Quantitative Aptitude Notes - Ratio and Proportions

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Quantitative Aptitude Notes - Ratio and Proportions

Introduction:
Ratio is a comparison of two quantities by division. Ratio represents the relation that one quantity bears to the other. It is represented as a:b. In any ratio a:b, a is called Antecedent and B is called Consequent. It is an abstract (without units) quantity.
A ratio remains unaltered if its numerator and denominator are multiplied or divided by the same number, e.g. 4:3 is the same as the (4 x 10) : (3 x 10) ie 40:30.
Different Types of Ratios:
Duplicate Ratio: a2 : b2 is called duplicate ratio of a : b.
Triplicate Ratio: a3 : b3 is called triplicate ratio of a : b.
Sub – Duplicate Ratio: √a :√b is called sub-duplicate ratio of a : b.
Sub – Triplicate Ratio: 3√a :3√b is called sub-triplicate ratio of a : b
Compound Ratio: ab : cd is the compound ratio of a : c and b : d. It is the ratio of the product of the antecedents to that of the consequents of two or more given ratios.
Inverse Ratio: 1/a : 1/b is the inverse ratio of a : b.
Componendo and Dividendo: If a/b = c/d, then (a + b)/(a – b) = (c + d)/(c – d)
Proportion:
Proportion is a statement that two ratios are similar. When two ratios are equal, they make a proportion, i.e. if a/b = c/d, then a, b, c and d are in proportion. This is represented as a:b :: c:d. When a, b, c and d are in proportion, then a and d are called the Extremes and b and c are called the Means, also Product of the Means = Product of the Extremes i.e. bc = ad.
Continued Proportion: If these quantities a, b and c are such that a:b :: b:c, then b2 = ac and a, b and c are in continued proportion. Also the quantity c is called the third proportion of a and b.
Fourth Proportion: If four quantities a, b, c and x are such that a:b :: c:x, then ax = bc and x is called the fourth proportion of a, b and c.
Mean or second Proportion: If three quantities a, b and x are such that a:x ::x:b, then x2 = ab and x is called the mean of a and b. Also, if a:b = c:d, then the following properties holds good.
i) b:a = d:c (Invertendo)
ii) a:c = b:d (Alternendo)
iii) (a + b) : b = (c + d) : d (Componendo)
iv) (a – b) : b = (c – d) : d (Divendendo)
v) (a + b)/(a – b) = (c + d)/(c – d) (Componendo – Divendendo)
Variation:
If two quantities x and y are related in such a way that as the quantity x changes it also brings a change in the second quantity y, then the two quantities are in variation.
Direct Variation: The quantity x is in direct variation to y, if an increase in x makes y to increase proportionally. Also decrease in x males y to decrease proportionally it can be expressed as x = ky, where k is called the constant of proportionality.
Eg: Cost is directly proportional to the number of articles brought.
Inverse Variation: The quantity x is in inverse variation to y, if an increase in x makes y o decrease proportionally. Also a decrease in x makes y to increase proportionally. It can be expressed as x = k/y, where k is a constant of proportionality.
Eg: The time taken by a vehicle in covering a certain distance is inversely proportional to the speed of the vehicle.
Joint Variation: If there are more than two quantities x, y and z and x varies with both y and z, then x is in joint variation to y and z. It can be expressed as kyz, where k is constant of proportionality.
Eg: Men doing a work in some number of days working certain hours a day.
Distribution of Amount: If an amount A is distributed in the ratio a:b, then
First part =a/(a+b) x A; Second Part =b/(a + b) x A
Some Important Formulae to be remembered:
1) If a:b :: b:c, then a/b = b/c => c = b2/a
2) If a:b :: c:d, then a/b = c/d => d = bc/a
3) If a:x :: x:b, then x = √ab (x is mean proportional)
4) If x/y = 1, then (x + a)/(y + a) = 1 and (x – a)/(y – a) = 1
5) If x/y > 1, then (x + a)/(y + a) < x/y and (x –a)/(y – a) > x/y
6) If x/y < 1, then (x + a)/(y + a) > x/y and (x –a)/(y – a)< x/y
7) If a/b = c/d = e/f = …………… = k (constant), then (a + c + e + ……..)/(b + d + f + ……..) = k


MEMORY BASED EXAMPLE PROBLEMS

Based on - PERCENTAGE:
1. Two numbers are respectively 25% and 60% more than third number. What would be the ratio of the two numbers?
Solution:
Let the third number be x.
Then first number is 125% of x = 125x/100 = 5x/4 and second number is 160% of x = 160x/100 = 8x/5.
Now the ratio of first to second number is 5x/4 : 8x/5 = 5x x 5 : 8x x 4 = 25 : 32.
2. The salary of three friends Vinod, Kamal, Krishna are divided into the ratio 5 : 6 : 8. If the increment has given of 10%, 20%, 25%, find the new ratio of three friend’s salary?
Solution:
We can assume ratio as 5x, 6x, 8x
Now, the increment of new salary of Vinod is 10% = 110/100, Kamal is 20% = 120/100 and Krishna is 25% = 125/100
Vinod’s new salary = 110 x 5x/100 = 55x/10; Kamal’s new salary = 36x/5 and Krishna’s new salary = 10x.
New ratio = 55x/10 : 36x/5 : 10x = 11/5 : 36/5 : 50/5 = 11 : 36 : 50
3. In a class of 125, 20% students can dance. 2/5 of the total students can sing and 2/5 of the remaining students are good at sports. What is the respective ratio of students who can dance to students who are good at sport?
Solution:
Total No. of Students in a class is 125.
Students who can dance (20% of 125) is = 25
Students who can sing(2/5th of 125) is = 50
Students who are good at sports {2/5th of (125-75)} is =20
Dance : Sports = 25:20 =5: 4
4. One-fifth of Ram's expense is equal to one third of his savings. If his monthly income is 3,200, what is his savings?
Solution:
Expenses: Savings = 5: 3
Savings = 3/8 * 3200 = 1200
5. Rate of income tax is increased from 4% to 5%. However, the total tax liability of a person remains the same as was in the last year. If his income for the last year was Rs. 10000, find his present income?
Solution:
Ratio of tax = 4: 5 ==> Ratio of income = 5 : 4
New Income = 10000 * 4/5 = 8000
6. Mohan distributed his assets to his wife, three sons, two daughters and five grand children in such a way that each grandchild got one-eighth of each son and one tenth of each daughter. His wife got 40% of the total share of his sons and daughter together. If each daughter receives asset of worth Rs 1.25 lakhs. What is the salary of his wife?
Solution:
Share of 1 grand child = 1/10 * 1.25 lakhs = 0.125 lakhs
Share of 1 son = 8 * 0.125 lakhs = 1 lakh ==> Share of 3 sons = 3 * 1 lakhs = 3 lakhs
Share of 2 daughters = 2 * 1.25 lakhs = 2.5 lakhs
Total share of two sons and daughters = (3 + 2.5 lakhs ) = 5.5 lakhs
Share of wife = 4/10 * 5.5 lakhs = 2.2 lakhs
COIN BASED PROBLEMS
1. A money bag contains 50p, 25p and 10p coins in the ratio 5 : 9 : 4, and the total amounting to Rs.206. Find the individual number of coins of each type.
Solution:
Let the number of 50p, 25p, 10p coins be 5x, 9x, 4x respectively
We can say two 50 paise, four 25 paise, ten 10 Paise make one rupee respectively.
Then 5x/2 + 9x/4 + 4x/10 = 206 => x = 40
So Number of 50 paise coins = 5x = 5 x 40 = 200 coins
Number of 25 paise coins = 9x = 9 x 40 = 360 coins
Number of 10 paise coins = 4x = 4 x 40 = 160 coins
2. A sum of Rs.11.70 consists of rupees, 50 paisa and 5 paisa coins in the ratio 3:5:7. Find the number of each kind of coins.
Solution:
Let the number of rupees, 50 paisa and 5 paisa coins be 3x, 5x, 7x respectively
3x *100+5x*50+7x*5=11.70*100 (converting of rupees to paise)
585x=1170 → x=2
So Number of 1 rupee coins = 3x = 3 x 2 = 6 coins
Number of 50 paise coins = 5x = 5 x 2 = 10 coins
Number of 5 paise coins = 7x = 7 x 2 = 14 coins
BASED ON PARTNERSHIP
1. Share of Rs.4200 among Rahul, Vijay and Mahidar in the ratio 2:4:6. Find the amount received by Mahidar.
Solution:
Amount received by Mahidar = (related ratio / sum ratio) x Total amount = 6/12 x 4200 = 2100.
So the amount received by Mahidar = 2100.
2. Rs. 385 were divided among A, B, C in such a way that A has Rs.20 more than B and C has Rs.15 more than A. How much was C’s share?
Solution:
Let B gets Rs.x. Then we can say A gets Rs.(x + 20) and C gets Rs.(x + 35)
x + 20 + x + x + 35 = 385 => 3x + 55 = 385 => 3x = 330 => x = 110.
C’s share = Rs.(110 + 35) = Rs.145.
3. An amount of money to be divided between A, B and C in the ratio 2 : 3 : 5 respectively. If the amount received by C is Rs.6000/- more than the amount received by B. Then the total amount of money received by A and B together is
Solution:
A : B : C = 2 : 3 : 5 and 5x – 3x = 6000 => x = 3000
A receives 3000 x 2 = 6000/- and B receives 3000 x 3 = 9000/-
Then total amount received by both A and B = 6000 + 9000 = Rs.15000/-
4. A started a business with a capital of Rs 1,00,000. One year later, B joined him with a capital of Rs.2,00,000. At the end of 3 years from the start of the business, the profit earned was 84,000. The share of B in the profit exceeded the share of A is
Solution:
A’s investment = Rs 1,00,000 for 3 years
B’s investment = Rs 2,00,000 for 2 years
Ratio of profit share = 100000 x 3 : 200000 x 2 = 3 : 4
B’s share in profit = (4/7) x 84000 = 48000
A’s share in profit = 84000 – 48000 = 36000
Difference = 48000 – 36000 = 12000
5. The ratio of the incomes of four persons A, B, C and D is 5:3:9:4.The sum of the incomes of A and C is 84000; find the difference of the incomes of B and D?
Solution:
Let the incomes of four persons A, B, C and D be 5x,3x,9x,4x respectively.
Sum of the incomes of A and C is 84000
5x+9x= 84000 → 14x=84000
Therefore, x = 6000
The difference of the incomes of B and D will be (4x-3x = x) = 6000.
6. The salary of Amit is 3/4th of the salary of Raju; while the salary of Raju is 25% less than the salary of Sohan. The difference between the salaries of Amit and Sohan is Rs.9800; find 14 2/7% of the sum of the salaries of Raju and Sohan.
Solution:
Let the salary of Raju, Sohan and Amit be x, y and z respectively.
z = 3/4 * x ------ (1)
x = 3/4 * y ------- (2) (Sine x is 25 % less than y. So x’s salary is equal to 75% salary of y)
Sub (2) in (1)  z = 3/4 * 3/4 * y ------ (3)
The difference between the salaries of Amit and Sohan is Rs.9800  y –z = 9800
From that, we get z = y – 9800 ---- (4)
By substituting (4) in (3) we will get, y = 224; By substituting ‘y’ in (2) we will get, x = 168;
Now, 14 2/7% of (x + y) = 100/7 * 1/100 * (224+168) = 392/7 = 56.
7. Rs.15600 is divided among three persons A, B and C in such a way that; A received 2/3rd of the total share of B and C together while B received 3/7th of the total share of A and C together. Find the share of C.
Solution:
A +B + C = 15600 ----- (1); A= 2/3(B+ C) ----- (2); B = 3/7 (A+ C) ------- (3);
Sub (3) in (2)  A = 2/3 (3/7(A+ C) +C);
A = 2/3 ((3A+ 10C)/7)
By solving above equn., A = 4/3 * C ---- (4)
Sub (4) in (3)  B = 3/7 (4/3*C +C) ---- (5)
Sub (4) & (5) in (1)  C = 4680.
8. The salary of two friends Ramu and Raju are in the ratio 4:5. If the salary of each one increases by Rs.6000, then the new ratio becomes 48:55. What is Raju's present salary?
Solution:
Ratio their salary is 4:5
Let the original salary of Ramu and Raju be 4k and 5k respectively.
After increasing Rs.6000, the ratio becomes 48:55
That is, (4k+6000)/(5k+6000) = 48/55
55(4k+6000) = 48(5k+6000)
220k+330000 = 240k+288000
20k= 42000
We have to find the original salary of Raju; that is, 5k.
If 20k = 42000 then 5k = 10500.
Hence the required answer is Rs.10500
Based on - RATIOS OF RATIOS
1. In a school, the ratio to the number of boys and girls is 4:9, after inclusion of new girls, the ratio become 4: 17. How many boys were present at the starting in this school?
Solution:
4x/(9x + 32) = 4/17 =>68x = 36x + 128 => x = 4.
So the number of boys in the school is (4 x 4) = 16.
2. In an examination, the number of those who passed and the number of those who failed were in the ratio 25:4. If five more had appeared and the number of failures was 2 less than earlier, the ratio of passers to failures would have been 22:3. The number of students who appeared at the examination, is
Solution:
Ratio of Pass to fail = 25:4 = 25x : 4x
New ratio = 25x + 7 : 4x -2 = 22 : 3
No of students passed increased by 7 because 5 more appeared and 2 less failed.
75x + 21 = 88x – 44
13x = 65 => x = 5
No of students appeared initially = 25x + 4x = 25 x 5 + 4 x 5 = 125 + 20 = 145
3. The students in the three classes are in the ratio 2:3:5. If 20students are increased in each class the ratio changes to 4:5:7. What was the total number of students in the three classes before the increase?
Solution:
Let the students in the three classes be 2x, 3x and 5x respectively.
Then, 2x+20+3x+20+5x+20 = 4x+5x+7x
10x + 60 = 16x => 6x = 60 => x= 10.
Therefore, total number of students in the three classes before the increase will be 2x+3x+5x=10x=100.
4. At a start of a seminar, the ratio of the number of male participants to the number of female participants was 3:1. During the tea break 16 participants left and 6 more participants registered. The ratio of the male to the female participants now becomes 2:1. What was the total number of participants at the start of the seminar?
Solution:
Let the number of male participants and the number of female participants be 3x and 1x respectively.
Now, 3x + x – 16 + 6 = 2x + x => x = 10; ( since 16 participants left and 6 participants registered).
Therefore, the total number of participants at the start of the seminar will be (3x +x = 4x) 40.
5. The numerator and denominator of a fraction is in the ratio 2: 3. If 6 are subtracted from the numerator the value of the fraction becomes 2/3 of the original fraction. The numerator of the original fraction is
Solution:
Let the numerator and denominator of a fraction be 2x, 3x respectively.
2𝑥−6/3𝑥 = 2/3 × 2𝑥/3𝑥 => 6x – 18 = 4x => x = 9.
6. The ratio of the first and second class train fares between two stations is 3: 1 and that of the number of passengers travelling between the two stations by first and second classes is 1:50 . If on a particular day, Rs 1325 are collected from passengers travelling between the two stations, then the amount collected from the second class passenger is?
Solution:
Ratio of Fares : 3:1
Ratio of Passenger : 1:50
Ratio of Money : 3:50
Required Amount = 50 /53 * 1325 = 1250
TWISTED PROBLEMS
1. A and B have 78 marbles with them. When A gives to B half the number of marbles which B has, both have equal number of marbles. How many marbles did B have initially?
Solution:
A+B=78 --------------- equation 1
A gives B/2 marbles to B. Now both have equal number of marbles.
A-B/2= B + B/2 = 3B/2 (here gives means subtract from A and B receives so add B/2 for both)
=>A=2B--------------- equation 2
From equations 1 and 2 we have B = 26
2. In a zoo, there are rabbits and pigeons. If heads are counted, there are 340 heads and if legs are counted there are 1060 legs. How many pigeons are there?
Solution:
Suppose there are all the pigeons then total no of heads are 340 and total no of legs are 680.
Now, since 380 (1060-680) legs are extra, it means there will be 190 (380/2) rabbits. As we know a rabbit has two extra legs than that of a pigeon.
Therefore, number of rabbits =190 and number of pigeons = 340- 190 = 150.
3. The ratio of any two angles of a triangle is 5:9. If the third angle is measured to be 110 degree, then find the difference of the other two angles.
Solution:
Let the two angles of a triangle be 5x and 9x.
We know that the sum of the angle of triangle = 180 degree
So, 110 + 9x + 5x = 180 → 14x=70 →x = 5
The difference of the other two angles will be (9x – 5x = 4x) = 20 degree.