d Word Problem 3 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs
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Word Problem 3

Problem 3: Divide Rs 1050 among A, B and C so that B’s share may be 2/3 of C’s share and A may get half as much as B and C together.



Solution:

Here,

1st Condition:

B’s share is equal to two third of C’s share;

i-e

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

2nd Condition:

A receives share equal to half of what  B and C get together;

i-e

A = \frac{1}{2}(B+C) ----(2)

From 2nd condition we can say;

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

\Rightarrow A : (B+C)= 1:2

Sum Of Ratio=1+2=3

Applying share formula;

A's Share = \frac{Total Amount}{Sum Of ratio}\times A's ratio

\Rightarrow A's Share = \frac{1050}{3}\times 1

\Rightarrow A's Share = 350

Again,

(B+C)'s Share=\frac{Total Amount}{Sum Of Ratios}\times (B+C)'s ratio

(B+C)'s Share=\frac{1050}{3}\times 2

(B+C)'s Share=700

Or we can write;

B+C=700 ----(2)

Put value of B from eq(1) in eq(2)

\Rightarrow \frac{2}{3}C+C=700

\Rightarrow \frac{2C+3C}{3}=700

\Rightarrow 5C=2100

\Rightarrow C=420

Put this value of C in eq (1)

\Rightarrow B= \frac{2}{3} \times 420

\Rightarrow B= 280

So , we calculated shares of A, B and C as 350, 280 and 420 respectively !

 

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