Ratio and Proportion - Aptitude

Chapter 8: RATIO AND PROPORTION

Introduction

Ratio means two quantities in the same units like Hari's height is 100 cm and Ram's height is 200 cm so the ratio of their heights is 100/200 or 1:2. the ratio has no units.

Proportion is equality of two ratio's a:b = c:d we write as a:b::c:d and a,d are extremes and b,c are means. The product of means = product of extremes.
a * d = b * c

When we have to divide a quantity 'x' into a:b we need to break it into two parts one is a/a+b and second is b/b+a.

Componendo - Dividendo rule:

if a:b = c:d then (a+b)/(a-b) = (c+d)/(c-d)

Proportionality:

If x ∝ y i.e. x is directly proportional to y which means when x increases y too increases and when x decreases y to decreases.

If x ∝ 1/y i.e. x is inversely proportional to y which means if x increases y decreases and when x decreases y increases.

x ∝ y then x = k*y where k = constant

x ∝ 1/ y then x * y= k where k = constant

If we multiply the numerator and the denominator of a ratio by the same number, the ratio remains unchanged. \( \frac{a}{b} = \frac{ma}{mb} \)

If we divide the numerator and the denominator of a ratio by the same number, the ratio remains unchanged. Thus \( \frac{a/d}{b/d} = \frac{a}{b} \)

The magnitudes of two ratios can be compared by equating the denominators of the two ratios and then checking for the value of the numerator. Thus, if we have to check for \( \frac{8}{3} = \frac{11}{4} \). We convert 8/3 into \( \frac{8 * 1.33}{3 * 1.33} = \frac{10.66}{4} \)Now denominators are equal and so we can compare the numerators and find which are greater and lesser.

If a/b = c/d = e/f = g/h = k then \( k = \frac{a + e + c +g}{b + d + f + h} \)

If a₁ / b₁ , a₂/b₂, a₃/b₃... a_n/b_n are unequal fractions then we have \( \frac{a_1 + a_2 + a_3 + .. + a_n}{b_1 + b_2 + b_3 + .. + b_n} \) lies between the lowest and the highest of these fractions.

If we have two equations containing three unknowns as a₁x + b₁y + c₁z = 0 and a₂x + b₂y + c₂z = 0 then we can find the proportion x : y : z. This will be given by \( b_1 * c_2 - b_2 * c_1 : a_2 * c_1 - c_2 * a_1 : a_1 * b_2 - a_2 * b_1 \)

If the ratio \( \frac{a}{b} > 1 \) (called a ratio of greater inequality) and if k is a positive number: \( \frac{a + k}{b + k} < \frac{a}{b} \) and \( \frac{a - k}{b - k} > \frac{a}{b} \) Similarly if \( \frac{a}{b} < 1 \) then \( \frac{a + k}{b + k} > \frac{a}{b} \) and \( \frac{a - k}{b - k} < \frac{a}{b} \)

Maintenance of equality when numbers are added in both the numerator and the denominators. This if best illustrated through an example: \( \frac{20}{30} = \frac{20 + 2}{30 + 3} \) i.e. \( \frac{a}{b} = \frac{a + c}{b + d} \) if and only if \( \frac{c}{d} = \frac{a}{b} \)

Consequently, if \( \frac{c}{d} > \frac{a}{b} \) then \( \frac{a + c}{b + d} > \frac{a}{b} \) and if \( \frac{c}{d} < \frac{a}{b} \) then \( \frac{a + c}{b + d} < \frac{a}{b} \)

To get the consolidated ratio A:B:C:D:E when individual ratios A:B, B:C, C:D, D:E are given

A:B = 2:3, B:C = 4:5, C:D = 6:11, D:E = 12:17

In order to create one consolidated ratio for this situation using the LCM process becomes too long. The short cut goes as follows:

A will correspond to the product of all numerators (2 × 4 × 6 × 12) while B will take the first denominator and the last 3 numerators (3 × 4 × 6 × 12). C on the other hand takes the first two denominators and the last 2 numerators (3 × 5 × 6 × 12), D takes the first 3 denominators and the last numerator (3 × 5 × 11 × 12) and E takes all the four denominators (3 × 5 × 11 × 17).

In mathematical terms this can be written as:

\( \frac{a}{b} = \frac{N_1}{D_1}, \frac{b}{c} = \frac{N_2}{D_2}, \frac{c}{d} = \frac{N_3}{D_3} , \frac{d}{e} = \frac{N_4}{D_4} \text{ then } a : b : c : d : e = N_1N_2N_3N_4 : D_1N_2N_3N_4 : D_1D_2N_3N_4 : D_1D_2D_3N_4 : D_1D_2D_3D_4 \)

Cross multiplication method : Compare two ratios

Two ratios can be compared using the cross multiplication method as follows. Suppose you have to compare 12/17 with 15/19. Then, to test which ratio is higher cross multiply and compare 12 × 19 and 15 × 17.

If 12 × 19 is bigger the Ratio 12/17 will be bigger. If 15 × 17 is higher, the ratio 15/19 will be higher. In this case, 15 × 17 being higher, the Ratio 15/19 is higher

Method for calculating the value of a percentage change in the ratio:

if 20/40 becomes 22/50 then Effect of numerator = 20 -> 22(10% increase) Effect of denominator = 50 -> 40(25% decrease) (reverse fashion)

Overall effect on the ratio:

100 increased by 10% gives 110 which when reduced by 25% gives 82.5.

Hence the overall change in the ratio is 17.5%

When two ratios are equal, the four quantities composing them are said to be proportionals. Thus if a/b = c/d, then a, b, c, d are proportionals. This is expressed by saying that a is to b as c is to d, and the proportion is written as a : b : : c : d OR a : b = c : d

The terms a and d are called the extremes while the terms b and c are called the means.

If four quantities are in proportion, the product of the extremes is equal to the product of the means. Let a, b, c, d be the proportionals. Then by definition a/b = c/d we get ad = bc Hence if any three terms of proportion are given, the fourth may be found. Thus if a, c, d are given, then b = ad/c

If three quantities a, b and c are in continued proportion, then a : b = b : c then \( ac = b^2 \) also a : c = \( a^2 : b^2 \)

If four quantities a, b, c and d form a proportion, many other proportions may be deduced by the properties of fractions. The results of these operations are very useful. These operations are

Invertendo: If a/b = c/d then b/a = d/c

Alternando: If a/b = c/d, then a/c = b/d

Componendo: If a/b = c/d, then \( \frac{a+b}{b} = \frac{c+d}{d} \)

Dividendo: If a/b = c/d, then \( \frac{a-b}{b} = \frac{c-d}{d} \)

Componendo and Dividendo: If a/b = c/d, then \( \frac{a+b}{a-b} = \frac{c+d}{c-d} \)

5783 is divided among Sherry, Berry, and Cherry in such a way that if t` 28, t` 37 and t` 18 be deducted from their respective shares, they have money in the ratio 4 : 6 : 9. Find Sherry’s shar

1256

1228

1456

1084

Ans .

Explanation :

The problem clearly states that when we reduce 28, 37 and 18 rupees respectively from
Sherry’s, Berry’s and Cherry’s shares, the resultant ratio is: 4 : 6 : 9.
Thus, if we assume the reduced values as
4x, 6x and 9x, we will have Æ
Sherry’s share is 4x + 28, Berry’s share is 6x + 37 and Cherry’s share is 9x + 18 and thus we have
(4x + 28) + (6x + 37) + (9x + 18) = 5783
so 19x = 5783 – 83 = 5700
Hence, x = 300.
Hence, Sherry’s share is 1228.

Divide 1870 into three parts in such a way that half of the first part, one-third of the second part and one-sixth of the third part are equal.

241, 343, 245

400, 800, 670

470, 640, 1160

None of these

Ans .

None of these

Explanation :

Solve this question using options. 1/2 of the first part should equal 1/3
rd of the second part and of the third part. This means that the first part should be divisible by 2, the second one by 3, and
the third one by 6. Looking at the options, none of the first 3 options has its third number divisible
by 6. Thus, option (d) is correct.

Divide 500 among A, B, C and D so that A and B together get thrice as much as C and D together, B gets four times of what C gets and C gets 1.5 times as much as D. Now the value of what B gets is

300

125

150

Ans .

Explanation :

(A + B) = 3 (C + D) so A + B = 375 and C + D = 125.
Also, since C gets 1.5 times D we have
C = 75 and D = 50, and B = 4 * C = 300.

If \( \frac{a}{b+c} = \frac{b}{a+c} = \frac{c}{b+a} \), then each fraction is equal to

(a + b + c)²

1/2

1/4

Ans .

1/2

Explanation :

The given condition has a, b and c symmetrically placed. Thus, if we use a = b = c = 2 (say) we get each fraction as 1/2

If 6x² + 6y² = 13xy, what is the ratio of x to y?

1 : 4

3 : 2

4 : 5

1 : 2

Ans .

3 : 2

Explanation :

Solve using options. Since x > y > 0 it is clear that a ratio of x:y as 3:2 fits the equation

If a : b = c : d then the value of \( \frac{a^2 + b^2}{c^2 + d^2} \)

1/2

a+b / c+d

a-b / c-d

ab/cd

Ans .

ab/cd

Explanation :

1 : 2 = 3 : 6 So, (a² + b²)/(c² + d²) = 5/45 = 1/9; From the given options, only ab/cd gives us this value.

If 4 examiners can examine a certain number of answer books in 8 days by working 5 hours a day, for how many hours a day would 2 examiners have to work in order to examine twice the number of answer books in 20 days.

7.5

Ans .

Explanation :

4 × 8 × 5 = 160 man-hours are required for ‘x’ no. of answer sheets. So, for ‘2x’ answer sheets we would require 320 man-hours = 2 × 20 × n Æ n = 8 Hours a day.

In a mixture of 40 litres, the ratio of milk and water is 4 : 1. How much water must be added to this mixture so that the ratio of milk and water becomes 2 : 3.

Ans .

Explanation :

In 40 litres, milk = 32 and water = 8. We want to create 2 : 3 milk to water mixture, for this we would need: 32 milk and 48 water. (Since milk is not increasing). Thus, we need to add 40 litres
of water

If three numbers are in the ratio of 1 : 2 : 3 and half the sum is 18, then the ratio of squares of the numbers is:

6 : 12 : 13

1 : 2 : 4

36 : 144 : 324

3 : 5 : 7

Ans .

6 : 144 : 324

Explanation :

1 : 2 : 3 then x, 2x and 3x add up to 36.
So the numbers are: 6, 12 and 18.
Ratio of squares = 36 : 144 : 324

The ratio between two numbers is 3 : 4 and their LCM is 180. The first number is

Ans .

Explanation :

The numbers would be 3x and 4x and their LCM would be 12x. This gives us the values as 45 and 60. The first number is 45.

If a, b, c, d are proportional, then (a – b) (a – c)/a =

a + c + d

a + d – b – c

a + b + c + d

a + c – b – d

Ans .

a + d – b – c

Explanation :

Assume a set of values for a, b, c, d such that they are proportional i.e. a/b = c/d. Suppose we take
a:b as 1:4 and c:d as 3:12 we get the given expression:
(a – b)(a – c)/a = –3 × –2/1 = 6. This value is also given by a + d – b – c and hence option (b) is
correct

A and B are two alloys of argentum and brass prepared by mixing metals in proportions 7 : 2 and 7 : 11 respectively. If equal quantities of the two alloys are melted to form a third alloy C, the proportion of argentum and brass in C will be:

5 : 9

5 : 7

7 : 5

9 : 5

Ans .

7 : 5

Explanation :

Since equal quantities are being mixed, assume that both alloys have 18 kgs (18 being a number which is the LCM of 9 and 18).
The third alloy will get, 14 kg of argentum from the first alloy and 7 kg of argentum from the
second alloy. Hence, the required ratio: 21:15 = 7:5

If 30 men working 7 hours a day can do a piece of work in 18 days, in how many days will 21 men working 8 hours a day do the same work?

22.5

Ans .

22.5

Explanation :

The total number of manhours required = 30 × 7 × 18 = 3780
21 × 8 × no. of days Æ 3780/168 = 22.5 days
Note, you could have done this directly by: (30 × 7 × 18)/(21 × 8)

The incomes of A and B are in the ratio 3 : 2 and their expenditures are in the ratio 5 : 3. If each saves 1000, then, A’s income can be:

3000

4000

6000

9000

Ans .

Explanation :

Solve using options. Option (c) fits the situation as if you take A’s income as t` 6,000, B’s income
will become t` 4,000 and if they each save t` 1,000, their expenditures would be t` 5,000 , t` 3,000
respectively. This gives the required 5: 3 ratio.

Divide t` 680 among A, B and C such that A gets 2/3 of what B gets and B gets 1/4th of what C gets. Now the share of C is:

480

300

420

360

Ans .

Explanation :

Solve by options. Option (a) C = 480 fits perfectly because if C = 480, B = 120 and A = 80

A, B, C enter into a partnership. A contributes one-third of the whole capital while B contributes as much as A and C together contribute. If the profit at the end of the year is t` 84,000, how much would each receive

24,000, 20,000, 40,000

28,000, 42,000, 14,000

28,000, 42,000, 10,000

28,000, 14,000, 42,000

Ans .

28,000, 42,000, 14,000

Explanation :

A’s contribution = 33.33%
B’s contribution = 50%
C’s contribution = 16.66%
Ratio of profit sharing = Ratio of contribution
= 2 : 3 : 1
Thus, profit would be shared as : 28000 : 42000 : 14000

The students in three batches at Mindworkzz are in the ratio 2 : 3 : 5. If 20 students are increased in each batch, the ratio changes to 4 : 5 : 7. The total number of students in the three batches before the increases were

100

150

Ans .

Explanation :

2x + 20 : 3x + 20 : 5x + 20 = 4 : 5 : 7 so x = 10 and initially the number of students would be 20, 30 and 50 and a total of 100.

The speeds of three cars are in the ratio 2 : 3 : 4. The ratio between the times taken by these cars to travel the same distance is

2 : 3 : 4

4 : 3 : 2

4 : 3 : 6

6 : 4 : 3

Ans .

6 : 4 : 3

Explanation :

The ratio of time would be such that speed × time would be constant for all three. Thus if you take the speeds as 2x, 3x and 4x respectively, the times would be 6y, 4y and 3y respectively

If a, b, c and d are proportional then the mean proportion between a² + c² and b² + d² is

ac/bd

ab + cd

a/b + d/c

a²/b² + c²/d²

Ans .

ab + cd

Explanation :

Let us say we assume a = 1, b = 4, c = 3 and d = 12 we get:
a² + c² = 10 and b² + d² = 160. The mean proportional between 10 and 160 is 40. ab + cd gives us
this value and can be checked by taking another set of values to see that it still works.

A number z lies between 0 and 1. Which of the following is true?

z > \( \sqrt{z} \)

z > 1/z

\( z^3 > z^2 \)

1/z > \( \sqrt{z} \)

Ans .

1/z > \( \sqrt{z} \)

Explanation :

Option (d) is true since 1/z will be greater than 1 and square root of z would be less then 1.

2250 is divided among three friends Amar, Bijoy and Chandra in such a way that 1/6th of Amar’s share, 1/4th of Bijoy’s share and 2/5th of Chandra’s share are equal. Find Amar’s share

720

1080

450

1240

Ans .

Explanation :

Amar’s share should be divisible by 6. Option d gets rejected by this logic.
Further: A + B + C = 2250. If Amar’s share is 720 (acc. To option a) Bijoy’s share should be 480
& Chandra’s share should be 300. (Gives us a total of 720 + 480 + 300 = 1500).
But the required total is 2250 (50% more than 1500). Since all relationships are linear, 1500 will
increase to 2250 if we increase all values by 50%. Hence, Amar’s share should be 1080.

After an increment of 7 in both the numerator and denominator, a fraction changes to 3/4. Find the original fraction.

5/12

7/9

2/5

3/8

Ans .

2/5

Explanation :

Trial and error would give us 2/5 as the original fraction

The difference between two positive numbers is 10 and the ratio between them is 5 : 3. Find the product of the two numbers.

375

175

275

125

Ans .

Explanation :

Their ratio being 5:3, the difference according to the ratio is 2. But this difference is 10. To get the values, expand the ratio 5 times. This gives 25 and 15 as the required values. Hence, the product
is 375.

If 30 oxen can plough 1/7th of a field in 2 days, how many days 18 oxen will take to do the remaining work?

Ans .

Explanation :

60 oxen days = 1/7 of the field so 420 oxen days are required to plough the field. Thus, the remaining work would be 360 oxen days. With 18 oxen, it would take 20 days

A cat takes 5 leaps for every 4 leaps of a dog, but 3 leaps of the dog are equal to 4 leaps of the cat. What is the ratio of the speed of the cat to that of the dog?

11 : 15

15 : 11

16 : 15

15 : 16

Ans .

15 : 16

Explanation :

Assume that 1 cat leap is equal to 3 metres and 1 dog leap is equal to 4 metres.
Then the speed of the cat in one unit time = 3 × 5 = 15 meters.
Also, the speed of the dog in one unit time = 4 × 4 = 16 meters.
The required ratio is 15:16

The present ratio of ages of A and B is 4 : 5. 18 years ago, this ratio was 11 : 16. Find the sum total of their present ages.

105

110

Ans .

Explanation :

4x, and 5x are their current ages. According to the problem, 4x – 18 : 5x – 18 = 11:16 so x = 10 and hence the sum total of their present ages is 90 years (40 + 50).

Four numbers in the ratio 1 : 3 : 4 : 7 add up to give a sum of 105. Find the value of the biggest number.

Ans .

Explanation :

x + 3x + 4x + 7x = 105 Æ x = 7
Thus, 7x = 49

Three men rent a farm for t` 7000 per annum. A puts 110 cows in the farm for 3 months, B puts 110 cows for 6 months and C puts 440 cows for 3 months. What percentage of the total expenditure should A pay?

14.68

16.66

11.01

Ans .

14.68

Explanation :

The share of the rent is on the basis of the ratio of the number of cow months. A uses 330 cow months (110 × 3), B uses 660(110 × 6) and C uses 1320 cow months (440 × 3)
Hence, the required ratio is: 330:660:1320 = 1:2:4

10 students can do a job in 8 days, but on the starting day, two of them informed that they are not coming. By what fraction will the number of days required for doing the whole work get increased?

4/5

3/8

3/4

1/4

Ans .

1/4

Explanation :

10 × 8 = 80 man days is required for the job. If only 8 students turn up, they would require 10 days to complete the task. The number of days is increasing by 1/4

A cask contains a mixture of 49 litres of wine and water in the proportion 5 : 2. How much water must be added to it so that the ratio of wine to water may be 7 : 4?

3.5

none

Ans .

Explanation :

Initial wine = 35 litres
Initial water = 14 litres
Since, we want to create 7 : 4 mixture of wine and water by adding only water, it mean that the
amount of wine is constant at 35 litres. Thus 7 : 4 = 35 : 20. So, we need 6 litres of water

A cask contains 12 gallons of mixture of wine and water in the ratio 3 : 1. How much of the mixture must be drawn off and water substituted, so that wine and water in the cask may become half and half.

none

Ans .

none

Explanation :

If we were to draw out 4 litres of wine and substitute it with plain water, the ratio of wine to water would become 1:1. Hence, option (d) is correct.

The total number of pupils in three classes of a school is 333. The number of pupils in classes I and II are in the ratio 3 : 5 and those in classes II and III are in the ratio 7 : 11. Find the number of pupils in the class that had the highest number of pupils.

105

165

180

Ans .

Explanation :

The overall ratio is: 21:35:55. Dividing 333 in 111 parts (21 + 35 + 55) each part will be 3 and Class III will have the highest number of pupils are 55 × 3 = 165

If two numbers are in the ratio of 5 : 8 and if 9 be added to each, the ratio becomes 8 : 11. Now find the lower number

none

Ans .

Explanation :

Solve using options. 15/24 becomes 24/33 gives 8/11

Vijay has coins of the denomination of Re. 1, 50 p and 25 p in the ratio of 12 : 10 : 7. The total worth of the coins he has is t` 75. Find the number of 25 p coins that Vijay has

none

Ans .

none

Explanation :

Ratio of no. of coins = 12 : 10 : 7
Ratio of individual values of coins = 1 : 0.5 : 0.25
Ratio of gross value of coins = 12 : 5 : 1.75
= 48 : 20 : 7 so 75 
Thus, he has rs 7 in 25 paisa coins. Which means that he would have 28 such coins.

A beggar had ten paise, twenty paise and one rupee coins in the ratio 10 : 17 : 7 respectively at the end of day. If that day he earned a total of rs 57, how many twenty paise coins did he have?

114

171

Ans .

Explanation :

The ratio of the values of the three coins are:
10 × 10 : 17 × 20 : 7 × 100 = 100:340:700
= 5:17:35 is the ratio of division of value of coins.
Thus, 20 paise coins correspond to t` 17. Hence, there will be 85 coins.

A mixture contains milk and water in the ratio 5 : 1. On adding 5 litres of water, the ratio of milk to water becomes 5 : 2. The quantity of milk in the mixture is:

32.5

22.75

Ans .

Explanation :

Let the values of milk and water be 5x and x respectively. Then when we add 5 liters of water to this mixture, water would become x + 5.
Now: 5x/(x + 5) = 5:2 so x = 5. Thus, 5x is 25.

If 391 bananas were distributed among three monkeys in the ratio 1/2 : 2/3 : 3/4, how many bananas did the first monkey get?

102

108

112

104

Ans .

Explanation :

The ratio : 1/2 : 2/3 : 3/4
Converts to 6 : 8 : 9 (on multiplying by 12)
Thus, the first monkey would get (391/23) × 6 = 102 bananas

If rs 58 is divided among 150 children such that each girl and each boy gets 25 p and 50 p respectively. Then how many girls are there?

Ans .

Explanation :

For option (c), 68 girls. Hence, 82 boys
Amount with Girls = 68 × 0.25 = 17
Amount with Boys = 82 × 0.5 = 41.
Total of t` 58.
Thus, option (c) fits the conditions.

At constant temperature, pressure of a definite mass of gas is inversely proportional to the volume. If the pressure is reduced by 20%, find the respective change in volume.

–16.66%

+25%

–25%

+16.66%

Ans .

+25%

Explanation :

Since pressure and volume are inversely proportional, we get that if one is reduced by 20% the other would grow by 25%.

If the ratio of the ages of Maya and Chhaya is 6 : 5 at present, and fifteen years from now, the ratio will get changed to 9 : 8, then find Maya’s present age

Ans .

Explanation :

6x + 15 : 5x + 15 = 9 : 8
so 45x + 135 = 48x + 120
3x = 15 so x = 5
Maya’s present age = 6x = 30

3000 is distributed among A, B and C such that A gets 2/3rd of what B and C together get and C gets 1/2 of what A and B together get. Find C’s share

750

1000

800

1200

Ans .

Explanation :

From the first statement A = 1200 and B + C = 1800. From the second statement C = 1000 and A + B = 2000

A dishonest milkman mixed 1 litre of water for every 3 litres of milk and thus made up 36 litres of milk . If he now adds 15 litres of milk to the mixture, find the ratio of milk and water in the new mixture

12 : 5

14 : 3

7 : 2

9 : 4

Ans .

14 : 3

Explanation :

The initial amount of water is 9 liters and milk is 27 liters. By adding 15 liters of milk the mixture
becomes 42 milk and 9 water Æ 14:3 the required ratio.

If x varies as y, and y = 7 when x = 18, find x when y = 21

Ans .

Explanation :

x = ky gives 18 = 7k and k = 18/7
Hence, x = 18/7 × y
When y = 21, x = 54.

A varies jointly as B and C; and A = 6 when B = 3, C = 2; find A when B = 5, C = 7.

17.5

105

Ans .

Explanation :

A = K × B × C so It is known that when A = 6, B = 3 and C = 2. Thus we get 6 = 6K gives K = 1. Thus, our relationship between A, B and C becomes A = B × C. Thus, when B = 5 and C = 7 we get A = 35

If x varies as y directly, and as z inversely, and x = 14 when y = 10; find z when x = 49, y = 45.

14/10

10/14

Cannot be determined

Ans .

 Cannot be determined

Explanation :

x = ky/z
We cannot determine the value of k from the given information and hence cannot answer the
question

Questions & Answers:

Q. If a:b = 3:4 and b:c is 5:4 then find a:b:c

A. The value common to both is b so b has to made same in both ratios. The LCM of 4,5 is 20 so we convert the ratios by multiplication

a:b = (3*5):(4*5) and b:c = (5*4)/(4*4)

a:b = 15:20 and b:c = 20:16 now b value is same in both sides and so we can merge on basis of 'b'.

a:b:c = 15:20:16

Q.Find fourth proportional between 4,9,12

A. Assume fourth proportional as 'x'. so we get 4:9::12:x .

As we know from above formula: 4*x = 9*12 so we can get value of x.

Q. Third proportional to 16, 36.

A. Assume third proportional is 'x'. so we get 16:36::36:x

As we know from formula that 16*x = 36*36 we can get 'x'.

Q. Mean proportional between 0.08 and 0.18

A. mean proportional = square-root(0.08*0.18)

= square-root(8/100 * 18/100)

= square-root(144/10000)

=12/100

=0.12

Q. If x : y = 3 : 4, find (4x + 5y) : (5x - 2y)

A. 4x+5y : 5x-2y = 4(x/y)+5 / 5(x/y) - 2 we get this by dividing by y to both numerator and denominator.

substituting x/y in equation we can solve it.

Q. Divide Rs. 600 in the ratio 4:2

A. the ratio 4:2 means first value shall be 4/4+2 and second value shall be 2/4+2 i.e. 600 has to be divided into 4/6 and 2/6 parts. so we get 400,200.

Q. A bag contains 50 p, 25 P and 10 p coins in the ratio 5: 9: 4, amounting to Rs. 206. Find the number of coins of each type.

A. Since ratio is 5:9:4 this means actual number of coins can be 5x, 9x and 4x.

We know that, 5x*(1/2) + 9x*(1/4) + 4x*(1/10) = 206 as we converted the paisa values to rupee.

Q. A mixture contains alcohol and water in the ratio 4 : 3. If 5 litres of water is added to the mixture, the ratio becomes 4: 5. Find the quantity of alcohol in the given mixture.

A. Alcohol and water were 4x and 3x and now 4x / 3x+5 = 4 / 5.

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Chapter 8: RATIO AND PROPORTION

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Proportions

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