 # Tricky questions of (Number system.) for SSC CGL Tier-II 2016

Q1. If the number 97215X6 is completely divisible by 11, then the smallest whole number in place of X will be:
(a) 3
(b) 2
(c) 1
(d) 5
S1. Ans.(a)
Sol. Given number = 97215×6
(6 + 5 + 2 + 9) – (x + 1 + 7) = (14 – x), which must be divisible by 11.

∴ x = 3.

Q2. If the number 91876X2 is completely divisibly by 8, then the smallest whole number in place of X will be:
(a) 1
(b) 2
(c) 3
(d) 4
S2. Ans.(c)
Sol. The number 6×2 must be divisible by 8.

∴ x = 3, as 632 is divisible by 8.

Q3. If x and y are the two digits of the number 653 xy such that this number is divisible by 80, then  x + y = ?
(a) 2
(b) 3
(c) 4
(d) 6
S3. Ans.(a)
Sol. 80 = 2 × 5 × 8
Since 653 xy is divisible by 2 and 5 both, so y = 0
Now, 653 x0 is divisible by 8, so 3×0 should be divisible by 8. This happens when x = 2

∴ x + y = (2 + 0) = 2

Q4. If the product 4864 × 9 P 2 is divisible by 12, the value of P is:
(a) 2
(b) 5
(c) 6
(d) 1
S4. Ans.(d)
Sol. Clearly, 4864 is divisibly by 4.
So, 9P2 must be divisible by 3. So, (9 + P + 2) must be divisible by 3.
∴ P = 1.

Q5. On dividing a number by 5, we get 3 as remainder. What will be the remainder when the square of this number is divided by 5?
(a) 0
(b) 1
(c) 2
(d) 4
S5. Ans.(d)
Sol. Let the number is=5x+3
Square of the number=(5x+3)^2

Remainder=9/5=4
Q6. The difference between a positive proper fraction and its reciprocal is 9/20. The fraction is:
(a) 3/5
(b) 3/10
(c) 4/5
(d) 5/4

Q8. A 3-digit number 4a3 is added to another 3-digit number 984 to give a 4-digit number 13b7, which is divisible by 11. Then, (a + b) = ?

(a) 10
(b) 11
(c) 12
(d) 15

Q9. When a number is divided by 13, the remainder is 11. When the same number is divided by 17, the remainder is 9. What is the number?
(a) 339
(b) 349
(c) 369
S9. Ans.(b)
Sol.
x = 13p + 11 and x = 17q + 9
∴13p+11=17q+9⇒17q-13p=2⇒q=(2 + 13p)/17
The least value of p for which q=(2 + 13p)/17 is a whole number is p = 26
∴ x=(13×26+11)=(338+11)=349

Q10. A boy multiplied 987 by a certain number and obtained 559981 as his answer. If in the answer both 9 are wrong and the other digits are correct, then the correct answer would be:
(a) 553681
(b) 555181
(c) 555681
(d) 556581
S10. Ans.(c)
Sol. 987 = 3 × 7 × 47
So, the required number must be divisible by each one of 3, 7, 47
553681 → (Sum of digits = 28, not divisible by 3)
555181 → (Sum of digits = 25, not divisible by 3)

555681 is divisible by each one of 3, 7, 47.

Q12. Three numbers are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:
(a) 75
(b) 81
(c) 85
(d) 89
S12. Ans.(c)
Sol. Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common
So, middle number = H.C.F of 551 and 1073 = 29;
First number =(551/29)=19; Third number =(1073/29)=37

∴ Required sum = (19 + 29 + 37) = 85.

Q13. Three different containers contain 496 litres, 403 litres and 713 litres of mixtures of milk and water respectively. What biggest measure can measure all the different quantities exactly?
(a) 1 litre
(b) 7 litres
(c) 31 litres
(d) 41 litres
S13. Ans.(c)

Sol. Required measurement = (H.C.F of 496, 403, 713) litres = 31 litres

Q14. The greatest number which can divide 1356, 1868 and 2764 leaving the same remainder 12 in each case, is:
(a) 64
(b) 124
(c) 156
(d) 260
S14. Ans.(a)
Sol.  Required number = H.C.F of (1356 – 12), (1868 – 12) and (2764 – 12)

= H.C.F of 1344, 1856 and 2752 = 64.

Q15. The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is:
(a) 11115
(b) 15110
(c) 15120
(d) 15210
S15. Ans.(b)
Sol. Here (48 – 38) = 10, (60 – 50) = 10, (72 – 62) = 10, (108 – 98) = 10 & (140 – 130) = 10

∴ Required number = (L.C.M of 48, 60, 72, 108, 140) – 10 = 15120 – 10 = 15110. Join India largest learning distination

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