Ratio and Proportion


Ratio and proportion is the heart of arithmetic.  If you understand this chapter properly you can solve virtually any problem in arithmetic.
If two numbers are in the ratio 2:3 means for every two units of the first number, second has 3 units.  This is a mere comparison between numbers, and actual numbers may be way bigger than these numbers.  If you multiply or divide a ratio the comparison does not change. i.e., 2:3 is same as 4:6.
If two numbers are in the ratio a:b then this ration has to be multiplied with a number K, to get actual numbers.  This K is called multiplication factor (MF)
If two ratios are equal then we say they are in proportion.  then $a:b::c:d \Rightarrow a \times d = b \times c$
 or $\displaystyle\frac{a}{b} = \frac{c}{d} \Rightarrow a \times d = b \times c$.

Chain Rule:

Chain rule is comes in handy when there are many variables need to compare with the given variable.  We can understand this rule by observing a practice problem.

Solved Example:
If 12 carpenters working 6 hours a day can make 460 chairs in 24 days, how many chairs will 18 carpenters make in 36 days, each working 8 hours a day?
Let us prepare small table to understand the problem.
Now with repect to the Chairs we need to understand how each variable is related.
If the number of men got increased (i.e., 12 to 18), do they manufacture more chairs or less chairs is the question we have to ask ourselves. If the answer is "more" then the higher number between 12, 18 will go to the numerator and other will go to denominator and vice versa. Here answer is "more"
So $460 \times \displaystyle\frac{{18}}{{12}}$
Next we go to Hours. If the number of hours they work each day got increases then do they manufactures more chairs or less chairs? Answer is more
So $460 \times \displaystyle\frac{{18}}{{12}} \times \frac{8}{6}$
Last, If the number of day they work increases then .... answer is more.
So $460 \times \displaystyle\frac{{18}}{{12}} \times \frac{8}{6} \times \frac{{36}}{{24}} = 1380$

MCQ's (Solved Examples)

1. The ratio between two numbers is 5 : 8. If 8 is subtracted from both the numbers, the ratio becomes 1 : 2. The smaller number is:
a. 10
b. 15
c. 20
d. 25
Correct Option: C
Let the numbers be 5x and 8x.
$ \Rightarrow \dfrac{{5x - 8}}{{8x - 8}} = \dfrac{1}{2}$
⇒ 10x - 16 = 8x - 8
⇒ x = 4.
Therefore, The numbers are 4 x 5 and 4 x 8 i.e. 20 and 32.

2. The ratio between Sumit's and Prakash's age at present is 2:3. Sumit is 6 years younger than Prakash. The ratio of Sumit's age to Prakash's age after 6 years will be :
a. 1 : 2
b. 2 : 3
c. 3 : 4
d. 3 : 8
Correct Option: C
Explanation:
Let their ages be 2x and 3x years.
3x-2x=6 or x=6
Sumit's age = 12 years, Prakash's age = 24 years
Ratio of their ages = 18 : 24 = 3 :4.

3. The ratio between the ages of Kamala and Savitri is 6:5 and the sum of their ages is 44 years.  The ratio of their ages after 8 years will be :
a. 5 : 6
b. 7 : 8
c. 8 : 7
d. 14:13
Correct Option: C
Explanation:
Let their ages be 6x and 5x years
6x+5x=44 or x = 4
So their present ages are 24 years & 20 years
Ratio of their ages after 8 years = 32 : 28 = 8: 7

4. In a mixture of 60 litres, the ratio of milk and water is 2:1.  What amount of water must be added to make the ratio 1:2?
a. 42 litres
b. 56 litres
c. 60 litres
d. 77 litres
Correct Option: C
Explanation:
Milk = $(60 \times \displaystyle\frac{2}{3})$ Litres = 40 litres
Water = (60-40)litres = 20 litres
Let x liters of water to be added to make it 1 : 2
$\displaystyle\frac{{40}}{{20 + x}} = \displaystyle\frac{1}{2} \Rightarrow 20 + x = 80$ or x = 60
Hence, water to be added = 60 litres

5. A's money is to B's money as 4:5 and B's money is to C's money as 2:3. If A has Rs.800, C has 
a. Rs.1000
b. Rs.1200
c. Rs.1500
d. Rs.2000
Correct Option: C
Explanation:
A:B = (4:5)×2 = 8 : 10
 B:C = (2:3)×5 = 10 : 15
($\because$ As B is common in both the ratios, we have to multiply these two ratios with suitable numbers to make B equal)
A:B:C =8:10:15
Let A, B, C has 8x, 10x, 15x.
Given 8x = 800 $\Rightarrow $ x = 100.
C = 15x = 15×100 = 1500

6. 15 litres of a mixture contains 20% alcohol and the rest water. If 3 litres of water is mixed in it, the percentage of alcohol in the new mixture will be :
a. 17
b. $16\displaystyle\frac{2}{3}$
c. $18\displaystyle\frac{1}{2}$
d. 15
Correct Option: B
Explanation:
Alcohol = $\left[ {\displaystyle\frac{{20}}{{100}} \times 15} \right]$ litres = 3 litres, Therefore water = 12 litres.  If 3 litres of water is mixed, new mixture contains alcohol = 3 litres, water = 15 litres.
Percentage of alcohol in new mix
=$\left[ {\displaystyle\frac{3}{{18}} \times 100} \right]$% = $16\displaystyle\frac{2}{3}$%

7. Vinay got thrice as many marks in Maths as in English.  The proportion of his marks in Maths and History is 4:3. If his total marks in Maths, English and History are 250, what are his marks in English ?
a. 120
b. 90
c. 40
d. 80
Correct Option: C
Explanation:
M=3E and $\displaystyle\frac{M}{H} = \displaystyle\frac{4}{3}$
H=$\displaystyle\frac{3}{4}M = \displaystyle\frac{3}{4} \times 3E = \displaystyle\frac{9}{4}E$
Now, M+E+H=250 $ \Rightarrow $ 3E+E+$\displaystyle\frac{9}{4}E$=250
25E = 1000 or E = 40

8. One-fourth of the boys and three-eighth of the girls in a school participated in the sports. What fractional part of the total student population of the school participated in the annual sports ?
a. $\displaystyle\frac{4}{{12}}$
b. $\displaystyle\frac{5}{8}$
c. $\displaystyle\frac{8}{{12}}$
d. Data inadequate
Correct Option:d
Explanation:
Boys and girls ratio was not given.

9. Gold is 19 times as heavy as water and copper 9 times as heavy as water.  The ratio in which these two metals be mixed so that the mixture is 15 times as heavy as water is:
a. 1 : 2
b. 2 : 3
c. 3 : 2
d. 19: 135
Correct Option: C
Explanation:
This question can be solved using weighted average formula. If two quantities of weights m, n have concentrations x, y are mixed then, final concentration = $\dfrac{{mx + ny}}{{m + n}}$
Take 1 unit of gold and x units of copper.
$\displaystyle\frac{{1 \times 19 + x \times 9}}{{1 + x}} = 15 \Rightarrow 19 + 9x = 15 + 15x$
$ \Rightarrow x = \displaystyle\frac{2}{3}$
So they are to be mixed in the ratio 1 : x =  $1:\displaystyle\frac{2}{3}$ or 3 : 2

10. If a:b=c:d, then $\displaystyle\frac{{ma + nc}}{{mb + nd}}$ is equal to
a. m : n
b. na:mb
c. a : b
d. md:nc
Correct Option: C
Explanation:
Let $\displaystyle\frac{a}{b} = \displaystyle\frac{c}{d} = k$. Then a = bk and c=dk
$\displaystyle\frac{{ma + nc}}{{mb + nd}} = \displaystyle\frac{{mbk + ndk}}{{mb + nd}} = k\left[ {\displaystyle\frac{{mb + nd}}{{mb + nd}}} \right]$ = k
But $k = \displaystyle\frac{a}{b}$ So the required ratio = a : b

11. Rs.1050 is divided among P, Q and R. The share of P is $\displaystyle\frac{2}{5}$ of the combined share of Q and R. Thus, P gets:
a. Rs.200
b. Rs.300
c. Rs.320
d. Rs.420
Correct Option: B
Explanation:
Let Q + R got 5 units then P gets 2 units.
P : (Q + R)= 2:5
But total P + Q + R = 7 units. So,
P's share =Rs.$\left( {1050 \times \displaystyle\frac{2}{7}} \right)$=Rs.300

12. Divided Rs.600 among A,B and C so that Rs.40 more than $\displaystyle\frac{2}{5}$th of A's share.  Rs.20 more than $\displaystyle\frac{2}{7}$ of B's share and Rs.10 more than $\displaystyle\frac{9}{{17}}$th of C's share may all be equal.  What is A's share ?
a. Rs.280
b. Rs.150
c. Rs.170
d. Rs.200
Correct Option: B
Explanation:
$\displaystyle\frac{2}{5}A + 40 = \displaystyle\frac{2}{7}B + 20 = \displaystyle\frac{9}{{17}}C + 10 = x$
A=$\displaystyle\frac{5}{2}(x - 40),B = \displaystyle\frac{7}{2}(x - 20)$ and c = $\displaystyle\frac{{17}}{9}(x - 10)$
Given that $\displaystyle\frac{5}{2}(x - 40) + \displaystyle\frac{7}{2}(x - 20)$ + $\displaystyle\frac{{17}}{9}(x - 10) = 600$
$\displaystyle\frac{{45x - 1800 + 63x - 1260 + 34x - 340}}{{18}} = 600$
45x - 1800+63x -1260 +34x-340=10800
142x=14200 or x = $\displaystyle\frac{{14200}}{{142}} = 100$
Hence, A's share = $\displaystyle\frac{5}{2}(100 - 40) = Rs.150$

13. 729 ml.of a mixture contains milk and water in the ratio of 7:2. How much more water is to be added to get a new mixture containing milk and water in the ratio of 7:3 ?
a. 60 ml
b. 70 ml
c. 81 ml
d. 90 ml
Correct Option: C
Explanation:
Milk = $\left( {729 \times \displaystyle\frac{7}{9}} \right) = 567ml$
Water= $\left( {729 \times \displaystyle\frac{2}{9}} \right) = 162ml$
$\displaystyle\frac{{567}}{{162 + x}} = \displaystyle\frac{7}{3} \Rightarrow 7(162 + x) = 3 \times 567$
$7x = 1701 - 1134$ or x = $\displaystyle\frac{{567}}{7} = 81$ml

14. A and B are two alloys of gold and copper prepared by mixing metals in proportions 7:2 and 7:11 respectively.  If equal quantities of the alloys are melted to form a third alloy C, the proportion of gold and copper in C will be :
a. 5 : 9
b. 5 : 7
c. 7 : 5
d. 9 : 5
Correct Option: C
Explanation:
Gold in C = $\left( {\displaystyle\frac{7}{9} + \displaystyle\frac{7}{{18}}} \right) = \displaystyle\frac{{21}}{{18}} = \displaystyle\frac{7}{6}$
Copper in C = $\left( {\displaystyle\frac{2}{9} + \displaystyle\frac{{11}}{{18}}} \right) = \displaystyle\frac{{15}}{{18}} = \displaystyle\frac{5}{6}$
Gold : Copper = $\displaystyle\frac{7}{6}:\displaystyle\frac{5}{6} = 7:5$

15. 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. The total number of students before the increase were :
a. 10
b. 90
c. 100
d. None of these
Correct Option: C
Explanation:
Let the number of students be 2x, 3x and 5x.
Then (2x + 20):(3x + 20):(5x + 20)
= 4 : 5 : 7
So, $\displaystyle\frac{{2x + 20}}{4} = \displaystyle\frac{{3x + 20}}{5} = \displaystyle\frac{{5x + 20}}{7}$
5(2x + 20)=4(3x + 20) or x = 10
Hence, total number of students before increase =10x = 100

16. The ratio of money with Ram and Gopal is 7:17 and that with Gopal and Krishna is 7:17 . If Ram has Rs.490, Krishna has :
a. Rs.2890
b. Rs.2330
c. Rs.1190
d. Rs.2680
Correct Option: A
Explanation:
Ram: Gopal = 7 : 17 = 49 : 119
Gopal : Krishna = 7 : 17 = 119 : 289
Ram : Gopal : Krishna = 49 : 119 : 289
or Ram : Krishna = 49 : 289
Thus, 49 : 289 = 490 : x
x = $\displaystyle\frac{{289 \times 490}}{{49}} = 2890$

17. Rs.5625 is divided among A, B and C so that A may receive $\displaystyle\frac{1}{2}$ as much as B and C together receive,  B receives $\displaystyle\frac{1}{4}$ of what A and C together receive.
The share of A is more than that of B by :
a. Rs.750
b. Rs.775
c. Rs.1500
d. Rs.1600
Correct Option: A
Explanation:
A=$\displaystyle\frac{1}{2}$ (B+c) or B+C = 2A
$ \Rightarrow $ A + B + C=3A
Thus , 3A = 5625 or A =1875
Again, B = $\displaystyle\frac{1}{4}$ (A+C)$ \Rightarrow $ A+C=4B
$ \Rightarrow $A+B+C=5B
5B=5625 or B = 1125
Then, A's share is more than that of B by
Rs.(1875-1125) i.e. Rs.750

18. A certain amount was divided between Kavita and Reena in the ratio of 4:3. If Reena's share was Rs.2400, the amount is :
a. Rs. 5600
b. Rs. 3200
c. Rs. 9600
d. None of these
Correct Option: A
Explanation:
Let their shares be Rs.4x and Rs.3x.
Thus, 3x = 2400 $ \Rightarrow $ x = 800
Total amount = 7x = Rs.5600

19.The prices of a scooter and a television set are in the ratio 3:2 . If a scooter costs Rs.6000 more than the television set, the price of the television set is :
a. Rs.6000
b. Rs.10,000
c. Rs.12,000
d. Rs.18,000
Correct Option: C
Explanation:
Let the price of scooter be Rs.3x and that of a television set be Rs.2x.
Then 3x-2x=6000 or x = 6000
Cost of a television set = 2x = Rs.12000

20.If 18:x = x:8, then x is equal to :
a. 144
c. 72
c. 26
d. 12
Correct Option: D
Explanation:
$18 \times 8 = {x^2}$ or x = $\sqrt {144}  = 12$

21. A right cylinder and a right circular cone have the same radius and the same volume.  The ratio of the height of the cylinder to that of the cone is :
a. 3:5
b. 2:5
c. 3:1
d. 1:3
Correct Option: D
Explanation:
Let the heights of the cylinder and cone be h and H respectively. Then,
$\pi {r^2}h = \displaystyle\frac{1}{3}\pi {r^2}H$ or $\displaystyle\frac{h}{H} = \displaystyle\frac{1}{3}$
So, their heights are in the ratio 1 : 3

22. A circle and square have same area.  Therefore, the ratio of the side of the square and the radius of the circle is :
a. $\sqrt \pi  :1$
b. $1:\sqrt \pi  $
c. $1:\pi $
d. $\pi :1$
Correct Option: A
Explanation:
Let the side of the square be x and let the radius of the circle be y.
Then, ${x^2} = \pi {y^2} \Rightarrow \displaystyle\frac{{{x^2}}}{{{y^2}}}\pi $ or $\displaystyle\frac{x}{y} = \sqrt \pi  $
x : y  = $\sqrt \pi  :1$

23. In a class, the number of boys is more than the number of girls by 12% of the total strength. The ratio of boys to girls is :
a. 11:4
b. 14:11
c. 25:28
d. 28:25
Correct Option: B
Explanation:
Let the number of boys and girls be x and y respectively. Then (x-y) = 12% of (x+y)
or x-y = $\displaystyle\frac{3}{{25}}$ (x+y)
25x-25y=3x+3y or 22x = 28y
or $\displaystyle\frac{x}{y} = \displaystyle\frac{{28}}{{22}} = \displaystyle\frac{{14}}{{11}} = 14:11$

24. A, B and C do a work in 20, 25 and 30 days respectively. They undertook to finish the work together for Rs.2220, then the share of A exceeds that of B by :
a. Rs.120
b. Rs.180
c. Rs.300
d. Rs.600
Correct Option: B
Explanation:
Remember remuneration is inversely proportional to the days taken to complete the work.
Ratio of shares of A,B & C = $\dfrac{1}{{20}}:\dfrac{1}{{25}}:\dfrac{1}{{30}} = \dfrac{{15:12:10}}{{300}}$ ($\because $ by taking LCM of 20, 25, 30 and multiplying the given ratios)
Sum of the ratios = 15 + 12 + 10 = 37.
A's share = Rs.$\left( {2220 \times \displaystyle\frac{{15}}{{37}}} \right) = Rs.900$
B's share = Rs.$\left( {2220 \times \displaystyle\frac{{12}}{{37}}} \right) = Rs.720$
Thus, the share of A exceeds that of B by Rs.
(900-720)=Rs.180

25. Three friends divide Rs.624 among themselves in the ratio $\displaystyle\frac{1}{2}:\displaystyle\frac{1}{3}:\displaystyle\frac{1}{4}$. The share of the third friend is :
a. Rs.288
b. Rs.192
c. Rs.148
d. Rs.144
Correct Option: D
Explanation:
Multiplying the entire ratio by 12 we get,
Ratio = $\displaystyle\frac{1}{2}:\displaystyle\frac{1}{3}:\displaystyle\frac{1}{4}$ = 6:4:3
Share of third friend = Rs.$\left( {624 \times \displaystyle\frac{3}{{13}}} \right)$
= Rs.144

26. One year ago the ratio between Laxman's and Gopal's salary was 3:4.  The ratio's of their individual salaries between last year's and this year's salaries are 4:5 and 2:3 respectively.  At present the total of their salary is Rs.4290.  The salary of Laxman now is :
a. Rs.1040
b. Rs.1650
c. Rs.2560
d. Rs.3120
Correct Option: B
Explanation:
Let the salaries of Laxman and Gopal one yer before be 12x and 16x.
Given that laxman's last year and present year salary are in the ratio 4 : 5 so his present salary = 5/4(12x) = 15x
Also Gopal's last year and present year salary are in the ratio 2 : 3 so his present salary = 3/2(16x) = 24x
But given that sum of the salaries 15x + 24x = 39x = 4290 $ \Rightarrow $ x = 110
Laxman's present salary = 15x = 15 x 110 = 1650

27.  Students in Class I, II and III of a school are in the ratio of 3 : 5 : 8. Had 15 more students admitted to each class, the ratio would have become 6 : 8 : 11. How many total students were there in the beginning?
a. 112
b. 64
c. 96
d. 80
Correct Option: D
Explanation:
Increase in ratio for 3 classes is 6 - 3 = 8 - 5 = 11 - 8 = 3.
Given, 15 more students are admitted to each class.
Therefore, 3 :: 15 ⇒ 1 :: 5
Therefore, 3 + 5 + 8 = 16 :: 16 x 5 = 80.
Hence, total students in the beginning were 80.

28. A spends 90% of his salary and B spends 85% of his salary. But savings of both are equal. Find the income of B, if sum of their incomes is Rs. 5000.
a. 2000
b. 2400
c. 2125
d. 2400
Correct Option: A
Explanation:
Let the incomes of A and B are x, y respectively.
Savings of A = (100 - 90)% (x) = 10% (x)
Savings of B = (100 - 85)% (y) = 15% (y)
Given, both saves equal amount.
Therefore, 10% (x) = 15% (y) $ \Rightarrow \dfrac{x}{y} = \frac{{15\% }}{{10\% }} = \dfrac{3}{2}$
Therefore, x : y = 3 : 2
Hence, B's salary = $\dfrac{{\rm{2}}}{{(2 + 3)}}$ x 5000 = Rs. 2000.

29. A man has some hens and some cows. If the number of heads is 50 and number of feet is 142. The number of cows is:
a. 26
b. 24
c. 21
d. 20
Correct Option: C

Explanation:
Let the mans has $x$ hens and $y$ cows.
Then total heads = $ x + y = 50 $ - - - (1)
total feet = $2x + 4y = 142$ - - - (2)
Solving above two equations, we get x = 29, y = 21
So cows are 21.
Alternative method:
Let, the man has hens only.
Then total heads = 50 x 1 = 50.
And, legs = 50 x 2 = 100 which is short by 42 from the actual legs i.e. = 142.
Now, replacement of one cow with one hen means same number of heads and two more legs.
Therefore, Hens replaced with cows = $\displaystyle\frac{{{\rm{42}}}}{{\rm{2}}}$ = 21
Therefore, Cows = 21

30. A sum of Rs. 350 made up of 110 coins, which are of either Rs. 1 or Rs. 5 denomination. How many coins are of Rs. 5?
a. 52
b. 60
c. 62
d. 72
Correct Option: B

Explanation:
Let, the number of Rs.1 coins are x and Rs.5 coins are y.
Then x + y = 110
x + 5y = 350
Solving above  two equations, we get y = 60.  So number of Rs.5 coins are 60.
Alternative method:
Let, all the coins are of Re. 1 denomination.
Then, total value of 110 coins = 110 x 1 = Rs. 110 which is short from Rs. 350 by Rs. 350 - Rs. 110 = Rs. 240.
Now, replacing 1 one-rupee coin with five-rupee coin mean Rs. 4 extra.
Therefore, Five rupee coins = $\displaystyle\frac{{{\rm{240}}}}{{\rm{4}}}$ = 60 coins

31. The population of a village is 10000. In one year, male population increase by 6% and female population by 4%. If population at the end of the year is 10520, find size of male population in the village (originally).
a. 6000
b. 4800
c. 5200
d. 5600
Correct Option: A
Explanation:
Let the males are x.  Then females are 10000 - x.
Now  $x \times 106\%  + (10000 - x)104\%  = 10520$
$ \Rightarrow x \times \dfrac{{106}}{{100}} + (10000 - x)\dfrac{{104}}{{100}} = 10520$
$ \Rightarrow x \times \dfrac{{106}}{{100}} - x \times \dfrac{{104}}{{100}} + \dfrac{{104}}{{100}}(10000) = 10520$
$ \Rightarrow x \times \dfrac{2}{{100}} + 10400 = 10520$
$ \Rightarrow x = 120 \times \dfrac{{100}}{2}$
$ \Rightarrow x = 6000$
Alternative Method:
Let, the population consists of females only.
Then, increase in the population is 4% of 10000 = 400
Therefore, Increased population = 10000 + 400 = 10400
But, actual increased population = 10520
Difference = 10520 - 10400 = 120
We know that for every 100 people, males grow at the rate of 6 and females at the rate of 4. So Males grow 2 people more than females. But we need 120 people extra. So
Therefore, Male population = $\displaystyle\frac{{{\rm{120}}}}{{\rm{2}}}$ x 100 = 6000

32. A mixture of 55 litres contains milk and water in the ratio of 7 : 4. How many litres of milk and water each must be added to the mixture to make the ratio 3 : 2?
a. 8
b. 9
c. 10
d. 11
Correct Option: C
Explanation:
Let milk and water in the mixture  are 7x and the 4x litre respectively.
Then, 7x + 4x = 55
Therefore, x = $\displaystyle\frac{{{\rm{55}}}}{{{\rm{11}}}}$ = 5
So milk = 35, water = 20
Let k liters of milk and water to be added to the mixture to make to 7 : 4.
$\dfrac{{35 + k}}{{20 + k}} = \dfrac{3}{2}$
⇒ (35 + k) × 2 = (20 + k) × 3
⇒ k = 10 liters