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 non-zero 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 (a2 : b2).
- Sub-duplicate ratio of (a : b) is (a1/2 : b1/2).
- Triplicate ratio of (a : b) is (a3 : b3).
- Sub-triplicate ratio of (a : b) is (a1/3 : b1/3).
- 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
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If the area of two circles are in the ratio 169:196 then the ratio of their radii is?
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10:11
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11:12
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12:13
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13:14
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Question 1 Explanation:
If r and R are radii if two circles, then
pi * r^2 /pi * R^2 = 169 / 196
r^2 / R^2 = 169 / 196 = (13/14)^2
so r/R = 13/14
pi * r^2 /pi * R^2 = 169 / 196
r^2 / R^2 = 169 / 196 = (13/14)^2
so r/R = 13/14
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Question 2
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The salaries of A, B, C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries?
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3 : 3 : 10
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10 : 11 : 20
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23 : 33 : 60
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Cannot be determined
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Question 2 Explanation:
Let A = 2k, B = 3k and C = 5k
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
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
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Question 3
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If 40% of a number is equal to two-third of another number, what is the ratio of first number to the second number?
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2:5
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3:7
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5:3
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7:3
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Question 3 Explanation:
Let 40% of A = 2 / 3 B
Then, 40A / 100 = 2B / 3
2A / 5 = 2B / 3
A / B = (2 / 3 * 5 / 2) = 5 / 3
A : B = 5 : 3
Then, 40A / 100 = 2B / 3
2A / 5 = 2B / 3
A / B = (2 / 3 * 5 / 2) = 5 / 3
A : B = 5 : 3
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Question 4
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The fourth proportional to 5, 8, 15 is?
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18
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24
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19
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20
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Question 4 Explanation:
Let the fourth proportional to 5, 8, 15 be x
Then, 5 : 8 : 15 : x
5x = (8 x 15)
x = (8 x 15) / 5 = 24
Then, 5 : 8 : 15 : x
5x = (8 x 15)
x = (8 x 15) / 5 = 24
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Question 5
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Two number are in the ratio 3 : 5. If 9 is subtracted from each, the new numbers are in the ratio 12 : 23. The smaller number is?
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27
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33
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49
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55
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Question 5 Explanation:
Let the numbers be 3x and 5x
Then, 3x - 9 / 5x - 9 = 12 / 23
23 (3x - 9) = 12 (5x - 9)
9x = 99
x = 11
The smaller number = (3 x 11) = 33
Then, 3x - 9 / 5x - 9 = 12 / 23
23 (3x - 9) = 12 (5x - 9)
9x = 99
x = 11
The smaller number = (3 x 11) = 33
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Question 6
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In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1 : 2 : 3. If there is Rs. 30 in all, how many 5 p coins are there?
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50
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100
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150
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200
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Question 6 Explanation:
Then, sum of their values = Rs. {(25x / 100) +[(10 * 2x) / 100)]+ (5 * 3x /100)} = Rs. 60x / 100
(60x / 100) = 30
x = (30 x 100) / 60 = 50
Hence, the number of 5 p coins = (3 x 50) = 150
(60x / 100) = 30
x = (30 x 100) / 60 = 50
Hence, the number of 5 p coins = (3 x 50) = 150
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Question 7
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The ratio of the number of boys and girls in a college is 7 : 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?
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8:9
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17:18
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21:22
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Cannot be determined
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Question 7 Explanation:
Originally, let the number of boys and girls in the college be 7x and 8x respectively
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
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
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Question 8
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Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40 : 57. What is Sumit's salary?
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17,000
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20,000
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25,500
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38000
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Question 8 Explanation:
Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively
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
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
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Question 9
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If 0.75 : x :: 5 : 8, then x is equal to?
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1.12
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1.20
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1.25
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1.30
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Question 9 Explanation:
(x * 5) = (0.75 * 8) x = 6 / 5 = 1.20
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Question 10
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The sum of three numbers is 98. If the ratio of the first to second is 2 : 3 and that of the second to the third is 5 : 8, then the second number is?
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20
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30
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48
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58
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Question 10 Explanation:
Let the three parts be A, B, C. Then,
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
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
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Question 11
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A and B together have Rs. 1210. If 4/15 of A's amount is equal to 2/5 of B's amount, how much amount does B have?
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460
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484
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550
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664
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Question 11 Explanation:
4 /15A = 2 /5B
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
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
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Question 12
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Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is?
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2 : 5
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3 : 5
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4 : 5
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6 : 7
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Question 12 Explanation:
Let the third number be x
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
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
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Question 13
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A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B's share?
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500
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1500
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2000
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None of these
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Question 13 Explanation:
Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively
Then, 4x - 3x = 1000
x = 1000
B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000
Then, 4x - 3x = 1000
x = 1000
B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000
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Question 14
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Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?
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2 : 3: 4
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6 : 7 : 8
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6 : 8 : 9
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None of these
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Question 14 Explanation:
Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively
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
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
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Question 15
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In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2,
then the quantity of water to be further added is?
then the quantity of water to be further added is?
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20
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30
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40
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60
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Question 15 Explanation:
Quantity of milk = (60 * 2/3)litres = 40 litres
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
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
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Question 16
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A sum of Rs.312 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?
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35
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40
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45
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50
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Question 16 Explanation:
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
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
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
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Question 17
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In a mixture 60 litres, the ratio of milk and water 2:1. If the this ratio is to be 1:2, then the quantity of water to be further added is?
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20
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30
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40
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60
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Question 17 Explanation:
Quantity of milk = 60 × 23 = 40 litres
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
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
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Question 18
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Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is?
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2 : 5
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3 : 5
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4 : 5
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5 : 4
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Question 18 Explanation:
Let the third number be x
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
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
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Question 19
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A sum of Rs. 36.90 is made up of 180 coins which are either 10 paise coins or 25 p coins. The number of 10 p coins is?
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48
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54
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56
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60
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Question 19 Explanation:
Total number of coins = 180
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
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
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Question 20
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The salaries A, B, C are in the ratio 2:3:5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries?
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3 : 3 : 10
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10 : 11 : 20
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23 : 33 : 60
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Cannot be determined
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Question 20 Explanation:
Let A = 2k, B = 3k and C = 5k
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
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
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Question 21
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Salaries of Ravi and Sumit are in the ratio 2:3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40:57. What is Sumit's salary?
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Rs. 17,000
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Rs. 20,000
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Rs. 34,000
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Rs. 38,000
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Question 21 Explanation:
Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively
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
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
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Question 22
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The sum of three numbers is 98. If the ratio of the first to second is 2:3 and that of the second to the third is 5:8, then the second number is?
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20
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30
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48
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58
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Question 22 Explanation:
Let the three parts be A, B, C. Then,
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
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
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Question 23
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Rs.432 is divided amongst 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?
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Rs.192
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Rs. 133
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Rs. 144
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Rs. 128
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Question 23 Explanation:
8 times A’s share = 12 times B’s share = 6 times C’s share
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
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
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Question 24
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If 20 men or 24 women or 40 boys can do a job in 12 days working for 8 hours a day, how many men working with 6 women and 2 boys take to do a job four times as big working for 5 hours a day for 12 days?
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120 men
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122 men
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128 men
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134 men
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Question 24 Explanation:
Amount of work done by 20 men = 24 women = 40 boys or 1 man = 1.2 woman = 2 boys
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
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
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Question 25
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P,Q and R enter into a partnership with capitals in the ratio 3:2:1. After 4 months, P leaves the business and after 4 more months Q also leaves the business and R continues till the end of the year. If R takes 10 of the profit for managing the business, then what part of the profit does R get?
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37%
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36%
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27%
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30%
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Question 25 Explanation:
Let Rs 100 be the profit
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%
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%
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Question 26
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An outgoing batch of students wants to gift a PA system worth Rs 4,200 to their school. If the teachers, offer to pay 50% more than the students and an external benefactor gives three times the teacher's contribution, then how much should the teachers donate?
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Rs 600
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Rs 840
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Rs 900
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Rs 1,200
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Question 26 Explanation:
The ratio of the share students : teacher : benefactor = 1 : 1.5 : 4.5
So the proportion to teacher's share = 1.57
Hence, the teachers would donate 1.57 × 4200 = Rs 900
So the proportion to teacher's share = 1.57
Hence, the teachers would donate 1.57 × 4200 = Rs 900
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Question 27
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IBM and KTC quote for a tender. On the tender opening day, IBM realizes that their quotations are in the ratio 7:4 and hence decreases its price during negotiations to make it Rs 1 Lakh lower than KTC's quoted price. KTC realizes that the final quotes of the two were in the ratio 3:4. What was the price at which IBM won the bid?
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Rs 7 Lakh
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Rs 4 Lakh
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Rs 3 Lakh
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Rs 1.5 Lakh
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Question 27 Explanation:
IBM initially quoted Rs 7x lakh. KTC quoted 4x 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
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
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Question 28
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The proportion of milk and water in 3 samples is 2:1, 3:2 and 5:3. A mixture comprising of equal quantities of all 3 samples is made. The proportion of milk and water in the mixture is?
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2 : 1
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5 : 1
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99 : 61
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227 : 133
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Question 28 Explanation:
Proportion of milk in 3 samples is 23,35,58
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
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
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Question 29
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Rs 4830 is divided among Abhishek, Dishant and Prashant such that if Abhishek's share diminishes by Rs 5, Dishant's share diminishes by Rs 10 and Prashant's share diminishes by Rs 15, their shares will be in the ratio 5:4:3. Find the Dishant's original share?
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1610
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2010
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2410
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1590
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Question 29 Explanation:
Let actual share of Abhishek, Dishant and Prashant be A, D , P respectively
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
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
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Question 30
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The ratio of squares of first n natural numbers to square of sum of first n natural numbers is 17:325. The value of n is?
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15
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25
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35
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40
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Question 30 Explanation:
Sum of squares of first n natural numbers
= ∑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
= ∑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
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Question 31
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Two men, sitting on the dining table. One man has 7 eggs and other had 5 eggs. A third man passing by requested them to share their food in return for money. The three of them shared the eggs equally and the third traveler paid the other two a total of Rs 24. Find the difference between the amounts received by first two men?
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18
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12
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16
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15
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Question 31 Explanation:
As the two men had a total of 12 = ( 7 + 5) eggs, also they agreed to share eggs in such a manner that
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
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
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Question 32
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A, B and C play cricket. A's runs are to B's runs and B's runs are to C's as 3:2. They get altogether 342 runs. How many runs did A make?
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162
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108
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72
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None
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Question 32 Explanation:
A : B = 3 : 2 = 9 : 6
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
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
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Question 33
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The most economical speed for this journey is?
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20 km/hr
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30 km/hr
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35 km/hr
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40 km/hr
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Question 33 Explanation:
Cost = K2 ×speed2[ Where k is constant]
64 = k × 16 × 16
k = 1/4
Cost = speed2/4
Total cost = (speed2 / 4 + 400 ) x 100 / speed
using options, putting different values of speed, we find speed = 40km/hr to be most economical
64 = k × 16 × 16
k = 1/4
Cost = speed2/4
Total cost = (speed2 / 4 + 400 ) x 100 / speed
using options, putting different values of speed, we find speed = 40km/hr to be most economical
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Question 34
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The total cost for the journey at this most economical speed is?
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Rs. 8000
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Rs. 6000
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Rs. 10,000
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Rs. 11,000
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Question 34 Explanation:
E=ks2
⇒ E= 64 and S = 16
K=1/4
Total cost =1/4 × s2t + 400t
=14 × s 2t + 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
⇒ E= 64 and S = 16
K=1/4
Total cost =1/4 × s2t + 400t
=14 × s 2t + 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
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Question 35
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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?
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70
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75
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80
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85
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Question 35 Explanation:
Let PA = 2x and AB = 3x
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
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
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Question 36
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Three beakers have capacity of 250ml, 650ml and 200ml. 682 ml of Juice is poured into them so that the same fraction of each is filled. The volume filled in the largest beaker will be?
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415 ml
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403 ml
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400 ml
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424 ml
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Question 36 Explanation:
As the beaker filled by the same fractions,
p250 = q650
= r200
= p + q + r250 + 650 + 200
Thus p + q + r = 682
⇒q = 403 ml
p250 = q650
= r200
= p + q + r250 + 650 + 200
Thus p + q + r = 682
⇒q = 403 ml
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Question 37
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The total surface area of a solid iron cube and a solid aluminium cuboid are the same. The length, breadth and height of the cuboid are in the ratio 1:2:4. Both are melted together in a vessel. What is the ratio of iron and aluminium in the resultant mixture?
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(14/3)3/2:8
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8:(14/3)3/2
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(3/14)3/2:8
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8:(3/14)3/2
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Question 37 Explanation:
lbh = 2 (lb + bh + hl)=1 : 2 : 4 = 2l , h = 4l = 2×(2l2 + 8l2 + 4l2)=28l2
= (14/3)3/2l3 = (14/3)3/2:8
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Question 38
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The monthly salaries of two persons are in the ratio of 4:7. If each receives an increase of Rs.25 in the salary, the ratio is altered to 3:5. Find their respective salaries?
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120 and 210
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80 and 140
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180 and 300
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200 and 350
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Question 38 Explanation:
Let the salaries be 4x and 7x
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
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
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Question 39
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On a certain day, the ratio of the passenger in the 1st class and the second class travelling by train is 1:3. The ratio of the fares collected from each first class and second class passengers is 30:1. If the total amount collected from all the passengers is Rs 1,320. Find the amount in Rs, collected from the second class passengers?
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240
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360
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480
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120
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Question 39 Explanation:
Let the number of passengers travelling by first class and second class be x and 3x respectively
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
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
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Question 40
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By mistake, instead of dividing Rs 117 among three persons P, Q and R in the ratio (12,13,14), it was divided in the ratio 2:3:4. Who gains the most and how much?
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28
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35
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25
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27
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Question 40 Explanation:
Ratio 12,13,14 is equivalent to 6 : 4 : 3
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
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
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Question 41
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20 of the students in a class failed in an examination. Out of the students who failed, 75 were males. Male students who failed constitute 90 of the economically poor students in the class. What is the ratio of the number of economically poor students to the number of students in the class?
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1 : 4
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1 : 5
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5 : 6
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1 : 6
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Question 41 Explanation:
Let the total number of students in the class be 't'
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
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
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Question 42
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The present ages of A and B are as 6:4. Five years ago their ages were in the ratio 5:3. Find their present ages?
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42, 28
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36, 24
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30, 20
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25, 15
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Question 42 Explanation:
Go from the choices
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
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
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Question 43
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A, B and C enter into a partnership by investing Rs.3600, Rs.4400 and Rs.2800. A is a working partner and gets a fourth of the profit for his services and the remaining profit is divided amongst the three in the rate of their investments. What is the amount of profit that B gets if A gets a total of Rs. 8000?
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4888.88
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9333.33
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4000
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3666.66
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Question 43 Explanation:
Let x be the profit
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
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
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Question 44
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A noodles merchant buys two varieties of noodles the price of the first being twice that of the second. He sells the mixture at Rs 17.50 per kilogram thereby making a profit of 25. If the ratio of the amounts of the first noodles and the second noodles in the mixture is 2:3, then the respective costs of each noodles are?
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Rs 20, Rs 10
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Rs 24, Rs 12
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Rs 16, Rs 8
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Rs 26, Rs 13
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Question 44 Explanation:
Let the price of one noodles = k
⇒ 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
⇒ 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
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Question 45
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In a house, there are dogs, cats and parrot in the ratio 3:7:5. If the number of cats was more than the number of dogs by a multiple of both 9 and 7, what is the minimum of pets in the house?
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945
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630
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252
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238
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Question 45 Explanation:
If three kinds of pets are taken be 3k,7k and 5k respectively,
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
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
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Question 46
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There are N numbers of apples in the house, in which four people are lived. If the first men woke up and divided the apples into 5 equal piles and found one extra apple. He took one of those piles along with the extra apple and hid them. He then gathered the 4 remaining piles into a big pile, woke up the second person and went to sleep. Each of the other 3 persons did the same one by one i.e. divided the big pile into 5 equal piles and found one extra apple. Each hid one of the piles along with the extra apple and gathered the remaining 4 piles into a big pile. If N>1000, what could be the least value of N?
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1249
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1023
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1202
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1246
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Question 46 Explanation:
Suppose N=5x + 1
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
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
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