# Ch.17 :- Arithmetical Reasoning : Exercise 1

1. There are 50 students admitted 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

Ans.(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 (ii) and (iii), x = 11
No. of students who can speak English only = 11
No. of students who can speak Hindi = (y + z)

From (i), we get
y + z = 50 – 11 = 39 … (iv)

Therefore, no. of students who can speak Hindi only is z.
From (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.

2. Consider the Venn diagram given below:

The numbers in Venn diagram indicates the number of persons reading the newspapers. The diagram is drawn after surveying 50 persons. In a population of 10,000, how many can be expected to read at least two newspapers?
(a) 5,000
(b) 5,400
(c) 6,000
(d) 6,250

Ans.(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 two newspapers.

No. of such persons per 10000
= (2750 x 10000)
=5400

3. 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 any other language. If 2 persons in the group 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

Ans.(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 any two languages = B + D + F = 2 …. (iv)
Persons who speak all the three language = E = 1 …. (v)

Substituting the value of E in equations (i), (ii) and (iii),we get
A + B + D = 5….. (vi)
B + C + F = 14….. (vii)
D + F + G = 5 …..(viii)

Subtracting (iv) from (vi), we get
A – F = 3 ….(ix)

Adding (vii) and (viii), we get
B + C + 2F + D + G = 19 ….. (x)

Adding (ix) and (x), we get
A + B + C + D + F + G = 22 ….. (xi)

Adding (v) and (xi), we get
A + B + C + D + E + F + G = 23

Therefore, total no. of persons = 23

4. Out of a total of 120 musicians in a club, 5% can play all the three instruments-guitar, violin and flute. It so happens that the number of musicians who can play any two and only two of the above instruments is 30. The number of musicians who can play the guitar alone is 40. What is the total number of those who can play violin alone or flute alone?
(a) 30
(b) 38
(c) 44
(d) 45

Ans.(c)
Let the circles x, y, z represent the musicians, who can play guitar, violin and flute, respectively.

P + Q + R + S + T + U + V = 120 … (i)

No. of musicians who can play all the three instruments
= S = 5% of 120 = 6
S = 6 … ….(ii)

No. of musicians who can play guitar only
= P = 40 … (iii)

No. of musicians who can play any two and only two of the instruments = Q + T + U = 30 … (iv)

Now, from Eqs. (i), (ii), (iii) and (iv), we get
P + Q + R + S + T + U + V = 120
P + (Q + T + U) + S + R + V = 120
40 t 30 + 6 + R + V = 120
R + V = 120 – 76
= 44

Total no. of musicians who can play violin alone or flute alone
= (R + V)
v = 44.

Hence, option (c) is the correct answer.

✦Directions for Q. No.5 to 7 : Head the following information carefully and then answer the question 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%.

5. The percentage of readers who read ail the three newspapers is
(a) 1
(b) 4
(c) 5
(d) 6

Ans.(c)
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 + U = 9 ….(vi)
Again, R = 0 …… (vii)

From (iii), (vi) and (vii), we have
(T + U) + V + R = 14
9 + V + 0 = 14        ∵ V = 14- 9 = 5
=> V = 5

From (v), U + V = 10 but V = 5       ∵ U = 5

From (iii), again
T + U + V + R = 14
T + 5 + 5 + 0 = 14           ∵ T = 4

From (iv), we get
S = 11 – 5 = 6                 ∵ S = 6

From (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

newspapers = U = 5

6. The percentage of readers who read Newspapers A and B but not C, is
(a) 2
(b) 4
(c) 5
(d) 6

Ans.(d)
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 + U = 9 ….(vi)
Again, R = 0 …… (vii)

From (iii), (vi) and (vii), we have
(T + U) + V + R = 14
9 + V + 0 = 14                   ∵ V = 14- 9 = 5
=> V = 5

From (v), U + V = 10 but V = 5        ∵ U = 5

From (iii), again
T + U + V + R = 14
T + 5 + 5 + 0 = 14              ∵ T = 4

From (iv), we get
S = 11 – 5 = 6              ∵ S = 6

From (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

Percentage of readers who read newspapers A and B but not C = S = 6

7. The percentage of readers who read at least one of the three newspapers is
(a) 40
(b) 50
(c) 60
(d) 65

Ans.(a)
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 + U = 9 ….(vi)
Again, R = 0 …… (vii)

From (iii), (vi) and (vii), we have
(T + U) + V + R = 14
9 + V + 0 = 14              ∵ V = 14- 9 = 5
=> V = 5

From (v), U + V = 10 but V = 5        ∵ U = 5

From (iii), again
T + U + V + R = 14
T + 5 + 5 + 0 = 14               ∵ T = 4

From (iv), we get
S = 11 – 5 = 6                ∵ S = 6

From (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

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

✦Directions for Q. No.8 to 10: 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:

8. What is the percentage of students who got distinction in two subjects?
(a) 8%
(b) 9%
(c) 10%
(d) 12%

Ans.(a)
No. of students who got distinction in two subjects =
(15 + 13 + 12) = 40
∴ Required percentage = (40500 x 100) = 8%

9. What is the percentage of students who got distinction?
(a) 28%
(b) 18.6%
(c) 38%
(d) 15%

Ans.(c)
No. of students who got distinction
= (50 + 47 + 42 + 12 + 11 + 13 + 15)
= 190
∴ Required percentage = (190500 x 100 = 38%)

10. The percentage of students with distinction marks in mathematics is :
(a) 17.8%
(b) 18.6%
(c) 19.2%
(d) 20.6%

Ans.(a)
No. of students with distinction marks in Mathematics
= (50 + 13 + 11 + 15)
= 89

∴ Required percentage = (89500 x 100) = 17.8%

✦Directions for Q. No. 11 to 13 : 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 Chemistry only 147
(J) Failed in Biology only 221

11. How many candidates failed in ail the subjects?
(a) 4815
(b) 3317
(c) 2867
(d) 362

Ans.(d)
Candidates failed in all the subjects
= (Candidates appeared) – (Candidates passed in one, two, three or five subjects + Candidates failed in any subject)

= 10500 – (5685 + 1498 + 1250 + 835 + 78 + 275 + 149 + 147 + 221)
= 10500 – 10138
= 362.

12. How many candidates passed in at least four subjects?
(a) 6555
(b) 5685
(c) 1705
(d) 870

Ans.(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 + 14 + 221) + 5685
= 870 + 5685
= 6555

13. How many candidates failed because of having failed in four or less subjects?
(a) 4815
(b) 4453
(c) 3618
(d) 2368

Ans.(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) + (Candidates failed in only four subjects)

= (Candidates failed in only one subject) + (Candidates passed in only three subjects) + (Candidates passed in only two subjects) + (Candidates passed in only one subject)

= 2 (78 + 275 + 149 + 147 +221) + 1498 + 1250 + 835
= 4453

✦Directions for Q. No. 14 to 18 : These questions are based on the information given below:
Data of 450 candidates, who took part in an examination in social sciences, Mathematics and Science is given as below:
(A) Passed in all the subjects 167
(B) Failed in all the subjects 60
(C) Failed in Social sciences 175
(D) Failed in Mathematics 199
(E) Failed in Science 191
(F) Passed in Social sciences only 62
(G) Passed in Mathematics only 481
(H) Passed in Science only 52

14 How many candidates failed in Social science only?
(a) 15
(b) 21
(c) 30
(d) 42

Ans.(a)
Candidates failed in Social science only
= (Candidates failed in Social science) – (Candidates failed in all subjects + Candidates passed in Science only + Candidates passed in Mathematics only)

= 175 – (60 + 52 + 48)
= 175 -160
= 15

15. How many candidates failed in one subject only?
(a) 152
(b) 144
(c) 61
(d) 56

Ans.(c)
Candidates failed in one subject only
= (Total no. of candidates) – (candidates passed in all the subjects + candidates failed in all the subjects + candidates passed in one subject only)

= 450 – (167 + 60 + 62 + 48 + 52)
= 450-389
= 61

16. How many candidates passed in Mathematics and at least in one more subject?
(a) 210
(b) 203
(c) 170
(d) 94

Ans.(b)
Candidates failed in Science only
= 191 – (62 + 60 + 48)
= 21

Candidates failed in Social science only = 15
Therefore, candidates passed in Mathematics and at least one more subject
= (21 + 15 + 167)
= 203

17. How many candidates failed in two subjects only?
(a) 56
(b) 61
(c) 152
(d) 162

Ans.(d)
Candidates failed in two subjects only
= candidates passed in one subject only
= 62 + 48 + 52
= 162

18. How many candidates passed in at least one subject?
(a) 450
(b) 390
(c) 304
(d) 167

Ans.(b)
Candidates passed in at least one subject
= (candidates passed in only one subject) + (candidates passed in only two subjects) + (candidates passed in all the subjects)

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

= 162 + 61 + 167
= 390

19. Five years ago Vinay’s age was one-third of the age of Vikas and now Vinay’s age is 17 years. What is the present age of Vikas ?
(a) 9 years
(c) 41 years
(b) 36 years
(d) 51 years

Ans.(c)
Vinay’s present age is 17 years.
Five years ago, Vinay’s age was 12 years and that of Vikas was 36 years.
Present age of Vikas
= 36 + 5
= 41 years.

20. Pushpa is twice as old as Rita was two years ago. If difference between their ages be 2 years, how old is Pushpa today?
(a) 6 years
(b) 8 years
(c) 10 years
(d) 12 years

Ans.(b)
Let the present age of Pushpa be x years.
Then, the present age of Rita would be (x- 2) years.

Now, as given is the question, x = 2(x – 2 – 2 )
or                                                   x = 2x-8
or                                                   x = 8 years

Therefore, present age of Pushpa is 8 years.

21. 10 years ago, Chandravati’s mother was 4 times older than her daughter. After 10 years, the mother will be twice older than the daughter. The present age of Chandravati is :
(a) 5 years
(c) 20 years
(b) 10 years
(d) 30 years

Ans.(c)
Let Chandravati’s age, 10 years ago, be x years.
Then, mother’s age, 10 years ago, was 4x years.
After 10 years, or after 20 years when Chandravati’s age would be 10 years,
we have
(4x + 20) = 2 (x + 20)
4x + 20 = 2x + 40
2x = 20                  or    x = 10 years

Present age of Chandravati = 20 years

22. The sum of ages of a father and son is 45 years. Five years 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 years, 10 years
(b) 36 years, 9 years
(c) 39 years, 6 years
(d) none of these