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Selina Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 17 Special Types of Quadrilaterals

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 17 Special Types of Quadrilaterals. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Special Types of Quadrilaterals Exercise 17 – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
In parallelogram ABCD, ∠A = 3 times ∠B. Find all the angles of the parallelogram. In the same parallelogram, if AB = 5x – 7 and CD = 3x +1 ; find the length of CD.
Solution:

Let ∠B = x
∠A = 3 ∠B = 3x
AD||BC
∠A + ∠B = 180°
3x + x = 180°
⇒ 4x = 180°
⇒ x = 45°
∠B = 45°
∠A = 3x = 3 x 45 = 135°
and ∠B = ∠D = 45°
opposite angles of || gm are equal.
∠A = ∠C = 135°
opposite sides of //gm are equal.
AB = CD
5x – 7 = 3x + 1
⇒ 5x – 3x = 1+7
⇒ 2x = 8
⇒ x = 4
CD = 3 x 4+1 = 13
Hence 135°, 45°, 135° and 45° ; 13

Question 2.
In parallelogram PQRS, ∠Q = (4x – 5)° and ∠S = (3x + 10)°. Calculate : ∠Q and ∠R.
Solution:
In parallelogram PQRS,
∠Q = (4x – 5)° and ∠S = (3x + 10)°

opposite ∠s of //gm are equal.
∠Q = ∠S
4x – 5 = 3x + 10
4x – 3x = 10+5
x = 15
∠Q = 4x – 5 =4 x 15 – 5 = 55°
Also ∠Q + ∠R = 180°
55° + ∠R = 180°
∠R = 180°-55° = 125°
∠Q = 55° ; ∠R = 125°

Question 3.
In rhombus ABCD ;
(i) if ∠A = 74° ; find ∠B and ∠C.
(ii) if AD = 7.5 cm ; find BC and CD.
Solution:
AD || BC
∠A + ∠B = 180°
74° + ∠B = 180°
∠B =180° – 74°= 106°

opposite angles of Rhombus are equal.
∠A = ∠C = 74°
Sides of Rhombus are equal.
BC = CD = AD = 7.5 cm
(i) ∠B = 106° ; ∠C = 74°
(ii) BC = 7.5 cm and CD = 7.5 cm Ans.

Question 4.
In square PQRS :
(i) if PQ = 3x – 7 and QR = x + 3 ; find PS
(ii) if PR = 5x and QR = 9x – 8. Find QS
Solution:
(i) sides of square are equal.

PQ = QR
=> 3x – 7 = x + 3
=> 3x – x = 3 + 7
=> 2x = 10
x = 5
PS=PQ = 3x – 7 = 3 x 5 – 7 =8
(ii) PR = 5x and QS = 9x – 8

As diagonals of square are equal.
PR = QS
5x = 9x – 8
=> 5x – 9x = -8
=> -4x = -8
=> x = 2
QS = 9x – 8 = 9 x 2 – 8 =10

Question 5.
ABCD is a rectangle, if ∠BPC = 124°
Calculate : (i) ∠BAP (ii) ∠ADP

Solution:
Diagonals of rectangle are equal and bisect each other.
∠PBC = ∠PCB = x (say)
But ∠BPC + ∠PBC + ∠PCB = 180°
124° + x + x = 180°
2x = 180° – 124°
2x = 56°
=> x = 28°
∠PBC = 28°
But ∠PBC = ∠ADP [Alternate ∠s]
∠ADP = 28°
Again ∠APB = 180° – 124° = 56°
Also PA = PB
∠BAP = \(\frac { 1 }{ 2 }\) (180° – ∠APB)
= \(\frac { 1 }{ 2 }\) x (180°- 56°) = \(\frac { 1 }{ 2 }\) x 124° = 62°
Hence (i) ∠BAP = 62° (ii) ∠ADP =28°

Question 6.
ABCD is a rhombus. If ∠BAC = 38°, find :
(i) ∠ACB
(ii) ∠DAC
(iii) ∠ADC.

Solution:
ABCD is Rhombus (Given)
AB = BC
∠BAC = ∠ACB (∠s opp. to equal sides)
But ∠BAC = 38° (Given)
∠ACB = 38°
In ∆ABC,
∠ABC + ∠BAC + ∠ACB = 180°
∠ABC + 38°+ 38° = 180°
∠ABC = 180° – 76° = 104°
But ∠ABC = ∠ADC (opp. ∠s of rhombus)
∠ADC = 104°
∠DAC = ∠DCA ( AD = CD)
∠DAC = \(\frac { 1 }{ 2 }\) [180° – 104°]
∠DAC = \(\frac { 1 }{ 2 }\) x 76° = 38°
Hence (i) ∠ACB = 38° (ii) ∠DAC = 38° (iii) ∠ADC = 104° Ans.

Question 7.
ABCD is a rhombus. If ∠BCA = 35°. find ∠ADC.
Solution:
Given : Rhombus ABCD in which ∠BCA = 35°

To find : ∠ADC
Proof : AD || BC
∠DAC = ∠BCA (Alternate ∠s)
But ∠BCA = 35° (Given)
∠DAC = 35°
But ∠DAC = ∠ACD ( AD = CD) & ∠DAC +∠ACD + ∠ADC = 180°
35°+ 35° + ∠ADC = 180°
∠ADC = 180° – 70° = 110°
Hence ∠ADC = 110°

Question 8.
PQRS is a parallelogram whose diagonals intersect at M.
If ∠PMS = 54°, ∠QSR = 25° and ∠SQR = 30° ; find :
(i) ∠RPS
(ii) ∠PRS
(iii) ∠PSR.
Solution:
Given : ||gm PQRS in which diagonals PR & QS intersect at M.
∠PMS = 54° ; ∠QSR = 25° and ∠SQR=30°

To find : (i) ∠RPS (ii) ∠PRS (iii) ∠PSR
Proof : QR || PS
=> ∠PSQ = ∠SQR (Alternate ∠s)
But ∠SQR = 30° (Given)
∠PSQ = 30°
In ∆SMP,
∠PMS + ∠ PSM +∠MPS = 180° or 54° + 30° + ∠RPS = 180°
∠RPS = 180°- 84° = 96°
Now ∠PRS + ∠RSQ = ∠PMS
∠PRS + 25° =54°
∠PRS = 54° – 25° = 29°
∠PSR = ∠PSQ + ∠RSQ = 30°+25° = 55°
Hence (i) ∠RPS = 96° (ii) ∠PRS = 29° (iii) ∠PSR = 55°

Question 9.
Given : Parallelogram ABCD in which diagonals AC and BD intersect at M.
Prove : M is mid-point of LN.
Solution:

Proof : Diagonals of //gm bisect each other.
MD = MB
Also ∠ADB = ∠DBN (Alternate ∠s)
& ∠DML = ∠BMN (Vert. opp. ∠s)
∆DML = ∆BMN
LM = MN
M is mid-point of LN.
Hence proved.

Question 10.
In an Isosceles-trapezium, show that the opposite angles are supplementary.
Solution:

Given : ABCD is isosceles trapezium in which AD = BC
To Prove : (i) ∠A + ∠C = 180°
(ii) ∠B + ∠D = 180°
Proof : AB || CD.
=> ∠A + ∠D = 180°
But ∠A = ∠B [Trapezium is isosceles)]
∠B + ∠D = 180°
Similarly ∠A + ∠C = 180°
Hence the result.

Question 11.
ABCD is a parallelogram. What kind of quadrilateral is it if :
(i) AC = BD and AC is perpendicular to BD?
(ii) AC is perpendicular to BD but is not equal to it ?
(iii) AC = BD but AC is not perpendicular to BD ?
Solution:

Question 12.
Prove that the diagonals of a parallelogram bisect each other.
Solution:

Given : ||gm ABCD in which diagonals AC and BD bisect each other.
To Prove : OA = OC and OB = OD
Proof : AB || CD (Given)
∠1 = ∠2 (alternate ∠s)
∠3 = ∠4 = (alternate ∠s)
and AB = CD (opposite sides of //gm)
∆COD = ∆AOB (A.S.A. rule)
OA = OC and OB = OD
Hence the result.

Question 13.
If the diagonals of a parallelogram are of equal lengths, the parallelogram is a rectangle. Prove it.
Solution:

Given : //gm ABCD in which AC = BD
To Prove : ABCD is rectangle.
Proof : In ∆ABC and ∆ABD
AB = AB (Common)
AC = BD (Given)
BC = AD (opposite sides of ||gm)
∆ABC = ∆ABD (S.S.S. Rule)
∠A = ∠B
But AD // BC (opp. sides of ||gm are ||)
∠A + ∠B = 180°
∠A = ∠B = 90°
Similarly ∠D = ∠C = 90°
Hence ABCD is a rectangle.

Question 14.
In parallelogram ABCD, E is the mid-point of AD and F is the mid-point of BC. Prove that BFDE is a parallelogram.
Solution:

Given : //gm ABCD in which E and F are mid-points of AD and BC respectively.
To Prove : BFDE is a ||gm.
Proof : E is mid-point of AD. (Given)
DE = \(\frac { 1 }{ 2 }\) AD
Also F is mid-point of BC (Given)
BF = \(\frac { 1 }{ 2 }\) BC
But AD = BC (opp. sides of ||gm)
BF = DE
Again AD || BC
=> DE || BF
Now DE || BF and DE = BF
Hence BFDE is a ||gm.

Question 15.
In parallelogram ABCD, E is the mid-point of side AB and CE bisects angle BCD. Prove that :
(i) AE = AD,
(ii) DE bisects and ∠ADC and
(iii) Angle DEC is a right angle.
Solution:

Given : ||gm ABCD in which E is mid-point of AB and CE bisects ZBCD.
To Prove : (i) AE = AD
(ii) DE bisects ∠ADC
(iii) ∠DEC = 90°
Const. Join DE
Proof : (i) AB || CD (Given)
and CE bisects it.
∠1 = ∠3 (alternate ∠s) ……… (i)
But ∠1 = ∠2 (Given) …………. (ii)
From (i) & (ii)
∠2 = ∠3
BC = BE (sides opp. to equal angles)
But BC = AD (opp. sides of ||gm)
and BE = AE (Given)
AD = AE
∠4 = ∠5 (∠s opp. to equal sides)
But ∠5 = ∠6 (alternate ∠s)
=> ∠4 = ∠6
DE bisects ∠ADC.
Now AD // BC
=> ∠D + ∠C = 180°
2∠6+2∠1 = 180°
DE and CE are bisectors.
∠6 + ∠1 = \(\frac { { 180 }^{ 0 } }{ 2 }\)
∠6 + ∠1 = 90°
But ∠DEC + ∠6 + ∠1 = 180°
∠DEC + 90° = 180°
∠DEC = 180° – 90°
∠DEC = 90°
Hence the result.

Question 16.
In the following diagram, the bisectors of interior angles of the parallelogram PQRS enclose a quadrilateral ABCD.

Show that:
(i) ∠PSB + ∠SPB = 90°
(ii) ∠PBS = 90°
(iii) ∠ABC = 90°
(iv) ∠ADC = 90°
(v) ∠A = 90°
(vi) ABCD is a rectangle
Thus, the bisectors of the angles of a parallelogram enclose a rectangle.
Solution:
Given : In parallelogram ABCD bisector of angles P and Q, meet at A, bisectors of ∠R and ∠S meet at C. Forming a quadrilateral ABCD as shown in the figure.
To prove :
(i) ∠PSB + ∠SPB = 90°
(ii) ∠PBS = 90°
(iii) ∠ABC = 90°
(iv) ∠ADC = 90°
(v) ∠A = 9°
(vi) ABCD is a rectangle
Proof : In parallelogram PQRS,
PS || QR (opposite sides)
∠P +∠Q = 180°
and AP and AQ are the bisectors of consecutive angles ∠P and ∠Q of the parallelogram
∠APQ + ∠AQP = \(\frac { 1 }{ 2 }\) x 180° = 90°
But in ∆APQ,
∠A + ∠APQ + ∠AQP = 180° (Angles of a triangle)
∠A + 90° = 180°
∠A = 180° – 90°
(v) ∠A = 90°
Similarly PQ || SR
∠PSB + SPB = 90°
(ii) and ∠PBS = 90°
But, ∠ABC = ∠PBS (Vertically opposite angles)
(iii) ∠ABC = 90°
Similarly we can prove that
(iv) ∠ADC = 90° and ∠C = 90°
(vi) ABCD is a rectangle (Each angle of a quadrilateral is 90°)
Hence proved.

Question 17.
In parallelogram ABCD, X and Y are midpoints of opposite sides AB and DC respectively. Prove that:
(i) AX = YC
(ii) AX is parallel to YC
(iii) AXCY is a parallelogram.
Solution:
Given : In parallelogram ABCD, X and Y are the mid-points of sides AB and DC respectively AY and CX are joined

To prove :
(i) AX = YC
(ii) AX is parallel to YC
(iii) AXCY is a parallelogram
Proof : AB || DC and X and Y are the mid-points of the sides AB and DC respectively
(i) AX = YC ( \(\frac { 1 }{ 2 }\) of opposite sides of a parallelogram)
(ii) and AX || YC
(iii) AXCY is a parallelogram (A pair of opposite sides are equal and parallel)
Hence proved.

Question 18.
The given figure shows parallelogram ABCD. Points M and N lie in diagonal BD such that DM = BN.

Prove that:
(i) ∆DMC = ∆BNA and so CM = AN
(ii) ∆AMD = ∆CNB and so AM CN
(iii) ANCM is a parallelogram.
Solution:
Given : In parallelogram ABCD, points M and N lie on the diagonal BD such that DM = BN
AN, NC, CM and MA are joined
To prove :
(i) ∆DMC = ∆BNA and so CM = AN
(ii) ∆AMD = ∆CNB and so AM = CN
(iii) ANCM is a parallelogram
Proof :
(i) In ∆DMC and ∆BNA.
CD = AB (opposite sides of ||gm ABCD)
DM = BN (given)
∠CDM = ∠ABN (alternate angles)
∆DMC = ∆BNA (SAS axiom)
CM =AN (c.p.c.t.)
Similarly, in ∆AMD and ∆CNB
AD = BC (opposite sides of ||gm)
DM = BN (given)
∠ADM = ∠CBN – (alternate angles)
∆AMD = ∆CNB (SAS axiom)
AM = CN (c.p.c.t.)
(iii) CM = AN and AM = CN (proved)
ANCM is a parallelogram (opposite sides are equal)
Hence proved.

Question 19.
The given figure shows a rhombus ABCD in which angle BCD = 80°. Find angles x and y.

Solution:
In rhombus ABCD, diagonals AC and BD bisect each other at 90°
∠BCD = 80°
Diagonals bisect the opposite angles also ∠BCD = ∠BAD (Opposite angles of rhombus)
∠BAD = 80° and ∠ABC = ∠ADC = 180° – 80° = 100°
Diagonals bisect opposite angles
∠OCB or ∠PCB = \(\frac { { 80 }^{ 0 } }{ 2 }\) = 40°
In ∆PCM,
Ext. CPD = ∠OCB + ∠PMC
110° = 40° + x
=> x = 110° – 40° = 70°
and ∠ADO = \(\frac { 1 }{ 2 }\) ∠ADC = \(\frac { 1 }{ 2 }\) x 100° = 50°
Hence x = 70° and y = 50°

Question 20.
Use the information given in the alongside diagram to find the value of x, y and z.

Solution:
ABCD is a parallelogram and AC is its diagonal which bisects the opposite angle
Opposite sides of a parallelogram are equal
3x + 14 = 2x + 25
=> 3x – 2x = 25 – 14
=> x = 11
∴ x = 11 cm
∠DCA = ∠CAB (Alternate angles)
y + 9° = 24
y = 24° – 9° = 15°
∠DAB = 3y° + 5° + 24° = 3 x 15 + 5 + 24° = 50° + 24° = 74°
∠ABC =180°- ∠DAB = 180° – 74° = 106°
z = 106°
Hence x = 11 cm, y = 15°, z = 106°

Question 21.
The following figure is a rectangle in which x : y = 3 : 7; find the values of x and y.

Solution:
ABCD is a rectangle,
x : y = 3 : 1
In ∆BCE, ∠B = 90°
x + y = 90°
But x : y = 3 : 7
Sum of ratios = 3 + 7=10

Hence x = 27°, y = 63°

Question 22.
In the given figure, AB // EC, AB = AC and AE bisects ∠DAC. Prove that:

(i) ∠EAC = ∠ACB
(ii) ABCE is a parallelogram.
Solution:
ABCE is a quadrilateral in which AC is its diagonal and AB || EC, AB = AC
BA is produced to D
AE bisects ∠DAC
To prove:
(i) ∠EAC = ∠ACB
(ii) ABCE is a parallelogram
Proof:
(i) In ∆ABC and ∆ZAEC
AC=AC (common)
AB = CE (given)
∠BAC = ∠ACE (Alternate angle)
∆ABC = ∆AEC (SAS Axiom)
(ii) ∠BCA = ∠CAE (c.p.c.t.)
But these are alternate angles
AE || BC
But AB || EC (given)
∴ ABCD is a parallelogram

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 8 Profit, Loss and Discount

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 8 Profit, Loss and Discount. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 8 Maths SolutionsPhysicsChemistryBiologyGeographyHistory & Civics

Profit, Loss and Discount Exercise 8A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Megha bought 10 note-books for Rs.40 and sold them at Rs.4.75 per note-book. Find, her gain percent.
Solution:

Question 2.
A fruit-seller buys oranges at 4 for Rs.3 and sells them at 3 for Rs.4 Find his profit percent.
Solution:

Question 3.
A man buys a certain number of articles at 15 for Rs. 112.50 and sells them at 12 for Rs.108. Find ;
(i) his gain as percent;
(ii) the number of articles sold to make a profit of Rs.75.
Solution:

Question 4.
A boy buys an old bicycle for Rs. 162 and spends Rs. 18 on its repairs before selling the bicycles for Rs. 207. Find his gain or loss percent.
Solution:
Buying price of the old bicycle = Rs.162
Money spent on repairs = Rs. 18
Real C.P. of the bicycle= 162+18 = Rs.180
S.P. of the bicycle = Rs.207
Profit = S.P. – C.P. = 207 – 162 = Rs. 45

Question 5.
An article is bought from Jaipur for Rs. 4,800 and is sold in Delhi for Rs. 5,820. If Rs. 1,200 is spent on its transportations, etc. ; find he loss or the gain as percent.
Solution:
Cost price = Rs. 4,800
Selling Price = Rs. 5,820
Transport etc. charges = Rs. 1,200
Total cost price = Rs, 4,800 + Rs. 1,200 = Rs. 6,000
Loss = Rs. 6000 – Rs. 5820 = Rs. 180

Question 6.
Mohit sold a T.V. for Rs. 3,600 ; gaining one-sixth of its selling price. Find :
(i) the gain
(ii) the cost price of the T.V.
(iii) the gain percent.
Solution:
S.P. of T.V. = Rs. 3,600

Question 7.
By selling a certain number of goods for Rs. 5,500; a shopkeeper loses equal to one-tenth of their selling price. Find :
(i) the loss incured
(ii) the cost price of the goods
(iii) the loss as percent.
Solution:

Question 8.
The selling price of a sofa-set is \(\frac { 4 }{ 5 }\) times of its cost price. Find the gain or the loss as percent.
Solution:

Question 9.
The cost price of an article is \(\frac { 4 }{ 5 }\) times of its selling price. Find the loss or the gain as percent.
Solution:

Question 10.
A shopkeeper sells his goods at 80% of their cost price. Find the percent gain or loses ?
Solution:

Question 11.
The cost price of an article is 90% of its selling price. What is the profit or the loss as percent ?
Solution:

Question 12.
The cost price of an article is 30 percent less than its selling price. Find, the profit or loss as percent.
Solution:

Question 13.
A shop-keeper bought 300 eggs at 80 paisa each. 30 eggs were broken in transaction and then he sold the remaining eggs at one rupee each. Find, his gain or loss as percent.
Solution:

Question 14.
A man sold his bicycle for Rs.405 losing one-tenth of its cost price, find :
(i) its cc price;
(ii) the loss percent.
Solution:
(i) Let C.P. of the bicycle = Rs. x

Question 15.
A man sold a radio-set for Rs.250 and gained one-ninth of its cost price. Find ;
(i) its cost price;
(ii) the profit percent.
Solution:

Profit, Loss and Discount Exercise 8B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the selling price, if:
(i) C.P. = Rs. 950 and profit = 8%
(ii) C.P. = Rs. 1,300 and loss = 13%
Solution:

Question 2.
Find the cost price, if :
(i) S.P. = Rs. 1,680 and profit = 12%
(ii) S.P. = Rs. 1,128 and loss = 6%
Solution:

Question 3.
By selling an article for Rs.900; a man gains 20%. Find his cost price and the gain.
Solution:

Question 4.
By selling an article for Rs.704; a person loses 12%. Find his cost price and the loss
Solution:

Question 5.
Find the selling price, if :
(i) C.P. = Rs.352; overheads = Rs.28 and profit = 20
(ii) C.P. = Rs.576; overheads = Rs.44 and loss = 16%
Solution:

Question 6.
If John sells his bicycle for Rs. 637, he will suffer a loss of 9%. For how much should it be sold, if he desires a profit of 5% ?
Solution:

Question 7.
A man sells a radio-set for Rs.605 and gains 10%. At what price should he sell another radio of the same kind, in order to gain 16% ?
Solution:

Question 8.
By selling a sofa-set for Rs.2,500; the shopkeeper loses 20%. Find his loss percent or profit percent ; if he sells the same sofa-set for Rs.3150.
Solution:

Question 9.
Mr. Sinha sold two tape-recorders for Rs.990 each; gaining 10% on one and losing 10% on the other. Find his total loss or gain as percent on the whole transaction.
Solution:

Question 10.
A tape-recorder is sold for Rs. 2,760 at a gain of 15% and a C.D. player is sold for Rs. 3,240 at a loss of 10% Find :
(i) the C.P. of the tape-recorder
(ii) the C.P. of the C.D. player.
(iii) the total C.P. of both.
(iv) the total S.P. of both
(v) the gain % or the loss % on the whole
Solution:

Question 11.
Rajesh sold his scooter to Rahim at 8% loss and Rahim, in turn, sold the same scooter to Prem at 5% gain. If Prem paid Rs. 14,490 for the scooter ; find :
(i) the S.P. and the C.P. of the scooter for Rahim
(ii) the S.P. and the C.P. of the scooter for Rajesh
Solution:

Question 12.
John sold an article to Peter at 20% profit and Peter sold it to Mohan at 5% loss. If Mohan paid Rs.912 for the article; find how much did John pay for it ?
Solution:

Profit, Loss and Discount Exercise 8C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A stationer buys pens at 5 for Rs.28 and sells them at a profit of 25 %. How much should a customer pay; if he buys
(i) only one pen ;
(ii) three pens ?
Solution:

Question 2.
A fruit-seller sells 4 oranges for Rs. 3, gaining 50%. Find :
(i) C.P. of 4 oranges,
(ii) C.P. of one orange.
(iii) S.P. of one orange.
(iv) profit made by selling one orange.
(v) number of oranges brought and sold in order to gain Rs. 24.
Solution:

Question 3.
A man sells 12 articles for Rs. 80 gaining 33\(\frac { 1 }{ 3 }\) %. Find the number of articles bought by the man for Rs. 90.
Solution:

Question 4.
The cost price of 20 articles is same as the selling price of 16 articles. Find the gain percent.
Solution:
C.P. of 20 articles = S.P. of 16 articles.
Let C.P. of 1 article = Re. 1
C.P. of 20 articles = Rs.20
and C.P. of 16 articles = Rs.16
S.P. of 16 articles = Rs.20
[S.P. of 16 articles = C.P. of 20 articles]
Gain = Rs.20 – Rs.16 = Rs.4
Gain% = \(\frac { 4 }{ 16 }\) x 100
= \(\frac { 4 x 100 }{ 16 }\)
= 25%

Question 5.
The selling price of 15 articles is equal to the cost price of 12 articles. Find the gain or loss as percent.
Solution:
S.P. of 15 articles = C.P. of 12 articles
Let C.P. of 1 article = Re.1
C.P. of 12 article = Rs.12
and C.P. of 15 articles = Rs.15
S.P. of 15 articles = Rs.12
[S.P. of 15 articles = C.P. of 12 articles]
Loss = Rs.15 – Rs.12 = Rs.3
Loss% = \(\frac { 3 }{ 15 }\) x 100 = 20%

Question 6.
By selling 8 pens, Shyam loses equal to the cost price of 2 pens. Find his loss percent.
Solution:
Let C.P. of 1 pen = Re.1
C.P. of 2 pens = Rs.2
and C.P. of 8 pens = Rs.8
Loss = Rs.2 ….[Loss = C.P. of 2 Pens]
Loss% = \(\frac { 2 }{ 8 }\) x 100 = 25%

Question 7.
A shop-keeper bought rice worth Rs.4,500. He sold one-third of it at 10% profit.
If he desires a profit of 12% on the whole ; find :
(i) the selling price of the rest of the rice ;
(ii) the percentage profit on the rest of the rice.
Solution:

Question 8.
Mohan bought a certain number of note-books for Rs.600. He sold \(\frac { 1 }{ 4 }\) of them at 5 percent loss. At what price should he sell the remaining note-books so as to gain 10% on the whole ?
Solution:

Question 9.
Raju sells a watch at 5% profit. Had he sold it for Rs.24 more ; he would have gained 11%. Find the cost price of the watch.
Solution:
Let C.P. of the watch = Rs.100
When profit = 5%; S.P. = Rs.(100+5) = Rs.105
When profit = 11%;
S.P. = Rs.(100 + 11) = Rs .111
Difference of two selling prices = Rs. 111 – Rs. 105 = Rs.6
When watch sold for Rs.6 more; then C.P. of the watch = Rs.100
When watch sold for Re. 1 more; then C.P. of the watch = Rs. \(\frac { 100 }{ 6 }\)
When watch sold for Rs.24 more; then C.P. of the watch = Rs. \(\frac { 100 }{ 6 }\) x 24 = Rs.400

Question 10.
A man sold a bicycle at 5% profit. If the cost had been 30% less and the selling price Rs.63 less, he would have made a profit of 30%. What is the cost price of the bicycle ?
Solution:
Let C.P. of the bicycle = Rs.100
In the I case :
When Profit = 5% ;

Question 11.
Renu sold an article at a loss of 8 percent. Had she bought it at 10% less and sold for Rs.36 more; she would have gained 20%. Find the cost price of the article.
Solution:

Profit, Loss and Discount Exercise 8D – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
An article is marked for Rs. 1,300 and is sold for Rs. 1,144 ; find the discount percent.
Solution:

Question 2.
The marked price of a dinning table is Rs. 23,600 and is available at a discount of 8%. Find its selling price.
Solution:

Question 3.
A wrist-watch is available at a discount of 9%. If the list-price of the watch is Rs. 1,400 ; find the discount given and the selling price of the watch.
Solution:
List price of the watch = Rs. 1,400
Discount = 9%
Discount = \(\frac { 1400 x 9 }{ 100 }\) = 14 x 9 = Rs. 126
S.P. = (List price – Discount) = Rs. (1400 – 126) = Rs. 1274

Question 4.
A shopkeeper sells an article for Rs. 248.50 after allowing a discount of 10%. Find the list price of the article.
Solution:

Question 5.
A shop-keeper buys an article for Rs.450. He marks it at 20% above the cost price. Find :
(i) the marked price of the article.
(ii) the selling price, if he sells the articles at 10 percent discount.
(iii) the percentage discount given by him, if he sells the article for Rs.496.80
Solution:

Question 6.
The list price of an article is Rs.800 and is available at a discount of 15 percent. Find :
(i) selling price of the article ;
(ii) cost price of the article, if a profit of 13\(\frac { 1 }{ 3 }\) % is made on selling it.
Solution:

Question 7.
An article is marked at Rs. 2,250. By selling it at a discount of 12%, the dealer makes a profit of 10%. Find :
(i) the selling price of the article.
(ii) the cost price of the article for the dealer.
Solution:

Question 8.
By selling an article at 20% discount, a shopkeeper gains 25%. If the selling price of the article is Rs. 1,440 ; find :
(i) the marked price of the article.
(ii) the cost price of the article.
Solution:

Question 9.
A shop-keeper marks his goods at 30 percent above the cost price and then gives a discount of 10 percent. Find his gain percent.
Solution:

Question 10.
A ready-made garments shop in Delhi, allows 20 percent discount on its garments and still makes a profit of 20 percent. Find the marked price of a dress which is bought by the shop-keeper for Rs.400.
Solution:

Question 11.
At 12% discount, the selling price of a pen is Rs. 13.20. Find its marked price. Also, find the new selling price of the pen, if it is sold at 5% discount.
Solution:

Question 12.
The cost price of an article is Rs. 2,400 and it is marked at 25% above the cost price. Find the profit and the profit percent, if the article is sold at 15% discount.
Solution:

Question 13.
Thirty articles are bought at Rs. 450 each. If one-third of these articles be sold at 6% loss; at what price must each of the remaining articles be sold in order to make a profit of 10% on the whole?
Solution:

Question 14.
The cost price of an article is 25% below the marked price. If the article is available at 15% discount and its cost price is Rs. 2,400; find:
(i) Its marked price
(ii) its selling price
(iii) the profit percent.
Solution:

Question 15.
Find a single discount (as percent) equivalent to following successive discounts:
(i) 20% and 12%
(ii) 10%, 20% and 20%
(iii) 20%, 10% and 5%
Solution:

Question 16.
Find the single discount (as percent) equivalent to successive discounts of:
(i) 80% and 80%
(ii) 60% and 60%
(iii) 60% and 80%
Solution:
(i) Successive discounts = 80% and 80%

Profit, Loss and Discount Exercise 8E – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Rajat purchases a wrist-watch costing Rs. 540. The rate of Sales Tax is 8%. Find the total amount paid by Rajat for the watch.
Solution:

Total Amount of Watch = ₹ 540 + ₹ 43.20 = ₹ 583.20

Question 2.
Ramesh paid ₹ 345.60 as Sales Tax on a purchase of ₹ 3,840. Find the rate of Sales Tax.
Solution:

Question 3.
The price of a washing machine, inclusive of sales tax is ₹ 13,530/-. If the Sales Tax is 10%, find its basic (cost) price.
Solution:

Question 4.
Sarita purchases biscuits costing ₹ 158 on which the rate of Sales Tax is 6%. She also purchases some cosmetic goods costing ₹ 354 on which rate of Sales Tax is 9%. Find the total amount to be paid by Sarita.
Solution:

Total cost of cosmetic goods = ₹ 354 + ₹ 31.86 = ₹ 385.86
Total amount paid by Sarita = ₹ 167. 48 + 385.86 = ₹ 553. 34

Question 5.
The price of a T.V. set inclusive of sales tax of 9% is ₹ 13,407. Find its marked price. If Sales Tax is increased to 13%, how much more does the customer has to pay for the T.V. ?
Solution:

Question 6.
The price of an article is ₹ 8,250 which includes Sales Tax at 10%. Find how much more or less does a customer pay for the article, if the Sales Tax on the article:
(i) increases to 15%
(ii) decreases to 6%
(iii) increases by 2%
(iv) decreases by 3%
Solution:
Price of an article = ₹ 8,250
Rate of Sales Tax = 10%
Let the list price = x

Question 7.
A bicycle is available for ₹ 1,664 including Sales Tax. If the list price of the bicycle is ₹ 1,600, find :
(i) the rate of Sales Tax
(ii) the price a customer will pay for the bicycle if the Sales Tax is increased by 6%.
Solution:

Question 8.
When the rate of sale-tax is decreased from 9% to 6% for a coloured T.V. ; Mrs Geeta will save ₹ 780 in buying this T.V. Find the list price of the T.V.
Solution:

Question 9.
A shopkeeper sells an article for ₹ 21,384 including 10% sales-tax. However, the actual rate of sales-tax is 8%. Find the extra profit made by the dealer.
Solution:
Sale Price of an article including S.T. = ₹ 21384

Profit, Loss and Discount Exercise 8F – Selina Concise Mathematics Class 8 ICSE Solutions

[In this exercise, all the prices are excluding tax/VAT unless specified]
Question 1.
A shopkeeper buys an article for ₹ 8,000 and sells it for ₹ 10,000. If the rate of tax under VAT is 10%, find :
(i) tax paid by the shopkeeper
(ii) tax charged by the shopkeeper
(iii) VAT paid by the shopkeeper
Solution:

(iii) VAT paid by the shopkeeper = ₹ 1000 – ₹ 800 = ₹ 200

Question 2.
A trader buys some goods for ₹ 12,000 and sells them for ₹ 15,000. If the rate of tax under VAT is 12%, find the VAT paid by the trader?
Solution:

Question 3.
The marked price of an article is ₹ 7,000 and is available at 20% discount. Manoj buys this article and then sold it at its marked price. If the rate of tax at each state is 10%, find the VAT paid by Manoj.
Solution:

Question 4.
A buys some goods for ₹ 4,000 and sold them to B for ₹ 5,000. B sold these goods to C for ₹ 6,000. If the rate of tax (under VAT) at each stage is 5%, find :
(i) VAT paid by A
(ii) VAT paid by B
Solution:
C.P. of some goods for A = ₹ 4000
C.P. of some goods for B = ₹ 5000
and C.P. for C = ₹ 6000

Question 5.
A buys an article for ₹ 8,000 and sold it to B at 20% profit. If the rate of tax under VAT is 8%, find :
(i) tax paid by A
(ii) tax charged by A
(iii) VAT paid by A
Solution:

Question 6.
A shopkeeper purchases an article for ₹ 12,400 and sells it to a customer for ₹ 17,000. If the tax under VAT is 8%, find the VAT paid by the shopkeeper.
Solution:
C.P. of article = ₹ 12,400
Rate of VAT =8%
Total VAT = ₹ 12,400 x \(\frac { 8 }{ 100 }\) = ₹ 992
S.P. of the article = ₹ 17000
VAT charge 8% = ₹ 17000 x \(\frac { 8 }{ 100 }\) = ₹ 1360
Amount of VAT paid by the shopkeeper = ₹ 1360 – ₹ 992 = ₹ 368

Question 7.
A purchases an article for ₹ 7,200 and sells it to B for ₹ 9,600. B, in turn, sells the article to C for ₹ 11,000. If the tax (under VAT) is 10%, find the VAT paid by A and B.
Solution:

Question 8.
A manufacturer buys some goods for ₹ 60,000 and pays 5% tax. He sells these goods for ₹ 80,000 and charges tax at the rate of 12%. Find the VAT paid by the manufacturer.
Solution:

VAT paid by the manufacturer = ₹ 9600 – ₹ 3000 = ₹ 6600

Question 9.
The cost of an article is ₹ 6,000 to a distributor, he sells it to a trader for ₹ 7,500 and the trader sells it further to a customer for ₹ 8,000. If the rate of tax under VAT is 8%; find the VAT paid by the:
(i) distributor
(ii) trader
Solution:

Question 10.
The marked price of an article is ₹ 10,000. A buys it at 30% discount on the marked price and sells it at 10% discount on the marked price. If the rate of tax under VAT is 5%, find the amount of VAT paid by A.
Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 22 Data Handling

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 22 Data Handling. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Data Handling Exercise 22A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Arrange the following data as an array (in ascending order):
(i) 7, 5, 15, 12, 10, 11, 16
(ii) 6.3, 5.9, 9.8, 12.3, 5.6, 4.7
Solution:
(i) Ascending order = 5, 7, 10, 11, 12, 15, 16
(ii) Ascending order = 4.7, 5.6, 5.9, 6.3, 9.8, 12.3

Question 2.
Arrange the following data as an array (descending order):
(i) 0 2, 0, 3, 4, 1, 2, 3, 5
(ii) 9.1, 3.7, 5.6, 8.3, 11.5, 10.6
Solution:
(i) Descending order = 5, 4, 3, 3, 2, 2, 1, 0
(ii) Descending order = 11.5, 10.6, 9.1, 8.3, 5.6, 3.7

Question 3.
Construct a frequency table for the following data:
(i) 6, 7, 5, 6, 8, 9, 5, 5, 6, 7, 8, 9, 8, 10, 10, 9, 8, 10, 5, 7, 6, 8.
(ii) 3, 2, 1, 5, 4, 3, 2, 5, 5, 4, 2, 2, 2, 1, 4, 1, 5, 4.
Solution:

Question 4.
Following are the marks obtained by 30 students in an examinations.

Taking class intervals 0-10, 10-20, ……… 40-50 ; construct a frequency table.
Solution:

Question 5.
Construct a frequency distribution table for the following data ; taking class-intervals 4-6, 6-8, ……… 14-16.

Solution:

Question 6.
Fill in the blanks:
(i) Lower class limit of 15-18 is ………
(ii) Upper class limit of 24-30 is ……..
(iii) Upper limit of 5-12.5 is ………
(iv) If the upper and the lower limits of a class interval are 16 and 10 ; the class-interval is ……..
(v) If the lower and the upper limits of a class interval are 7.5 and 12.5 ; the class interval is ……..
Solution:
(i) Lower class limit of 15 – 18 is 15.
(ii) Upper class limit of 24 – 30 is 30.
(iii) Upper limit of 5 – 12.5 is 12.5
(iv) If the upper and lower limits of a class interval are 16 and 10 ; the class interval is 10 – 16
(v) If the lower and upper limits of a class interval are 7.5 and 12.5 ; the class interval is 7.5 – 12.5

Data Handling Exercise 22B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Hundred students from a certain locality use different modes of travelling to school as given below. Draw a bar graph.

Solution:

Question 2.
Mr. Mirza’s monthly income is Rs. 7,200. He spends Rs. 1,800 on rent, Rs. 2,700 on food, Rs. 900 on education of his children ; Rs. 1,200 on Other things and saves the rest.
Draw a pie-chart to represent it.
Solution:

Question 3.
The percentage of marks obtained, in different subjects by Ashok Sharma (in an examination) are given below. Draw a bar graph to represent it.

Solution:

Question 4.
The following table shows the market position of different brand of tea-leaves.

Draw it-pie-chart to represent the above information.
Solution:

Question 5.
Students of a small school use different modes of travel to school as shown below:

Draw a suitable bar graph.
Solution:

Question 6.
For the following table, draw a bar-graph

Solution:

Question 7.
Manoj appeared for ICSE examination 2018 and secured percentage of marks as shown in the following table:

Represent the above data by drawing a suitable bar graph.
Solution:

Question 8.
For the data given above in question number 7, draw a suitable pie-graph.
Solution:
∵ 60 + 45 + 42 + 48 + 75 = 270

Question 9.
Mr. Kapoor compares the prices (in Rs.) of different items at two different shops A and B. Examine the following table carefully and represent the data by a double bar graph.

Solution:

Question 10.
The following tables shows the mode of transport used by boys and girls for going to the same school.

Draw a double bar graph representing the above data.
Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 23 Probability. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Probability Exercise 23 – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A die is thrown, find the probability of getting:
(i) a prime number
(ii) a number greater than 4
(iii) a number not greater than 4.
Solution:
A die has six numbers : 1, 2, 3, 4, 5, 6
∴ Number of possible outcomes = 6

Question 2.
A coin is tossed. What is the probability of getting:
(i) a tail? (ii) ahead?
Solution:
On tossing a coin once,
Number of possible outcome = 2

Question 3.
A coin is tossed twice. Find the probability of getting:
(i) exactly one head (ii) exactly one tail
(iii) two tails (iv) two heads
Solution:

Question 4.
A letter is chosen from the word ‘PENCIL’ what is the probability that the letter chosen is a consonant?
Solution:
Total no. of letters in the word ‘PENCIL = 6
Total Number of Consonant = ‘PNCL’ i.e. 4

Question 5.
A bag contains a black ball, a red ball and a green ball, all the balls are identical in shape and size. A ball is drawn from the bag without looking into it. What is the probability that the ball drawn is:
(i) a red ball
(ii) not a red ball
(iii) a white ball.
Solution:
Total number of possible outcomes = 3

Question 6.
6. In a single throw of a die, find the probability of getting a number
(i) greater than 2
(ii) less than or equal to 2
(iii) not greater than 2.
Solution:
A die has six numbers = 1, 2, 3, 4, 5, 6
∴ Number of possible outcomes = 6

Question 7.
A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size.
A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:
(i) a black ball.
(ii) a red ball.
(iii) a white ball.
(iv) not a red ball.
(v) not a black ball.
Solution:

Question 8.
In a single throw of a die, find the probability that the number:
(i) will be an even number.
(ii) will be an odd number.
(iii) will not be an even number.
Solution:

Question 9.
In a single throw of a die, find the probability of getting :
(i) 8
(ii) a number greater than 8
(iii) a number less than 8
Solution:
On a die the numbers are 1, 2, 3, 4, 5, 6 i.e. six.
∴ Number of possible outcome = 6

Question 10.
Which of the following can not be the probability of an event?

Solution:
The probability of an event cannot be
(ii) 3.8 i.e. the probability of an even cannot exceed 1.
(iv) i.e. -0.8 and
(vi) -2/5, This is because probability of an even can never be less than 1.

Question 11.
A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out:
(i) a white ball,
(ii) a black ball
Solution:

Question 12.
Three identical coins are tossed together. What is the probability of obtaining:
all heads?
exactly two heads?
exactly one head?
no head?
Solution:

Question 13.
A book contains 92 pages. A page is chosen at random. What is the probability that the sum of the digits in the page number is 9?
Solution:

Question 14.
Two coins are tossed together. What is the probability of getting:
(i) at least one head
(ii) both heads or both tails.
Solution:

Question 15.
From 10 identical cards, numbered 1, 2, 3, …… , 10, one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:
(i) 2 (ii) 3
(iii) 2 and 3 (iv) 2 or 3
Solution:

Question 16.
Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 0
(ii) 12
(iii) less than 12
(iv) less than or equal to 12
Solution:

Question 17.
A die is thrown once. Find the probability of getting:
(i) a prime number
(ii) a number greater than 3
(iii) a number other than 3 and 5
(iv) a number less than 6
(v) a number greater than 6.
Solution:

Question 18.
Two coins are tossed together. Find the probability of getting:
(i) exactly one tail
(ii) at least one head
(iii) no head
(iv) at most one head
Solution:

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable (With Problems Based on Linear equations)

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 14 Linear Equations in one Variable (With Problems Based on Linear equations)

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 14 Linear Equations in one Variable (With Problems Based on Linear equations). You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Linear Equations in one Variable Exercise 14A – Selina Concise Mathematics Class 8 ICSE Solutions

Solve the following equations:
Question 1.
20 = 6 + 2x
Solution:
20 = 6 + 2x
20 – 6 = 2x
14 = 2x
7 = x
x = 7

Question 2.
15 + x = 5x + 3
Solution:
15 – 3 = 5x – x
12 = 4x
3 = x
x = 3

Question 3.
\(\frac { 3x+2 }{ x-6 }\) = -7
Solution:
3x + 2 = -7 (x – 6) (by cross multiplying)
3x + 2 = -7x + 42
3x + 7x = 42 – 2
10x = 40
x = 4

Question 4.
3a – 4 = 2(4 – a)
Solution:
3a – 4 = 8 – 2a
3a + 2a = 8+4
5a = 12
a = 2.4

Question 5.
3(b – 4) = 2(4 – b)
Solution:

Question 6.
\(\frac { x+2 }{ 9 } =\frac { x+4 }{ 11 }\)
Solution:

Question 7.
\(\frac { x-8 }{ 5 } =\frac { x-12 }{ 9 }\)
Solution:

Question 8.
5(8x + 3) = 9(4x + 7)
Solution:

Question 9.
3(x +1) = 12 + 4(x – 1)
Solution:
3(x + 1) = 12 + 4(x – 1)
3x + 3 = 12 + 4x – 4
3x – 4x = 12 – 4 – 3
-x = 5
x = -5

Question 10.
\(\frac { 3x }{ 4 } -\frac { 1 }{ 4 } \left( x-20 \right) =\frac { x }{ 4 } +32\)
Solution:

Question 11.
\(3a-\frac { 1 }{ 5 } =\frac { a }{ 5 } +5\frac { 2 }{ 5 }\)
Solution:

Question 12.
\(\frac { x }{ 3 } -2\frac { 1 }{ 2 } =\frac { 4x }{ 9 } -\frac { 2x }{ 3 }\)
Solution:

Question 13.
\(\frac { 4\left( y+2 \right) }{ 5 } =7+\frac { 5y }{ 13 }\)
Solution:

Question 14.
\(\frac { a+5 }{ 6 } -\frac { a+1 }{ 9 } =\frac { a+3 }{ 4 }\)
Solution:

Question 15.
\(\frac { 2x-13 }{ 5 } -\frac { x-3 }{ 11 } =\frac { x-9 }{ 5 } +1\)
Solution:

Question 16.
6(6x – 5) – 5 (7x – 8) = 12 (4 – x) + 1
Solution:

Question 17.
(x – 5) (x + 3) = (x – 7) (x + 4)
Solution:

Question 18.
(x – 5)2 – (x + 2)2 = -2
Solution:

Question 19.
(x – 1) (x + 6) – (x – 2) (x – 3) = 3
Solution:

Question 20.
\(\frac { 3x }{ x+6 } -\frac { x }{ x+5 } =2\)
Solution:

Question 21.
\(\frac { 1 }{ x-1 } +\frac { 2 }{ x-2 } =\frac { 3 }{ x-3 }\)
Solution:

Question 22.
\(\frac { x-1 }{ 7x-14 } =\frac { x-3 }{ 7x-26 }\)
Solution:

Question 23.
\(\frac { 1 }{ x-1 } -\frac { 1 }{ x } =\frac { 1 }{ x+3 } -\frac { 1 }{ x+4 }\)
Solution:

Question 24.
Solve: \(\frac { 2x }{ 3 } -\frac { x-1 }{ 6 } +\frac { 7x-1 }{ 4 } =2\frac { 1 }{ 6 }\)
Hence, find the value of ‘a’, if \(\frac { 1 }{ a }\) + 5x = 8.
Solution:

Question 25.
Solve: \(\frac { 4-3x }{ 5 } +\frac { 7-x }{ 3 } +4\frac { 1 }{ 3 } =0\)
Hence find the value of ‘p’ if 2p – 2x + 1 = 0
Solution:

Hence x = 8
Now, 3p – 2x + 1=0
⇒ 3p – 2 x 8 + 1 = 0
⇒ 3p – 16 + 1 =0
⇒ 3p – 15 = 0.
⇒ 3p=15
⇒ p = 5

Question 26.
Solve: \(0.25+\frac { 1.95 }{ x } =0.9\)
Solution:

Question 27.
Solve: \(5x-\left( 4x+\frac { 5x-4 }{ 7 } \right) =\frac { 4x-14 }{ 3 }\)
Solution:

Linear Equations in one Variable Exercise 14B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Fifteen less than 4 times a number is 9. Find the number.
Solution:
Let the required number be x
4 times the number = 4x
15 less than 4 times the number = 4x-15
According to the statement :
4x – 15 = 9
⇒ 4x = 9 + 15
⇒ 4x = 24
⇒ x = 6

Question 2.
If Megha’s age is increased by three times her age, the result is 60 years. Find her age
Solution:
Let Megha’s age = x years
Three times Megha’s age = 3x years
According to the statement :
x + 3x = 60
=> 4x = 60
=> x = 15
Megha’s age = 15 years

Question 3.
28 is 12 less than 4 times a number. Find the number.
Solution:
Let the required number be x
4 times the number = 4x
12 less than 4 times the number = 4x – 12
According to the statement
4x – 12 = 28
=> 4x = 28 + 12
=> 4x = 40
x = 10
Required number = 10

Question 4.
Five less than 3 times a number is -20. Find the number.
Solution:
Let the required number = x
3 times the number = 3x
5 less than 3 times the number = 3x – 5
According to statement :
3x – 5 = -20
=> 3x = -20 + 5
=> 3x = -15
=> x = -5
Required number = -5

Question 5.
Fifteen more than 3 times Neetu’s age is the same as 4 times her age. How old is she ?
Solution:
Let Neetu’s age = x years
3 times Neetu’s age = 3x years
Fifteen more than 3 times Neetu’s age = (3x + 15) years
4 times Neetu’s age = 4x
According to the statement :
4x = 3x + 15
=> 4x – 3x = 15
=> x = 15
Neetu’s age = 15 years

Question 6.
A number decreased by 30 is the same as 14 decreased by 3 times the number; Find the number.
Solution:
Let the required number = x
The number decreased by 30 = x – 30
14 decreased by 3 times the number = 14 – 3x
According to the statement :
x – 30 = 14 – 3x
=> x + 3x = 14 + 30
=> 4x = 44
x = 11
Required number =11

Question 7.
A’s salary is same as 4 times B’s salary. If together they earn Rs.3,750 a month, find the salary of each.
Solution:
Let B’s salary = Rs. x
A’s salary = Rs. 4x
According to the statement :
x + 4x = 3750
=> 5x = 3750
=> x = 750
4x = 750 x 4 = 3000
A’s salary = Rs. 3000
B’s salary = Rs. 750

Question 8.
Separate 178 into two parts so that the first part is 8 less than twice the second part.
Solution:
Let first part = x
Second part = 178 – x
According to the problem :
First Part = 8 less than twice the second part
x = 2(178 – x) – 8
=> x = 356 – 2x – 8
=> x+2x = 356 – 8
=> 3x = 348
=> x = 116
First Part = 116
=> Second Part = 178 – x = 178 – 116 = 62
First Part = 116
=> Second Part = 62

Alternative Method :
Let Second part = x
First part = 2x – 8
According to the problem :
x + 2x – 8 = 178
=> x + 2x = 178 + 8
=> 3x = 186
=> x = 62
First part = 2x – 8 = 2 x 62 – 8 = 124 – 8 = 116
First part = 116
Second part = 62

Question 9.
Six more than one-fourth of a number is two-fifth of the number. Find the number.
Solution:
Let the required number = x

x = 40
Required number = 40

Question 10.
The length of a rectangle is twice its width. If its perimeter is 54 cm; find its length.
Solution:
Let width of the rectangle = x cm
Length of the rectangle = 2x cm
Perimeter of the rectangle = 2 [Length + Width] = 2 [2x + x] = 2 x 3x = 6x cm
Given perimeter = 54 cm
6x = 54
=> x = 9
Length = 2x = 2 x 9 = 18 cm

Question 11.
A rectangle’s length is 5 cm less than twice its width. If the length is decreased by 5 cm and width is increased by 2 cm; the perimeter of the resulting rectangle will be 74 cm. Find the length and the width of the origi¬nal rectangle.
Solution:
Let width of the original rectangle = x cm
Length of the original rectangle = (2x – 5)cm
Now, new length of the rectangle = 2x – 5 – 5 = (2x – 10) cm
New width of the rectangle = (x + 2) cm
New perimeter = 2[Length+Width] = 2[2x – 10 + x + 2] = 2[3x – 8] = (6x – 16) cm
Given; new perimeter = 74 cm
6x – 16 = 74
=> 6x = 74 + 16
=> 6x = 90
=>x = 15
Length of the original rectangle = 2x – 5 = 2 x 15 – 5 = 30 – 5 = 25 cm
Width of the original rectangle = x = 15 cm

Question 12.
The sum of three consecutive odd numbers is 57. Find the numbers.
Solution:
Let the three consecutive odd numbers be x, x+2, x+4.
According to the statement :
x + x + 2 + x + 4 = 57
=> x + x + x = 57 – 2 – 4
=> 3x = 51
=> x = 17
Three consecutive odd numbers are 17, 19, 21

Question 13.
A man’s age is three times that of his son, and in twelve years he will be twice as old as his son would be. What are their present ages.
Solution:
Let present age of the son = x years
present age of the man = 3x years
In 12 years :
Son’s age will be = (x + 12) years
The man’s age will be = (3x + 12) years
According to the statement :
3x + 12 = 2(x + 12)
=> 3x + 12 = 2x + 24
=> 3x – 2x = 24 – 12
=> x = 12
3x = 3 x 12 = 36
Hence, present age of the man = 36 years
Present age of the son = 12 years.

Question 14.
A man is 42 years old and his son is 12 years old. In how many years will the age of the son be half the age of the man at that time?
Solution:
Man’s age = 42 years
Son’s age = 12 years
Let after x years the age of the son will be half the age of the man.
Man’s age after x years = (42 + x) years
Son’s age after x years = (12 + x) years
According to the statement :

Hence after 18 years, the age of the son will be half the age of the man

Question 15.
A man completed a trip of 136 km in 8 hours. Some part of the trip was covered at 15 km/hr and the remaining at 18 km/hr. Find the part of the trip covered at 18 km/hr.
Solution:

Question 16.
The difference of two numbers is 3 and the difference of their squares is 69. Find the numbers.
Solution:
Let one number = x
Second number = x + 3 [Difference of two numbers is 3]
According to the statement :

One number = 10
Second number = x + 3 = 10 + 3 = 13

Question 17.
Two consecutive natural numbers are such that one-fourth of the smaller exceeds one-fifth of the greater by 1. Find the numbers.
Solution:
Let two consecutive natural numbers = x, x+1

Two consecutive numbers are 24 and 25

Question 18.
Three consecutive whole numbers are such that if they be divided by 5, 3 and 4 respectively; the sum of the quotients is 40. Find the numbers.
Solution:
Let the three consecutive whole numbers be x, x + 1 and x + 2
According to the statement:

x = 50
x + 1 = 50+1 = 51
x + 2 = 50 + 2 = 52
Three consecutive whole numbers are 50, 51 and 52

Question 19.
If the same number be added to the numbers 5, 11, 15 and 31, the resulting numbers are in proportion. Find the number.
Solution:
Let x be added to each number, then the numbers will be 5 + x, 11 + x, 15 + x and 31 + x
According to the condition

1 should be added

Question 20.
The present age of a man is twice that of his son. Eight years hence, their ages will be in the ratio 7 : 4. Find their present ages.
Solution:
Let present age of son = x year
Then age of his father = 2x
8 years hence,
Age of son = (x + 8) years and age of father = (2x + 8) years
According to the condition,
\(\frac { 2x+8 }{ x+8 } =\frac { 7 }{ 4 }\)
=> 8x + 32 = 7x + 56
=> 8x – 7x = 56 – 32
=> x = 24
Present age of son = 24 years
and age of father = 2x = 2 x 24 = 48 years
Hence age of man = 48 years and age of his son = 24 years

Linear Equations in one Variable Exercise 14C – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Solve:
(i) \(\frac { 1 }{ 3 } x-6=\frac { 5 }{ 2 }\)
(ii) \(\frac { 2x }{ 3 } -\frac { 3x }{ 8 } =\frac { 7 }{ 12 }\)
(iii) (x + 2)(x + 3) + (x – 3)(x – 2) – 2x(x + 1) = 0
(iv) \(\frac { 1 }{ 10 } -\frac { 7 }{ x } =35\)
(v) 13(x – 4) – 3(x – 9) – 5(x + 4) = 0
(vi) x + 7 – \(\frac { 8x }{ 3 } =\frac { 17x }{ 6 } -\frac { 5x }{ 8 }\)
(vii) \(\frac { 3x-2 }{ 4 } -\frac { 2x+3 }{ 3 } =\frac { 2 }{ 3 } -x\)
(viii) \(\frac { x+2 }{ 6 } -\left( \frac { 11-x }{ 3 } -\frac { 1 }{ 4 } \right) =\frac { 3x-4 }{ 12 }\)
(ix) \(\frac { 2 }{ 5x } -\frac { 5 }{ 3x } =\frac { 1 }{ 15 }\)
(x) \(\frac { x+2 }{ 3 } -\frac { x+1 }{ 5 } =\frac { x-3 }{ 4 } -1\)
(xi) \(\frac { 3x-2 }{ 3 } +\frac { 2x+3 }{ 2 } =x+\frac { 7 }{ 6 }\)
(xii) \(x-\frac { x-1 }{ 2 } =1-\frac { x-2 }{ 3 }\)
(xiii) \(\frac { 9x+7 }{ 2 } -\left( x-\frac { x-2 }{ 7 } \right) =36\)
(xiv) \(\frac { 6x+1 }{ 2 } +1=\frac { 7x-3 }{ 3 }\)
Solution:

Question 2.
After 12 years, I shall be 3 times as old as 1 was 4 years ago. Find my present age.
Solution:
Let present age = x years
According to question,
(x + 12) = 3(x – 4)
x + 12 = 3x – 12
2x = 24
=> x = 12 years
Present age = 12 years

Question 3.
A man sold an article for 7396 and gained 10% on it. Find the cost price of the article
Solution:
S.P. of article = ₹ 396
Gain = 10%
Let cost price = ₹ x

Cost price of an article = ₹ 360

Question 4.
The sum of two numbers is 4500. If 10% of one number is 12.5% of the other, find the numbers.
Solution:
Let the first number = x
and the second number = y
According to question,
x + y = 4500 ……(i)
and 10% x = 12.5% y
i.e. 10x = 12.5y

x = 2500
Hence, the numbers are 2500 and 2000

Question 5.
The sum of two numbers is 405 and their ratio is 8 : 7. Find the numbers.
Solution:
Let the first number = x
and the second number = 7
According to the question, x + y = 405 ……..(i)
and the numbers are in the ratio 8 : 7


x = 189
Hence, the numbers are 189 and 216

Question 6.
The ages of A and B are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.
Solution:
Ratio in the present ages of A and B = 7 : 5
Let age of A = 7x years
Let age of B = 5x years
10 years hence,
Then age of A = 7x + 10 years
and age of B = 5x + 10 years
According to the condition,
\(\frac { 7x+10 }{ 5x+10 } =\frac { 9 }{ 7 }\)
By crossing multiplication
7(7x + 10) = 9(5x + 10)
=> 49x + 70 = 45x + 90
=> 49x – 45x = 90 – 70
=> 4x = 20
=> x = 5
Present age of A = 7x = 7 x 5 = 35 years
and present age of B = 5x = 5 x 5 = 25 years

Question 7.
Find the number whose double is 45 greater than its half.
Solution:
Let the required number = x
Double of it = 2x

Required number = 30

Question 8.
The difference between the squares of two consecutive numbers is 31. Find the numbers.
Solution:
Let first number = x
and The second number = x + 1
According to the condition,

First number = 15
and second number = 15 + 1 = 16
Hence, the numbers are 15, 16

Question 9.
Find a number such that when 5 is subtracted from 5 times the number, the result is 4 more than twice the number.
Solution:
Let the required number = x
5 times of it = 5x
Twice of it = 2x
According to the condition,
5x – 5 = 2x + 4
=> 5x – 2x = 4 + 5
=> 3x = 9
=> x = 3
Required number = 3

Question 10.
The numerator of a fraction is 5 less than its denominator. If 3 is added to the numerator, and denominator both, the fraction becomes \(\frac { 2 }{ 3 }\). Find the original fraction.
Solution:
Let denominator of the original fraction = x
Then numerator = x – 5
.

Selina Concise Chemistry Class 8 ICSE Solutions – Physical and Chemical Changes

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 2 Physical and Chemical Changes. You can download the Selina Concise Chemistry ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Chemistry for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 2 Physical and Chemical Changes

Exercise

Question 1.
Define:
(a) a physical change, (b) a chemical change.
Answer:
(a) Physical Change: A physical change is a temporary change in which no new substance is formed and the chemical composition of the original substance remains the same, even though its physical properties like colour, state, shape, size etc. might change.
(b) Chemical Change: A chemical change is permanent change in which new substances are formed whose chemical composition and physical and chemical properties are different from those of in original substance.

Question 2.
Classify the following as a physical or a chemical change.
(a) Drying of wet clothes
(b) Manufacture of salt from sea water
(c) Butter getting rancid
(d) Boiling of water
(e) Burning of paper
(f) Melting of wax
(g) Burning of coal
(h) Formation of clouds
(i) Making of a sugar solution
(j) Glowing of an electric bulb
(k) Curdling of milk
Answer:
Physical change
(a) Drying of wet clothes
(b) Manufacture of salt from sea water
(d) Boiling of water
(f) Melting of wax
(h) Formation of clouds
(i) Making of a sugar solution
(j) Glowing of an electric bulb.
Chemical change
(c) Butter getting rancid
(e) Burning of paper
(g) Burning of coal
(k) Curdling of milk

Question 3.
Fill in the blanks.
Answer:
(a) The process of a liquid changing into a solid is called freezing.
(b) A change, which alters the composition of a substances, is known as a chemical change.
(c) There is no change in the composition of the substance during a physical change.
(d) The reaction in which energy is evolved is called exothermic reaction.

Question 4.
Given reason:
(a) Freezing of water to ice and evaporation of water are physical changes.
(b) Burning of a candle is both a physical and chemical change.
(e) Burning of paper is a chemical change.
(d) Cutting of a cloth piece is a physical change, though it cannot be reversed.
Answer:
(a) Freezing of water to ice and evaporation of water are physical change because water can be brought back to its original (liquid) form by

1. We can heat the ice to bring it back to water.
2. We can cool down the vapours to bring it back to water.

(b) When a candle is lighted, some of the solid wax first melts and turns into liquid, then it turns into vapours to produce a flame. New substances CO₂ and H₂O vapours are formed alongwith the evolution of light and heat energy. This shows a chemical change. When some of the molten wax drops to the floor, it again solidifies. Which shows a physical change. Thus the melting of candle wax is a physical change and the production of CO₂ and H₂O represents chemical change.

(c) When a piece of paper is burnt a new substance ash is produced. Even when the burning is stopped, the ash cannot be changed back into paper. This shows that the formation of the ash from paper is a permanent and irreversible change.

(d) Because it does not change chemical composition of cloth and the change is only in the state, size, shape, colour, texture or the smell of some or all of the substances that undergo physical change.

Question 5.
Give four difference between physical and chemical changes.
Answer:
The differences are Physical and Chemical Changes:
Physical change

1. In a physical change no new substance is formed and the chemical composition of substance remains same. There are changes only in physical properties and state.
2. Temporaiy change which can be reversed by simple physical methods.
3. Weight of original substance doesn’t change
4. Energy like heat, light etc. may or may not be absorbed or released

Chemical change

1. In a chemical change new substance with entirely different chemical composition and properties is formed.
2. Permanent change and irreversible
3. Weight of original substances may increase or decrease
4. Energy like heat, light etc. are given out or absorbed.

Selina Concise Chemistry Class 8 ICSE Solutions – Chemical Reactions

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 6 Chemical Reactions. You can download the Selina Concise Chemistry ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Chemistry for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 6 Chemical Reactions

Exercise – I

Question 1.
(a) Define a chemical reaction.
(b) What happens during a chemical reaction?
(c) What do you understand by a chemical bond?
Answer:
(a) Any chemical change in a matter which involves transformation into one or more substances with entirely different properties is called a chemical reaction.
(b) A chemical reaction involves breaking of chemical bonds between the atoms or groups of atoms of reacting substances and rearrangement of atoms making new bonds to form new substances.

(c) A chemical bond is the attractive force that holds the atoms of a molecule together, in a compound.

Question 2.
Give one example each of which illustrates the following characteristics of a chemical reaction:
(a) evolution of a gas
(b) change of colour
(c) change in state
Answer:
(a) When Zinc reacts with dil. sulphuric acid. Hydrogen gas is evolved, with an effervescence

(b) When blue coloured copper sulphate reacts with hydrogen sulphide gas, a black coloured substance copper sulphide is formed.

(c) The reaction between hydrogen sulphide and chlorine (both gases) produces sulphur (solid) and hydrogen chloride (gas).

Question 3.
How do the following help in bringing about a chemical change?
(a) pressure (b) light
(c) catalyst (d) heat.
Answer:
(a) Some chemical reactions take place when reactants are subjected to high pressure.
e.g: Nitrogen and hydrogen when subjected to high pressure produce ammonia gas.

(b) Some chemical reactions can take place in the presence of light. Ex. Photosynthesis.

(c) A catalyst can either increases or decreases the rate of a chemical reaction and some chemical reactions need a catalyst to change the rate of the reaction, in case it is too slow or too fast.

1. Positive catalyst: When a catalyst increases the rate of reaction finely divided iron is used as a positive catalyst in the manufacturing of ammonia from hydrogen and nitrogen.
   
2. Negative Catalyst: When a catalyst decreases the rate of reaction.
   Ex. Phosphoric acid act as a negative catalyst to decrease the rate of the decomposition of hydrogen peroxide.

(d) Some chemical reactions take place only in the presence of heat.
e.g. When lead nitrate is heated, it breaks into lead monoxide, nitrogen dioxide, and oxygen.

Question 4.
(a) Define catalyst.
(b) What are (i) positive catalysts and (ii) negative catalysts? Support your answer with one example for each of them.
(c) Name three biochemical catalysts found in the human body.
Answer:
(a) Catalyst: A catalyst is a substance that either increases or decreases the rate of a chemical reaction without itself undergoing any chemical change during the reaction.

(b) (i) Positive catalyst: When a catalyst increases the rate of chemical reaction, it is called a positive catalyst.
e.g. when potassium chlorate heated to 700°C decomposes to evolve oxygen gas, when MnO₂ has added the decomposition takes place at 300°C

(ii) Negative catalyst: When a catalyst decreases the rate of chemical reaction it is called a negative catalyst.
Example. Phosphoric acid acts as a negative catalyst to decrease the rate of the decomposition of hydrogen peroxide. Alcohol too acts as a negative catalyst in certain chemical reactions.

(c) Biochemical catalysts found in the human body:

1. Pepsin
2. Trypsin
3. lipase.

Question 5.
What do you observe when
(a) dilute sulphuric acid is added to granulated zinc?
(b) a few pieces of iron are dropped in a blue solution of copper sulphate?
(c) silver nitrate is added to a solution of sodium chloride?
(d) ferrous sulphate solution is added to an aqueous solution of sodium hydroxide.
(e) solid lead nitrate is heated?
(f) when dilute sulphuric acid is added to barium chloride solution?
Answer:
(a) When Zinc reacts with dilute sulphuric acid, hydrogen gas is evolved with effervescence.
Zn + dil. H₂SO₄ → Zn SO₄ + H₂.

(b) When a few pieces of iron are dropped into a blue coloured copper sulphate solution, the blue colour of the solution fades and eventually turns green.

(c) When a solution of silver nitrate is added to a solution of sodium chloride, white insoluble ppt. of silver chloride is formed.
AgNO₃ (aq) + NaCl (aq) → AgCl (ppt) + NaNO₃ (aq)

(d) When ferrous sulphate solution is added to sodium hydroxide solution, a dirty green ppt. of ferrous hydroxide is formed.
FeSO₄ (aq) + 2NaOH (aq) → Fe(OH)₂ ↓ + Na₂SO₄(aq)

(e) When solid lead nitrate is heated, it decomposes to produce light yellow solid lead monoxide, reddish-brown nitrogen dioxide gas and colourless oxygen gas.

(f) When few drops of dilute sulphuric acid is added to barium chloride solution, a white precipitate of barium sulphate is formed.

Question 6.
Complete and balance the following chemical equations:

Answer:

Exercise – II

Question 1.
1. Fill in the blanks.
(a) A reaction in which two or more substances combine to form a single substance is called a combination reaction.
(b) A catalyst is a substance which changes the rate of a chemical reaction without undergoing a chemical change.
(c) The formation of gas bubbles in a liquid during a reaction is called effervescence
(d) The reaction between an acid and a base is called a neutralization reaction.
(e) Soluble bases are called alkalis.
(f) The chemical change involving iron and hydrochloric acid illustrates a displacement reaction.
(g) In the type of reaction called double decomposition reaction, ions two compounds exchange their positive and negative radicals ions respectively.
(h) A catalyst either increases or decreases the rate of a chemical change but itself remains unchanged at the end of the reaction.
(i) The chemical reaction between hydrogen and chlorine is a combination reaction
(j) When a piece of copper is added to silver nitrate solution, it turns blue in colour.

Question 2.
Classify the following reactions as a combination, decomposition, displacement, precipitation, and neutralization. Also, balance the equations.


Answer:

Question 3.
Define:
(a) precipitation (b) neutralization (c) catalyst
Answer:
(a) Precipitation: A chemical reaction in which two compounds in their aqueous state react to form an insoluble salt as one of the product.
Acid + Base → Salt + Water
Example.

(b) Neutralization: A chemical reaction in which a base or an alkali reacts, with an acid to produce a salt and water only.

(c) Catalyst: A catalyst is a substance that either increases or decreases the rate of a chemical reaction without itself undergoing any chemical change.

here iron act as a catalyst and increases the rate of a chemical reaction.

Question 4.
Explain the following types of chemical reactions giving two examples for each of them.
(a) combination reaction
(b) decomposition reaction
(c) displacement reaction
(d) double decomposition reaction
Answer:
(a) Combination reaction: A reaction in which two or more substances combine to form a single substance is called a combination reaction.
A + B → AB
e.g (i) When iron and sulphur are heated together, they combine to form iron sulphide.

(ii) When carbon bums in oxygen to form a gaseous compound called carbon dioxide.

(b) Decomposition reaction: A reaction in which a compound breaks up due to the application of heat into two or more simple substances is called a decomposition reaction.

e.g. (i) Mercuric oxide when heated, decomposes to form two elements mercury and oxygen

(ii) CaCO₃ when heated decomposes to calcium oxide and carbon dioxide.

(c) Displacement reaction: A reaction in which a more active element displaces a less active element from a compound is called a displacement reaction.
AB + C → CB + A
e.g. (i) Zinc, displaces copper from copper sulphate solution.
Zn + CuSO₄ (aq) → ZnSO₄ (aq) + Cu
(ii) Iron piece when added to copper sulphate solution, copper is displaced.
Fe + CuSO₄ → FeSO₄ + Cu.

(d) Double decomposition reaction: A chemical reaction in which two compounds in their aqueous state exchange their ions to form new compounds is called a double decomposition reaction.
AB + CD → CB + AD
e.g. (i) AgNO₃ + HCl  AgCl + HNO₃(aq)
(ii) NaOH (aq) + HCl (aq)  NaCl (aq) + H₂O.

Question 5.
Write the missing reactants and products and balance the equations.

Answer:

Question 6.
How will you obtain it?
(a) Magnesium oxide from magnesium.
(b) Silver chloride from silver nitrate.
(c) Nitrogen dioxide from lead nitrate.
(d) Zinc chloride from zinc.
(e) Ammonia from nitrogen.
Also, give balanced equations for the reactions
Answer:
(a) Magnesium when burnt in air (oxygen) Magnesium oxide is formed

(b) When silver nitrate solution reacts with sodium chloride, silver chloride is formed.

(c) Lead nitrate when heated nitrogen oxide is obtained

(d) Zinc when reacts with hydrochloric acid zinc chloride and hydrogen (g) is formed.
Zn + 2HCl → ZnCl₂ + H₂

(e) Nitrogen when reacts with hydrogen at 450°C and under 200 atm, ammonia is formed.

Question 7.
What do you observe when
(a) Iron nail is kept in copper sulphate solution for some time.
(b) Phenolphthalein is added to sodium hydroxide solution.
(c) Blue litmus paper is dipped in dilute hydrochloric acid.
(d) Lead nitrate is heated.
(e) Magnesium ribbon is burnt in oxygen.
(f) Ammonia is brought in contact with hydrogen chloride. gas.
Answer:
(a) A brown layer of copper gets deposited on an iron nail. This is due to a chemical reaction.
Fe (s) + CuSO₄ (aq) → FeSO₄ (aq) + Cu (s)
(b) Solution turns pink.
(c) Blue litmus turns red in an acid solution.
(d) The pale yellow solid is lead monoxide, the reddish-brown gas is nitrogen dioxide and the colourless gas is oxygen.

(e) Magnesium ribbon bums with a dazzling white light and produces a white powder which is magnesium oxide.
The reaction can be represented as
2Mg + O₂ → 2MgO (white powder)
(f) Ammonia and hydrogen chloride, both compounds, combine to form a compound, ammonium chloride.

Question 8.
Give reason:
(a) A person suffering from acidity is advised to take an antacid.
(b) Acidic soil is treated with quick lime.
(c) Wasp sting is treated with vinegar.
Answer:
(a) An antacid neutralizes stomach acidity.
(b) If the soil is acidic it can be treated with a base like quick lime, to make it neutral.
(c) Wasp stings are alkaline and can be neutralized by vinegar which is a weak acid.

Question 9.
What is meant by the metal reactivity series? State its importance, (any two points).
Answer:
A list in which the metals are arranged in the decreasing order of their chemical reactivity is called the metal reactivity series.

Special features of the activity series:

1. The ease with which a metal in solution loses an electron(s) and forms a positive ion decreases down the series, i.e. from potassium to gold.
2. Hydrogen is included in the activity series because, as metals do, it too loses an electron and becomes positively charged (H⁺) in most chemical reactions.
3. The series facilitates the comparative study of metals in terms of the degree of their reactivity.
4. The compounds of the metals (oxides, carbonates, nitrates and hydroxides) too can be easily compared.

Question 10.
What are oxides? Give two examples of each of the following oxides.
(a) Basic oxide (b) Acidic oxide
(c) Amphoteric oxide (d) Neutral oxide
Answer:
An oxide is a compound that essentially contains oxygen in its molecule, chemically combined with a metal or a non-metal.

Question 11.
Define exothermic and endothermic reactions. Give two examples of each.
Answer:
Exothermic reactions: The chemical reaction in which heat is given out is called exothermic reactions. It causes rise in temperature. .
e.g. (i) When carbon bums in oxygen to form carbon dioxide, a lot of heat is produced.
C + O₂ → CO₂ + heat.
When water is added to quicklime a lot of heat is produced which boils the water.
CaO + H₂O → Ca (OH)₂ + Heat.

Endothermic reaction: A chemical reaction in which heat is absorbed is called endothermic reaction. It causes fall in temperature.
e.g. (i) When nitrogen and oxygen together are heated to a temperature of about 3000°C, nitric oxide gas is formed.
N₂ + O₂ + heat → 2NO (g)
(ii) Decomposition of calcium carbonate into carbon dioxide and calcium oxide when heated to a 1000°C.
CaCO₃ + Heat → CaO (s) + CO₂ (g)

Question 12.
State the effect of:
(a) an endothermic reaction
(b) an exothermic reaction on the surroundings.
Answer:
(a) Carbon dioxide present in the atmosphere is trapped by infrared radiations, gives rise to temperature which is exothermic reaction.
(b) The melting of glaciers by global warming.

Question 13.
What do you observe when
(a) an acid is added to a basic solution.
(b) ammonium chloride is dissolved in water.
Answer:
(a) A chemical reaction in which a base or an alkali reacts with an acid to produce a salt and water.
Acid + Base → Salt + Water
(b) Dissolution of ammonium chloride in water is an endothemic reaction in which heat energy is absorbed.

Selina Concise Chemistry Class 8 ICSE Solutions – Matter

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 1 Matter. You can download the Selina Concise Chemistry ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Chemistry for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 1 Matter

Exercise

Question 1.
Define:
(a) matter
(b) intermolecular force of attraction.
Answer:
(a) Matter is anything which has mass, occupies space and can be percieved by our senses.
Example: Air, Book.
(b) The molecules of matter are always in motion and attract each other with a force called intermolecular force of attraction due to which they are held together.

Question 2.
What are the three states of matter ? Define each of them with two examples.
Answer:
The three states of matter are:
solids, liquids and gases

• Solids — A solid has a definite shape and definite volume.
  Example – wood, stone, iron, ice etc.
• Liquid — A liquid has a definite volume but not definite shape.
  Example — water, juice, milk, oil, etc.
• Gases — A gas neither has definite shape nor a definite volume.
  Example – air, hydrogen, oxygen, watervapour etc.

Question 3.
Define interconversion of states of matter. What are the two factors responsible for the change of states of matter?
Answer:
The process by which matter changes from one state to another and back to original state, without any change in its chemical composition is called interconversion state of matter.
Two factors responsible for change of state of matter are: change in
(i) Temperature (ii) Pressure

Question 4.
State the main postulates of kinetic theory of matter.
Answer:
The main postulates of the theory are:

1. Matter is composed of very small particles called atoms and molecules.
2. The constituent particles of a kind of matter are identical in all respects.
3. These particles have space or gaps between them which is known as interparticular or intermolecular space.
4. There exists a force of attraction between the particles of matter which holds them together. This force of attraction is known as interparticular or intermolecular force of attraction.
5. Particles of matter are always in a state of random motion and possess kinetic energy, which increases with increase in temperature and vice-versa.

Question 5.
What happens to water if
(a) it is kept in a deep freezer
(b) it is heated
Explain the phenomenon of change of state of water.
Answer:
(a) When water is kept in a deep freezer, it gets cooled and change into ice at 0°C ice.

(b) Water on heating changes into steam at 100°C

Phenomenon of change of state of water:
Water is a liquid under ordinary conditions but, when it is kept in a deep freezer, it changes into ice at 0°C and when ice is kept at room temperature again changes back into liquid water.
Similarly, water on heating change into steam at 100°C, which on cooling changes back into liquid water. But there is no change in the chemical composition of water. When its state changes from liquid to solid or liquid to gaseous state.

Question 6.
(a) State the law of conversation of mass.
(b) What do you observe when barium chloride solution is mixed with sodium sulphate solution?
Answer:
(a) “Matter can neither be created nor be destroyed in a chemical reaction”. However, it may change from one form to another in the process.
It can also be stated as, “In a chemical reaction, the total mass of the reactants is equal to the total mass of the products”.

(b)

We will observe that a white insoluble solid (precipitate) of barium sulphate is formed along with a solution of sodium chloride. Wait for ten minutes to complete the reaction and the solid formed to settle down.
Weigh the content again and note the reading.
We will observe that,
total mass of the apparatus + reactants = total mass of apparatus + products
Hence the law of conservation of mass is verified.

Question 7.
Give reasons:
(a) A gas can fill the whole vessel in which it is enclosed.
(b) Solids cannot be compressed.
(c) Liquids can flow.
(d) When magnesium is burnt in air, there is an increase in mass after the reaction.
Answer:
(a) Because, in gases, the molecules are free to move.
They are not stuck to each other and the intermolecular force of attraction is least in the gases. So the gas almost filled the whole vessel in which it is enclosed.

(b) In solids, particles are closely packed. There is a strong force of attraction and the intermolecular space is almost zero. Therefore the molecules are not free to move, which makes them hard and rigid. So solids can not be compressed.

(c) In liquids intermolecular force is weaker because the particles are not closely packed and hence there is large intermolecular space. So molecules in a liquids can move randomely and hence liquids can flow easily.

(d) When magnesium ribbon is burnt in air, a white solid, magnesium oxide is formed. The mass of magnesium oxide is more than the mass of magnesium. This is because mass of oxygen used is not taken. If that is considered, the total mass of the reactants and the products is found to be almost equal.

Question 8.
Fill in the blanks:

(a) The change of a solid into a liquid is called melting or fusion.
(b) The process in which a solid directly changes into a gas is called sublimation.
(c) The change of water vapour into water is called condensation.
(d) The temperature at which a liquid starts changing into its vapour state is evaporation or vaporisation.

Question 9.
Give two examples for each of the following:
(a) The substances which sublime.
(b) The substances which do not change their state on heating.
Answer:
(a) Camphor, iodine, naphthalene, ammonium chloride, dry ice (solid carbon dioxide), etc.
(b) Gases do not change their state on heating.
Example: O₂.

Question 10.
Define:
(a) Diffusion.
(b) Brownian motion.
Answer:
(a) Diffusion: The intermixing of two or more substances due to the motion of their particles in order to get a uniform mixture is called ‘diffusion’.
(b) Brownian motion: The haphazard, random motion of suspended particles on the surface of a liquid or in air is called ‘Brownian motion’.

Question 11.
When sodium chloride is added to a definite volume of water and stirred well, a solution is formed, but there is no increase in the level of water. Why?
Answer:
This is because there is some space between the particles of water in which the salt particles get accomodated when dissolved.

Question 12.
What do you observe when a gas jar which appears empty is inverted over a gas jar containing Bromine vapours? Name the phenomenon.
Answer:
When a gas jar full of bromine vapours (reddish brown) is inverted over a gas jar containing air over it. It is observed that after sometime, the reddish brown vapours of bromine also spread out into the upper jar. This mixing is called diffusion. The rate of diffusion is the fastest in gases and the slowest in solids. It increases with an increase in temperature.

Question 13.
Why can a piece of chalk be broken easily into smaller pieces while a coal piece cannot be broken easily?
Answer:
The particles of matter have force acting between them. This force keeps the particles together. The strength of this force of attraction is lesser in chalk, hence it could be broken easily into smaller pieces.
But the strength of inter-molecular force of attraction is very strong in coal, therefore it is not possible to break them into small pieces.

Selina Concise Physics Class 8 ICSE Solutions – Light Energy

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Physics Chapter 5 Light Energy. You can download the Selina Concise Physics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Physics for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 8 Physics Chapter 5 Light Energy

• Light is a form of energy which produces in us the sensation of sight i.e. we can see objects only when light falls on them and then reflected into our eye.
•  Velocity of light in air or in vacuum is 300000 km per second.
  Or
  3 x 10⁸ ms^-1
• As light passes into different mediums its speed changes and depends upon the density of medium i.e. it decreases with increase in density i.e. it is 2.25 × 10⁸ m/s in water and 2 x 10⁸ ms^-1 in glass as water is
  denser than air ( \(_{ w }^{ a }{ \mu }\) = 1.33 ) and glass is still optically denser than water
  ( \(_{ g }^{ a }{ \mu }\) =1.5 ) i.e. slower in water and still slower in glass.
• Light travels in a straight line.
• As light travels from one transparent medium to other transparent medium and falls oblique at another medium, its path changes and this change in path is called REFRACTION OF LIGHT.
•  When ray of light travels from RARER (less-denser) to DENSER medium, it bends TOWARD the normal AND when it travels from a DENSER to a RARER medium it bends away from NORMAL
•  ANGLE of INCIDENCE : “The angle which incident ray makes with normal”. “∠i”
•  ANGLE OF REFRACTION: “The angle which refracted ray makes with normal” “ ∠r ”
  ∠i is not equal to ∠r
•  LAWS OF REFRACTION or SNELL’S LAWS OF REFRACTION:
  (i) Incident ray, normal at the point of incidence and Refracted ray all lie in the same plane.
  (ii) Ratio of sine of angle of incidence to the sine of angle of refraction is constant.
• 
•  EFFECTS OF REFRACTION :
  (i) A coin placed in water appears to be raised.
  (ii) Swimming pool seen from above appears SHADOW.
  (iii) A pencil in water appears to be bent.
  (iv) MIRAGE in desert, EARLY Sunrise, LATE SUN set are all due to REFRACTION of light.
• White light is a band of seven colours-VIBGYOR. Speed of all colours of the white light in AIR or VACUUM is same, but different different transparent mediums.
•  In glass or water Speed of VIOLET colour is MINIMUM and speed of RED light is MAXIMUM
• Refractive index of medium is minimum for VIOLET light and R.I. of medium is maximum for red light.
• DISPERSION: “The splitting (breaking) of white light into seven colours is called DISPERSION OF LIGHT.
•  CAUSE OF DISPERSION: Speed of different colours is different in glass or water and different colours get separated from each other on refraction at second surface of glass prism.

Test yourself

A. Objective Questions

1. Write true or false for each statement

(a) Water is optically denser than glass.
Answer. False.
Water is optically denser than air.

(b) A ray of light when passes from glass to air, bends towards the normal.
Answer. False.

(c) The speed of light is more in glass than in water.
Answer. False.

(d) The depth of a pond when seen from above appears to be less.
Answer. True.

(e) Light travels at a lower speed in water than in air.
Answer. True.

(f) Light travels in the same straight line path while passing through different media.
Answer. False.

(g) The angle formed between the normal and the refracted ray is known as the angle of incidence.
Answer. False.

(h) At the point of incidence, a line drawn at right angles to the surface, separating the two media, is called the normal.
Answer. True.

(i) Image is formed by a mirror due to refraction of light.
Answer. False.

(j) Rays of light incident parallel to the principal axis pass through the focus after reflection from a concave mirror.
Answer. True.

(k) A convex mirror is used as a shaving mirror.
Answer. False.

(l) The focal length of a convex mirror is equal to its radius of curvature.
Answer. False.

(m) A concave mirror converges the light-rays, but a convex mirror diverges them.
Answer. True.

(n) A virtual image formed by a spherical mirror is always erect and situated behind the mirror.
Answer. True.

2. Fill in the blanks

(a) Water is optically denser than air.
(b) Air is optically rarer than glass.
(c) When a ray of light travels from water to air, it bends away from the normal.
(d) When a ray of light travels from air to glass, it bends towards the normal.
(e) When white light passes through a prism, it disperses
(f) The splitting of white light into its constituent colours is called dispersion.
(g) A concave mirror is obtained on silvering the outer surface of a part of a hollow glass sphere.
(h) Radius of curvature of a spherical mirror is two times its focal length.
(i) The angle of incidence for a ray of light passing through the centre of curvature of a spherical mirror is 0°
(j) A convex mirror always forms a virtual image.
(k) A concave mirror forms a virtual image for an object placed between pole and focus.

 

3. Match the following

4. Select the correct alternative

(a) The speed of light in air or vacuum is

1. 3 × 10⁸ M s^-1
2.  2.25 × 10⁸ m s^-1
   
3.  332 ms^-1
4.  2.0 × 10⁸ ms^-1

(b) A ray of light moving from an optically rarer to a denser medium

1.  bends away from the normal
2.  bends towards the normal
3.  remains undeviated
4.  none of the above

(c) The angle between the normal and refracted ray is called

1.  angle of deviation
2.  angle of incidence
3.  angle of refraction
4.  angle of emergence.

(d) The property of splitting of white light into its seven constituent colours is known as

1.  rectilinear propagation
2.  refraction
3.  reflection
4.  dispersion

(e) The seven colours in the spectrum of sunlight in order, are represented as :

1.  VIBGYOR
2.  VIGYBOR
3.  BIVGYOR
4.  RYOBIVG

(f) A ray of light passing through centre of curvature of a spherical mirror, after reflection

1. passes through the focus
2.  passes through the pole
3.  becomes parallel to the principal axis
4.  retraces its own path.

(g) If the radius of curvature of a concave mirror is 20 cm, its focal length is:

1.  10 cm
2.  20 cm
3.  40 cm
4.  80 cm

(h) The image formed by a convex mirror is

1.  erect and diminished
2.  erect and enlarged
3.  inverted and diminished
4.  inverted and enlarged.

(i) The image formed by a concave mirror is of the same size as the object, if the object is placed

1. at the focus
2. between the pole and focus
3.  between the focus and centre of curvature
4.  at the centre of curvature.

(j) A convex mirror is used

1.  as a shaving mirror
2.  as a head mirror by a dentist
3.  as a rear view mirror by a driver
4.  as a reflector in torch.

Selina Concise Chemistry Class 8 ICSE Solutions – Elements, Compounds and Mixtures

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ICSESolutions.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 3 Elements, Compounds and Mixtures. You can download the Selina Concise Chemistry ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Chemistry for Class 8 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 8 Chemistry Chapter 3 Elements, Compounds and Mixtures

Exercise 3(A)

Question 1.
Define: (a) Elements (b) Compounds
Answer:
(a) Elements: Element is a substance which cannot be broken further into simpler substances and has a definite set of properties. Elements are made up of only one kind of atoms.
(b) Compounds: Compounds are pure substances composed of two or more elements in definite proportion by mass and has properties, entirely different from those of its constituents elements.
Compound, are made up of different types of atoms combined chemically.

Question 2.
Give two examples for each of the following:
(a) Metals (b) Non-metals
(c) Metalloids (d) Inert gases
Answer:
(a) Metals: Iron, silver, gold.
(b) Non-metals: Carbon, sulphur, oxygen.
(c) Metalloids: Antimony, silicon, boron.
(d) Inert gases: Helium, argon, neon.

Question 3.
Differentiate between:
(a) Pure and impure substances
(b) Homogenous and heterogenous substances
Answer:
(a) Pure substances —

1. Pure substances have definite composition and definite physical and chemical properties.
2. They are all homogeneous i.e. their composition is uniform throughout the bulk.
3. Examples: Elements and compounds.

Impure substances —

1. Impure substances are made up of two or more pure substances mixed together in any proportion.
2. They may be homogeneous or hetergeneous i.e. their composition is not uniform throughout the bulk.
3. They are all mixtures.
   Examples: air, sea water, petroleum, a solution of sugar in water are all impure substances.

(b) Homogeneous mixture — is a mixture where the components that make up the mixture are uniformly distributed throughout the mixture.
Example — air, sugar water, rain water.
Heterogeneous mixture — is a mixture, where the components of the mixture are not uniform or have localized regios with different properties.
Example—Cereal in milk, vegetable soup.

Question 4.
Write the chemical name of the following and also give their molecular formulae:
(a) Baking soda (b) Vinegar
(c) Marble (d) Sand
Answer:
(a) Sodium bicarbonate (Baking soda) — NaHCO₃
(b) Acetic acid (Vinegar) — CH₃COOH
(c) Calcium carbonate (Marble) — CaCO₃
(d) Silicon dioxide (Sand) — SiO₂

Question 5.
Name:
(a) a soft metal
(b) a metal which is brittle
(c) a non-metal which is lustrous
(d) a liquid metal
(e) a metal which is a poor conductor of electricity.
(f) a non-metal which is a good conductor of electricity.
(g) a liquid non-metal
(h) the hardest naturally occurring substance
(i) an inert gas
Answer:
(a) Gold
(b) Zinc
(c) Iodine
(d) Mercury
(e) Tungsten
(f) Graphite
(g) Bromine
(h) Diamond
(i) Neon, helium

Question 6.
How is sodium chloride different from its constituent elements ?
Answer:
The properties of sodium chloride are completely different from those of sodium and chlorine. Sodium is a soft, highly reactive metal. Chlorine is a poisonous non-metallic gas while sodium chloride is a very useful non poisonous compound which is added to our food to get minerals and also to add taste to it.

Question 7.
Why is iron sulphide a compound ?
Answer:
Iron sulphide is a compound which can be broken into the elements iron and sulphur they both have different properties. The properties of compound are entirely different from there of its constituents elements.

Exercise 3(B)

Question 1.
Classify the following substances into compounds and mixtures:
Answer:
Carbon dioxide, air, water, milk, common, salt, blood, fruit juice, iron sulphide.
Carbon dioxide — (Compound)
air — (Mixture)
water — (Compound)
milk — (Mixture)
common salt — (Compound)
blood — (Mixture)
fruit juice — (Mixture)
iron sulphide — (Compound)

Question 2.
Give one example for each of the following types of mixtures
(a) solid-solid homogenous mixture
(b) solid-liquid heterogenous mixture
(c) misicible liquids
(d) liquid-gas homogenous mixture
Answer:
(a) Solid-solid homogenous mixture — Alloys of metals e.g. brass, bronze stainless steel etc.
(b) Solid-liquid heterogenous mixture — Sand and water, mud and water, sugar and oil.
(c) Misicible liquids — water and ethanol.
(d) Liquid-gas homogenous mixture — Air

Question 3.
Suggest a suitable technique to separate the constituents of the following mixtures. Also give the reason for selecting the particular method.
(a) Salt from sea water
(b) Ammonium chloride from sand
(c) Chalk powder from water
(d) Iron from sulphur
(e) Water and alcohol
(f) Sodium chloride and potassium nitrate
(g) Calcium carbonate and sodium chloride
Answer:

(a) The technique used to separate the salt from seawater is Evaporation.
Reason – Because this method is used to separate the components of the homogeneous solid-liquid mixture. In this method, sea water is collected in a shallow bed and allowed to evaporate in the sun. When all the water is evaporated, salt is left behind. By this method, we only get solid and liquid is evaporated in its vapour form.

(b) Technique used to separate Ammonium chloride from sand is sublimation.
Because this method is used for solid mixtures in which one of the components can sublime on heating. In this method, Ammonium chloride changes into vapours on heating and salt is left behind.

(c) Technique used to separate chalk powder from water is filtration.
Reason – Because this process is used to separate the components of a heterogeneous solid-liquid mixture in which solids are lights and insoluble in liquids. Substances used as filters are sand filter paper at C. These filters allows the liquid to pass through them, but not solids.

(d) Technique to separate iron from sulpher is magnetic separation.
Because, this method is used when one of the component of mixture is Iron. Iron gets attracted towards the magnet and hence get separated.

(e) Technique used to separate water and Alcohol is Fractional Distillation.
Because in this method, the vapours of water is left behind in the original vessel as the alcohol boils at lower temperature than water. Thus these two liquids can be separated.

(f) Technique used is Fractional-crystallisation.
Because: This method is used when solubility of solid components of mixture and different in the same solvent. Here, sodium chloride and potassium nitrate. Both are soluble in water but solubility of potassium nitrate is more.

(g) Technique used is Solvent Extraction Method: Because, by this method, salts get dissolve in water while calcium carbonate being insoluble in water settles down in the container. And hence get separated about.

Question 4.
(a) Define mixture.
(b) Why is it necessary to separate the constituents of a mixture.
(c) State four differences between compounds and mixtures.
Answer:
(a) “Mixtures can be defined as. a kind of matter which is formed by mixing two or more pure substances (elements and compounds) in any proportion, such that they do not undergo any chemical change and retain their individual properties. Therefore they are impure substances.

(b) Because: The mixtures contain unwanted substances which may be harmful and may degrade the properties of mixtures. So we, need to separated them and extract useful substances.
This is necessary because
(i) It removes unwanted and harmful substances
(ii) to obtain pure and useful substances them.
Example: Sea water is rich in common salt which is an important ingredient of our food to add taste and nutrients. But sea water, cannot be directly used to get the salt.
Hence, it is necessary to separate both.

(c) Compound

1. A compound is formed from its constituent elements as a result of chemical reaction.
2. A compound is always homogeneous in nature.
3. In a compound the elements are present in a fixed ratio by weight.
4. The components of a compound can’t be separated by physical methods but can be separated by chemical methods only.
5. The properties of a compound are different from those of its elements.
6. The formation of a compound from its elements is accompanied by energy changes.

Mixture

1. A mixture is obtained form its (elements, compounds) components as a result of physical change.
2. The mixtures can be homogeneous or heterogeneous.
3. In a mixture the components can be present in any ratio.
4. The components of a mixture can be separated by physical methods only.
5. The properties of a mixture lie between those of-its components.
6. The formation of a mixture from its constituents is not accompanied by energy changes.

Question 5.
(a) What is chromatography ? For which type of mixture is it used ?
(b) What are the advantages of chromatography.
Answer:
(a) This is one of the latest techniques to separate the coloured components of a mixture when all the components are very similar in their properties. Example: Components of ink are separated by this method. Ink is a mixture of different dyes, which are separated by chromatography because some of the dyes are less soluble and some are more soluble in a solvent.

(b)

1. A very small quantity of the substance can be separated.
2. Components with very similar physical and chemical properties can be separated.
3. It identifies the different constitutes of a mixture.
4. It also helps in quantitive estimation of components of a mixture.

6. Choose the most appropriate answer from the options given below:

(a) a mixture of sand and ammonium chloride can be separated by

1. filtration
2. distillation
3. sublimation
4. crystallisation

(b) A pair of metalloids are

1. Na and Mg
2. B and Si
3. C and P
4. HeandAr

(c) Which of the following property is not shown by compounds?

1. They are heterogeneous.
2. They are homogeneous.
3. They have definite molecular formulae.
4. They have fixed melting and boiling points.

(d) A solvent of Iodine is

1. Water
2. Kerosene oil
3. Alcohol
4. Petrol

(e) Which of the gas is highly soluble in water ?

1. Ammonia
2. Nitrogen
3. Carbon monoxide
4. Oxygen

Selina Concise Physics Class 7 ICSE Solutions – Heat

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Physics. You can download the Selina Concise Physics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Physics for Class 7 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina Concise ICSE Solutions for Class 7 Physics Chapter 5 Heat

• Points to Remember
•  Heat is a form of energy that leads to the sensations of hotness or coldness.
•  Temperature is the degree of hotness and coldness of a body.
•  Thermometer is used to measure temperature.
•  The S.I. unit of temperature is °C.
•  The most common liquid for a thermometer is mercury.
•  The main sources of heat are (i) Fire (ii) Sun (iii) Electricity.
•  Those substances which can easily catch fire are called inflammable substances.
•  Those substances which are fire resistant are called non-inflammable substances.
•  The fixed temperature at which freezing of liquid occurs is known as freezing point.
•  The temperature at which vapourisation occurs is known as the boiling point.
•  Substances through which heat is easily conducted are called good conductors e.g. silver, gold, copper etc.
•  Substances through which heat is not easily conducted are called Insultors.
•  Radiation is the process of transfer of heat from a hot body to a cold body without affecting the intermediate medium.

Test Yourself

A. Objective Questions 

1. Write true or false for each statement

(a) On touching a lump of ice, we feel cold because some heat passes from our body to the ice.
Answer. True.

(b) Heat flows from a body at a high temperature to a body at a low temperature when they are kept in contact. .
Answer. True.

(c) All solids expand by the same amount when heated to the same rise in temperature.
Answer. False.

(d) Telephone wires are kept tight between the two poles in summer.
Answer. False.

(e) Equal volumes of different liquids expand by different amounts when they are heated to the same rise in temperature.
Answer. True.

(f) Solids expand the least and gases expand the most on being heated.
Answer. True.

(g) A mercury thermometer makes use of the property of expansion of liquids on heating.
Answer. True.

(h) Kerosene contracts on heating.
Answer. False.

(i) Water is a bad conductor of heat.
Answer. True.

(j) Medium is necessary for the transfer of heat by radiation.
Answer. False.

(k) Land and sea breezes are convection currents of cold and warm air.
Answer. True.

(l) Liquids are heated by conduction and radiation.
Answer. False.

(m) Black surfaces are the poor absorbers of heat radiations.
Answer. False.

2. Fill in the blanks

(a) Heat is a form of energy.
(b) Temperature determines the degree of hotness or coldness of a body.
(c) On heating a body, its temperature rises.
(d) We use a thermometer for measuring the temperature of a body.
(e) The S.I. unit of temperature is kelvin.
(f) In a thermometer, the commonly used liquid is mercury.
(g) The temperature of a normal human body is 37 °C.
(h) A person is said to have fever if his body temperature is more than 98.6
(i) A hot metallic piece is placed in tap water contained in a bucket. Heat will flow from metallic piece to water.
(j) The temperature of boiling water is 100°C.
(k) Liquids expand more than the solids.
(l) Gases expand more than the liquids.
(m) Heat transfer in solids is by conduction.
(n) Heat transfer in liquids and gases is by convection.
(o)Metals are conductors of heat.
(p) Still air is an insulator of heat.
(q) Black and dull surfaces are good absorbers of heat.

3. Match the following

4. Select the correct alternative

(a) If we add a lump of ice to a tumbler containing water,

1. heat flows from water to ice
2.  heat flows from ice to water
3.  heat flows from water to ice if water is more
4.  heat flows from ice to water if ice is more

(b) The temperature of pure melting ice is

1.  0°C
2.  100°C
3.  95°C
4.  98.6°F

(c) A thermometer uses

1.  water
2. mercury
3.  air
4.  none of the above

(d) Which of the statement is correct

1.  Iron rims are cooled before they are placed on cart wheels
2.  A glass stopper gets tight on warming the neck of the bottle
3.  Telephone wires sag in winter, but become tight in summer
4. A little space is left between two rails on a railway track

(e) Heat in a liquid is transferred by

1.  conduction
2. convection
3.  radiation
4.  conduction and radiation

(f) In the process of convection, heat travels

1.  sideways
2.  downwards
3. upwards
4.  in all directions

(g) The vacuum kept in between the walls of a thermos flask reduces the heat transfer by

1.  conduction only
2.  convection only
3.  radiation only
4. conduction and convection

B. Short/Long Answer Questions

Question 1.
What is heat ? State its S.I. unit.
Answer:
Heat is a form of energy which flows. It is the energy of motion of molecules constituting the body.
The unit of heat is same as that of energy, The S.I. unit of heat is joule (abbreviated as J) and other common units of heat are calorie and kilo calorie, where 1 k cal = 1000 cal.

Question 2.
What is meant by the term temperature.
Answer:
Temperature is a quantity which tells the thermal state of a body (i.e. the degree of hotness or coldness). It determines the direction of flow of heat when the two bodies at different temperatures are placed in contact.

Question 3.
State the three units of temperature.
Answer:
The S.I. unit of temperature is kelvin or K. The other most common unit of temperature is degree Celsius (°C) and degree Fahrenheit (°F).

Question 4.
Name the instrument used to measure the temperature of a body.
Answer:
To measure the temperature of a body with the help of a thermometer.

Question 5.
Name two scales of temperature. How are they inter-related?
Answer:
Two scales of temperature are
(i) Celsius (ii) Fahrenheit
Relation:

Question 6.
How is the size of a degree defined on a Celsius scale ?
Answer:
The interval between the ice point and steam point divided by 100 (hundred) equal parts is called a degree on the Celsius scale.

Question 7.
How is the size of a degree defined on a Fahrenheit scale?
Answer:
The interval between the ice point and steam point divided into 180 equal parts is called a degree on the Fahrenheit scale.

Question 8.
State the temperature of (i) ice point and (ii) steam point, on the Celsius scale.
Answer:
(i) Ice point. Is the the mark on Celsius scale at which ice melts. Ice point on the Celsius scale is 0°C.
(ii) Steam point. On the Celsius scale is the mark at which water changes into steam at normal atmospheric pressure. On Celsius scale it is 100°C.

Question 9.
Write down the temperature of (i) lower fixed point, and (ii) upper fixed point, on the Fahrenheit scale.
Answer:
Lower fixed point: On the Falirenheit scale is the mark at which pure ice melts. It is 32°F on Fahrenheit scale.
Upper fixed point: On the Fahrenheit scale is the mark at which water starts changing into steam at normal atmospheric pressure. It is 212°F.

Question 10.
What is the Celsius scale of temperature ?
Answer:
Celsius scale is that which has ice point as 0°C and steam point marked as 100°C.

Question 11.
What is the Fahrenheit scale of temperature ?
Answer:
Fahrenheit scale is that which has ice point as 32°F and the steam point marked as 100°C.

Question 12.
What is the Kelvin scale of temperature ?
Answer:
On Kelvin scale of temperature zero mark is when no molecular motion occurs. Ice point is at 273 and steam point is at 373 K. Thus 0 K = – 273°C and one degree on Kelvin scale is same as one degree on Celsius scale.

Question 13.
The fig. shows a glass tumbler containing hot milk which is placed in a tub of cold water. State the direction in which heat will flow.

Answer:
When we bring two objects of different temperature together, energy will always be transferred from hotter to the cooler object.
Here, also heat will flow from hot milk tumbler to tub of cold water.

Question 14.
Draw a neat labelled diagram of a laboratory thermometer.
Answer:

Question 15.
Write down the body temperature of a healthy person.
Answer:
The temperature of a healthy persons is 98.6 degrees fahrenheit or 37.0 degree Celsius or 310 k.

Question 16.
What do you understand by thermal expansion of a substance ?
Answer:
The expansion of a substance when, heated, is called thermal expansion.
Or
Thermal expansion is the tendency of matter to change .in shape, area and volume in response to a change in temperature.

Question 17.
Name two substances which expand on heating.
Answer:
Mercury and Aluminium wire.

Question 18.
Why do telephone wires sag in summer ?
Answer:
The telephone wires will sag in summer due to expansions and will break in winter due to contraction.
Therefore, while putting up the wires between the poles, care is taken that in summer they are kept slightly loose so that they may not break in winter due to contraction.
While in winter they are kept light so that they may not sag too much in summer due to expansion.

Question 19.
Iron rims are heated before they are fixed on the wooden wheels. Explain the reason.
Answer:
The wooden wheels of a bullock-cart are fitted with iron tyres. To ensure a tight fit, the tyre is made slightly smaller in diameter than the wheel. The tyre is first heated due to which it expands. The heated tyre is then fitted on the wheel. When the tyre cools, it contracts and makes a tight fit on the wheel.

Question 20.
Why are gaps left between successive rails on a railway track ?
Answer:
The rails of railway track are made of steel. While laying the railway track, a small gap is left between the two successive length of rails. The reason is that the rails expand in summer. The gap is provided to allow for this expansion. If no gap is left, the expansion in summer will cause the rails to bend sideways. This may result in a train accidents.

Question 21.
A glass stopper stuck in the neck of a bottle can be removed by pouring hot water on the neck of the bottle. Explain why ?
Answer:
When hot water is poured over the neck of the bottle, it expands. As a result the stopper gets loosened and can be removed easily.

Question 22.
Why is a cement floor laid in small pieces with gaps in between?
Answer:
The floor is laid in small pieces with gaps in between to allow for the expansion during summer. However glass strips can be placed in the gaps.

Question 23.
One end of a steel girder in a bridge is not fixed, but is kept on roliers. Give the reason.
Answer:
In the construction of a bridge, steel girders are used. One end of the girder is fixed into the concrete or brick pillars and its other end is not fixed, but it is placed on rollers. The reason is that if there is any rise (or fall) in temperature of atmosphere, the girder can freely expand (or contract) without affecting the pillars.

Question 24.
Describe one experiment to show that liquids expand on heating.
Answer:
(i) Take an empty bottle with a tight fitting cork having a hole drilled in its middle, a drinking straw, two bricks, a wire guaze and a burner.
(ii) Fill the bottle completely with water and add few drops of ink in it to make it coloured.
(iii) Fix the cork in the mouth of the bottle and pass the drinking straw through the cork. Put some molten wax around the hole so as to avoid the leakage of water.
(iv) Pour some more water into the drinking straw so that water level in the straw can be seen. Mark the water level in the straw as shown in Figure.

(v) Place the bottle on the wire gauze kept over the two bricks as shown in Figure. Then heat the bottle by means of a burner.
(vi) Look at the level of water in the straw.
You will notice that as the water is heated more and more, the level of water in the drinking straw rises. This shows that water expands on heating.

Question 25.
State one application of thermal expansion of liquids.
Answer:
Mercury is a metal found in liquid state. It expands more and uniformly over a wide range of temperature. So mercury is used as thermometric liquid.

Question 26.
Describe an experiment to show that air expands on heating.
Answer:
(i) Take an empty bottle. Actually the empty bottle contains air. Attach a rubber balloon to its neck as shown in Figure. Initially, the balloon is deflated.
(ii) Place the bottle in a water bath containing boiling water. After some time you will notice that the balloon gets inflated as shown in Figure. The reason is that the air inside the bottle expands on heating and it fills the balloon.

(iii) Take the bottle out of the water bath and 7 allow it to cool by itself. We will notice that the balloon gets deflated and it collapses. This is because the air inside the balloon and the bottle, has contracted on cooling. The air from balloon passes to the bottle, so the balloon gets deflated.

Question 27.
An empty glass bottle is fitted with a narrow tube at its mouth. The open end of the tube is kept in a beaker containing water. When the bottle is heated, bubbles of air are seen escaping into the water. Explain the reason.
Answer:
When the bottle is heated, bubbles of air are seen escaping into the water. This happens because the air present in glass bottle expands on heating and tries to escape out through the tube into the water.

Question 28.
State which expands more, when heated to the same temperature : solid, liquid or gas ?
Answer:
Gases expand much more than the liquids and the solids. Like liquids, the gases do not have a definite shape, so they also have only the cubical expansion.

Question 29.
Name the three modes of transfer of heat.
Answer:
There are three modes of transfer of heat (i) Conduction (ii) Convection (iii) Radiation.
(i) Conduction “is that mode of transfer of heat, when heat travels from hot end to cold end from particle to particle of the medium, without actual movement of particles.”
(ii) Convection. “Is a process of transfer of heat by actual move-ment of the medium particles.”
(iii) Radiation. “Is that mode of transfer of heat in which heat directly passes from one body to the other body without heating the medium.”

Question 30.
Name the mode of transfer of heat in the following :
(a) solid,
(b) liquid,
(c) gas
(d) vacuum
Answer:

Question 31.
What are the good and bad conductors of heat ? Give two examples of each.
Answer:
Good conductors. “The substances through which heat is easily conducted are called good conductors of heat.”
Example : Copper, iron.
Bad conductors. “The substances through which heat is not conducted easily are called bad conductors of heat or poor conductors of heat.”
Example : Wood, cloth.

Question 32.
Name a liquid which is a good conductor of heat.
Answer:
Mercury is good conductor of heat.

Question 33.
Name a solid which is a good conductor of heat.
Answer:
Aluminium is a good conductor of heat.

Question 34.
Select good and bad conductors of heat from the following :
copper, mercury, wood, iron, air, saw-dust, cardboard, silver, plastic, wool.
Answer:
Good conductors — Mercury, copper, silver, iron.
Bad conductors — Wood, air, saw dust, plastic, wool, cardboard.

Question 35.
Why is an oven made of double walls with the space in between filled with cork ?
Answer:
An oven is made of double walls and the space between them is filled with wool, cork etc. because the wool and cork are the insulator of heat. They prevent the heat of the oven to escape.

Question 36.
Why do we use cooking utensils made up of copper.
Answer:
Cooking utensils are made of metals such as copper, aluminium, brass, steel etc., so that heat is easily conducted through the base to their contents. But they are provided with handles of bad conductors (such as ebonite or wood) to hold them easily as handles will not conduct heat from the utensil to our hand.

Question 37.
Why is a tea kettle provided with an ebonite handle ?
Answer:
Tea kettles are provided with wooden or ebonite handles. The wood or the ebonite being the insulators of heat, does not pass heat from the utensils to our hand. Thus, we can hold the hot utensils or pans comfortably by their handles.

Question 38.
In summer, ice is kept wrapped in a gunny bag. Explain the reason.
Answer:
In summer, the ice is kept wrapped in a gunny bag or it is covered with saw dust. The air filled in the fine pores of the gunny bag or saw dust, is the insulator of heat. The air does not allow heat from outside to pass through it to the ice. Thus, the ice is prevented from melting rapidly.

Question 39.
Explain why
(a) we wear woolen clothes in winter.
(b) the water pipes are covered with cotton during very cold weather.
Answer:
(a) Woolen clothes have fine pores filled with air. Wool and air both are bad conductors of heat. Therefore in winter, we wear woolen clothes as they check the conduction of heat from the body to the surroundings and thus keeps the body warm.
(b) During very cold weather, the water pipes are covered with cotton. The cotton has air trapped in its fine pores. The cotton and air are the insulators of heat. They do not pass heat from water inside the pipes to the outside atmosphere. Thus, cotton prevents the water in the pipes from freezing.

Question 40.
Why are quilts filled with fluffy cotton ?
Answer:
Quilts are filled with fluffy cotton. Air is trapped in the fine pores of cotton. Cotton and air are the insulators of heat. They check heat from our body to escape and thus keep us warm.
The newly made quilts are warmer than the old ones because in the old quilts, there is no air trapped in the cotton.

Question 41.
State the direction of heat transfer by way of convection.
Answer:
By the process of convection, heat is always transferred vertically upwards. The reason is that the medium particles near the source of heat absorb heat from the source and they start moving faster. As a result, the medium at this place becomes less dense so it rises up and the medium from above being denser, moves down to take its place. Thus, current is set up in the medium which is called a convection current. The current continues till the entire medium acquires the same temperature.

Question 42.
Why is a ventilator provided in a room ?
Answer:
Ventilators and windows are provided in rooms for proper ventilation. The reason is that when we breathe out in a room, the air in the room becomes warm and impure. The warm air is less dense i.e. lighter, so it rises up and moves out through the ventilators. Then the cold fresh air comes in the room through the windows to take its place. Thus the continuous circulation of fresh air keeps the air in the room fresh.

Question 43.
Why are chimneys provided over furnace in factories ?
Answer:
Chimneys are provided over the furnace in factories. This is because the hot gases coming out of the furnace are less dense than the air. They rise up through the chimney. The smoke, fumes etc. around the furnace rush in so as to take their place and they are sucked out. Thus, the chimney helps to remove the undesired fumes, smoke etc. from the premises.

Question 44.
What are the land and sea breezes ? Explain their formation.
Answer:
LAND BREEZE : Blowing of breeze (air) from land towards sea is called land breeze.
During night land and sea water both lose heat. Specific heat capacity of land being very low as compared to that of sea water, land loses heat energy fast and cools more rapidly as compared to sea. Sea water being at higher temperature, the air above it becomes lighter and rise up. Air from land being at higher pressure. So air from land starts blowing towards sea and gives rise to Land Breeze.

SEA BREEZE : Blowing of breeze (cold air) from sea towards land during the day is called the SEA BREEZE. During day time land and sea both are heated equally by the sun, but land has very low specific heat capacity as compared to sea, is heated up more quickly. Thus air above land due to heat becomes lighter and rises up. Thus pressure decreases and cold and humid air above the sea starts blowing towards land, thereby giving rise to SEABREEZE.

Question 45.
Why is the freezing chest in a refrigerator fitted near its top?
Answer:
Freezing chest in a refrigerator is fitted near the top, because it cools the remaining space of the refrigerator by convection current. Air near the top comes in contact with the freezing chest gets cooled, becomes denser and therefore descends while the hot air from the lower part rises and hence convection currents produced cool the whole space inside.

Question 46.
Explain briefly the process of heat transfer by radiation.
Answer:
RADIATION. “The transfer of heat energy from a hot body to cold body directly, without heating the medium between two bodies is called RADIATION.”
The radiant heat or thermal radiation is of the form of ELECTROMAGNETIC WAVES. These waves can travel even in vacuum in all directions in straight line with the speed of light. They do not heat the medium through which they pass. Heat radiations are also called INFRA-RED RADIATIONS because the wavelength of heat radiations is longer than that of visible light. These radiations can cause heating effect only if they are absorbed by some material.

Question 47.
Give one example of heat transfer by radiation.
Answer:
When we sit in the sun, we feel warm. We cannot get heat from the sun by the process of conduction or convection because most of the space between the sun and the earth is a vacuum and both of these modes of heat transfer require medium. Hence, one must be getting heat from the sun by the mode of radiation.

Question 48.
Why do we prefer to wear white or light coloured clothes in summer and black or dark coloured clothes in winter ?
Answer:
We prefer to wear white clothes in summer. The reason is that the white clothes reflects most of the sun’s heat and absorb very little of the sun’s heat, thus they keep our body cool.
We prefer to wear black and dark coloured clothes in winter. The reason is that the black or dark colour clothes absorb most of the sun’s heat and keep our body warm.

Question 49.
The bottom of a cooking utensil is painted black. Give the reason.
Answer:
The bottom part of the cooking utensil or pan is painted black. The reason is that the black surface absorbs more heat and so the contents of utensil or pan get cooked rapidly if its bottom part is painted black.

Question 50.
Draw a labelled diagram of a thermo flask. Explain how the transfer of heat by conduction, convection and radiation is reduced to a minimum in it.
Answer:
Heat transfer is minimised because of:
(1) The vacuum between the two walls, rubber, glass, cork and air do not allow the loss of heat by conduction.
(2) Cork in the neck of flask and the cup over it prevent loss of heat by convection.
(3) Heat cannot be lost by conduction or convection because of vacuum between the two walls.
(4) Heat loss is also minimised by radiation, by making outer surface of inner wall and inner surface of outer wall silvered. The inner wall is a BAD RADIATOR and the outer wall is a GOOD REFLECTOR of radiation.

C. Numericals

Question 1.
The temperature of a body rises by 1°C. What is the corresponding rise on the (a) Fahrenheit scale (b) Kelvin scale?
Answer:
(a) Since 100 divisions on Celsius scale =180 divisions on the Fahrenheit scale 1 division on Celsius scale
∴ 1 division on Celsius scale
= 1.80 / 1.00 × 1
= 1.8 divisions in the Fahrenheit scale.
For 1°C rise corresponding rise in Fahrenheit = 1.8°F
(b) Since 100 divisions in the Celsius scale = 100 divisions in the Kelvin scale
1 division on Celsius scale = 100 / 100 × 1
= 1 division on Kelvin scale
For 1°C rise corresponding rise in Kelvin is 1 K.

Question 2.
The temperature rises by 18°F. What is the rise on the Celsius scale ?
Answer:
Since 100 divisions on the Celsius scale =180 divisions on the Fahrenheit scale
∴ 18 divisions on Fahrenheit scale.

Question 3.
Convert 5°F to the Celsius scale.
Answer:

Question 4.
Convert 40°C to the (a) Fahrenheit scale (b) Kelvin Scale.
Answer:
(a) Fahrenheit scale
C = 40°C
Substitute value of C = 40° in below equation

Question 5.
Convert – 40°F to the Celsius scale.
Answer:

Selina Concise Chemistry Class 7 ICSE Solutions – Elements, Compounds and Mixtures

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Selina Concise ICSE Solutions for Class 7 Chemistry Chapter 3 Elements, Compounds and Mixtures

Points to Remember :

1. Every substance is made up of very tiny particles, called molecules. Molecules are formed from even smaller particles called atoms.
2. Element— (a) Element is the simplest pure substance. It cannot be divided further into simpler substances by any chemical method, e.g. oxygen, hydrogen, sulphur, etc.
   (b) At present 116 elements are known, of which 92 are natural elements.
3. Based on their properties, elements are classified into : metals, non-metals, metalloids, noble gases.
4. Metals are ductile, malleable, good conductors of heat and electricity, high melting and boiling points. Metals are sonorous, e.g., Iron, Gold, Silver, etc.
5. Non-metals are solids and brittle in nature, bad conductor of heat and electricity (exception Graphite) low melting and boiling points, e.g. sulphur, carbon, hydrogen, etc.
6. Metalloids— These elements show properties of both metals and non-metals. They are hard solids, e.g. Boron, Silicon, Arsenic.
7. Inert or noble gases— These elements do not react chemically with other elements or compounds are called noble (Inert) gases, e.g., helium, neon, argon, etc.
8. Symbols of Elements— Each element is denoted by a symbol usually to first letter.
   Examples : Oxygen by O Hydrogen by H.
9. Atom— “An Atom is the smallest particle of an element that can take part in a chemical reaction but may or may not have independent existence.”
   The atom of an element exhibits all the properties of that element.
10. Molecule— A molecule is the smallest particle of a pure substance of element or compound which has independent existence. It exhibits all the properties of pure substance.
11. Atomicity— The number of atoms of an element that join together to form a molecule of that element is known as the atomicity.
12. Molecular Formula— of an element is the symbolic representation of its molecule. It indicates the number of atoms present in it. e.g. Magnesium oxide – MgO.

EXERCISE – I

Question 1.
Write the symbols of helium, silver, krypton, antimony, barium.
Answer:

Element                   Symbol
Helium                         He
Silver                            Ag
Krypton                       Kr
Antimony                    Sb
Barium                        Ba

Question 2.
Write the names of following elements Na, C, Kr, U, Ra, Fe, Co.
Answer:

Symbol           Element
Na                        Sodium
C                          Carbon
Kr                        Krypton
U                         Uranium
Ra                        Radium
Fe                           Iron
Co                        Cobalt

Question 3.
Define :

1. Elements : An element is the basic form of matter that cannot be broken down into simpler substances by chemical reactions.
2. Compounds : A compound is a pure substance formed by the chemical combination of two or more elements in a fixed ratio by mass.

Question 4.
Name the main metal present in the following :
Answer:

(a) Haemoglobin                                    Iron
(b) Chalk                                               Calcium
(c) Chlorophyll                                   Magnesium
(d) Chocolate wrappers                    Aluminium

Question 5.
Give four examples of non-metallic elements.
Answer:
Examples : Hydrogen, oxygen, nitrogen, carbon, chlorine, sulphur, phosphorus, etc.

Question 6.
What do you understand by :
Answer:

1. Metalloids : Metalloids are those substances which have some properties of metals and some of non-metals e.g. boron, silicon.
2. Noble gases : Noble gases are those which do not react chemically with other elements or compounds e.g. helium, neon, etc.

Question 7.
Select elements and compounds from the following list: Iron, plaster of paris, chalk, common salt, copper, aluminium, calcium oxide, cane sugar, carbon, silica, sodium sulphate, uranium, potassium carbonate, silver, carbon dioxide.
Answer:

EXERCISE – II

Question 1.
State four difference between compounds and mixtures.
Answer:

Compound	Mixture
1. A compound is a pure substance.	1. A mixture is an impure substance.
2. Compounds are always homogeneous.	2. Mixtures may be homogeneous or heterogeneous.
3. A compound has a fixed composition, i.e., it is formed when two or more pure substances chemically combine in a definite ratio by mass.	3. A mixture has no fixed composition, i.e., it is formed by mixing two or more substances in any ratio without any chemical reaction.
4. Formation of a compound involves change in energy.	4. Formation of a mixture does not involve any change in energy.
5. Compounds have specific set of properties.	5. Mixtures do not have any specific set of properties.
6. Components of compounds can be separated only by complex chemical processes.	6. Components of mixtures can be separated by simple physical methods.

 Question 2.
What are the characteristic properties of a pure substance? Why do we need them?
Answer:
Pure substance : Pure substances have a definite set of properties such as boiling point, melting point, density, etc. They are all homogeneous i.e., their composition is uniform throughout the bulk. Both elements and compounds are pure substances.
Pure substances are needed to :

1. Manufacture medicines.
2. To prepare chemicals in industry.
3. For scientific purposes.
4. To maintain the good health of human beings.

Question 3.
Give two examples for each of the following :
(a) Solid + Solid mixture
(b) Solid + Liquid mixture
(c) Liquid + Liquid mixture
Answer:
(a) Solid + Solid mixture :Sand and sugar,

• Sand and stone,
• sand and sugar.

(b) Solid + Liquid mixture :

• Sand and water,
• Charcoal and water.

(c) Liquid + Liquid mixture :

• Oil in water,
• Alcohol and water.

Question 4.
Define :

1. Evaporation : Is the process ~of converting a liquid into its vapours state either by exposing it to air or by heating.
2. Filtration : The process of separating solid particles from liquid by allowing it to pass through a filter paper is called filtration.
3. Sublimation : The process in which a solid changes directly into its vapours on heating is called sublimation.
4. Distillation : Distillation is the method of getting a pure liquid from a solution by evaporating and then condensing the vapours.
5. Miscible liquids : Homogeneous liquid-liquid mixtures are called miscible liquids.
6. Immiscible liquids : Heterogeneous liquid-liquid mixtures are called immiscible liquids.

Question 5.
Name the process by which the components of following mixtures can be separated.

1. Iron and sulphur
2. Ammonium chloride and sand
3. Common salt from sea water
4. Chaff and grain
5. Water and mustard oil
6. Sugar and water
7. Cream from milk

Answer:

1. Magnetic separation.
2. Sublimation.
3. Evaporation.
4. Winnowing separates chaff (lighter) from heavier grains in two different heaps.
5. Mustard oil and water is liquid-liquid immiscible mixture and is separated by separating funnel. Water being the heavier forms the lower layer.
6. By evaporation in this process of converting a liquid into its vapour state by heating. Liquid is heated and water evaporate and sugar is obtained.
7. Centrifugation.

Question 6.
How will you separate a mixture of common salt, chalk powder and powdered camphor? Explain.
Answer:
Comphor with sublimation. Chalk powder by Alteration then the residual left is common salt.

Question 7.
How is distillation more advantageous than evaporation?
Answer:
The advantage of distillation is that both components of the
solid and liquid mixture are obtained. Whereas in evaporation only solid is obtained.

Question 8.

1. What is chromatography?
2. Why is it named so?
3. What are the advantages of chromatography?
4. Name the simplest type of chromatography?
5. On what principle is this method based?
6. What is meant by stationary phase and mobile phase in chromatography?

Answer:

1. The process of separating different dissolved constituents of a mixture by their absorption on an appropriate material is called chromatography.
2. It is named so, because earlier it was used to separate mixtures containing coloured components only but these days this technique is applied to colourless substances too.
3. Advantages of chromatography :
   (i) A very small quantity of the substance can be separated.
   (ii) Components with very similar physical and chemical properties can be separated.
   (iii) It identifies the different constituents of a mixture.
   (iv) lt also helps in quantitative estimation of components of a mixture.
4. The simplest type of chromatography is “Paper chromatography”.
5. Chromotography is based on differential affinities of compounds towards two phases i.e. stationary and mobile phase.
6. The filter paper acts as “stationary phase” while the solvent act as “mobile phase”.

Question 9.
On what principle are the following methods of separation based? Give one example of a mixture for each of the methods mentioned in which they are used
Answer:

1. Sublimation : Change of solid into vapours directly on heating and change of vapours into solid again on
   Example : Salt from ammonium chloride.
2. Filtration : The process of separating insoluble solid particles from a liquid by allowing it to pass through a filter is called Alteration. These filters allow liquids to pass through them but not solids. The insoluble solid left on the filter is called the residue, while the liquid which passes through the filter is called the filtrate. Mixtures like chalk and water, clay and water, tea and tea leaves, sawdust and water, etc., are separated by this method.
3. Sedimentation and decantation : The settling down of suspended, insoluble, heavy, solid particles in a solid- liquid mixture when left undistrubed is called sedimentation.
   The solid which settles at the bottom is called sediment while the clear liquid above it is called supernatant liquid.
   The process of pouring out the clear liquid, without disturbing the sediment, is called decantation.
   Example : A mixture of sand and water.
4. Solvent extraction method : This method is used when one of the solid components is soluble in a liquid.
   Example : A mixture of sand and salt can be separated by this method. Salt gets dissolved in water while sand settles down in the container. The salt solution is then decanted. Salt is separated from the solution by evaporation. In this way, they can be separated.
5. Magnetic separation : This method is used when one of the components of the mixture is iron. Iron gets attracted towards a magnet and hence can be separated. Mixtures of iron and sulphur, iron and sand, etc., can be separated by moving a magnet over them. Iron gets attached to the magnet and is separated.
6. By using a separating funnel : It is a simple device used to separate the components of a liquid-liquid heterogeneous mixture.
   Example : Kerosene oil and water. The mixture is placed in a separating funnel and allowed to stand for sometime. The components form two clear layers. Water being heavier forms the lower layer and oil being lighter forms the upper layer. When the stopper of the funnel is opened, the heavier liquid trickles out slowly and is collected in a vessel. The stopper is closed when the bottom layer is entirely removed the funnel. In this way, the two liquids are separated.
7. Fractional distillation : The process of distillation is used for separating the components of a homogeneous liquid-liquid mixture, like water and alcohol. This is based on the fact that alcohol boils at a lower temperature than water. The vapour of alcohol are collected and cooled while water is left behind in the original vessel. Thus, two liquids having different boiling points can be separated by distillation provided that difference in their boiling points must be 25 °C or more.

OBJECTIVE TYPE QUESTIONS

Question 1.
Fill in the blanks:
Answer:

1. Elements are made up of same kind of atoms.
2. Elements and compounds are pure substances.
3. In a mixture the substances are not combined chemically.
4. Clay is separated from water by the method called loading and decantation.
5. Crystallisation is a process to obtain a very pure form of a solid dissolved in a liquid.
6. Camphor and ammonium chloride can sublimate.

Question 2.
Give one word answers for the following :
Answer:

1. The solid particles which remain on the filter paper after the filtration residue.
2. The liquid which evaporates and then condenses during the process of distillation distillate.
3. The process of transferring the clean liquid after the solid settles at the bottom of the container decantation.
4. The process by which two miscible liquids are separated fractional distillation.

MULTIPLE CHOICE QUESTIONS

Select the correct alterative from the choices given for the following statements:
Question 1.
A pure liquid is obtained from a solution by :
Answer:

1. evaporation
2. distillation
3. Alteration
4.  crystallisation

Question 2.
Components of crude petroleum can be separated by :
Answer:

1. distillation
2. evaporation
3. filtration
4. fractional distillation

Question 3.
Example of a homogeneous mixture is :
Answer:

1. tap water 
2. distilled water
3.  sand and water
4. water and oil

Question 4.
In chromatography the filter paper is :
Answer:

1.  stationary phase 
2. mobile phase
3. mixture
4. none of the above

Question 5.
A set of mixture is :
Answer:

1.  ink, honey, icecream, milk
2. tapwater, gold, common salt, alloy
3.  milk, brass, silver, honey
4. butter, petroleum, tapwater, iron