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# Ratio and Proportion – Concepts, Properties, and CAT Questions Quant Notes | EduRev

## Quant : Ratio and Proportion – Concepts, Properties, and CAT Questions Quant Notes | EduRev

The document Ratio and Proportion – Concepts, Properties, and CAT Questions Quant Notes | EduRev is a part of the Quant Course Quantitative Aptitude for SSC CGL.
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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 trianglesmixtures and allegations. Hence fundamentals of this concept are important not just from a stand-alone perspective, but also to answer questions from other concepts.

1. Ratio (Concepts and Properties)
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.

1.1 Properties of Ratio:
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)

2. Proportion:
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).

2.1 Properties of proportion
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 b2 = 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 = 495/9
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
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.

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
Ans. (a)
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. 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

Example 9. Three travellers are sitting around a fire, and are about to eat a meal. One of them has 5 small loaves of bread, the second has 3 small loaves of bread. The third has no food, but has 8 coins. He offers to pay for some bread. They agree to share the 8 loaves equally among the three travellers, and the third traveller will pay 8 coins for his share of the 8 loaves. All loaves were of the same size. The second traveller (who had 3 loaves) suggests that he will be paid 3 coins, and that the first traveller be paid 5 coins. The first traveller says that he should get more than 5 coins. How much should the first traveller get?(a) 5
(b) 7
(c) 1
(d) None of these
Ans. (b)
Solution.
Each traveler had 8/3 loaves after the distribution of the loaves.
First traveler has given (5 - 8/3) = 7/3 loaves to the third one.
Second traveler had given (3 - 8/3) = 1/3 loaves to the third one.
Since the first traveler had given 7 times the loaves of the second traveler , the coin should be distributed in the ratio of 7: 1
Hence, first traveler should get 7 coins and second traveler should get 1 coin.

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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