A, B and C enter a partnership with initial investments in the ratio $$\frac{11}{5}:\frac{7}{2}:\frac{15}{8}$$​. After 4 months A raises her share of investment by p%. If the total profit after 12 months since the beginning of the partnership is Rs. 138584, and A gets Rs. 50864 as her share of profit, what is the value of p?

Correct option is B

Given:

Investment ratios: A : B : C = $$\frac{11}{5} : \frac{7}{2} : \frac{15}{8}$$​​

Duration: 12 months total.

A increases investment by p % after 4 months.

A’s share of profit = ₹50864

Total profit = ₹138584

Formula Used:

Profit Share = $$\frac{\text{(Investment × Time)}}{\text{Total of all (Investment × Time)}}$$​​

Solution:

LCM of 5, 2, 8 = 40
A = $$\frac{11}{5} = \frac{88}{40}$$

​
B = $$\frac{7}{2} = \frac{140}{40}$$

​
C = $$\frac{15}{8} = \frac{75}{40}$$​​

A : B : C = 88 : 140 : 75

Let’s denote A’s increased investment as $$88 \times \left(1 + \frac{p}{100} \right)$$​​

A’s total investment units:
​$$88 \times 4 + 88 \times \left(1 + \frac{p}{100} \right) \times 8 = 352 + 704 \left(1 + \frac{p}{100} \right)$$​​

B: $$140 \times 12 = 1680$$​​

C: $$75 \times 12 = 900$$​​

A’s share in total profit:

Total profit = ₹138584, A’s share = ₹50864

​$$\frac{352 + 704 \left(1 + \frac{p}{100} \right)}{352 + 704 \left(1 + \frac{p}{100} \right) + 1680 + 900} = \frac{50864}{138584}$$​​

​$$\frac{50864}{138584} = \frac{12716}{34646} = \frac{6358}{17323}$$​​

Let:

​$$x = 352 + 704 \left(1 + \frac{p}{100} \right)$$​​

​$$\Rightarrow \frac{x}{x + 2580} = \frac{6358}{17323}$$​​

​$$\Rightarrow 17323x = 6358x + 16376400$$​​

​$$\Rightarrow 10965x = 16376400$$​​

​$$\Rightarrow x = \frac{16376400}{10965} = 1493.4$$​​

​$$352 + 704 \left(1 + \frac{p}{100} \right) = 1493.4$$​​

​$$\Rightarrow 704 \left(1 + \frac{p}{100} \right) = 1141.4$$​​

​$$\Rightarrow 1 + \frac{p}{100} = \frac{1141.4}{704} = 1.6213$$​​

​$$\Rightarrow \frac{p}{100} = 0.6213$$​​

​$$\Rightarrow p \approx 62.13\%$$​​

Thus, the correct option is (b) 62.5%