Practice MCQ Questions and Answer on Number System
21.
Three numbers are in Arithmetic progression (AP) whose sum is 30 and the product is 910. Then the greatest number in the AP is:
(A) 17
(B) 15
(C) 13
(D) 10
Solution:
Let the three number is a - d, a, a + d a is first term, d is common difference a + d + a + a - d = 30 (Given) 3a = 30 a = 10 (a + d)(a)(a - d) = 910 (10 + d)(10)(10 - d) = 91 × 10 (10 + d)(10 - d) = 91 Put d = 3 So, d = 3 So, greater number is = a + b = 10 + 3 = 13
22.
What number multiplied by 48 will give the same product as 173 multiplied by 240 ?
(A) 495
(B) 545
(C) 685
(D) 865
Solution:
Let x × 48 = 173 × 240Then, x = x = 173 × 5 x = 865
23.
111, 111, 111, 111 is divisible by :
(A) 3 and 37 only
(B) 3, 11 and 37 only
(C) 3, 11, 37 and 111 only
(D) 3, 11, 37, 111 and 1001
Solution:
Sum of the digits = 12, which is divisible by 3. So, the given number is divisible by 3 (Sum of digits at odd places) - (Sum of digits at even places) = 6 - 6 = 0 So, the given number is divisible by 11 The given number when divided by 37 gives 3003003003 So, the given number is divisible by 37 The given number when divided by 111 gives 1001001001 Clearly, it is divisible by 111 as well as by 1001 Hence, the given number is divisible by each one of 3, 11, 37, 111 and 1001
24.
The sum of 10 terms of the arithmetic series is 390. If the third term of the series is 19. Find the first term?
(A) 3
(B) 5
(C) 7
(D) 8
Solution:
25.
Given n = 1 + x and x is the product of four consecutive integers. Then which of the following us true ?
I. n is an odd integer. II. n is prime. III. n is a perfect square
(A) Only I is correct
(B) Only III is correct
(C) Both I and II are correct
(D) Both I and III are correct
Solution:
Out of four consecutive integers two are even and therefore, their product is even and on adding 1 to it, we get an odd integer. So, n is odd. Some possible values of n are as under : n = 1 + (1 × 2 × 3 × 4) = (1 + 24) = 25 = 52 n = 1 + (2 × 3 × 4 × 5) = (1 + 120) = 121 = 112 n = 1 + (3 × 4 × 5 × 6) = (1 + 360) = 361 = 192 n = 1 + (4 × 5 × 6 × 7) = (1 + 840) = 841 = 292 Ans so on..... Hence, n is odd and a perfect square.
26.
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
27.
A, B, C and D purchase a gift worth Rs. 60. A pays of what others are paying. B pays of what others are paying and C pays of what others are paying. What is the amount paid by D ?
(A) 13
(B) 15
(C) 12
(D) 14
Solution:
A + B + C + D = 60 -----------> (1) Now According to Question, A = (B + C + D) 2A = B + C + D Put the value of B + C + D in Equation 1 A + 2A = 60 3A = 60 A = 20 Now again according to Question B = (A + C + D) A + C + D = 3B By Putting the Value of A + C + D in Equation 1 we get, B + 3B = 60 4B = 60 B = 15 Now again according to Question C = (A + B + D) A + B + D = 4C By Putting the Value of A + B + D in equation 1 We get C + 4C = 60 5C = 60 C = 12 Since A + B + C + D = 60 20 + 15 + 12 + D = 60 D = 60 - 47 = 13 Amount of D = Rs. 13
28.
There are 2 teams-A and B. If 3 people are shifted from Team A to Team B, then Team B has thrice the number of members than Team A. If 2 people are shifted from Team B to Team A, then Team B has double the number of members than Team A. How many members does Team B have originally?
(A) 15
(B) 18
(C) 42
(D) 45
Solution:
29.
In a certain series, each number except the first and second is obtained by adding the previous two numbers. If the first no is 2 and sixth no is 26, then the seventh number is: