Question

A and B started a business with initial investments in the ratio 14:15 and their annual profits were in the ratio 7:6. If A invested the money for 10 months, for how many months did B invest his money?
(a) 6
(b) 7
(c) 8
(d) 9

Answer 1

Hint: To solve the question given above, we will assume that the investment of A is x, and the investment of B is y. Then we will assume that the profit of A is p and the profit of B is q. Then we will make use of the fact that the profit achieved by them will be proportional to the product of the number of months they did investment and investment for one month. From there, we will find the number of months of investment of B by taking the ratios of profits.

Complete step-by-step solution -
To start with, we will assume that the investment of A is x, and the investment of B is y for one month. Now, it is given that the ratio of investment for one month is equal to 14:15. Thus, we will get the following equation,
\[\dfrac{x}{y}=\dfrac{14}{15}\]
\[\Rightarrow 15x=14y\]
\[\Rightarrow x=\dfrac{14}{15}y......\left( i \right)\]
Now, we will assume that the annual profit made by A is ‘p’ and the annual profit made by B is ‘q’. It is given in the question that the ratio of annual profits is 7:6. Thus, we will get the following equation,
\[\dfrac{p}{q}=\dfrac{7}{6}\]
\[\Rightarrow 6p=7q\]
\[\Rightarrow p=\dfrac{7q}{6}......\left( ii \right)\]
Another information given in the question is that A invested for 10 months. We have to find the number of months B did investment. Let it be ‘n’ months.
Now, we will make use of the fact that the profits achieved by A and B are proportional to the product of the number of months of investment and investment for one month. Thus, we will get,
\[p\propto x\times 10\]
\[\Rightarrow p=k\left( 10x \right).....\left( iii \right)\]
Similarly,
\[q\propto y\times n\]
\[\Rightarrow q=k\left( yn \right).....\left( iv \right)\]
We will now divide p by q. Thus, we will get,
\[\dfrac{p}{q}=\dfrac{k\left( 10x \right)}{k\left( yn \right)}\]
\[\Rightarrow \dfrac{p}{q}=\dfrac{10x}{yn}\]
\[\Rightarrow p=q\left( \dfrac{10x}{yn} \right)....\left( v \right)\]
From (ii) and (v), we have,
\[\dfrac{7q}{6}=q\left( \dfrac{10x}{yn} \right)\]
\[\Rightarrow \dfrac{7}{6}=\dfrac{10x}{yn}\]
\[\Rightarrow x=\dfrac{7yn}{60}.....\left( vi \right)\]
From (i) and (vi), we have,
\[\dfrac{14}{15}y=\dfrac{7yn}{60}\]
\[\Rightarrow \dfrac{14}{15}=\dfrac{7n}{60}\]
\[\Rightarrow n=\dfrac{60\times 14}{15\times 7}\]
\[\Rightarrow n=8\]
Thus, B invested for 8 months.
Hence, the option (c) is the right answer.

Note: Here, we have assumed the ideal condition that the profit is proportional to the investment they did. It may be possible that the proportionality constant for A and B are not the same. In these conditions, we will need more data to calculate the number of months of investment of B.