## Quant Question Of The Day: 53

[latexpage]

# Numbers

How many three digit numbers are increased by 99 when there digits are reversed ?

A. 4

B. 8

C. 10

D. 80

E. 90

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[latexpage]

How many three digit numbers are increased by 99 when there digits are reversed ?

A. 4

B. 8

C. 10

D. 80

E. 90

We need to find a positive integer of the form ABC such that CBA = ABC + 99

so , 100C + 10B + A = 100A + 10B + C + 99

that is 99C = 99A + 99

C – A = 1 .

A and C takes values (1 , 2 ), ( 2 , 3) , …….., ( 8, 9) => 8 ways

and B can take any values from 0 to 9 => 10 ways

Hence , 10 ×8 = 80 cases

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Ans. D=80

100a + 10b+c +99= 100c +10b+a

99a-99c = – 99

c-a =1

With this condition (a,c) will take values (1,2),(2,3),(3,4)….(8,9)

b can take any value from 0-9

Numbers will be 102,203,304……112,213……122,223…….192,293,394…899.

Answer :D=80

100a+10b+c+99= 100c+10b+a

99a-99c+99=0

c-a=1

Values for (a,c) will be (1,2),(2,3)..(8,9)

b can take any value from 0-9

D. 80

90

D. 80

80