## Aptitude Test Shortcut Tricks

Cracking any Aptitude test requires-Knowledge, Accuracy & Speed. It’s the combination of all these three factors that rears a winner and this is exactly you will learn here. You can score easily on quantitative aptitude section if you understand the basics of number system. Since the questions on number systems are simple, importance lies in acquiring the right skills to tackle these problems with speed.

Example 1: If the side of a cube is increased by 100%, find by

What percentage the surface area of the cube is increased?

- a) 150%
- b) 200%
- c) 300%
- d) 350%

Speed Trick Example

**SOLUTION**

Let side of cube be a.

Increase of side by 100% would mean double of side i.e. 2a.

As we know surface area of a cube is 6a2.

With doubled side we can mentally calculate it to be 24a2

i.e. (6 x(2a) 2).

The increased amount for initial surface area would be 24a2- 6a2

= 18a2.Obviously 18a2 is 300% of 6a2.

** Regular Solution**

Surface area of cube is 6a .

If a = 1, then surface area would be 6.

If a = 2 then surface area would be 24.

24-6 = 18, so 24 is 300% of 6.

So the answer is c.

** Fast Solution**

It the sides are doubled you can spatially imagine forming three

additional squares on each face of the cube.

So the increase in surface area would be 300%.

**Shortcut tricks of Time and Work**

To solve these problems very quickly, you should understand the concept of Time and Work and some shortcut methods.

If a man can do a piece of work in 5 days, then he will finish 1/5th of the work in one day.

If a man can finish 1/5th of the work in one day then he will take 5 days to complete the work.

If a man 5/6th of work in one hour then he will take 6/5 hours to complete the full work.

If A works three times faster than B then A takes 1/3rd the time taken by B.

Here are some shortcut rules which can be very useful while solving Time and Work problems.

**Rule 1: Universal Rule **This rule can be used in almost every problem.

If M1 persons can do W1 work in D1 days and M2 persons can do W2 works in D2days then we can say M1D1W2 = M2D2W1

If the persons work T1 and T2 hours per day respectively then the equation gets modified to

M1D1T1W2 = M2D2T2W1

If the persons has efficiency of E1 and E2 respectively then, M1D1T1E1W2 = M2D2T2E2W1

**Rule 2: **If A can do a piece of work in n days, then The work done by A in one day = 1/n

**Rule 3: **If A can do a work in D1 days and B can do the same work in D2 days then A and B together can do the same work

**Rule 4: **If A is twice as good a workman as B, then A will take half of the time taken by B to complete a piece of work.

**Rule 5: **If A is thrice as good a workman as B, then A will take one third of the time taken by B to complete a piece of work.

**Rule 6: **If A and B together can do a piece of work in x days, B and C together can do in y days and C and A together can do in z days, then the same work can be done

**Rule 7: **If A can do a piece of work in D1 days, B can do in D2 days and C can do in D3 days then they together can do the same work in

**Rule 8: **If A and B together can do a piece of work in D1 days and A alone can do it in D2days, then B alone can do the work in

**Rule 9: **If the number of men are changed in the ratio of m:n, then the time taken to complete the work will change in the ratio n:m

**Formulas:**

No. of ways of selecting *r *objects out of *n *objects = *n*C*r*

No. of ways of arranging *n *objects = *n*!

No. of ways of arranging *n *objects with *a *identical & *b *identical =*n*!/ *a*! *b*!

No. of ways of arranging *r *objects out of *n *objects = *n*(*n*–1)(*n*2) … (*n *– (*r *– 1)) = *n*P*r *= *n*C*r r*!

No. of ways of arranging *n *objects in a circle = (*n *– 1)!

**Choosing from Different Types of People (e.g. Boys & Girls)**

Example: A team of 3 is to be chosen from a group of 3 boys and 4 girls. How many ways can this be done if

(i) there are no restrictions?

(ii) there must be exactly 1 boy?

(iii) there must be at least 1 boy?

(iv) there must be at least 1 boy and at least 1 girl?

** SOLUTION**

of ways of choosing 4 people = 7C3 = 35

of ways of choosing 1 boy and 2 girls = 3C1

4C2 = 18

of ways = Total no. of ways – no. of ways with no boys = 35 – 4C3 = 35 – 4 = 31

**Note: **It is wrong to say no. of ways = 3C1

6C2 = 45

- of ways = Total no. of ways – no. of ways with no boys – no. of ways with no girls

= 35 – 4C3 – 3C3 = 35 – 4 – 1 = 30

**Note: **It is wrong to say no. of ways = 3C1

4C1

5C1 = 60

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