"> Word Problem 2 | Job Exam Rare Mathematics (JEMrare)

Word Problem 2

Problem 2: Divide Rs 2160 among A, B and C so that A may get half as much as B, and B may get twice as much as C.

a) 1080, 720, 1080           b) 720, 1080, 360                c) 300, 720, 1080            d) 300, 780, 1080



(general math for jobs)

Solution:

1st Condition

A gets half of B and then again as much as B gets

\Rightarrow A=\frac{1}{2}B+B

\Rightarrow A=\frac{3}{2}B ----(1)

2nd Condition

B gets twice as much as C

\Rightarrow B=2C ----(2)

Now as the amount is being divided among A, B and C , so, sum of shares of all three should make 2160;

\Rightarrow A+B+C=2160 ------------(3)

Put values from eq (1) and (2) in eq (3)

\Rightarrow \frac{3}{2}B+2C+C=2160

\Rightarrow \frac{3}{2}B+3C=2160

Again put value of B from eq(2)

\Rightarrow \frac{3}{2}(2C)+3C=2160

\Rightarrow 3C+3C=2160

\Rightarrow 6C=2160

\Rightarrow C=\frac{2160}{6}

\Rightarrow C=360

This is C’s share. Now put this value of C in eq (2);

\Rightarrow B=2(360)=720

This is B’s share. Now put  value of B in eq (1);

\Rightarrow A = \frac{3}{2}\times 720

\Rightarrow A = 1080

So shares of A, B and C are 1080,  720 and 360 respectively