Question

Divide the amount Rs. 4260 into three parts proportional to $2\dfrac{1}{2}:1\dfrac{2}{3}:1\dfrac{3}{4}$.
A. Rs. 1800, Rs. 1200 and Rs. 1260
B. Rs. 1600, Rs. 1200 and Rs. 1260
C. Rs. 1800, Rs. 1200 and Rs. 1280
D. Rs. 1600, Rs. 1400 and Rs. 1260

Answer 1

Hint: To solve this question, we should know the concept related to ratios. Let us consider the three parts be x, y, z respectively. From the question, we can write that$x:y:z=2\dfrac{1}{2}:1\dfrac{2}{3}:1\dfrac{3}{4}$. We know a property that if there is a relation $a:b:c=d:e:f$, we can rewrite them using a constant k as $a=kd,b=ke,c=kf$. Likewise, we can assume a constant k and write the values of x, y, z. We know that the sum of the three parts is Rs. 4260. Using this, we can get the value of k and finally the values of x, y, z.

Complete step by step answer:
Let us assume the three required parts as x, y, z. We can write from the question that
$x:y:z=2\dfrac{1}{2}:1\dfrac{2}{3}:1\dfrac{3}{4}$.
We know a property that if there is a relation $a:b:c=d:e:f$, we can rewrite them using a constant k as $a=kd,b=ke,c=kf$.
Let us assume a constant k and we get the above ratio as
$\begin{align}
  & x=2\dfrac{1}{2}\times k=\dfrac{5k}{2} \\
 & y=1\dfrac{2}{3}\times k=\dfrac{5k}{3} \\
 & z=1\dfrac{3}{4}\times k=\dfrac{7k}{4} \\
\end{align}$
We know that the sum of the values of x, y, z is equal to the total amount that is Rs. 4260. We can write it as
$x+y+z=4260$
Using the values of x, y, z from the derived relations, we can write that
$\dfrac{5k}{2}+\dfrac{5k}{3}+\dfrac{7k}{4}=4260$
Taking k common and simplifying, we get
$k\left( \dfrac{5}{2}+\dfrac{5}{3}+\dfrac{7}{4} \right)=4260$
The L.C.M of the denominators is 12. Using this, we get
$\begin{align}
  & k\left( \dfrac{5}{2}+\dfrac{5}{3}+\dfrac{7}{4} \right)=k\left( \dfrac{5\times 6+5\times 4+7\times 3}{12} \right)=4260 \\
 & k\left( \dfrac{30+20+21}{12} \right)=4260 \\
 & k\dfrac{71}{12}=4260 \\
 & k=4260\times \dfrac{12}{71}=\dfrac{71\times 60\times 12}{71}=720 \\
\end{align}$
Using this value of k in the equations of x, y, z, we get
$\begin{align}
  & x=\dfrac{5k}{2}=\dfrac{5\times 720}{2}=5\times 360=1800 \\
 & y=\dfrac{5k}{3}=\dfrac{5\times 720}{3}=5\times 240=1200 \\
 & z=\dfrac{7k}{4}=\dfrac{7\times 720}{4}=7\times 180=1260 \\
\end{align}$
$\therefore $The values of x, y, z are Rs. 1800, Rs. 1200, Rs. 1260 respectively. The answer is option-A.

Note:
 We should observe that the values of the three parts should be equal to the value Rs. 4260. If we carefully observe the options, we have only option-A, which satisfies the condition of the sum equalling Rs. 4260. SO, without any calculations, we can conclude that the answer is option-A. Generally, these types of questions are given to test the lateral thinking capability of the student. Students who don’t observe this will do the whole procedure and fall behind those who observed this and opted for the answer right away. So, we should first check which are the options satisfying the basic rules.