Question

The sum of three numbers is 68. if the ratio between first and second be 2 : 3 and that between second and third be 5 : 3, then the second number is

Answer 1

x + y+z=68
x/z=2/3
y/z=5/3
x=2/3z
y=5/3z
hence,x+y+z=68
2/3z+5/3z+z= 68
z=204/10
y=5/3z= 5/3*204/10=34

Answer 2

Answer:

Step-by-step explanation:

Lets assume those three numbers to be a, b, c.

a:b = 3x : 2x,. ........ (Step1)

b:c = 5y : 3y. ........ (Step2)

a(q)+b(q)+c(q) = 68 ................ (Step3)

where 'q' is an integer to be found.

From "b" of Step1 & Step2, we get,

b = 2x = 5y

To get a single value for "b", let's put

x=5 and y=2. ......... (Step4)

therefore, b = 2(5) = 5(2) = 10 ..... (Step5)

Applying Step4 in Step1 & Step2, we get,

a:b = 3(5) : 2(5) = 15 : 10

b:c = 5(2) : 3(2) = 10 : 6

therefore, a:b:c = 15 : 10 : 6 .........Step6

Applying Step6 in Step3, we get,

15(q) + 10(q) + 6(q) = 68. ......Step7

let's put q = 2,

15(2) + 10(2) + 6(2) = 30 + 20 + 12 = 62.

.....................................................(Step8)

I'm sorry, the above question is wrong as 68 is not a possible sum of

a(q)+b(q)+c(q) with the given ratios of a:b & b:c

However if the question expects

a(q)+b(q)+c(q) = 62,

then

## the second number = 20 ## as shown in (Step8)

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