Sum, product and proportion of ratios

Friday, June 28, 2013
Problems on Ratio and proportion

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Ratio

The ratio of a is to b is defined as a/b, where a and b are two non zero quantities.

Important Notes about ratio
1. In the ratio a: b, a is called antecedent and the b is called consequent.
2. Both a and b should be of the same kind.
3. If both antecedent and the consequent of the ratio a:b are multiplied or divided by
the same non zero quantity then the ratio is not changed. It remains the same.
i.e. a:b = na:nb or a:b = ma:mb

1. a:b is said to be in lowest form when a and b are co-primes to each other or in other
   words there is no common factor between a and b except for 1.

How to compare two ratios a:b and p:q

Let a:b and p:q are the two ratios to be compared then,
1. a:b > p:q if aq > pb
2. a:b = p:q if aq=pb
3. a:b < p:q if aq < pb

Compound ratio

Compounded ratio is obtained by multiplying all the antecedents of all the ratios involved and multiplying all the consequent of all the ratios involved and expressing them a numerator and denominator of a fraction.
Example:

For example the compounded ratio of a:b, c:d, e:f, g:h is
(a x c x e x g)/(b x d x f x h)

Proportion

4 quantities are said to be in proportion if a:b = c:d.
This proportion is expressed as a:b::c:d

Terms related to proportion

  1.    In the proportion a:b::c:d ,

a is the first term of the proportion
b is the second term of the proportion
c is third term of the proportion and
d is the fourth term of the proportion

 2.    In the proportion a:b::c:d,
a and d are called the extremes and b and c are the means

 3.    Continued proportion
Three quantities a, b and c are said to be in continued proportion if
a:b::b:c
or
a/b = b/c
or
(ac) = (b)2

In continued proportion,
a is the first proportional, b is the second proportional or the mean proportional and c is the third proportional. There is no fourth proportional in continued proportion.

Difference between ratio and the proportion

1.In ratio two quantities of the same kind are involved whereas in the proportion, there are four quantities.

1. Ratio a:b is read as a is to b. means if quantity 'a' doubled then quantity 'b' is to be doubled.
2. In proportion a:b::c:d means what a is to b is exactly what c is to d means,
   i. if we double the quantity represented by 'a' then either we have to double the quantity
   'b' or halve the quantity 'd' to maintain the same proportion. The same is the with the quantity 'b' and 'c'.

Solved Problems

1. If A:B = 3:4 B:C =8: 10 and C:D =15:17 then find A:B.C.D

(a) 9:12:15:17 (b) 6:12:20:32
(c) 9:22:32:34 (d) 7:19:27:42

Answer: (a) 9:12:15:17
Solution:
If there are three ratio a:b and c:d and e:f
Then the ratio among these 4 quantities is given by ace: bce : bde : bdf
Using this the required ratio is:
3 x 8 x 15:4 x 8 x 15: 4 x 10 x 15 : 4 x 10 x 17
= 9 : 12 : 15 : 17

1. Find the ratio compounded of the four ration:
   4:3, 9:13, 26:5, and 2:15

Solution:
The ratio compounded = (4 x 9 x 26 x 2)/(3 x 13 x 5 x 15) = 16/25

1. An amount of Rs 750 is distributed among A,B and C in the ratio of 4:5:6 what is the share of B ?
   (a) 500 (b) 250
   (c) 1050 (d) 370

Answer: (b) 250
Solution:
In such question, the first we add all the part. we get 4 + 5 + 6 = 15. Out these, B got 5 parts.
Thus, B got (5/15) x 750 = 250

1. Two number are in the ratio of 9: 14 if the larger number is 55 more than the smaller number.Find the numbers.
   (a) 99,154 (b) 48:88
   (c) 17:96 (d) 59:74

Answer: (a) 99,154
Solution:
The smaller number be x, then the larger number be (x+ 55) . Then according to the question
x / (x+55) = 9/14
Thus,
14x = 9(x + 55) ( multiplying the LHS and RHS of the above equation by (x+55)/14 }
=> 14x = 9x + 495
=> 5x = 495
=> x = 99
Thus the number are 55 and 154.

5.The sum of the three is 98 if the ratio between the first and the second be 2:3 and that between the second and the third be 5:8, then find the second number .
(a) 90 (b) 45
(c) 77 (d)30

Answer: (d)30
Solution:
Let the three numbers be A , B and C.
Then the ratio of A to B is = 2:3 or 10:15
And the ratio of B to C is = 5 : 8 15:24
Thus, the ratio of A:B:C is 10:15:24.
Again, Given that A + B + C = 98.
Therefore, A = (10/49) x 98 = 20
B = (15/49) x 98 = 30

1. A sum of money is divided between two people in the ratio of 3:5. If the share of one person is Rs 20 less than that of other,find the sum .
   (a) 88 Rs (b) 76 Rs
   (c) 85 Rs (d) 80 Rs

Answer: (d) 80 Rs
Solution:
Let the share of one person be x then the share of other person is x + 20.
It is given that
(x / x +20) = 3:5
=> 5x = 3(x + 20)
=> 5x = 3x + 60
=> 2x = 60
=> x = 30
And ( x +20) = 30 + 20 = 50
And therefore the sum is 50 + 30 = Rs . 80

1. The ratio between two number is 3:4 If each number be increased by 6 the ratio becomes 4:5 find the number.
   (a) 43,16 (b) 18,24
   (c) 22,12 (d) 32, 22

Answer: (b) 18,24
Solution:
Let the number be 3x and 4x. Then (3x/4x) = 3/4
Again, (3x+6/4x+6) = 4/5 ( given that when both numbers are increased by 6 the ratio becomes 4/5)
=> 5(3x + 6 ) = 4(4x + 6)
=> 15x + 30 = 16x + 24
=> 15x - 16x = 24 - 30
=> -x = -6
=> x = 6

Thus, the first number is 6 x 3 = 18
and the seconder number is 4 x 6 = 24

1. The ratio between two number is 3: 4. If each number be increased by 2, the ratio becomes 7:9. Find the number.
   (a) 43, 16 (b) 22,10
   (c) 12, 16 (d) 18,9

Answer: (c) 12, 16
Solution:
Let the number be 3x and 4x. Then (3x/4x) = 3/4
Again, (3x + 2/4x+2)=7/9 (given that when both numbers are increased by 2 the ratio becomes 7/9)
=> 9(3x + 2 ) = 7(4x + 2)
=> 27x + 18 = 28x + 14
=> 27x - 28x = 14 -18
=> - x = -4
=> x = 4

Thus, the first number is 4 x 3 =12
and the seconder number is 4 x 4 =16

1. The students in three classes are in the ratio 2:3:5. If 20 students are increased in each class the ratio changes to 4:5:7. what was the total number of students in the three classes before the increases?
   (a) 120 (b) 85
   (c) 72 (d) 100

Answer: (d) 100
Solution:
Let the number be 2x,3x and 5x.
It is given that if 20 students in each class are increased then the ratio becomes 4:5:7
It can written as (2x +20) : (3x +20) : (5x + 20) = 4 : 5 : 7
Taking the first two ratio
we have,
(2x +20) / (3x +20) = 4/5
=> 5(2x +20) ) = 4(3x +20)
=> 10x + 100 = 12x + 80
=> 10x - 12x = 80 - 100
=> -2x = -20
=> x =10
Therefore, the student in each class before the increase are 2x10, 3x10, 5x10 = 20,30,50
And total student = 100

1. The ratio between two number is 3:4 ,if each number be increased by 9, the ratio becomes 18:23 find the sum of the number
   (a) 135 (b) 105
   (c) 155 (d) 165

Answer: (b) 105

Solution: Let the two numbers be 3x and 4x.

When they are increased by 9 they become 3x + 9 and 4x + 9.

It is given that the ratio is 18:23

Thus, 3x + 9/4x + 9 = 18:23

23(3x + 9) = 18(4x + 9)

69x + 207 = 72x + 162

69x – 72x = 162- 207

-3x = -45

X = 15

Thus two numbers are 3x15 = 45 and 4 x 15 = 60
And the sum is 60+45 = 105

1. Find the number which, when added to the terms of the ratio 11:23 makes it equal to the ratio 4:7
   (a) 15 (b) 5
   (c) 25 (d) 20

Answer: (b) 5
Solution:

Let the number be x which when added to each terms of the ratio 11:23 makes it equal to 4:7.

Thus, (11 + x)/(23 + x ) = 4/7

7(11 + x ) = 4(23 + x)

77 + 7x = 92 + 4x

7x-4x = 92 – 77

3x = 15

X = 5

Thus, 5 should be added.

1. Find the number which, when subtracted from the terms of ratio 11:23 makes it equal to the ratio 3:7
   (a) 2 (b) 8
   (c) 16 (d) 6

Answer: (a) 2
Solution:

Let the number be x which when subtracted from each terms of the ratio 11:23 makes it equal to 3:7.

Thus, (11 - x)/(23 - x ) = 3/7

7(11 - x ) = 3(23 - x)

77 - 7x = 69 - 3x

-7x+3x = 69 – 77

-4x = -8

X = 2

Thus, 2 should be subtracted

1. In 40 liters mixture of milk and water the ratio of milk and water is 3: 1 how much water should be added in the mixture so that the ratio of milk to water becomes 2:1 ?

(a) 500ml (b) 20 liters

(c) 5 liters (d) 10 liters

Answer:
Solution:

Let x liters of water be added to the mixture to make the ratio of milk to water to become 2:1.

Thus,

   30/(10 + x) = 2/1

30(1) = 2(10+x )

30 = 20 + 2x

10 = 2x

X = 5 liters

1. A mixture contains milk and water in the ratio of 3:2 liter of water is added to the mixture, milk, milk and water in the mixture become equal find the quantities of milk and water in the mixture .

(a) 12, 8 liter (b) 4,3 liter

(c) 12, 6 liters (d) 10,8 liters

Answer: (a) 12, 8 liter

Solution:

Let quantities of milk and water in the mixture be 3x and 2x. Then if 4 liters of water is added to the mixture the ratio of milk and water become 1:1.

It can be written as (3x): (2x + 4) = 1/1

Thus, 3x = 2x +4

        x = 4

Therefore, the milk in the mixture is 4x3 = 12 litres and quantity of water = 4x2 = 8 liters

1. A bag contain equal number of one rupee, 50 paisa and 25 paisa coins respectively if the total value of rs 35, how many coins of each type are three?
   (a) 40 (b)80
   (c) 160 (d) 20

Answer: (d) 20
Solution:

50 paisa = .5 rupee. .25 paisa = .25 rupees.

Now, given that X (1 + 0.50 + 0.25) = 35

                         => x = 35/(1.75)   = 20

Thus, there are 20 coins of each type.

1. The speed of these cars is in the ratio of 2:3:4. what is the ratio among the time taken by these cars to travel the same distance ?
   (a) 7:8:12 (b) 3:6:9
   (c) 9:3:6 (d) 6:4:3

Answer: (d) 6:4:3
Solution:

Let the distance be 2x3x4 = 24 km.

Then, ratio of time taken by each car is 24/2: 24/3: 24/4

12:8:6

or 6:4:3

1. The incomes of A of B are in the ratio 3:2 and their expenditure are in the ratio 5:3 if each saves rs 2000, what is their income ?
   (a) 32000 (b) 20000
   (c) 1190 (d) 8000

Answer: (d) 8000
Solution:
Let the income be 3x and 2x. It is given that the saving of each is Rs. 2000.

Then, their expenditures are 3x – 2000 and 2x – 2000

Again, (3x – 2000)/(2x – 2000) = 5/3

 => 3(3x – 2000) = 5(2x – 2000)

 =>  9x – 6000    = 10x – 10000

 => 9x -10x        = -10000+ 6000

 =>                -x = -4000

 =>                 x = 4000

Therefore, their salaries are 3 x 4000 = 12000 and 2 x 4000 = 8000

1. The contents of two vessels containing water and milk are in the ratio 1:2 and 2:5 are mixed in the ratio 1:4, the resulting mixture will have water and milk in the ratio.
   (a) 11:22 (b) 71:58
   (c) 31:74 (d) 91:49

Answer:(c) 31:74
Solution:
In vessel 1
The fraction of water = 1/3
The fraction of milk = 2/3
In vessel 2
The fraction of water = 2/7
The fraction of milk = 5/7

Now, From vessel 1, 1/5 of the mixture is taken out. Thus water taken out = (1/5)(1/3) = 1/15 and milk taken out = (1/5)(2/3) = 2/15
And from vessel 2, 4/5 of the mixture is taken out. Thus water taken out = (4/5)(2/7) = 8/35 and milk taken out = (4/5)(5/7) = 20/35

Thus Ratio of water and milk in resulting solution is (1/15 + 8/35) : (2/15+20/35) = 31/105 : 74/105
= 31:74

1. Find the mean proportion of 9, 16
   (a)12 (b) 144
   (c) 25 (d) 4/3

Answer:(a)12
Solution:
Mean proportion of 9 and 16 = √ (9 x 16) = 3 x 4 = 12

1. Find the third proportional to 15 and 20
   (a)80/3 (b) 55/6
   (c) 37/2 (d) 71/2

Answer:(a)80/3
Solution:
Here, given one ratio
15:20 and 20:x . We have to find the value of x.
Now 15/20 = 20/x
=> x = (20x20)/ 15 = 400/15 = 80/3

1. Find the fourth proportional to the number 6,8 and 15
   (a) 48 (b) 48
   (c) 41 (d) 20

Solution:
Let the fourth proportional be F.
Then,
6/8 = 15/F
=> F = (15 x 8 )/6 = 120/6 = 20

1. The ratio of three number is 3:4:5 and the sum of their squares is 1250. The sum of their number is :
   (a) 30 (b) 50
   (c) 60 (d) 90

Answer: (c) 60
Solution:Let the numbers be 3x, 4x and 5x.
Then,
Given that sum of their squares is 1250
=> (3x)² + (4x ) ² + (5x)² = 1250
=> 9x² + 16 x² + 25x² =1250
=> 50x² = 1250
=> x² = 25
=> x = 5
Thus, the number are 3 x 5 = 15, 4 x 5 = 20 and 5 x 5 = 25
Ad therefore their sum = 60

1. The ratio of three number is 3:4:7 and product 18144 the number are :
   (a) 9,12,21 (b) 15,20,25
   (c) 18,24,42 (d) None of the

Answer:(c) 18,24,42
Solution: Let numbers are 3x, 4x and 7x
Given that the product of the numbers is 18144
Therefore, (3x)(4x)(7x) = 18144
=> 84x³ = 18144
=> x ³ = 216
=> x = 6
Therefore the numbers are : 3 x 6 = 18
4 x 6 = 24
and 7 x 6 = 42

1. The value of K that must be added to 7,16,43,79 so that they are in proportion is.
   (a) 7 (b) 5
   (c) 9 (d) None of the

Answer: (b) 5
Solution:
k is added to 7,16,43,79 to make them in proportion then
(7 + k)/(16 + k) = (43 + k)/(79 + k)
=> (7 + k)(79 + k) = (43 + k)(16 + k)
=> 553 + 7k +79k + k² = 688 + 43k + 16k + k²
=> 553 + 86k = 688 + 59k
=> 553 - 688 = -86k + 59k
=> -135 = -27k
=> k = 5

1. What should be subtracted from 15,28,20 and 38 so that the remaining number may by proportional?
   (a) 2 (b) 4
   (c) 6 (d) None of the

Answer: (a) 2
solution:
Let k be subtracted from 15,28,20,38 such that yhey may be proportional. then
(15 - k)/(28-k) = (20 - k)/(38 - k)
=> (15 - k)( 38-k) = (20-k)(28-k)
=> 570 - 53k + k = 560 - 48k +k
=> 570 - 53k = 560 - 48k
=> 570- 560 = 53k - 48k
=> 10 = 5k
=> k = 2

1. Two number are respectively 20% and 50% more than a third number. The ratio of two numbers is :
   (a) 2:5 (b) 3:5
   (c) 4:5 (d) 6:7

Answer:(c) 4:5
Solution:
Let the third number be x.
Then the first number = (1.2) x
And the second number = (1.5)x
Therefore their ration = (1.2x)/(1.5x) = 12/15=4/5 or 4 : 5

1. A and B together have Rs 1210, if 4/15 of A's amount is equal to (2/5) of B's amount. How much amount does B have ?
   (a) Rs.460 (b) Rs.484
   (c) Rs.550 (d) Rs.664

Answer: (b) Rs.484
Solution:
Let A's amount to be a.
And B's amount be b.
Given that (4/15)a = (2/5)b => a = (2/5)(15/4)b = (3/2)b
Thus, a + b = 1210
=> (3/2)b + b = 1210
=> (5/2)b = 1210
=> b = (1210)(2/5) = (242)x 2 = 484

1. Rs. 900 is to be distributed among A,B,C and in the proportion 2:3:4 how much C get?
   (a) Rs.400 (b) Rs.450
   (c) Rs.540 (d) Rs.None of the

Answer:(a) Rs.400
Solution:
Share of C = (4/9)x 900 = 4 x 100 = 400

1. If x:y is 9:7 then (x+y):(x-y) is ......
   (a)8:1 (b) 1:8
   (c)4:1 (d) 1:4

Answer:(a)8:1
Solution:
(x+y) :(x-y) = {(x/y) + 1 } /{(x/y) -1} = {9/7 +1}/{9/7 - 1} = 16/2 = 8/1 or 8:1

1. The ratio of the present ages of Sunita and Vinita is 4:5. Six years hence, the ratio of their ages will be 14:17. What will be their ages 12 years hence.
   (a) 13:19 (b)16:19
   (c) 17:19 (d) 15:19

Answers: (b)16:19
Solution:
The word to understand here is 'hence'. 'Hence' in such questions means 'after' the years mentioned.
Let the present ages of Sunita and Vinita be 4x and 5x. Now, according to the question,
(4x + 6)/(5x + 6 ) = 14/17
=> 68x + 102 = 70x + 84
=> 102-84 = 70x - 68x
=> 18 = 2x
=> x = 9
Therefore, Sunita's age = 4x9 = 36 years
Vinita's age = 5x9 = 45
After, 12 years Sunita's age = 36+ 12 = 48 years
12 years Vinita's age = 45+ 12 = 57 years
And ratio of their ages = 48/57 = 16/19 or 16:19

1. The ratio of the two numbers is 3:4 and their LCM is 180. The second number is
   (a) 45 (b) 90
   (c) 30 (d) 60

Answer: (d) 60
Solution:
The point to understand here is that,
when we find ratio of two numbers then the HCF of the numbers is eliminated while reducing the ratio to the lowest term. Thus, the two numbers are 3 x HCF and 4 x HCF
Now, Let the HCF be x. Then using the formula
LCM x HCF = First Number and Second Number
180 x HCF = (3xHCF)(4xHCF)
=> 180/(3x4) = HCF
=> 15 = HCF

Therefore, the second number is 4 x15 = 60

1. From each of the two given unequal numbers, half the smaller number is subtracted. Then, of the resulting numbers, the larger one is five times than the smaller one. Then the ratio of the larger to smaller one is
   (a) 2 : 1 (b) 3 : 2
   (c) 3 : 1 (d) 1 : 4

Answer:(c) 3 : 1
Solution:
Let the numbers be X and Y and X >Y.
Then according to the conditions given in the problem, we have
(X-Y/2)/(Y/2) = 5/1
=> (2X-Y/(Y)) = 5/1
=> 2X-Y = 5Y
=> 2X = 6Y
=> X/Y = 6/2 = 3/1
or X:Y = 3:1

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