Introduction: Ratio, Proportion & Variation | General Aptitude for GATE - Mechanical Engineering PDF Download

Understanding ratios, proportions, and variations is crucial for aptitude exams. These topics frequently show up in tests like CAT, XLRI, CMAT, NMIMS, SNAP, NIFT, IRMA, Bank PO, etc. Questions related to these concepts are common in aptitude tests, and they also play a significant role in Data Interpretation, particularly in questions involving changes and comparisons in ratios. Questions from this topic are based on conceptual clarity and their different applications.

Ratio

The ratio of a number x to a number y is defined as the quotient of the numbers x and y.

• Ratios are used to compare similar quantities. In simple words, the ratio is the number that can be used to express one quantity as a fraction of the other ones.
• The ratio of a number x to a number y is defined as the quotient of the numbers x and y. The numbers that form the ratio are called the terms of the ratio. The numerator of the ratio is called the antecedent and the denominator is called the consequent of the ratio.
• The ratio of any number of quantities is expressed after removing any common factors that ALL the terms of the ratio have. For example, if there are two quantities having values of 4 and 2, their ratio is 4: 2, i.e., 2: 1 after taking the common factor 1 between them out.
  
• Ratios can be expressed as percentages. To express the value of a ratio as a percentage, we multiply the ratio by 100. Thus, 4/5 = 0.8 = 80%

Properties of Ratios

• If two quantities whose values are A and B respectively are in the ratio a : b, since we know that some common factor k(>0) would have been removed from A and B to get the ratio a: b, we can write the original values of the two quantities (i.e., A and B) as ak and bk respectively.
  Example: If the salaries of two persons are in the ratio 7: 5, we can write their individual salaries as 7k and 5k respectively.
• A ratio a: b can also be expressed as a/b. 
  So if two items are in the ratio 2 : 3, we can say that their ratio is 2/3. 
• A ratio is said to be a ratio of greater or less inequality or of equality according to as antecedent is greater than, less than or equal to consequent.
  In other words:
  (i) The ratio a: b where a > b is called a ratio of greater inequality (for example, 3: 2)
  (ii) The ratio a: b where a < b is called a ratio of less inequality (for example, 3: 5)
  (iii) The ratio a: b where a = b is called a ratio of equality (for example, 1: 1)

From this, we can find that a ratio of greater inequality is diminished and a ratio of less inequality is increased by adding the same quantity to both terms, i.e., in the ratio a : b.

• When we add the same quantity x (positive) to both the terms of the ratio, we have the following results:
  (i) if a < b then (a + x) : (b + x) > a : b
  (ii) if a > b then (a + x) : (b + x) < a : b
  (iii) if a = b then (a + x) : (b + x) = a : b
• If we multiply the numerator and the denominator of a ratio by the same number, the ratio remains unchanged.

  i.e  a/b = ma/mb

• If we divide the numerator and the denominator of a ratio by the same number, the ratio remains unchanged. Thus, 
  a/b = (a/d) / (b/d)

• If the ratio a/b>1 and if k is a positive integer then,
  (a+k)/(b+k) < a/b  and (a-k)/ (b-k)> a/b

• If the ratio a/b<1 and if k is a positive integer then,
  (a+k)/(b+k) > a/b  and (a-k)/ (b-k)< a/b

• If either or both the terms of a ratio are a surd quantity, then the ratio will never evolve into integral numbers unless the surd quantities are equal. 
  For example, √3 /√2  can never be expressed as an integer 
  Until and unless both the surds are same like √3 /√3 = 1

• If a₁/b₁ , a₂/b₂,.....,a_n/b_n  are unequal fractions then,
  (a₁ + a₂ + a₃ +----+ a_n) / (b₁+b₂+b₃+----+b_n) lies between the lowest and the highest fraction of these fractions.

Types of Problems in Ratios

Type 1: Scaling ratios

• Suppose you have a ratio relationship given between the salaries of two individuals A and B. Further, if there is another ratio relationship between B and C. 
• Then, by combining the two ratios, you can come up with a single consolidated ratio between A, B and C. 
• This ratio will give you the relationship between A and C.

Example 1: The Ratio of A’s salary to B’s salary is 2:3. The ratio of B’s salary to C’s salary is 4:5. What is the ratio of A’s salary to C’s salary? 

Sol: Using the conventional process in this case:

Take the LCM of 3 and 4 (the two values representing B’s amount). The LCM is 12.  

Convert B’s value in each ratio to 12. 

Thus, Ratio 1 = 8/12 and Ratio 2 = 12/15 

Thus, A:B:C = 8:12:15 Hence, A:C = 8:15 
Further, if it were given that A's salary was 800, you could derive the values of C's salary (as 1500). 

Shortcut for this Process:

• The LCM process gets very cumbersome, especially if you are trying to create a bridge between more than 3 quantities.
• Suppose, you have the ratio train as follows: A:B = 2:3 B:C = 4:5 C:D = 6:11 D:E = 12:17
• In order to create one consolidated ratio for this situation, using the LCM process becomes too long.
• The shortcut goes as follows:

  A : B : C : D : E can be written directly as:

  2 * 4 * 6 * 12 : 3 * 4 * 6 * 12 : 3 * 5 * 6 * 12 : 3 * 5 * 11 * 12 : 3 * 5 * 11 * 17

Type 2: Comparison of Ratios

• Two ratios at a time:
  Consider two ratios a : b and c : d. To compare two ratios, cross-multiply to get ad and bc
  If ad > bc, then a : b > c : d
  If ad < bc, then a : b > c : d
  If ad = bc, then a : b = c : d
  Here, the necessary condition is that b > 0 and d > 0
• Multiple ratios at a time:
  If you have to compare more than two ratios, either make all the numerators equal or all the denominators equal.
  When denominators are equal, the ratio with the largest numerator has the largest value and the ratio with the smallest numerator has the smallest value.
  When numerators are equal, the ratio with the largest denominator has the smallest value and the ratio with the smallest denominator has the largest value.

Example 2: 10 persons can cut 8 trees in 12 days. How many days will 8 persons take to cut 6 trees?

Sol: Let us see this question from a changed perspective.
Suppose if the question is: 10 persons can cut 8 trees in 12 days. How many days will 10 persons take to cut 4 trees? 

The answer to this question is: Since the amount of work is getting halved, so the number of days will also get halved.
There are three factors, namely:
(i) The number of men
(ii) The number of days
(iii) The number of trees
which are responsible for the final answer.

Since the number of men is less in the final situation, so more days will be required. 

Hence, multiplier = 10/8 (had there been 12 persons, the multiplier would have been 10/12.) 
The number of trees is less in the final situation, so fewer number of days will be required. So, multiplier = 6/8 
Hence, the total number of days = 12 x 10/8 x 6/8 = 90/8 = 11.25 days

Example 3: Two numbers are in the ratio 3: 4. What part of the larger number must be added to each number so that their ratio becomes 5: 6?

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

3x + k/4x + k = 5/6

18x + 6k = 20x + 5k

k = 2x

∴ Half of the larger number must be added to each number. 

Example 4: There are 2 classes A and B. If 10 students leave class A and join class B, then the ratio of the number of students in class A and class B would reverse. Find the difference in the numbers of students in class A and class B.

Sol: Let the numbers of students in class A and class B be ax and bx, respectively.

Given:
⇒ ax - 10/bx + 10 = b/a
⇒ a²x − 10a = b²x + 10b
⇒ a²x – b²x – 10a – 10b = 0
⇒ (ax – bx – 10) (a + b) = 0

∴ ax – bx = 10

The difference in the number of students in class A and class B is 10.

Methods to Compare Ratios

1. The Cross Multiplication Method

• Two ratios can be compared using the cross multiplication method.
• Suppose we have to compare 12/17 with 15/19
  Then to test which ratio is higher, cross multiply and compare 12 x 19 and 15 x 17.
• In this case, 15 x 17 is higher hence 15/19 is higher.

2. Percentage Value Comparison

• Suppose you have to compare: 173/212 with 181/241
• In such a case, just by estimating the 10% ranges for each ratio you can clearly see that -
• The first ratio is > 80% while the second ratio is < 80%. Hence, the first ratio is obviously greater.
• This method is extremely convenient if the two ratios have their values in different 10% ranges.
  However, this problem will become slightly more difficult, if the two ratios fall in the same 10% range. 
• Thus, if you had to compare 173/212 with 181 / 225 both the values would give values between 80 and 90%. The next step would be to calculate the 1% range.
• The first ratio here is 81-82% while the second ratio lies between 80 and 81%
  Hence, the first ratio is the larger of the two.

3. Numerator Denominator Percentage Change Method

• There is another way in which you can compare close ratios like 173/212 and 181/225. 
• For this method, you need to calculate the percentage changes in the numerator and the denominator.
• Thus, 173→181 is  a%  increase of 4-5% 
  while 212→225 is  a%  increase of 6-7% 
• In this case denominator is increasing more than the numerator, the second ratio is smaller. 
• This method is the most powerful to compare two ratios.

4. Ratio of the Changes Method

• To effectively compare ratios, it's best to look at how the increases relate to the original ratios.For example, if you need to find out which is larger between 181/225 and 173/212, you can use this method.By examining the changes in each ratio, you can determine their sizes more clearly. This approach helps in understanding the differences without just looking at the numbers alone.
• Step 1: place them in the order such that there are increases in both the numerator and the denominator. Thus, view them as: 173/212 (first ratio) and 181/225 (second ratio).
• Step 2: Instead of comparing 173/212 with 181/225, first find out the ratio of the increases in the numerator and denominator. 
• In this case, we can see that the numerator is going from 173 to 181 (an increase of 8); the denominator is increasing from 212 to 225 (an increase of 13). Thus, we get the ratio of increases in the numerator and denominator as 8/13.
• Step 3: Compare the first ratio with the ratio of the increases. In this case, we will compare 173/212 with 8/13. The following cases might happen in general:
• Case 1: The ratio of increases is larger than the first ratio 
• → Conclusion: The second ratio is larger.
• Case 2: The ratio of increases is equal to the first ratio 
• → Conclusion: The first and second ratios are equal.
• Case 3: The ratio of increases is smaller than the first ratio 
• Conclusion: The second ratio is smaller. In this case, 8/13 (just larger than 60%) is smaller than 173/212 (definitely over 80%). Hence, the second ratio is smaller. 
• We conclude that 173/212>181/225.

Proportion  

When two ratios are equal, the four quantities composing them are said to be proportional.
Thus if a/b = c/d, then a,b, c, d are proportionals. This is expressed by saying that a is to b as c is to d, and the proportion is written as
a:b :: c :d
or
a: b = c : d

• The terms a and d are called the extremes while the terms b and c are called the means.
  

• If four quantities are in proportion, the product of the extremes is equal to the product of the means.
Let a, b, c, d be the proportional.
Then by definition a/b = c/d
Therefore, ad = bc
Hence if any three terms of proportion are given, the fourth may be found.
Thus if a, c, d are given, then b = ad/c.

• If three quantities a, b and c are in continued proportion, then
a : b = b : c
Therefore, ac = b²
In this case, b is said to be a mean proportional between a and c; and c is said to be a third proportional to a and b.

• If three quantities are proportional the first is to the third is the duplicate ratio of the first to the second.
That is: for a:b :: b: c
a: c = a² :b²

• If four quantities a, b, c and d form a proportion, many other proportions may be deduced by the properties of fractions. The results of these operations are very useful. These operations are:

Invertendo: If a/b = c/d then b/a = d/c 

Alternando: If a/b = c/d, then a/c = b/d 

Componendo: If a/b=c/d, then (a+b)/b=(c+d)/d 

Dividendo: If a/b=c/d, then (a-b)/b=(c-d)/d

Componendo and Dividendo: if a/b = c/d, then (a + b)/(a - b) = (c + d)/(c - d) 

Tips and Tricks on Proportion

1. a/b = c/d ⇒ ad = bc

2. a/b = c/d ⇒ b/a = d/c

3. a/b = c/d ⇒ a/c = b/d

4. a/b = c/d ⇒ (a + b)/b = (c + d)/d

5. a/b = c/d ⇒ (a - b/b = (c - d)/d

6. a/(b + c) = b/(c + a) = c/(a + b) and a + b + c ≠0, then a = b = c.

7. a/b = c/d ⇒ (a + b)/(a - b) = (c + d)/(c - d), which is known as componendo -dividendo rule

8. If both the numbers a and b are multiplied or divided by the same number in the ratio a:b, then the resulting ratio remains the same as the original ratio.

Types of Problems in Proportion

Example 1: Given that both x and y vary directly from each other. If x = 10 and y = 15, which of the following pairs is not possible with respect to the value of x and y?
x = 2 and y = 3
x = 8 and y = 12
x = 15 and y = 20
x = 25 and y = 37.5

Sol: Given that x and y are directly proportional.
Hence, x/y = k(constant)
So, x/y = 10/15 = 2/3 …(1)
Now, check with the options provided here.
(a) x = 2 and y = 3
x/y = 2/3 …(2)
Hence, (1) = (2)
(b) x = 8 and y = 12
x/y = 8/12 = 2/3 …(3)
Hence, (1) = (3)
(c) x = 15 and y = 20
x/y = 15/20 = 3/4 …(4)
Hence, (1) ≠ (4)

(d) x = 25 and y = 37.5
x/y = 25/37.5 = 2/3 …(5)
Hence, (1) = (5)

Therefore, option (c) x = 15 and y = 20 should not be a possible pair with respect to the values x and y.

Example 2: Ramya purchased 97 meters of cloth that cost Rs. 242.50. What will the length of the cloth be if she purchased it for Rs. 302.50?

Sol: As we know, the length of the cloth and its costs are directly proportional. Because if we purchase more, the cost will be higher. Similarly, if we purchase less, the cost will decrease.

Hence, we get

Length (in Meters)	97	x
Cost (in Rs)	242.50	302.50

Now, we have to find the value of “x”.
Since the length and cost of cloth are directly proportional, we can write
97/242.50 = x/302.50
Now, cross multiply the above equation, we get
242.50x = 97(302.50)
242.50x = 29342.5
Hence, x = 29342.5 / 242.50 = 121.
Hence, the length of the cloth is 121 meters, if she purchased it for Rs. 302.50.
Therefore,

Length (in Meters)	97	121
Cost (in Rs)	242.50	302.50

Variation

➢ Some Typical Examples of Variation

• Area (A) of a circle = π.R², where R is the radius of the circle. Area of a circle depends upon the value of the radius of a circle, or, in other words, we can say that the area of a circle varies as the square of the radius of a circle.
• At a constant temperature, the pressure is inversely proportional to the volume.
• If the speed of any vehicle is constant, then the distance traversed is proportional to the time taken to cover the distance.

Types of Variation

1. Direct Variation

When it is said that A varies directly as B, you should understand the following implications:

(a) Logical Implication: When A increases B increases.
(b) Calculation Implication: If A increases by 10%, B will also increase by 10.
(c) Graphical Implications: The following graph is representative of this situation.

2. Inverse Variation

(d) Equation Implication: The product A X B is constant. 

Example : The height of a tree varies as the square root of its age (between 5 and 17 years). When the age of a tree is 9 years, its height is 4 feet. What will be the height of the tree at the age of 16?

Sol: Let us assume the height of the tree is H and its age is A years. 

So, H ∝ √A, or, H = K x √A Now, 4 = K x √9 K = 4/3 

Height at the age of 16 years = H = K x √A 
= 4/3 x 4 
= 16/3 = 5 feet 4 inches.

Note: For problems based on this chapter we are always confronted with ratios and proportions between different number of variables. For the above problem we had three variables which were in the ratio of 4 : 6 : 9. When we have such a situation we normally assume the values in the same proportion, using one unknown ‘x’ only (in this example we could take the three values as 4x, 6x and 9x, respectively).Then, the total value is represented by the addition of the three giving rise to a linear equation, which on a solution, will result in the answer to the value of the unknown ‘x’.

Solved Examples 

Q 1: A student gets an aggregate of 60% marks in five subject in the ratio 10 : 9 : 8 : 7 : 6. If the passing marks are 50% of the maximum marks and each subject has the same maximum marks, in how many subjects did he pass the examination?
(a) 2
(b) 3
(c) 4
(d) 5

Ans: Option (c) is correct

Sol: Let his marks be 10, 9, 8, 7 and 6 in the five subjects.
Hence, totally he has scored 40 marks. This constitutes only 60% of the total marks. Hence, total marks 40/0.6 = 66.7 or 67 approx. , which is the maximum marks in all 5 subjects.
Since the total marks in each subject is the same, hence maximum marks in each subject will be 67/5 = 13 approx. 
Out of this 50% is the passing marks . 
In other words to pass in a subject, he needs to score 6.5 marks. 
We can see that only in 1 subject, he scored less than this viz. 6. Hence, he passed in 4 subject.

Q 2: Total salary of A, B & C is Rs.350. If they spend 75%, 80% & 56% of their salaries respectively their savings are as 10 : 12 : 33. Find the salary of C?
(a) 80
(b) 150
(c) 180
(d) None of These
Ans: Option (b) is correct

Sol: A’s saving = 100 – 75 = 25% of his salary. B’s saving = 100 – 80 = 20% of his salary C’s saving = 100 – 56 = 44% of his salary 25/100 of A’s salary : 20/100 of B’s salary : 44/100 of C’s salary = 10 : 12 : 33
or 25 × A’s salary : 20 × B’s salary : 44 × C’s salary = 10 : 12 : 33
or 25 × A’s salary / 20 × B’s salary = 10/12
or A’s salary : B’s salary = 2 : 3,
B’s salary : C’s salary = 4 : 5
Thus A : B = 2 : 3, B : C = 4 : 5 Now making B common we have
A : B = 8 : 12, B : C = 12 : 15, or A : B : C = 8 : 12 : 15
Total salary = 350 Þ A’s salary = 8 / (8 + 12 + 15) × 350 = 80
B’s salary = 12 / (8 + 12 + 15) = 120, and C’s Salary = 150

Q 3:  Arvind Singh purchased a 40 seater bus. He started his services on route number 2 (from Mahu Naka to Dewas Naka with route length of 50 km). His profit (P) from the bus depends upon the number of passengers over a certain minimum number of passengers ‘n’ and upon the distance travelled by bus. His profit is Rs.3600 with 29 passengers in the bus for a journey of 36 km and Rs.6300 with 36 passengers in the bus for a journey of 42 km. What is the minimum number of passengers are required so that he will not suffer any loss?
(a) 12
(b) 20
(c) 18
(d) 15
Ans: Option (d) is correct.

Sol: The minimum number of passengers n, at which there is no loss and number of passengers travelling = m
and let the distance travelled is d, Then

or p = k(m – n)d  k is a constant.

When P = 3600, m = 29 and d = 36, then

3600 = k(29 – n) × 36 ...(1)

Again, when p = 6300, m = 36, d = 42, then

6300 = k(36 – n) × 42 ...(2)

Dividing equation (2) by (1)

Sol: Work backwards from the options. If the cylinder has a capacity of 1200 litre, then the conical vessel shall have a capacity of 700 litres. Once 200 litres have been taken out from the same, the remaining holding of each of them shall be 1000 & 500.
Alternate: Let the cylinder has a capacity of X litre, then the conical vessel shall have a capacity of
(x – 500) litres.
(x – 200) = 2 (x – 700) = x = 1200.

Q 5:  Anil mixes cocoa with sugar in the ratio 3 : 2 to prepare mixture A, and coffee with sugar in the ratio 7 : 3 to prepare mixture B. He combines mixtures A and B in the ratio 2 : 3 to make a new mixture C. If he mixes C with an equal amount of milk to make a drink, then the percentage of sugar in this drink will be
a. 21
b. 17
c. 16
d. 28
Ans: Option 'b' is  correct .

Sol: Let the volume of mixture A be 200 ml, which implies the quantity of cocoa in the mixture is 120 ml, and the quantity of sugar In the mixture 80 ml.
Similarly, let the volume of the mixture be 300 ml, which implies the quantity of coffee, and sugar in the mixture is 210, and 90 ml, respectively.
Now we combine mixture A, and B in the ratio of 2:3 (if 200 ml mixture A, then 300 ml of mixture B).
Hence, the volume of the mixture C is (200+300) = 500 ml, and the quantity of the sugar is (90+80) = 170 ml.
Now he mixes C with an equal amount of milk to make a drink, which implies the quantity of the final mixture is (500+500) 1000 ml.
The quantity of sugar in the final mixture is 170 ml.
Hence, the percentage is 17%

Q 6: In a company, 20% of the employees work in the manufacturing department. If the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company, then the ratio of the average salary obtained by the manufacturing employees to the average salary obtained by the nonmanufacturing employees is
a. 6:5
b. 4:5
c. 5:4
d. 3:2

Ans: Option 'b' is correct.

Sol: Let the number of total employees in the company be 100x, and the total salary of all the employees be 100y.
It is given that 20% of the employees work in the manufacturing department, and the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company.
Hence, the total number of employees in the manufacturing department is 20x, and the total salary received by them is (100y/6)
Average salary in the manufacturing department = (100y/6*20x) = 5y/6x
Similarly, the total number of employees in the nonmanufacturing department is 80x, and the total salary received by them is (500y/6)
Hence, the average salary in the nonmanufacturing department (500y/6*80x) = 25y/24x
Hence, the ratio =  (5y/6x): (25y/24x)
                                     =120: 150 = 4:5

Q 7: Ankita buys 4 kg cashews, 14 kg peanuts and 6 kg almonds when the cost of 7 kg cashews is the same as that of 30 kg peanuts or 9 kg almonds. She mixes all the three nuts and marks a price for the mixture in order to make a profit of 1752. She sells 4 kg of the mixture at this marked price and the remaining at a 20% discount on the marked price, thus making a total profit of ₹744. Then the amount, in rupees, that she had spent in buying almonds is
a. 1680
b.1176
c. 2520
d. 1440

Ans: Option 'a' is correct

Sol: It is given,
7C = 30P = 9A 
And Ankita bought 4C, 14P and 6A. 
Let 7C = 30P = 9A = 630k C = 90k P = 21k and A = 70k 
Cost price of 4C, 14P and 6A = 4(90k) + 14(21k) + 6(70k) = 1074k 
Marked up price = 1074k + 1752 
S. P = 1/6 * (1074k + 1752) + (4/5)(5/6)(1074k + 1752) 
         = 5/6 * (1074k + 1752) 
S.P-C.P = profit 
1460 - (1074k)/6 = 744 (1074k)/6 =716 k = 4 
Money spent on buying almonds = 420k = 420*4 = Rs 1680