 # Quantitative Aptitude: Data Interpretation Set 17 (Bar Graph)

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
Option E
Solution:

In Karnal, valid votes are 45+36 = 81%, so invalid votes are 100-81 = 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. 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
Option B
Solution:

Total number of votes from Sirsa = 30,000
Difference between votes = 30000/100 * (55-36) = 5700
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
Option D
Solution:

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. 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%, thhen 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
Option B
Solution:

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. 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
Option E
Solution:

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

Directions (6 – 10): Study the following bar graph and answer the questions that follow.
The bar graph shows the percentage of students passing in various standards in a school. Maximum marks for each subject of every class is same. 1. If the number of boys and girls passing in class 5 are same, then what is the ratio between the number of boys and girls in class 5?
A) 4 : 3
B) 5 : 3
C) 4 : 9
D) 3 : 4
E) 3 : 8
Option D
Solution:

In class 5, let there are x boys and y girls
So number of boys passing = 0.8x, and girls = 0.6y
and 0.8x = 0.6y
S0 x/y = 3/4
2. In class 7, 44% of the total students got passed. If total number of boys in class 7 is 200, then what is the total number of girls in class 7?
A) 50
B) 52
C) 59
D) 66
E) 60
Option A
Solution:

Let total number of girls in class 7 = x
total number of boys passing = 200 * 0.4 = 80
Total number of girls passing = x*0.6 = 0.6x
So passed percentage of students = (80+0.6x)/(200+x)
So 44/100 = (80+0.6x)/(200+x)
Solve, x = 50
3. If the total number of boys and girls in each class is 150 and 120 respectively, then what is overall pass percentage of the school?
A) 64%
B) 69%
C) 72%
D) 81%
E) 54%
Option B
Solution:

Ther are 6 classes in all, total number of students = 6*(150+120) = 1620
Boys passing = 150 * (0.8+0.8+0.4+0.9+0.7+0.7) = 645
Girls passing = 120 * (0.6+0.7+0.6+0.6+0.8+0.6) = 468
So overall pass% = (645+468)/1620 * 100 = 69%
4. If the ratio between the number of boys and the number of girls in class 9 is 4 : 1, then what is the ratio between number of boys passed and number of girls passed in class 9?
A) 9 : 4
B) 9 : 5
C) 4 : 1
D) 6 : 7
E) 7 : 2
Option E
Solution:

Let there are 400 boys and 100 girls in class 9,
Number of boys passing = 70/100 * 400 = 280
Girls = 80/100 * 100 = 80
So ratio = 280 : 80 = 7 : 2
5. Assuming the data of question 8, if the overall average marks of boys is 50% and that of girls is 60%, then what is the average marks of the students in the school?
A) 42%
B) 54%
C) 58%
D) 65%
E) 63%
Option B
Solution:

Required average = (50/100 * 150 + 60/100 * 120)/[150+120] * 100= 14700/270 = 54%

## 27 Thoughts to “Quantitative Aptitude: Data Interpretation Set 17 (Bar Graph)”

1. Dipankar

mam 10q explan plz

1. Shubhra

see the explanation now

100 gets cancelled so did not use before,
now u will get it

2. ǟʍɮɨӄǟ

i skip que no 1 :p
itna long calculation 🙁

3. Oliver Queen

question 10 ka explanation kuch alg hi hai que se..

1. ǟʍɮɨӄǟ

how ??? i think it is correct

2. Shubhra

its correct
I dint use 100 for percent – cz it gets cancelled
see the explanation now

1. Oliver Queen

ye 100 ki baat nahi hai ma’am
que ne boy ka 50% avg bola hai per ye nahi clear kia ki total boy ki strength ka 50% same as in girl

1. Shubhra

Question 8 me given hai ki in each class 150 boys n 120 girls.
So ek class ka lo ya sari classes ka. Answer same hi aega.
Sari classes ka lenge to 6 se multiply hga in numerator n denominator and that 6 will be cancelled.

Overall Average of boys is 50%.
Means total jitne b boys hai

3. Ankit Saxena

bhai yaarr allegation se karloo would be easy

1. dream girl ;;;;/@ :)

allegation se kaise hoga?

4. Ankit Saxena

thank you mam

1. dream girl ;;;;/@ :)

bro q.no. 4 btao

5. !!angry bird !!

ty 🙂

6. sTrike Like a Puma

In Q5. Total no. of valid votes asked, why the options are in %?

1. Shubhra

1. karthik

in the bar graph of 1st one, how to measure % of the cities easily?… because its too difficult..plz help me

7. sTrike Like a Puma

Q7. B—————G
——40————–60
————–44———–
—–16——————-4
—-4————————1
4——>200
1——->50

8. jaga

thank u…

9. dream girl ;;;;/@ :)

1. Shubhra

From 1st q add the voters from constituencies
Hisar k to 1st q me find nhi ho sakte
That iis 207000

1. dream girl ;;;;/@ :)

bt mam 10% se increase kar rahe voters toh hum purana kyu lenge
different constituencies k bhi increase honge na

2. MiMi

mam 3 no. qstn pls explain kar do……..
What is the total number of valid votes received by winning candidates from Panipat, Ambala and Karnal together?
aisa kyun nhi hua mam?

3. karthik

mam, in the 7th question:
Let total number of girls in class 7 = x
total number of boys passing = 200 * 0.4 = 80
Total number of girls passing = x*0.6 = 0.6x

=>So passed percentage of students = (80+0.6x)/(200+x) here in this step
//” (80+.6x)/(200+x) * 100″// but without taking 100, % is possible if i am wrong plz correct me…
So 44/100 = (80+0.6x)/(200+x)

10. ronnie (honey)

ty mam

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