Word Problems on Ratio and Proportion 2 :
This is the continuity of our web content given on "Word Problems on Ratio and Proportion 1".
Before look at the problems, if you would like to know the shortcuts on ratio and proportion,
Problem 1 :
Two numbers are respectively 20% and 50% are more than a third number, Find the ratio of the two numbers.
Let "x" be the third number.
Then, the first number is
= (100+20)% of x
= 120% of x
The second number is
= (100+50)% of x
= 150% of x
The ratio between the first number and second number is
= 1.2x : 1.5x
= 1.2 : 1.5
= 12 : 15
= 4 : 5
Hence, the ratio of two numbers is 4 : 5.
Problem 2 :
The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio, the liquids in both the vessels be mixed to obtain a new mixture in vessel C consisting half milk and half water ?
Vessel A :
[4:3 ------> 4+3 = 7, M----> 4/7, W----> 3/7]
Let "x" be the quantity of mixture taken from vessel A to obtain a new mixture in vessel C.
Quantity of milk in "x" = (4/7)x = 4x/7
Quantity of water in "x" = (3/7)x = 3x/7
Vessel B :
[ 2:3 ------> 2+3 = 5, M----> 2/5, W----> 3/5 ]
Let "y" be the quantity of mixture taken from vessel B to obtain a new mixture in vessel C.
Quantity of milk in "y" = (2/5)y = 2y/5
Quantity of water in "y" = (3/5)y = 3y/5
Vessel A and B :
Quantity of milk from A and B is
= (4x/7) + (2y/5)
= (20x + 14y) / 35
Quantity of water from A and B is
= (3x/7) + (3y/5)
= (15x + 21y) / 35
According to the question, vessel C must consist half of the milk and half of the water.
That is, in vessel C, quantity of milk and water must be same.
Quantity of milk in (A+B) = Quantity of water in (A+B)
(20x+14y) / 35 = (15x+21y) / 35
20x + 14y = 15x+21y
5x = 7y
x/y = 7/5
x : y = 7 : 5
Hence, the required ratio is 7 : 5.
Problem 3 :
A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. From the vessel, 10 liters of the mixture is removed and replaced with an equal quantity of pure milk. Find the ratio of milk and water in the final mixture obtained.
[ 3 : 2 ------> 3+2 = 5, M----> 3/5, W----> 2/5 ]
In 20 liters of mixture,
no. of liters of milk = 20 ⋅ 3/5 = 12
no. of liters of water = 20 ⋅ 2/5 = 8
Now, 10 liters of mixture removed.
In this 10 liters of mixture, milk and water will be in the ratio 3:2.
No. of liters of milk in this 10 liters = 10 ⋅ 3/5 = 6
No. of liters of water in this 10 liters = 10 ⋅ 2/5 = 4
After removing 10 liters (1st time),
No. of liters of milk in the vessel = 12 - 6 = 6
No. of liters of water in the vessel = 8 - 4 = 4
Now, we add 10 liters of pure milk in the vessel,
After adding 10 liters of pure milk in the vessel,
No. of liters of milk in the vessel = 6 + 10 = 16
No. of liters of water in the vessel = 4 + 0 = 4
After removing 10 liters of mixture and adding 10 liters of pure milk, the ratio of milk and water is
= 16 : 4
= 4 : 1
Hence, the required ratio is 4 : 1.
Problem 4 :
If $782 is divided among three persons A, B and C in the ratio 1/2 : 2/3 : 3/4, then find the share of A.
Given ratio ---> 1/2 : 2/3 : 3/4
First let us convert the terms of the ratio into integers.
L.C.M of denominators (2, 3, 4) = 12
When we multiply each term of the ratio by 12, we get
12 ⋅ 1/2 : 12 ⋅ 2/3 : 12 ⋅ 3/4 ------> 6 : 8 : 9
From the ratio 6 : 8 : 9,
Share of A = 6x
Share of B = 8x
Share of C = 9x
We know that,
Share of A + Share of B + Share of C = 782
6x + 8x + 9x = 782
23x = 782
x = 34
Share of A is
= 6 ⋅ 34
Hence, the share of A = $ 204.
Problem 5 :
An amount of money is to be divided among P, Q and R in the ratio 3 : 7 : 12. The difference between the shares of P and Q is $2400. What will be the difference between the shares of Q and R ?
From the given ratio 3 : 7 : 12,
Share of P = 3x
Share of Q = 7x
Share of R = 12x
Difference between the shares of P and Q is $ 2400
Share of Q - Share of P = 2400
7x - 3x = 2400
4x = 2400
x = 600
Difference between the shares of Q and R is
= 12x - 7x
= 5 ⋅ 600
Hence, the difference between the shares of Q and R is $3000.
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