Introduction:

The Reasoning section of every competitive exam includes questions from the topic“Arithmatical Reasoning”. This topic is considered to be quiet important and every year number of questions are asked from this topic. We are providing you different types of questions were asked which will surely help you in the upcoming Exams.

This section deals with questions on simple Arithmetical operations. Here, the fundamental operations

• Time and work
• Time, speed, and distance
• Simple interest
• Compound interest
• Percentage
• Profit and loss
• Number system
• Average
• Ratio and proportion

So, it is one of the most interesting chapters in reasoning, because it contains both aptitude and reasoning.

Let’s explain a little bit of each type arithmetic reasoning

Time and work − Problems on time and work will be of normal men work and men women work type problems. In such type of questions, we have to bring the number to 1 always. If it is given that 5 men can do a certain work in 10 days, and after this data it is mentioned that 10 men can take how many days to do the work then at first we have to find that, 1 man can do the job in how many days and then we can proceed further.

Time speed and distance − For problems regarding this chapter, there is one formula which we can use in this context i.e. distance = time x speed.

Simple interest − If P is a principal, R is taken as rate of interest, T is taken as time, and I is taken as interest then the relationship between them is

I = (P x T x R) / 100

Compound Interest − If P is principal, R is rate, amount is A and time is n years then if interest is −

Compounded annually : A = P (1 + R/100)ⁿ

Compounded half yearly : A = P [1 + (R/2)/100]²ⁿ

Compounded quarterly : A = P [1+(R/4)/100]⁴ⁿ

Percentage − If it is mentioned that at a certain percent, it will be meant that many hundredths. Thus if we say a percent it means a hundredths, and will be written as a %.

Profit and loss − Profit = sale price – cost price and %profit = (profit x 100) / cost price

Average − The average is a measure of central point of a set of numbers. It is an estimation of where the center point or weight of a set of number lies.

Number system − It is very important in arithmetical reasoning to know about the numbers. It is considered as backbone of mathematics.

• Natural Numbers − Natural numbers are called as counting numbers and are represented as 1, 2 , 3, 4, 5, 6,…
• Whole Numbers − Whole numbers are those numbers which start from 0 to infinity. i.e. 0, 1, 2 …
• 0 is not a natural number.
• Integers − If we connect positive numbers and negative numbers with zero then we got integers. Also we can define integers as negative numbers + whole numbers. i.e. {..., - 3, - 2, - 1, 0, 1, 2, 3, …}

There are also even numbers and odd numbers. An even number is that number which can be divided by 2 and an odd number is that number which cannot be divided by 2.

A prime number is that number which can be divided by only two numbers that is 1 and the number itself. The smallest prime number is 2. Other prime numbers under 50 are, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47.

Number system − It is very important in arithmetical reasoning to know about the numbers. It is considered as backbone of mathematics.

• Natural Numbers − Natural numbers are called as counting numbers and are represented as 1, 2 , 3, 4, 5, 6,…
• Whole Numbers − Whole numbers are those numbers which start from 0 to infinity. i.e. 0, 1, 2 …
• 0 is not a natural number.
• Integers − If we connect positive numbers and negative numbers with zero then we got integers. Also we can define integers as negative numbers + whole numbers. i.e. {..., - 3, - 2, - 1, 0, 1, 2, 3, …}

There are also even numbers and odd numbers. An even number is that number which can be divided by 2 and an odd number is that number which cannot be divided by 2.

A prime number is that number which can be divided by only two numbers that is 1 and the number itself. The smallest prime number is 2. Other prime numbers under 50 are, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47.

SI Example-

Example 1: If the principal is 100 Rs. Difference of Simple Interest for 4yrs and 6yrs is Rs 8. Calculate the rate of simple interest.

Solution: In simple interest questions, interest always remains same for a year if the principal, rate of interest is constant for the same.

Let Interest for 4 yrs is I then interest for 6 yrs is (I+8)

interest for 2 yrs is Rs. 8

interest for 1 yr = 4

rate of interest = (4/100) × 100 = 4%

Example 2: If the amount is (10/9) times of Principal and rate of interest and time both are numerically equal. Then, what is the rate of interest per annum?

Solution: Let Principal is P. Given, numerically R = T

Interest = Amount – principal

I = (10/9)P – P

I = P/9 (Interest is in the multiples of Principal)

Now, I =[(P×R×T)/100]

P/9 = (P× R× T)/100

R2 = 100/9 (using, R=T)

R = (10/3)%

We can also say the time period is (10/3)years.

Annual Instalments for Simple Interest:

Let's discuss a real example to understand instalment concepts:

A person deposit Rs.140 to bank every year up to 5 yrs . The bank gives him 5% rate of interest simple annually. And at the end of 5 yrs he get total amount of Rs.770

So, 140 is the instalment, time is 5 years rate of interest is 5% and the amount or debt is Rs.770

This Installment is also known as annual payment. Debt is total amount, so don’t confuse between these two terms.

Some Important examples based on Simple Interest.

Example3: A sum amounts to Rs. 702 in 2 years and Rs. 783 in 3 years. Calculate the sum, rate of interest and the amount after 5 years?

Solution:

Amount for 2 years(A2) = 702

Amount for 3 years (A3)= 783

Interest for 1 year (I) = 783-702 = 81

So Sum = A2 – 2I = 702 – 2×81

= 702-162 = 540

rate of interest = (81/540)×100

= 15%

Amount after 5 years = Sum+5I

= 540+ 5×81

= 945

Example 4: A sum of money doubles itself in 3 yrs at a simple interest. In how many yrs will it amount to 8 times itself?

Solution: Doubles in 3 yrs

3 times in 3× 2 = 6yrs

4 times in 3× 3 = 9yrs

8 times in 3× 7 = 21yrs

Example5: Atul and Vijay are friends. Atul borrowed a sum of Rs.400 at 5% per annum simple interest from Vijay. He returns the amount with interest after 2 yrs. Vijay returns to Atul 2% of the total amount returned. How much did Atul receive?

Solution: After 2 yrs, amount returned to Vijay = 400+ (400*5*2)/100 = Rs 440

Amount returned to Atul = 2% of 440 = 8.8

Example6: Rs.4000 is divided into two parts such that if one part be invested at 3% and the other at 5%, the annual interest from both the investments is Rs. 144. Find each part.

Solution: Let the amount lent at 3% rate be Rs.X, then amount lent at 5% rate is 4000-X

So, 3% of X + 5% of (4000-X) = 144

5% of 4000 – 2% of X = 144

200 – 2% of X = 144

2% of X = 56

X = (56/2)×100

X = 2800

And 4000 -X = 1200.

C.I. Example

we will discuss the concepts of Compound Interest and also talk about how to solve and approach the questions based on this topic.

Now, Let’s discuss the basic difference between Simple Interest and Compound Interest.

Principal = 1000, rate of interest (r) = 10%, time = 3yrs

Simple Interest

SI for 1st yr = (1000×10×1)/100 = 100,

SI for 2nd yr = 100 (In SI it will be the same as 1st yr)

SI for 3rd yr = 100

Compound Interest:

CI for 1st yr = 100

CI for 2nd yr will not be same as 1st yr because principal for 2nd yr is the amount of 1st yr.

So, CI (2nd yr) = (1100×10×1)/100 = 110

CI for 3rd yr will also not be the same as 1st yr and 2nd yr because principal for 3rd yr is the amount of 2nd yr.

principal (3rd yr) = Amount (2nd yr) = Principal(2nd yr)+Interest(2nd yr) = 1100+110 = 1210

CI (3rd yr) = (1210×10×1)/100 = 121

Hence total CI for 3yrs = 100+110+121 = 331

Amount after 3 yrs = 1331

Interest is always calculated on the Principal. But in case of CI, Principal is get changed every year.

If we calculate it by net rate concept then the Principal will remain same.

Concept1: How to calculate net CI rate for 2 years?

Let rate is r% per annum for 2 years

Net CI rate for 2yrs can be calculated by = 2r+(r2/100)

If rate is 1%, net CI rate for 2yrs = 2×1+(12/100) = 2.01%

If rate is 3%, net CI rate for 2yrs = 2×3+(32/100) = 6.09%

If rate is 14%, net CI rate for 2yrs = 2×14+(142/100) = 29.96%

Concept2: How to calculate net CI rate for 3 years?

Let rate is r% per annum for 3 years

Net CI rate for 3yrs can be calculated = 3r+3(r2/100)+1(r3/10000)

If rate is 3% p.a., net CI rate for 3 yrs

= 3×3+3(9/100)+1(27/10000)

= 9+.27+.0027 = 9.2727

If rate is 12% p.a., net CI rate for 3 yrs

= 3×12+3(144/100)+1(1728/10000)

= 36+4.32+.1728

= 40.4928

Representation while calculating net rate %.

Let’s calculate it for the rate 3% p.a.

write, r/r2/r3 = 3/9/27

then,3r/3r2/1r3 = 9/27/27

= 9.2727

Concept3: If the r% p.a. is in fraction:

Example: if rate is 16(2/3) % and principal is 216, then calculate CI for 2yrs and 3yrs.

Solution: We can write 16(2/3)% = 1/6 (Discussed in percentage study notes)

For 2 years

216×(1/6)= 36, Now multiply 36 by 2 = 72

36× (1/6) = 6 , multiply 6 by 1 = 6

Add both the above value = 72+6 = 78

CI for 2yrs = 78

For 3 years

216×(1/6) = 36, Now multiply 36 by 3 = 108

36× (1/6) = 6, multiply 6 by 3 = 18

6× (1/6) = 1, multiply 1 by 1 = 1

Add all the above values = (108+18+1)= 127

CI for 3yrs = 127

Concept4: When different rates are given for 2 years.

If a% is given for 1st year and b% is given for 2nd year.

Net rate of CI for 2 yrs = (a+b+ab/100) % (discussed in percentage study notes)

Note: The net CI rate will be the same if b% is given for 1st year and a% is given for 2nd year.

Example: If principal is 1000 Rs and r(1st yr) = 4% and r(2nd yr) = 6%. Calculate the CI after 2yrs.

Solution:

Net CI rate = 4+6+(4×6)/100

= 10.24%

Now CI = 1000×10.24% = 102.4 Rs

Concept 5: When difference between CI and SI is given.

We know, net CI for 2yrs = 2r+(r2/100) %,

net SI for 2 yrs = 2r%

So, difference = (r2/100)%

Example: If difference between CI and SI is Rs.10 and the principal is Rs.1000.Calculate the rate % per annum.

Solution: difference = 10 Rs.

So difference% = (10/1000)×100 = 1%

We know that, if rate of interest is 10%

then, net CI rate (2yrs) = 21%

net SI rate (2yrs) = 20%

difference = 1%

Definitely we can say r% per annum is 10%.

Example: Calculate the difference between CI and SI for 3 yrs if Principal = 8000 and r = 6% p.a.

Solution: Net rate CI(3yrs) = 19.1016%

Net rate SI (3yrs) = 18%

Difference = 1.1016%

So, difference = 1.1016% of 8000 = 88.128

Example: If difference between CI and SI is Rs.64 and r = 8% p.a.. Calculate the Principal and Amount?

Solution: If r = 8% p.a.

then, net rate CI (2yrs) – net rate SI (2yrs)

= 16.64% -16% = 0.64%

Given, difference is Rs. 64

So, 0.64% = 64

100% = 10000

Hence, Principal is 10000 Rs.

Amount = principal× (116.64%)= 10000× 116.64% = Rs.11664

Samples:

Samples

1 - Govt. has decided to connect Tripura and Delhi via a train service which is called ‘Tripura Sundari Express’ Two trains are running from Tripura and Delhi towards each other. Train from Tripura in covering a distance of 60 km takes 2 hours more than that of the train from Delhi. If Tripura train doubles its speed, then it would take 1 hour less than that of Delhi. Tripura train’s speed is?

Options −

A - 5

B - 10

C - 7

D - 8

Answer − Option B

Explanation − Let Tripura train's speed be X km/hr.

Then, 60/x - 60/2x = 3

6x = 60

x = 10 km/hr.

2 - Creative constructors has hired some workers from Bihar. From those newly appointed workers if 10 men working 6 hours a day can do a work in 20 days. Then 8 men working 10 hours a day can do it in how many days?

Options −

A - 15

B - 14

C - 17

D - 18

Answer − Option A

Explanation − 10 men work for 6 hours so total 60 hours and work is done in 20 days. 8 men working 10 hours means total 80 hours and the work will be completed in = (60 x 20)/80 = 15 days.

3 - Riyaz and Saqlain are two workers and they work for GPR pumps and pipes. Riyaz is twice as good a workman as Saqlain and together Riyaz and Saqlain finish a piece of work in 20 days. In how many days will Riyaz alone finish the work?

Options −

A - 90

B - 66

C - 30

D - 29

Answer − Option C

Explanation − If Riyaz takes x days to do a work then Saqlain takes 2x days to do the same work.

1/x + 1/2x = 1/20

3/2x = 1/20

x = 30 days

Hence, Riyaz alone can finish the work in 30 days.

Questions and Answers:

Q 1 - In DAVV school, number of boys in a class are four times the number of girls. Which of the following represents the total number of students in the class?

Options :

A - 40

B - 42

C - 46

D - 52

Answer : A

Explanation

Let the number of girls be x. From question, boys are 4x.Total number of students = x + 4x = 5x. Hence total number of students must be divisible by 5.Hence correct answer is option (A).

Q 2 - A college organised a function in which 1/3^rd of boys and ½ of girls are present. Total number of students present is

Options :

A - 90

B - 110

C - 125

D - Data inadequate

Answer : D

Explanation

Number of boys and number of girls are not given.

Q 3 - In between two file ends in the rack are six paper weights. If you decide to arrange the six paper weights in every possible combination and moved just one paper weight every minute, how long would it take to you?

Options :

A - 12

B - 16

C - 15

D - 9

Answer : A

Explanation

Clearly, we can get the answer by taking the factorial.

= 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.

So total time taken = 720 minutes = 12 hours.

Q 4 - Mr. Sharma has a cattle farm of his own. And in his farm there are goats and hens. In that group of goats and hens, the number of legs are more than twice the number of heads. The number of goats is

Options :

A - 25

B - 36

C - 17

D - Data Inadequate

Answer : D

Explanation

Data is inadequate to get to any conclusion.

Q 5 - In the annual exam Purusottam Meher has secured a particular number in Mathematics and that number consists of two digits whose sum is 11. If 27 is added to the number, exchange in the places of the digits took place. Find out the number that Purusottam has scored in Mathematics in the annual exam?

Options :

A - 47

B - 28

C - 24

D - 37

Answer : A

Explanation

Purusottam Meher has scored 47 marks in mathematics, hence, option A is correct.

Q 6 - P and Q can finish a piece of work in 15 and 10 days, Q starts the work and leaves it after 5 days. The number of days in which P can complete the work is

Options :

A - 15/2

B - 22/2

C - 11/7

D - 13/2

Answer : A

Explanation

Q’s 1 day work = 1/10

Q’s 5 day work = 1/2

Remaining work = 1 - 1/2 = 1/2

Let P completes remaining work in x days, so

1/15 × X = 1/2

X = 15/2

Q 7 - In Tertiary club Bhubaneswar different sports took place in winter season. It may be tennis or cricket or football. On one year 30 fit members of the club decided to play a tennis singles tournament. Every time a member loses a game he is out of the tournament. In that year there are no ties in the tournament. Find out the minimum number of matches that must be played to determine the winner?

Options :

A - 10

B - 15

C - 29

D - 61

Answer : C

Explanation

To determine the winner of the tournament, the minimum number of matches to be played is 30 - 1 = 29.

Q 8 - P is able to complete a certain task in 6 days and Q can do it within 8 days. P and Q agreed to do the task together for Rs. 4000. R joined them and they finished the task within 3 days. How much is to be paid to R?

Options :

A - Rs.300

B - Rs.700

C - Rs.400

D - Rs.500

Answer : D

Explanation

$Work \: done \: by \: R \: in \: one \: day \: = \: \frac{1}{3} - \lgroup\frac{1}{6} + \frac{1}{8}\rgroup = \frac{1}{3} - \frac{7}{24} = \frac{1}{24}$

$P's share \: : \: Q's share \: : \: R's Share \: = \: \frac{1}{6} : \frac{1}{8} : \frac{1}{24} = 4 \: : \: 3 \: : \: 1$

$\therefore \: R's \: share \: for \: 3 \: days \:= \: Rs.\lgroup3*\frac{1}{24}*4000\rgroup \: = \: Rs.500.$

Q 9 - Mehendi plucked 175 flowers and Pulak plucked 225 flowers. If flowers are to be placed in 5 threads. Then each thread will contain how many flowers in it?

Options :

A - 75

B - 80

C - 85

D - 90

Answer : B

Explanation

175 + 225 = 400. 400/5 = 80.

Q 10 - A data is given regarding the number of pens of a local company sold in a fortnight.

15 14 16 12 15 12 18 19 16 12 15 18 17 71 19

Find the mode.

Options :

A - 12 and 15

B - 12

C - 15

D - 16

Answer : A

Explanation

As 12 and 15 both repeats for 3 times hence, option A is correct.