## RS Aggarwal Class 8 Chapter 5 Assertion Reason Playing with Numbers Solutions

**Directions (Questions 1-8): Each question consists of two-statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:**

**(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) : The generalized from of a three-digit number abc is 100a + 10b+c

Reason (R): 100a +10b + c denotes the sum of the place values of the digits of the number.

2.) Assertion (A): If 23x is divisible by 9 then x=6.

Reason (R): A number is divisible by 9 if the sum of its digits is divisible by 9.

3.) Assertion (A): For 4y5 to be visible by 3, y can have at the most 3 possible values.

Reason (R): Every number divisible by 9 must be divisible by 3.

4.) Assertion (A): Every three digit number consisting of three identical digits is divisible by each one of 3, 37 and 111.

Reason (R): 100a + 10a + a= 111a, which is clearly divisible by each one of 3, 37 and 111.

5.) Assertion (A): The sum of a two- digit number and the number obtained by reversing the digits is always divisible by 11.

Reason (R): The difference of a two- digit number and the number obtained by reversing the digits is always divisible by 9.

6.) Assertion (A): If N ÷ 5 leaves remainder 3 and N ÷ 2 leaves remainder 0 then N ÷ 10 leaves remainder 4.

Reason (R): N is a multiple of 2 and leaves remainder 3 when divided by 5.

7.) Assertion (A): If 31a2 is divisible by 6 then a has 3 possible values.

Reason (R): For a number to be divisible by 6 it must be divisible by both 2 and 3.

8.) Assertion (A): If a number is divisible by 18 it is divisible by each one of 2, 3, 6 and 9.

Reason (R): A number divisible by n is divisible by each of the factors of n.

**CASE-BASED QUESTIONS**

**Directions (Questions 9 to 12): Study the following information carefully and answer the given questions.**

a, b, c are three different nonzero digits. Using these digits, different numbers are formed.

**9.) If a>b>c then the difference of three-digit number abc and the number obtained by reversing the digits is always divisible by**

(a.) 9 only

(b.) 11 only

(c.) both 9 and 11

(d.) none of 9 and 11

**10.) How many different three-digit numbers can be formed using the given digits it being given that repetition of digits is not allowed?**

(a.) 3

(b.) 4

(c.) 6

(d.) 9

**11.)The sum of all possible three-digit numbers formed above, is always divisible by**

(a) 2 only

(b) 37 only

(c) 111 only

(d) each one of 2. 37 and 111

**12.) Which of the following is not true? **

(a) The number aabb is divisible by 11

(b) The number abab is divisible by 101

(c) The number aabbcc is divisible by 11

(d) The number abcabc is divisible by 101

**SOLUTIONS**

1.) (a)

2.) (d): 23x is divisible by 9=> (2+3+x) is divisible by 9

=>x=4.

So. A is false. Clearly, R is true.

3.) (d): 4y5 is divisible by 3 =>(4+y+5) is divisible by 3

=>(9+y) is divisible by 3

=>y=0, 3, 6 or 9.

So, y has 4 possible values.

Thus, A is false. Clearly, R is true.

4.) (a)

5.) (b): Let the given two-digit number be 10a + b.

Then, number obtained by reversing the digits = 10b + a.

Now, (10a+b)+(10b+a)=11a+11b = 11(a + b), which is clearly divisible by 11.

So. A is true.

And. (10a+b)-(10b+a)=9a- 9b= 9(a-b), which is clearly divisible by 9.

So, R is true.

6.) (d): N÷2 leaves remainder 2 → N is even.

N÷ 5 leaves remainder 3 -> N is 3 more than a multiple of 5.

So, units digit of N is 8.

Hence, N÷10 leaves remainder 8.

So. A is false but R is true.

7.) (d): 31a2 is divisible by 6

→31a2 is divisible by both 2 and 3

->31a2 is divisible by 2 for all values of a.

But, 31a2 is divisible by 3 for a = 0, 3, 6 or 9.

Thus, a has 4 possible values.

So. A is false. But R is true.

8.) (a)

9.) (c): Difference = (100a+10b+c)-(100c+10b+a)=99a-99c = 99(a-c), which is clearly divisible by 99 and hence by both 9 and 11.

10.) (c): The different three-digit numbers that can be formed using the given digits are abc, acb, bac, bca, cab, cba, i.e., 6 in all.

11.)d): Required sum can be calculated as under: …

abc ->100a + 10b + c

acb -> 100a + b + 10c

bac -> 10a + 100b + c

bca -> a + 100b + 10c

cab -> 10a + b + 100c

cba ->__a + 10b + 100c__

222a + 222b + 222c

= 222 (a + b + c), which is clearly divisible by 2, 37 and 111.

12.) (d): aabb->1000a + 100a + 10b + b

= 1100a + 11b = 11(100a + b), which is clearly divisible by 11.

abab->1000a + 100b + 10a + b

= 1010a + 101b = 101(10a + b), which is clearly divisible by 101.

aabbcc->100000a + 10000a + 1000b + 100b + 10c + c

= 110000a + 1100b + 11c

= 11(10000a + 100b + c), which is clearly divisible by 11.

abcabc->100000a + 10000b + 1000c + 100a + 10b + c

= 100100a + 10010b + 1001c

= 1001(100a + 10b + c), which is clearly divisible by 1001 and not by 101.