Quant Question Of The Day: 53
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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
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
See our previous ‘Questions of the Day’:
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