## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 25 Value Added Tax Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test.

Question 1.
A shopkeeper bought a washing machine at a discount of 20% from a wholesaler, the printed price of the washing machine being ₹ 18000. The shopkeeper sells it to a consumer at a discount of 10% on the printed price. If the rate of sales tax is 8%, find:
(i) the VAT paid by the shopkeeper.
(ii) the total amount that the consumer pays for the washing machine.
Solution:
(i) SP of washing machine

Question 2.
A manufacturing company sold an article to its distributor for ₹22000 including VAT. The distributor sold the article to a dealer for ₹22000 excluding tax and the dealer sold it to a consumer for ₹25000 plus tax (under VAT). If the rate of sales tax (under VAT) at each stage is 10%, find :
(i) the sale price of the article for the manufacturing company.
(ii) the amount of VAT paid by the dealer.
Solution:

Question 3.
The marked price of an article is ₹7500. A shopkeeper sells the article to a consumer at the marked prices and charges sales tax at . the rate of 7%. If the shopkeeper pays a VAT of ₹105, find the price inclusive of sales tax of the article which the shopkeeper paid to the wholesaler.
Solution:

Question 4.
A shopkeeper buys an article at a discount of 30% and pays sales tax at the rate of 6%. The shopkeeper sells the article to a consumer at 10% discount on the list price and charges sales tax at the’ same rate. If the list price of the article is ₹3000, find the price inclusive of sales tax paid by the shopkeeper.
Solution:

Question 5.
Mukerjee purchased a movie camera for ₹27468. which includes 10% rebate on the list price and then 9% sales tax (under VAT) on the remaining price. Find the list price of the movie camera.
Solution:

Question 6.
A retailer buys an article at a discount of 15% on the printed price from a wholesaler. He marks up the price by 10%. Due to competition in the market, he allows a discount of 5% to a buyer. If the buyer pays ₹451.44 for the article inclusive of sales tax (under VAT) at 8%, find :
(i) the printed price of the article
(ii) the profit percentage of the retailer.
Solution:

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 1 Value Added Tax Chapter Test are helpful to complete your math homework.

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## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 24 Probability Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 24 Probability Chapter Test.

Question 1.
A game consists of spinning an arrow which comes to rest at one of the regions 1, 2 or 3 (shown in the given figure). Are the outcomes 1, 2 and 3 equally likely to occur? Give reasons.

Solution:

Question 2.
In a single throw of a die, find the probability of getting
(i) a number greater than 5
(ii) an odd prime number
(iii) a number which is multiple of 3 or 4.
Solution:

Question 3.
A lot consists of 144 ball pens of which 20 are defective and the others are good. Rohana will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that :
(i) She will buy it ?
(ii) She will not buy it ?
Solution:
In a lot, there are 144 ball pens in which defective ball pens are = 20
and good ball pens are = 144 – 20 = 124
Rohana buys a pen which is good only.
(i) Now the number of possible outcomes = 144
and the number of favourable outcomes = 124

Question 4.
A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is
(i) acceptable to Varnika?
Solution:

Question 5.
A bag contains 6 red, 5 black and 4 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is
(i) white
(ii) red
(iii) not black
(iv) red or white.
Solution:

Question 6.
A bag contains 5 red, 8 white and 7 black balls. A ball is drawn from the bag at random. Find the probability that the drawn ball is:
(i) red or white
(ii) not black
(iii) neither white nor black
Solution:

Question 7.
A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is :
(i) white or blue
(ii) red or black
(iii) not white
(iv) neither white nor black ?
Solution:

Question 8.
A box contains 20 balls bearing numbers 1, 2, 3, 4,……, 20. A ball is drawn at random from the box. What is the probability that the number on the ball is
(i) an odd number
(ii) divisible by 2 or 3
(iii) prime number
(iv) not divisible by 10 ?
Solution:

Question 9.
Find the probability that a number selected at random from the numbers 1, 2, 3,……35 is a
(i) prime number
(ii) multiple of 7
(iii) multiple of 3 or 5.
Solution:

Question 10.
Cards marked with numbers 13, 14, 15,…..60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card is
(i) divisible by 5
(ii) a number which is a perfect square.
Solution:

Question 11.
The box has cards numbered 14 to 99. Cards are mixed thoroughly and a card is drawn at random from the box. Find the probability that the card drawn from the box has
(i) an odd number
(ii) a perfect square number.
Solution:

Question 12.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is four times that of a red ball, find the number of balls in the bags.
Solution:

Question 13.
A bag contains 18 balls out of which x balls are white.
(i) If one ball is drawn at random from the bag, what is the probability that it is white ball?
(ii) If 2 more white balls are put in the bag, the probability of drawing a white ball will be $$\\ \frac { 9 }{ 8 }$$ times that of probability of white ball coming in part (i). Find the value of x.
Solution:
Total numbers of balls in a bag = 18
No. of white balls = x
(i) One ball is drawn a random
Probability of being a white ball = $$\\ \frac { x }{ 18 }$$
(ii) If 2 more white balls an put, then number of white balls = x + 2
and probability is $$\\ \frac { 9 }{ 8 }$$ times

Question 14.
A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card drawn is :
(i) a red face card
(ii) neither a club nor a spade
(iii) neither an ace nor a king of red colour
(iv) neither a red card nor a queen
(v) neither a red card nor a black king.
Solution:

Question 15.
From pack of 52 playing cards, blackjacks, black kings and black aces are removed and then the remaining pack is well-shuffled. A card is drawn at random from the remaining pack. Find the probability of getting
(i) a red card
(ii) a face card
(iii) a diamond or a club
(iv) a queen or a spade.
Solution:

Question 16.
Two different dice are thrown simultaneously. Find the probability of getting:
(i) sum 7
(ii) sum ≤ 3
(iii) sum ≤ 10
Solution:

Question 17.
Two dice are thrown together. Find the probability that the product of the numbers on the top of two dice is
(i) 6
(ii) 12
(iii) 7
Solution:

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 24 Probability Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 24 Probability Chapter Test, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 23 Measures of Central Tendency Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 23 Measures of Central Tendency Chapter Test.

Question 1.
Arun scored 36 marks in English, 44 marks in Civics, 75 marks in Mathematics and x marks in Science. If he has scored an average of 50 marks, find x.
Solution:

Question 2.
The mean of 20 numbers is 18. If 3 is added to each of the first ten numbers, find the mean of new set of 20 numbers.
Solution:

Question 3.
The average height of 30 students is 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for computation of mean. Find the correct mean.
Solution:

Question 4.
There are 50 students in a class of which 40 are boys and the rest girls. The average weight of the students in the class is 44 kg and average weight of the girls is 40 kg. Find the average weight of boys.
Solution:

Question 5.
The contents of 50 boxes of matches were counted giving the following results

Calculate the mean number of matches per box.
Solution:

Question 6.
The heights of 50 children were measured (correct to the nearest cm) giving the following results :

Solution:

Question 7.
Find the value of p for the following distribution whose mean is 20.6 :

Solution:

Question 8.
Find the value of p if the mean of the following distribution is 18.

Solution:

Question 9.
Find the mean age in years from the frequency distribution given below:

Solution:

Question 10.
Calculate the Arithmetic mean, correct to one decimal place, for the following frequency distribution :

Solution:

Question 11.
The mean of the following frequency distribution is 62.8. Find the value of p.

Solution:

Hence p = 10

Question 12.
The daily expenditure of 100 families are given below. Calculate f1, and f2, if the mean daily expenditure is Rs 188.

Solution:

Question 13.
The measures of the diameter of the heads of 150 screw is given in the following table. If the mean diameter of the heads of the screws is 51.2 mm, find the values of p and q

Solution:
Mean = 51.2
No. of screws = 150

Question 14.
The median of the following numbers, arranged in ascending order is 25. Find x, 11, 13, 15, 19, x + 2, x + 4, 30, 35, 39, 46
Solution:
Here, n = 10, which is even

Question 15.
If the median of 5, 9, 11, 3, 4, x, 8 is 6, find the value of x.
Solution:

Question 16.
Find the median of: 17, 26, 60, 45, 33, 32, 29, 34, 56 If 26 is replaced by 62, find the new median.
Solution:

Question 17.
The marks scored by 16 students in a class test are : 3, 6, 8, 13, 15, 5, 21, 23, 17, 10, 9, 1, 20, 21, 18, 12
Find
(i) the median
(ii) lower quartile
(iii) upper quartile
Solution:
Arranging the given data in ascending order:
1, 3, 5, 6, 8, 9, 10, 12, 13, 15, 17, 18, 20, 21, 21, 23
Here n = 16 which is even.

Question 18.
Find the median and mode for the set of numbers : 2, 2, 3, 5, 5, 5, 6, 8, 9
Solution:

Question 19.
Calculate the mean, the median and the mode of the following distribution :

Solution:

Question 20.
The daily wages of 30 employees in an establishment are distributed as follows :

Estimate the modal daily wages for this distribution by a graphical method.
Solution:

Question 21.
Using the data given below, construct the cumulative frequency table and draw the ogive. From the ogive, estimate ;
(i) the median
(ii) the inter quartile range.

Also state the median class
Solution:

Question 22.
Draw a cumulative frequency curve for the following data :

Hence determine:
(i) the median
(ii) the pass marks if 85% of the students pass.
(iii) the marks which 45% of the students exceed.
Solution:

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 23 Measures of Central Tendency Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 23 Measures of Central Tendency Chapter Test, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Heights and Distances Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Heights and Distances Chapter Test.

Question 1.
The angle of elevation of the top of a tower from a point A (on the ground) is 30°. On walking 50 m towards the tower, the angle of elevation is found to be 60°. Calculate
(i) the height of the tower (correct to one decimal place).
(ii) the distance of the tower from A.
Solution:

Question 2.
An aeroplane 3000 m high passes vertically above another aeroplane at an instant when the angles of elevation of the two aeroplanes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the two planes.
Solution:

Question 3.
A 7m long flagstaff is fixed on the top of a tower. From a point on the ground, the angles of elevation of the top and bottom of the flagstaff are 45° and 36° respectively. Find the height of the tower correct to one place of demical.
Solution:

Question 4.
A boy 1.6 m tall is 20 m away from a tower and observes that the angle of elevation of the top of the tower is 60°. Find the height of the tower.
Solution:

Question 5.
A boy 1.54 m tall can just see the sun over a wall 3.64 m high which is 2.1 m away from him. Find the angle of elevation of the sun.
Solution:

Question 6.
In the adjoining figure, the angle of elevation of the top P of a vertical tower from a point X is 60° ; at a point Y, 40 m vertically above X, the angle of elevation is 45°. Find
(i) the height of the tower PQ
(ii) the distance XQ

Solution:

Question 7.
An aeroplane is flying horizontally 1 km above the ground is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°. Find the speed of the aeroplane in km/hr.
Solution:

Question 8.
A man on the deck of a ship is 16 m above the water level. He observes that the angle of elevation of the top of a cliff is 45° and the angle of depression of the base is 30°. Calculate the distance of the cliff from the ship and the height of the cliff.
Solution:

Question 9.
There is a small island in between a river 100 metres wide. A tall tree stands on the island. P and Q are points directly opposite to each other on the two banks and in the line with the tree. If the angles of elevation of the top of the tree from P and Q are 30° and 45° respectively, find the height of the tree.
Solution:

Question 10.
A man standing on the deck of the ship which is 20 m above the sea-level, observes the angle of elevation of a bird as 30° and the angle of depression of its reflection in the sea as 60°. Find the height of the bird
Solution:

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Heights and Distances Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 21 Heights and Distances Chapter Test, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Trigonometric Tables Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Tables Chapter Test.

Question 1.
Using trigonometrical tables, find the values of :
(i) sin 48° 52′
(ii) cos 37° 34′
(iii) tan 18° 21′.
Solution:

Question 2.
Use tables to find the acute angle θ, given that
(i) sin θ = 0.5766
(ii) cos θ = 0.2495
(iii) tan θ = 2.4523.
Solution:

Question 3.
If θ is acute and cos θ = 0.53, find the value of tan θ.
Solution:

Question 4.
Find the value of: sin 22° 11′ + cos 57° 20′ – 2 tan 9° 9′.
Solution:

Question 5.
If θ is acute and sin θ = 0.7547, find the value of: (i) θ (ii) cos θ (iii) 2 cos θ – 3 tan θ.
Solution:

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Trigonometric Tables Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 20 Trigonometric Tables Chapter Test, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test.

Question 1.
(i) If θ is an acute angle and cosec θ = √5 find the value of cot θ – cos θ.
(ii) If θ is an acute angle and tan θ = $$\\ \frac { 8 }{ 15 }$$, find the value of sec θ + cosec θ.
Solution:
(i) θ is an acute angle.
cosec θ = √5

Question 2.
Evaluate the following:
(i) $$2\times \left( \frac { { cos }^{ 2 }{ 20 }^{ O }+{ cos }^{ 2 }{ 70 }^{ O } }{ { sin }^{ 2 }{ 25 }^{ O }+{ sin }^{ 2 }{ 65 }^{ O } } \right)$$ – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°
(ii) $$\frac { { sec }29^{ O } }{ { cosec }61^{ O } }$$ + 2 cot 8° cot 17° cot 45° cot 73°0 cot 82° – 3(sin2 38° + sin2 52°)
(iii) $$\frac { { sin }^{ 2 }{ 22 }^{ O }+{ sin }^{ 2 }{ 68 }^{ O } }{ { cos }^{ 2 }{ 22 }^{ O }+{ cos }^{ 2 }{ 68 }^{ O } }$$ + sin2 63° + cos 63° sin 27°
Solution:
(i) $$2\times \left( \frac { { cos }^{ 2 }{ 20 }^{ O }+{ cos }^{ 2 }{ 70 }^{ O } }{ { sin }^{ 2 }{ 25 }^{ O }+{ sin }^{ 2 }{ 65 }^{ O } } \right)$$ – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°

Question 3.
If $$\\ \frac { 4 }{ 3 }$$ (sec2 59° – cot2 31°) – $$\\ \frac { 2 }{ 2 }$$ sin 90° + 3tan2 56° tan2 34° = $$\\ \frac { x }{ 2 }$$, then find the value of x.
Solution:
Given
$$\\ \frac { 4 }{ 3 }$$ (sec2 59° – cot2 31°) – $$\\ \frac { 2 }{ 2 }$$ sin 90° + 3tan2 56° tan2 34° = $$\\ \frac { x }{ 2 }$$

Question 4.
(i) $$\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA$$
(ii) $$\frac { cosA }{ cosecA+1 } +\frac { cosA }{ cosecA-1 } =2tanA$$
Solution:
(i) $$\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA$$
L.H.S = $$\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA }$$

Question 5.
(i) $$\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } =sec\theta$$
(ii) (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1.
Solution:
(i) $$\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } =sec\theta$$
L.H.S = $$\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 }$$

Question 6.
(i) sin2 θ + cos4 θ = cos2 θ + sin4 θ
(ii) $$\frac { cot\theta }{ cosec\theta +1 } +\frac { cosec\theta +1 }{ cot\theta } =2sec\theta$$
Solution:
L.H.S. = sin2 θ + cos4 θ
= (1 – cos2 θ + cos4 θ
= 1 – cos2 θ + cos4 θ
= 1 – cos2 θ (1 – cos2 θ)

Question 7.
(i) sec4 A (1 – sin4 A) – 2 tan2 A = 1
(ii) $$\frac { 1 }{ sinA+cosA+1 } +\frac { 1 }{ sinA+cosA-1 } =secA+cosecA$$
Solution:
(i) sec4 A (1 – sin4 A) – 2 tan2 A = 1
L.H.S = sec4 A (1 – sin4 A) – 2 tan2 A

Question 8.
(i) $$\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1$$
(ii) (sec A – tan A)2 (1 + sin A) = 1 – sin A.
Solution:
(i) $$\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1$$
L.H.S = $$\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta$$

Question 9.
(i) $$\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA$$
(ii) (sec A – cosec A) (1 + tan A + cot A) = tan A sec A – cot A cosec A
(iii) $$\frac { { tan }^{ 2 }\theta }{ { tan }^{ 2 }\theta -1 } -\frac { { cosec }^{ 2 }\theta }{ { sec }^{ 2 }\theta -{ cosec }^{ 2 }\theta } =\frac { 1 }{ { sin }^{ 2 }\theta -{ cos }^{ 2 }\theta }$$
Solution:
(i) $$\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA$$
L.H.S = $$\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA }$$

Question 10.
$$\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 }$$
Solution:
$$\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 }$$
L.H.S = $$\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA }$$

Question 11.
2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = θ
Solution:
2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = θ
L.H.S = 2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1

Question 12.
If cot θ + cos θ = m, cot θ – cos θ = n, then prove that (m2 – n2)2 = 16 run.
Solution:
cot θ + cos θ = m…..(i)
cot θ – cos θ = n……(ii)

Question 13.
If sec θ + tan θ = p, prove that sin θ = $$\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 }$$
Solution:
sec θ + tan θ = p,
prove that sin θ = $$\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 }$$
$$\frac { 1 }{ cos\theta } +\frac { sin\theta }{ cos\theta } =p$$

Question 14.
If tan A = n tan B and sin A = m sin B, prove that cos2 A = $$\frac { { m }^{ 2 }-1 }{ { n }^{ 2 }-1 }$$
Solution:
m = $$\\ \frac { sinA }{ sinB }$$
n = $$\\ \frac { tanA }{ tanB }$$

Question 15.
If sec A = $$x+ \frac { 1 }{ 4x }$$, then prove that sec A + tan A = 2x or $$\\ \frac { 1 }{ 2x }$$
Solution:
sec A = $$x+ \frac { 1 }{ 4x }$$
To prove that sec A + tan A = 2x or $$\\ \frac { 1 }{ 2x }$$

Question 16.
When 0° < θ < 90°, solve the following equations:
(i) 2 cos2 θ + sin θ – 2 = 0
(ii) 3 cos θ = 2 sin2 θ
(iii) sec2 θ – 2 tan θ = 0
(iv) tan2 θ = 3 (sec θ – 1).
Solution:
0° < θ < 90°
(i) 2 cos2 θ + sin θ – 2 = 0

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Mensuration Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Mensuration Chapter Test.

Question 1.
A cylindrical container is to be made of tin sheet. The height of the container is 1 m and its diameter is 70 cm. If the container is open at the top and the tin sheet costs Rs 300 per m2, find the cost of the tin for making the container.
Solution:
Height of container opened at the top (h) = 1 m = 100 cm
and diameter = 70 cm
∴Radius (r) = $$\\ \frac { 70 }{ 2 }$$ = 35 cm
∴Total surface area = 2πrh + πr2

Question 2.
A cylinder of maximum volume is cut out from a wooden cuboid of length 30 cm and cross-section of square of side 14 cm. Find the volume of the cylinder and the volume of wood wasted.
Solution:
Dimensions of the wooden cuboid = 30 cm × 14 cm × 14 cm
Volume = 30 × 14 × 14 = 5880 cm3

Question 3.
Find the volume and the total surface area of a cone having slant height 17 cm and base diameter 30 cm. Take π = 3.14.
Solution:
Slant height of a cone (l) = 17 cm
Diameter of base = 30 cm
Radius (r) = $$\\ \frac { 30 }{ 2 }$$ = 15 cm

Question 4.
Find the volume of a cone given that its height is 8 cm and the area of base 156 cm2.
Solution:
Height of a cone = 8 cm
Area of base = 156 cm
.’. Volume = $$\\ \frac { 1 }{ 3 }$$ × area of base × height

Question 5.
The circumference of the edge of a hemispherical bowl is 132 cm. Find the capacity of the bowl.
Solution:
Circumference of the edge of bowl = 132 cm

Question 6.
The volume of a hemisphere is $$2425 \frac { 1 }{ 2 }$$ cm2. Find the curved surface area.
Solution:
Volume of a hemisphere = $$2425 \frac { 1 }{ 2 }$$ cm3
= $$\\ \frac { 4851 }{ 2 }$$ cm3

Question 7.
A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volume of the toy
Solution:
A wooden solid toy is of a shape of a right circular cone
mounted on a hemisphere.
Radius of hemisphere (r) = 4.2 cm
Total height = 10.2 cm

Question 8.
A medicine capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of the entire capsule is 2 cm. Find the capacity of the capsule.
Solution:
Diameter of cylindrical part = 0.5 cm
Total length of the capsule = 2 cm

Question 9.
A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and the total surface area of the solid.
Solution:
Radius of cylinder = $$\\ \frac { 7 }{ 2 }$$cm
and height of cylinder = 19 – 2 × $$\\ \frac { 7 }{ 2 }$$ cm
= 19 – 7 = 12 cm
and radius of hemisphere = $$\\ \frac { 7 }{ 2 }$$ cm

Question 10.
The radius and height of a right circular cone are in the ratio 5 : 12. If its volume is 2512 cm , find its slant height. (Take π = 3.14).
Solution:
Let radius of cone (r) = 5x
then height (h) = 12x

Question 11.
A cone and a cylinder are of the same height. If diameters of their bases are in the ratio 3 : 2, find the ratio of their volumes.
Solution:
Let height of cone and cylinder = h
Diameter of the base of cone = 3x
Diameter of base of cylinder = 2x

Question 12.
A solid cone of base radius 9 cm and height 10 cm is lowered into a cylindrical jar of radius 10 cm, which contains water sufficient to submerge the cone completely. Find the rise in water level in the jar.
Solution:
Radius of the cone (r) = 9 cm
Height of the cone (h) = 10 cm
Volume of water filled in cone

Question 13.
An iron pillar has some part in the form of a right circular cylinder and the remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if one cu. cm of iron weighs 7.8 grams.
Solution:
Radius of the base of cone = 8 cm

Question 14.
A circus tent is made of canvas and is in the form of right circular cylinder and a right circular cone above it. The diameter and height of the cylindrical part of the tent are 126 m and 5 m respectively. The total height of the tent is 21 m. Find the total cost of the tent if the canvas used costs Rs 36 per square metre.
Solution:

Question 15.
The entire surface of a solid cone of base radius 3 cm and height 4 cm is equal to the entire surface of a solid right circular cylinder of diameter 4 cm. Find the ratio of their
(i) curved surfaces
(ii) volumes.
Solution:
Radius of the base of a cone (r) = 3 cm
Height (h) = 4 cm

Question 16.
A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. Find the radius of the sphere.
Solution:
Radius of base of a cone (r) = 2. 1 cm
and height (h) = 8.4 cm

Question 17.
How many lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimensions 66 cm, 42 cm and 21 cm.
Solution:
Dimensions of a solid rectangular lead piece
= 66 cm × 42 cm × 21 cm
.’. Volume = 66 × 42 × 21 cm3
Diameter of a lead shot = 4.2 cm

Question 18.
Find the least number of coins of diameter 2.5 cm and height 3 mm which are to be melted to form a solid cylinder of radius 3 cm and height 5 cm.
Solution:
Radius of a cylinder (r) = 3 cm
Height (h) = 5 cm

Question 19.
A hemisphere of lead of radius 8 cm is cast into a right circular cone of base radius 6 cm. Determine the height of the cone correct to 2 places of decimal.
Solution:
Radius of hemisphere = 8 cm
Volume = $$\frac { 2 }{ 3 } \pi { r }^{ 3 }{ cm }^{ 3 }$$

Question 20.
A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of the water in the cylinder.
Solution:
Radius of hemispherical bowl = 6 cm
.’. Volume of the water in the bowl

Question 21.
The diameter of a metallic sphere is 42 cm. It is metled and drawn into a cylindrical wire of 28 cm diameter. Find the length of the wire.
Solution:
Diameter of sphere = 42 cm
Radius of sphere =$$\\ \frac { 42 }{ 2 }$$ = 21 cm

Question 22.
A sphere of diameter 6 cm is dropped into a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?
Solution:
Radius of sphere = $$\\ \frac { 6 }{ 2 }$$ = 3 cm

Question 23.
A solid sphere of radius 6 cm is metled into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of the cylinder.
Solution:
Radius of solid sphere = 6 cm
Volume of solid sphere = $$\frac { 4 }{ 3 } \pi { r }^{ 3 }$$

Question 24.
A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 3.5 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical vessel, full of water, in such a Way that the whole solid is submerged in water. If the radius of the cylindrical vessel is 5 cm and its height is 10.5 cm, find the volume of water left in the cylindrical vessel.
Solution:
Radius of hemisphere (r) = 3.5 cm
Height of cone (h1) = 4 cm

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Mensuration Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 18 Mensuration Chapter Test, drop a comment below and we will get back to you at the earliest.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Constructions Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Constructions Chapter Test.

Question 1.
Draw a circle of radius 3 cm. Mark its centre as C and mark a point P such that CP = 7 cm. Using ruler and compasses only, Construct two tangents from P to the circle.
Solution:

Steps of Construction :

1. Draw a circle with centre C and radius 3 cm.
2. Mark a point P such that CP = 7 cm.
3. With CP as diameter, draw a circle intersecting the given circle at T and S.
4. Join PT and PS.
5. Draw a tangent at Q to the circle given. Which intersects PT at D.
6. Draw the angle bisector of ∠PDQ intersecting CP at E.
7. With centre E and radius EQ, draw a circle.
It will touch the tangent T and PS and the given circle at Q.
This is the required circle.

Question 2.
Draw a line AQ = 7 cm. Mark a point P on AQ such that AP = 4 cm. Using ruler and compasses only, construct :
(i) a circle with AP as diameter.
(ii) two tangents to the above circle from the point Q.
Solution:

Steps of construction :

1. Draw a line segment AQ = 7 cm.
2. From AQ,cut off AP = 4cm
3. With AP as diameter draw a circle with centre O.
4. Draw bisector of OQ which intersect OQ at M.
5. With centre M and draw a circle with radius MQ
which intersects the first circle at T and S.
6. Join QT and QS.
QT and QS are the tangents to the first circle.

Question 3.
Using ruler and compasses only, construct a triangle ABC having given c = 6 cm, b = 1 cm and ∠A = 30°. Measure side a. Draw carefully the circumcircle of the triangle.
Solution:

Steps of Construction :

1. Draw a line segment AC = 7 cm.
2. At C, draw a ray CX making an angle of 30°
3. With centre A and radius 6 cm draw an arc
which intersects the ray CX at B.
4. Join BA.
5. Draw perpendicular bisectors of AB and AC intersecting each other at O.
6. With centre O and radius OA or OB or OC,
draw a circle which will pass through A, B and C.
This is the required circumcircle of ∆ABC

Question 4.
Using ruler and compasses only, construct an equilateral triangle of height 4 cm and draw its circumcircle.
Solution:

Steps of Construction :

1. Draw a line XY and take a point D on it.
2. At D, draw perpendicular and cut off DA = 4 cm.
3. From A, draw rays making an angle of 30°
on each side of AD meeting the line XY at B and C.
4. Now draw perpendicular bisector of AC intersecting AD at O.
5. With centre O and radius OA or OB or OC
draw a circle which will pass through A, B and C.
This is the required circumcircle of ∆ABC.

Question 5.
Using ruler and compasses only :
(i) Construct a triangle ABC with the following data: BC = 7 cm, AB = 5 cm and ∠ABC = 45°.
(ii) Draw the inscribed circle to ∆ABC drawn in part (i).
Solution:
Steps of construction :

1. Draw a line segment BC = 7 cm.
2. At B, draw a ray BX making an angle of 45° and cut off BA = 5 cm.
3. Join AC.
4. Draw the angle bisectors of ∠B and ∠C intersecting each other at I.
5. From I, draw a perpendicular ID on BC.
6. With centre, I and radius ID, draw a circle
which touches the sides of ∆ABC at D, E and F respectively.
This is the required inscribed circle.

Question 6.
Draw a triangle ABC, given that BC = 4cm, ∠C = 75° and that radius of circumcircle of ∆ABC is 3 cm.
Solution:
Steps of Construction:

1. Draw a line segment BC = 4 cm
2. Draw the perpendicular bisector of BC.
3. From B draw an arc of 3 cm radius which intersects the perpendicular bisector at O.
4. Draw a ray CX making art angle of 75°
5. With centre O and radius 3 cm draw a circle which intersects the ray CX at A.
6. Join AB.
∆ABC is the required triangle

Question 7.
Draw a regular hexagon of side 3.5 cm construct its circumcircle and measure its radius.
Solution:
Steps of construction:

1. Draw a regular hexagon ABCDEF whose each side is 3.5 cm.
2. Draw the perpendicular bisector of AB and BC
which intersect each other at O.
3. Join OA and OB.
4. With centre O and radius OA or OB, draw a circle
which passes through A, B, C, D, E and P.
Then this is the required circumcircle.

Question 8.
Construct a triangle ABC with the following data: AB = 5 cm, BC = 6 cm and ∠ABC = 90°.
(i) Find a point P which is equidistant from B and C and is 5 cm from A. How many such points are there ?
(ii) Construct a circle touching the sides AB and BC, and whose centre is equidistant from B and C.
Solution:
Steps of Construction :

1. Draw a line segment BC = 6 cm.
2. At B, draw a ray BX making an angle of 90° and cut off BA = 5 cm.
3. Join AC.
4. Draw the perpendicular bisector of BC.
5. From A with 5 cm radius draw arc which intersects the perpendicular bisector of BC at P and P’.
There are two points.
6. Draw the angle bisectors of ∠B and ∠C intersecting at 0.
7. From O, draw OD ⊥ BC.
8. With centre O and radius OD, draw a circle which will touch the sides AB and BC.
This is the required circle.

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 17 Constructions Chapter Test are helpful to complete your math homework.

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## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 10 Reflection Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 10 Reflection Chapter Test.

Question 1.
The point P (4, – 7) on reflection in x-axis is mapped onto P’. Then P’ on reflection in the y-axis is mapped onto P”. Find the co-ordinates of P’ and P”. Write down a single transformation that maps P onto P”.
Solution:
P’ is the image of P (4, – 7) reflected in x-axis
∴ Co-ordinates of P’ are (4, 7)
Again P” is the image of P’ reflected in y-axis
∴ Co-ordinates of P” are ( – 4, 7)
∴ Single transformation that maps P and P” is in the origin.

Question 2.
The point P (a, b) is first reflected in the origin and then reflected in the y-axis to P’. If P’ has co-ordinates (3, – 4), evaluate a, b
Solution:
The co-ordinates of image of P(a, b) reflected in origin are ( – a, – b).
Again the co-ordinates of P’, image of the above point ( – a, – b) reflected in the y-axis are (a, – b).
But co-ordinates of P’ are (3, – 4)
∴a = 3 and – b = – 4
=>b = 4 Hence a = 3, b = 4.

Question 3.
A point P (a, b) becomes ( – 2, c) after reflection in the x-axis, and P becomes (d, 5) after reflection in the origin. Find the values of a, b, c and d.
Solution:
If the image of P (a, b) after reflected in the x-axis be (a, – b) but it Is given ( – 2, c).
a = – 2, c = – b
If P is reflected in the origin, then its co-ordinates will be ( – a, – b), but it is given (d, 5)
∴ – b = 5 =>b = – 5
d = – a = – ( – 2) = 2, c = – b = – ( – 5) = 5
Hence a = – 2, b = – 5, c = 5, d = 2 Ans.

Question 4.
A (4, – 1), B (0, 7) and C ( – 2, 5) are the vertices of a triangle. ∆ ABC is reflected in the y-axis and then reflected in the origin. Find the co-ordinates of the final images of the vertices.
Solution:
A (4, – 1), B (0, 7) and C ( – 2, 5) are the vertices of ∆ ABC.
After reflecting in y-axis, the co-ordinates of points will be A’ ( – 4, – 1), B’ (0, 7), C’ (2, 5). Again reflecting in origin, the co-ordinates of the images of the vertices will be
A” (4, 1), B” (0, – 7), C” ( – 2, – 5) Ans.

Question 5.
The points A (4, – 11), B (5, 3), C (2, 15), and D (1, 1) are the vertices of a parallelogram. If the parallelogram is reflected in the y-axis and then in the origin, find the co-ordinates of the final images. Check whether it remains a parallelogram. Write down a single transformation that brings the above change.
Solution:
The points A (4, – 11), B (5, 3), C (2, 15) and D (1, 1) are the vertices of a parallelogram. After reflecting in/-axis, the images of these points will be
A’ ( – 4, 11), B’ ( – 5, 3), C ( – 2, 15) and D’ ( – 1, 1).
Again reflecting these points in origin, the image of these points will be
A” (4, – 11), B” (5, – 3), C” (2, – 15), D” (0, – 1)
Yes, the reflection of single transformation is in x-axis. Ans.

Question 6.
Use a graph paper for this question (take 2 cm = 1 unit on both x and y axes).
(i) Plot the following points:
A (0, 4), B (2, 3), C (1, 1) and D (2, 0).
(ii) Reflect points B, C, D on 7-axis and write down their coordinates. Name the images as B’, C’, D’ respectively.
(iii) Join points A, B, C, D, D’, C’, B’ and A in order, so as to form a closed figure. Write down the equation of line of symmetry of the figure formed. (2017)
Solution:
(i) On graph A (0, 4), B (2, 3), C (1, 1) and D (2, 0)
(ii) B’ = ( – 2, 3), C’ = ( – 1, 1), D’ = ( – 2, 0)

The equation of the line of symmetry is x = 0

Question 7.
The triangle OAB is reflected in the origin O to triangle OA’B’. A’ and B’ have coordinates ( – 3, – 4) and (0, – 5) respectively.
(i) Find the co-ordinates of A and B.
(ii) Draw a diagram to represent the given information.
(iii) What kind of figure is the quadrilateral ABA’B’?
(iv) Find the coordinates of A”, the reflection of A in the origin followed by reflection in the y-axis.
(v) Find the co-ordinates of B”, the reflection of B in the x-axis followed by reflection in the origin.
Solution:
∆ OAB is reflected in the origin O to ∆ OA’B’, Co-ordinates of A’ = ( – 3, – 4), B’ (0, – 5).
.’. Co-ordinates of A will be (3, 4) and of B will be (0, 5).
(ii) The diagram representing the given information has been drawn here.
(iii) The figure in the diagram is a rectangle.
(iv) The co-ordinates of B’, the reflection of B is the x-axis are (0, – 5) and co-ordinates of B”, the reflection in origin of the point (0, – 5) will be (0, 5).
(v) The co-ordinates of the points, the reflection of A in the origin are ( – 3, – 4) and coordinates of A”, the reflected in y-axis of the point ( – 3, – 4) are (3, – 4) Ans

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 10 Reflection Chapter Test are helpful to complete your math homework.

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## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test.

Question 1.
(a) In the figure (i) given below, triangle ABC is equilateral. Find ∠BDC and ∠BEC.
(b) In the figure (ii) given below, AB is a diameter of a circle with centre O. OD is perpendicular to AB and C is a point on the arc DB. Find ∠BAD and ∠ACD

Solution:
(a) ∆ABC is an equilateral triangle.

Question 2.
(a) In the figure given below, AB is a diameter of the circle. If AE = BE and ∠ADC = 118°, find
(i) ∠BDC (ii) ∠CAE.

(b) In the figure given below, AB is the diameter of the semi-circle ABCDE with centre O. If AE = ED and ∠BCD = 140°, find ∠AED and ∠EBD. Also Prove that OE is parallel to BD.

Solution:
(a) Join DB, CA and CB.

Question 3.
(a) In the figure (i) given below, O is the centre of the circle. Prove that ∠AOC = 2 (∠ACB + ∠BAC).
(b) In the figure (ii) given below, O is the centre of the circle. Prove that x + y = z.

Solution:
(a) Given : O is the centre of the circle.
To Prove : ∠AOC = 2 (∠ACB + ∠BAC).
Proof: In ∆ABC,
∠ACB + ∠BAC + ∠ABC = 180°
(Angles of a triangle)

Question 4.
(a) In the figure (i) given below, AB is diameter of a circle. If DC is parallel to AB and ∠CAB = 25°, find :
(b) In the figure (ii) given below, the centre O of the smaller circle lies on the circumference of the bigger circle. If ∠APB = 70° and ∠BCD = 60°, find :
(i) ∠AOB (ii) ∠ACB

Solution:
(a) AB is diameter and DC || AB,

Question 5.
(a) In the figure (i) given below, ABCD is a cyclic quadrilateral. If AB = CD, Prove that AD = BC.
(b) In the figure (ii) given below, ABC is an isosceles triangle with AB = AC. If ∠ABC = 50°, find ∠BDC and ∠BEC.

Solution:
(a) Given : ABDC is a cyclic quadrilateral AB = CD.

Question 6.
A point P is 13 cm from the centre of a circle. The length of the tangent drawn from P to the circle is 12 cm. Find the distance of P from the nearest point of the circle.
Solution:
Join OT, OP = 13 cm and TP = 12 cm

Question 7.
Two circles touch each other internally. Prove that the tangents drawn to the two circles from any point on the common tangent are equal in length.
Solution:
Given : Two circles with centre O and O’ touch each other internally at P.

Question 8.
From a point outside a circle, with centre O, tangents PA and PB are drawn. Prove that
(i) ∠AOP = ∠BOP.
(ii) OP is the perpendicular bisector of the chord AB.
Solution:
Given : From a point P, outside the circle with centre O. PA and PB are the tangents to the circle, OA, OB and OP are joined.

Question 9.
(a) The figure given below shows two circles with centres A, B and a transverse common tangent to these circles meet the straight line AB in C. Prove that:
AP : BQ = PC : CQ.

(b) In the figure (ii) given below, PQ is a tangent to the circle with centre O and AB is a diameter of the circle. If QA is parallel to PO, prove that PB is tangent to the circle.

Solution:
(a) Given : Two circles with centres A and B and a transverse common tangent to these circles meet AB at C.

Question 10.
In the figure given below, two circles with centres A and B touch externally. PM is a tangent to the circle with centre A and QN is a tangent to the circle with centre B. If PM = 15 cm, QN = 12 cm, PA = 17 cm and QB = 13 cm, then find the distance between the centres A and B of the circles.

Solution:
In the given figure, two chords with centre A and B touch externally. PM is a tangent to the circle with centre A and QN is tangent to the circle with centre B. PM = 15 cm, QN = 12 cm, PA = 17 cm, QB = 13 cm. We have to find AB.

Question 11.
Two chords AB, CD of a circle intersect externally at a point P. If PB = 7 cm, AB = 9 cm and PD = 6 cm, find CD.
Solution:
∵ AB and CD are two chords of a circle which intersect each other at P, outside the circle.

Question 12.
(a) In the figure (i) given below, chord AB and diameter CD of a circle with centre O meet at P. PT is tangent to the circle at T. If AP = 16 cm, AB = 12 cm and DP = 2 cm, find the length of PT and the radius of the circle

(b) In the figure (ii) given below, chord AB and diameter CD of a circle meet at P. If AB = 8 cm, BP = 6 cm and PD = 4 cm, find the radius of the circle. Also find the length of the tangent drawn from P to the circle. .

Solution:
Given : (a) AB is chord of a circle with centre O and PT is tangent and CD is the diameter of the circle which meet at P.
AP = 16 cm, AB = 12 cm, OP = 2 cm
∴PB = PA – AB = 16 – 12 = 4 cm
∵ABP is a secant and PT is tangent.
∴PT² = PA x PB .

Question 13.
In the figure given below, chord AB and diameter PQ of a circle with centre O meet at X. If BX = 5 cm, OX = 10 cm and.the radius of the circle is 6 cm, compute the length of AB. Also find the length of tangent drawn from X to the circle.

Solution:
Chord AB and diameter PQ meet at X on producing outside the circle

Question 14.
(a) In the figure (i) given below, ∠CBP = 40°, ∠CPB = q° and ∠DAB = p°. Obtain an equation connecting p and q. If AC and BD meet at Q so that ∠AQD = 2 q° and the points C, P, B and Q are concyclic, find the values of p and q.
(b) In the figure (ii) given below, AC is a diameter of the circle with centre O. If CD || BE, ∠AOB = 130° and ∠ACE = 20°, find:
(i)∠BEC (ii) ∠ACB
(iii) ∠BCD (iv) ∠CED.

Solution:
(a) (i) Given : ABCD is a cyclic quadrilateral.

Question 15.
(a) In the figure (i) given below, APC, AQB and BPD are straight lines.
(i) Prove that ∠ADB + ∠ACB = 180°.
(ii) If a circle can be drawn through A, B, C and D, Prove that it has AB as diameter

(b) In the figure (ii) given below, AQB is a straight line. Sides AC and BC of ∆ABC cut the circles at E and D respectively. Prove that the points C, E, P and D are concyclic.

Solution:
(a) Given : In the figure, APC, AQB and BPD are straight lines.

Question 16.
(a) In the figure (i) given below, chords AB, BC and CD of a circle with centre O are equal. If ∠BCD = 120°, find
(i) ∠BDC (ii) ∠BEC
(iii) ∠AEC (iv) ∠AOB.
Hence Prove that AOAB is equilateral.
(b) In the figure (ii) given below, AB is a diameter of a circle with centre O. The chord BC of the circle is parallel to the radius OD and the lines OC and BD meet at E. Prove that
(i) ∠CED = 3 ∠CBD (ii) CD = DA.

Solution:
(a) In ∆BCD, BC = CD
∠CBD = ∠CDB
But ∠BCD + ∠CBD + ∠CDB = 180°
(∵ Angles of a triangle)

Question 17.
(a) In the adjoining figure, (i) given below AB and XY are diameters of a circle with centre O. If ∠APX = 30°, find
(i) ∠AOX (ii) ∠APY (iii) ∠BPY (iv) ∠OAX.

(b) In the figure (ii) given below, AP and BP are tangents to the circle with centre O. If ∠CBP = 25° and ∠CAP = 40°, find :
(i) ∠ADB (ii) ∠AOB (iii) ∠ACB (iv) ∠APB.

Solution:
(a) AB and XY are diameters of a circle with centre O.
∠APX = 30°.
To find :
(i) ∠AOX (ii) ∠APY
(iii) ∠BPY (iv) ∠OAX

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test are helpful to complete your math homework.

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## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Locus Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Locus Chapter Test.

Question 1.
Draw a straight line AB of length 8 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.
Solution:
(i) Draw a line segment AB = 8 cm.

(ii) Draw the perpendicular bisector of AB intersecting AB at D.
∴ Every point P on it will be equidistant from A and B.
(iii) Take a point P on the perpendicular bisector.
(iv) Join PA and PB.
Proof : In ∆PAD and ∆PBD
PD = PD (common)
AD = BD (D is mid-point of AB)
∠PDA = ∠PDB (each 90°)
∴ ∆ PAD ≅ ∆ PBD (SAS axiom of congruency)
∴PA = PB (c.p.c.t.)
Similarly we can prove any other point on the perpendicular bisector of AB is equidistant from A and B.
Hence Proved.

Question 2.
A point P is allowed to travel in space. State the locus of P so that it always remains at a constant distance from a fixed point C.
Solution:
The point P is moving in the space and it is at a constant distance from a fixed point C.
∴ Its locus is a sphere.

Question 3.
Draw a line segment AB of length 7 cm. Construct the locus of a point P such that area of triangle PAB is 14 cm².
Solution:
Base of ∆PAB = 7 cm
and its area = 14 cm²

Now draw a line XY parallel to AB at a distance of 4 cm.
Now take any point P on XY
Join PA and PB
area of ∆PAB = 14 cm.
Hence locus of P is the line XY which is parallel to AB at distance of 4 cm.

Question 4.
Draw a line segment AB of length 12 cm. Mark M, the mid-point of AB. Draw and describe the locus of a point which is
(i) at a distance of 3 cm from AB.
(ii) at a distance of 5 cm from the point M. Mark the points P, Q, R, S which satisfy both the above conditions. What kind of quadrilateral is PQRS ? Compute the area of the quadrilateral PQRS.
Solution:
Steps of Construction :

(i) Take a line AB = 12 cm
(ii) Take M, the mid point of AB.
(iii) Draw straight lines CD and EF parallel to AB at a distance of 3 cm.
(iv) With centre M and radius 5 cm, draw areas which intersects CD at P and Q and EF at R and S.
(v) Join QR and PS.
PQRS is a rectangle where length
PQ = 8 cm.
Area of rectangle PQRS = PQ x RS = 8 x 6 = 48 cm²

Question 5.
AB and CD are two intersecting lines. Find the position of a point which is at a distance of 2 cm from AB and 1.6 cm from CD.
Solution:
(i) AB and CD are the intersecting lines which intersect each other at O.

(ii) Draw a line EF parallel to AB and GH parallel to CD intersecting each other at P
P is the required point.

Question 6.
Two straight lines PQ and PK cross each other at P at an angle of 75°. S is a stone on the road PQ, 800 m from P towards Q. By drawing a figure to scale 1 cm = 100 m, locate the position of a flag staff X, which is equidistant from P and S, and is also equidistant from the road.
Solution:
1 cm = 100 cm
800 m = 8 cm.
Steps of Construction :
(i) Draw the lines PQ and PK intersecting each other at P making an angle of 75°.

(ii) Take a point S on PQ such that PS = 8 cm.
(iii) Draw the perpendicular bisector of PS.
(iv) Draw the angle bisector of ∠KPS intersecting the perpendicular bisector at X. X is the required point which is equidistant from P and S and also from PQ and PK.

Question 7.
Construct a rhombus PQRS whose diagonals PR, QS are 8 cm and 6 cm respectively. Find by construction a point X equidistant from PQ, PS and equidistant from R, S. Measure XR.
Solution:
Steps of Construction :

(i) Take PR = 8 cm and draw the perpendicular bisector of PR intersecting it at O.
(ii) From O, out. off OS = OQ = 3 cm
(iii) Join PQ, QR, RS and SP.
PQRS is a rhombus. Whose diagonal are PR and QS.
(iv) PR is the bisector of ∠SPQ.
(v) Draw the perpendicular bisector of SR intersecting PR at X
∴ X is equidistant from PQ and PS and also from S and R.
On measuring length of XR = 3.2 cm (approx)

Question 8.
Without using set square or protractor, construct the parallelogram ABCD in which AB = 5.1 cm. the diagonal AC = 5.6 cm and the diagonal BD = 7 cm. Locate the point P on DC, which is equidistant from AB and BC.
Solution:
Steps of Construction :

(i) Take AB = 5.1 cm
(ii) At A, with readius $$\\ \frac { 5.6 }{ 2 }$$ = 2.8 cm and at B with radius $$\\ \frac { 7.0 }{ 2 }$$ = 3.5 cm, draw two arcs
intersecting each other at O.
(iii) Join AO and produce it to C such that OC = AD = 2.8 cm and join BO and produce it to D such that BO = OD = 3.5 cm
(iv) Join BC, CD, DA
ABCD is a parallelogram.
(v) Draw the angle bisector of ∠ABC intersecting CD at P. P is the required point which is equidistant from AB and BC.

Question 9.
By using ruler and compass only, construct a quadrilateral ABCD in which AB = 6.5 cm, AD = 4cm and ∠DAB = 75°. C is equidistant from the sides AB and AD, also C is equidistant from the points A and B.
Solution:
Steps of Construction :

(i) Draw a line segment AB = 6.5 cm.
(ii) At A, draw a ray making an angle of 75° and cut off AD = 4 cm.
(iii) Draw the bisector of ∠DAB.
(iv) Draw perpendicular bisector of AB intersecting the angle bisector at C.
(v) Join CB and CD.

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Locus Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. APlusTopper try to provide online math tutoring for you.

## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test.

Question 1.
In the given figure, ∠1 = ∠2 and ∠3 = ∠4. Show that PT x QR = PR x ST.
Solution:
Given : In the given figure,
∠1 = ∠1 and ∠3 = ∠4

To prove : PT x QR = PR x ST
Proof: ∠1 = ∠2
∠1 + ∠6 = ∠2 + ∠6
∠SPT = ∠QPR
In ∆PQR and ∆PST

Question 2.
In the adjoining figure, AB = AC. If PM ⊥ AB and PN ⊥ AC, show that PM x PC = PN x PB.

Solution:
Given : In the given figure,
AB = AC, PM ⊥ AB and PN ⊥ AC
To prove : PM x PC = PN x PB
Proof: In ∆ABC, AB = AC
∠B = ∠C
Now in ∆CPN and ∆BPM,

Question 3.
(a) In the figure (1) given below. ∠AED = ∠ABC. Find the values of x and y.
(b) In the fig. (2) given below, CD = $$\\ \frac { 1 }{ 2 }$$ AC, B is mid-point of AC and E is mid-point of DF. If BF || AG, prove that :
(i) CE || AG
(ii) 3 ED = GD.

Solution:
(a) Given : In following figure, ∠AED = ∠ABC
Required : The values of x and y.
∠AED = ∠ABC (given)
∠A = ∠A (common)
(By A.A. axiom of similarity)

Question 4.
In the given figure, 2 AD = BD, E is mid-point of BD and F is mid-point of AC and EC || BH. Prove that:
(i) DF || BH
(ii) AH = 3 AF.

Solution:
Given : E is the mid-point of BD and F is mid-point of AC also 2 AD = BD and EC || BH
To Prove : (i) DF || BH
(ii) AH = 3 AF

Question 5.
In a ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE.
Solution:
Given : In ∆ABC, D and E are the points on the sides AB and AC respectively
DE || BC
AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm, BC = 5 cm

Question 6.
In a ∆ABC, D and E are points on the sides AB and AC respectively such that AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and AC = 8.8cm. Is DE || BC? Justify your answer.
Solution:
In ∆ABC, D and E are points on the sides AB and AC respectively
AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and AC = 8.8 cm

Question 7.
In a ∆ABC, DE is parallel to the base BC, with D on AB and E on AC. If $$\frac { AD }{ DB } =\frac { 2 }{ 3 } ,\frac { BC }{ DE }$$
Solution:
In ∆ABC, DE || BC
D is on AB and E is on AC

Question 8.
If the area of two similar triangles are 360 cm² and 250 cm² and if one side of the first triangle is 8 cm, find the length of the corresponding side of the second triangle.
Solution:
Let ∆ABC and ∆DEF are similar and area of
∆ABC = 360 cm²

Question 9.
In the adjoining figure, D is a point on BC such that ∠ABD = ∠CAD. If AB = 5 cm, AC = 3 cm and AD = 4 cm, find
(i) BC
(ii) DC
(iii) area of ∆ACD : area of ∆BCA.

Solution:
In ∆ABC and ∆ACD
∠C = ∠C (Common)
∴ ∆ABC ~ ∆ACD

Question 10.
In the adjoining figure the diagonals of a parallelogram intersect at O. OE is drawn parallel to CB to meet AB at E, find area of DAOE : area of ||gm ABCD.

Solution:
(a) In the figure
Diagonals of parallelogram ABCD are AC and BD which intersect each other at O.
OE is drawn parallel to CB to meet AB in E.

Question 11.
In the given figure, ABCD is a trapezium in which AB || DC. If 2AB = 3DC, find the ratio of the areas of ∆AOB and ∆COD.

Solution:
In the given figure, ABCD is trapezium in
which AB || DC, 2AB = 3DC

Question 12.
In the adjoining figure, ABCD is a parallelogram. E is mid-point of BC. DE meets the diagonal AC at O and meet AB (produced) at F. Prove that .

(i) DO : OE = 2 : 1
(ii) area of ∆OEC : area of ∆OAD = 1 : 4.
Solution:
Given : In || gm ABCD,
E is mid point of BC and DE meets the diagonal AC at O and meet AB produced at F.
To prove : (i) DO : OE = 2 : 1
(ii) area of ∆OEC : area of ∆OAD = 1 : 4
Proof: In ∆AOD and ∆EDC
∠AOD = ∠EOC (vertically opposite angle)
∆AOD ~ ∆EOC (AA postulate)

Question 13.
A model of a ship is made to a scale of 1 : 250. Calculate :
(i) the length of the ship, if the length of model is 1.6 m.
(ii) the area of the deck of the ship, if the area of the deck of model is 2.4 m².
(iii) the volume of the model, if the volume of the ship is 1 km³.
Solution:
Scale factor (k) of the model of the ship = $$\\ \frac { 1 }{ 250 }$$
(i) Length of model = 1.6 m

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. APlusTopper try to provide online math tutoring for you.