Quentin put an amount of money into each of two new investments, X and

Question

Quentin put an amount of money into each of two new investments, X and Y, that pay simple annual interest. Investment X pays x% simple annual interest and investment Y pays y% simple annual interest. After 1 year, the average (arithmetic mean) interest earned by investments X and Y was 7.6%. If x and y are consecutive positive integers, in that order, what fraction of the total amount was put in investment X?

A. 1/5
B. 1/3
C. 2/5
D. 3/5
E. 2/3

Bunuel · Accepted Answer

Quentin deposited an amount of money into each of two new investments, X and Y. Investment X pays  x\%  simple annual interest, and investment Y pays  y\%  simple annual interest. After 1 year, the total interest earned was  7.6\%  of the total investment. If  x  and  y  are consecutive positive integers, in that order, what fraction of the total amount was deposited into investment X?

A.  \frac{1}{5} 
B.  \frac{1}{3} 
C.  \frac{2}{5} 
D.  \frac{3}{5} 
E.  \frac{2}{3}

Since  x  and  y  are consecutive positive integers, in that order, and the total interest earned was  7.6\%  of the total investment, then  x=7\%  and  y=8\% .
 
 \frac{7}{100}*a+\frac{8}{100}*b=\frac{7.6}{100}(a+b) , where  a  and  b  are amounts invested in X and Y, respectively.
 
 \frac{a}{b}=\frac{2}{3} ;
 
 \frac{a}{a+b}=\frac{2}{2+3}=\frac{2}{5} .

Answer: C

ankitsaroha · Answer

x and y are consecutive integers where x<y and their weighted mean is 7.6%. So, x=7% and y= 8%.
Let amount invested in X = a
Let amount invested in Y = b
Simple Interest = ((amount invested)*(rate of interest)*(time))/100

Here, time = 1 year.
Interest from X = 0.07a
Interest from Y = 0.08b
Average Interest = 0.076(a+b)

0.076(a+b) = 0.07a + 0.08b
7.6a + 7.6b = 7a + 8b
0.6a = 0.4b
1.5a = b

We need to find a/(a+b)
= a/(a+1.5a)
= 2/5

Therefore, answer is C

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TimeTraveller · Answer

Since the mean of  x  and  y  is  7.6 , the only possible choice for  x  and  y  is  7  and  8 , respectively.

Interest earned from investments  X  and  Y  after a year are  0.07X  and  0.08Y , respectively. Interest earned on the whole equals  0.076(X+Y) .

So, we have:  0.076(X+Y) = 0.07X + 0.08Y

0.076X - 0.07X = 0.08Y - 0.076Y

0.6X = 0.4Y

3X = 2Y

\frac{X}{X+Y} = \frac{X}{X+(3X/2)} = \frac{2X}{5X} = \frac{2}{5}.

Ans - C.

chetan2u · Answer

Hi..
We can also use weighted average method...
Since average interest is 7.6 and two interest x and y are consecutive integers, they are 7 and 8 respectively..

Now use weighted average..
Amount of x/ TOTAL=  \frac{8-7.6}{8-7}=\frac{0.4}{1}=\frac{2}{5} 
C

theharvardafricaner · Answer

chetan2u

How do you find X and Y? Can you explain?

TimeTraveller · Answer

Line up the values in increasing order, with average value in the middle:  7, 7.6, 8.

Now the ratio of  \frac{X}{Y} = \frac{8 - 7.6}{7.6 - 7} = \frac{0.4}{0.6} = \frac{2}{3} .

The question asks for ratio of  \frac{X}{Total} = \frac{2}{2+3} = \frac{2}{5}.

Bunuel · Answer

Par of  GMAT CLUB'S New Year's Quantitative Challenge Set

Kinshook · Answer

Given: Quentin put an amount of money into each of two new investments, X and Y. Investment X pays x% simple annual interest and investment Y pays y% simple annual interest. After 1 year, the average (arithmetic mean) interest earned by investments X and Y was 7.6%.

Asked: If x and y are consecutive positive integers, in that order, what fraction of the total amount was put in investment X?

Let the fraction of total amount X put in Investment X be k.

x = 7%
y = 8%

{k 7% + (1-k) 8%} = 7.6%
7k + 8(1-k) = 7.6%
8 - k = 7.6
k = .4 = 2/5

IMO C

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Quentin put an amount of money into each of two new investments, X and

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