APTITUDE PREPARATION: Ratio and Proportion

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It is the relation with respect to magnitude between two similar quantities.
The ratio ‘a’ with respect to ‘b’ may be represented by (a/b) or (a:b).
As a general practice, (a/b) is kept such that ‘a’ and ’b’ are PRIME NUMBERS.
The value of the fraction (a/b) does not change even if you multiply (or) divide BOTH the numerator and denominator by the same number.
Example: (a/b)=(3a)/(3b)=(a/5)/(b/5)…
But, the value changes if you add (or) subtract BOTH the numerator and denominator by the same number.
Example: (a/b)≠(a+3)/(b+3)≠(a-5)/(b-5)…
If the ratio is to be represented in terms of percentage, multiply (a/b) by 100.


It is an equality of two ratios.
Example: 10:30=1:3 (or) (10/30)=(1/3)
Each term, 10, 30, 1 and 3 is called a proportional.
If we have four proportionals in order then, “the product of end proportionals equals the product of middle two proportionals”.
That is, (10×3)=(30×1).

Example 1:

If 12 men can do a piece of work in 45 days, in how many days will 27 men do it?


In such problems, the rate of doing work is assumed to be same and constant for each man and for each day.
The work remains the same and equal to “12×45=540 man days”.
Hence, 27x =540, where, x=the number of days required to complete the same work by 27 men.
x=540/27=20 days.

Other Method:

As the number of men increases, the number of days will decrease which means ‘the direct ratio of number of men is equal to the inverse ratio of the number of days’.
12:27=x:45 (or) 12/27=x/45
Note that in the Left Hand Side (LHS), we keep the ratio of the number of men (the known data) and at the Right Hand Side (RHS), we keep the INVERSE ratio of number of days (with the un-known ‘x’).
By cross multiplying, 27x=12×45 and x=20 days.

Example 2:

If 8 boys or 12 girls can do a piece of work in 25 days, in how many days can the same work be done by 6 boys and 11 girls working together?


Work done by 8 boys = work done by 12 girls (or)
1 boy = (12/8) = (3/2) girls
Therefore, 6 boys = 6 x (3/2) girls = 9 girls
Number of persons employed = 6 boys + 11 girls = 9 girls +11girls = 20 girls
The work remains the same and equal to “12×25=300 girl days”.
Hence, 20x =300, where, x=the number of days required to complete the same work by 20 girls.
x=300/20=15 days.

Example 3:

Rs 210 is divided among A, B and C so that if A gets Rs 2, B gets Rs 3 and if B gets Rs 4, C gets Rs 5. What is the share of A, B and C?


Such problems can be tackled by the following method.
Share 2 : 3
Share 4 : 5
Since B is common, make the element same by taking LCM of 3 and 4 = 12.
Now, share of A = 2x(12/3)=8 ….. 3 is the share of B with respect to A
Share of C = 5x(12/4)=15 …… 4 is the share of B with respect to C
Therefore the related shares:
Shares 8 : 12 : 15
Total number of shares =(8+12+15) = 35
Total amount = 210
Share of A = 8x(210/35) = Rs. 48
Share of B = 12x(210/35) = Rs. 72
Share of C = 15x(210/35) = Rs. 90.

Example 4:

4 men can earn as much as 6 women, 2 women as much as 3 boys and 4 boys as much as 5 girls. If a girl earns Rs. 100 a day, what is the earning of a man?


We have to relate the earnings of a man and a girl.
x = 1 man’s earning
4 men = 6 women
2 women = 3 boys
4 boys = 5 girls
1 girl’s earning = 100 Rs.
Relating all:
x x 4 x 2 x 4 x 1 = 1 x 6 x 3 x 5 x 100 and,
x = (9000/32)=Rs. 281.25.

Other Method:

We have to relate the earnings of a man and a girl.
6 women = 4 men and one woman = (4/6) = (2/3) man
3 boys = 2 women and one boy = (2/3) woman
5 girls = 4 boys and one girl =(4/5) boy.
Therefore, one girl = (4/5) x (2/3) x (2/3) man = (16/45) man
That is, one man = (45/16) girls
Earning of one girl = 100
Earning of one man = 100 x (45/16) = Rs. 281.25.

Example 5:

A cyclist travels for 10 hours, the first half at a speed of 12 km/hr and the second half at 18 km/hr. Find the total distance travelled.


Distance travelled at 12 km/hr = distance travelled at 18 km/hr
Average speed = (2.X.Y)/(X+Y) = (2x12x18)/(12+18) = 14.4 km/hr
Total distance travelled in 10 hours = (10 x 14.4) = 144 km.

Example 6:

Rs 664 to be divided among three persons A, B and C. Four times of share of A is equal to 5 times of B and 7 times of C. Find the share of A.


4A=5B=7C and (A+B+C)=664.
As we have to estimate A, convert others in terms of A.
B=A(4/5) and C=A(4/7)
A+A(4/5)+A(4/7)=664 (or) (35A+28A+20A)/35=664 …. LCM of 5 and 7=35
Share of A=280.

Example 7:

A boat moves upstream at a speed of 5 km/hr and downstream at 8 km/hr. What will be its speed in still water? Also find the speed of the stream.


Let the speed of the stream (river) is v km/hr and speed of the boat in still water is x km/hr.
Upstream: the speeds will oppose each other and resultant speed = (x-v) = 5
Downstream: the speeds will get added and resultant speed = (x+v) = 8
Adding both we get, 2x=13 and x=6.5 km/hr
Speed of the stream, v=(x-5)=(6.5-5)=1.5 km/hr.

Example 8:

The ratio of apples and oranges in a basket is 5:9. If 10 apples are removed, the ratio becomes 1:3. Find the number of oranges in the basket.


Let the number of apples = 5x and number of oranges = 9x.
After 10 apples are removed, ratio of apples to oranges = (5x-10):(9x) = (1:3) or
(5x-10)/(9x) = (1/3)
Cross multiplying, 3(5x-10)=9x
6x=30 and x=5.
Number of oranges in the basket = 9x = 9×5 =45.

Example 9:

Five years ago, the ratio of Anitha’s age to Beena’s age was 2:3. After 10 years from now, the ratio of their ages will be 5:6. Find their present ages.


Let the present age of Anitha is A and that of Beena is B.
Five years ago:
(A-5):(B-5)=(2:3) or (A-5)/(B-5)=(2/3) or 3(A-5)=2(B-5) or 3A-15=2B-10
3A-2B=5 … (i)
After ten years from now:
(A+10):(B+10)=(5:6) or (A+10)/(B+10)=(5/6) or 6(A+10)=5(B+10) or 6A+60=5B+50
5B-6A=10 … (ii)
Multiplying equation (i) by 2 and adding to equation (ii):
B=20 which is the age of Beena.
3A-(2×20)=5 and age of Anitha, A=15.

Example 10:

The price of a pen is Rs 7 and scale is Rs 4. If Anand spent 30 rupees in purchasing both the items, find the number of pens and scales purchased.


For such problems, we can form a simple equation.
Let the number of pens purchased is P and scale is S.
The problem now is we have only one equation but, two unknowns.
However, we have one additional information; P and S are only integers (whole numbers).
P = 1, 2, 3, 4… and 7P=7, 14, 21, 28…
S=1, 2, 3, 4, 5, 6, 7… and 4S=4, 8, 12, 16, 20, 24, 28…
To have a combination of 7P+4S=30, the only possible combination is: 7P=14 and 4S=16.
That is P=2 and S=4.

About Author


a former Research Scientist Government of India, presently working as a Professor in the Dept of Mechanical Engineering, Authored books on Kinematics and Dynamics of Machinery, Also Published Numerous Research papers moreover a Geek in Mathematics

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