# The sum of three numbers is 98. The ratio of the first to the second is 2/3, and the ratio of the second to the third is 5/8. The second number is:

(a) 15, (b) 20, (c) 30, (d) 32, (e) 33

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(a) 15, (b) 20, (c) 30, (d) 32, (e) 33

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x : y = 2 / 3

y : z = 5 / 8

z : y = 8 / 5

Now,

2y + y + 8y = 98

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3 5

10y + 15y + 24y = 98

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15

49y = 98 * 15

y = 98 * 15

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49

y = 2 * 15

Now,

Its given that their sum is 98.

So, x + y + z = 98 --Eq.1

Again, given ratios –

x : y = 2 : 3 --Eq.2

y : z = 5 : 8 --Eq.3

So, now we have to find the second number i.e. y

Finding the value of x in terms of y.

Using Eq.2, we can write

x/y = 2/3

x = 2y/3 --Eq.4

Now, again finding the value of z in terms of y.

Using Eq.3, we can write

y/z = 5/8

z = 8y/5 --Eq.5

Putting the values of x and z according to Eq.4 and Eq.5 in Eq.1

x + y + z = 98

2y/3 + y + 8y/5 = 98

Making the denominators common

10y/15 + 15y/15 + 24y/15 = 98

49y/15 = 98

Transposing 15 and 49 to RHS

y = 98×15/49

y = 2 × 15

y = 30

So, the answer is (C)30