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

Word Problem 3

Problem 3: On a part of  amount of Rs 1500,  simple interest is applied @ 5% for one year. On the other part interest is applied @ 6% annually. The sum of both interests is Rs 85. Find the ration between two parts of the amount.


simple arithmetic questions and answers



Let First Part=y

\Rightarrow 2nd Part = 1500-y

Using formula for Interest;

I= \frac{P\times R\times T}{100}

For First Case;

P = y, R=5, T= 1

So, I= \frac{y\times 5\times 1}{100}

I= \frac{ 5y}{100}

So. we take it first Interest,

I1= \frac{ 5y}{100}

For 2nd case;

P= 1500-y, R = 6, T =1

Again using;

I= \frac{P\times R\times T}{100}

I= \frac{(1500-y)\times 6\times 1}{100}

I= \frac{6(1500-y)}{100}

We take it 2nd interest,

I2= \frac{6(1500-y)}{100}

According to statement;


\Rightarrow \frac{ 5y}{100} + \frac{ 6(1500-y)}{100}=85

\Rightarrow \frac{ 5y}{100} + \frac{ 9000-6y}{100}=85

\Rightarrow \frac{ 5y+9000-6y}{100}=85

\Rightarrow 9000-y=85\times 100

\Rightarrow 9000-y=8500

\Rightarrow y=9000-8500

\Rightarrow y=500

So, First Part = 500

\Rightarrow 2nd Part = 1500-500=1000


1st Part : 2nd Part

\Rightarrow 500:1000

\Rightarrow 1:2

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