​The sum of 2 numbers is 184. If one-third of the one exceeds one-seventh of the other by 8, find the smaller number.

Correct option is D

Given:

Sum of two numbers = 184

One-third of one exceeds one-seventh of the other by 8

Solution:

Let the numbers be x and y

From condition 1;

​$$x + y = 184 \quad\tag{1}$$​​

From condition 2:

​$$\frac{1}{3}x - \frac{1}{7}y = 8$$​​

​$$7x - 3y = 168\quad \tag{3}$$​​

From equation (1):

y = 184 - x

Substitute into equation (3):

7x - 3(184 - x) = 168

7x - 552 + 3x = 168

10x = 720

x = 72

Then, y = 184 - 72 = 112

Smaller number is 72