Directions (1 – 5): Study
the following bar graph and answer the questions that follow.
Votes are polled in 10 constituencies
of Haryana. After the polls, some votes were declared invalid.
The bar chart shows the % of votes
received by the winning and the losing candidates (Suppose there are only 2
candidates) out of the “valid votes”, which is “total votes” polled minus the
“invalid votes”. The one which got the highest number of votes from the “valid
votes” was declared the winner.
The total number of invalid votes in
each constituency – Karnal, Faridabad, Gurgaon, Hisar, Sirsa, Bhiwani, Ambala,
Rohtak, Panipat and Kurukshetra are 3800, 2000, 11400, 0, 2700, 150, 4200, 360,
320, 6800 respectively.
1. What is the total number of voters in the given 10
constituencies of Haryana?
a) 217000
b) 234000
c) 276000
d) 211000
e) cannot be determined
2. What is the difference between the number of votes received
by the winning and the losing candidate from Sirsa?
a) 5400
b) 5700
c) 6600
d) 4500
e) 5900
3. What is the total number of valid votes received by winning
candidates from Panipat, Ambala and Karnal together?
a) 26540
b) 21760
c) 23780
d) 26340
e) Cannot be determined
4. In the last elections, there was a total of 3,00,000 voters
in these 10 constituencies. If this time, there is an increase by 10%, then
what is eth total number of voters from Hisar?
a) 1,28,000
b) 1,23,000
c) 1,32,000
d) 1,34,000
e) 1,25,000
5. What is the total number of valid votes from Bhiwani and
Kurukshetra together?
a) 32,030
b) 38,400
c) 23,760
d) 45,550
e) 48,050
Direction (6 – 10): The table given below shows the scorecard
of India during a test match. Some values are missing. Find the answers based
on information in table and respective questions.
Player

Runs

Balls Faced

4’s

6’s

V Kohli

148

104

13

–

S Raina

40

55

3

–

R Ashwin

37

42

1

1

C Pujara

–

21

2

0

R Jadeja

13

5

0

1

MS Dhoni

81

–

11

3

Total

–

–

–

10

Further information is:
(i) Total runs scored by V Kohli from
those scored by 1’s and 2’s are in the ratio of 1 : 5
(ii) Number of white balls faced by C
Pujara is 6.
(iii) A white ball is defined as the
ball on which no run is scored.
(iv) During the entire match only
1’s, 2’s, 4’s and 6’s were taken by the batsmen.
(v) V Kohli hits the maximum number
of 6’s among all the batsmen.
(vi) Beside C Pujara every player has
hitted at least a six.
(vii) Every player has taken at least
a 1 and a 2.
6. What is the total number of white balls face by V Kohli?
a) 42
b) 53
c) 49
d) 45
e) 38
7. What is the minimum number of balls faced by MS Dhoni?
a) 24
b) 32
c) 29
d) 36
e) 20
8. If the number of balls faced by C Pujara to take 1’s is
greater than that for 2’s, then he can score a maximum of how many runs?
a) 24
b) 27
c) 32
d) 21
e) 34
9. Assume if only the given players score during the match for
the team, then what is the minimum score of the team?
a) 333
b) 309
c) 403
d) 358
e) 341
10. What was the maximum possible new run rate of the team?
a) 9.32
b) 8.48
c) 7.56
d) 4.45
e) 12.23
Solutions:
1. E) In Karnal,
valid votes are 45+36 = 81%, so invalid votes are 10081 = 19%
Same as with other constituencies. So
Total number of votes polled in
Karnal = (3800/19) * 100 = 20,000
Total number of votes polled in
Faridabad = (2000/10) * 100 = 20,000
Total number of votes polled in Gurgaon
= (11400/38) * 100 = 30,000
Total number of votes polled in Hisar
cannot be found because there is no invalid vote here.
Total number of votes polled in Sirsa
= (2700/9) * 100 = 30,000
Total number of votes polled in
Bhiwani = (150/1) * 100 = 15,000
Total number of votes polled in
Ambala = (4200/14) * 100 = 30,000
Total number of votes polled in
Rohtak = (360/2) * 100 = 18,000
Total number of votes polled in
Panipat = (320/8) * 100 = 4,000
Total number of votes polled in
Kurukshetra = (6800/17) * 100 = 40,000
2. B) Total number of votes from Sirsa =
30,000
Difference between votes = 30000/100
* (5536) = 5700
3. D) Total number of votes of winners from
Panipat, Ambala and Karnal =
4000*81/100 + 20000*45/100 +
30000*47/100 = 3240 +9000+ 14100 = 26340
4. B) Total number of votes this time =
110/100 * 300000 = 3,30,000
Total number of voters from other
constituencies this time = 2,07,000
So total number of votes from Hisar =
3,30,000 – 2,07,000 = 1,23,000
5. E) Total number of valid votes from
Bhiwani = 15000 – 150 = 14,850
Total number of valid from
Kurukshetra = 40,000 – 6800 = 33,200
So total number of valid votes =
14850+33200 = 48,050
6. D) Runs scored by 4’s = 13*4 = 52
Total 6’s = 10, Each player except
Pujara hit at least a 6. And Kohli hit the maximum 6’s. So 6’s by Kohli = 4. So
runs scored by 6’s = 4*6 = 24
Total runs scored by 4’s and 6’s =
52+24 = 76
For these runs balls played are 13+4
= 17 …….(1)
So runs by 1’2 and 2’s = 14876 = 72
1’s and 2’s runs ratio = 1 : 5
So 12 runs by 1’2 and 60(2*30) by 2’s
So balls for 1’s and 2’s = 12+30 = 42
……..(2)
So total balls = 17+42 = 59
So number of white balls = 104 – 59 =
45
7. A) Runs scored by 4’s = 11*4 = 44 ….(1)
by 6’s = 3*6 = 18 ……(2)
Total runs by 4’s and 6’s = 44+18 =
62
remaining runs = 8162 = 19
By taking 1’s and 2’s he has scored
these 19 runs. To minimize the number of balls Dhoni has to score more runs
taking 2’s. So he can score 18 runs by taking 2’s and 1 by taking a 1.
So balls for 1’s and 2’s = 9+1 = 10
…….(3)
So total balls (minimum) = 11+3+10 =
24
8. B) Runs by 4’s = 2*4 = 8 ….(1)
Number of white balls = 6
So remaining balls = 21 – (2+6) = 13
So, to maximize his score in 13
balls, he can take 7 one’s and 6 two’s. ……(2)
So a maximum of 8 + 7*1 + 6*2 = 27
runs
9. E) In order to find the minimum score of
team, we have to find the minimum runs scored by Pujara.
he takes 2 four’s so 2*4 = 8 runs
So now 212 = 19 balls left. Now
given that he faced 6 white balls, So now balls left = 196 = 13 balls. Now for
minimum runs, he should score more 1’s than 2’s
Since at least one 1’s and one 2’s is
necessary so minimum runs on 13 balls is 1*12 + 2*1
So total min runs by Pujara = 2*4 + 1*12 + 2*1 = 22
So minimum score of team =
148+40+37+22+13+81 = 341
10. B) First find the maximum runs scored
and minimum balls faced.
Minimum number of balls faced by
Dhoni = 24
Maximum runs scored by Pujara = 2*12
+ 1*1 + 4*2 = 33
Maximum runs scored by team =
148+40+37+33+13+81 = 352
Minimum number of balls faced =
104+55+42+21+5+24 = 251 balls or 251/6 = 41.5 overs
So maximum run rate = 352/41.5 = 8.48