Let monthly salary for each of the 11 months except December was x, then

11x*1/10 + Nx*1/5 = 1/8(11x + Nx);

11/10 + N/5 = 1/8(11 + N)

N = 11/3.

Answer: D.
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Hi,
Apart from the method of [b]Bunuel, we can do it by weighted average method, by taking this too as MIXTURE.
so the quantities are
1) 11 units of 1/10
2) N units of 1/5
average= 1/8 of (N+11)[/b]

so\({(\frac{1}{8}-\frac{1}{10})}/{(\frac{1}{5}-\frac{1}{10})}=\frac{N}{(N+11)}\)..
\({\frac{1}{40}/\frac{1}{10}}=\frac{N}{(N+11)}\)..
\(\frac{1}{4}=\frac{N}{(N+11)}\)
N+11=4N..
3N=11..
N=11/3
D
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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

We can let the normal salary = k, so his total salary for the first 11 months was 11k. Each month k/10 went to charity, so for the first 11 months, his total charitable contribution was 11k/10.

The December salary was kN, and kN/5 was donated to charity.

Since the worker's charitable contributions totaled one-eighth of his earnings for the entire year:

kN/5 + 11k/10 = (kN + 11k)/8

Multiplying the equation by 40, we have:

8kN + 44k = 5kN + 55k

Dividing the equation by k, we have:

8N + 44 = 5N + 55

3N = 11

N = 11/3

Answer: D
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