Online Quiz in Ratio and Proportions For Quantitative Aptitude for Competitive exams like IBPS BANK PO/Clerical,SBI,RRB,SSC,LIC,UPSC-CSAT,SCRA.MAT,CMAT,MBA,SNAP,CAT,NTSE,CLAT
Ratio and Proportions-Test 2
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Subject :- Quantitative Aptitude
Chapter :- Ratio and Proportion- Test 2
Questions :- 25
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Two alloys contain platinum and gold in the ratio of 1:2 and 1:3 respectively. A third
alloy C is formed by mixing alloys one and alloy two in the ratio of 3:4. Find the percentage of gold in the mixture
Correct
Platinum = 1/3 and 1/4
gold = 2/3 and 3/4
Alloy one and two are mixed in the ratio of 3:4, so ratio of platinum and gold in final ratio – 2:5
So gold % = (5/7)*100=71.3/7%
Incorrect
Platinum = 1/3 and 1/4
gold = 2/3 and 3/4
Alloy one and two are mixed in the ratio of 3:4, so ratio of platinum and gold in final ratio – 2:5
So gold % = (5/7)*100=71.3/7%
Question 2 of 25
2. Question
1 points
The sum of three numbers is 980. If the ratio between first and second number is 3:4 and
that of second and third is 3:7. Find the difference between first and last number.
Correct
ratio between three numbers – 9:12:28
49x = 980, x = 20 difference between number = 19*20 = 380
Incorrect
ratio between three numbers – 9:12:28
49x = 980, x = 20 difference between number = 19*20 = 380
Question 3 of 25
3. Question
1 points
The ratio between number of girls and boys in a school is 5: 6. If 40 percent of the boys
and 20 percent of the girls are scholarship holders, what percentage of the students does
not get scholarship?
Correct
Girls = 5x and boys = 6x
Girls that don’t get scholarship = 5x * 80/100 = 4x and boys that don’t get scholarship = 6x *
60/100 = 3.6x
Percent students that didn’t get scholarship = (7.6x/11x)*100 = 69 (approx.)
Incorrect
Girls = 5x and boys = 6x
Girls that don’t get scholarship = 5x * 80/100 = 4x and boys that don’t get scholarship = 6x *
60/100 = 3.6x
Percent students that didn’t get scholarship = (7.6x/11x)*100 = 69 (approx.)
Question 4 of 25
4. Question
1 points
A bag contains 25p coins, 50p coins and 1 rupee coins whose values are in the ratio of
8:4:2. The total values of coins are 840. Then find the total amount in rupees.
Correct
Value is given in the ratio 8:4:2.
(8x/0.25) + (4x/0.5) + (2x/1) = 840.
X = 20. Total amount = 14*20 = 280
Incorrect
Value is given in the ratio 8:4:2.
(8x/0.25) + (4x/0.5) + (2x/1) = 840.
X = 20. Total amount = 14*20 = 280
Question 5 of 25
5. Question
1 points
An amount is to be divided between A, B and C in the ratio 2:3:5 respectively. If C gives
200 of his share to B the ratio among A, B and C becomes 3:5:4. What is the total sum?
Correct
s2x, 3x + 200, 5x – 200
2x/(3x + 200) = 3/5, we will get x = 600, so total amount = 10*600 = 6000
Incorrect
s2x, 3x + 200, 5x – 200
2x/(3x + 200) = 3/5, we will get x = 600, so total amount = 10*600 = 6000
Question 6 of 25
6. Question
1 points
A bag contains 25p coins, 50p coins and 1 rupee coins whose values are in the ratio of 8:4:2.The total values of coins are 840. Then find the total number of coins
Correct
Value is given in the ratio 8:4:2.
(8x/0.25) + (4x/0.5) + (2x/1) = 840.
X = 20. Total amount = 14*20 = 280
Incorrect
Value is given in the ratio 8:4:2.
(8x/0.25) + (4x/0.5) + (2x/1) = 840.
X = 20. Total amount = 14*20 = 280
Question 7 of 25
7. Question
1 points
Two vessels contains equal quantity of solution contains milk and water in the ratio of
7:2 and 4:5 respectively. Now the solutions are mixed with each other then find the ratio of milk and water in the final solution?
Correct
milk = 7/9 and water = 2/9 – in 1st vessel
milk = 4/9 and water = 5/9 – in 2nd vessel
(7/9 + 4/9)/ (2/9 + 5/9) = 11:7
Incorrect
milk = 7/9 and water = 2/9 – in 1st vessel
milk = 4/9 and water = 5/9 – in 2nd vessel
(7/9 + 4/9)/ (2/9 + 5/9) = 11:7
Question 8 of 25
8. Question
1 points
Two alloys contain gold and silver in the ratio of 3:7 and 7:3 respectively. In what ratio
these alloys must be mixed with each other so that we get a alloy of gold and silver in the
ratio of 2:3?
Correct
Gold = 3/10 and silver = 7/10 – in 1st vessel
gold = 7/10 and silver = 3/10 – in 2nd vessel
let the alloy mix in K:1, then
(3k/10 + 7/10)/ (7k/10 + 3/10) = 2/3. Solve this equation , u will get K = 3
Incorrect
Gold = 3/10 and silver = 7/10 – in 1st vessel
gold = 7/10 and silver = 3/10 – in 2nd vessel
let the alloy mix in K:1, then
(3k/10 + 7/10)/ (7k/10 + 3/10) = 2/3. Solve this equation , u will get K = 3
Question 9 of 25
9. Question
1 points
The sum of three numbers is 123. If the ratio between first and second numbers is 2:5
and that of between second and third is 3:4, then find the difference between second and
the third number.
Correct
a:b = 2:5 and b:c = 3:4 so a:b:c = 6:15:20
41x = 123, X = 3. And 5x = 15
Incorrect
a:b = 2:5 and b:c = 3:4 so a:b:c = 6:15:20
41x = 123, X = 3. And 5x = 15
Question 10 of 25
10. Question
1 points
If 40 percent of a number is subtracted from the second number then the second number
is reduced to its 3/5. Find the ratio between the first number and the second number.
Correct
Incorrect
Question 11 of 25
11. Question
1 points
The ratio between the number of boys and girls in a school is 4:5. If the number of boys
are increased by 30 % and the number of girls increased by 40 %, then what will the new
ratio of boys and girls in the school.
Correct
boys = 4x and girls = 5x.
Required ratio = [(130/100)*4x]/ [(140/100)*5x]
Incorrect
boys = 4x and girls = 5x.
Required ratio = [(130/100)*4x]/ [(140/100)*5x]
Question 12 of 25
12. Question
1 points
One year ago the ratio between rahul salary and rohit salary is 4:5. The ratio between
their individual salary of the last year and current year is 2:3 and 3:5 respectively. If the
total current salary of rahul and rohit is 4300. Then find the current salary of rahul.
Correct
4x and 5x is the last year salry of rahul and rohit respectively
Rahul last year to rahul current year = 2/3
Rohit last year to rohit current year = 3/5
Current of rahul + current of rohit = 4300
(3/2)*4x + (5/3)*5x = 4300.
X = 300.
So rahul current salary = 3/2 * 4* 300 = 1800
Incorrect
4x and 5x is the last year salry of rahul and rohit respectively
Rahul last year to rahul current year = 2/3
Rohit last year to rohit current year = 3/5
Current of rahul + current of rohit = 4300
(3/2)*4x + (5/3)*5x = 4300.
X = 300.
So rahul current salary = 3/2 * 4* 300 = 1800
Question 13 of 25
13. Question
1 points
A sum of 12600 is to be distributed between A, B and C. For every rupee A gets, B gets
80p and for every rupee B gets, C get 90 paise. Find the amount get by C.
Correct
Ratio of money between A and B – 100:80 and that of B and C – 100:90
so the ratio between A : B :C – 100:80:72
so 252x = 12600, x = 50. So C get = 50*72 = 3600
Incorrect
Ratio of money between A and B – 100:80 and that of B and C – 100:90
so the ratio between A : B :C – 100:80:72
so 252x = 12600, x = 50. So C get = 50*72 = 3600
Question 14 of 25
14. Question
1 points
The sum of the squares between three numbers is 5000. The ratio between the first and
the second number is 3:4 and that of second and third number is 4:5. Find the difference
between first and the third number.
The ratio between two numbers is 7:5. If 5 is subtracted from each of them, the new ratio
becomes 3:5. Find the numbers.
Correct
(7x – 5)/(5x – 5) = 3/5
X = 1/2 so the numbers are 7/2 and 5/2
Incorrect
(7x – 5)/(5x – 5) = 3/5
X = 1/2 so the numbers are 7/2 and 5/2
Question 16 of 25
16. Question
1 points
Three cars travel same distance with speeds in the ratio 2 : 4 : 7. What is the ratio of the
times taken by them to cover the distance?
Correct
s = d/t
Since distance is same, so ratio of times:
1/2 : 1/4 : 1/7 = 14 : 7 : 4
Incorrect
s = d/t
Since distance is same, so ratio of times:
1/2 : 1/4 : 1/7 = 14 : 7 : 4
Question 17 of 25
17. Question
1 points
Section A and section B of 7th class in a school contains total 285 students. Which of the
following can be a ratio of the ratio of the number of boys and number of girls in the class?
Correct
The number of boys and girls cannot be in decimal values, so the denominator should completely
divide number of students (285).
Check each option:
6+5 = 11, and 11 does not divide 285 completely.
10+9 = 19, and only 19 divides 285 completely among all.
Incorrect
The number of boys and girls cannot be in decimal values, so the denominator should completely
divide number of students (285).
Check each option:
6+5 = 11, and 11 does not divide 285 completely.
10+9 = 19, and only 19 divides 285 completely among all.
Question 18 of 25
18. Question
1 points
180 sweets are divided among friends A, B, C and D in which B and C are brothers also
such that sweets divided between A and B are in the ratio 2 : 3, between B and C in the
ratio 2 : 5 and between C and D in ratio 3 : 4. What is the number of sweets received by the brothers together?
Correct
A/B = N1/D1 B/C = N2/D2 C/D = N3/D3
A : B : C : D = N1*N2*N3 : D1*N2*N3 : D1*D2*N3 : D1*D2*D3
A/B = 2/3 B/C = 2/5 C/D = 3/4
A : B : C : D
2*2*3 : 3*2*3 : 3*5*3 : 3*5*4
4 : 6 : 15 : 20
B and C together = [(6+15)/(4+6+15+20)] * 180
Incorrect
A/B = N1/D1 B/C = N2/D2 C/D = N3/D3
A : B : C : D = N1*N2*N3 : D1*N2*N3 : D1*D2*N3 : D1*D2*D3
A/B = 2/3 B/C = 2/5 C/D = 3/4
A : B : C : D
2*2*3 : 3*2*3 : 3*5*3 : 3*5*4
4 : 6 : 15 : 20
B and C together = [(6+15)/(4+6+15+20)] * 180
Question 19 of 25
19. Question
1 points
Number of students in 4th and 5th class is in the ratio 6 : 11. 40% in class 4 are girls and
48% in class 5 are girls. What percentage of students in both the classes are boys?
Correct
Total students in both = 6x+11x = 17x
Boys in class 4 = (60/100)*6x = 360x/100
Boys in class 5 = (52/100)*11x = 572x/100
So total boys = 360x/100 + 572x/100 = 932x/100 = 9.32x
% of boys = [9.32x/17x] * 100
Incorrect
Total students in both = 6x+11x = 17x
Boys in class 4 = (60/100)*6x = 360x/100
Boys in class 5 = (52/100)*11x = 572x/100
So total boys = 360x/100 + 572x/100 = 932x/100 = 9.32x
% of boys = [9.32x/17x] * 100
Question 20 of 25
20. Question
1 points
Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of alloy B. A contains
brass and copper in the ratio 3 : 2, and B contains them in the ratio 4 : 3 respectively. What is the ratio of copper to brass in the mixture?
Correct
Brass in A = 3/5 * 50 = 30 kg, Brass in B = 4/7 * 70 = 40 kg
Total brass = 30+40 = 70 kg
So copper in mixture is (50+70) – 70 = 50 kg
So copper to brass = 50 : 70
Incorrect
Brass in A = 3/5 * 50 = 30 kg, Brass in B = 4/7 * 70 = 40 kg
Total brass = 30+40 = 70 kg
So copper in mixture is (50+70) – 70 = 50 kg
So copper to brass = 50 : 70
Question 21 of 25
21. Question
1 points
Ratio of A and B is in the ratio 5 : 8. After 6 years, the ratio of ages of A and B will be in
the ratio 17 : 26. Find the present age of B.
Correct
A/B = 5/8 , A+6/B+6 = 17/26
Solve both, B = 72
Incorrect
A/B = 5/8 , A+6/B+6 = 17/26
Solve both, B = 72
Question 22 of 25
22. Question
1 points
A bag contains 25p, 50p and 1Re coins in the ratio of 2 : 4 : 5 respectively. If the total
money in the bag is Rs 75, find the number of 50p coins in the bag.
Correct
2x, 4x, 5x
(25/100)*2x + (50/100)*4x + 1*5x = 75
x = 10, so 50 p coins = 4x = 40
Incorrect
2x, 4x, 5x
(25/100)*2x + (50/100)*4x + 1*5x = 75
x = 10, so 50 p coins = 4x = 40
Question 23 of 25
23. Question
1 points
A is directly proportional to B and also directly proportional to C. When B = 6 and C =
2, A = 24. Find the value of A when B = 8 and C = 3.
Correct
A directly proportional B, A directly proportional to C:
A = kB, A = kC
Or A = kBC
When B = 6 and C = 2, A = 24:
24 = k*6*2
k = 2
Now when B = 8 and C = 3:
A = 2*8*3
Incorrect
A directly proportional B, A directly proportional to C:
A = kB, A = kC
Or A = kBC
When B = 6 and C = 2, A = 24:
24 = k*6*2
k = 2
Now when B = 8 and C = 3:
A = 2*8*3
Question 24 of 25
24. Question
1 points
A is directly proportional to B and also inversely proportional to the square of C. When
B = 16 and C = 2, A = 36. Find the value of A when B = 32 and C = 4.
Correct
Incorrect
Question 25 of 25
25. Question
1 points
A is directly proportional to the inverse of B and also inversely proportional to C. When
B = 36 and C = 9, A = 42. Find the value of A when B = 64 and C = 21.
Correct
A = k√B, A = k/C
Or A = k√B/C
When B = 36 and C = 9, A = 42:
42 = k√36/9
k = 63
Now when B = 64 and C = 21:
A = 63*√64/21
Incorrect
A = k√B, A = k/C
Or A = k√B/C
When B = 36 and C = 9, A = 42:
42 = k√36/9
