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08 Jun 2021, 03:45
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Difficulty:
35%
(medium)
Question Stats:
73% (02:07) correct
27%
(02:24)
wrong
based on 150
sessions
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The sum of a two-digit number and the number obtained by reversing the digits is a multiple of 88. If the difference of the digits at the ten’s place and unit’s place is 6, find the digit at the ten’s place of the number.
(A) 2
(B) 3
(C) 5
(D) 6
(E) 7
(A) 2
(B) 3
(C) 5
(D) 6
(E) 7
08 Jun 2021, 05:04
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Solution:
Let the number xy have a value of y + 10x
=>The reversed number yx will have a value of x + 10 y
=>The sum of these numbers = 11(x + y) =88
=> x + y = 8 ..........(1)
Also the difference of digits at the ten’s place and unit’s place is 6
=> x - y = 6............(2)
Adding both equations we have 2x = 14 => x = 7 (tens place digit) (option e)
Happy Studying !
Devmitra Sen
GMAT SME
Let the number xy have a value of y + 10x
=>The reversed number yx will have a value of x + 10 y
=>The sum of these numbers = 11(x + y) =88
=> x + y = 8 ..........(1)
Also the difference of digits at the ten’s place and unit’s place is 6
=> x - y = 6............(2)
Adding both equations we have 2x = 14 => x = 7 (tens place digit) (option e)
Happy Studying !
Devmitra Sen
GMAT SME
