Question

The sum of three numbers is 123. If the ratio between first and second number is 2:5 and that of between second and third is 3:4 then find the difference between second and third number.​

Answer 1

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Let the three numbers be a, b, and c.

Given that

a:b = 2:5

We can also write it as

a:b=6:15

Alos second given condition is

b:c = 3:4

We can also write it as

b:c = 15:20

Hence from above equations

a:b:c = 6:15:20

Let us assume that a = 6x, b = 15x and c = 20x

Given sum of all the three numbers = 123

On adding all the three numbers we get,

a+b+c =123

6x+15x+20x =123

41x = 123

x = 3

Hence a = 18, b = 45 and c = 60.

The difference between the second and the third number is 60 – 45 =15

Answer 2

Answer:

▪︎Let the three numbers be A, B and C.

Let the three numbers be A, B and C.A:B = 2:5, or

Let the three numbers be A, B and C.A:B = 2:5, orA:B=6:15

Let the three numbers be A, B and C.A:B = 2:5, orA:B=6:15B:C = 3:4, or

Let the three numbers be A, B and C.A:B = 2:5, orA:B=6:15B:C = 3:4, orB:C = 15:20

Let the three numbers be A, B and C.A:B = 2:5, orA:B=6:15B:C = 3:4, orB:C = 15:20So A:B:C = 6:15:20.

Let the three numbers be A, B and C.A:B = 2:5, orA:B=6:15B:C = 3:4, orB:C = 15:20So A:B:C = 6:15:20.Let A = 6x, B = 15x and C = 20x.

Let the three numbers be A, B and C.A:B = 2:5, orA:B=6:15B:C = 3:4, orB:C = 15:20So A:B:C = 6:15:20.Let A = 6x, B = 15x and C = 20x.A+B+C = 6x+15x+20x = 41x = 123, or x = 3.

Let the three numbers be A, B and C.A:B = 2:5, orA:B=6:15B:C = 3:4, orB:C = 15:20So A:B:C = 6:15:20.Let A = 6x, B = 15x and C = 20x.A+B+C = 6x+15x+20x = 41x = 123, or x = 3.Hence A = 18, B = 45 and C = 60.

Let the three numbers be A, B and C.A:B = 2:5, orA:B=6:15B:C = 3:4, orB:C = 15:20So A:B:C = 6:15:20.Let A = 6x, B = 15x and C = 20x.A+B+C = 6x+15x+20x = 41x = 123, or x = 3.Hence A = 18, B = 45 and C = 60.The difference between the second and the third number is 5x or 15.

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