Quantitative Aptitude DI
Sep 09 2020
Here we are providing new series of Quantitative Aptitude Questions for upcoming exams, so the aspirants can practice it on a daily basis.
Study the information carefully to answer the following questions.
In the following table there are five colleges in which total student and percentage of arts students and the ratio of civil and mechanical engineering students are given. Calculate the missing data if necessary:
1) If the ratio of boys and girls in college A for civil engineering students are 4:1 and the civil engineering students are 50% more than the mechanical engineering students. Then find the difference of boys and girls in civil department?
a) 120
b) 430
c) 351
d) 320
e) 270
2) If the total engineering student in college E is 525 and students in civil department are 12 ½% less than the students in mechanical department and the engineering student in college D is 735. Then find ratio of civil engineering student in college D and E?
a) 445:247
b) 441:243
c) 453:247
d) 441:245
e) 441:249
3) If arts student in college A is 375 less than arts student in college B. Then the total student in college B is what percent more or less than the total students in college D?
a) 5 1/7%
b) 8 1/7%
c) 2 1/7%
d) 7 1/7%
e) 1 1/7%
4) If total student in college C is 1380 and total arts student in college C is equal to the total students in engineering. And the ratio of boys and girls in college C in arts is 5:1. If 20% of boys are transferred to college E, then find the total students in college E?
a) 1546
b) 1456
c) 1585
d) 1865
e) 1687
5) Suppose there is another college X in which arts students are 2/5th of arts student in college A and engineering student in college X is 40% of total students of college D then what is the total students in X?
a) 1050
b) 1205
c) 1640
d) 1550
e) 4520
Answers :
1) Answer: C
Total number of engineering student in college A= (100-35) % of 1500
= (65/100)*1500=975
Given that the civil engineering students are 50% more than the mechanical engineering students.
Let x be the number of mechanical engineering students in college A.
Then number of civil engineering students in college A=x+50% of x= 150% of x
So, x+(150/100)x=975
(250/100)x=975
x=390
So, the no of mechanical engineering students= 390
The number of civil engineering students=585
Given that the ratio of boys and girls in college A for civil engineering students=4:1
So, the difference of boys and girls in mechanical students= (3/5)*585=351
2) Answer: D
Given that, the total engineering student in college E = 525
Let the number of mechanical engineering students in college E be X.
Then the number of civil engineering student in college E= X- 12 ½% of X= 87 ½% of X
Therefore, X + 87 ½% of X= 525
187 ½% of X=525
X=280
So, the number of mechanical engineering students in college E=280
And the number of civil engineering students in college E=525-280=245
The number of engineering student in college D = 735
The number of civil engineering in college D= (3/5)*735=441
Required ratio=441:245
3) Answer: D
Number of arts student in college A=(35/100)*1500=525
Number of arts student in college B = 525+375= 900
Let the total number of students in college B be X.
We know that, there are 40% of students in college B are arts.
So, (40/100)*X=900
X=2250
(i.e) Total number of students in college B=2250
Required percent= [(2250-2100)/2100]*100=7 1/7%
4) Answer: D
Given that, total student in college C = 1380 and total arts student in college C = total engineering students in college C.
So, total arts student + total engineering student =1380
Total arts student + total arts student=1380
Therefore, total arts student=690
Number of arts boys in college C= (5/6)*690=575
If 20% of boys are transferred to college E, then the total students in college E=1750 + (20/100)*575=1865
5) Answer: A
Total students in X= [(2/5)*(35/100)*1500] + [(40/100)*2100]
=210+840
=1050