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Aptitude & Reasoning Solutions and Tricks, Cuttack (2026)

05/10/2018

Hey friends,sorry for a long gap

A common sense question for you about the mirror effect....

If a analog clock shows time as 9.30 in its mirror image then what will be the real time?????
A.6.30
B.9.30
C.3.30
D.2.30

Note - please don't try to use a mirror to solve it because in the exam hall there will be no mirror......try to answer and if possible explain the process which you used to solve it in order to help others.......

20/03/2018

All students who want to score better in the quantitative aptitude and reasoning part,I personally suggest that they have to do some particular practices before going to the syllabus based training.....

Initially it seems awkward,but if you are regular with it......then you will see that the problems will be easy for you than the earlier scenario.

1.Practise & memorize the squares up to 100
and the reverse square root also.(up to 50 is mandatory.and who will practise and memorize up to 100 is sufficient for the exams)
Means 8^2 = 64 and √9801 = 99 etc

2. Practise & memorize the cubes up to 30 and the reverse cube root also ( up to 15 is very important and very regular in the exams....and who will go for 30 is excellent......totally sufficient for the examinations)
Means 8^3 = 512 and 3√3375 = 15

3. Compare between the 1st and 2nd point that which numbers are perfect square as well as perfect cube and memorize them.
Means 64 is such no i.e 4^3 = 8^2 = 64

4.Memorize the comparison between the fraction and the percentage resultants.....It is the most important practice you have to do to get good score in the section.... i.e
Memorize the comparison up to 1/20.(up to 1/10 is is very important....who will go for 1/20 is sufficient for most of the exams)
Means 1/2 = 50%, 1/3=33.33%,1/7 =14.29% etc up to 1/20 = 5%......

5.Memorize the divisibility formulae of the Nos. As much possible.....like divisibility formula of 2,3 ,4 .....12,15,18 etc.

Remember these five points and practise with respect to them and then see the result.

Look Friends, the word "memorization" is quite awkward for the Math section.😅😅😅

The question will definitely arise in your mind why to memorize so many things because there are many rules to obtain them......just memorize the fundamental rules and if needed then obtain the answer by applying them.

I will also agree with you because I also follow the basic concepts of math to solve the problems......

But I suggest you to memorize the resultants according to these five points specially.

Because in the exams we have not sufficient time for the solutions.

As per my knowledge we have only 30 second to 1min time to solve a question varying from exam to exam.
And in most of the exams the time limit is about 30 to 40 second.......

So by memorizing the above things as I said and also with the basic concepts with respect to the syllabus your speed limit to solve a question will increase rapidly......

So if you don't believe,just try it and also share your experience with me😊😊😊.

If any doubt,then message me

Thank You.......

05/09/2017

Who can answer?????????????

05/09/2017

Ans of the last qstn asked is the loss value of the shop keeper is in between 800 and 1100....that,if it is represented by x, then....... 800 < x < 1100

cause......the 2nd shopkeeper has no role as the first shopkeeper exchange the amount.......that is indirectly his loss......not the loss of the others............and as we don't know in what price does the shop keeper buy the perfume......so we can't say the exact value.......so as he returns 800 rupees,the loss amount is 800 + the cost of the perfume with respect to the shop keeper......i.e varies from 0

29/08/2017

Let's try a qstn from profit and loss

14/06/2017

SIMPLE INTEREST (part 2)

Formula 4 ÷

If an amount becomes "x" times of it self in " t " years at the rate of simple interest "r"....then the time "T" taken by the amount to be "y" time of it self at the same rate of interest is

T = {(y - 1) t} / (x-1) yr...

Ex- An amount will be thrice in 20yr at a rate of simple interest 10%...then at same rate of simple interest,time taken by the amount to twice it self is ??????
x = 3 y = 2 t = 20 then T = (2-1)20 / 3 - 1 = 20/2=10yr...

Formula 5 ÷

If an amount becomes x time of it self in some time at rate of r% of simple interest.....then in the same time period the rate of interest R% to make the amount y times of itself is

R = {(y-1))r} / (x-1) %

Ex -
An amount becomes twice in a time period of 10yrs....at the rate of simple interest 5%..then in the same time the rate to make the amount 4times of it self is?????

x = 2 y = 4 r = 5 then R = (4 - 1) 5 / (2-1) = 15%

Formula 6 ÷

If the difference of interest on an amount P obtained from two souces in time t yr will be "x"....then the difference in the rates of interest will be
Difference in the rate of difference
= (x × 100) / (P × t)

Ex -
If the difference of interest obtained on Rs 20000 in 10 yr is Rs2000...then the difference in their rates will be?????
= (2000 × 100) / (20000 × 5) = 2 %

12/04/2017

SIMPLE INTEREST (part 1)

The used terms are Principal - P
Rate - r
Time - t
Simple Interest - SI
and. Amount - A

Amount is equal to Principal + SI

➡ A= P + SI

General formula÷

SI = ( Ptr)/100

Derived formulae ÷

1⃣. A = P【 1 + {(rt)/100}】

2⃣. P = (100 × A) / (100 × rt)

3⃣. SI = (Art) / (100 + rt)

SHORT CUT TRICKS÷÷÷÷÷÷÷

FORMULA - 1

If rate of interest becomes r2% from r1% and rs "G" more interest are earned in " t " yr then

Principal can be calculated by

P = (G × 100) / { (r2 - r1) × t }

Example÷

A person deposited some amount in bank at a rate of 5% interest ,,but suddenly bank increased its rate of interest to 7% and it was found that the increased interest was 4000 more than the previous interest in 5 yrs....then what amount did the person deposit in the bank?????

Ans÷
Here r1 = 5% , r2 = 7% , t = 5yrs , increasing SI , G = Rs 4000/-

Then P = (4000×100) / { ( 7 - 5 ) × 5 }

= 400000/10 = 40000

The man deposited 40000 in the bank.........

.............................................................................

FORMULA - 2

The time taken by an amount to become " n " times of itself at r% rate of simple interest is

t = { (n-1) × 100 } / r

Example÷

Ram deposited some money in a bank at a rate of 20% of simple interest...after some yr he found that his money was increased by 5 times than the amount he deposited...then find after how many yrs the amount increased by 5 times?

Ans ÷

Here n = 5 times , r = 20%

Then the time period is calculated by

t = { ( 5 - 1) × 100 } / 20

= 400 / 20 = 20 yrs.
.............................................................................

FORMULA - 3

An amount will be " n " times of itself in " t " yrs at the rate of interest

r = { ( n - 1 ) × 100 } / t %

Example÷

Ram deposited some money in the bank in simple interest and he found that it became 7 times in 12 yrs.... then what was the rate of the simple interest?

Ans÷

Here n = 7 times , t = 12 yrs then

r = { ( 7 - 1 ) × 100 } / 12 %

= 600 / 12 %

= 50%
..............to be continued for part 2.................

19/03/2017

PERFECT SQUARE÷

What do we mean by a perfect square??????

A perfect square is a natural no. ,that gives a integer in result when square rooted

Ex÷ 25, √25 = +5 or -5.

How to identify a perfect square ÷

1.A perfect square generally ends with 00,1,4,5,6 and 9, and except these digits in the unit place ,the rest i.e unit palce with 2,3,7,8 are not perfect square.

2.If we add the digits of a perfect square,then the final resultant root digits are 1,4,7 and 9 except these root digits,the number having other digits can't be a perfect square.

3.when a perfect square is devided by 7,it can't leave a remainder of 3,5 or 6.

4.In a perfect square if the unit place digit is 6,then the tenth place digit must be a odd digit.

5.In a perfect square if the unit place digit is 5 then the tenth place digit must be 2

6.in a perfect square even no. Of 0 are present at the end side

7.in a perfect square if the unit digit are 1,4 and 9,then the tenth place digit must be even

If a number satisfies all the above rules then it may be a perfect square,but these rules are right in 90% case ,the rest 10% numbers are exceptional case.

By using these rules we can identify 90% of the perfect square numbers

Example 1 ÷

Let's take a number "65,536"

We have to check whether it is aperfect square or not
1.it ends with 6,so may be
2.root digit is = 6+5+5+3+6=25=2+5=7,so it may be
3.65,536÷7 = leaves a remainder of 2,so it may be
4.unit place 6,tenth palce odd,satisfies

So it is a perfect square.

Example 2 ÷

Let 216
1.unit place 6 ,may be
2.tenth place odd, may be
3.root digit 9,may be
4.when divided by 7 leaves a remainder of 6 ,that doesn't satisfy.

So it is not a perfect square

Example 3 ÷

Exceptional case (only 10% of the total no.)

Let take 184
1.unit place 4 satisfies
2.tenth place even satisfies
3.root digit 4 satisfies
4.when divided by 7 leaves a ramainder of 2 that also satisfies

So it may be a perfect square,but actually it is not a perfect square

So in some cases there are some numbers who satisfy the rules,but they are not actually perfect squares,And the cases are limited to just 10% and the rest 90% must be perfect squares.

30/12/2016

APTITUDE

NUMBER SYSTEM

Let's have some basic ideas about squaring the numbers.I hope it is helpful to us if we know some square of two digit numbers.

✔For Example, if we know the squares up to 20 then we can easily calculate the squares up to 200 by this formula.

Look

✔If the given sum is (x)^2
Then
●Rule 1-
Find the nearest no. ending with zero
●Rule 2-
Find the difference of x with respect to the nearest no. ending with zero.

Then it's game time.

✔Let the differnce is remarked as 'n'

◆Case 1- If 'x' is greater than the nearest no ending with zero.

●Then the formula is

➡X^2={(square of the nearest no. ending with zero) + ( n × the nearest no ending with zero) + ( n × x)}

✔Example-

(123)^2=???
nearest no ending with zero = 120
Difference = 3
So
{(120)^2 +(3×120) + (3×123)}
= 14400+360+369
= 15129(ans)

◆Case 2- If 'x' is smaller than the nearest no. ending with zero

●Then the formula is-

➡X^2=[{square of the nearest no. ending with zero} - {( n × the nearest no ending with zero) + ( n × x)}]

✔Example-

(296)^2 =???
Nearest no.ending with zero = 300
Difference = 4
So
[(300)^2 - {(4 × 300) + (4 × 296)}]
= 90000 - (1200+1184)
= 90000 - 2384
= 87616 (ans)
Here x^2 means square of 'x'

25/11/2016

APTITUDE

TIME AND WORK

How many person can complete the task with in how many days or how long it will take to complete the given work or how much work one person can do etc.

BASIC FORMULA

●If a person can do a piece of work in x days, then person one day work is ( 1 / x )

●If a person 1 day work is ( 1 / x ) then the person will complete the work in x days.

●While solving the problems assume the work done by person to be equal to 1.

EXAMPLE

1. A alone can do a piece of work in 8 days, B alone can do the same work in 12 days. In how many days can A and B together completes the same work?
A′s 1 day′s work = ( 1 / 8 )
B′s 1 day′s work = ( 1 / 12 )
( A + B )′s 1 day′s work = ( 1 / 8 ) + ( 1 / 12 )
= ( 5/ 24 )
( A + B ) will complete the work in 24/5 Days=4.8~5 days

BASIC FORMULA

If a person A Works x times more that of second person B ,then
●Ratio of work done by A and B is x : 1
●The ratio of time taken by A and B is 1: x

EXAMPLE

A work can be finished in 16 days by twenty women. The same work can be finished in fifteen days by sixteen men. Find the ratio between the capacity of a man and a woman ?

Work done by 20 women in 1 day = 1 / 16

Work done by 1 woman in 1 day = 1 / ( 16 × 20 )

Work done by 16 men in 1 day = 1 / 15

Work done by 1 man in 1 day = 1 / ( 15 × 16 )

Ratio of the capacity of a man and woman = { 1 / ( 15 × 16 ) } : { 1 / ( 16 × 20 ) }
= ( 1 / 15 ) : ( 1 / 20 )
= ( 1 / 3 ) : ( 1 / 4 )
= 4 : 3

18/11/2016

APTITUDE

🔹🔷PARTNERSHIP🔷🔹

More than one person agree to invest their money to run a business or firm then this kind of agreement is called partnership. The persons involved in the partnership are called partners.

➡TWO TYPES OF PARTNERSHIP

➡SIMPLE PARTNERSHIP

In simple partnership, capitals of partners are invested for the same period of time.

➡Basic Formulas

🔷( Capital of A / Capital of B ) = ( Profit of A / Profit of B )

Example÷

A, B and C started a business by investing Rs. 120000, Rs. 135000 and Rs. 150000 respectively. Find the share of each, out of an annual profit of Rs. 56700?

Ans

Ratio of shares of A, B and C = Ratio of their investments.
= 120000 : 135000 : 150000
= 8 : 9 : 10
Total ratio = 27

A′s share = Rs. [ 56700 × ( 8 / 27 ) ]
= Rs. ( 2100 × 8 )
= Rs. 16800

B′s share = Rs. [ 56700 × ( 9 / 27 ) ]
= Rs. ( 2100 × 9 )
= Rs. 18900

C′s share = Rs. [ 56700 × ( 10 / 27 ) ]
= Rs. ( 2100 × 10 )
= Rs. 21000

➡COMPOUND PARTNERSHIP

In compound partnership, capitals of partners are invested for the different period of time.

➡Basic Formulas

🔷( Capital of A × Time period of A ) / ( Capital of B × Time Period of B ) = ( Profit of A / Profit of B )

Example÷

In a business, Lucky invests Rs. 35,000 for 8 months and manju invests Rs 42,000 for 10 months. Out of a profit of Rs. 31,570. Find manju′s share?

Ans

Lucky : Manju = ( 35000 × 8 ) : ( 42000 × 10 )
= 280000 : 420000
= 2 : 3
Total ratio = 5

Manju′s share = Rs. [ ( 3 / 5 ) × 31570 ]
= Rs. ( 3 × 6314 )
= Rs. 18942

➡RATIO OF DIVISIONS

When investments of all the partners are for the same time, the gain or loss is distributed among the partners in the ratio of their investments.

Example÷

Ramya starts a business with Rs. 45000. Janani joins in the business after 3 months with Rs. 30000. What will be the ratio in which they should share the profit at the end of the year?

Ans

Ramya and Janani share profit = Ratio of the investments multiplied by the time period

= ( 45000 × 12 ) : ( 30000 × 9 )
= 540000 : 270000
= 54 : 27
= 2 : 1

09/11/2016

APTITUDE

🔷PERCENTAGE🔷

Percentage is a number or ratio expressed as a fraction of 100.

➡PROPERTIES OF PERCENTAGE

1. If we have to convert percentage into fraction than it is divided by 100.
2. If we have to convert fraction into percentage we have to multiply with 100.

➡LOSS / DECREASE CONDITION

If there is increase of X% and subsequently X% decrease then there is always loss / decrease in the condition.

Example

If rohan salary is increase by 50% and subsequently decrease by 50%. How much percentage loss?

= ( 50 × 50 ) / 100
= ( 5 × 5 ) %
= 25 % decrease

➡PERCENTAGE INCREASES

If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure

🔹 [ { R / ( 100 + R ) } × 100 ] %

Example

If radha earning is 25% more than sita. Then sita earning is how many percentage less then by radha?

= [ { 25 / ( 100 + 25 )} × 100 ] %
= 20 %

➡PERCENTAGE DECREASES

If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is :

🔹[ { R / ( 100 - R ) } × 100 ] %

Example

If golu age is 20% less than gita than gita age is how many percentage more than golu?

= {20 / ( 100 - 20 ) } × 100 %
= 25 %

➡RESULTS ON POPULATION

Let the population of a town be P now and suppose it increases at the rate of R% per annum,

1. Population after n years =

🔹 P × { 1 + ( R / 100 ) }^n

2. Population n years ago =

🔹 P / { 1 + ( R / 100 ) }^n

Example

The population of a town is 176400. If it increases at the rate of 5% per annum, what will be its population 2 years hence, What was it 2 years ago ?

Population after 2 years =

[ 176400 × { 1 + ( 5 / 100 ) }^2 ]
= 176400 × ( 105 / 100 )^2
= 176400 × ( 21/ 20 )^2
= ( 176400 × 21 × 21 ) / ( 20 × 20 )
= 194481

Population 2 years ago =

176400 / { 1 + ( 5 / 100 ) }^2
= 176400 / (21/20)^2
= 160000

RESULTS ON DEPRECIATION

Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum.

1. Value of the machine after n years =

🔹P × { 1 - ( R / 100 ) }^n

2. Value of the machine n years ago =

🔹 P / { 1 - ( R / 100 ) }^n

Example

The value of a machine depreciates at the rate of 10% per annum. If its present value is Rs. 162000, what will be its worth after 2 years, What was the value of the machine 2 years ago ?

Value of the machine after 2 years =

Rs. [ 162000 × { 1 - ( 10 / 100 ) }^2 ]
= Rs. [ 162000 × ( 9 / 10 )^2 ]
= Rs. [ 162000 × ( 9 / 10 ) × ( 9 / 10 ) ]
= Rs. 131220

Value of the machine 2 years ago =

Rs. [ 162000 / { 1 - ( 10 / 100 ) }^2 ]
= Rs. [ 162000 / ( 9 / 10 )^2 ]
= Rs. ( 162000 × ( 10 / 9 ) × ( 10 / 9 ) )
= Rs. 200000

Note- x^n means x to the power n.

Aptitude & Reasoning Solutions and Tricks

05/10/2018

20/03/2018

05/09/2017

05/09/2017

29/08/2017

14/06/2017

12/04/2017

19/03/2017

30/12/2016

25/11/2016

18/11/2016

09/11/2016

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