Quantitative Aptitude Discussion Day4 (Ratio& Proportion) with Exercise Questions:
Dear Readers, We all knew a famous saying of the Legend Aristotle,“Well begun is half done” and we also knew that this statement is 100% true. To Crack the Bank Exams you need to know where to start and how to start. Now this is the absolute correct time to kick start the preparation for upcoming IBPS Exams 2017. To help you in this aspect and to be a part of your preparation here we, IBPS Guide Team providing Topic wise Discussion on Quantitative Aptitude, this session will be conducted regularly on daily basis. This will provide a complete overview on the topics along with exercise questions. Kindly Make use of it.
RATIO AND PROPORTION
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:a^{2}: b^{2} is called duplicate ratio of a : b.
Triplicate Ratio:a^{3}: b^{3} is called triplicate ratio of a : b.
Sub – Duplicate Ratio: √a :√b is called subduplicate ratio of a : b.
Sub – Triplicate Ratio:^{3}√a :^{3}√b is called subtriplicate 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 b^{2}= 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 x^{2}= 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
Formulae:
1) If a:b :: b:c, then a/b = b/c => c = b^{2}/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
Solved examples:
Type 1:
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.
Type 2:
2). 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.
Type 3:
3). 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.
Type 4:
4). 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
Type 5:
5). 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
More Types on Ratio& Proportion Problems will be discussed in the next Session, Kindly follow us daily.
Dear Aspirants, Here below we have given exercise questions on Ratio& Proportion based on the above types, solve these questions by yourself and comment your answers below. Correct Answers with explanation will be updated in the end of the day.
Exercise problems:
1). If two numbers are respectively 35% and 75% more than a third number, what percent is the first of the second?
a) 92.14%
b) 84.14%
c) 80.24%
d) 65.14%
e) None of these
2). The monthly expenses of P and Q are in the ratio 5 : 6, their incomes are in the ratio 4 : 5. If ‘P’ saves Rs.50 per month and ‘Q’ saves Rs.75 per month, what are their respective incomes?
a) Rs.300 and Rs.375
b) Rs.240 and Rs.300
c) Rs.320 and Rs.475
d) Rs.425 and Rs.525
e) None of these
3). Rs. 2970 has been divided into three parts in such a way that half of the first parts, onethird of the second parts, and onesixth of the third parts are equals. The third parts is ?
a) Rs. 510
b) Rs. 680
c) Rs. 850
d) Rs. 1620
e) None of these
4). The ratio of money with Raju and Govind is 7 : 17 and that with Govind and Krish is 7: 17. If Raju has Rs. 490, Krish has ?
a) Rs. 2890
b) Rs. 2330
c) Rs. 1190
d) Rs. 2680
e) None of these
5). Points A and B are both in the line segment PQ and on the same side of its midpoint. A divides PQ in the ratio 2:3, and B divides PQ in the ratio 3: 4. If AB = 2, then the length of PQ is:
a) 70
b) 75
c) 80
d) 85
e) None of these
6). In a college, the number of boys is more than the numbers of girls by 12% of the total strength. The ratio of boys to girls is ?
a) 11 : 14
b) 14 : 11
c) 25 : 28
d) 28 : 25
e) None of these
7). The ratio of water and milk in 85 kg of contaminated milk is 7 : 27. The amount of water which must be added to make the ratio 3 : 1 is ?
a) 5 kg
b) 6.5 kg
c) 7.25 kg
d) 8 kg
e) 9.5 kg
8). A bag contains 25 paise, 10 paise and 5 paise coins in the ratio 1 : 2 : 3 . If their total value is Rs. 360, then the number of 5 paise coins is ?
a) 140
b) 150
c) 180
d) 240
e) 160
9). One year ago, the ratio between A’s and B’s salary was 4:5. The ratio of their individual salaries of last year and present year are 3:5 and 2:3 respectively. If their total salaries for the present year are Rs 13600, the present salary of A is?
a) Rs. 4080
b) Rs. 6400
c) Rs. 4533.40
d) Rs. 2720
e) None of these
10). A sum of Rs.324 was divided among 100 boys and girls in such a way that the boy gets Rs.3.60 and each girl Rs. 2.40 the number of girls is
a) 35
b) 40
c) 45
d) 50
e) None of these
Answers with Explanation:
1). Answer: e)
Let the third number be 100. Then,
1^{st} number = 135
2^{nd} number = 175
% 1^{st} to the 2^{nd} number = (135/175)*100 = 77.14%.
1^{st} number = 135
2^{nd} number = 175
% 1^{st} to the 2^{nd} number = (135/175)*100 = 77.14%.
2). Answer: a)
Let P’s income be = 4x
P’s expenses, therefore = 4x – 50
Let Q’s income be = 5x
Q’s expenses, therefore = 5x – 75
We know that the ratio of their expenses = 5 : 6
(4x – 50) / (5x – 75) = 5/6
24x – 300 = 25x – 375
Therefore, x = 75
P’s income = 4x = 4 x 75 = 300 and Q’s income = 5x = 5 x 75 = 375.
3). Answer: d)
(1/2)A = (1/3)B =(1/6)C = x
⇒ A = 2x, B = 3x, C =6x
∴ A : B : C = 2 : 3 : 6
Third part = Rs. (2970 x 6/11) = Rs. 1620
⇒ A = 2x, B = 3x, C =6x
∴ A : B : C = 2 : 3 : 6
Third part = Rs. (2970 x 6/11) = Rs. 1620
4). Answer: a)
Raju : Govind = 7 : 17 = 49 : 119
Govind : Krish = 7 : 17 = 119 : 289
Govind : Krish = 7 : 17 = 119 : 289
Raju = 7*7 =49; Govind = 17 * 7 = 119; Krish = 17 * 17 = 289]
∴ Raju : Govind : Krish = 49 : 119 : 289
Raju : Krish = 49 : 289
Thus, 49 : 289 = 490 : N
N = 289 x 490 / 49
N = Rs. 2890
∴ Raju : Govind : Krish = 49 : 119 : 289
Raju : Krish = 49 : 289
Thus, 49 : 289 = 490 : N
N = 289 x 490 / 49
N = Rs. 2890
5). Answer: a)
Let PA = 2x and AB = 3x
and PB = 3y and BQ = 4y
PB : BQ = 3 : 4
PA : QA = 2 : 3
PQ = 5x = 7y
X = 7y/5 …..(i)
From equation (i) and (ii),
Now, AB = PQ – PA − BQ = 7y− 4y− 2x
3y − 2x = 2 ————– (ii)
y = 10 and Hence, PQ = 70.
6). Answer: b)
Let the number of boys and girls be x and y respectively. Then
From question, (x – y) = 12% of (x + y)
x – y = 3/25 (x + y )
25x – 25y = 3x + 3y
22x = 28y
∴ x / y = 28 / 22 = 14 / 11
From question, (x – y) = 12% of (x + y)
x – y = 3/25 (x + y )
25x – 25y = 3x + 3y
22x = 28y
∴ x / y = 28 / 22 = 14 / 11
7). Answer: a)
Milk = 85 x (27/34) kg = 135/2 kg = 67.5 kg
Water = 85 x (7/34) kg = 35/2 kg = 17.5 kg
Let Amount of water to be added is x,
∵ 67.5 / (17.5 + x) = 3/1
3 (17.5 + x) = 67.5
∴ x = 5
∴ Water to be added = 5 kg.
Water = 85 x (7/34) kg = 35/2 kg = 17.5 kg
Let Amount of water to be added is x,
∵ 67.5 / (17.5 + x) = 3/1
3 (17.5 + x) = 67.5
∴ x = 5
∴ Water to be added = 5 kg.
8). Answer: c)
Let the number of 25 paise, 10 paise and 5 paise coins be N, 2N and 3N, respectively. Then ,
N + 2N + 3N = 360
N = 60
∴ The number of 5 paise coins = 3 x 60 = 180.
N + 2N + 3N = 360
N = 60
∴ The number of 5 paise coins = 3 x 60 = 180.
9). Answer: b)
Ratio of A’s last year and present year salary = 3:5
Let salary be 3x and 5x respectively.
Ratio of B’s in last year and present year salary = 2:3
Let salary be 2y and 3y respectively.
Given that, 3x / 2y = 4 / 5
15x = 8y ———– (i)
Also, given that, 5x + 3y = 13600 ———– (ii)
From equation (i) and (ii), we get
y = 2400 and x = 1280.
A’s present salary =5x = 5 × 1280 = 6400
10). Answer: e)
Let x be the number of boys and y be the number of girls.
Given total number of boys and girls = 100
x + y=100 —— (i)
A boy gets Rs. 3.60 and a girl gets Rs. 2.40
The amount given to 100 boys and girls = Rs. 324
3.60x + 2.40y = 324 ——– (ii)
Solving (i) and (ii)
3.60x + 3.60y = 360 ——— Multiply (i) by 3.60
3.60x + 2.40y = 324 ——— (ii)
1.20y = 36
y = 36 / 1.20
= 30–>Number of girls = 30
Given total number of boys and girls = 100
x + y=100 —— (i)
A boy gets Rs. 3.60 and a girl gets Rs. 2.40
The amount given to 100 boys and girls = Rs. 324
3.60x + 2.40y = 324 ——– (ii)
Solving (i) and (ii)
3.60x + 3.60y = 360 ——— Multiply (i) by 3.60
3.60x + 2.40y = 324 ——— (ii)
1.20y = 36
y = 36 / 1.20
= 30–>Number of girls = 30
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