*Wednesday, June 6th, 2018*

Ratio and Proportions is one of the easiest concepts in CAT. It is just an extension of high school mathematics. Questions from this concept are mostly asked in conjunction with other concepts like similar triangles, mixtures and allegations. Hence fundamentals of this concept are important not just from a stand-alone perspective, but also to answer questions from other concepts

Many a times we compare two data values of the same type. One way to do this is to find out the difference (a-b). Other method of comparison could be by division or finding out ratios. (For example, a/b also written as a: b).

For example: Cost of a car is Rs. 250,000 and cost of a bike is Rs. 50,000. If we compare the two numbers using difference method, difference is Rs. 200,000. If we compare by division or ratios, they would be in the ratio: 250,000/50,000 = 5/1.

Thus, we can say that cost of the car is five times the cost of the bike.

- A ratio remains the same if both antecedent and consequent are multiplied or divided by the same non-zero number,
- a/b = pa/pb = qa/qb , p, q ≠0
- a/b = (a/p) / (b/p) = (a/q) / (b/q) , p, q ≠0

- Two ratios in their fraction notation can be compared just as we compare real numbers.
- a/b = p/q ⟺ aq = bp
- a/b > p/q ⟺ aq > bp
- a/b < p/q ⟺ aq < bp

- If two ratios a/b and c/d are equal
- a/b = c/d ⟹ b/a = d/c (Invertendo)
- a/b = c/d ⟹ a/c = b/d (Alternendo)
- a/b = c/d ⟹ (a+b)/b = (c+d)/d (Componendo)
- a/b = c/d ⟹ (a-b)/b = (c-d)/d (Dividendo)

If two ratios are equal, we say that they are in proportion and use the symbol “::” or “=” to equate the two ratios.

** For example: **Consider two ratios, a:b and c:d. These ratios are said to be in proportion if a/b=c/d and we write a:b::c:d or a,b,c and d are in proportion.

Here first and fourth terms are known as extreme terms (a and d). Second and third terms are known as middle terms (b and c).

If a:b = c:d is a proportion, then

- Product of extremes = product of means i.e., ad = bc
- a, b, c, d,…. are in continued proportion means, a:b = b:c = c:d
- a:b = b:c then b is called mean proportional and b
^{2}= ac - The third proportional of two numbers, a and b, is c, such that, a:b = b:c
- d is fourth proportional to numbers a, b, c if a:b = c:d

**Example 1:** Divide Rs. 60 in the ratio 1:2 between Mike and John.

**Solution:** Let Mike’s part be x.

Then John’s part is 2x.

Thus, x+2x = 60

=> 3x = 60

=> x = (60/3)

=> x = 20.

Therefore, Mike’s part = x = Rs. 20

John’s part = 3x = Rs. (2*20) = Rs. 40

** **

**Example 2: **If there are Rs. 495 in a bag in denominations of one-rupee, 50-paisa and 25-paisa coins which are in the ratio 1:8:16. How many 50 paisa coins are there in bag?

**Solution: **Assume, you have x numbers of one rupee coin. Now coins are in the ratio 1:8:16. This means that if we have x number of one rupee coins, we have 8x number of 50 paisa coins and 16x number of 25 paisa coins. Here order in which ratios are mentioned in the question is very important. In this case order is one rupee, 50 paisa and 25 paisa and ratio is 1:8:16.Thus,

Number of 50-paisa coins = 8x

Number of 25- paisa coins = 16x

Now,

Total money in the bag = Rs. 495

- x+ (8x/2) + (16x/4) = 495

(50 paisa coins divided by 2 to convert into rupee and 25 paisa coins divided by 4 to convert into rupee)

- 9x = 495
- x = 495/5
- x= 55

Thus, number of 50 paisa coins = 55*8 = 440

** **

**Example 3: **Three Jars contain alcohol to water in the ratios 3:5, 1:3 and 1:1. If all the three solutions are mixed, what will be the ratio of alcohol to water in final solution?

**Solution:** Here we are not given the quantities of the solution in three jars. Only the ratio of alcohol to water is given. If the ratio of quantity of solution would have been there, we could determine the ration of alcohol to water in final solution. Hence, the answer here will be **cannot be determined**.

**Example 4**: A lump of two metals weighing 18gm is worth Rs. 87. If their weights are interchanged, it would be worth Rs. 78.6. If the price of one metal is Rs. 6.7 per gram, find the weight of the other metal in the mixture.

**Solution:** Let the weight of first metal be x. Then the weight for other metal would be (18-x).

Also, let the price of other metal be Rs. K per gram. We know that price for the first metal is Rs. 6.7. Thus

Now, Price for first metal = Rs. 6.7 per gram

Price for second metal = Rs. K per gram

Originally, Weight of first metal = x gm and weight of second metal = (18-x) gm

Also, originally worth of the lump = Rs. 87

Therefore, **6.7x + K (18-x) = 87**

When the weights are interchanged, Weight of the first metal = (18-x) gm and Weight of second metal = x gm

Therefore, **6.7(18-x) + Kx = 78.6**

Adding the two equations above:

6.7x + 18K – Kx + (6.7*18) -6.7x +Kx = 87+ 78.6

=>18K + 120.6 = 165.6

=>18k = 45

=>K= 45/18 = 5/2

Substituting value of K= 5/2 in equation 1

6.7x + (5/2 * (18 – X)) = 87

=> x = 10

Thus, weight of other metal in the mixture = (18-10) gm = 8 gm

**Example 5**: A bottle of whisky contains ¾ of whisky and the rest is water. How much of the mixture must be taken away and substituted by equal quantity of water so as to have half whisky and half water?

**Solution:** We will use the formula below for answering this. This is a standard formula for mixtures and very useful when questions concerning mixture and ratio comes in.

F = I (1- X/V)

F= Fraction of whisky in final mixture = 1/2

I = Fraction of whisky in Initial mixture = 3/4

X= Quantity of mixture replaced = ??

V= Total Volume of mixture

Substituting the values in above formula

=>½=3/4 (1-X/V)

=>(1-X/V) = (1/2)*(4/3)

=>(1-X/V) = 2/3

=>X/V = 1/3

So, the answer is Quantity of mixture replaced would be **1/3 of the original mixture.**

You can also see: Divisibility Rules for CAT Quantitative Aptitude Preparation

**Example 6**: A man covered a total distance of 1,000 km in 16 hr, partly in a taxi at 36 km/hr and partly in a bus at 80 km/hr. The distance covered by the bus is approx.?

a.) 770 km b.)640 km c.) 720 km d.) 680 km e.) None of the above

**Solution: **This is a classic example of how ratios questions comes combined with the concept of speed, time and distance. Now let’s look at how we can solve this particular question.

We know, Speed = distance / time.

Now, Let the distance covered by bus be x km

Then, distance covered by taxi = (1000-x) km

Also, Time taken by the bus to cover x km = Distance/Speed = x km / 80 km/hr

= x/80 hr.

Similarly, time taken by the taxi to cover (1000-x) km = (1000-x)/36 hr.

We know as per question that total time taken is 16 hr.

So, (x/80) + ((1000-x)/36) = 16

=>X =~ 770

Thus, answer would be, Distance covered by the bus = 770 km

** **

**Example 7**: In a mixture of 35 L, the ratio of milk to water is 4:1. If 7 L water is added to the mixture, the ratio of milk to water changes to a new ratio. If we want ratio of milk and water to change back to the original value, how much milk is to be added now?

**Solution: ** Initial Ratio of milk: water = 4:1

Let the quantity of water be x. Then, quantity of milk would be 4x.

As per question,

x+4x = 35

=>x = 7

In original mixture,

Quantity of water = 7 L

Quantity of milk = 28 L

After 7 L of water is added to the mixture,

Quantity of water = 14 L

Quantity of milk = 28 L

New ratio of milk: water = (28:14) = (2:1)

Original ratio = 4:1

Let the quantity of milk to be added = x L

Thus, New quantity of milk / New quantity of water = 4/1

=>(28+x) / 14 = 4/1

=>28+x = 56

=>x = 56-28 = 28

Hence, quantity of milk to be added = 28 L

**Example 8**** : **Two tanks of similar volume are full of a mixture of oil and water. In the first, the ratio of oil and water is 5:8 and in the second, it is 7:19. If both these tanks are poured in a larger tank, what would be the resultant ratio of oil and water?

**Solution: ** In this type of questions, take one of the components and find its proportion in the overall mixture.

In first tank, Oil/Water = 5/8 therefore, Oil/Mixture = 5/13.

In second tank, Oil/Water = 7/19 therefore, Oil/Mixture = 7/26

Let us assume that the volume of each tank is x.

Thus, quantity of oil in first tank = 5x / 13

Quantity to oil in second tank = 7x / 26

Total quantity of oil in the new tank = (5x/13) + (7x/26)

= (17x/26)

Now, total quantity of the new mixture = x + x = 2x

Therefore ratio of oil to the overall mixture in the tank would be or oil: Mixture = (17x/26) / 2x

= 17x / 52

This means in 52 parts mixture, 17 parts is oil.

Therefore in 52 parts mixture, 35 parts would be water.

So, answer is ratio of oil: water = 17:35

**Example 9:** The ratio of a two-digit natural number to a number formed by reversing its digits is 4 : 7. Which of the following is the sum of all the numbers of all such pairs?

a. 99 b. 198 c. 330 d. 132

**Solution:** Let the two digit number be 10a + b and the number formed by reversing its digits be 10b + a.

(10a + b) / (10b + a) = 4/7

70a + 7b = 40b + 4a

66a = 33b

Therefore,

a/b = 1/2

So, let us list down all possible values for a and b.

a b Number Reversed Number

1 2 12 21

2 4 24 42

3 6 36 63

4 8 48 84

Hence, the sum of all the numbers would be,

12 + 21 + 24 + 42 + 36 + 63 + 48 + 84 = 330.

Answer choice (c)

**Example 10: **Rs.432 is divided among three workers A, B and C such that 8 times A’s share is equal to 12 times B’s share which is equal to 6 times C’s share. How much did A get?

**Solution: **

**As per question,**

8A = 12B = 6C

Expressing B’s and C’s share in terms of A’s share

B = (8/12) A = (2/3) A (Since 8A = 12 B)

C = (8/6) A = (4/3) A (Since 8A = 6C)

Now A+B+C = 432

=>A + (2/3)A + (4/3)A = 432

=>3A = 432

=>A = 144

Thus, A’s share is Rs. 144

**Conclusion:** Ratio and proportion is a very easy topic in itself, but questions are mostly asked from ratio and proportion by combining this with various other topics. Thus, In CAT preparation, it becomes very essential that we have a familiarity of all the topics – if not at expert level in all, surely the basics.

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