# Arithmetical Reasoning | arithmetical reasoning questions and answers pdf | test meaning

By   June 6, 2020

arithmetical reasoning questions and answers pdf  , Arithmetical Reasoning | test meaning ?

This chapter of reasoning deals with general arithmetical problems common in nature. solutions of these problems require common sense with slight amount of logical reasoning. these questions are invariably asked in competitive exams. we have dealt with the questions on arithmetical reasoning in detail in the following examples.

Example 1. The number of boys in a class are three times times the number of girls. which one of the following number can’t represent the total number of children in the class?

(a) 48

(b) 44

(c) 42

(d) 40

Sol. Let the number of girls be x, then from the question it is clear that number of boys are 3x.

Therefore, total no. of students

=Number of boys + Number of girls

=3x+ x =4x

Now, the total number of children in the class must be a multiple of 4. Out of the four options given (c) does not qualify this condition. therefore, 42 does not represent the total number of children in the class.

Hence, the correct answer is (c).

Example 2. In 1o yr, a will be twice as old as B was 10 yr ago. if at present a is 9 yr older than B,the present age of B is

(a) 19 yr

(b) 29 yr

(c) 39 yr

(d) 49 yr

Sol. Let the present age of B be x yr.

then, the present age of A would be (x + 9) yr

After 10 yr, the age of A would be (x + 9+ 10)

=(x+19) yr

and before 10 yr, the age of B was (x – 10) yr

Now, from the information given in the question,

(x + 19) =2 (x-10)

or               x+ 19 = 2 x-20

or             x=19 +20 =39 yr

Therefore, the present age of B is 39 yr

Hence, the correct answer is (c).

Example 3. The 1 st bunch of bananas if the 2nd bunch has 3 bananas less than the 1 st bunch, then  the number of bananas in 1 st bunch is

(a) 9

(b) 10

(c) 12

(d) 15

sol. Let the number of bananas in 2nd bunch be x.

then, the number of bananas in 1 st bunch =x + x/4 = 5x/4.

therefore, 5x/4 – x = 3 > 5x – 4x =12 > x= 12.

then, the number of bananas in 1 st bunch = 5/4 x 12 =15

hence, the correct answer is (d) .

Example 4. In a town, 65% people watch the news on television, 40% read a newspaper and 25% read a newspaper and watch the news on television also. what percentage of the people neither watch the news on television nor read a newspaper?

(a) 5%

(b) 10%

(c) 15%

(d) 20%

Sol. Let the total number of people be 100.

Let circle A represents people who watched television and B represents people who read newspaper.

Then,      x + y =65,     y + z = 40,       y = 25

We get,   x = 40,      y = 25,   z = 15

Then, the number of people who neither watched television nor read newspaper

= 100 – (x + y +z )

= 100 – ( 40 + 25 +15 )

= 100 – 80 = 20

Therefore, the required percentage is 20%.

Hence, the correct answer is (d).

Example 5. In a chess tournament each of six players will play with all other players exactly once. how many matches will be played during the tournament?

(a) 12

(b) 15

(c) 30

(d) 36

Sol. The situation in which the matches will be played would be as follows

(1) 1 st player will play matches with other 5 players.

(2) 2nd player will play matches with 4 players other than the 1st player.

(3) 3rd player will play matches with 3 players others than 1st and 2nd players.

(4) 4th player will play matches with 2 players other than  1st, 2nd and 3rd players.

(5) 5th player will play match with 6th player only.

Therefore, the number of matches played during the tournament is

= 5 + 4 +3 +2 +1 =15.

Example 6. Consider the diagram given below :

Five hundred candidates appeared in the examination conducted for the tests in english, Hindi and mathematics. the diagram gives the number of candidates who failed in different tests. what is the percentage of candidates who failed in at least two subjects?

(a) 0.078%

(b) 1.0%

(c) 6.8%

(d) 7.8%

Sol. From the diagram it is clear that number of candidates who failed in at least two subjects = number of candidates who failed in two or more subjects.

= (10 + 12 + 12 + 5 ) =39.

Therefore, the required percentage = {39/500 x100}% = 7.8%

Therefore, the option (d) is the correct answer.

Unit Exercise

1. There are 50 students admitted to a nursery to a nursery class. some students can speak only english and some can speak only hindi. ten students can speak both english and hindi. if the number of students who can speak english is 21, then how many students can speak hindi, only hindi and only english?

(a) 39, 29 and 11, respectively

(b) 37, 27 and 13, respectively

(c) 28, 18 and 22, respectively

(d) 21, 11 and 29 respectively

1. Consider the venn diagram given below :

The numblers in venn diagram indicates the number of persons reading the newspapers. the diagram is drawn after surveying 50 persons. in a population of 10000, how many can be expected to read at least two newspapers?

(a) 5000

(b) 5400

(c) 6000

(d) 6250

1. In a group of persons travelling in a bus, 6 persons can speak tamil, 15 can speak hindi and 6 can speak gujarati. in that group, none can speak two languages and one person can speak all the three languages, then how many persons are there in the group?

(a) 21

(b) 22

(c) 23

(d) 24

Directions (q. Nos. 4-6) read the following information carefully and then answer the questions that follow

A publishing firm publishes newspapers A, B and C. in an effort to persuade advertisers to insert advertisements in these newspapers, the firm sends out the following statements to possible advertisers

A survey of representative sample of the whole population shows that

Newspaper A is read by 26%.

Newspaper B is read by 25%.

Newspaper C is read by 14%

Newspaper A and B are read by 11%

Newspaper B and C are read by 10%

Newspaper C and A are read by 9%

Newspaper C only is read by 0%

1. The percentage of readers who read all the three newspapers is

(a) 1

(b) 4

(c) 5

(d) 6

1. The percentage of readers who read newspapers A and B but not C is

(a) 2

(b) 4

(c) 5

(d) 6

1. The percentage of readers who read at least one of the three newspapers is

(a) 40

(b) 50

(c) 60

(d) 65

Directions (Q. Nos. 7-9) the following diagram shows the number of students who got distinction in three subjects in a total of 500 students. study the diagram carefully and then answer the questions that follow

1. What is the percentage of students who got distinction in two subjects?

(a) 8%

(b) 9%

(c) 10%

(d) 12%

1. What is the percentage of students Who got distinction?

(a) 28%

(b) 18.6%

(c) 38%

(d) 15%

1. The percentage of students with distinction marks in mathematics is

(a) 17.8%

(b) 18. 6%

(c) 19.2%

(d) 20.6%

Directions (Q. Nos. 10-12) these questions are based on the following information for an examination

(a) candidates appeared          10500

(b) passed in all the five subjects     5685

(c) passed in three subjects only      1498

(d) passed in two subjects only        1250

(e) passed in one subject only       835

(f)failed in english only                    78

(g) failed in mathematics only       275

(h) failed in physics only                149

(i) failed in chemistey only           147

(j) failed in biology only                221

1. How many candidates failed in all the subjects?

(a) 4815

(b) 3317

(c) 2867

(d) 362

1. How many candidates passed in at least four subjects?

(a) 6555

(b) 5685

(c) 1705

(d) 870

1. How many candidates failed because of having failed in four or less subjects?

(a) 4815

(b) 4453

(c) 3618

(d) 2368

1. 10 yr ago chandravati’s mother was 4 times older then her daughter. after 10 yr the mother will be twice older then the daughter. the present age of chandravati is

(a) 5 yr

(b) 10 yr

(c) 20 yr

(d) 30 yr

1. The sum of ages of a father and son is 45 yr. 5 yr ago the product of their ages was 4 times the father’s age at that time. the present age of the father and son respectively are

(a) 25 yr 10 yr

(b) 36 yr 9yr

(c) 39 yr 6 yr

( d) None of these

1. jayesh is as much younger to anil as he is older to prashant. if the sum of the ages of anil and prashant be 48 yr , what is the age of jayesh?

(a) 20 yr

(b) 24 yr

(c) 30 yr

(d) can not be determined

1. Rajan’s age is 3 times than that of ashok in 12 yr rajan’s age will be double the age of ashok. rajan’s present age is

(a) 27 yr

(b) 32 yr

(c) 36 yr

(d) 40 yr

1. A number of friends decided to go on a picnic and planned to spend 96 on eatables. four of them, however, did not tum up, as a result, the remaining ones had to contribute 4each extra. the number of those who attended the picnic was

(a) 8

(b) 12

(c) 16

(d) 24

1. A, B, C, D and E play a game of cards. a says to B, if you give me three cards, you will have as many as E has and if I give you three cards, you will have as many as D has. Aand Btogether have 10 cards more then what D and E together have. if B has two cards more than what C has and the total number of cards be 133, how many cards does B have?

(a) 22

(b) 23

(c) 25

(d) 35

1. A worker may claim rs 15 for each km which he travels by taxi and rs 5 for each km which he drives his own car. if in one week he chaimed re 500 for travalling 80 km , how many km did he travel by taxi?

(a) 10

(b) 20

(c) 30

(d) 40

1. A bus starts from city X. the number of women in the bus is half the number of men. in the city Y, 10 men left the bus and five women boarded in to it. Now, number of men and women becomes equal. in the beginning, how many passengers entered the bus?

(a) 15

(b) 30

(c) 36

(d) 45

1. A, B, C, D and E play a game of cards. A says to B, if you give me 3 cards, you will have as many I have at this moment while if d takes 5 cards from you , he will have as many as E has’ , A and c together have twice as many cards as E has B and D together also have the same number of cards as A and C have taken together. if all together they have 150 cards, how many cards has C got?

(a) 28

(b) 29

(c) 31

(d) 35

1. Between two book-ends in your study are displayed your five favourite puzzle books. if you decide to arrange the five books in every possible combination and moved just one book every minute, how long would it take?

(a) 1 h

(b)2 h

(c) 3 h

(d) 4 h

1. At the end of a business conference, ten people present shake hands with each other once. how many handshakes will be there all together?

(a) 20

(b)45

(c) 55

(d) 90

1. I have a few sweets to be distributed. if i keep 2, 3 or 4 in a pack i am left with one sweet. if i keep 5 in a pack, i am left with none. what is the minimum number of sweets i can have to pack and distribute?

(a) 25

(b) 37

(c) 54

(d) 65

1. In a family , a couple has a son and a daughter. the age of the fathe is three times that of his daughter and the age of the son is half of his mother. the wife is 9 yr younger to her husband and the brother is 7 yr older than his sister. what is the age of the mother?

(a) 40 yr

(b) 50 yr

(c) 45 yr

(d) 60 yr

1. (a) circles A and B represent the students who speak english and hindi, respectively.

Now, according to the given information total no. of students

= x + y + z =50      ………………… (i)

also        y = 10 and      …………………….(ii)

x + y =21    …………………………………….(iii)

from eqs. (ii) and (iii) , x =11

No. of students who can speak english only = 11

No. of students who can speak hindi = (y + z)

From eq. (i)  we get

y + z =50 – 11 =39  ………………(iv)

therefore, no of students who can speak hindi only is z.

From eq. (iv), y + z =39 > z =39 – y = 39 – 10 =29

Thus, no. of students speaking hindi = 39, no. of

students speaking hindi only = 29 and no. of

students speaking english only = 11.

1. (b) No. of persons who read at least two newspapers

= no. of persons who read two newspapers and more

= 12 + 2 + 8 + 5 = 27

It means out of 50 persons 27 read at least at two newspapers.

No. of such persons per  10000 = {27/50 x 10000}

= 5400

1. (c) Let circles x, y and z represent persons who can speak tamil, hindi and gujarati, respectively.

persons who speak tamil = A + B + D + E = 6 …………..(i)

persons who speak hindi = B +C +E +F = 15 ……………(ii)

persons who speak gujrati = D + E + F +G =6 ………………….(iii)

persons who speak and two languages = B +D + F = 2 ………… (iv)

persons who speak all the three languege = E = 1 ……………..(v)

substituting the value of E in equations Eqs. (i) , (ii) and (iii) we get

A + B + C =5 ……………….(vi)

B +C +F = 14 …………..(vii)

D + F + G = 5 ……………(viii)

subracting eq (iv) from eq. (vi) we get

A- F = 3 ……………… (ix)

adding eqs. (vii) and (viii) we get

B + C + 2F + D + G = 19 …………………….. (X)

adding eqs. (ix) and (x) we get

A + B + C + D + F + G = 22 …………….(xi)

adding eqs. (v) and (xi) we get

A + B + C + D + E + F + G = 23

therefore, total no. of persons= 23

For Q. No. 4 to 6

Let the no. of persons be 100.

from the information given in the question,we have

P + S + T + U =26 ……………..(i)

S + U + Q + V = 25………………… (ii)

T + U + V + R = 14 ………………… (iii)

S + u = 11………………………(iv)

u + v = 10 ………………….. (v)

T + V = 9 …………………… (vi)

Again,                      R = 0 …………………..(vii)

Form eqs. (iii) , (vi) and (vii), we have

(T + U +V + R = 14

9 + V + 0 = 14

V = 14 – 19 = 5

>               V = 5

From eq. (v) , u + V = 10but V = 5

U = 5

From eq. (iii), again

T + U + V + R = 14

T + 5 + 5 + 0 =14

T = 4

From eq. (iv), we get

S = 11 – 5 = 6  s = 6

From eq. (ii), we get

Q + 6 + 5 + 5 = 25

Q  = 25 – 16 = 9    Q = 9

Therefore, we get

P = 11, Q = 9, R = 0, S = 6, T = 4 U = 5 and V = 5

1. (c) Percentage of readers who read all the three newspapers = u = 5
2. (d) percentage of readers who read newspapers A and B but not C = S = 6.
3. (a) percentage of readers who read at least one of the three newspapers = P + Q + R + S + T + U + V = (11 + 9 + 0 + 6 + 4 + 5 + 5) = 40.
4. (a) No. of students who got distinction in two subjects = (15 + 13 + 12) = 40

Required percentage = [40/500×100] = 8%

1. (c) No. of students who got distinction

=(50 + 47 + 42 + 12 + 11 + 13 + 15) =190

Required percentage = [190/500×100 = 38%]

1. (a) No. of students with distinction marks in mathematics

= (50 + 13 + 11 + 5) = 89

Required percentage =[89/500×100]= 17.8%

1. (d) Candidates failed in all the subjects

=(candidates appeared) – (candidates failed in any subject)

= 10500 – (5685 + 1498 + 1250 + 835 + 78 + 275 + 149 + 147 + 221)

= 10500 – 10138

= 362

1. (a) Candidates passed in at least four subjects

=(candidates passed in four subjects) + (candidates passed in all the five subjects)

= (candidates failed in only one subject) + (candidates passed in all the five subjects)

= (78 + 275 + 149 + 147 +221)+ 5685

= 870 + 5685

=6555

1. (b) Candidates failed in four or less subjects

=(candidates failed in only one subject )

+ (candidates failed in only two subjects)

+ (candidates failed in only three subjects)

+ (candiddates failed in only four subjects)

= (camdidates failed in only one subject)

+ (candidates passed in only three subjects)

+ (candidates passed in only two subjects)

+(candidates passed in only one subjects)

= 2 (78 + 275 + 149 + 147 + 221) + 1498 + 1250 + 835

= 4453

1. (c) Lat chandravati’s age, 10 yr ago, be x yr

Then, mother’s age, 10 yr ago, was 4x yr

After 10 yr, or after 20 yr when chandravati’s age would be 10 yr, we have

(4x + 20) = 2 (x + 20)

4x + 20 = 2x + 40

2x = 20

x =10 yr

present age of chandravati = 20 yr

1. (b) Lat the present age of father be x yr and that of the son be (45 – x ) yr

Five yr ago the age of son was (45 – x – 5)

=(40 – x) and that of father was (x – 5) yr

Now, (x – 5) (40 – x) = 4 (x -5)

or                  x – 41x + 180 =0

or     40 x – x -200 + 5x = 4x -20

or           x – (36 + 5) x + 180 = 0

or           41x – x – 180 = 0

or      x -36x – 5x + 180 = 0

or    (x – 36) (x – 5) = 0

or    x (x – 36) – 5 (x -36) = 0

or    x = 36 and     x = 36                         x = 5

unacceptable

Therefore, the present age of father and son are 36 yr and 9 yr respectively.

1. (b) Let the age of anil and prashant be x and (48 – x) yr, respectively.

Let the age of jayesh be y yr, then

(x – y) = y – (48 – x)

x – y = y – 48 + x

or                   2y = 48, or    y = 24 yr

Age of jayesh is 24 yr.

1. (c) Let the present age of ashok be z yr.

> the age of rajan = 3x yr.

In 12 yr hence, the age of ashok would be (x + 12)

yr and that of rajan be (3x + 12) yr

As given in the question,

(3x + 12) = 2 (x + 12)

or                     3x + 12 = 2x + 24

or                          x = 12 yr

Therefore, the age of rajan = 3 x 12 = 36 yr.

1. (a) Let the number of friends who decided to go on a picnic be x.

Then, the contribution of each member = 96x

In case four of them do not turn up, the contribution of each member = 96/(x – 4)

Now as per question, 96/(x – 4) = 96/x + 4

or      1/(x – 4) -1/x = 4/96     or      x – 4x – 96 = 0

or        (x – 12) (x + 8) = 0       or x =12

Therefore, the number of persons who attended the picnic is (12 – 4 ) = 8.

1. (c) From the question, we heve

B – 3  = E ……………… (i)

B + 3   = D …………….(ii)

MA + B = D + E + 10 ……………(iii)

B = C + 2 ……………………………(iv)

A + B +C + D + E = 133…………………….(v)

From eqs. (i) and(ii),

B – E = D – B > E + D = 2B ………………(vi)

From eqs. (iii) and (vi)

A + B = 2B + 10 > B = A – 10

Or           A = B + 10 ……………………(vii)

From (v), (vi) , (iv) and (vii) , we have

(B + 10 ) (B) + (B – 2) + 2B = 133

5B = 133 – 8 = 125

Or                   B = 25

1. (a) Let the distance covered by taxi be x km

Then, distance covered by car = (80 – x) km

15x + 5 (80 – x) = 500

or  15x + 400 – 5x = 500

or            10x = 100 > x = 10

Therefore, the distance covered by taxi is 10 km

1. (d) In the city X let the number of men be x

Then the number of women would be x/2

In the city y, x – 10 = x/2 + 5

or x/2 = 15        or x = 30   and  y = 15

Therefore, total number of passenger in the beginning was 30 + 15 = 45

1. (a) From the question, it is clear that

B – 3 = A …………….(i)

D + 5 = E ……………(ii)

A + C = 2E ……………..(iii)

A + C = B = D …………..(iv)

A + B + C + D + E = 150 ……..(v)

From eqs. (iii) , (iv) and (v) ,we get

5E = 150           E = 30

Now from eq. (ii), D = 30 – 5 = 25

And from eq. (iv),    A = 32

Now from eq. (iii), 32 + C = 2 x 30

C = 60 – 32 = 28

1. (b) No. of ways in which 5 books can be arranged = 5

= 5 x 4 x 3 x 2 x 1 = 120

so, total time taken = 120 min = 2 h

1. (b) Total number of handshakes

= (9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1) = 45

1. (a) The required number will be obtained by taking LCM of 2, 3 and 4 and adding 1 into it. the number so obtained is 25.
2. (d) Let the age of the mother be x yr

Age of the son = x/2 yr

and the age of daugher = [x/2 – 7]yr

and the age of the father = 3 [x/2 – 7]yr

3 [x/2 – 7 ] – x = 9

>                3x/2 – x – 21 = 9

>              x/2 = 9 + 21 = 30

x = 30 x 2 = 60yr