# Quantitative Aptitude Question With Answers – Make Maths Simple

In this post lets discuss some of the questions for simplification. Try to solve by yourself then scroll down for answers.

1. Find the value of 40000-40000÷100

2. Simplify the following 515*515*515-221*221*221/(515*515+515*221*221+221)

3. Simplify the following (578291*578291-457428*457428)/120863

4. Find the value of ((598+194)^{2}-(598-194)^{2})/(598*194)

5. If the value of a-b=6 and a^{2}+b^{2}=40, then find the value of ab

**Answers**

1. Most of you might have probably got the answer as 0 if you’re calculating in hurry. But its wrong.

We mostly forget about the BODMAS while doing these kind of simple calculations.

The answer is 39600 (1st divide 40000÷100 = 400, then subtract 40000 with 400)

2. How long you too for this calculation or used calculator to do this sum.

Here is the simple way. This follows the formula as (a^{3}-b^{3}) /(a^{2}+a*b+b^{2})

=(a-b)(a^{2}+a*b+b^{2})/(a^{2}+a*b+b^{2})

=(a-b)

=(515-221)

=294

3. This is of the form a^{2}-b^{2}/(a-b)

=(a+b)(a-b)/(a-b)

=(a+b)

=578291+457428

=1035719

4. This is of the form ((a+b)^{2}-(a-b)^{2})/(a*b)

= ((a^{2}+2*a*b+b^{2})-(a^{2}-2*a*b+b^{2}))/(a*b)

=4ab/ab

=4

5. (a-b)^{2}-(a-b)^{2}=0

=>(a^{2}-2ab+b^{2})-(a-b)^{2}=0

=>(a^{2}+b^{2})-(a-b)^{2}=2ab

=>40-6^{2}=2ab

=>40-36=2ab

=>2ab=4

=>ab=2

Make your calculation as simple as possible 🙂 Share your comments with us 🙂

Algebra makes horrific calculations look simple…really amazing!!

nice one… now i got some idea about this like problems… 🙂

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