Question

A man divided two sums of money among 4 sons R, S, T, U the first in the ratio 4 : 3 : 2 : 1 and the second in the ratio 5 : 6 : 7 : 8. If the second sum of money is twice the first sum, which son receives the largest part?
(a) R
(b) S
(c) T
(d) U

Answer 1

Hint: We will make use of the constant of proportionality to solve this question. We will assume the constants of first ratio and second ratio as K and C respectively. Then, with the information on the relation between the sums, we find the relation between K and C. With this relation, we will be able to find the son who gets the largest part.

Complete step by step solution:
The first sum is divided in the ratio as R : S : T : U = 4 : 3 : 2 : 1. Let us assume the constant of ratio be K. Thus, the portion of the sum received by the sons is
 R = 4K, S = 3K, T = 2K and U = 1K.
And the first sum, therefore, is 4K + 3K + 2K + K = 10K.
The second sum is divided in the ratio as R : S : T : U = 5 : 6 : 7 : 8. Let us assume the constant of ratio be C. Thus, the portion of the sum received by the sons is
 R = 5C, S = 6C, T = 7C and U = 8C.
And the second sum, therefore, is 5C + 6C + 7C + 8C = 26C.
So, total money received by various sons is as follows:
R = 4K + 5C
S = 3K + 6C
T = 2K + 7C
U = 1K + 8C
Based on the given information, we find the relation between C and K.
twice that of first sum = second sum
$\begin{align}
  & \Rightarrow 2\times 10K=26C \\
 & \Rightarrow 20K=26C \\
 & \Rightarrow 10K=13C \\
 & \Rightarrow K=\dfrac{13}{10}C \\
\end{align}$
Now, we will replace K with $\dfrac{13}{10}C$ and find the part received by all the sons.
R = $4\left( \dfrac{13C}{10} \right)+5C=\dfrac{52}{10}C+5C=\dfrac{102}{10}C$
S = $3\left( \dfrac{13}{10}C \right)+6C=\dfrac{39}{10}C+6C=\dfrac{99}{10}C$
T = $2\left( \dfrac{13}{10}C \right)+7C=\dfrac{26}{10}C+7C=\dfrac{96}{10}C$
U = $\dfrac{13}{10}C+8C=\dfrac{93}{10}C$
As we can clearly see, R got the largest part of the money.
Hence, option (a) is the correct option.

Note: It is advisable to be careful while taking LCM to solve the equations. It is not necessary to keep the denominator common while comparing, but the same denominator makes the comparison easy.