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How many pairs of \( X\) and \( Y\) are possible in the number \( 763X4Y2\), if the number is divisible by 9?

A) 8
B) 9
C) 10
D) 11

Correct Answer:




D) 11

Description for Correct answer:
The given number is 763 x 4yz and divisible by 9

7 + 6 + 3 + 4 + 2 + x + y = 22 + x + y

For divisibility by 9,

x + y = 5

x + y = 14

So the possible values of x and y are

For x + y = 5,

x = 0, y = 5

x = 1, y = 4

x = 2, y = 3

x = 3, y = 2

x = 4, y = 1

x = 5, y = 0

For x + y = 14

x = 5, y = 9

x = 6, y = 8

x = 7, y = 7

x = 8, y = 6

x = 9, y = 5

Hence total number of possible pairs of x and y are 11

Option (D)is correct.

Part of solved CDS Maths(1) questions and answers : Exams >> CDSE >> CDS Maths(1)

Comments

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