### Find solutions for all the questions at the end of each set:

### Set A

1. If A:B = 2:3, B:C = 4:5 and C:D = 6:7, then find the value of A:B:C:D

A. 15:24:30:35

B. 16:24:30:35

C. 17:24:30:35

D. 18:24:30:35

2. 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

3.The salaries of A, B and C are of ratio 2:3:5. If the increments of 15%, 10% and 20% are done to their respective salaries, then find the new ratio of their salaries.

A. 20:33:60

B. 21:33:60

C. 22:33:60

D. 23:33:60

4. 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

5. A dog takes 3 leaps for every 5 leaps of a hare. If one leap of the dog is equal to 3 leaps of the hare, the ratio of the speed of the dog to that of the hare is

A. 4:5

B. 9:5

C. 8:5

D. 9:2

6. In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quanity of water to be further added is:

A. 20 litres

B. 30 litres

C. 40 litres

D. 60 litres

7. 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 B have.

A. Rs 484

B. Rs. 480

C. Rs 478

D. Rs 470

8. 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:

A. 20

B. 30

C. 48

D. 58

9. If Rs. 782 be divided into three parts, proportional to 1/2 : 2/3 : 3/4, then the first part is:

A. Rs. 182

B. Rs. 190

C. Rs. 196

D. Rs. 204

10. 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

ANSWER AND SOLUTION:

1(B)Explanation:

A : B = 2 : 3

B : C = 4 : 5 = (4*3/4 : 5*3/4)

=3 : 15/4

C:D=6:7=(6*15/24:7*15/24)

=15/4 : 35/8

A : B : C : D = 2 : 3 : 15/4 : 35/8

= 16 : 24 : 30 : 35

2(C)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.

3(D)Explanation:

Let A salary be 2k

B salary be 3k and C salary be 5k

A’s new salary = 115/100*2k

=23/10k

B’s new salary = 110/100*3k

=33/10k

C’s new salary = 120/100*5k

=6k

New ratio = 23k/10 : 33k/10 : 6k

= 23 : 33 : 60

4(A)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 x 5x ) , (150/100 x 7x) and (175/100 x 8x)

=> 7x , 21x/2 and 14x.

so The required ratio = 7x : 21x/2 : 14x

=> 14x : 21x : 28x

=> 2 : 3 : 4.

5(B)Explanation:

Dog : Hare = (3*3) leaps of hare : 5 leaps of hare

= 9 : 5

6(D)Explanation:

Quantity of milk = (60 x 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.

so Quantity of water to be added = 60 litres.

7(A)Explanation:

In this type of question, we will first try to calculate the ratio of two persons. Once we get ratio then we can easily get our answer. So lets solve this.

4/15 A = 2/5 B

A = (2/5*14/4) B

A = 3/2B

A/B = 3/2

A : B = 3 : 2

B’s Share = 2/5*1210 = 484

8(B)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 5/49) = 30.

9(D)Explanation:

Given ratio = 1/2 : 2/3 : 3/4 = 6 : 8 : 9.

so 1st part = Rs.(782 x 6/23) = Rs. 204

10(C)Explanation:

Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.

Then, sum of their values = Rs.(25x/100 + (10 x 2x)/100 + (5 x 3x)/100)

= Rs. 60x/100

So 60x/100 = 30 <=> x = (30 x 100)/60 = 50

Hence, the number of 5 p coins = (3 x 50) = 150.

### Find solutions for all the questions at the end of each set:

### Set B

1. How much percent more than the cost price should a shopkeeper mark his goods so that after allowing a discount of 25% on the marked price, he gains 20%?

A. 60%

B. 55%

C. 70%

D. 50%

2. A dishonest dealer professes to sell his goods at the cost price but uses a false weight of 850 g instead of 1 kg. His gain percent is

A. 71 11/17%

B. 11 11/17%

C. 17 12/17%

D. 17 11/17%

3. An article is sold at 10% loss. If the selling price is Rs. 40 more, there will be a gain of 15%. The cost price of the article is:

A. Rs. 140

B. Rs. 120

C. Rs. 175

D. Rs. 160

4. The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, find out the value of x

A. 15

B. 25

C. 18

D. 16

5.In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?

A. 30%

B. 70%

C. 100%

D. 250%

6.The percentage profit earned by selling an item for Rs. 1920 is equal to the percentage loss incurred by selling the same item for Rs. 1280. At what price should the item be sold to make 25% profit?

A. Insufficient Data

B. Rs. 3000

C. Rs. 2000

D. Rs. 2200

7. Some articles were bought at 6 articles for Rs. 5 and sold at 5 articles for Rs. 6. Gain percent is:

A. 30%

B. 33 1/3%

C. 35%

D. 44%

8. 100 oranges are bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen. The percentage of profit or loss is:

A. 14 2/7% gain

B. 15% gain

C. 14 2/7 % loss

D. 15 % loss

9.A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. What is his profit percentage?

A. 6%

B. 5%

C. 4%

D. 7%

10. A trader gives 12% additional discount on the discounted price, after giving an initial discount of 20% on the labeled price of an item. The final sale price of the item is Rs.704. Find out the labeled price?

A. 1000

B. 2000

C. 1200

D. 920

ANSWER AND SOLUTION :

1(A)Explanation :

Let cost price of goods be Rs 100.

Gain = 20%

Therefore, Selling price = Rs 120

Discount = 25%

Marked Price = (100/100-25)x120

= Rs. 160

i .e. 60% more

2(D)Explanation :

If a trader professes to sell his goods at cost price, but uses false weights, then

Gain% = {Error/(True value – Error) x 100}%

In the given question, Error = 1000 – 850 = 150

Thus, Gain% = {150/(1000 – 150) x 100}%

= 17 11/17%

3(D)Explanation :

Let the cost price be Rs. x.

Selling Price at 10% loss = 90x/100

Selling Price at 15% gain = 115x/100

Thus, according to the problem,

115x/100 – 90x/100 = 40

x = Rs.160

4(D)Explanation :

Let the Cost Price (CP) of one article = 1

=> CP of x articles = x ——————————(Equation 1)

CP of 20 articles = 20

Given that cost price of 20 articles is the same as the selling price of x articles

=> Selling price (SP) of x articles = 20————–(Equation 2)

Given that Profit = 25%

(SP-CP/CP)=25/100=1/4————( Equation 3)

Substituting equations 1 and 2 in equation 3,

(20-x)/x=1/4

80-4x=x

5x=80

x=80/5=16

5(B)Explanation:

Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs. 420.

Profit = Rs. (420 – 125) = Rs. 295.

Required percentage = (295/420 x 100)% = 1475/21 % = 70% (approximately).

6(C)Explanation :

Let CP = x

Percentage profit earned by selling an item for Rs. 1920

= (SP-CP/CP0 ×100

=(1920-x)/x×100

Percentage loss incurred by selling the same item for Rs. 1280

= (CP-SP)/CP×100

= (x-1280)/x ×100

Given that Percentage profit earned by selling an item for Rs. 1920=Percentage loss incurred by selling the same item for Rs. 1280

(1920-x)/x ×100 = x-1280/x ×100

(1920-x)/x = (x-1280)/x

1920–x = x–1280

2x=1920+1280=3200

x=3200/2

=1600

Required Selling Price = CP×125/100

=1600×125/100 =1600×5/4

=400×5=2000

7(D)Explanation:

Suppose, number of articles bought = L.C.M. of 6 and 5 = 30.

C.P. of 30 articles = Rs.(5/6 x 30) = Rs. 25.

S.P. of 30 articles = Rs.(6/5 x 30) = Rs. 36.

Gain % = (11/25 x 100) % = 44%.

8(A)Explanation:

C.P. of 1 orange = Rs.(350/100) = Rs. 3.50

S.P. of 1 orange = Rs (48/12) = Rs. 4

Gain% = (0.50/3.50 x 100) % = 100/7 % = 14 2/7%

9(B)Explanation :

CP of 1st variety rice=20

CP of 2nd variety rice=36

CP of the 56 kg rice mixture=(26×20+30×36)=520+1080=1600

SP of the 1 kg rice mixture=30

SP of the 56 kg rice mixture=30×56=1680

Gain=SP-CP=1680-1600=80

Gain%=Gain/CP×100=80/1600×100=100/20=5%

10(A)Explanation :

Let the labeled price=x

SP=704

Initial Discount=20%

Price after initial discount=x×80/100

Additional discount=12%

Price after additional discount=x×80/100 × 88/100

But Price after additional discount=SP=704

=> x×80/100 × 88100=704

=>x×4/5 × 22/25=704

=>x=704×25/22 × 5/4=176×25/22×5

=8×25×5=40×25=1000

### Find solutions for all the questions at the end of each set:

### Set C

1.A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

A. 26 litres

B. 29.16 litres

C. 28 litres

D. 28.2 litres

2. A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

A. 1/3

B. 1/4

C. 1/5

D. 1/7

3. Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety of tea in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, what is the price of the third variety per kg ?

A. Rs.182.50

B. Rs.170.5

C. Rs.175.50

D. Rs.180

4.A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?

A. 10

B. 20

C. 21

D. 25

5. Two vessels A and B contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively. Find the ratio in which these mixture be mixed to obtain a new mixture in vessel C containing spirit and water in the ration 8 : 5 ?

A. 3: 4

B. 4 : 3

C. 9 : 7

D. 7 : 9

6. In what ratio must a grocer mix two varieties of pulses costing Rs. 15 and Rs. 20 per kg respectively so as to get a mixture worth Rs. 16.50 kg?

A. 3 : 7

B. 5 : 7

C. 7 : 3

D. 7 : 5

7.8 litres are drawn from a cask full of 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 water is 16 : 65. How much wine did the cask originally hold?

A. 30 litres

B. 26 litres

C. 24 litres

D. 32 litres

8. A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is:

A. 4%

B. 6 1/4%

C. 20%

D. 25%

9. A jar full of whiskey contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohols and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is

A. 4/3

B. 3/4

C. 3/2

D. 2/3

10. A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is:

A. 1/3

B. 2/3

C. 2/5

D. 3/5

ANSWER AND SOLUTION :

1( B)Explanation :

Assume that a container contains x of liquid from which y units are taken out and replaced by water. After n operations, the quantity of pure liquid

=x(1-y/x)^n

Hence milk now contained by the container = 40(1-4/40)^3

=40(1-1/10)^3

=40×9/10×9/10×9/10 =(4×9×9×9)/100 =29.16

2(C)Explanation:

Suppose the vessel initially contains 8 litres of liquid.

Let x litres of this liquid be replaced with water.

Quantity of water in new mixture = (3 – 3x/8 + x) litre

Quantity of syrup in new mixture = (5 – 5x/8) litres

So (3 – 3x/8 + x) = (5 – 5x/8) litres

=> 5x + 24 = 40 – 5x

=>10x = 16

=> x = 8/5 .

So, part of the mixture replaced = (8/5 x 1/8) = 1/5

3(C)Explanation :

Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed in the ratio 1 : 1

So their average price = (126+135)/2=130.5

Hence let’s consider that the mixture is formed by mixing two varieties of tea.

one at Rs. 130.50 per kg and the other at Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. Now let’s find out x.

By the rule of alligation, we can write as

Cost of 1 kg of 1st kind of tea Cost of 1 kg of 2nd kind of tea

(130.50) (x)

Mean Price

(153)

(x – 153) (22.50)

=>(x – 153) : 22.5 = 1 : 1

=>x – 153 = 22.50

=> x = 153 + 22.5 = 175.5

4(C)Explanation:

Suppose the can initially contains 7x and 5x of mixtures A and B respectively.

Quantity of A in mixture left = (7x – 7/12 x 9)litres

= (7x – 21/4) litres.

Quantity of B in mixture left = (5x – 5/12 x 9) litres

= (5x – 15/4) litres.

So (7x – 21/4)/((5x – 15/4) +9) = 7/9

=> (28x – 21)/(20x + 21) = 7/9

=> 252x – 189 = 140x + 147

=> 112x = 336

=> x = 3.

So, the can contained 21 litres of A.

.

5(D)Explanation :

Let Cost Price(CP) of 1 litre spirit be Rs.1

Quantity of spirit in 1 litre mixture from vessel A = 5/7

Cost Price(CP) of 1 litre mixture from vessel A = Rs. 5/7

Quantity of spirit in 1 litre mixture from vessel B = 7/13

Cost Price(CP) of 1 litre mixture from vessel B = Rs. 7/13

Quantity of spirit to be obtained in 1 litre mixture from vessel C = 8/13

Cost Price(CP) of 1 litre mixture from vessel C = Rs. 8/13 = Mean Price

By the rule of alligation, we can write as

CP of 1 litre mixture CP of 1 litre mixture

from vessel A (5/7) from vessel B (7/13)

Mean Price

(8/13)

8/13 – 7/13 = 1/13 5/7 – 8/13 = 9/91

=> Mixture from Vessel A : Mixture from Vessel B = 1/13 : 9/91 = 7 : 9 = Required Ratio

6(C)Explanation:

By the rule of alligation:

Cost of 1 kg pulses of Cost of 1 kg pulses of

1st kind Rs. (15) 2nd kindRs. (20)

Mean Price

Rs. (16.50)

(3.50) (1.50)

Required rate = 3.50 : 1.50 = 7 : 3.

7(C)Explanation :

Let initial quantity of wine = x litre

After a total of 4 operations, quantity of wine = x(1-y/x)^n=x(1-8/x)^4

Given that after a total of 4 operations, the ratio of the quantity of wine left in cask to that of water = 16 : 65

Hence we can write as (x(1-8/x)^4)/x =16/81

(1-8/x)^4 = (2/3)^4

(1-8/x) = 2/3

(x-8/x) = 2/3

3x-24=2x

x=24

8(C)Explanation:

Let C.P. of 1 litre milk be Re. 1

Then, S.P. of 1 litre of mixture = Re. 1, Gain = 25%.

C.P. of 1 litre mixture = Re.(100/125 x 1) = 4/5

By the rule of alligation, we have:

C.P. of 1 litre of milk C.P. of 1 litre of water

Re. (1) (0)

Mean Price

Re. 4/5

4/5 1/5

Ratio of milk to water = 4/5 : 1/5 = 4 : 1.

Hence, percentage of water in the mixture = (1/5 x 100)% = 20%

9(D)Explanation :

Concentration of alcohol in 1st Jar = 40%

Concentration of alcohol in 2nd Jar = 19%

After the mixing, Concentration of alcohol in the mixture = 26%

By the rule of alligation,

Concentration of alcohol in Concentration of alcohol in

1st Jar(40%) 2nd Jar (19%)

Mean

(26%)

(7) (14)

Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2

10(B)Explanation:

By the rule of alligation, we have:

Strength of first jar Strength of 2nd jar

(40%) (19%)

Mean

Strength

(26%)

(7) (14)

So, ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2

Required quantity replaced = 2/3