Ratio and Proportion
Numbers can be used to make comparisons in day-to-day situations. When comparing any two numbers, sometimes it is necessary to find out how many times one number is greater than or less than the other. In other words, we often need to express one number as a fraction of the other.
For example,
Akshay scored 76 runs and Harsh scored 19 runs in the finals of the CB series.
Then, we can say that Akshay scored 57 runs more than Harsh or alternatively Akshay scored 4 times as many runs as Harsh. The latter way of comparing is by finding ratios between the runs scored by the two batsmen.
Ratios are useful when making comparisons. They represent a relationship between two quantities of the same unit. One of the values is divided by the other to find the value of one quantity in terms of the other. However, a ratio does not have any units. A ratio can also be expressed as a fraction that one quantity is of the other. The ratio of two terms a and b is denoted by a : b and is equal to a/b where a is called the antecedent and b is called the consequent.
For example,
5000 g of type A rice is mixed with 3 kg of type B rice. The ratio of type A rice to type B rice can be calculated by dividing the amount of type A rice by the amount of type B rice. To calculate a ratio, the two quantities should be of the same unit.
Ratio of type A rice to type B rice = 5000/3000 = 5/3
This ratio can be represented as 5 : 3.
The fraction 5/3 represents the amount of type A rice to the amount of type B rice.
However, the fraction 5/8 represents the amount of type A rice in the mixture.
REMEMBER:
- The order of the terms in a ratio is important.
a : b is not the same as b : a.
- The two quantities should be of the same unit.
- For example, 30 marks can be compared with 45
marks but not with Rs. 45.
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70 shares have to be distributed among brokers A and B in the ratio 2 : 3. How many shares will each of them get?
Solution: The shares have to be distributed in the ratio 2 : 3. A will get 2/5th of the shares and B will get 3/5th of the shares.
Number of shares A gets = 2/5
Number of shares B gets = 3/5 |
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Asha, Altheda and Amata had a total of Rs. 2750 with them. They decided to divide this money between themselves such that 1/4th of Asha’s share was equal to 1/5th of Altheda’s share, which in turn was equal to half of Amata’s share. How much money did Amata receive?
Solution: Let the ratio of money received by Asha, Altheda and Amata be x : y : z. Hence,
Thus, the ratio of Asha, Altheda and Amata becomes 1 : 5/4 : 1/2; which is equivalent to 4 : 5 : 2.
Hence, Amata received 2/11 |
Ratios are usually reduced to the lowest form for simplicity. Multiplying or dividing the terms in a ratio by the same number does not change it.
For example, if there are 5000 students in college A, 4000 students in college B and 3500 students in college C, the ratio of the students in the three colleges is 5000 : 4000 : 3500.
To simplify the ratio, divide every term by 500. The simplified ratio is 10 : 8 : 7.
5000 : 4000 : 3500 and 10 : 8 : 7 are equivalent ratios.
Ratios can also be expressed in percentages. To express the value of a ratio as a percentage, multiply the ratio by 100.
There may be situations where there are more than two quantities and they are not in the same ratio. The ratios can be scaled to find a common ratio.
For example, if the ratio of red marbles to blue marbles is 2 : 5 and the ratio of blue marbles to yellow marbles is 6 : 7, then we can find a common ratio using the scaling ratio method.
| Red | Blue | Yellow |
| 2 | 5 | |
| 6 | 7 |
Blue is common in both the ratios, so find the LCM of 5 and 6.
LCM of 5 and 6 = 30
The value of 5 corresponds to 30. So, any other value in the same ratio should
be multiplied by 6.
The value of 2 will correspond to 12. So, 2 : 5 and 12 : 30 are equivalent
ratios.
Similarly, the value of 6 corresponds to 30. So, any other value in the same ratio should be multiplied by 5.
The value of 7 will correspond to 35. So, 6 : 7 and 30 : 35 are equivalent
ratios.
The ratio of red marbles to blue marbles to yellow marbles is 12 : 30 : 35.
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$9000 is divided among Harry, Ron and Hermione. The ratio of the amount Hermione got to that Ron got is 1 : 3. The ratio of the amount Ron got to that Harry got is 1 : 2. Find the amount that each of them should get.
Solution: The amount Ron gets is common. The LCM of 3 and 1 is 3. So, multiply the second ratio by 3. The two ratios are 1 : 3 and 3 : 6.
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Finding the common ratio when more than two ratios are involved
While it is quite easy to find the common ratio when only 2 different ratios are given, in cases that involve more ratios it would be difficult to find the common ratio using the LCM method. For example, consider that the given ratios are:
Solving the above using the LCM method, you would have got the common ratio as follows:
LCM of 7 and 2 is 14; hence multiply 3/7 by 2 and 2/11 by 7 to get 6/14 and 14/77 respectively.
LCM of 77 and 4 is 308; so multiply 14/77 by 4 and 4/5 by 77 to get 56/308 and 308/385 respectively.
LCM of 385 and 9 is 3465; so multiply 308/385 by 9 and 9/13 by 385 to get 2772/3465 and 3465/5005 respectively.
LCM of 308 and 2772 is 2772; so multiply 56/308 by 9 to get 504/2772.
LCM of 14 and 504 is 504; so multiply 6/14 by 36 to get 216/504.
a : b : c : d : e = 216 : 504 : 2772 : 3465 :
5005
However, this would have required a lot of time and effort. Fortunately, there exists an easier and quicker method of solving the same.
Then,
So, solving the above example using this method, we get,
a 
3 
2
4
9 = 216
b
7
2
4
9 = 504
c
7
11
4
9 = 2772
d
7
11
5
9 = 3465
e
7
11
5
13 = 5005
a : b : c : d : e = 216 : 504 : 2772 : 3465 :
5005
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Let a, b, c, d and e be integers such that a = 6b = 12c, and 2b = 9d = 12e. Then which of the following pairs contains a number that is not an integer? [CAT 2003 Re-test]
Solution: Since a = 6b = 12c and 2b = 9d = 12e
Evaluating the given options:
Hence, option 4. |
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There are four empty jars, to be filled with milk, water, honey and wine respectively. The ratio of the quantity of milk to water is 13 : 7; that of water to honey is 5 : 19; and that of honey to wine is 7 : 3. Also, the ratio of the capacity of the milk jar to the water jar is 3 : 5; that of the water jar to the honey jar is 7 : 9; and that of the honey jar to the wine jar is 8 : 11. A man decides that he will either pour out the entire amount of a liquid into its corresponding jar, or if the capacity of the jar isn’t enough, then he won’t transfer any of the liquid into the jar. The capacity of all four jars combined is 9121 units, and the amount of water is such that when poured into its jar, it fills exactly to the brim. How many liquids does the man transfer into jars?
Solution: First let us deal with the capacity of the jars:
Capacity of milk jar : water jar = 3 : 5 Capacity of water jar : honey jar = 7 : 9 Capacity of honey jar : wine jar = 8 : 11
= (3 = 168 : 280 : 360 : 495
168x + 280x + 360x + 495x = 9121
Thus, the capacities of the milk, water, honey and wine jars are 1176, 1960, 2520 and 3465 respectively.
Now, let us deal with the quantity of the liquids:
Quantity of milk : water = 13 : 7 Quantity of water : honey = 5 : 19 Quantity of honey : wine = 7 : 3
= (13 = 455 : 245 : 931 : 399
i.e. 245y = 1960
Thus, the quantities of milk, water, honey and wine are 3640, 1960, 7448 and 3192 respectively.
Comparing these values with those of the corresponding jar capacities, we see that only water and wine can be poured into their respective jars without any spilling out.
Hence, the man transfers 2 liquids into jars. |
Let us consider two ratios a : b and c : d.
Now, a : b is greater than c : d if
Multiplying both sides by bd we get,
ad > bc
Hence, a : b is greater than c : d if ad > bc and vice versa.
Thus, to determine which of the two given
ratios a : b and c : d is greater, we compare
a
d and b
c where b > 0 and d > 0.
For example, to compare 4 : 5 and 3 : 4,
compare 4
4 with 3
5.
Since 16 > 15, 4 : 5 is greater than 3 : 4.
- When a ratio, say a : b, is multiplied with itself, then the new ratio formed, i.e. a2 : b2, is known as the duplicate ratio. Also, a3 : b3 is called the triplicate ratio,
.
Moreover, b : a is called the reciprocal ratio of a : b.
- Consider that we are given two simultaneous equations with three unknown variables (say x, y and z). Although we require a third equation to find all three unknowns, two equations are enough to determine the ratio of the variables, i.e. x : y : z. This can be done as follows:
If the given simultaneous equations are:
p1x + q1y + r1z = 0 and p2x + q2y + r2z = 0
Then,
x : y : z =
q1r2 
q2r1 :
r1p2
r2p1 : p1q2
p2q1
- Multiplying or dividing the same number (say x) to both the numerator and the denominator of a ratio (say a : b) will not change the value of the ratio:-
- Effect of adding or subtracting a number (say x) from the numerator and denominator of a ratio a : b :-
i)If a < b or (a/b) < 1, then for a positive quantity x,
Explanation:
Consider a < b
b (a + x) > a(b + x)
i.e. if ab + bx > ab + ax
i.e. if bx > ax which is true since b > a and x is positive.
Similarly,
ii)If a > b or (a/b) > 1, then for a positive quantity x,
- If the numerator and denominator of the ratio a : b are increased by, say, c and d respectively, then the new ratio formed will be equal to the original ratio only if the ratios a : b and c : d are equal.
i.e. a : b = (a + c) : (b + d) only if a : b = c : d
Thus,
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Two numbers are in the ratio 3 : 5 and the difference of their squares is 64. Find the numbers.
Solution: Let the numbers be 3x and 5x.
Hence, the numbers are 6 and 10 or |
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In an election for college president, Mehul received 5 votes for every 7 votes Harish got. If Harish got 140 votes, then how many students participated in the elections? Consider that every student has voted for either one of them.
Solution: Let the number of votes that Mehul received be x. Then, 5/7 = x/140
Hence, the total no of students who voted = 140 + 100 = 240 |
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Weekly incomes of two persons are in the ratio of 7 : 3 and their weekly expenses are in the ratio of 5 : 2. If each of them saves Rs. 300 per week, then the weekly income of the first person is [SNAP 2009]
(1) Rs. 7500(2) Rs. 4500 (3) Rs. 6300(4) Rs. 5400
Solution: Let the incomes be 7x and 3x and let the expenses be 5y and 2y.
and 3x – 2y = 300… (ii) Solving (i) and (ii) simultaneously, we get
Hence, option 3. |
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Two numbers are such that their difference, their sum, and their product are to one another as 1 : 7 : 24. The product of the two numbers is:
[FMS 2010]
(1) 6(2) 12 (3) 24(4) 48
Solution: Let the numbers be a and b
ab = 24x … (iii)
Solving (i) and (ii) simultaneously, we get a = 4x and b = 3x Substituting the values of a and b in (iii) we get, x = 2
Hence, option 4. |
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Solution: Substitute the value of a = 3 and b = 4
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Four high school football teams, Gryffindor, Ravenclaw, Hufflepuff and Slytherin, had qualified to play in the international tournament. Each team, before the start of the tournament, was assigned some amount of points based on their performance in the preliminary rounds. During this tournament, the points only shift from one team to another depending on their performance; i.e. the total number of points remains the same throughout.
In the first round, the teams Ravenclaw, Hufflepuff and Slytherin all beat Gryffindor and Gryffindor had to give away (from its initial points’ bank) that many number of points to each team as they already had. In the second round, Slytherin lost all its matches, and it had to give away that many number of points to each of the other 3 teams as they already had. In the third round, Hufflepuff lost all its games and a similar fate as that of Gryffindor in the first round and Slytherin in the second round befell upon it. At the end of the three rounds, all four teams were tied with equal number of points.
What was the ratio of the points of the teams Gryffindor and Slytherin at the start of the tournament?
Solution: Let the ratio of Gryffindor, Ravenclaw, Hufflepuff and Slytherin at the end of the three rounds be 4x : 4x : 4x : 4x. Thus, at the start of the third round (in which Hufflepuff lost), their ratio must have been 2x : 2x : 10x : 2x. This is because, the winning teams doubled their points from 2x to 4x in this round; and Hufflepuff after giving away 6x points still had 4x. So, it must have had 10x at the start of the round.
Similarly, at the start of the second round (in which Slytherin lost), their ratio must have been x : x : 5x : 9x.
At the start of the first round (in which Gryffindor lost), their ratio must have been
Hence, the ratio of the points of the teams Gryffindor and Slytherin at the start of the tournament was 17 : 9. |
The equality of two ratios is called proportion. A proportion is an equation that has two equivalent ratios on either side.
In other words, if a/b = c/d, then a, b, c and d are said to be in proportion. This equality of ratios is denoted as a : b :: c : d.
When a, b, c and d are in proportion, they are called the first, second, third and fourth proportional respectively. a and d are called the extremes and b and c are called the means. When four numbers are in proportion, the product of the extremes is equal to the product of the means.
where, a and d are the extremes and b and c are the means.
Since the product of the extremes = product of the means,
80
3 = 5x
Hence, 3 : 5 :: 48 : 80
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A 4 inch long and 6 inch wide photo is scaled proportionally. Find the width of the new scaled photo if it is 6 inch long.
Solution: Let the width of the new photo = x
Hence, the scaled photo will be 9 inches wide. |
The concept of proportion is not restricted to only two equal ratios. It can be extended to more than two equal ratios.
If a/b = c/d = e/f = g/h, then a, b, c, d, e, f, g and h are said to be in proportion.
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Find the fraction which bears the same ratio to 1/27 that 3/11 does to 5/9
Solution: Let x be the fraction.
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If a/b = b/c, then a, b, and c are said to be in continued proportion. In this case, b is called the mean proportional and it is also the geometric mean of a and c, as b2 = ac
Also, in the case of a continued proportion, the ratio of the first and third proportional is equal to the duplicate ratio of the first and second proportional.
Explanation:
Since a, b and c are in continued proportion, hence a/b = b/c
b2 = ac
Dividing both sides of the equation by c2, we get,
However, b/c = a/b
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Three numbers are in continued proportion. Their mean proportional is 10 and the sum of the other two is 29. Find the numbers.
Solution: Let a, b and c be the numbers which are in continued proportion. Then, b = 10
b2 = ac = 100 and a + c = 29
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If (a + 2b + c), (a – c) and (a – 2b + c) are in continued proportion, find the mean proportional between a and c.
Solution: Since, (a + 2b + c), (a – c) and (a – 2b + c) are in continued proportion,
= [(a + c) + 2b][(a + c) – 2b]
= a2 + 2ac + c2 – 4b2
Hence, the mean proportional of a and c is b. |
1. If a : b :: c : d
or a/b = c/d, then
lowest of the given
fractions.
Explanation:
Componendo and Dividendo law:
a = bx and c = dx
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Solve the following equation:
Solution: Using the Componendo and Dividendo law, we get
Taking square roots,
7x + 5 = 49x – 35 or 7x + 5 = –49x + 35
42x = 40 or 56x = 30
x = 20/21 or x = 15/28 |
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then r cannot take any value except __________. [CAT 2004]
Solution:
By property of equal ratios,
Assuming (a + b + c ≠ 0),
If a + b + c = 0, a = –(b + c)
Hence, option 3. |
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Solution:
Now we solve using the rule:
Add the first two ratios and subtract the third ratio from it.
Similarly, add the second and third ratio and subtract the first ratio from it.
Similarly, add the first and third ratio and subtract the second ratio from it.
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Solution:
Using the Componendo law, we have,
Dividing all four ratios by (w + x + y + z), we have
The same logic applies to the remaining 3 ratios as well. Hence, the sum of the four ratios = 4 |
If P and Q invest amounts
a1 and a2 respectively for time period
t1 and t2 respectively then the ratio in
which the total profit earned will be divided amongst P and Q will be
(a1
t1) : (a2
t2)
Note: The unit of measurement of the time period t1 and t2 should be same.
If n people invest amounts
a1, a2, a3,
a4,... for time periods t1,
t2, t3, t4, ... the ratio
of the profit earned will be divided in the ratio (a1
t1) : (a2
t2) : (a3
t3) : (a4
t4) : ...
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Amber Chew opened a departmental store at Great India Palace in Noida by investing Rs. 20 million. After a few months, her brother Sheesh Chew joined the business and invested Rs. 30 million. At the end of the year, the profit was shared in the ratio of 3 : 2. After how many months did Amber’s brother join the business? [SNAP 2008]
(1) 4 months (2) 6 months (3) 7 months (4) 8 months
Solution: Amber Chew invested Rs. 20 million for a total period of 12 months. Let Sheesh Chew join the business after x months.
Then, Sheesh Chew invested Rs. 30 million, but only for a period of (12 – x) months. Amber and Sheesh shared profits in the ratio 3 : 2. Hence,
Thus, Amber’s brother joins the business after 6 months (and 20 days).
Hence, option 2. |
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A, B and C started a business by investing 1/2, 1/3rd and 1/6th of the capital respectively. After 1/3rd of the total time, A withdrew his capital completely and after 1/4th of the total time B withdrew his capital. C kept his capital for the full period. The ratio in which total profit is to be divided amongst the partners is [SNAP 2008]
(1) 1 : 2 : 1 (2) 4 : 1 : 4 (3) 2 : 1 : 2 (4) 1 : 2 : 2
Solution: The table below shows the investment together with the time frame for which the amount is invested. For convenience, the total amount of time is assumed to be 12 months and the total capital is assumed to be 6x.
The required ratio is 2 : 1 : 2 Hence, option 3. |
