## RS Aggarwal Class 8 Chapter 3 Assertion Reason Square and Square Roots Solutions

**Squares and Square Roots**

**Assertion – Reason Questions**

**Directions (Questions 1-7): Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:**

**(a.) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).**

**(b.) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).**

**(c.) Assertion (A) is true but Reason (R) is false.**

**(d.) Assertion (A) is false but Reason (R) is true.**

**1.)** **Assertion (A):** Each of the numbers 400, 900, 1600 and 2500 is a perfect square.

**Reason (R):** A number ending in an even number of zeroes is always a perfect square.

Ans: Option (c) is the correct answer as assertion A is true and reason R is false.

√400=20; √900=30; √1600=40; √2500=50. So, A is true.

None of the numbers 200, 300, 500, 600, 700,… ending with even number of zeroes is a perfect square. So, R is false.

**2.) Assertion (A):** √1(11/25)=1(1/5)

**Reason (R):** 25 is a perfect square, but 11 is not.

Ans: Option (b) is correct answer as the assertion A and reason R are both correct however R does not explain A.

√1(11/25)= √36/25=6/5=1(1/5) So, A is true.

√25=5, but 11 is not a perfect square.

So, we can say that R is true but R doesn’t explain A.

**3.) Assertion (A):** A number ending in 2, 3, 7 or 8 is never a perfect square.

**Reason (R):** None of the numbers from 1 to 9 when squared ends in 2, 3, 7 or 8.

Ans: Option (a) is the correct option as the assertion A is correct and the reason R is also correct and explains the assertion accurately.

Units digit in x |
1 | 2 | 3 | 4 | 5 | 6 | 7 |
8 |
9 |

Units digit in x^{2} |
1 | 4 | 9 | 6 | 5 | 6 | 9 | 4 |
1 |

From this we can see that the numbers from 1 to 9 when squared does not end with the numbers 2, 3, 7 and 8.

So we can say that R is true and correctly explains A.

**4.) Assertion (A):** (3/4)^{2} <3/4

**Reason (R):** The square of a proper fraction is smaller than the fraction.

Ans: Options (a) is the correct answer as the assertion A is correct and the reason R is also correct. R correctly explains A.

(3/4)^{2}= 9/16 <3/4. So, A is true.

So we can say that R is true and correctly explains A.

**5.) Assertion (A):** the Square of a prime number may be prime or composite.

**Reason (R):** The Square of a number can never be prime.

Ans: option (d) is correct answer as the assertion A is false and the reason R is true.

The square of a number has at least 3 factors. So, It can never be prime.

Hence , A is false but R is true.

**6.) Assertion (A):** Since 9 is a perfect square, so each one of 0.9, 0.09, 0.009, 0.0009 is a perfect square.

**Reason (R):**0.9=9/10, 0.09=9/100, 0.009=9/1000, 0.0009= 9/10000

Ans: Option (d) is the correct answer as the assertion A is false and the reason R is true.

√0.09=√9/100=3/10=0.3;

√0.0009= √9/10000=3/100=0.03.

But 0.9 and 0.009 are not perfect squares.

So we can say that A is false.

From the above observations it is clear that R is true.

**7.) Assertion (A):** For positive values of a rational number n, n^{2}>n only if n>1.

**Reason (R):** For 0<n<1, we have n^{2}<n.

Ans: Option (a) is the correct answer as R is true and correctly explains A.

**Case – Based Questions**

**Directions (Questions 8 to 12): Examine the following case study carefully and answer the questions that follow.**

**There are m boys and n girls in a class. Each boy donated twice as many rupees as is the total number of boys and each girl donated thrice as many rupees as is the total number of girls. The boys collected ₹ 1152 in all while the girls collected ₹ 768 in all.**

**8.) What is the value of m?**

(a.) 22

(b.) 24

(c.) 26

(d.) 28

Ans: option (b) is the correct answer.

Amount donated by each boy= ₹ (2m)

Total amount collected by boys= ₹ (m×2m)= ₹ (2m^{2}).

∴ 2m^{2} = 1152=>m^{2}=576=>m=√576=24.

The value of m or the number of boys in the class is 24.

**9.) What is the value of n?**

(a.) 12

(b.) 14

(c.) 16

(d.) 18

Ans:

Option (c) is the correct answer.

Amount donated by each girl= ₹ (3m).

Total amount collected by girls= ₹ (n ×3n)= ₹ (3n^{2}).

∴ 3n^{2}= 768=>n^{2}=256=>n= √256=16.

∴ The value of n or the number of girls is 16.

**10.) Had there been (m+ 3) boys and (n-3) girls in the class, the total amount collected would be**

(a.) ₹ 1925

(b.) ₹ 1940

(c.) ₹ 1965

(d.) ₹ 2000

Ans: option (c) is the answer

According to the problem

Number of boys= 27 and number of girls= 13.

Total amount collected= ₹ [27× (2×27)+ 13 × (3 × 13)]

= ₹ (2 × 729 +3 ×169)= ₹ (1458 + 507)

= ₹ 1965.

The total amount collected according to the question is ₹1965.

**11.) Let p= m ^{2}+n^{2}. Then, which of the following is true for p?**

(a.) p is a perfect square.

(b.) (p+9) is a perfect square.

(c.) (p+ 2) is a perfect square.

(d.) (p-32) is a perfect square.

Ans:

Option (b) is the answer

p=m^{2}+n^{2}= (24)^{2} +(16)^{2}= 576 +256= 832.

Now, (28)^{2}< 832 < (29)^{2}.

So, next perfect square after 832= (29)^{2}= 841.

Now, 841- 832= 9.

∴ (p+9) is a perfect square.

**12.) Let q=m ^{2}-n^{2}. Then, the least number to be added to q to make it a perfect square is **

(a.) 4

(b.) 6

(c.) 8

(d.) 12

Solutions

Ans: Option (a) is the correct answer

q = m^{2}-n^{2} = (24)^{2}– (16)^{2}= 576- 256 = 320.

Now, (17)^{2} <320 <(18)^{2}.

∴ Least number to be added= (18)^{2}– 320= 324- 320= 4.