Explanation / Important formulas:
Ratio: The ratio of two quantities a and b in the same units is the fraction a/b and we write it as a : b.
 In the ratio a : b, we call a as the first term or antecedent and b, the second term or consequent.
 Example: The ratio 5 : 9 represents 5/9 With antecedent = 5, consequent = 9.
[Rule: The multiplication or division of each term of a ratio by the same nonzero number does not affect the ratio.]
Example: 4 : 5 = 8 : 10 = 12 : 15. Also, 4 : 6 = 2 : 3.
Proportion: The equality of two ratios is called proportion.
 If a : b = c : d, we write a : b :: c : d and we say that a, b, c, d are in proportion.
 Here a and d are called extremes, while b and c are called mean terms.
 Product of means = Product of extremes.
 Thus, a : b :: c : d (b x c) = (a x d).
Fourth Proportional: If a : b = c : d, then d is called the fourth proportional to a, b, c.
Third Proportional: a : b = c : d, then c is called the third proportion to a and b.
Mean Proportional: Mean proportional between a and b is ab.
Comparison of Ratios: We say that (a : b) > (c : d) <=> a / b > c / d
Compounded Ratio: The compounded ratio of the ratios: (a : b), (c : d), (e : f) is (ace : bdf).
Duplicate Ratios: Duplicate ratio of (a : b) is (a^{2} : b^{2}).
 Subduplicate ratio of (a : b) is (a^{1/2} : b^{1/2}).
 Triplicate ratio of (a : b) is (a^{3} : b^{3}).
 Subtriplicate ratio of (a : b) is (a^{1/3} : b^{1/3}).
 If a/b = c/d, then a + b / a – b =c + d / c – d [Componendo and dividendo]
Variations
 We say that x is directly proportional to y, if x = ky for some constant k and we write, x ∝ y.
 We say that x is inversely proportional to y, if xy = k for some constant k and we write, x ∝ 1/ y
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Test  Ratio and Proportion
Ratio and proportion  Question and Answers
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Question 1

10:11


11:12


12:13


13:14

pi * r^2 /pi * R^2 = 169 / 196
r^2 / R^2 = 169 / 196 = (13/14)^2
so r/R = 13/14
Question 2

3 : 3 : 10


10 : 11 : 20


23 : 33 : 60


Cannot be determined

A's new salary = (115 / 100) of 2k = (115 / 100) x 2k = 23k / 10
B's new salary = (110 / 100) of 3k = (110 / 100) x 3k = 33k / 10
C's new salary = (120 / 100) of 5k = (120 / 100) x 5k = 6k
New ratio = [23k / 10 : 33k / 10 : 6k]
= 23 : 33 : 60
Question 3

2:5


3:7


5:3


7:3

Then, 40A / 100 = 2B / 3
2A / 5 = 2B / 3
A / B = (2 / 3 * 5 / 2) = 5 / 3
A : B = 5 : 3
Question 4

18


24


19


20

Then, 5 : 8 : 15 : x
5x = (8 x 15)
x = (8 x 15) / 5 = 24
Question 5

27


33


49


55

Then, 3x  9 / 5x  9 = 12 / 23
23 (3x  9) = 12 (5x  9)
9x = 99
x = 11
The smaller number = (3 x 11) = 33
Question 6

50


100


150


200

(60x / 100) = 30
x = (30 x 100) / 60 = 50
Hence, the number of 5 p coins = (3 x 50) = 150
Question 7

8:9


17:18


21:22


Cannot be determined

Their increased number is (120% of 7x) and (110% of 8x)
(120 / 100 * 7x) And (110 / 100 * 8x)
42x / 5 and 44x / 5
The required ratio = 42x / 5 : 44x / 5 = 21 : 22
Question 8

17,000


20,000


25,500


38000

Then, (2x + 4000) / (3x + 4000) = 40 / 57
57(2x + 4000) = 40(3x + 4000)
6x = 68,000
3x = 34,000
Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000
Question 9

1.12


1.20


1.25


1.30

Question 10

20


30


48


58

A : B = 2 : 3 and B : C = 5 : 8 = (5 x 3 / 5 : 8 x 3 / 5) = 3 : (24 / 5)
A : B : C = 2 : 3 : (24 / 5) = 10 : 15 : 24
B = 98 x (15 / 49) = 30
Question 11

460


484


550


664

A = (2/5 * 15/4)B
A = 3 / 2B
A / B = 3 / 2
A : B = 3 : 2
B’s share = Rs. (1210 * (2 / 5)) = Rs. 484
Question 12

2 : 5


3 : 5


4 : 5


6 : 7

Then the first number = 120% of x = 120x / 100 = 6x /5
Second number = 150% of x = 150x / 100 = 3x /2
Ratio of first two numbers = 6x / 5 : 3x / 2 = 12x : 15x = 4 : 5
Question 13

500


1500


2000


None of these

Then, 4x  3x = 1000
x = 1000
B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000
Question 14

2 : 3: 4


6 : 7 : 8


6 : 8 : 9


None of these

Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x)
= (140/100 * 5x) ,(150/100 * 7x) and (175/100 * 8x)
= 7x, 21x /2 and 14x
The required ratio = 7x : 21x /2 : 14x
= 14x : 21x : 28x
= 2 : 3 : 4
Question 15

then the quantity of water to be further added is?
20


30


40


60

Quantity of water in it = (60  40) litres = 20 litres
New ratio = 1 : 2
Let quantity of water to be added further be x litres
Then, milk : water = (40 / 20 + x )
Now, (40 / (20 + x)) = 1 / 2
20 + x = 80
x = 60
Quantity of water to be added = 60 litres
Question 16

35


40


45


50

Given total number of boys and girls = 100
x + y = 100
A boy gets Rs. 3.60 and a girl gets Rs. 2.40
The amount given to 100 boys and girls = Rs. 312
3.6x + 2.4y = 312
By Solving
3.6x + 3.6y = 360 multiplying by 3.6
3.6x + 2.4y1.20yy = 312 = 48 = 40
The number of girls is 40
Question 17

20


30


40


60

Quantity of water in it = (60 − 40) litres = 20 litres
New ratio = 1 : 2
Let, the quantity of water to be added further be x litres
Then milk : water = 4020 + x
Now, 4020 + x20 + x ⇒x =12 = 80 = 60
Question 18

2 : 5


3 : 5


4 : 5


5 : 4

Then, first number = 120% of x = 120 x 100=6 x 5
Second number = 150% of x = 150 x 100 = 3 x 2
Ratio of first two numbers = 6 x 5 : 3 x 2 = 4 : 5
Question 19

48


54


56


60

Let x be number of 10p coins and y be number of 25p coins
x + y = 180 (i)
Step (ii) Given 10p coins and 25p coins make the sum = Rs. 36.90
10x 100 + 25y 100 ⇒ 10x + 25y = 36.9 = 3690(ii)
Step (iii)
Solving (i) and (ii),
10x + 10y10x + 25y − 15yy = 1800 = 3690 = 1890 = 126
Substitute y value in equation (i)
x = 180 – 126 = 54
Number of 10p coins = 54
Question 20

3 : 3 : 10


10 : 11 : 20


23 : 33 : 60


Cannot be determined

A's new salary = (115100 × 2k) = 23k10
B's new salary = (110100 × 3k) = 33k10
C's new salary = (120100 × 5k) = 6k
⇒New ratio = (23k10 : 33k10 : 6k) = 23 : 33 : 60
Question 21

Rs. 17,000


Rs. 20,000


Rs. 34,000


Rs. 38,000

then, 2x + 40003x + 4000 ⇒57 × (2x + 4000) 6x3x = 4057 = 40 × (3x + 4000) = 68,000 = 34,000
Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000
Question 22

20


30


48


58

A:B=2:3 and B:C=5:8
=> 5 × 35 : 8 × 35 ⇒ A : B : C = 2 : 3 : 245 = 10 : 15 : 24
=> B = 98 × 1549 = 30 = 3 : 245
Question 23

Rs.192


Rs. 133


Rs. 144


Rs. 128

Note that this is not the same as the ratio of their wages being 8 : 12 : 6
In this case, find out the L.C.M of 8, 12 and 6 and divide the L.C.M by each of the above numbers
to get the ratio of their respective shares
The L.C.M of 8, 12 and 6 is 24
Therefore, the ratio A : B : C is
248 : 2412 : 246 ⇒ A : B : C :: 3 : 2 : 4
The sum of the total wages =3x + 2x + 4x = 432
9x = 432 or x = 48
Hence A gets 3 × 48 = Rs 144
Question 24

120 men


122 men


128 men


134 men

Let us therefore, find out the amount of men required
If only men were working on the job, to complete the new job under the new conditions and then make adjustments for the women and children working with the men
The man hours required to complete the new job = 4 times the man hours required to complete the old job
(As the new job is 4 times as big as the old job)
Let n be the number of men required
n × 5 × 12 = 20 × 8 × 12 × 4 = or n = 128
i.e. 128 men working on the job will be able to complete the given job
However, the problem states that 6 women and 2 boys are working on the job
6 women $ = \dfrac{6}{1.2} = 5 men and 2 boys = 1 man
∴ The equivalent of 5 + 1 = 6 men are already working
Thus, final number of men working
= 128 − 6 = 122 men
Question 25

37%


36%


27%


30%

Rs 90 is to be divided in the ratio 12 : 16 : 12. i.e 3 : 4 : 3
R gets 310 × 90 = 27 and 10 for managing
Thus 27 + 10 = 37%
Question 26

Rs 600


Rs 840


Rs 900


Rs 1,200

So the proportion to teacher's share = 1.57
Hence, the teachers would donate 1.57 × 4200 = Rs 900
Question 27

Rs 7 Lakh


Rs 4 Lakh


Rs 3 Lakh


Rs 1.5 Lakh

IBM's final quote = (4x1) Lakh
Thus, 4x − 14x ⇒ x = 34 = 1
IBM's bid winning price = Rs 3. Lakh
So IBM wins the bid at 4x1= Rs. 3 lakh
Question 28

2 : 1


5 : 1


99 : 61


227 : 133

Proportion of water in 3 samples is 13,25,38
Since equal quantities are taken,
Total proportion of milk is 23 + 35 + 58 = 227120
Total proportion of water is 13 + 25 + 38 = 133120
Proportion of milk and water in the solution is = 227 : 133
Question 29

1610


2010


2410


1590

A + D + P = 4830
Hence, A's, D's and P's share are diminished by Rs 5, Rs 10 and Rs 15, their net share will be Rs.4800
Dishant's diminished share = 412 × 4800 = Rs 1600
Hence, Dishant actual share = Rs 1600 + Rs 10 = Rs 1610
Question 30

15


25


35


40

= ∑n2 = n(n + 1)(2n + 1)6
Squares of sum of first n natural numbers
= (∑n)2 = n(n + 1)2 × n (n + 1)2
Now the ratio is
= n(n +1)(2n + 1) 6 : n (n+1) 2 × n (n+1) 2= 17 : 325
⇒ (2n + 13)(n (n + 1) 2) = 17325
⇒2(2n + 1)3n(n + 1) = 17325
⇒ 650 (2n + 1) = 51n (n + 1)
⇒ 1300n + 650 = 51n2 + 51n
⇒ 51n2 − 1300n + 51n − 650 = 0
⇒ 51n2 − 1249n − 650 = 0
Upon solving the above equation, we get,
n = 25, − 0.5098
Out of these two,
n = 25 is there in the solution
Question 31

18


12


16


15

all three of them got an equal number of eggs
So each of them must have got 4 = (123) eggs
The first and the second man must have given 3 = (7 − 4) eggs and 1= (5 − 4) egg to the third man
Hence, the ratio of their share is 3 : 1
Now, money has to be distributed in the ratio of their contribution. As total money paid by the stranger is Rs. 24
Hence the first and the second man get Rs 18 = (24 × 33 + 1) and Rs 6 = (24 × 13 + 1) respectively
The 1st man gets Rs 12 = (18 − 6) more than the second
Question 32

162


108


72


None

B : C = 3 : 2 = 6 : 4 (making B equal)
Therefore, A : B : C = 9 : 6 : 4
Therefore, the runs made by A = 919 × 342 = 162
Question 33

20 km/hr


30 km/hr


35 km/hr


40 km/hr

64 = k × 16 × 16
k = 1/4
Cost = ^{ }speed^{2}/4
Total cost = (speed^{2 }/ 4 + 400 ) x 100 / speed
using options, putting different values of speed, we find speed = 40km/hr to be most economical
Question 34

Rs. 8000


Rs. 6000


Rs. 10,000


Rs. 11,000

⇒ E= 64 and S = 16
K=1/4
Total cost =1/4 × s^{2}t + 400t
=14 × s ^{2}t + 400t
Most economical speed, checking options we get most economical speed at 40km/hr
Total cost at 40km/hr
= ¼ x 40 ^{2} + 400 ( 400 / 40)
= Rs. 8000
Question 35

70


75


80


85

and PB = 3y and BQ = 4y
PB : BQ = 3 : 4
PA : QA = 2 : 3
PQ = 5x = 7y
x = 7 / 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
Question 36

415 ml


403 ml


400 ml


424 ml

p250 = q650
= r200
= p + q + r250 + 650 + 200
Thus p + q + r = 682
⇒q = 403 ml
Question 37

(14/3)3/2:8


8:(14/3)3/2


(3/14)3/2:8


8:(3/14)3/2

= (14/3)3/2l3 = (14/3)3/2:8
Question 38

120 and 210


80 and 140


180 and 300


200 and 350

Therefore, 4x + 257x + 255(4x + 25)20x + 125x = 35 = 3(7x + 25) = 21x + 75 = 50
Therefore, their salaries are 4 × 50 and 7 × 50 i.e., 200 and 350
Question 39

240


360


480


120

Lets the fares collected from each of the first class and second class passengers be 30y and y respectively
Hence
x(30y) + 3x(y) = 30xy + 3xy = 33xy = 1320
xy = 40
Total amount collected from the second class
= 3xy = 3 × 40 = Rs 120
Question 40

28


35


25


27

So, in case of correct distribution, P would have got,
=> 117 × 66 + 4 + 3 = Rs. 54
Q would have got, = 117 × 46 + 4 + 3 = Rs. 36
and R would have got, = 117 × 36 + 4 + 3 = Rs. 27
But actually the money was divided in the ratio 2 : 3 : 4 and shares of P, Q and R in this case would be
For P, = 117 × 22 + 3 + 4 = Rs. 26
For Q, = 117 × 32 + 3 + 4 = Rs. 39,
and for R, = 117 × 42 + 3 + 4 = Rs. 52
Thus P gains, = 54 − 26 = 28,
Q gains, = 36 − 39 = −3 (loss)
and R gains, = 27 − 52 = −25 (loss)
Thus, P gains the most Rs. 28
Question 41

1 : 4


1 : 5


5 : 6


1 : 6

Let total number of students in the class who are economically poor be 'p'
Total number of students who failed = t5
Total number of male students who failed
= 34 × t5 = 9p10
⇒ pt = 16
⇒ Number of economically poor studentsNumber of Students = 16 = 1 : 6
Question 42

42, 28


36, 24


30, 20


25, 15

Choice (C) 30 and 20 are in the ratio of 6 : 4
Five years ago their ages would be 25 and 15 which are in the ratio 5 : 3
Hence choice (C) is the right answer
Question 43

4888.88


9333.33


4000


3666.66

Their investment ratio = 3600 : 4400 : 2800 = 9 : 11 : 7
A's profit of Rs. 8000 = 14 × x + 13 × 34 × x =12 × x
x = Rs 16,000
Therefore B's profit = 1127 × 34 × 16,000 = Rs 4888.88
Question 44

Rs 20, Rs 10


Rs 24, Rs 12


Rs 16, Rs 8


Rs 26, Rs 13

⇒ the price of other noodle = k2
Price of 1 kg = 2k5 + 35 × k2 = 7k10
But CP = 17.50 × 100125 = 14
⇒ 7k10 = 14
⇒ k = 20
So price of the noodles's are 20 and 10
Question 45

945


630


252


238

Then 7k − 3k = 63p (where p is any positive integer)
As the number is a multiple of both 9 and 7, it has to be multiple of 63
⇒ k = 63p4
Minimum value of p for which k is a natural number is 4
Thus, k=63
Hence, the number of pets = 15k = 945
Question 46

1249


1023


1202


1246

A took (x + 1) biscuit
Now 4x is of the form 5y + 1 then x must be in the form 5z + 4
⇒4(5z + 4) = 5y + 1
⇒ y = 4z + 3 and x = 5z + 4
The ratio of number of biscuits that A and B took is
[(5z + 4) +1] : [(4z + 3) + 1] = 5 : 4
So, we can say that any two successive persons A, B, C and D take coins in the ratio of 5 : 4
Let the number of biscuits that A, B, C and D took be a, b, c and d respectively
a : b = b : c = c : d = 5 : 4
a : b : c : d = 125 : 100 : 80 : 64
⇒ a = 125k
⇒ x = 125k − 1 and N = 5x + 1 = 625k − 4
As, N>1000, the least value of N is when k = 2
N = 1246