1. Total number of men, women and children working in a factory is 18. They earn Rs. 4000 in a day. If the sum of the wages of all men, all women and all children is in ratio of 18 : 10 : 12 and if the wages of an individual man, woman and child is in ratio 6 : 5 : 3, then how much a woman earn in a day?
a) Rs. 120
b) Rs. 150
c) Rs. 250
d) Rs. 4000
Discussion
Explanation: Ratio of number of men, women and children,
$$\eqalign{ & = \frac{{18}}{6}:\frac{{10}}{5}:\frac{{12}}{3} \cr & = 3:2:4 \cr} $$
Total (Men + Women + Children) = 18
3X + 2X + 4X = 18
9X = 18
X = 2
Hence, number of women = 2X = 2 × 2 = 4
Share of all women = $$\frac{{10 \times 4000}}{{40}}$$ = Rs. 1000 [18 + 10 + 12 = 40]
Share of each woman = $$\frac{{1000}}{4}$$ = Rs. 250
2. A and B are two alloys in which ratios of gold and copper are 5 : 3 and 5 : 11 respectively. If these equally amount of two alloys are melted and made alloy C. What will be the ratio of gold and copper in alloy C?
a) 25 : 23
b) 33 : 25
c) 15 : 17
d) 17 : 15
Discussion
Explanation : Ratio of Gold and Copper in Alloy A = 5 : 3
Ratio of Gold and Copper in Alloy B = 5 : 11
Amount of Gold in Alloy A = $$\frac{5}{8}$$
Amount of Gold In Alloy B = $$\frac{5}{{16}}$$
Amount of Copper in A = $$\frac{3}{8}$$
Amount of Copper in B = $$\frac{{11}}{{16}}$$
Amount of Gold In C,
= (Amount of gold in A + Amount of gold in B) = $$\frac{5}{8}$$ + $$\frac{5}{{16}}$$ = $$\frac{{10 + 5}}{{16}}$$ = $$\frac{{15}}{{16}}$$
Amount of Copper in C,
= Amount of Copper in A + Amount of Copper in B = $$\frac{3}{8}$$ + $$\frac{{11}}{{16}}$$ = $$\frac{{17}}{{16}}$$
Ratio of Gold and Copper in C,
$$ = \frac{{15}}{{16}}:\frac{{17}}{{16}} = 15:17$$
3. A bag contains an equal number of one rupee, 50 paise and 25 paise coins. If the total value is Rs. 35, how many coins of each type are there?
a) 15
b)18
c) 20
d) 27
Discussion
Explanation: Let X coins of each type of there
Total Value = Rs. 35
X + $$\frac{{\text{X}}}{2}$$ + $$\frac{{\text{X}}}{4}$$ = 35
4X + 2X + X = 140
7X = 140
X = 20
4. A bucket contains a mixture of two liquids A and B in the proportion 7 : 5. If 9 litres of mixture is replaced by 9 liters of liquid B, then the ratio of the two liquids becomes 7 : 9. How much of the liquid A was there in the bucket?
a) 21 liters
b) 23 liters
c) 24 liters
d) 27 liters
Discussion
Explanation: Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively. When 9 litres of mixture are drawn off, quantity of A in mixture left:
$$\eqalign{ & = \left[ {7x - {\frac{7}{{12}}} \times 9} \right]\, \text{litres} \cr & = \left[ {7x - {\frac{{21}}{4}} } \right]\,{\text{litres}} \cr & {\text{Similarly quantity of B in mixture left}}, \cr & = \left[ {5x - {\frac{5}{{12}}} \times 9} \right]\, \text{litres} \cr & = \left[ {5x - {\frac{{15}}{4}} } \right]\,{\text{litres}} \cr & \therefore \,{\text{ratio becomes}}, \cr & \frac{{ {7x - {\frac{{21}}{4}} } }}{{ {\left(5x - {\frac{{15}}{4}}\right)+9 } }} = \frac{7}{9} \cr & \frac{{ {28x - 21} }}{{ {20x + 21} }} = \frac{7}{9} \cr & {252x - 189} = 140x + 147 \cr & 112x = 336 \cr & x = 3 \cr & {\text{So the can contained}}, \cr & = 7 \times x \cr & = 7 \times 3 \cr & = 21\,{\text{litres of A initially}}{\text{.}} \cr} $$
5. A bucket contains a mixture of two liquids A & B in the proportion 5 : 3. If 16 litres of the mixture is replaced by 16 litres of liquid B, then the ratio of the two liquids becomes 3 : 5. How much of the liquid B was there in the bucket?:
a) 25 litres
b) 15 litres
c) 18 litres
d) None of these
Discussion
Explanation: Let bucket contains 5x and 3x of liquids A and B respectively.
When 16 litres of mixture are drawn off, quantity of A in mixture left:
$$\eqalign{ & {5x - {\frac{5}{8}} \times 16} = {5x - 10} \cr & {\text{Similarly quantity of B in mixture left}}, \cr & {3x - {\frac{3}{8}} \times 16} = {3x - 6} \cr & {\text{Now the ratio becomes,}} \cr & \frac{{ {5x - 10} }}{{ {3x - 6} }} = \frac{3}{5} \cr & 25x - 50 = 9x - 18 \cr & 16x = 32 \cr & x = 2 \cr & {\text{Quantity of liquid B initially}}, \cr & = 3 \times 2 = 6\,{\text{litres}} \cr} $$
6. The salaries of A, B and C are in the ratio 1 : 3 : 4. If the salaries are increased by 5%, 10% and 15% respectively, then the increased salaries will be in the ratio
a) 20 : 66 : 95
b) 21 : 66 : 96
c) 21 : 66 : 92
d) 19 : 66 : 92
Discussion
Explanation: Let A's Salary = Rs. 100
Then, B's Salary = Rs. 300
And, C's Salary = Rs. 400
Salary has given in 1 : 3 : 4 ratio
Now,
5% increase in A's Salary,
A's new Salary = (100 + 5% of 100) = Rs. 105
B's Salary increases by 10%, Then,
B's new Salary = (300 + 10% of 300) = Rs. 330
C's Salary increases by 15%,
C's new Salary = (400 + 15% of 400) = Rs. 460
Then, ratio of increased Salary,
A : B : C = 105 : 330 : 460 = 21 : 66 : 92
7.If A : B = 2 : 3 and B : C = 4 : 5 then A : B : C is
a) 2 : 3 : 5
b) 5 : 4 : 6
c) 8 : 12 : 15
d) 6 : 4 : 5
Discussion
Explanation:
$$\eqalign{ & \frac{{\text{A}}}{{\text{B}}} = \frac{2}{3} \cr & \frac{{\text{B}}}{{\text{C}}} = \frac{4}{5} \cr} $$
A : B : C = 2 × 4 : 3 × 4 : 3 × 5 = 8 : 12 : 15
8.If two times A is equal to three times of B and also equal to four times of C, then A : B : C is
a) 2 : 3 : 4
b) 3 : 4 : 2
c) 4 : 6 : 3
d) 6 : 4 : 3
Discussion
Explanation:
$$\eqalign{ & 2A = 3B \cr & Or,\,B = \left( {\frac{2}{3}} \right)A;\,{\text{and}} \cr & 2A = 4C \cr & Or,\,C = \left( {\frac{1}{2}} \right)A; \cr & {\text{Hence}}, \cr & A:B:C = A:\frac{{2A}}{3}:\frac{A}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1:\frac{2}{3}:\frac{1}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6:4:3 \cr} $$
9.In a school having roll strength 286, the ratio of boys and girls is 8 : 5. If 22 more girls get admitted into the school, the ratio of boys and girls becomes
a) 12 : 7
b) 10 : 7
c) 8 : 7
d) 4 : 3
Discussion
Explanation: Boys : girls = 8 : 5 (let the boys = 8x, girl = 5x)
Total strength = 286
8x + 5x = 286
13x = 286
x = $$\frac{{286}}{{13}}$$ = 22
Boys = 176 and girls = 110
22 more girls get admitted then number of girls become
(5x + 22) = 110 + 22 = 132
New ratio of boys and girls = 176 : 132 = 4 : 3.
10. Two numbers are in ratio 4 : 5 and their LCM is 180. The smaller number is
a) 9
b) 15
c) 36
d) 45
Discussion
Explanation: Let two numbers be 4x and 5x
Their LCM = 180 and HCF = x
Now,
1st number × 2nd number = LCM × HCF
4x × 5x = 180 × x
20x = 180
x = 9
The smaller number = 4 × 9 = 36
11.The ratio of water and milk in a 30 liter mixture is 7 : 3. Find the quantity of water to be added to the mixture in order to make this ratio 6 : 1.
a) 30
b) 32
c) 33
d) 35
Discussion
Explanation: Here,Let water = 7x and milk = 3x
7x + 3x = 30
x = 3
So, water = 7x = 7 × 3 = 21 liter
Milk = 3x = 3 × 3 = 9 liter
Now, we keep milk constant and add water to mixture to get ratio 6 : 1
Let water in this mixture = 6y and milk = y
We have, milk = 9 liter, so y = 9 liter
Water = 6y = 6 × 9 = 54 liter
Then extra water to be added is 33 liter
12. The incomes of A and B are in the ratio 3 : 2 and their expenditure are in ratio 5 : 3. If each saves Rs. 1000, then, A's income can be:
a) Rs. 3000
b) Rs. 4000
c) Rs. 6000
d) Rs. 9000
Discussion
Explanation: Let income of A and B be 3x and 2x respectively. Also, their expenditure is 5y and 3y.
3x - 5y = 1000 ------- (i) × 3
2x - 3y = 1000 ---------- (ii) × 5
9x - 15y - 10x + 15y = 3000 - 5000
-x = -2000
x = 2000
Then, income of A = 3x = 3 × 2000 = Rs. 6000
13. The difference between two positive numbers is 10 and the ratio between them is 5 : 3. Find the product of the two numbers.
a) 375
b) 175
c) 275
d) 125
Discussion
Explanation: Let the two positive numbers be 5x and 3x respectively
5x - 3x = 10
x = 5
Then numbers are 25 and 15
Thus, their product = 25 × 15 = 375
14. If 30 oxen can plough $$\frac{1}{7}$$ th of a field in 2 days, how many days 18 oxen will take to do the remaining work?
a) 30 days
b) 20 days
c) 15 days
d) 18 days
Discussion
Explanation: We will use work equivalence method,
$$\eqalign{ & \frac{{30}}{{18}} = \frac{{ {\frac{1}{7}} }}{{ {\frac{6}{7}} }} \times \frac{x}{2} \cr & \frac{5}{3} = {\frac{1}{6}} \times \frac{x}{2} \cr & \,x = \frac{{60}}{3} = 20\,{\text{days}} \cr} $$
15. A cat leaps 5 leaps for every 4 leaps of a dog, but 3 leaps of the dog are equal to 4 leaps of the cat. What is the ratio of the speed of the cat to that of the dog?
a) 11 : 15
b) 15 : 11
c) 16 : 15
d) 15 : 16
Discussion
Explanation: 3dog = 4 cat
$$\frac{{{\text{dog}}}}{{{\text{cat}}}} = \frac{4}{3}$$
Let cat's 1 leap = 3 meter and dogs 1 leap = 4 meter
Then, ratio of speed of cat and dog = $$\frac{{3 \times 5}}{{4 \times 4}}$$ = 15 : 16
16. Brindavan Express leave Chennai Central Station every day at 07.50 am and goes to Bangalore City Railway station. This train is very popular among the travelers. On 25th July 2012 number of passengers traveling by I class and II class was in the ratio 1 : 4. The fare for this travel is in the ratio 3 : 1. The total fare collected was Rs. 224000. (Rs. Two lakhs twenty four thousand only). What was the fare collected from I class passengers on that day?
a) Rs. 32,000
b) Rs. 96,000
c) Rs.1,28,000
d) Rs. 5,00,000
Discussion
Explanation: Let the number of passenger traveling by first class be x
Then, number of passenger traveling by second class will be 4x
But the fare is in the ratio 3 : 1
In other words, if 3y fare is collected per I class passenger, y would be collected per II class passenger
Fares of I class passengers : Fares of II class passengers
= x × 3y : 4x × y
= 3 : 4
If total fare is 3 + 4 = 7, then I class passengers should pay Rs. 3
Similarly, we can calculate the fare of I class passengers when total was 224000
| Total Fare | Class Fare |
| 7 | 3 |
| 224000 | ? |
= $$224000 \times \frac{3}{7}$$
= Rs. 96000
17. A vessel of capacity 2 litre has 25% alcohol and another vessel of capacity 6 litre had 40% alcohol. The total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water. what is the new concentration of mixture ?
a) 29%
b) 49%
c) 31%
d) 71%
Discussion
Explanation: Amount of alcohol in first vessel,
= 25% of 2 litre
= 0.25 × 2 = 0.5 litre
Amount of alcohol in second vessel,
= 40% of 6 litre
= 0.4 × 6 = 2.4 litre
Total amount of alcohol out of 10 litres of mixture is
0.5 + 2.4 = 2.9 litre
Concentration of the mixture is,
$$\frac{{2.9 \times 100}}{{10}}$$
= 29%
18. The number of oranges in three basket are in the ratio 3 : 4 : 5. In which ratio the no. of oranges in first two basket must be increased so that the new ratio becomes 5 : 4 : 3 ?
a) 3 : 4
b) 2 : 3
c) 1 : 3
d) 2 : 1
Discussion
Explanation: Let, B1 : B2 : B3 = 3x : 4x : 5x and
B1 : B2 : B3 = 5y : 4y : 3y
Number of oranges remain constant in third basket as increase in oranges takes place only in first two baskets.
Hence, 5x = 3y
x = $$\frac{3y}{5}$$ and,
∴ 3x : 4x : 5x (putting the vale of x)
= $$\frac{{9{\text{y}}}}{5}:\frac{{{\text{12y}}}}{5}:\frac{{{\text{15y}}}}{5}$$
= 9y : 12y : 15y
5y : 4y : 3y (multiple by 5) → 25y : 20y : 15y
Increment in first basket = 16
Increment in second basket = 8
Required ratio = $$\frac{{16}}{8}$$ = 2 : 1
19. At a casino in Mumbai, there are 3 tables A, B and C. The payoffs at A is 10 : 1, at B is 20 : 1 and C is 30 :1. If a man bets Rs. 200 at each table and win at two of the tables, what is the maximum and minimum difference between his earnings can be ?
a) Rs. 4000
b) Rs. 4500
c) Rs. 2500
d) Rs. 2000
Discussion
Explanation: Maximum earning will be only possible when he will won on the maximum yielding table
A → 10 : 1
B → 20 : 1
C → 30 : 1
i.e., he won B and C but lost on A
20 × 200 + 30 × 200 - 1 × 200 = 9800
Minimum earnings will be when he won on table A and B and lose on table 3
10 × 200 + 20 × 200 - 1 × 200 = 5800
Therefore, difference = 9800 - 5800 = Rs. 4000
20. A track covers a distance of 550 metres in 1 minute whereas a bus covers a distance of 33 kms in 45 minute. The ratio of their speeds is:
a) 4 : 3
b) 3 : 5
c) 3 : 4
d) 50 : 3
Discussion
Explanation: Speed of track = 550 per minute.
Speed of bus = $$\frac{{33\,{\text{kms}}}}{{45}}$$ = $$\frac{{33000}}{{45}}$$ = 733.33 m/minute
Ratio of their speeds = $$\frac{{550}}{{733.33}}$$ = 3 : 4
21. If a : b : c = 3 : 4 : 7, then the ratio (a + b + c) : c is equal to
a) 2 : 1
b) 14 : 3
c) 7 : 2
d) 1 : 2
Discussion
Explanation:
$$\eqalign{ &\,\,\,\, {\text{ }}a{\text{ }}:{\text{ }}b{\text{ }}:{\text{ }}c \cr &\,\,\,\, {\text{ }}3{\text{ }}:{\text{ }}4{\text{ }}:{\text{ }}7{\text{ }} \cr & \boxed{\,\,3x:4x:7x\,\,\,\,} \Rightarrow 14x \cr & a + b + c = 14x \cr & c = 7x \cr & \left( {a + b + c} \right):c = 14x:7x \cr & = 2:1 \cr} $$
22. The number of students in 3 classes is in the ratio 2 : 3 : 4. If 12 students are increased in each class this ratio changes to 8 : 11 : 14. The total number of students in the three classes in the beginning was
a) 162
b) 108
c) 96
d) 54
Discussion
Explanation: Let the number of students in the classes be 2x, 3x and 4x respectively;
Total students = 2x + 3x + 4x = 9x
$$\eqalign{ & \frac{{ {2x + 12} }}{{ {3x + 12} }} = \frac{8}{{11}} \cr & \,24x + 96 = 22x + 132 \cr & \,2x = 132 - 96 \cr & \,x = \frac{{36}}{2} = 18 \cr & {\text{Original}}\,{\text{number}}\,{\text{of}}\,{\text{students}}, \cr & 9x = 9 \times 18 \cr & \,\,\,\,\,\,\,\, = 162 \cr} $$
23. A box has 210 coins of denominations one-rupee and fifty paise only. The ratio of their respective values is 13 : 11. The number of one-rupee coin is
a) 65
b) 66
c) 77
d) 78
Discussion
Explanation: Respective ratio of the NUMBER of coins;
= 13 : 11 × 2 = 13 : 22
Number of 1 rupee coins;
= $$\frac{{13 \times 210}}{{13 + 22}}$$ = 78
24.If $$\frac{2}{3}$$ of A=75% of B = 0.6 of C, then A : B : C is
a) 2 : 3 : 3
b) 3 : 4 : 5
c) 4 : 5 : 6
d) 9 : 8 : 10
Discussion
Explanation:
$$\eqalign{ & {\frac{{2A}}{3}} = {\frac{{75B}}{{100}}} = {\frac{{C \times 6}}{{10}}} \cr & {\text{Above}}\,{\text{relation}}\,{\text{gives}}; \cr & \frac{{A \times 2}}{3} = \frac{{B \times 3}}{4} \cr & \to \frac{A}{B} = \frac{9}{8} \cr & {\text{And}}, \cr & \frac{{B \times 3}}{4} = \frac{{C \times 3}}{5} \cr & \to \frac{B}{C} = 4:5 \cr & \to \frac{B}{C} = 8:10 {\text{ (multiple by 2)}} \cr & A:B:C = 9:8:10 \cr} $$
25. If A and B are in the ratio 3 : 4, and B and C in the ratio 12 : 13, then A and C will be in the ratio
a) 3 : 13
b) 9 : 13
c) 36 : 13
d) 13 : 9
Discussion
Explanation:
$$\eqalign{ & {\frac{A}{B}} \times {\frac{B}{C}} = {\frac{3}{4}} \times {\frac{{12}}{{13}}} \cr & \,\frac{A}{C} = \frac{{36}}{{52}} = 9:13 \cr} $$
26. 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?
a) 50
b) 100
c) 150
d) 200
Discussion
Explanation: Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively
Then, sum of their values
$$\eqalign{ & = Rs.\,\left( {\frac{{25x}}{{100}} + \frac{{10 \times 2x}}{{100}} + \frac{{5 \times 3x}}{{100}}} \right) \cr & = Rs.\,\frac{{60x}}{{100}} \cr & \frac{{60x}}{{100}} = 30 \Leftrightarrow x = \frac{{30 \times 100}}{{60}} = 50 \cr & {\text{Hence,}}\,{\text{the}}\,{\text{number}}\,{\text{of}}\,{\text{5p}}\,{\text{coins}} = \left( {3 \times 50} \right) \cr & = 150 \cr} $$
27. What is the ratio in Rs. 2.80 and 40 paise?
a) 1 : 7
b) 2 : 7
c) 7 : 1
d) 1 : 14
Discussion
Explanation: Rs. 2.80 = 280 paise
Required ration = 280 : 40
= 7 : 1
28. A person spends Rs. 8100 in buying some tables at Rs. 1200 each and some chairs at Rs. 300 each. The ratio of the number of chairs to that of tables when the maximum possible number of tables is purchased,
a) 1 : 2
b) 1 : 4
c) 2 : 1
d) 5 : 7
Discussion
Explanation: Maximum possible number of tables = 6
[1200 × 6 = 7200]
Number of chairs purchased
$$\eqalign{ & {\text{ = }}\frac{{{\text{8100}} - {\text{7200}}}}{{{\text{300}}}}{\text{ = }}\frac{{{\text{900}}}}{{{\text{300}}}}{\text{ = 3}}{\text{}} \cr & {\text{Required ratio}} = {\text{3}}:{\text{6}} \cr & {\text{ = 1}}:{\text{2}} \cr} $$
29. If $$x = \frac{1}{3}y$$ and $$y = \frac{1}{2}z{\text{,}}$$ then x : y : z is equal to = ?
a) 3 : 2 : 1
b) 1 : 2 : 6
c) 1 : 3 : 6
d) 2 : 4 : 6
Discussion
Explanation:
$$\eqalign{ & x = \frac{1}{3}y\,\,{\text{and }}y = \frac{1}{2}z \cr & \frac{x}{y} = \frac{1}{3}\,{\text{and }}\frac{y}{z} = \frac{1}{2} \cr & x:y = 1:3 \cr & y:z = 1:2 \times 3\,({\text{multiply}}) \cr & i.e.\,y:z = 3:6 \cr & x:y:z = 1:3:6 \cr} $$
30. If x : y = 3 : 1, then x3 - y3 : x3 + y3 = ?
a) 13 : 14
b) 14 : 13
c) 10 : 11
d) 11 : 10
Discussion
Explanation:
$$\eqalign{ & x:y = 3:1 \cr & \frac{x}{y} = \frac{3}{1} \cr & \frac{{{x^3} - {y^3}}}{{{x^3} + {y^3}}} \cr & \Rightarrow \frac{{{y^3}\left( {\frac{{{x^3}}}{{{y^3}}} - 1} \right)}}{{{y^3}\left( {\frac{{{x^3}}}{{{y^3}}} + 1} \right)}} \cr & {\text{taking }}{{\text{y}}^3}{\text{ common}} \cr & {\text{ = }}\frac{{\frac{{{x^3}}}{{{y^3}}} - 1}}{{\frac{{{x^3}}}{{{y^3}}} + 1}} \cr & \Rightarrow \frac{{27 - 1}}{{27 + 1}} \cr & \Rightarrow \frac{{26}}{{28}} \cr & \Rightarrow \frac{{13}}{{14}} \cr} $$
31. Number of students in institutes A and B were in the ratio of 7 : 15 respectively in 2012. In 2013, the number of students in institute A increased by 25% and number of students in institutes B increased by 26%, then what was the respective ratio between number of students in institutes A and B?
a) 25 : 56
b) 24 : 55
c) 24 : 53
d) 25 : 54
Discussion
Explanation: Ratio of students in 2012 in institutes A and B = 7 : 15
Let number of students in institute A in 2012 = 700
And Number of students in institutes B in 2012 = 1500
25% increase in the number of students in 2013
Now, number of students in Institute A = 700 + 25% of 700 = 875
Number of students in B in 2013 as 26% students increased in B
= 1500 + 26% of 1500 = 1890
Current Ratio of the students,
$$ = \frac{{875}}{{1890}} = 25:54$$
32. A and B together have Rs. 1210. If $$\frac{4}{{15}}$$ of A's amount is equal to $$\frac{2}{{5}}$$ of B's amount, how much amount does B have?
a) Rs. 460
b) Rs. 484
c) Rs. 550
d) Rs. 664
Discussion
Explanation:
$$\eqalign{ & \frac{4}{{15}}A = \frac{2}{5}B \cr & A = \left( {\frac{2}{5} \times \frac{{15}}{4}} \right)B \cr & A = \frac{3}{2}B \cr & \frac{A}{B} = \frac{3}{2} \cr & A:B = 3:2. \cr & {\text{B's}}\,{\text{share}} = Rs.\left( {1210 \times \frac{2}{5}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = Rs.\,484 \cr} $$
33. Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:
a) 2 : 5
b) 3 : 5
c) 4 : 5
d) 6 : 7
Discussion
Explanation:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{third}}\,{\text{number}}\,{\text{be}}\,x \cr & {\text{Then,}}\,{\text{first}}\,{\text{number}} \cr & = 120\% \,{\text{of}}\,x = \frac{{120x}}{{100}} = \frac{{6x}}{5} \cr & {\text{Second}}\,{\text{number}} \cr & = 150\% \,{\text{of}}\,x = \frac{{150x}}{{100}} = \frac{{3x}}{2} \cr & {\text{Ratio}}\,{\text{of}}\,{\text{first}}\,{\text{two}}\,{\text{numbers}} \cr & = {\frac{{6x}}{5}:\frac{{3x}}{2}} = 12x:15x = 4:5 \cr} $$
34.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?
a) Rs. 500
b) Rs. 1500
c) Rs. 2000
d) None of these
Discussion
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
35. 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?
a) 2 : 3 : 4
b) 6 : 7 : 8
c) 6 : 8 : 9
d) None of these
Discussion
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)
$$ = \left( {\frac{{140}}{{100}} \times 5x} \right),$$ $$\left( {\frac{{150}}{{100}} \times 7x} \right)$$ and $$\left( {\frac{{175}}{{100}} \times 8x} \right)$$
$$\eqalign{ & = 7x,\,\,\frac{{21x}}{2}{\text{ and }}14x \cr & {\text{The required ratio}} = 7x:\frac{{21x}}{2}:14x \cr & = 14x:21x:28x \cr & = 2:3:4 \cr} $$
36. 8 litres are drawn from a cask filled with wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the total solution is 16 : 81. How much wine did the cask hold originally?
a) 24 litres
b) 45 litres
c) 49 litres
d) 44 litres
Discussion
Explanation: Let the quantity of the wine in the cask originally be x litres.
Final Amount of solute that is not replaced = Initial Amount × $${\left( {\frac{{{\text{Vol}}{\text{. after removal}}}}{{{\text{Vol}}{\text{. after replacing}}}}} \right)^{\text{N}}}$$
Where N = No. of operation done.
Then ratio of wine to total solution in cask after 4 operations,
$$\eqalign{ & 1 \times { {\left( {\frac{{x - 8}}{x}} \right)} ^4} = \frac{{16}}{{81}} \cr & 1 \times {\left\{ {\frac{{ {x - 8} }}{x}} \right\}^4} = {\left( {\frac{2}{3}} \right)^4} \cr & \frac{{ {x - 8} }}{x} = \frac{2}{3} \cr & 3x - 24 = 2x \cr & x = 24\,{\text{litres}} \cr} $$
37. The milk and water in a mixture are in the ratio 7 : 5. When 15 liters of water are added to it, the ratio of milk and water in the new mixture becomes 7 : 8. The total quantity of water in the new mixture is:
a) 35 litres
b) 40 litres
c) 60 litres
d) 96 litres
Discussion
Explanation:
| Milk | : | Water |
| 7 | : | 5 |
| 7 | : | 8 |
| 3 unit |
Remember water is added and not milk, so make milk equal but here milk is already equal
3 units = 15 litres
1 units = 5 litres
8 units = 40 litres
Total quantity of water in the new mixture = 40 litres
38. If x : y = 5 : 2, then (8x + 9y) : (8x + 2y) is :
a) 22 : 29
b) 26 : 61
c) 29 : 22
d) 61 : 26
Discussion
Explanation:
$$\frac{{\text{x}}}{{\text{y}}} = \frac{5}{2}$$
Means x = 5, y = 2
Putting value of x and y in expression
8 × 5 + 9 × 2 = 58
8 × 5 + 2 × 2 = 44
58 : 44 = 29 : 22
39. Tom is chasing Jerry. In the same interval of time Tom jumps 8 times while Jerry jumps 6 times. But the distance covered by Tom in 7 Jumps is equal to the distance covered by Jerry in 5 Jumps. The ratio of speed of Tom and Jerry is:
a) 48 : 35
b) 28 : 15
c) 24 : 20
d) 20 : 21
Discussion
Explanation: 7 jumps of Tom = 5 jumps of Jerry
$$\frac{{{\text{Tom}}}}{{{\text{Jerry}}}} = \frac{5}{7}$$
Let Jerry's 1 leap = 7 meter and Tom's 1 leap = 5 meter
Then, ratio of speed of Tom and Jerry
= $$\frac{{8 \times 5}}{{6 \times 7}}$$
= $$\frac{{40}}{{42}}$$
= 20 : 21
40. The ratio of ducks and frogs in a pond is 37 : 39 respectively. The average number of ducks and frogs in the pond is 152. What is the number of frogs in the pond ?
a) 148
b) 152
c) 156
d) 144
Discussion
Explanation: Ratio of Ducks and Frogs in Pond = 37 : 39
Average of Ducks and Frogs in Pond = 152
So, total number of Ducks and Frogs in the Pond = 2 × 152 = 304
Number of Frogs = $$\frac{{304 \times 39}}{{76}}$$ = 156
41. If 60% A = $$\frac{3}{4}$$ of B, then A : B is
a) 4 : 5
b) 5 : 4
c) 9 : 20
d) 20 : 9
Discussion
Explanation:
$$\eqalign{ & {\text{60}}\% \,{\text{of}}\,{\text{A}} = \frac{{\text{3}}}{{\text{4}}}\,{\text{of}}\,{\text{ B}} \cr & \Rightarrow \frac{{60}}{{100}}{\text{A}} = \frac{3}{4}\,{\text{B}} \cr & \frac{{\text{3}}}{{\text{5}}}{\text{A = }}\frac{{\text{3}}}{{\text{4}}}\,{\text{B}} \cr & \frac{{\text{A}}}{{\text{B}}}{\text{ = }}\frac{{\text{3}}}{{\text{4}}} \times \frac{{\text{5}}}{{\text{3}}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{5}{4} \cr} $$
A : B = 5 : 4
42. Which of the following represents ab = 64?
a) 8 : a = 8 : b
b) a : 16 = 6 : 4
c) a : 8 = b : 8
d) 32 : a = b : 2
Discussion
Explanation: A = 8 : a = 8 : b ⇒ 8a = 8b ⇒ a = b.
B = a : 16 = b : 4 ⇒ 4a = 16b ⇒ a = 4b
C = a : 8 = b : 8 ⇒ 8a = 8b ⇒ b = a
D = 32 : a = b : 2 ⇒ ab =64
43. The ratio of boys and girls in a club is 3 : 2. Which of the following could be the actual number of members ?
a) 16
b) 18
c) 24
d) 25
Discussion
Explanation: The total number of members must be a multiple of the sum ratio terms.
3 + 2 = 5
and 25 is a multiple of 5
44. If m : n = 3 : 2, then (4m + 5n) : (4m - 5n) is equal to = ?
a) 4 : 9
b) 9 : 4
c) 11 : 1
d) 9 : 1
Discussion
Explanation:
$$\eqalign{ & m:n = 3:2 \cr & \frac{m}{n} = \frac{3}{2} \cr & \frac{{4m + 5n}}{{4m - 5n}} \cr & \frac{{n\left( {4\frac{m}{n} + 5} \right)}}{{n\left( {4\frac{m}{n} - 5} \right)}} \cr & \frac{{4 \times \frac{3}{2} + 5}}{{4 \times \frac{3}{2} - 5}} \cr & \frac{{6 + 5}}{{6 - 5}} \cr & = \frac{{11}}{1} \cr & = 11:1 \cr} $$
45. The sum of two numbers is 40 and their difference is 4. The ratio of the numbers is = ?
a) 21 : 19
b) 22 : 9
c) 11 : 9
d) 11 : 18
Discussion
Explanation : A + B = 40
A - B = 4
A = 22
B = 18
A : B = 22 : 18
= 11 : 9
46. If A : B = 2 : 3, B : C = 4 : 5 and C : D = 5 : 9 then A : D is equal to:
a) 11 : 17
b) 8 : 27
c) 5 : 9
d) 2 : 9
Discussion
Explanation:
$$\eqalign{ & \frac{A}{D} = {\frac{A}{B}} \times {\frac{B}{C}} \times {\frac{C}{D}} \cr & \,\,\,\,\,\,\,\,\, = {\frac{2}{3}} \times {\frac{4}{5}} \times {\frac{5}{9}} \cr & \,\,\,\,\,\,\,\,\, = \frac{{ {2 \times 4 \times 5} }}{{ {3 \times 5 \times 9} }} \cr & \,\,\,\,\,\,\,\,\, = \frac{8}{{27}} \cr & \,\,\,\,\,\,\,\,\, = 8:27 \cr} $$
47. In a class, the number of girls is 20% more than that of the boys. The strength of the class is 66. If 4 more girls are admitted to the class, the ratio of the number of boys to that of the girls is
a) 1 : 2
b) 3 : 4
c) 1 : 4
d) 3 : 5
Discussion
Explanation: Girls : boys = 6 : 5
Hence, girls = $$6 \times \frac{{66}}{{11}}$$ = 36
Boys = 30
New ratio, 30 : (36 + 4) = 3 : 4
48. What must be added to each term of the ratio 7 : 11, So as to make it equal to 3 : 4?
a) 8
b) 7.5
c) 6.5
d) 5
Discussion
Explanation:
$$\eqalign{ & {\text{Let }}x{\text{ be added to each term}}. \cr & \frac{{ {7 + x} }}{{11 + x}} = \frac{3}{4} \cr & \,33 + 3x = 28 + 4x \cr & \,x = 5 \cr} $$
49. Two numbers are in ratio 7 : 11. If 7 is added to each of the numbers, the ratio becomes 2 : 3. The smaller number is
a) 39
b) 49
c) 66
d) 77
Discussion
Explanation: Let the numbers be 7x and 11x.
$$\eqalign{ & \frac{{ {7x + 7} }}{{ {11x + 7} }} = \frac{2}{3} \cr & \,22x + 14 = 21x + 21 \cr & \,x = 7 \cr & {\text{Smaller}}\,{\text{number}} = 49 \cr} $$
50. Two numbers are in ratio P : Q. when 1 is added to both the numerator and the denominator, the ratio gets changed to $$\frac{{\text{R}}}{{\text{S}}}$$. again, when 1 is added to both the numerator and denominator, it becomes $$\frac{1}{2}$$. Find the sum of P and Q.
a) 3
b) 4
c) 5
d) 6
Discussion
Explanation: We will solve this question through options
Taking Option A:
It has P + Q = 3.
The possible value of $$\frac{{\text{P}}}{{\text{Q}}}$$ is $$\frac{1}{2}$$ or $$\frac{2}{1}$$
Using $$\frac{1}{2}$$, we see that on adding 2 in both the numerator and denominator we get $$\frac{3}{4}$$ (not required value)
Similarly we check for $$\frac{2}{1}$$, this will also not give the required value
Option B:
We have $$\frac{1}{3}$$ possible ratio
Then, we get the final value as $$\frac{3}{5}$$ (not = to $$\frac{1}{2}$$)
Hence, rejected
Option C:
Here we have $$\frac{1}{4}$$ or $$\frac{2}{3}$$
Checking for $$\frac{1}{4}$$ we get $$\frac{3}{6}$$ = $$\frac{1}{2}$$
Hence, the option c is correct
51. If A : B = 3 : 4 and B : C = 8 : 9, then A : B : C is = ?
a) 8 : 6 : 9
b) 9 : 8 : 6
c) 6 : 8 : 9
d) 3 : 32 : 9
Discussion
Explanation: A : B = 3 : 4
B : C = 8 : 9
A : B = 3 : 4 (multiply with 2)
i.e. A : B = 6 : 8 and B : C = 8 : 9
A : B : C = 6 : 8 : 9
52. If 2A = 3B = 4C, the A : B : C is = ?
a) 2 : 3 : 4
b) 4 : 3 : 2
c) 6 : 4 : 3
d) 3 : 4 : 6
Discussion
Explanation:
$$\eqalign{ & \,\,\,{\text{A}}\,\,\,\,\,\,\,{\text{:}}\,\,\,\,\,\,{\text{B}}\,\,\,\,\,\,{\text{:}}\,\,\,\,\,{\text{C}} \cr & 3 \times 4:2 \times 4:2 \times 3 \cr & \,\,\,12\,\,\,\,:\,\,\,\,\,{\text{8}}\,\,\,\,\,:\,\,\,\,{\text{6}} \cr & \,\,\,\boxed{\,\,6\,\,\,\,:\,\,\,\,4\,\,\,\,\,\,:\,\,\,\,3\,\,} \cr} $$
53. x, y, z, u are real numbers such that x : y = y : z = z : u and x : u = 64 : 27. the value of x : z is -
a) 64 : 27
b) 16 : 9
c) 4 : 3
d) 3 : 4
Discussion
Explanation:
$$\eqalign{ & {\text{Let}}\,\frac{x}{y} = \frac{y}{z} = \frac{z}{u} = k. \cr & {\text{Now, }}\frac{x}{u} = \frac{{64}}{{27}} \cr & \frac{x}{y} \times \frac{y}{z} \times \frac{z}{u} = \frac{{64}}{{27}} \cr & {k^3} = {\left( {\frac{4}{3}} \right)^3} \cr & k = \frac{4}{3}. \cr & {\text{So}},\,x:y = y:z = z:u = 4:3. \cr & \frac{x}{z} = \frac{x}{y} \times \frac{y}{z} = \frac{4}{3} \times \frac{4}{3} = \frac{{16}}{{9.}} \cr} $$
54. If p : q : r = 1 : 2 : 4, then $$\sqrt {5{p^2} + {q^2} + {r^2}} $$ is equal to -
a) 5
b) 2q
c) 5p
d) 4r
Discussion
Explanation:
$$\eqalign{ & p = k,\,q = 2k,\,r = 4k \cr & \sqrt {5{p^2} + {q^2} + {r^2}} \cr & = \sqrt {5{k^2} + {{\left( {2k} \right)}^2} + {{\left( {4k} \right)}^2}} \cr & = \sqrt {5{k^2} + 4{k^2} + 16{k^2}} \cr & = \sqrt {25{k^2}} \cr & = 5k \cr & = 5p. \cr} $$
55. If A : B : C = 2 : 3 : 4, then the ratio $$\frac{{\text{A}}}{{\text{B}}}$$ : $$\frac{{\text{B}}}{{\text{C}}}$$ : $$\frac{{\text{C}}}{{\text{A}}}$$ is equal to -
a) 4 : 9 : 16
b) 8 : 9 : 24
c) 8 : 9 : 12
d) 8 : 9 : 16
Discussion
Explanation:
$$\eqalign{ & {\text{let A}} = 2{\text{k}},\,{\text{B}} = 3{\text{k,}}\,{\text{C}} = 4{\text{k}} \cr & \Rightarrow \frac{{\text{A}}}{{\text{B}}} = \frac{{2{\text{k}}}}{{3{\text{k}}}} = \frac{2}{3}, \cr & \Rightarrow \frac{{\text{B}}}{{\text{C}}} = \frac{{3{\text{k}}}}{{4{\text{k}}}} = \frac{3}{4}, \cr & \Rightarrow \frac{{\text{C}}}{{\text{A}}} = \frac{{4{\text{k}}}}{{2{\text{k}}}} = 2 \cr & \frac{{\text{A}}}{{\text{B}}}:\frac{{\text{B}}}{{\text{C}}}:\frac{{\text{C}}}{{\text{A}}} = \frac{2}{3}:\frac{3}{4}:2 \cr & = 8:9:24 \cr} $$
56. If 8a = 9b then the ratio of $$\frac{{\text{a}}}{9}$$ to $$\frac{{\text{b}}}{8}$$ is
a) 1 : 1
b) 1 : 2
c) 2 : 1
d) 64 : 81
Discussion
Explanation:
$$\eqalign{ & = {\text{8a}} = {\text{9b}} \Rightarrow a = \frac{9}{8}b \cr & \therefore \frac{{\text{a}}}{{\text{9}}}:\frac{{\text{b}}}{{\text{8}}} = \frac{{\left( {\frac{9}{8}b} \right)}}{9}:\frac{{\text{b}}}{{\text{8}}} \cr & = \frac{{\text{b}}}{{\text{8}}}:\frac{{\text{b}}}{{\text{8}}} = 1:1 \cr} $$
57. If x : y = 3 : 4, then (2x + 3y) : (3y - 2x) would be equal to
a) 2 : 1
b) 3 : 1
c) 3 : 2
d) 21 : 1
Discussion
Explanation:
$$\eqalign{ & = \frac{x}{y} = \frac{3}{4} \cr & = \frac{{2x + 3y}}{{3y - 2x}} \cr & = \frac{{2\left( {\frac{x}{y}} \right) + 3}}{{3 - 2\left( {\frac{x}{y}} \right)}} \cr & = \frac{{2 \times \frac{3}{4} + 3}}{{3 - 2 \times \frac{3}{4}}} \cr & = \frac{9}{2} \times \frac{2}{3} = 3 \cr & = \left( {2x + 3y} \right):\left( {3y - 2x} \right) \cr & = 3:1 \cr} $$
58. Rs. 33630 are divided among A, B and C in such a manner that the ratio of the amount of A to that of B is 3 : 7 and the ratio of the amount of B to that of C is 6 : 5. The amount of money received by B is = ?
a) Rs. 14868
b) Rs. 16257
c) Rs. 13290
d) Rs. 12390
Discussion
Explanation: A : B = 3 : 7
B : C = 6 : 5
A : B = 3 : 7 (multiply with 6) and
B : C = 6 : 5 (multiply with 7)
i.e. A : B = 18 : 42 and
B : C = 42 : 35
A : B : C = 18 : 42 : 35
18x + 42x + 35x = 33630
x = 354
Money received by B = 42 × 354 = Rs. 14868
59. 200 litres of a mixture contains milk and water in the ratio 17 : 3. After the addition of some more milk to it, the ratio of milk to water in the resulting mixture becomes 7 : 1. The quantity of milk added to it was =?
a) 20 Litres
b) 40 Litres
c) 60 Litres
d) 80 Litres
Discussion
Explanation: Milk : water = 17 : 3 = 17x : 3x
17x + 3x = 200
⇒ x =10 litre
Milk = 170 litre and water = 30 litre in initial mixture.
Let 'y' litre of milk added in mixture
i.e. 170 + y : 30 = 7 : 1
$$\frac{{170 + y}}{{30}} = \frac{7}{1}$$
y = 210 - 170 = 40 litre
60. In an innings of a cricket match,three players A, B and C scored a total of 361 runs. If the ratio of the number of runs scored by A to that scored by B and also number of runs scored by B to that scored by C be 3 : 2, the number of runs scored by A was = ?
a) 171
b) 181
c) 185
d) 161
Discussion
Explanation: Given total runs of A, B, C
(A + B + C = 361)
$$\eqalign{ & \Rightarrow {\text{A}}:{\text{B}}:{\text{C}} \cr & \,\,\,\,\,\,\,\,\,3:2 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3:2 \cr & \underline {\overline {\,\,\,\,\,\,\,\,\,\,9:6:4{\text{ }}} } \cr & 9x + 6x + 4x = 361 \cr & \Rightarrow 19x = 361 \cr & \Rightarrow x = 19 \cr & {\text{Runs scored by A }} = 9x \cr & = 9 \times 19 \cr & = 171 \cr} $$
61. If $$\frac{x}{2} = \frac{y}{3} = \frac{z}{4} = $$ $$\frac{{2x - 3y + 5z}}{k}{\text{,}}$$ then the value of k is -
a) 12
b) 15
c) 16
d) 18
Discussion
Explanation:
$$\eqalign{ & {\text{Let }}\frac{x}{2} = \frac{y}{3} = \frac{z}{4} = l \cr & {\text{Then,}} \cr & = x = 2l,y = 3l,z = 4l \cr & \frac{x}{2} = \frac{{2x - 3y + 5z}}{k} \cr & \frac{{2l}}{2} = \frac{{2 \times 2l - 3 \times 3l + 5 \times 4l}}{k} \cr & k = 4 - 9 + 20 = 15 \cr} $$
62. A mixture contains wine and water in the ratio 3 : 2 and another mixture contains them in the ratio 4 : 5. How many litres of the latter must be mixed with 3 litres of the former so that the resulting mixture may contain equal quantities of wine and water = ?
a) $$5\frac{2}{5}{\text{ litres}}$$
b) $$5\frac{2}{3}{\text{ litres}}$$
c) $$4\frac{1}{2}{\text{ litres}}$$
d) $$3\frac{3}{4}{\text{ litres}}$$
Discussion
Explanation: The first mixture contains wine and water in the ratio of 3 : 2
Wine in 3 liters of mixture = $$\frac{3}{5} \times 3 = \frac{9}{5}$$
Water in 3 liters of mixture = $$\frac{2}{5} \times 3 = \frac{6}{5}$$
Let us consider that the first mixture is mixed with the second mixture that has quantity as 9x liters (4x liters of wine and 5x liters of water).
After mixing,
The total quantity of wine = Total quantity of water
$$\eqalign{ & \frac{9}{5} + 4x = \frac{6}{5} + 5x \cr & x = \frac{9}{5} - \frac{6}{5} \cr & x = \frac{3}{5} \cr} $$
Second Mixture required = 9x = $$9 \times \frac{3}{5}$$ = $$5\frac{2}{5}{\text{ litres}}$$
63. The monthly salaries of A, B and C are in the ratio 2 : 3 : 5. If C's monthly salary is Rs. 12000 more than that of A, then B's annual salary is = ?
a) Rs. 120000
b) Rs. 144000
c) Rs. 180000
d) Rs. 240000
Discussion
Explanation:
| A | : | B | : | C | |
| 2 | : | 3 | : | 5 | |
| Let | 2x | : | 3x | : | 5x |
5x - 2x = 12000
3x = 12000
x = 4000
Monthly salary of B is = 3x = Rs. 12000
Annual salary of B is = 12 × 1200
= Rs. 144000
64.The ratio of the numbers of boys and girls in a school was 5 : 3. Some new boys and girls were admitted to the school, in the ratio 5 : 7. At this, the total number of students in the school became 1200 and the ratio of boys to girls changed to 7 : 5. The number of students in the school before new admissions was = ?
a) 700
b) 720
c) 900
d) 960
Discussion
Explanation:
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,{\text{A}}:{\text{B}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,5:3 \cr & {\text{Let }}5x:3x = 8x \cr & \Rightarrow {\text{New comers}} \cr & 5y:7y = 12y \cr & 8x + 12 = 1200 \cr & \Rightarrow 2x + 3y = 300......(i) \cr & {\text{Again,}}\frac{{5x + 5y}}{{3x + 7y}} = \frac{7}{5} \cr & 25x + 25y = 21x + 49y \cr & \Rightarrow 4x - 24y = 0 \cr & \Rightarrow 4x = 24y \cr & \Rightarrow x = 6y...........(ii) \cr & {\text{From equation }}.......(i) \cr & 12y + 3y = 300 \cr & y = 20 \cr & x = 6 \times 20 = 120 \cr} $$
The number of students initially
8x = 8 × 120 = 960
65. If a : b = c : d, then $$\frac{{{\text{ma}} + {\text{nc}}}}{{{\text{mb}} + {\text{nd}}}}$$ is equal to -
a) m : n
b) dm : cn
c) an : mb
d) a : b
Discussion
Explanation:
$$\eqalign{ & {\text{Let }}\frac{a}{b} = \frac{c}{d} = k \cr & \Rightarrow a = bk,\,c = dk \cr & \frac{{ma + nc}}{{mb + nd}} \cr & = \frac{{mbk + ndk}}{{mb + nd}} \cr & = \frac{{k\left( {mb + nd} \right)}}{{\left( {mb + nd} \right)}} \cr & \Rightarrow k = \frac{a}{b} \cr & \Rightarrow a:b \cr} $$
66. If a : b = 5 : 7 and c : d = 2a : 3b then ac : bd is = ?
a)20 : 38
b) 50 : 147
c) 10 : 21
d) 50 : 151
Discussion
Explanation:
$$\eqalign{ & {\text{a}}:{\text{b}}\,\,\,\,\,\,\,\,\,\,\,\,{\text{c}}:{\text{d}} \cr & 5:7\,\,\,\,\,\,\,\,\,\,\,\,\,2{\text{a}}:3{\text{b}} \cr & \frac{a}{b} = \frac{5}{7},\,\frac{c}{d} = \frac{{2a}}{{3b}} \cr & = \frac{2}{3} \times \frac{5}{7} \cr & = \frac{{10}}{{21}} \cr & ac:bd \cr & = \frac{{{\text{ac}}}}{{{\text{bd}}}} \cr & = \frac{5}{7} \times \frac{{10}}{{21}} \cr & = \frac{{50}}{{147}} \cr & = 50:147 \cr} $$
67.If 2A = 3B = 4C, then A : B : C is equal to -
a) 2 : 3 : 4
b) 3 : 4 : 6
c) 4 : 3 : 2
d) 6 : 4 : 3
Discussion
Explanation: Let 2A = 3B = 4C = k
$$\eqalign{ & A = \frac{k}{2}, \cr & B = \frac{k}{3}, \cr & C = \frac{k}{{4}} \cr & {\text{A}}:{\text{B}}:{\text{C}} = \frac{k}{2}:\frac{k}{3}:\frac{k}{4} \cr & = \frac{1}{2}:\frac{1}{3}:\frac{1}{4} {\text{ (Multiply by 12)}} \cr & = 6:4:3 \cr} $$
68. If x2 + 4y2 = 4xy, then x : y is -
a) 1 : 1
b) 1 : 2
c) 2 : 1
d) 1 : 4
Discussion
Explanation:
$$\eqalign{ & = {x^2} + 4{y^2} = 4xy \cr & \Rightarrow {x^2} - 4xy + 4{y^2} = 0 \cr & \left( {x - 2{y^2}} \right) = 0 \cr & x = 2y \cr & \frac{x}{y} = 2 \cr & \Rightarrow x:y = 2:1 \cr} $$
69. If W1 : W2 = 2 : 3 and W1 : W3 = 1 : 2, then W2 : W3 is -
a) 3 : 4
b) 4 : 3
c) 2 : 3
d) 4 : 5
Discussion
Explanation:
$$\eqalign{ & = \frac{{{{\text{W}}_{\text{2}}}}}{{{{\text{W}}_{\text{1}}}}} = \frac{3}{2}{\text{and}}\frac{{{{\text{W}}_{\text{1}}}}}{{{{\text{W}}_{\text{3}}}}} = \frac{1}{2} \cr & \frac{{{W_2}}}{{{W_3}}} = \left( {\frac{{{{\text{W}}_{\text{2}}}}}{{{{\text{W}}_{\text{1}}}}} \times \frac{{{{\text{W}}_{\text{1}}}}}{{{{\text{W}}_{\text{3}}}}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4} \cr & {W_2}:{W_3} = 3:4 \cr} $$
70.If a : b = b : c, then a4 : b4 is equal to = ?
a) ac : b2
b) a2 : c2
c) c2 : a2
d) b2 : ac
Discussion
Explanation:
$$\eqalign{ & a:b = b:c \cr & \frac{a}{b} = \frac{b}{c} \cr & {b^2} = ac \cr & {b^4} = {a^2}{c^2} \cr & {a^4}:{b^4} = \frac{{{a^4}}}{{{b^4}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{a^4}}}{{{a^2}{c^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{a^2}}}{{{c^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {a^2}:{c^2} \cr} $$
71. A sum of Rs. 1240 is distributed among A, B and C such that the ratio of amount received by A and B is 6 : 5 and that of B and C is 10 : 9 respectively. Find the share of C ?
a) Rs. 480
b) Rs. 360
c) Rs. 400
d) Rs. 630
Discussion
Explanation: A : B = 6 : 5
B : C = 10 : 9
A : B = 6 : 5 (multiply with 2)
i.e. A : B = 12 : 10
A : B : C = 12 : 10 : 9
12x + 10x + 9x = 1280
x = 40
Share of C = 9 × 40 = 360
72. If x : 7.5 = 7 : 17.5, then the value of x is -
a) 1
b) 2.5
c) 3
d) 3.5
Discussion
Explanation:
$$\eqalign{ & = x:7.5 = 7:17.5 \cr & 17.5x = 7.5 \times 7 \cr & x = \frac{{7.5 \times 7}}{{17.5}} \cr & = 3 \cr} $$
73. The value of $$x$$ where $$x$$ : $$2\frac{1}{3}$$ :: $$21$$ : $$50$$ is -
a) $$1\frac{1}{{49}}$$
b) $$1\frac{1}{{50}}$$
c) $$ \frac{{49}}{{50}}$$
d) $$\frac{{27}}{{50}}$$
Discussion
Explanation:
$$\eqalign{ & = x:2\frac{1}{3}::21:50 \cr & 50x = \frac{7}{3} \times 21 \cr & 50x = 49 \cr & x = \frac{{49}}{{50}} \cr} $$
74. If $$\sqrt 2 $$ : $$\left( {1 + \sqrt 3 } \right)$$ :: $$\sqrt 6 $$ : $$x$$, then $$x$$ is equal to -
a) $$\sqrt 3 + 3$$
b) $$1 - \sqrt 3 $$
c) $$1 + \sqrt 3 $$
d) $$\sqrt 3 - 3$$
Discussion
Explanation:
$$\eqalign{ & = \sqrt 2 :\left( {1 + \sqrt 3 } \right)::\sqrt 6 :x \cr & \sqrt {2}x = \sqrt 6 \left( {1 + \sqrt 3 } \right) \cr & x = \frac{{\sqrt 6 \left( {1 + \sqrt 3 } \right)}}{{\sqrt 2 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 3 \left( {1 + \sqrt 3 } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 3 + 3 \cr} $$
75. In an alloy, the ratio of copper and zinc is 5 : 2. If 1.250 kg of zinc is mixed in 17 kg 500 gm of alloy, then the ratio of copper and zinc will be = ?
a) 2 : 1
b) 2 : 3
c) 3 : 2
d) 1 : 2
Discussion
Explanation: The ratio of copper and zinc is 5 : 2
i.e 5x : 2x
Quantity of initial mixture = 17.5 kg
5x + 2x = $$\frac{35}{2}$$
⇒ x = $$\frac{5}{2}$$
Quantity of Copper in initial mixture
$$\eqalign{ & = \frac{5}{2} \times 5 \cr & = \frac{{25}}{2}kg \cr} $$
Quantity of Zinc in initial mixture
$$\eqalign{ & = \frac{5}{2} \times 2 \cr & = 5\,kg \cr} $$
Now after adding 1.250 kg of zinc
= 5 + 1.250
= 6.250 kg
= $$\frac{{25}}{4}kg$$
New Ratio of copper : zinc
$$\eqalign{ & = \frac{{25}}{2}:\frac{{25}}{4} \cr & = 2:1 \cr} $$
76. If A : B = $$\frac{1}{2}$$ : $$\frac{3}{8},$$ B : C = $$\frac{1}{3}$$ : $$\frac{5}{9}$$ and C : D = $$\frac{5}{6}$$ : $$\frac{3}{4},$$ then the ratio A : B : C : D is
a) 4 : 6 : 8 : 10
b) 8 : 6 : 10 : 9
c) 6 : 8 : 9 : 10
d) 6 : 4 : 8 : 10
Discussion
Explanation:
$$\eqalign{ & {\text{A}}:{\text{B}} = \frac{1}{2}:\frac{3}{8} = 4:3, \cr & {\text{B}}:{\text{C}} = \frac{1}{3}:\frac{5}{9} = 3:5, \cr & {\text{C}}:{\text{D}} = \frac{5}{6}:\frac{3}{4} = 10:9 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5:\frac{9}{2}. \cr & {\text{A}}:{\text{B}}:{\text{C}}:{\text{D}} \cr & = 4:3:5:\frac{9}{2} \cr & = 8:6:10:9 \cr} $$
77. If p : q : r = 1 : 2 : 4, then $$\sqrt {5{p^2} + {q^2} + {r^2}} $$ is equal to
a) 5
b) 2q
c) 5p
d) 4r
Discussion
Explanation:
$$\eqalign{ & {\text{p}}:{\text{q}}:{\text{r}} \cr & 1:2:4 \cr & x:2x:4x \cr & \sqrt {5{p^2} + {q^2} + {r^2}} \cr & = \sqrt {5{x^2} + 4{x^2} + 16{x^2}} \cr & = \sqrt {25{x^2}} \cr & = 5x \cr & = 5p \cr} $$
78. The mean proportion between $$\left( {3 + \sqrt 2 } \right)$$ and $$\left( {12 - \sqrt {32} } \right)$$ is = ?
a) $$\sqrt 7 $$
b) $$2\sqrt 7 $$
c) 6
d) $$\frac{{15 - 3\sqrt 2 }}{2}$$
Discussion
Explanation:
$$\eqalign{ & \left( {3 + \sqrt 2 } \right):x:\left( {12 - \sqrt {32} } \right) \cr & a:b:c \cr & {\text{mean proportion}} \cr & {b^2} = a \times c \cr & {x^2} = \left( {3 + \sqrt 2 } \right) \times \left( {12 - \sqrt {32} } \right) \cr & {x^2} = \left( {3 + \sqrt 2 } \right) \times \left( {12 - 4\sqrt 2 } \right) \cr & {x^2} = \left( {3 + \sqrt 2 } \right) \times 4\left( {3 - \sqrt 2 } \right) \cr & {x^2} = 4 \times \left\{ {{{\left( 3 \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}} \right\} \cr & {x^2} = 28 \cr & x = 2\sqrt 7 \cr} $$
79. a : b : c = 2 : 3 : 4 and 2a - 3b + 4c = 33, then the value of c is = ?
a) 6
b) 9
c) 12
d) $$\frac{{66}}{7}$$
Discussion
Explanation:
$$\eqalign{ & a:b:c \cr & 2:3:4 \cr & {\text{Let }}2x:3x:4x \cr & 2a - 3b + 4c = 33 \cr & 2 \times 2x - 3 \times 3x + 4 \times 4x = 33 \cr & 4x - 9x + 16x = 33 \cr & 11x = 33 \cr & x = 3 \cr & {\text{c}} \Rightarrow 4 \times 3 = 12 \cr} $$
80. In a ratio which is equal to 7 : 8, if the antecedent is 35, what is the consequent?
a) 35
b) 40
c) 56
d) 64
Discussion
Explanation: Let the consequent be x
$$\eqalign{ & \frac{7}{8} = \frac{{35}}{x} \Rightarrow 35 \times 8 \cr & x = \frac{{35 \times 8}}{7} = 40 \cr} $$
81. The ratio of the first and second class fares between two railway stations is 4 : 1 and that of the number of passengers travelling by first and second classes is 1 : 40. If on a day Rs. 1100 are collected as total fare, the amount collected from the first class passengers is = ?
a) Rs. 315
b) Rs. 275
c) Rs. 137.50
d) Rs. 100
Discussion
Explanation:
| 1st | : | 2nd | |
| Fare | 4x | : | x |
| Passengers | 1 | : | 40 |
Total fare = 4x : 40x = 44x
44x = 1100
x = $$\frac{1100}{44}$$ = 25
Fare = 1st class amount received per day
4x = 4 × 25
4x = 100
82. If a : (b + c) = 1 : 3 and c : (a + b) = 5 : 7, then b : (a + c) is equal to.
a) 1 : 2
b) 2 : 3
c) 1 : 3
d) 2 : 1
Discussion
Explanation:
$$\eqalign{ & = \frac{a}{{b + c}} = \frac{1}{3} \cr & a = \frac{{b + c}}{3} \cr & \frac{c}{{a + b}} = \frac{5}{7} \cr & 7c = 5a + 5b \cr & 7c = \frac{{5\left( {b + c} \right)}}{3} + 5b \cr & 7c - \frac{5}{3}c = 5b + \frac{5}{3}b \cr & \frac{{16c}}{3} = \frac{{20b}}{3} \cr & 16c = 20b \cr & b = \frac{4}{5}c. \cr & a = \frac{{b + c}}{3} = \frac{{\frac{4}{5}c + c}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{9c}}{5} \times \frac{1}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{5}c. \cr & \frac{b}{{a + c}} = \frac{{\left( {\frac{4}{5}c} \right)}}{{\left( {\frac{3}{5}c + c} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{4c}}{5} \times \frac{5}{{8c}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1:2 \cr} $$
83.If a : b = 3 : 4, b : c = 4 : 7, then $$\frac{{a + b + c}}{c}$$ is equal to -
a) 1
b) 2
c) 3
d) 7
Discussion
Explanation:
$$\eqalign{ & = a:b = 3:4 \cr & = b:c = 4:7 \cr & \Rightarrow a:b:c = 3:4:7 \cr & {\text{Let }}a = 3k, \cr & \,\,\,\,\,\,\,\,\,\,b = 4k, \cr & \,\,\,\,\,\,\,\,\,\,c = 7k. \cr & \frac{{a + b + c}}{c} \cr & = \frac{{3k + 4k + 7k}}{{7k}} \cr & = \frac{{14k}}{{7k}} \cr & = 2 \cr} $$
84.If A : B : C = 2 : 3 : 5 and A = x% of (B + C), then x is equal to -
a) 20
b) 24
c) 25
d) 28
Discussion
Explanation: Let A = 2k, B = 3k, C = 5k
A = x% of (B + C)
⇒ 2k = x% of (3k + 5k) = x% of 8k
$$\eqalign{ & \Rightarrow \frac{x}{{100}} = \frac{{{\text{2k}}}}{{{\text{8k}}}} = \frac{1}{4} \cr & x = \frac{{100}}{4} = 25 \cr} $$
85. If a and b are rational numbers and $$a + b\sqrt 3 $$ $$ = $$ $$\frac{1}{{2 - \sqrt 3 }}{\text{,}}$$ then a : b is equal to = ?
a) 2 : 1
b) 2 : 3
c) $$\sqrt 3 $$ : 1
d) - $$\sqrt 3 $$ : 1
Discussion
Explanation:
$$\eqalign{ & a + b\sqrt 3 = \frac{1}{{2 - \sqrt 3 }} \cr & \Rightarrow \frac{1}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} \cr & \Rightarrow \frac{{2 + \sqrt 3 }}{{4 - 3}} \cr & \Rightarrow 2 + \sqrt 3 \cr} $$
By rationalisation of denominator
⇒ a + b$$\sqrt 3 $$ = 2 + $$\sqrt 3 $$
⇒ Now compare the rational & irrational parts
a = 2
b = 1
So, a : b
2 : 1
86. If $${\text{A}}$$ : $${\text{B}}$$ = $$\frac{1}{2}$$ : $$\frac{3}{8}{\text{,}}$$ $${\text{B}}$$ : $${\text{C}}$$ = $$\frac{1}{3}$$ : $$\frac{5}{9}$$ and $${\text{C}}$$ : $${\text{D}}$$ = $$\frac{5}{6}$$ : $$\frac{3}{4}{\text{,}}$$ Then the ratio of A : B : C : D is ?
a) 6 : 4 : 8 : 10
b) 6 : 8 : 9 : 10
c) 8 : 6 : 10 : 9
d) 4 : 6 : 8 : 10
Discussion
Explanation:
$$\eqalign{ & {\text{A}}:{\text{B}} = \frac{1}{2}:\frac{3}{8}, \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{4}{3} \cr & {\text{B}}:{\text{C}} = \frac{1}{3}:\frac{5}{9}, \cr & \frac{{\text{B}}}{{\text{C}}} = \frac{1}{3} \times \frac{9}{5} = \frac{3}{5} \cr & {\text{C}}:{\text{D}} = \frac{5}{6}:\frac{3}{4}, \cr & \frac{{\text{C}}}{{\text{D}}} = \frac{{5 \times 4}}{{6 \times 3}} = \frac{{10}}{9} \cr & {\text{A}}:{\text{B}}:{\text{C}}:{\text{D}} \cr & {\text{4}}:{\text{3}} \cr & \,\,\,\,\,\,\,{\text{3}}:{\text{5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ 10}}:{\text{9}} \cr & \Rightarrow {\text{A}}:{\text{B}}:{\text{C}}:{\text{D}} \cr & \,\,\,\,\,\,\,\,\,8:6:10:9 \cr} $$
87.If A : B : C = 2 : 3 : 4, then ratio $$\frac{{\text{A}}}{{\text{B}}}$$ : $$\frac{{\text{B}}}{{\text{C}}}$$ : $$\frac{{\text{C}}}{{\text{A}}}$$ is equal to?
a) 8 : 9 : 16
b) 8 : 9 : 12
c) 8 : 9 : 24
d) 4 : 9 : 12
Discussion
Explanation:
$$\eqalign{ & {\text{A}}:{\text{B}}:{\text{C}} = 2:3:4 \cr & {\text{Let }} \cr & {\text{A}} = 2x,{\text{B}} = 3x,{\text{C}} = 4x \cr & \frac{{\text{A}}}{{\text{B}}}:\frac{{\text{B}}}{{\text{C}}}:\frac{{\text{C}}}{{\text{A}}} = \frac{{2x}}{{3x}}:\frac{{3x}}{{4x}}:\frac{{4x}}{{2x}} \cr & \Rightarrow \frac{2}{3}:\frac{3}{4}:\frac{2}{1} \cr} $$
Multiply by the L.C.M of denominator to remove fraction
$$\eqalign{ & {\text{L}}{\text{.C}}{\text{.M of }}\left( {3,4,1} \right) = 12 \cr & \frac{{\text{A}}}{{\text{B}}}\,\,\,\,\,:\,\,\,\,\,\frac{{\text{B}}}{{\text{C}}}\,\,\,\,\,\,:\,\,\,\,\,\frac{{\text{C}}}{{\text{A}}} \cr & \frac{2}{3} \times 12:\frac{3}{4} \times 12:\frac{2}{1} \times 12 \cr & \,\,\,\,\,\,\,8\,\,\,\,\,\,\,:\,\,\,\,\,\,\,9\,\,\,\,\,\,\,:\,\,\,\,\,\,\,24 \cr} $$
88. If A : B = 2 : 3, B : C = 2 : 4 and C : D = 2 : 5, then A : D is equal to -
a) 1 : 5
b) 2 : 5
c) 3 : 5
d) 2 : 15
Discussion
Explanation:
$$\eqalign{ & = \frac{{\text{A}}}{{\text{B}}} = \frac{2}{3},\,\frac{{\text{B}}}{{\text{C}}} = \frac{2}{4},\,\frac{{\text{C}}}{{\text{D}}} = \frac{2}{5} \cr & \frac{{\text{A}}}{{\text{D}}} = \left( {\frac{{\text{A}}}{{\text{B}}} \times \frac{{\text{B}}}{{\text{C}}} \times \frac{{\text{C}}}{{\text{D}}}} \right) \cr & \frac{2}{3} \times \frac{2}{4} \times \frac{2}{5} = \frac{2}{{15}} \cr & {\text{A}}:{\text{D}} = 2:15. \cr} $$
89.If 3A = 5B and 4B = 6C, then A : C equal to -
a) 2 : 5
b) 3 : 5
c) 4 : 5
d) 5 : 2
Discussion
Explanation:
$$\eqalign{ & = {\text{3A}} = {\text{5B and 4B}} = {\text{6C}} \cr & \frac{{\text{A}}}{{\text{B}}} = \frac{5}{3}{\text{ and }}\frac{{\text{B}}}{{\text{C}}} = \frac{6}{4} = \frac{3}{2} \cr & \frac{{\text{A}}}{{\text{C}}} = \left( {\frac{{\text{A}}}{{\text{B}}} \times \frac{{\text{B}}}{{\text{C}}}} \right) = \frac{5}{3} \times \frac{3}{2} = \frac{5}{2} \cr & {\text{A}}:{\text{C}} = {\text{5}}:{\text{2}} \cr} $$
90. In a business, the ratio of the capitals of A and B is 2 : 1, that of B and C is 4 : 3 and that of D and C is 6 : 5. Then the ratio of the capitals of A and D is -
a) 9 : 20
b) 3 : 5
c) 5 : 3
d) 20 : 9
Discussion
Explanation:
$$\eqalign{ & = \frac{{\text{A}}}{{\text{B}}} = \frac{2}{1},\,\,\frac{{\text{B}}}{{\text{C}}} = \frac{4}{3},\,\,\frac{{\text{C}}}{{\text{D}}} = \frac{5}{6} \cr & \Rightarrow \frac{{\text{A}}}{{\text{D}}} = \left( {\frac{{\text{A}}}{{\text{B}}} \times \frac{{\text{B}}}{{\text{C}}} \times \frac{{\text{C}}}{{\text{D}}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{2}{1} \times \frac{4}{3} \times \frac{5}{6}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20}}{9} \cr & {\text{A}}:{\text{D}} = {\text{20}}:{\text{9}} \cr} $$
91. If a : b = b : c, then a4 : b4 would be equal to
a) ac : b2
b) a2 : c2
c) c2 : a2
d) b2 : ac
Discussion
Explanation:
$$\eqalign{ & = \frac{a}{b} = \frac{b}{c} \cr & \Rightarrow {b^2} = ac \cr & \Rightarrow \frac{{{a^4}}}{{{b^4}}} = \frac{{{a^4}}}{{{{\left( {ac} \right)}^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{a^4}}}{{{a^2}{c^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{a^2}}}{{{c^2}}} \cr & {a^4}:{b^4} = {a^2}:{c^2} \cr} $$
92. If (4x2 - 3y2) : (2x2 + 5y2) = 12 : 19, then x : y is
a) 2 : 3
b) 1 : 2
c) 3 : 2
d) 2 : 1
Discussion
Explanation:
$$\eqalign{ & {\text{ = }}\frac{{4{x^2} - 3{y^2}}}{{2{x^2} + 5{y^2}}} = \frac{{12}}{{19}} \cr & 19\left( {4{x^2} - 3{y^2}} \right) = 12\left( {2{x^2} + 5{y^2}} \right) \cr & 76{x^2} - 57{y^2} = 24{x^2} + 60{y^2} \cr & 52{x^2} = 117{y^2} \cr & 4{x^2} = 9{y^2} \cr & \frac{{{x^2}}}{{{y^2}}} = \frac{9}{4} \cr & {\left( {\frac{x}{y}} \right)^2} = {\left( {\frac{3}{2}} \right)^2} \cr & \frac{x}{y} = \frac{3}{2} \cr & x:y = 3:2 \cr} $$
93. If x : y = 3 : 4 and a : b = 1 : 2, then the value of $$\frac{{2xa + yb}}{{3yb - 4xa}}$$ is
a) $$\frac{5}{6}$$
b) $$\frac{6}{5}$$
c) $$\frac{6}{7}$$
d) $$\frac{7}{6}$$
Discussion
Explanation:
$$\eqalign{ & = \frac{x}{y} = \frac{3}{4}{\text{ and }}\frac{a}{b} = \frac{1}{2} \cr & \Rightarrow \frac{{xa}}{{yb}} = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8} \cr & \frac{{2xa + yb}}{{3yb - 4xa}} \cr & = \frac{{2\left( {\frac{{xa}}{{yb}}} \right) + 1}}{{3 - 4\left( {\frac{{xa}}{{yb}}} \right)}} \cr & = \frac{{2 \times \frac{3}{8} + 1}}{{3 - 4 \times \frac{3}{8}}} \cr & = \frac{7}{4} \times \frac{2}{3} \cr & = \frac{7}{6} \cr} $$
94.Income of A and B are in the ratio 4 : 3 and their annual expenses are in the ratio 3 : 2. If each save Rs. 60000 at the end of the year, the annual income of A is = ?
a) Rs. 120000
b) Rs. 150000
c) Rs. 240000
d) Rs. 360000
Discussion
Explanation:
| A | : | B | |
| Income | 4x | : | 3x |
| Expenses | 3y | : | 2y |
| Saving | 6000 | : | 6000 |
Income = Expenses + Saving
$$ \frac{{4x - 60000}}{{3x - 60000}} = \frac{3}{2}$$
8x - 120000 = 9x - 180000
x = 60000
Income of A = 4 × 60000 = 240000
95.The weight of Mr. Gupta and Mrs. Gupta are in the ratio 7 : 8 and their total weight is 120 kg. After taking a dieting course Mr. Gupta reduces by 6 kg and the ratio between their weights change to 5 : 6, so Mrs. Gupta has reduced by = ?
a) 2 kg
b) 4 kg
c) 3 kg
d) 5 kg
Discussion
Explanation:
| Mr. | : | Mrs. | |
| Before | 7x | : | 8x |
| After | 5y | : | 6y |
Before 7x + 8x = 120
15x = 120
x = 8
Mr. Gupta = 7 × 8 = 56
Mrs. Gupta = 8 × 8 = 64
After losing 6kg by Mr. Gupta the ratio becomes 5 : 6
Let Mrs. Gupta loss x kg
$$\eqalign{ & \therefore \frac{{56 - 6}}{{64 - x}} \times \frac{5}{6} \cr & \left( {{\text{Cross multiplication}}} \right) \cr} $$
300 = 320 - 5x
5x = 20
x = 4 kg
96. The income of A, B and C are in the ratio 7 : 9 : 12 and their spending are in the ratio 8 : 9 : 15. If A saves $$\frac{1}{4}$$ th of his income then the savings of A, B and C are in the ratio of = ?
a) 56 : 99 : 69
b) 69 : 56 : 99
c) 99 : 56 : 69
d) 99 : 69 : 56
Discussion
Explanation: Let income of A, B and C are 7x, 9x and 12x respectively
and expenditure of A, B and C are 8y, 9y and 15y respectively
$$\eqalign{ & \Rightarrow {\text{Income}}\,{\text{of,}} \cr & A \times \frac{1}{4} = {\text{Saving}}\,{\text{of}}\,{\text{A}}\,\left( {{\text{given}}} \right) \cr & 7x - 8y = 7x \times \frac{1}{4} \cr & 28x - 32y = 7x \cr & 21x = 32y \cr & x:y = 32:21 \cr} $$
The ratio of savings of A, B and C
⇒ (7x - 8y) : (9x - 9y) : (12x - 15y)
⇒ (7 × 32 - 8 × 21) : (9 × 32 - 9 × 21) : (12 × 32 - 15 × 21)
⇒ (224 - 168) : (288 - 189) : (384 - 315)
⇒ 56 : 99 : 69
97. What will be the simplest form of the ratio 3 hours : 1 day?
a) 1 : 3
b) 1 : 6
c) 1 : 8
d) 1 : 25
Discussion
Explanation: 1 day = 24 hours.
Given ration = 3 : 24
= 1 : 8
98. In a proportion the product of 1st and 4th terms is 40 and that of 2nd and 3rd terms is 2.5x. Then the value of x is.
a) 16
b) 26
c) 75
d) 90
Discussion
Explanation: Product of 1st and 4th terms (extremes) = product of 2nd and 3rd terms (means)
$$\eqalign{ & {\text{2}}{\text{.5}}x = {\text{40}} \cr & x = \frac{{40}}{{2.5}} = 16 \cr} $$
99. If 20% of A = 30% of B = $$\frac{1}{6}$$ of C, then A : B : C is -
a) 2 : 3 : 6
b) 3 : 2 : 16
c) 10 : 15 : 18
d) 15 : 10 : 18
Discussion
Explanation:
$$\eqalign{ & 20\% \,{\text{of A}} = 30\% \,{\text{of B}} = \frac{1}{6}{\text{of C}} \cr & \Rightarrow \frac{{20{\text{A}}}}{{100}} = \frac{{30{\text{B}}}}{{100}} = \frac{{\text{C}}}{{\text{6}}} \cr & \Rightarrow \frac{{\text{A}}}{5} = \frac{{3{\text{B}}}}{{10}} = \frac{C}{6} = k(say) \cr & A = 5K,\,B = \frac{{10k}}{3},\,C = 6k \cr & {\text{A}}:{\text{B}}:{\text{C}} = 5k:\frac{{10k}}{3}:6k \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5:\frac{{10}}{3}:6 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 15:10:18 \cr} $$
100. Two numbers are in the ratio 17 : 45, One - third of the smaller is less than $$\frac{1}{5}$$ of the bigger by 15. The smaller number is = ?
a) $$25\frac{1}{2}$$
b) $$67\frac{1}{2}$$
c) $$76\frac{1}{2}$$
d) $$86\frac{1}{2}$$
Discussion
Explanation:
$$\eqalign{ & {\text{A}}:{\text{B}} \cr & 17:45 \cr & 17x:45x \cr & \Rightarrow 17x \times \frac{1}{3} + 15 = 45x \times \frac{1}{5} \cr & \Rightarrow \frac{{17x + 45}}{3} = 9x \cr & \Rightarrow 17x + 45 = 27x \cr & \Rightarrow 10x = 45 \cr & \Rightarrow x = \frac{{45}}{{10}} \cr & {\text{Smaller number is }} \cr & = \frac{{45}}{{10}} \times 17 \cr & = \frac{{765}}{{10}} \cr & = 76\frac{1}{2} \cr} $$