Ratio And Proportion Online Test 2020 - Aptitude Questions And Answers MCQ, Online Quiz

1. Two persons ages are in the ratio 1:A.When 1 is added to both the numerator and the denominator , the ratio gets changed to B/C. Again when 1 is added to both the numerator and the denominator it becomes ½. Find the sum of 1 and Q.

6
4
5
7

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Solution: Let the ratio of two original numbers be 1 : x.
So (1+1)/(x+1) =B/C
and 3/( x+2) = ½
=>x= 4
So, 1/x= 1/A = ¼
So required Sum = 5.

2. Three persons A , B and C entered into a partnership with Capital contribution with Rs. 50,000 ,Rs.60,000 and Rs.30,000 respectively. A is working partner and get 30% of profit for managing the business .The remaining profit is distributed in the respect of capital . At the end of a year A gets Rs. 200 more than that of B and C together .Then what is the total profit?

1840
1800
1700
2000

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Solution: A:B:C= 5:6:3 =14
Let the profit be 100
=>profit shared = 70
=>Difference of profit between A&B+C,
(30+70*5/14) –( 70* 6/14 +70 *3/14) =200
55-45=200
ie )10 ——– *20 = 200
so total profit = 100 *20 =Rs.2000

3. Rani and Jansi enter into a partnership with their initial investments of Rs. 48,000 and 80,000 respectively. They decided to distribute 40% of profit equally between them and rest is according to their investment ratio. If total profit after a year was Rs,16,800.Then find the profit of Rani?

7200
7140
7340
5420

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Solution: Ratio of investment = 48,000 : 80,000
=3:5
Profit of Ruhi ½ * 40/100 *16,800 +3/8 *60/100 *16,800
=3360 +3780
=Rs. 7,140

4. Three positive numbers first , second and third are in the ratio of 2:3:4 respectively. If 5 is added to first number then the product of first, second number becomes 21. What is the difference between first and third number?

2
5
6
3

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Solution: Let the three numbers be 2x,3x and 4x
Then ( 2x+5) *3x= 21
=>2×2 + 5x-7 = 0
=>(2x+7)( x-1) = 0
=>x=1, -7/2
So the numbers are 2, 3, 4
So difference = 4-2 = 2

5. Arun and Biju started a business after investing Rs. 12,000 and Rs. 14,400 respectively. At the end of the year Biju gets 12% of the total profit as management fees and the rest amount is distributed between both the partners in the ratio of their investments. If the total profit is Rs. 7, 500, then find the difference between their profits.

1500
1200
1600
1400

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Solution: Ratio of capitals = 5:6
Required difference = (6-5)/(6+5) * (88/100* 750 0) + 12/100 *7500
=600+900 = Rs.1500

6. Volume of a cylinder is 3888 cm3 , If the ratio between the radius and the height of cylinder is 6:4.find the radius of the cylinder

18
16
21
24

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Solution: Let the radius of the cylinder be r cm.
Given h = 4r/6 cm
Volume of cylinder = 3888 
 r2 * 4r/ 6 = 3888
=>r3= 5832
=> r= 18 cm

7. Ganesh has n sweets. He distributed them among 4 children in the ratio ¼ : 1/6: 1/10:1/16. If he gave them each one a complete sweet, Find the minimum number of sweets he brought?

159
148
156
165

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Solution: LCM ( 4, 6, 10,16) = 240
So ¼*240 = 80
1/6*240=40
1/10 *240 = 24
1/16*240=15
So total number of sweets= 80+ 40+24+15= 159

8. Two friends ages are 40% and 50% more than a third friend. What is the ratio of the first two friend’s age?

17:5
14:5
11:10
14:15

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Solution: Let the third friend age be x.
Then first friend age 140 % of x= 140x/100 = 7x/5
Second friend age = 150% of x= 150x/100 = 3x/2
So ratio= 7x/5: 3x/2 = 14x:15x= 14:15

9. Two triangles having the ratio s of area 5:3. Height ratio 3:2.Find the ratio of the base.

8:7
9:8
10:9
11:5

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Solution: Let the base be x and y of two triangles.
Heights are 3h and 2h.
(½ X x X 3h) /( ½ X y X 2h) = 5/3
x/ y = 5/ 3 x 2/ 3 = 10/9 =>x:y= 10: 9

10. Two glasses of equal capacity contains sand and color powder.The ratio of the mixing is 2:3 and 5:2 respectively. These mixture are transferred into a big bowl.Then what is the ratio of sand and color powder in the bowl?

7:5
1:5
4:5
6:5

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Solution: Ratio= 2 + 5 : 3+2 = 7: 5

11. Chintu, Pintu and Mintu are three friends. Chintu is twice as old as Pintu and Mintu is as old as Cintu and Pintu together. Two years later, the ratio of age of Chintu and Mintu would be 7:10 what was the age of Pintu five years ago?

2 years
4 years
5 years
1 years

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Solution: Let Pintu’s present age =x, Chintu = 2x and Mintu = x+2x =3x
Two years later [2x+2/3x+2] = (7/10)
20x+20=21x+14 = 6=x
Pintu’s present age =6
Five years ago, his age was 6-5= 1 year

12. Find the ratio of speed of A,B and C if for every 11 steps taken by A,B takes 12 steps and C takes 14 steps. Also 12 steps of A are equal to 14 steps of B and 16 steps of C.

154:144:147
132:168:224
25:24:23
161:123:113

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Solution: Ratio of speed is same as ratio of distance covered by them as speed and distance are directly proportional to each other. When time is constant.
A B C
(x/12)×11 (x/14)×12 (x/16)×14
Ratio is (11/12): (12/14): (14/16)
L.C.M of 12, 14, 16 is 336
=308:288:294 = 154:144:147
[What is x? Since distance covered by A in 12 steps is equal to distance covered by B in 14 steps and distance covered by C in 16 steps, hence let this distance =x. Distance covered in 1 step by A, B and C= (x/12), (x/14) and (x/16) respectively ]

13. The price of a ring varies jointly, with the cube root of number of diamonds and weight of gold used in it. The price of the ring was “10,000. When it had 64 diamonds and 15 gm. of gold. Find the no. of diamonds if the price of ring was~` 10,500 and weight of gold was 21 gm.

3
9
27
15

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Solution: Let price =P, number of diamonds = n and weight of gold = g
When p=10000, n=64 and g=15, k=?
P= kg3√n => 10000 = k(15) (3√64)
K=(10000/15×4)=(500/3)
P=10500, g=21, k=(500/3), n=?
P=(500/3) g3√n => 10500=(500/3).213√n
3√n=10500×(3/500)×(1/21)=3=>n=27

14. There are two containers A and B. Container A is filled with 30 liters of kerosene while container B is filled with 30 liters of water. Ten liters of kerosene is taken from container A and put in container B and then twelve liters of the mixture (of kerosene and water) is taken from container B and put in container A. Find the ratio of kerosene in A and B respectively.

23:9
21:5
23:7
1:3

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Solution: Container A Container B
Kerosene water Kerosene water
30 0 0 30
-10 –+10 –
20 0 10 30
When 10 litres of kerosene is taken out from A and put in B, then kerosene in A= 20 litres. Mixture in B= 40 litres in which ratio of Kerosene to water =10:30 = 1:3, when 12 litres of mixture is taken out from B, and put in A, then out of 12 litres
12×(1/4)=3 litres is kerosene and remaining 9 litres is water.
Container A Container B
Kerosene water Kerosene water
20 0 10 30
+3 +9-3 -9
23 9 -7 21
Ratio of Kerosene in A to B = 23:7

15. In a school, students of class I and class II are going for a picnic to Surajkund and Badkal lake respectively. The ratio f number of students in class I and II is 5:3. Also the ratio of the contribution made by each student of class I and II is 19:17. If the total contribution made by the all the students of both the class is Rs.29200, then find the total contribution made by class II students only.

Rs.10950
Rs.13789
Rs.10200
Rs.13272

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Solution: Let the no. of students in class II be 5x and 3x respectively and contribution made by each student of class I and class II be 19y and 17y respectively.
Hence, ratio of total contribution of class I and class II =5x×19y: 3x×17y = 95:51
Total contribution made by students of class II (51/(95+51))×29200=10200

16. A donkey moves at a speed of 8 kmph, when no load is put on him. Reduction in the speed of donkey varies directly to the square root of the kgs of load put on him. When only 4 kgs of load is put the speed of the donkey becomes 6 kmph. Find the minimum load that can be put on the donkey with which it cannot move.

64 kg
163.9 kg
65.2 kg
Cannot be determined

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Solution: Speed of the donkey, Without any load = 8 kmph. With 4 kgs of load, speed becomes 6 kmph, hence speed is reduced by 2 kmph.
Reduction in speed varies directly with the square root of the load.
Hence (8-6) = k√4=k±1
The donkey cannot move at zero speed. i.e. when his speed reduced by 8 kmph. So reduction in speed= 8 kmph
8=1√l=>√l=8 and l=64, when √l=-8, l=64
At 64 kg the donkey will stop.

17. A container has a mixture of kerosene and castor oil in the ratio of 7:5 and another container contains kerosene and castor oil in the ratio of 5:3. Find the proportion in which the mixtures from two containers should be mixed so that the resultant mixture has ratio of kerosene and castor oil of 3:2.

2:3
3:2
4:1
5:1

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Solution: Let quantity of mixture taken from first be x and second be y.
Amount of kerosene oil in the resultant mixture (x+y) is
(7/12)x+(5/8)y=(3/5)(x+y)
(7/12)x-(3/5)x=(3/5)y-(5/8)y-(1/60)x=-(1/40)y=>(x/y)=(6/4)=(3/2)=3:2

18. The no. of pens in three different pencil boxes in the ratio of 1:2:3. Find the ratio in which the number of pens in the first and the second boxes must be increased so that the new ratio becomes 3:2:1.

1:3
2:3
2:1
3:1

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Solution: let the no. of pens in 1st, 2nd and 3rd pencil box be x, 2x and 3x respectively and let the required no. be 3y, 2y and y.
The quantity f pens in the third pencil box would remain the same, hence
3x=y or x =(y/3)
Quantity of pens in the boxes originally is x, 2x and 3x
When x=(y/3), hence quantity is x=(y/3), 2x=(2y/3), 3x=y
i.e (y/3), (2y/3), y
The required number of pens is 3y, 2y and y in
Increase in 1st box=3y-(y/3)=(8/3)y&
Increase in 2nd box = 2y-(2y/3)=(4/3)y
Ratio of increase = (8y/3)L4/3)y=2:1

19. There are three vessels 1, 2 and 3. The ratio of the total capacity of vessels 1, 2 and 3 is 5:4:3 respectively. All the vessels are full of mixture of sugar syrup and water. In vessel 1, ratio of sugar syrup to water is 2:3. Similar ratio in case of vessel 2 and vessel 3 is 5:4 and 1:3 respectively. The mixture of all the three vessels is emptied into one bigger vessel. What is the resulting ratio of sugar syrup and water?

17:25
179:253
253:179
233:169

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Solution: Method-1: Let the total mixture in vessel 1,2 and 3 be 5 litres and 3 litres respectively. So, quantity of water in (5+4+3)=12 litres of mixture is
=(3/5)×5+(4/9) ×4+(3/4) ×3
=3+(16/9)+(9/4)=[(108+64+81)/36]=(253/36)
Quantity of sugar syrup is
12-(253/36)=(179/36)
Ratio of sugar syrup to water = 179:253

20. If x2+4y2= 4xy, then x:y is :

-1:2
-2:1
-2:3
-3:2

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Solution: (X-2Y)2=0
X-2Y=0
X=2Y
X:Y=2:1