# Mixed Quantitative Aptitude Questions Set 53

1. The number of students in 3 classes are in the ratio 4:5:6. If 15 students are increased in each class this ratio changes to 11:13:15. The total number of students in the three classes in the beginning was
A) 165
B) 150
C) 175
D) 180
E) None
Option B
Solution:

Let the number of students in the classes be 4x, 5x and 6x respectively;
Total students = 4x+5x+6x = 15x.
Given ,
(4x+15)/(5x+15) = 11/13
3x=30==>x=10.
Then Total no of students is 15x=15*10=150.
2. A, B and C have 40, x and y balls with them respectively. If B gives 20 balls to A, he is left with half as many balls as C. If together they had 60 more balls, each of them would have had 100 balls on an average. What is the ratio of x to y?
A) 4 : 3
B) 3 : 2
C) 2 : 3
D) 2 : 5
E) None
Option C
Solution:

Given,
40+x+y+60/3=100
X+y=200—1
x-20=y/2
2x-y=40—2
Solving 1 and 2
We get x=80, y=120.
Ratio of x:y=2:3
3. The incomes of A, B, C are in the ratio of 12 : 9 : 7 and their spending are in the ratio 15 : 9 : 8. If A saves 25% of his income. What is the ratio of the savings of A, B and C respectively?
A) 12:15:19
B) 11:15:18
C) 15: 18:11
D) 21:24:29
E) None
Option C
Solution:

Let the income be 12x, 9x, 7x and expenditure is 15y, 9y, 8y.
I-E=S
A 12x-15y=25%of 12x=3x
9x=15y==>y=3x/5
B Saving = 9x-9y
C Saving = 7x-8y
Substitute y value
Savings Ratio A:B:C
3x: 9x-9*3x/5 : 7x-8*3x/5
3x:18x/5:11x/5==>15:18:11Â­Â­Â­
4. A Student obtained equal marks in Maths and Science. The ratio of marks in Science and Social is 2:3 and the ratio of marks in Maths and English is 1 : 2. If he has scored an aggregate of 55% marks. The maximum marks in each subject is same. In how many subjects did he score greater than 50% marks?
A) 1
B) 2
C) 3
D) 4
E) None
Option B
Solution:
M:S=1:1, S:So= 2:3, M:E=1:2
Then M: S: So: E=2:2:3:4
Now 2x+2x+3x+4x/4=11x/4=55%
X=20.
So Marks, M=40, S=40, So=60, E=80.
Above 50 mark is in 2 subjects.
5. In a class, the number of girls is 30% more than that of the boys. The strength of the class is 92. If 8 more girls are admitted to the class, the ratio of the number of boys to that of the girls is
A) 4:5
B) 3:2
C) 2:3
D) 4:7
E) None
Option C
Solution:

G:B=130:100=13:10
Then (10+13)23 92
G 13 52
B 10 40
If 8 girls admitted then total girls is 52+8=60
Now ratio of B:G=40:60=2:3.
6. Rs 3440 is divided, among A, B, C and D such that Bâ€™s share is 6/11th of Aâ€™s; Câ€™s share is 1/4th of Bâ€™s and D has 2/5th as much as B and C together. Find Aâ€™s share
A) 1760
B) 1540
C) 1320
D) 1850
E) None
Option A
Solution:

Let Aâ€™s share be 1
Then Bâ€™s share is 6/11*1=6/11
Câ€™s share is 6/11*1/4=3/22
Dâ€™s share is 2/5*(6/11+3/22)=3/11
A:B:C:D=1:6/11:3/22:3/11=22:12:3:6
Total 43(22+12+6+3) 3440
Aâ€™s share 22 ?==>1760
7. When 20 is added to the numerator and denominator, then the new ratio of numerator to denominator becomes 7:8. What is the original ratio?
A) 3:4
B) 4:5
C) 4:3
D) Canâ€™t be determined
E) None
Option D
Solution:

Let the fraction be x/y.
Then (x+20)/(y+20) =7/8
We have two variable and only one equation so we canâ€™t find the solution.
8. The value of the diamond is in proportion to the square of its weight A diamond was broken into 3 parts in the ratio of 3: 4: 5, thus a loss of Rs.9.4 lakh is incurred. What is the actual value of diamond (in lakhs)?
A) 12
B) 13.5
C) 11
D) 14.4
E) None
Option D
Solution:

Ratio of broken parts is 3x : 4x: 5x
Value of broken parts of diamond is (3x)2Â + (4x) 2+ (5x) 2Â = 50x2
The value of original diamond (3x+4x+5x)2Â = 144x2
Then loss in value =144x2– 50x2=9.4 lakh
x2=10000.
The actual value of the diamond is 144 x2=14.4lakh
9. The ratio of the monthly salaries of P and Q is in the ratio 10 : 13 and that of Q and R is in the ratio 13 : 14. Find the monthly income (in Rupees) of R if the total of their monthly salary is Rs 1,85,000.
A) 70,000
B) 81,000
C) 55,000
D) 60,000
E) None
Option A
Solution:

P/Q = 10/13 and Q/R = 13/14
So P : Q : R = 10: 13: 14
Total (10+13+14) is 37 == 1,85,000
So Râ€™s salary 14 ? ==>70,000.
10. Two candles of the same height are lighted at the same time. The first is consumed in 8 hrs and second in 4 hrs. Assuming that each candle burns at a constant rate. In how many hour after being lighted, the rate between the first and second candles become 3 : 1?
A) 2hrs 45min
B) 3hrs 12min
C) 3hrs 20min
D) 2hrs 25min
E) None
Option B
Solution:

After x times ratio become 3:1.
Then (1-x/8)/(1-x/4)=3/1
8-x/2(4-x)=3/1
X=16/5hrs ie 3hrs 12min.

## 9 Thoughts to “Mixed Quantitative Aptitude Questions Set 53”

1. Barish!!

thanku mam ðŸ™‚

2. *********

10th q….. by using short method??

1. Always smile

8—4
16
16-2t/16-4t=3/1
t=3 hr 12 mint

explain 10..:)

1. Premashanthi

In ques it is given that After x hrs candle lightened the ratio become 3:1.
Before that the 1st candle consumed 8hrs and second 4hrs.
So we equate that to find x.
Then (1-x/8)/(1-x/4)=3/1
8-x/2(4-x)=3/1
X=16/5hrs ie 3hrs 12min.

shukriya:)

4. karthik

2nd question is confusing….

1. Premashanthi

In question it is given if they had 60 more balls, each has average 100. So 40+x+y+60/3=100 from this we got 1st eqn.

If B gives 20 balls to A, he is left with half as many balls as C. So x-20=y/2 from this we got 2nd eqn.
Solving both we get the ans.

5. Drishant

Sir mam please Ap daily update kiya Kijiye it’s a lovely site for preparation