## Quant Quiz On Ratio & Proportion Day 19 Bag

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• A school has 4 sections of class 12, such that half the number of students of 1st section, 1/3rd of 2nd section, 1/4th of 3rd section and 1/5th of the 4th section are equal. If total number of students in class 12 is 420, find the number of students in sections 1st and 2nd.
A) 180
B) 120
C) 240
D) 150
E) 260

D) 150
Explanation:

Let number of students in 4 sections be A, B, C, D respectively. Then
1/2 of A = 1/3 of b = 1/4 of C = 1/5 of D
So A : B : C : D = 2 : 3 : 4 : 5            [When A/2 = B/3 = C/4, then ratio A: B : C = 2 : 3 : 4] So students in 1st and 2nd section = [(2+3)/(2+3+4+5)] * 420 = 150
• The income of A, B, and C are in the ratio 3 : 4 : 7. If their incomes be changed such that the new income of A is 50% increased, 25% increased for B and 25% decrease for C. Find the ratio of their new incomes.
A) 18 : 40 : 23
B) 17 : 12 : 21
C) 18 : 20 : 21
D) 28 : 20 : 21
E) None of these

C) 18 : 20 : 21
Explanation:

Previous ratio = 3 : 4 : 7
New ratio = (150/100) * 3 : (125/100) * 4 : (75/100) * 7
• A, B and C divide Rs 4200 among themselves in the ratio 7 : 8 : 6. If Rs 200 is added to each of their shares, what is the new ratio in which they will receive the money?
A) 9 : 8 : 7
B) 8 : 9 : 7
C) 8 : 9 : 8
D) 9 : 10 : 8
E) 7 : 9 : 8

B) 8 : 9 : 7
Explanation:

A gets = [7/(7+8+6)] * 4200 = 1400
B gets = [8/(7+8+6)] * 4200 = 1600
C gets = [6/(7+8+6)] * 4200 = 1200
Rs 200 added to each share, so new ratio =
1400+200 : 1600+200 : 1200+200
1600 : 1800 : 1400
• The ratio of the monthly salaries of A and B is in the ratio 15 : 16 and that of B and C is in the ratio 17 : 18. Find the monthly income of C if the total of their monthly salary is Rs 1,87,450.
A) Rs 66,240
B) Rs 72,100
C) Rs 62,200
D) Rs 65,800
E) Rs 60,300

A) Rs 66,240
Explanation:

A/B = 15/16 and B/C = 17*18
So A : B : C = 15*17 : 16*17 : 16*18
= 255 : 272 : 288
So C’s salary = [288/(255+272+288)] * 1,87,450
• The ratio of the incomes of A and B last year was 9 : 13. Ratio of their incomes of last year to this year is 9 : 10 and 13 : 15 respectively. The sum of their present incomes is Rs 50,000. What is the present income of B?
A) Rs 32,000
B) Rs 24,000
C) Rs 20,000
D) Rs 30,000
E) None of these

D) Rs 30,000
Explanation:

Ratio of last year income to this year income of A is 9 : 10. So income of A last year is 9x and this year is 10x.
Ratio of last year income to this year income of B is 13 : 15. So income of B last year is 13y and this year is 15y.
So ratio of the incomes of A and B last year was 9x : 13y
Now given that ratio of the incomes of A and B last year was 9 : 13.
So 9x/13y = 9/13
This gives x = y
Total of incomes of A and B this year = 10x+15y = 10x+15x = 25 x              (because x=y)
So 25x = 50,000
This gives x = 2,000
So present income of B = 15y = 15x = 15*2000 = 30,000
• A sum of Rs 315 consists of 25 paise, 50 paise and 1 Re coins in the ratio 3 : 4 :6. What is the number of each kind of coin respectively?.
A) 216, 144, 27
B) 108, 144, 216
C) 27, 72, 216
D) 120, 35, 108
E) 102, 150, 210

B) 108, 144, 216
Explanation:

25 paise = 25/100 Rs, 50 paise = 50/100 Rs
So value ratio of these coins become = 3*(25/100) : 4*(50/100) : 6*(1)
= 3/4 : 2 : 6 = 3 : 8 : 24
So 25 paise coins value= [3/(3+8+24)] * 315 = Rs 27, so coins = 27 * (100/25) = 108
Similarly find others.
• Rs 650 was divided among 3 children in the ratio 2 : 4 : 7. Had it been divided in the ratio 1/2 : 1/4 : 1/7, who would have gained the most and by how much?
A) C, Rs 246
B) C, Rs 264
C) B, Rs 18
D) A, Rs 246
E) A, Rs 264

E) A, Rs 264
Explanation:

New ratio = 1/2 : 1/4 : 1/7 = 14 : 7 : 4
So both ratio suggests that C has not gained any money, rather he has lose the money.
For both ratio find the shares of A and B
With ratio 2 : 4 : 7, A gets = [2/(2+4+7)] * 650 = 100, B gets = [4/(2+4+7)] * 650 = 200
With ratio 14 : 7 : 4, A gets = [14/(14+7+4)] * 650 = 364, B gets = [7/(14+7+4)] * 650 = 182
B has also lose the money, A gain the money and = 364 – 100 = 264
• The ratio of the number of boys to the number of girls in a school is 6 : 5. If 20% of boys and 45% of girls come by bus to school, what percentage of students opt transport other than bus to come to school?
A) 68 9/11%
B) 68 7/11%
C) 72 7/11%
D) 73%
E) 73 5/11%

B) 68 7/11%
Explanation:

If 20% of boys and 45% of girls come by bus, then 80% of boys and 55% of girls opt transport other than bus.
Let total number of students in school = x
So boys who opt other transport are (80/100) * 6/(6+5) * x = 24x/55
And girls who opt other transport are (55/100) * 5/(6+5) * x = x/4
So total students who opt other transport = (24x/55) + (x/4) = 151x/220
So required % = [(151x/220)/x] * 100 = 755/11 %
• The incomes of A and B are in the ratio 1 : 2 and their expenditures are in the ratio 2 : 5. If A saves Rs 20,000 and B saves Rs 35,000, what is the total income of A and B?
A) Rs 30,000
B) Rs 90,000
C) Rs 90,000
D) Rs 60,000
E) Rs 80,000

C) Rs 90,000
Explanation:

Income of A = x, of B = 2x
Expenditure of A = 2y, of B = 5y
Savings is (income – expenditure). So
x – 2y = 20,000
2x – 5y = 35,000
Solve the equations, x = 30,000
So total = x+2x = 3x = 3*30,000 = 90,000
• Rs 5750 is divided among A, B, and C such that if their share be reduced by Rs 10, Rs 15 and Rs 25 respectively, the reminder amounts with them shall be in the ratio 4 : 6 : 9. What was C’s share then?
A) Rs 2700
B) Rs 2725
C) Rs 2750
D) Rs 2625
E) None of these

B) Rs 2725
Explanation:

When the shares reduce, the total amount will also reduce which is to be divided among them. So after reducing shares by Rs 10, Rs 15 and Rs 25 respectively, total amount is 5750 – (10+15+25) = 5700
So C’s share shall be [9/(4+6+9)] * 5700 = 2700
Actually C would have received = 2700 + 25

• The students in 3 classes are in the ratio 3:4:5.If 20 students added in each class, the ratio becomes 5:6:7.Find the total no of students in all the 3 classes now ?
A.160
B.170
C.180
D.200
E.None of these

Explanation :
5x+6x+7x+ 18x = 180
X=10
Check 3x+4x+5x = 12x = 120
120+60 = 180
• Two equal containers are filled with water and acid. The concentration of acid in each container is 20% and 30%. What is the ratio of water in both the containers respectively ?
A.7:8
B.8:7
C.6:4
D.3:2
E.None of these

Explanation :
Acid : 20    30
Water : 80  70 => 8:7
• A horse takes 8 steps for every 5 steps of a fox, but 6 steps of a fox are equal to the 3 steps of the horse. What is the ratio of the speed of horse to the fox ?
A.16:5
B.20:19
C.18:23
D.17:21
E.None of these

Explanation :
6 step fox=3 step horse = 2:1
16 step fox=8step horse=5 step fox
16 step fox=5 step fox
16:5
• The ratio of age of Krish  and her mother is 5:12 and difference of their ages is 21. What will be the ratio of their ages after 3 years ?
A.7:15
B.11:5
C.13:7
D.9:13
E.None of these

Explanation :
12x-5x =>7x = 21
X=21/7 =3
Present ratio = 15:36
After 3 years = 18:39 = 6:13
• A started a business with Rs.32,000 after 4 months B joins with the business by investing Rs.48,000.At the end of the year, in what ratio should share their profit ?
A.8:7
B.7:6
C.5:7
D.9:5
E.None of these

Explanation :
32000*12 : 48,000*8 = 32*12 : 42*8
384 : 384 = 1:1
• A working partner gets 25% as his commissions after his commissions paid  that is equal to Rs.7500, then what is the total profit ?
A.Rs.32,000
B.Rs.30,000
C.Rs.37,500
D.Rs.40,000
E.None of these

Explanation :
X=total profit
25/100*[x-7500] = 7500
x-7500 = 7500*100/25 =30,000
x = 37,500
• Equal quantities of 3 mixtures of milk and water are mixed in the ratio 1:3, 2:3 and 3:4.The ratio of water and milk in the new mixture is
A.45:76
B.151:269
C.123:154
D.145:245
E.None of these

Explanation :
Milk = 1/4 : 2/5 :3/7
= 35/140 :56/140 : 60/140
Quantity of milk in new mix = 35+56+60 = 151
Quantity of water in new mix = 140*3 = 420-151 = 269
M:W = 151:269
• The speed of P, Q and R are in the ratio of 2:3:6, what is the ratio of time taken by each one of them for the same distance ?
A.5:7:2
B.3:2:4
C.1:2:3
D.3:2:1
E.None of these

Explanation :
Speed = ½ : 1/3: 1/6
= 3/6:2/6:1/6
Time = 3:2:1
• The ratio of income of X and Y is 4:3. The sum of their expenditure is Rs.12,000 and the amount of savings is X is equal to the amount of expenditure of Y.What is the salary of  Y?
A.9000
B.7000
C.12000
D.15000
E.None of these

Explanation :
X’s saving = Expenditure of Y = S
4x-S + S = 12000
X = 3000
3x =3*3000 =  9000
• When 7 is added to the numerator and denominator of the fraction, then the new ratio of numerator and denominator becomes 13:19, what is the original ratio ?
A.11:13
B.7:9
C.4:7
D.Can’t be determined
E.None of these

Explanation :
X+7/y+7 = 13/19 x and y are different variable ,so original fraction cannot be determined

• 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
A.220
B.240
C.260
D.280
E.None of these

Explanation :
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
• 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?
A.11:7
B.11:6
C.11:5
D.11:9
E.None of these

Explanation :
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
• 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?
A.2:1
B.3:1
C.4:3
D.3:5
E.None of these

Explanation :
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
• 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.
A.12
B.14
C.15
D.17
E.None of these

Explanation :
a:b = 2:5 and b:c = 3:4 so a:b:c = 6:15:20
41x = 123, X = 3. And 5x = 15
• 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.
A.1:3
B.1:2
C.1:1
D.2:3
E.None of these

Explanation :
[ b – (40/100)a] = (3/5)b.
So we get a = b.
• 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.
A.13/35
B.26/35
C.26/41
D.23/13
E.None of these

Explanation :
boys = 4x and girls = 5x.
Required ratio = [(130/100)*4x]/ [(140/100)*5x]
• 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.
A.1200
B.1800
C.1600
D.2000
E.None of these

Explanation :
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
• 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.
A.3200
B.3600
C.4200
D.4600
E.None of these

Explanation :
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
• 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.
A.20
B.30
C.40
D.50
E.None of these

Explanation :
a^2 + b^2 + c^2 = 5000
a:b:c = 3:4:5
50x^2 = 5000.
X = 10.
5x – 3x = 2*10 = 20
• 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.
A.7/2, 5/2
B.3/2, 7/2
C.9/2, 7/2
D.11/2, 5/2
E.None of these

Explanation :
(7x – 5)/(5x – 5) = 3/5
X  = 1/2 so the numbers are 7/2 and 5/2

• 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?
A) 12 : 6 : 7
B) 14 : 7 : 4
C) 10 : 5 : 9
D) 7 : 4 : 14
E) 14 : 10 : 7

B) 14 : 7 : 4
Explanation:

s = d/t
Since distance is same, so ratio of times:
1/2 : 1/4 : 1/7 = 14 : 7 : 4
• 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?
A) 6 : 5
B) 10 : 9
C) 11 : 9
D) 13 : 12
E) Cannot be determined

B) 10 : 9
Explanation:

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.
• 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?
A) 78
B) 84
C) 92
D) 102
E) 88

B) 84
Explanation:

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
• 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?
A) 62.5%
B) 54.8%
C) 52.6%
D) 55.8%
E) 53.5%

B) 54.8%
Explanation:

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
• 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?
A) 8 : 5
B) 7 : 5
C) 5 : 11
D) 4 : 9
E) 5 : 7

E) 5 : 7
Explanation:

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
• 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.
A) 72
B) 65
C) 77
D) 60
E) None of these

A) 72
Explanation:

A/B = 5/8 , A+6/B+6 = 17/26
Solve both, B = 72
• 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.
A) 45
B) 50
C) 25
D) 40
E) None of these

D) 40
Explanation:

2x, 4x, 5x
(25/100)*2x + (50/100)*4x + 1*5x = 75
x = 10, so 50 p coins = 4x = 40
• 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.
A) 42
B) 40
C) 58
D) 48
E) None of these

D) 48
Explanation:

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
• 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.
A) 25
B) 20
C) 18
D) 32
E) None of these

C) 18
Explanation:

A = kB, A = k/C2
Or A = kB/ C2
When B = 16 and C = 2, A = 36:
36 = k*16/ 22
k = 9
Now when B = 32 and C = 4:
A = 9*32/ 42
• 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.
A) 24
B) 40
C) 32
D) 48
E) None of these

A) 24
Explanation:

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