Question

In an examination, the number of those that passed and the number of those
that failed were in the ratio of 3:1. Had 8 more appeared, and 6 less passed,
the ratio of passed to failures would have been 2:1. Find how many appeared?
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Answer 1

let passed = 3x ; failed=x

total = 3x+x= 4x

8 more appeared, (4x+8)

6 less passed = (3x-6)

now total failed = total appeared- total passed

(4x+8)-(3x-6)= x+14

now ratio of passed : failed = 2:1

(3x-6)/ (x+14) = 2:1

3x-6= 2x+28

x= 34

total appeared = 4x= 4×34= 136

Answer 2

Answer:

_________

Let the tens digit of the required number be x and the units digit be y. Then,

x+y=12 .........(1)

Required Number = (10x+y).

Number obtained on reversing the digits = (10y+x).

Therefore,

(10y+x)−(10x+y)=18

9y−9x=18

y−x=2 ..........(2)

On adding (1) and (2), we get,

2y=14⟹y=7

Therefore,

x=5

Hence, the required number is 57.