Question

The difference of two numbers is 16. If one-third of the smaller number is greater than one-seventh of the larger number by 4, then find the numbers.

Answer 1

Hint: Assume that one of the numbers is x and the other is y. Use the fact that the difference of the two numbers is 16 to form an equation in x and y. Use the fact that one third of the smaller number is greater than one-seventh of the larger number by 4 to form another equation in x and y. Solve the system of the equation using any of the known methods like elimination method, substitution method etc and hence find the value of x and y. Hence find the two numbers. Verify your answer.

Complete step by step answer:
Let one of the numbers is x and the other is y. Let x > y.
Since the difference of the two number is 16, we have
$x-y=16\text{ }\left( i \right)$
Also, since one-third of the smaller number is greater than one-seventh of the larger number by 4, we have
$\dfrac{y}{3}-\dfrac{x}{7}=4$
Multiplying both sides by 21, we get
$7y-3x=84\text{ }\left( ii \right)$
From equation (i), we have
$\begin{align}
  & x-y=16 \\
 & \Rightarrow x=y+16\text{ }\left( iii \right) \\
\end{align}$
Substituting the value of x in equation (ii), we get
$7y-3\left( y+16 \right)=84$
Using distributive property of multiplication over addition, i.e. a(b+c) = ab+ac, we get
$\begin{align}
  & 7y-3y-48=84 \\
 & \Rightarrow 4y-48=84 \\
\end{align}$
Dividing both sides by 4, we get
$y-12=21$
Adding 12 on both sides, we get
$y=33$
Substituting the value of y in equation (iii), we get
$x=y+16=33+16=49$
Hence the two numbers are 49 and 33.
Note:
[1] Verification:
We can verify the correctness of our solution by checking that the numbers satisfy the properties described in the questions.
We have $49-33=16$
Hence the difference of the two number is 16
We have $\dfrac{1}{3}\times 33-\dfrac{1}{7}\times 49=11-7=4$
Hence one-third of the smaller number is greater than one-seventh of the larger number by 4.
Hence our answer is verified to be correct.