Practice MCQ Questions and Answer on Number System
671.
Which one of the following is a prime number ?
(A) 161
(B) 221
(C) 373
(D) 437
Solution:
437 > 22 All prime numbers less than 22 are : 2, 3, 5, 7, 11, 13, 17, 19 161 is divisible by 7, and 221 is divisible by 13 373 is not divisible by any of the above prime numbers. Therefore 373 is prime
672.
Let x be the least number which when divided by 15,18, 20 and 27, the remainder in each case is 10 and x is a multiple of 31. What least number should be added to x to make it a perfect square?
(A) 39
(B) 37
(C) 43
(D) 36
Solution:
L.C.M. of 15,18, 20, 27 = 540 x = 540K + 10 x = 540 × 4 + 10 = 2170 462 2170 > 472 472 = 2209 To make perfect square = 2209 - 2170 = 39
673.
If the product of two positive numbers be 1575 and their ration is 7 : 9, then the greatest number is :
(A) 45
(B) 135
(C) 35
(D) 63
Solution:
Let the numbers are 7x and 9x According to the question, Then greater number = 9x= 9 × 5= 45
674.
Arrangement of the fractions ÃÂÃÂ ÃÂÃÂ into ascending order:
(A)
(B)
(C)
(D)
Solution:
675.
The product of two number is 48. If one number equals "The number of wings of a bird plus 2 times the number of figure on your hand divided by the number of wheels of a Tricycle." Then the other number is?
(A) 9
(B) 10
(C) 12
(D) 18
Solution:
676.
The real number to be added to 13851 to get a number which is divisible by 87 is :
(A) 18
(B) 43
(C) 54
(D) 69
Solution:
By hit and trial method = 13851 + 69 = 13920 Which is divisible by 87
677.
How many number are there between 1 to 200 which are divisible by 3 but not by 7?
(A) 38
(B) 45
(C) 57
(D) 66
Solution:
Number from 1 to 200 which is divisible by 3 3, 6, . . . . . . . . ., 198 Total number which divisible by 3 & 7 Total number which is divisible by 3 but not 7 = 66 - 9 = 57 Alternate solution Total number which is divisible by 3 & 7 both from 1 to 200 LCM (3 & 7) Total number which is divisible by only 3, from 1 to 200 ∴ Total number which is divisible by 3 but not 7 = 66 - 9 = 57
678.
Find the HCF of (3125-1) and (335-1).
(A) 34 - 1
(B) 35 - 1
(C) 312 - 1
(D) None of these
Solution:
The solution of this question is based on the rule, The HCF of (am - 1) and (an - 1) is given by (aHCF of m, n - 1) Thus for this question the answer is (35 - 1) Since, 5 is the HCF of 35 and 125
679.
A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. When it is successively divided by 5 and 4, then the respective remainders will be
(A) 1, 2
(B) 2, 3
(C) 3, 2
(D) 4, 1
Solution:
4 | x y = (5 × 1 + 4) = 9 ------------ 5 | y - 1 x = (4 × y + 1) = (4 × 9 + 1) = 37 ------------ | 1 - 4 Now, 37 when divided successively by 5 and 4, we get 5 | 37 ------------- 4 | 7 - 2 ------------- | 1 - 3 Respective remainders are 2 and 3
680.
A number when divided by 14 leaves reminder of 8, but when the same number is divided by 7, it will leave the remainder :
(A) 3
(B) 2
(C) 1
(D) 4
Solution:
When the number is divided by 14 it gives a remainder of 8, The number = 14N + 8 (14N is divisible by 14) When same number is divided by 7 it will give remainder 1