Question

the sum of three numbers is 68.if the ratio of first to second 3:2 and second to third is 5:3 then the second number

Answer 1

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)

Answer 2

Answer:

Step-by-step explanation:

=>>> 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)

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