Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots

Selina Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 3 Squares and Square Roots

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Squares and Square Roots Exercise 3A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the square of :
(i) 59
(ii) 63
(iii) 15
Solution:
(i) Square of 59= 59 x 59 = 3481
(ii) Square of 6.3 = 6.3 x 6.3 = 39.69
(iii) Square of 15 = 15 x 15 = 225

Question 2.
By splitting into prime factors, find the square root of :
(i) 11025
(if) 396900
(iii) 194481
Solution:

Question 3.
(i) Find the smallest number by which 2592 be multiplied so that the product is a perfect square.
(ii) Find the smallest number by which 12748 be mutliplied so that the product is a perfect square?
Solution:

On grouping the prime factors of 2592 as shown; on factor i.e. 2 is left which cannot be paired with equal factor.

The given number should be multiplied by 2 to make the given number a perfect square.

On grouping the prime factors of 12748 as shown; one factor i.e. 3187 is left which cannot be paired with equal factor.

The given number should be multiplied by 3187.

Question 4.
Find the smallest number by which 10368 be divided, so that the result is a perfect square. Also, find the square root of the resulting numbers.
Solution:

Question 5.
Find the square root of :
(i) 0.1764
(ii) \(96\frac { 1 }{ 25 }\)
(iii) 0.0169
Solution:

Question 6.
Evaluate


Solution:

Question 7.

Solution:

Question 8.
A man, after a tour, finds that he had spent every day as many rupees as the number of days he had been on tour. How long did his tour last, if he had spent in all ₹ 1,296
Solution:
Let the number of days he had spent = x
Number of rupees spent in each day = x
Total money spent = x x x = x² = 1,296 (given)

Question 9.
Out of 745 students, maximum are to be arranged in the school field for a P.T. display, such that the number of rows is equal to the number of columns. Find the number of rows if 16 students were left out after the arrangement.
Solution:
Total number of students = 745
Students left after standing in arrangement = 16
No. of students who were to be arranged = 745 – 16 = 729
The number of rows = no. of students in each row
No. of rows = √729

Question 10.
13 and 31 is a strange pair of numbers such that their squares 169 and 961 are also mirror images of each other. Find two more such pairs.
Solution:

Question 11.
Find the smallest perfect square divisible by 3, 4, 5 and 6.
Solution:
L.C.M. of 3, 4, 5, 6 = 2 x 2 x 3 x 5 = 60

in which 3 and 5 are not in pairs L.C.M. = 2 x 3 x 2 x 5 = 60
We should multiple it by 3 x 5 i.e. by 15
Required perfect square = 60 x 15 = 900

Question 12.
If √784 = 28, find the value of:
(i) √7.84 + √78400
(ii) √0.0784 + √0.000784
Solution:

Squares and Square Roots Exercise 3B – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Find the square root of:
(i) 4761
(ii) 7744
(iii) 15129
(iv) 0.2916
(v) 0.001225
(vi) 0.023104
(vii) 27.3529
Solution:

Question 2.
Find the square root of:
(i) 4.2025
(ii) 531.7636
(iii) 0.007225
Solution:

Question 3.
Find the square root of:
(i) 245 correct to two places of decimal.
(ii) 496 correct to three places of decimal.
(iii) 82.6 correct to two places of decimal.
(iv) 0.065 correct to three places of decimal.
(v) 5.2005 correct to two places of decimal.
(vi) 0.602 correct to two places of decimal
Solution:

Required square root = 0.78 upto two places of decimals.

Question 4.
Find the square root of each of the following correct to two decimal places:
(i) \(3\frac { 4 }{ 5 }\)
(ii) \(6\frac { 7 }{ 8 }\)
Solution:

Question 5.
For each of the following, find the least number that must be subtracted so that the resulting number is a perfect square.
(i) 796
(ii) 1886
(iii) 23497
Solution:

Question 6.
For each of the following, find the least number that must be added so that the resulting number is a perfect square.
(i) 511
(ii) 7172
(iii) 55078
Solution:
(i) 511
Taking square root of 511, we find that 27 has been left We see that 511 is greater than (22)²

On adding the required number to 511, we get (23)² i.e., 529
So, the required number = 529 – 511 = 18
(ii) 7172
Taking square root of 7172, we find that 116 has been left
We see that 7172 is greater than (84)²

Taking square root of 55078, we find that 322 has been left
We see that 55078 is greater than (234)²
On adding the required number to 55078, we get (235)² i.e., 55225
Required number = 55225 – 55078 = 147

Question 7.
Find the square root of 7 correct to two decimal places; then use it to find the value of \(\sqrt { \frac { 4+\sqrt { 7 } }{ 4-\sqrt { 7 } } }\) correct to three significant digits.
Solution:
√7 = 2.645 = 2.65

Question 8.
Find the value of √5 correct to 2 decimal places; then use it to find the square root of \(\sqrt { \frac { 3-\sqrt { 5 } }{ 3+\sqrt { 5 } } }\) correct to 2 significant digits.
Solution:

Question 9.
Find the square root of:
(i) \(\frac { 1764 }{ 2809 }\)
(ii) \(\frac { 507 }{ 4107 }\)
(iii) \(\sqrt { 108\times 2028 }\)
(iv) 0.01 + √0.0064
Solution:

Question 10.
Find the square root of 7.832 correct to :
(i) 2 decimal places
(ii) 2 significant digits.
Solution:
Square root of 7.832

√7.832 = 2.80 upto two decimal places
= 2.8 upto two significant places

Question 11.
Find the least number which must be subtracted from 1205 so that the resulting number is a perfect square.
Solution:
Clearly, if 49 is subtracted from 1205, the number will be a perfect square.

Question 12.
Find the least number which must be added to 1205 so that the resulting number is a perfect square.
Solution:

Question 13.
Find the least number which must be subtracted from 2037 so that the resulting number is a perfect square.
Solution:
Clearly; if 12 is subtracted from 2037, the remainder will be a perfect square.

Question 14.
Find the least number which must be added to 5483 so that the resulting number is a perfect square.
Solution:

Squares and Square Roots Exercise 3A – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
Seeing the value of the digit at unit’s place, state which of the following can be square of a number :
(i) 3051
(ii) 2332
(iii) 5684
(iv) 6908
(v) 50699
Solution:
We know that the ending digit (the digit at units place) of the square of a number is 0, 1, 4, 5, 6, or 9
So, the following numbers can be squares : 3051, 5684, and 50699 i.e., (i), (iii), and (v)

Question 2.
Squares of which of the following numbers will have 1 (one) at their unit’s place :
(i) 57
(ii) 81
(iii) 139
(iv) 73
(v) 64
Solution:
The square of the following numbers will have 1 at their units place as (1)² = 1, (9)² = 81
81 and 139 i.e., (i) and (iii)

Question 3.
Which of the following numbers will not have 1 (one) at their unit’s place :
(i) 322
(ii) 572
(iii) 692
(iv) 3212
(v) 2652
Solution:
The square of the following numbers will not have 1 at their units place : as only (1)² = 1, (9)² = 81 have 1 at then units place
322, 572, 2652 i.e., (i), (ii) and (v)

Question 4.
Square of which of the following numbers will not have 6 at their unit’s place :
(i) 35
(ii) 23
(iii) 64
(iv) 76
(v) 98
Solution:
The squares of the following numbers, Will not have 6 at their units place as only (4)² = 16, (6)² = 36 has but its units place 35, 23 and 98 i.e., (i), (ii), and (v)

Question 5.
Which of the following numbers will have 6 at their unit’s place :
(i) 262
(ii) 492
(iii) 342
(iv) 432
(v) 2442
Solution:
The following numbers have 6 at their units place as (4)² = 16, (6)² = 36 has 6 at their units place 262, 342, 2442 i.e., (i), (iii) and (v)

Question 6.
If a number ends with 3 zeroes, how many zeroes will its square have ?
Solution:
We know that if a number ends with n zeros, then its square will have 2n zeroes at their ends
A number ends with 3 zeroes, then its square will have 3 x 2 = 6 zeroes

Question 7.
If the square of a number ends with 10 zeroes, how many zeroes will the number have ?
Solution:
We know that if a number ends with n zeros Then its square will have 2n zeroes Conversely, if square of a number have 2n zeros at their ends then the number will have n zeroes
The square of a number ends 10 zeroes, then the number will have \(\frac { 10 }{ 2 }\) = 5 zeroes

Question 8.
Is it possible for the square of a number to end with 5 zeroes ? Give reason.
Solution:
No, it is not possible for the square of a number, to have 5 zeroes which is odd because the number of zeros of the square must be 2n zeroes i.e., even number of zeroes.

Question 9.
Give reason to show that none of the numbers, given below, is a perfect square.
(i) 2162
(ii) 6843
(iii) 9637
(iv) 6598
Solution:
A number having 2,3,7 or 8 at the unit place is never a perfect square.

Question 10.
State, whether the square of the following numbers is even or odd?
(i) 23
(ii) 54
(iii) 76
(iv) 75
Solution:
(i) 23 – odd
(ii) 54 – even
(iii) 76 – odd
(iv) 75 – even

Question 11.
Give reason to show that none of the numbers 640, 81000 and 3600000 is a perfect square.
Solution:
No, number has an even number of zeroes.

Question 12.
Evaluate:
(i) 37² – 36²
(ii) 85² – 84²
(iii) 101² – 100²
Solution:

Question 13.
Without doing the actual addition, find the sum of:
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
(ii) 1 + 3 + 5 + 7 + 9 + ……………… + 39 + 41
(iii) 1 + 3 + 5 + 7 + 9 + ………………… + 51 + 53
Solution:
(i) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 +23
= Sum of first 12 odd natural numbers = 122 = 144
(ii) 1+3 + 5 + 7 + 9 + ……….. + 39 + 41
= Sum of first 21 odd natural numbers = 212 = 441
(iii) 1 + 3 + 5 + 7 + 9 + ……………. + 51 + 53
= Sum of first 27 odd natural number = 272 = 729

Question 14.
Write three sets of Pythagorean triplets such that each set has numbers less than 30.
Solution:
The three sets of Pythagorean triplets such that each set has numbers less than 30 are 3, 4 and 5; 6, 8 and 10; 5, 12 and 13
Proof: