Practice MCQ Questions and Answer on Number System
91.
If p is a prime number greater than 3, then (p2 - 1) is always divisible by :
(A) 6 but not 12
(B) 12 but not 24
(C) 24
(D) None of these
Solution:
⇔ p = 5 ⇒ (p2 - 1) = (25 - 1) ⇒ (p2 - 1) = 24, which is divisible by 24 ⇔ p = 7 ⇒ (p2 - 1) = (49 - 1) ⇒ (p2 - 1) = 48, which is divisible by 24 ⇔ p = 11 ⇒ (p2 - 1) = (121 - 1) ⇒ (p2 - 1) = 120, which is divisible by 24 Hence, (p2 - 1) is always divisible by 24
92.
8888 + 848 + 88 - ? = 7337 + 737
(A) 1450
(B) 1550
(C) 1650
(D) 1750
Solution:
Let, 8888 + 848 +88 - x = 7337 + 737 Then, 9824 - x = 8074 x = 9824 - 8074 x = 1750
93.
Of the three numbers, the second is twice the first and is also thrice the third. If the average of these three numbers is 44, the largest number is -
(A) 24
(B) 36
(C) 72
(D) 108
Solution:
No. → I II III 3x 6x 2x = 11x
94.
The digit at Hundred's place value of 17! is:
(A) 1
(B) 0
(C) 2
(D) 3
Solution:
We know that after 5! we get one zero at the end of the number. We can say that in 17! we get minimum three zeros. The hundred's place value od 17! = 0
95.
Largest fraction among and is :
(A)
(B)
(C)
(D)
Solution: (105) (88) ⤧ So is largest
96.
The sum of three numbers is 252. If the first number is thrice the second and third number is two-third of the first, then the second number is :
(A) 41
(B) 21
(C) 42
(D) 84
Solution:
Let the number is a, b, c a = 3b = c = a = c a b 2 3 3 1 c : a : b = 2 : 3 : 1 So, (2 + 3 + 1) = 252 units 6 units → 252 1 unit → 42 3 units → 42 × 3 So, second number is = 42
97.
Which of the following numbers are completely divisible by 7 ?
I. 195195
II. 181181
III. 120120
IV. 891891
(A) Only I and II
(B) Only II and III
(C) Only I and IV
(D) Only II and IV
Solution:
Answer & Solution Answer: Option E Solution: I. We have (195 - 195) = 0 ∴ 195195 is divisible by 7 II. We have (181 - 181) = 0 ∴ 181181 is divisible by 7 III. We have (120 - 120) = 0 ∴ 120120 is divisible by 7 IV. We have (891 - 891) = 0 ∴ 891891 is divisible by 7 Hence, all are divisible by 7
98.
If 37 X 3 is a four-digit natural number divisible by 7, then the place marked as X must have the value :
(A) 0
(B) 3
(C) 5
(D) 9
Solution:
37 × 3 is divisible by 7 ⇒ (7 × 3 - 3) is either 0 or divisible by 7 ⇒ 7 × 0 is divisible by 7 ⇒ X = 0 or X = 7
99.
P is a prime number and (P2 + 3) is also a prime number . The no. of numbers that P can assume is:
(A) 3
(B) 2
(C) 1
(D) Can't determined
Solution:
Only, one value P can be assumed, which is at 2 at P = 2 P2 + 3 = 7 which is also prime Again at P = 3, 5, 7, 11 ....... P2 + 3 = an even number which can not be prime