d Word Problem 6 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs
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Word Problem 6

Word Problem 6

Problem 6: Divide Rs 290 among A, B and C persons, so that 1/2 of A’s share may be equal to 2/3 of B’s share, and 2/3 of B’s share may be equal to 1/4 of C’s share.



Solution:

Amount to be divided = 290 Rs

1st-Condition of division – half of A’s share should be equale to two third of B’s share;

\Rightarrow \frac{1}{2} A's=\frac{2}{3}B's

2nd-Condition of division – two third of B’s share should be equale to one fourth of C’s share;

\Rightarrow \frac{2}{3} B's=\frac{1}{4}C's

Also the amount divided to A,B and C should be 290 in total;

\Rightarrow A+B+C=290---(1)

From 1st condition we can write;

A = \frac{4}{3}B----(2)

From 2nd condition we can write;

C = \frac{8}{3}B----(3)

Put values of A and C from eqs  2 & 3 in eq 1;

\frac{4}{3}B+\frac{8}{3}B+B=290

\Rightarrow \frac{ 4B+8B+3B}{3}=290

\Rightarrow \frac{ 15B }{3}=290

\Rightarrow B= 58

Put this value of B in eq 2 and eq 3

\Rightarrow A= 77.33 and

C= 154.66

 

 

 

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