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DisciplinedPrep
\(\frac{1}{5}\) of a number is \(\frac{5}{8}\) times another number. If 35 is added to the first number, it becomes four times the second number. Find the second number.

A. 25
B. 39
C. 40
D. 70
E. 110

We can create the equations:

n/5 = 5y/8

8n = 25y

and

n + 35 = 4y

Multiplying the above equation by 8, we have:

8n + 280 = 32y

Substituting, we have:

25y + 280 = 32y

280 = 7y

40 = y

Answer: C
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(1/5)a = (5/8)b
=> 8a = 25b
=> 8a-25b = 0. ....(i)

a+35 = 4b
=> a-4b = -35
multiply the equation by '8'

8a-32b = -280 .....(ii)

(i) - (ii)

7b=280
=> b=40
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DisciplinedPrep
\(\frac{1}{5}\) of a number is \(\frac{5}{8}\) times another number. If 35 is added to the first number, it becomes four times the second number. Find the second number.

A. 25
B. 39
C. 40
D. 70
E. 110

x/5 = 5/8*y

x +35 = 4y

x = 4y - 35

Replacing x, 4y- 35/5 = 5y/8 or, 32y -280 = 25y or, 7y = 280 so, y =40
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