Two numbers are such that the sum of $$\frac{1}{3}$$​ of the first number and $$\frac{1}{2}$$​ of the second number is 8. The sum of $$\frac{1}{5}$$​ of the first number and $$\frac{1}{6}$$​ of the second number is 4. What is the largest of the two numbers?

Correct option is D

Solution:

Let the first number be x, and the second number be y.

$$\frac{1}{3}x + \frac{1}{2}y = 8 \\ \ \\2x + 3y = 48 \quad \text{(i)} \\ \ \\\frac{1}{5}x + \frac{1}{6}y = 4 \\ \ \\6x + 5y = 120 \quad \text{(ii)}$$​

Multiply (i) by 2:

​$$4x + 6y = 96 \quad \text{(iii)}$$​

Now subtract (iii) from (ii):

(6x + 5y) - (4x + 6y) = 120 - 96

​$$2x - y = 24 \quad \text{( iv)}$$​

​$$y = 2x - 24 \quad \text{(v)}$$​

Substitute in (i):

2x + 3(2x - 24) = 48

2x + 6x - 72 = 48

8x = 120

x = 15

Now from (v):

y = 2(15) - 24 = 30 - 24 = 6

Answer:15 (the larger of the two numbers)