DIRECTIONS for questions 1 to 10: In each of these questions, two quantities are mentioned, at column A and at column B.
To mark the answers to the questions, use the following instructions.
Choose 1: If the quantity in column A is greater. Choose 2: If the quantity in column B is greater. Choose 3: If the two quantities are equal. Choose 4: If the relationship cannot be determined.
Option: 3 Explanation: v = 10 x k , Let k = 325 , v = 3250,
The sum of the digits for both remains same which is = (3 + 2 + 5 + 0) = 10
m and n are integers. 0 < m < n < 10
Column A
The number of multiples of m between 1 and 100
Column B
The number of multiples of n between 1 and 100
Option: 1 Explanation: When m, n are integers between 1 and 10, and m < n, the number of multiples of m will be more than the number of multiples of n. E.g. for 8 , 9, the number of multiples are respectively 12 and 11
Column A
The sum of three different prime numbers if each number is less than 10
Column B
The sum of three different positive even integers if each integer is less than 10
Option: 4 Explanation: The max value of column A is 15 and min is 10,
The max value of column B is 18 and min is 12.
Hence, no relation can be found.
a is a positive integer.
Column A
The remainder when a is divided by 7
Column B
The remainder when a2 is divided by 7
Option: 4 Explanation: When a = 14, the remainder of a and a2 are respectively 0 and 0.
But when a = 16, the remainder of a and a2 are respectively 2 and 4.
S = (-3, -2, 2, 3). The members of a set T are the squares of numbers in set S.
Column A
The number of members of S
Column B
The number of members of T
Option: 1 Explanation: S = { -3 , -2 , 2 ,3 }, T = {9 , 4 }, n ( S ) = 4, n (T ) =2
The average (arithmetic mean) of 5 positive integers is 70.
Option: 4 Explanation: If the integers are (70 ,70 ,70, 70, 70) the median is 70, but if the integers are (60, 65, 65, 80, 80), the median is 65.
R, S and T are 3 consecutive odd integers and R < S < T
Column A
R + S + 1
Column B
S + T – 1
Option: 2 Explanation: Since, they are consecutive odd numbers and R < S < T, (S – R) = 2, (T – S) = 2, Adding (S + T) – (R + S) = 4
⇒(S + T) = ( R + S ) + 4,
⇒( S + T ) - 1 = ( R + S ) + 3 > ( R + S ) + 1
Option: 4 Explanation: For x = -2, y = -3, z = – 4, x × y × z = -24, x + y + z = -9
But for x = y = z = -1, x × y × z = -1, x + y + z = -3
x is an integer, and the remainder when 2x is divided by 4 is 0
Column A
The remainder when x is divided by 4
Column B
0
Option: 4 Explanation: If x = 4, 2x = 8, 8 / 4, the remainder is = 0. Also for 4/4, the remainder is = 0. But if x = 6 ,2x = 12 , 12 / 4, the remainder is = 0. But for 6 / 4 , the remainder is ≠ 0.
Each of w and x is less than 5 and greater than 2. Each of y and z is less than 2 and greater than 1.
Column A
w + x
Column B
y + z
Option: 1 Explanation: 2 < w < 5, 2 < x < 5. Adding, we get 4 < ( x + w ) < 10, 1 < y < 2, 1 < z < 2,
Adding, we get 2 < ( y + z ) < 4 < ( w + x ) < 10