MCQ on Playing with Numbers for Class 8
Class 8 Maths Chapter 16 Playing with Numbers MCQ
MCQ of Playing with Numbers Class 8
I. Choose the correct answer.
Question 1.
ab, ba are two, two digit number, then their sum is always divisible by __________ .
A) 9
B) 10
C) 11
D) none of the above
Answer:
C) 11
Question 2.
pq, qp are two, two digit number, then pq – qp = __________ .
A) 9(p – q)
B) (q – p)9
C) ‘A’ or ‘B’
D) none
Answer:
C) ‘A’ or ‘B’
Question 3.
pq is a two digit number, then (p – q) or (q – p) always divides __________ .
A) pq
B) qp
C) pq – qp
D) pq + pq
Answer:
C) pq – qp
Question 4.
pq is a two digit number, then (p + q) always divides _________ .
A) pq
B) qp
C) pq – qp
D) pq + qp
Answer:
D) pq + qp
Question 5.
klm is a 3 digit number, then the remainder when klm + mlk divided by 99 is __________ .
A) 98
B) 96
C) 11
D) 0
Answer:
D) 0
Question 6.
‘pqrs’ is a 4 – digit number. Then 999 divides __________ .
A) pqrs + srqp
B) pqrs – sqrp
C) pqrs + srqp
D) pqrs – psrq
Answer:
B) pqrs – sqrp
Question 7.
‘pqrs’ is a 4 – digit number and if (pqrs – sqrp) is divided by 99.9, then the quotient is __________ .
A) p – s
B) s – p ,
C) ‘A’ or ‘B’
D) p + s
Answer:
C) ‘A’ or ‘B’
Question 8.
‘pqr’ is a 3 – digit number, then (pqr + qrp + rpq) is always divisible by __________ .
A) 37
B) 3
C) 111
D) all the above
Answer:
D) all the above
Question 9.
Sum of all 5 possible 3 digit numbers of a 3 – digit number is always divisible by __________ .
A) 37
B) 3
C) 11
D) all the above
Answer:
D) all the above
Question 10.
If 41 p + 1p4 = 591, then possible value of ‘p’ = __________ .
A) 0
B) 7
C) 0 or 7
D) none
Answer:
B) 7
Question 11.
A + A + A + A + A = BA, then A – B = __________ .
A) 5
B) 7
C) 3
D) 25
Answer:
C) 3
Question 12.
, then the value of ‘A’ = __________ ; B = __________ .
A) A = 2, B – 5
B) A = 5, B = 3
C) A = 5, B = 2
D) A = 3,B = 5
Answer:
C) A = 5, B = 2
Question 13.
If ‘pq’ is a two digit number, then which of the following is true ?
A) pq + qp is always less than 198
B) pq + qp is always divisible by 11
C) pq + qp is always divisible by p + q
D) all the above
Answer:
D) all the above
Question 14.
3P + 42 = Q0, then values of ‘P’ and ‘Q’ are __________ .
A) equal and 8
B) not equal to 8
C) 7, 9
D) 2. 7
Answer:
A) equal and 8
Question 15.
, then Q + R = __________ .
A) 15
B) 6
C) 154
D) 65
Answer:
B) 6
Question 16.
9A – 62 = B5, then the values of A, B are __________ .
A) A = 3, B = 7
B) A = 3, B = 5
C) A = 7,B = 3
D) none
Answer:
C) A = 7,B = 3
Question 17.
, then the value of ‘P’and ‘R’ are __________ .
A) P = 5, R ≠ 1
B) P = 5, R = 2
C) P = 6, R = 5
D) P = 9, R = 9
Answer:
C) P = 6, R = 5
Question 18.
(pqr + qrp + _______???_____) ÷ 37 = 3(p + q + r) fill the blank
A) rpq
B) rrr
C) ppp
D) qqq
Answer:
A) rpq
Question 19.
(pqr + qrp + rpq) ÷ 111 __________ .
A) p + q
B) p + r
C) q + r
D) P + q + r
Answer:
D) P + q + r
Question 20.
1111 × 1111 = 1 ABCDE 1, then A + B + C + D + E = _________ .
A) 16
B) 15
C) 14
D) 13
Answer:
C) 14
Question 21.
4 5 6 P is divisible by 3, then the least possible value of ‘P’ is __________ .
A) 7
B) 6
C) 3
D) 0
Answer:
D) 0
Question 22.
If ‘ppp’ a 3 digit number is divisible by 9, then the possible value of p are _________ .
A) 3
B) 6
C) 9
D) all the above
Answer:
D) all the above
Question 23.
2p 46 5 is divisible by 3, then the maxi-mum possible value of ‘p’ = __________ .
A) 1
B) 4
C) 7
D) 9
Answer:
C) 7
Question 24.
1234ab is divisible by ‘6’. Then the least possible value of ‘a’ and ‘b’ are ________ .
A) 6 and 2
B) 2 and 6
C) 8 and 8
D) none
Answer:
B) 2 and 6
Question 25.
A number leaves a remainder 3, when divided by ‘5’. So, the possible value of one’s digit in given number are ___________
A) 3 or 8
B) 2 or 7
C) 1 or 6
D) 4 or 9
Answer:
A) 3 or 8
Question 26.
An 8-digit number 20101 Δ 17 is divisible by 9. What digit does Δ stand for ?
A) 0
B) 3
C) 6
D) 9
Answer:
C) 6
II. Fill in the blanks.
1. 5678p is divisible by 4. Then ihe least possible value of ‘p’ is __________ .
Answer:
0
2. pqr is a multiple of 11. Then p + r – q = __________ .
Answer:
0
3. A four digit number ‘pqrs’ is 3 divisible by 11, then p + r = _________ .
Answer:
q + s
4. 5p5 is divisible by 5. The least possible value of ‘p’ = _________ .
Answer:
0
5. P + P + P = Q7;Q + Q + Q = _________ .
Answer:
6
6. P × P × P = ABP, So, AB = _________ or _________ .
Answer:
12 or 21
7. (AB) × 3 = CAB, then A = _________, B = _________ ,C = _________
Answer:
A = 5, B = 0, C = 1
8. Sum of a two digit number and its reversed number is always divisible by _________ and _________ .
Answer:
11 and sum of its digits
9. Sum of a three digit number and its reversed number is always divisible by _________ and _________ .
Answer:
99, 9
10. PQ + QP is a maximum two digit number, then P = _________ , Q = _________ .
Answer:
3 and 6 or 6 and 3
III. Match the following.
Question 1.
A | B |
1) 100 P | a) always divisible by 11 |
2) ![]() |
b) divisible by 9 |
3) PQ + QP | c) P = 0 or 5 if divisible by 5 |
4) 21436587 | d) A = 6 |
5) (A)(A)(A) = PQA | e) A = 5 or 6 |
Answer:
(1, c) (2, d) (3, a) (4, b) (5, e)
A | B |
1) 100 P | c) P = 0 or 5 if divisible by 5 |
2) ![]() |
d) A = 6 |
3) PQ + QP | a) always divisible by 11 |
4) 21436587 | b) divisible by 9 |
5) (A)(A)(A) = PQA | e) A = 5 or 6 |
Question 2.
A | B |
1) 600 | a) divisible by 2, 3, 4, 5, 6, 8, 9, 10 |
2) 172800 | b) units digit of ‘p’ might be 1, 3, 5, 7, 9 |
3) If the division p ÷ 5 leaves a remainder 4, then | c) divisible by 3 and 6 but not by 9 |
4) If the division p ÷ 6 leaves a remainder 5, then | d) 7, 6 |
5) 4A × A = 2BA, then values of B, A are _________ | e) one’s digit of ‘p’ might be 4 or 9 |
Answer:
(1, c) (2, a) (3, e) (4, b) (5, d)
A | B |
1) 600 | c) divisible by 3 and 6 but not by 9 |
2) 172800 | a) divisible by 2, 3, 4, 5, 6, 8, 9, 10 |
3) If the division p ÷ 5 leaves a remainder 4, then | e) one’s digit of ‘p’ might be 4 or 9 |
4) If the division p ÷ 6 leaves a remainder 5, then | b) units digit of ‘p’ might be 1, 3, 5, 7, 9 |
5) 4A × A = 2BA, then values of B, A are _________ | d) 7, 6 |