Question

For each month of a given year except December, a worker earned the same monthly salary and donated one-tenth of that salary to charity. In December, the worker earned N times his usual monthly salary and donated one-fifth of his earnings to charity. If the worker's charitable contributions totaled one-eighth of his earnings for the entire year, what is the value of N?
A) 8/5

B) 5/2

C) 3

D) 11/3

E) 4

Answer 1

Answer:

D) 11/3

Step-by-step explanation:

Let x represent monthly salary.

We have been given that for each month of a given year except December, a worker earned the same monthly salary and donated one-tenth of that salary to charity.        

Salary earned in 11 months would be $11x$.

Money donated in 11 months would be $\frac{11x}{10}$.

Further we are told that in December, the worker earned N times his usual monthly salary and donated one-fifth of his earnings to charity.

Salary for December would be $Nx$.

Money donated in December would be $\frac{Nx}{5}$.  

The worker's charitable contributions totaled one-eighth of his earnings for the entire year that is $\frac{1}{8}\cdot (11x+Nx)$.

$\frac{11x}{10}+\frac{Nx}{5}=\frac{1}{8}\cdot (11x+Nx)$

Dividing whole equation by x, we will get:

$\frac{11x}{10x}+\frac{Nx}{5x}=\frac{1}{8x}\cdot x(11+N)$                

$\frac{11}{10}+\frac{N}{5}=\frac{1}{8}\cdot (11+N)$

$\frac{11}{10}+\frac{N}{5}=\frac{11}{8}+\frac{N}{8}$

Combine like terms:

$\frac{N}{5}-\frac{N}{8}=\frac{11}{8}-\frac{11}{10}$

$\frac{N}{5}*40-\frac{N}{8}*40=\frac{11}{8}*40-\frac{11}{10}*40$

$8N-5N=11*5-11*4$

$3N=55-44$

$3N=11$

$\frac{3N}{3}=\frac{11}{3}\\\\N=\frac{11}{3}$

Therefore, the value of N is 11/3 and option D is the correct choice.