வெல்லுங்கள் CSAT 2025: 7 - Ratio, Mixtures, Alligation & Partnership

Questions & Answers (Multiple Choice) - கொள்குறி வினா- விடைகள்
விகிதம் - Ratio, Mixtures, Alligation & Partnership

1. Find the third proportion to the numbers 0.81 and 0.09.
a. 1 b. 0.1 c. 0.01 d. 0.001

2. What number must be subtracted from 40, 74, 53 and 100 to leave the remainders are in proportion?
a. 1 b. 3 c. 4 d. 6

3. Krish takes 5 leaps for every 7 leaps of Lal. If two leaps of Krish is equal to 3 leaps of Lal, then find the ratio of speed of Lal to Krish.
a. 15: 14 b. 14 : 15
c. 10 :21 d. 21 : 10

4. Find the least of three numbers which are in the ratio 5 : 2 : 3 and the sum of the squares is 7448.
a. 14 b. 28 c. 42 d. 70

5. One bottle is half-full of oil and another bottle with twice the capacity is one-quarter full of oil. If water is added so that both the bottles are full, then the contents of both the bottles are poured into a third bottle which is empty but to hold the contents of both. What is the ratio of oil to water in the third bottle?
a. 1 : 2 b. 3 : 1 c. 2 : 3 d. 1: 1

6. The scale of a map
3 decimetre = 4 hectametre. If the distance between two villages in the map is 60decimetre, find the actual distsnce in km. between the two villages.
a. 16 b. 10 c. 8 d. 12

7. Ratio of water to soil in the earth is 2 : 1. Ratio water to soil in southern hemisphere is 3 : 5. Find the ratio of water to soil in the northern hemisphere.
a. 2 : 1 b. 5 : 6 c. 6 :5 d. 23 : 1

8. In what proportion should one variety of oil at ₹ 95 per kg be mixed with another variety of oil at ₹ 100 per kg to get a mixture worth ₹ 96 per kg.
a. 4 : 1 b. 5 : 1 c. 1 :5 d. 1 : 4

9. A container contained 40 kg of pure milk. From this container, 4kg of milk is taken out and replaced by water. This proccess was further repeated. Now how much kg of milk is in the container?
a. 32 b. 33 c. 32.4 d. 33.2

10. Three milkmen rented a pasture. A grazed 24 cows for 8 months, B grazed 33 cows for 6 months and C grazed 36 cows for 5 months. If C's share of rent is ₹ 1680, then find the rent of the field.
a. ₹3250 b. ₹5320
c. ₹4800 d. ₹5630

11. Three hikers A, B and C start on a trip with ₹ 750 each, and agree to share the expenses equally. If at the end of the trip, A has left ₹ 250 with him, but B ₹200 and C ₹ 300. In which way they settle their accounts?
a. A shoud give ₹ 50 to B
b. A shoud give ₹ 50 to C
c. B shoud give ₹ 50 to C
d. C shoud give ₹ 50 to B

12. Gold is 19 times as heavy as water and copper is 9 times as heavy as water. In what ratio should these be mixed to get an alloy of 16 times as water.
a. 7 : 3 b. 3 : 7 c. 4 :5 d. 5 : 4

13. Diamond's value is proportional to the square of its weight. When the diamond broke, the weights of the pieces are in the ratio
1 : 2 : 3 : 4. Total loss in value is
₹ 630000. Find the value of the original diamond before it broke.(in lakhs)
a. 12 b. 10 c. 9 d. 8

14. The ratio of volumes of two cones in the ratio is 2 : 3 and the ratio of the radii of their bases is
1 : 2. Find the ratio of their heights.
a. 2 : 3 b. 8 : 3 c. 1 :3 d. 5 : 6

15. The ratio of investments of partners A and B is 11 : 12 and the ratio of their profits is in the ratio
2 : 3. If B invested his money for 11 months, then find for how many months A invested.
a. 9 b. 11 c. 12 d. 8

Answers

1. c 2. d 3. b 4. b 5. a

6. c 7. d 8. a 9. c 10. b

11. d 12. a 13. c 14. b 15. d

Explanation to answers

1. If a, b and c are in proportion, then
a/b = b/c -> c = (b^2)/a
Here a = 0.81; b = 0.09
So c= 0.09×0.09 ÷ 0.81 = 0.01

2. Let 'n' be the number to be subtracted.
So, (40 - n)/(74 - n)
= (53 - n)/(100 - n)
–> (40 - n)(100 - n)=(53 - n)(74 - n)
-> 4000 - 140n = 3922 - 127n
-> 13n = 78 -> n = 6.

3. Let the distance of one leap of Krish and Lal be x and y units respectively.
Here 2x = 3y -> x = (3/2)y
The speeds of Lal and Krish is in the ratio 7y : 5x = 7y : 5(3/2)y
= 7 : (15/2).
Hence the required ratio is 14 : 15.

4. Let the numbers be 5x, 2x and 3x respectively.
So 25x^2 + 4x^2 + 9x^2 = 7448
-> 38x^2 = 7448
-> x^2 = 196 -> x = 14
The least number is 2x =2×14 =28

5. Let the capacities of both the bottles be 1 unit and 2 units respectively.
So the oil contents in both the bottles are 1/2 unit and 2(1/4) unit respectively.
i.e., oil in the third bottle is
1/2 + 1/2 = 1 unit.
When both bottles filled with water, the quantity of Water in both
(1 - 1/2) + (2 - 1/2) = 2 units.
Hence the ratio of oil to water in the third bottle is 1 : 2

6. We know 1 hectametre = 0.1 k.m.
1 decimetre = (4/3)(0.1) k.m.
= 4/30 k.m.
So 60 dm = (4/30)×60 = 8 k.m.

7. Parts of water and soil in the earth are 2/3 and 1/3. Half of the part of earth is Northern Hemisphere and other half is Southern Hemisphere. But in Southern Hemisphere the parts of water and soil are 3/8 and 5/8. In the whole earth the parts of water and soil in the Southern Hemisphere are 3/16 and 5/16.
So the ratio of water to soil in Northern Hemisphere are
(2/3 - 3/16) : (1/3 - 5/16)
= 23/48 : 1/48
= 23 : 1

8. I II
Price 95 100
Networth 96
Ratio 100 - 96 : 96 - 95 = 4 : 1
This method of solving is called Method of Alligation.

9. We can use the technique,
{1- (D/Q)}^n : 1 - {1- (D/Q)}^n where Q denotes the total quantity of Milk at the beginning, D denotes the quantity of milk taken out each time and 'n' denotes the number of times repeated. Here
D/Q = 4/40 = 1/10 and n =2
So the ratio of milk to water in the ratio
{1- (1/10)}^2 : 1 - {1- (1/10)}^2
= 81/100 : 1- (81/100) = 81 : 19
Hence at the end of 2 times replacement, the amount of milk in the mixture is
(81/100)×40 = 32.4 kg.

10. The rent's share ratio is
24×8 : 33×6 : 36×5 = 32 : 33 : 30
If C's share is ₹ 30, then the total rent is (32+33+30) = ₹95.
Hence the total rent is
(95/30) × 1680
= 95×56 = 5320

11. The total amount = 3 × 750
= ₹ 2250
The total amount left with them
= 250 + 200 + 300 = ₹ 750
The total expenses = 2250 - 750
= ₹1500
Each share of expenses = ₹ 500
The balance to be with them ₹250 each
But A,B and C had left with them ₹250, ₹200 and ₹300 respectively.
It is evident that C should give ₹ 50 to B.

12. Gold Copper
Weights 19 9
Net weight 16
Ratio 16 - 9 : 19 - 16
Required ratio = 7 : 3
This method of solving is called Method of Alligation.

13. Let the total weight of the diamond be
1 + 2 + 3 + 4 = 10w units.
So the value of the original diamond
(10w)^2 = 100w^2.
After broke, the value of the diamond
w^2 + 4w^2 + 9w^2 + 16w^2
= 30w^2
Loss value = 100w^2 - 30w^2
= 70w^2.
On equating, 70w^2 = 630000
So w^2 = 9000
Hence the value of the original diamond before broken is
100 × 9000 = 900000

14. Volume of cone =(1/3)π(r^2)h
Where 'r' is the radius of the base and 'h' is the height of the cone.
Here the ratio of squares of the radii is 1^2 : 2^2 = 1 : 4
Let the ratio of heights be H : h
So the ratio of their volumes is
1(H) : 4(h). But the given ratio is
2 : 3
i.e., H : 4h = 2 : 3 -> 3H = 8h
-> H : h = 8 : 3

15. Let the investment of A be used for 'x' months.
The compound ratios of investment and time is the profit ratio.
Here the compound ratios of 11: 12 and x : 11 is
11×(x) : 12×11 = 2 : 3
-> 3×11×(x) = 2×12×11 -> x = 8

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