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

### Set A

1. Three partners shared the profit in a business in the ratio 5 : 7 : 8. They had partnered for 14 months, 8 months and 7 months respectively. What was the ratio of their investments?

A. 5 : 7 : 8

B. 20 : 49 : 64

C. 38 : 28 : 21

D. None of these

2. P , Q, R enter into a partnership & their share are in the ratio 1/2 : 1/3 : 1/4 , after two months , P withdraws half of the capitals & after 10 months , a profit of Rs 378 is divided among them . What is Q’s share?

A. 114

B. 120

C. 134

D. 144

3. A, B, C subscribe Rs. 50,000 for a business. A subscribes Rs. 4000 more than B and B Rs. 5000 more than C. Out of a total profit of Rs. 35,000, A receives:

A. Rs. 8400

B. Rs. 11,900

C. Rs. 13,600

D. Rs. 14,700

4. 29. P, Q, R enter into a partnership. P initially invests 25 lakh & adds another 10 lakhs after one year. Q initially invests 35 lakh & withdrawal 10 lakh after 2 years and R invests Rs 30 Lakhs . In what ratio should the profit be divided at the end of 3 years?

A. 18:19:19

B. 18:18:19

C. 19:19:18

D. 18:19:19

5. A and B started a business in partnership investing Rs. 20,000 and Rs. 15,000 respectively. After six months, C joined them with Rs. 20,000. What will be B’s share in total profit of Rs. 25,000 earned at the end of 2 years from the starting of the business?

A. Rs. 7500

B. Rs. 9000

C. Rs. 9500

D. Rs. 10,000

6. In a business, A and C invested amounts in the ratio 2 : 1 , whereas the ratio between amounts invested by A and B was 3 : 2 . If Rs 157300 was their profit, how much amount did B receive?

A. 48000

B. 48200

C. 48400

D. 48600

6. A, B, C rent a pasture. A puts 10 oxen for 7 months, B puts 12 oxen for 5 months and C puts 15 oxen for 3 months for grazing. If the rent of the pasture is Rs. 175, how much must C pay as his share of rent?

A. Rs. 45

B. Rs. 50

C. Rs. 55

D. Rs. 60

7. If 4 (P’s Capital ) = 6 ( Q’s Capital ) = 10 ( R’s Capital ) , then out of the total profit of Rs 4650 , R will receive

A. 600

B. 700

C. 800

D. 900

8. Three partners A , B , C start a business . B’s Capital is four times C’s capital and twice A’s capital is equal to thrice B’s capital . If the total profit is Rs 16500 at the end of a year ,Find out B’s share in it.

A. 4000

B. 5000

C. 6000

D. 7000

9. P and Q invested in a business. The profit earned was divided in the ratio 2 : 3. If P invested Rs 40000, the amount invested by Q is

A. 40000

B. 50000

C. 60000

D. 70000

10. Kamal started a business investing Rs 9000. After five months, Sameer joined with a capital of Rs 8000. If at the end of the year, they earn a profit of Rs. 6970, then what will be the share of Sameer in the profit ?

A. Rs 2380

B. Rs 2300

C. Rs 2280

D. Rs 2260

ANSWERS AND SOLUTION :

1(B)Explanation:

Let their investments be Rs. x for 14 months, Rs. y for 8 months and Rs. z for 7 months respectively.

Then, 14x : 8y : 7z = 5 : 7 : 8.

Now, 14x/8y = 5/7 => 98x = 40y => y = 49/20 x

And, 14x/7z = 5/8 => 112x = 35z => z = 112/35 x = 16/5 .x.

So x : y : z = x : 49/20 x : 16/5 x = 20 : 49 : 64.

2(D)Explanation :

The ratio of their initial investment = 1/2 : 1/3 : 1/4

= 6 : 4: 3

Let’s take the initial investment of P, Q and R as 6x, 4x and 3x respectively

A:B:C = (6x * 2 + 3x * 10) : 4x*12 : 3x*12

= (12+30) : 4*12 : 3*12

=(4+10) : 4*4 : 12

= 14 : 16 : 12

= 7 : 8 : 6

B’s share = 378 * (8/21) = 18 * 8 = 144

3(D)Explanation:

Let C = x.

Then, B = x + 5000 and A = x + 5000 + 4000 = x + 9000.

So, x + x + 5000 + x + 9000 = 50000

=> 3x = 36000

=> x = 12000

A : B : C = 21000 : 17000 : 12000 = 21 : 17 : 12.

So A’s share = Rs. (35000 x 21/50) = Rs. 14,700.

4(C)Explanation :

P:Q:R = (25*1+35*2) : (35*2 : 25*1) : (30*3)

= 95 : 95 : 90

= 19 : 19: 18

5(A)Explanation:

A : B : C = (20,000 x 24) : (15,000 x 24) : (20,000 x 18) = 4 : 3 : 3.

So B’s share = Rs. (25000 x 3/10) = Rs. 7,500.

6(C)Explanation :

Assume that investment of C = x

Then, investment of A =2x

Investment of B = 4x/3

A:B:C = 2x : 4x/3 : x = 2 : 4/3 : 1 =6 : 4 : 3

B’s share = 157300 * 4/(6+4+3) = 157300*4/13

= 12100*4 = 48400

6(A)Explanation:

A : B : C = (10 x 7) : (12 x 5) : (15 x 3) = 70 : 60 : 45 = 14 : 12 : 9.

C’s rent = Rs.(175 x 9/35) = Rs. 45.

7(D)Explanation :

Let P’s capital = p, Q’s capital = q and R’s capital = r

Then

4p = 6q = 10r

=> 2p = 3q = 5r

=>q = 2p/3

r = 2p/5

P : Q : R = p : 2p/3 : 2p/5

= 15 : 10 : 6

R’s share = 4650 * (6/31) = 150*6 = 900

8(C)Explanation :

Suppose C’s capital = x then

B’s capital = 4x (Since B’s Capital is four times C’s capital)

A’s capital = 6x ( Since twice A’s capital is equal to thrice B’s capital)

A:B:C =6 x : 4x : x

= 6 : 4 : 1

B’s share = 16500 * (4/11) = 1500*4 = 6000

9(C)Explanation :

Let the amount invested by Q = q

40000 : q = 2 : 3

=> 40000/q = 2/3

=> q = 40000 * (3/2) = 60000

10(A)Explanation:

Now as per question, Kamal invested for 12 months and Sameer invested for 7 months.

So Kamal: Sameer = (9000*12):(8000*7)

= 108 : 56

= 27 : 14

Sameer Ratio in profit will be =(6970*14/41)=Rs2380

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

### Set B

1.Six years ago, the ratio of the ages of Kunal and Sagar was 6:5, Four years hence, the ratio of their ages will be 11:10. What is Sagar age at present

a.10 years

b.12 years

c.14 years

d.16 years

2. A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is

a.20 years

b.21 years

c.22 years

d.24 years

3. A man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. The ratio of the speed of the boat (in still water) and the stream is:

a. 3 : 1

b. 1 : 3

c. 1 : 2

d. 2 : 1

4.Tap ‘A’ can fill the tank completely in 6 hrs while tap ‘B’ can empty it by 12 hrs. By mistake, the person forgot to close the tap ‘B’, As a result, both the taps, remained open. After 4 hrs, the person realized the mistake and immediately closed the tap ‘B’. In how much time now onwards, would the tank be full?

a. 2 hours

b. 4 hours

c. 5 hours

d. 1 hour

5.A man can row at a speed of 12 km/hr in still water to a certain upstream point and back to the starting point in a river which flows at 3 km/hr. Find his average speed for total journey.

a. 12 3/4 km/hr

b. 11 3/4 km/hr

c. 12 1/4 km/hr

d. 11 1/4 km/hr

6.How many litres of water must be added to 16 liters of milk and water contains 10% water to make it 20% water in it?

a. 4 litre

b. 2 litre

c. 1 litre

d. 3 litre

7.John bought 20 kg of wheat at the rate of Rs.8.50 per kg and 35 kg at the rate of Rs.8.75 per kg. He mixed the two. Approximately at what price per kg should he sell the mixture to make 40% profit as the cost price?

a. Rs.12

b. Rs.8

c. Rs.16

d. Rs.20

8.Mr. Mani invested an amount of Rs. 12000 at the simple interest rate of 10% per annum and another amount at the simple interest rate of 20% per annum. The total interest earned at the end of one year on the total amount invested became 14% per annum. Find the total amount invested.

a. Rs. 25000

b. Rs. 15000

c. Rs. 10000

d. Rs. 20000

9.A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?

a. 3.46%

b. None of these

c. 4.5%

d. 5%

10. At what rate percent of simple interest will a sum of money double itself in 20 years?

a. 4%

b. 5%

c. 6%

d. 8%

Answers:

1.D

Let six years ago the age of Kunal and Sagar are 6x and 5x resp.

then,

((6x+6)+4)/((5x+6)+4)=11/10

10(6x+10)=11(5x+10)

5x=10=>x=2

So Sagar age is (5x+6) = 16

2.C

Let the son’s present age be x years. Then, man’s present age = (x + 24) years

(x + 24) + 2 = 2(x + 2)

x + 26 = 2x + 4

So, x = 22

3.A

Let speed upstream = x

Then, speed downstream = 2x

Speed in still water = (2x+x)/2=3x/2

Speed of the stream = (2x−x)/2=x/2

Speed in still water : Speed of the stream = 3x/2:x/2 = 3 : 1

4.B

Tap A can fill the tank completely in 6 hours

In 1 hour, Tap A can fill 1⁄6 of the tank

Tap B can empty the tank completely in 12 hours

In 1 hour, Tap B can empty 1⁄12 of the tank

i.e., In one hour, Tank A and B together can effectively fill 1⁄6 – 1⁄12 = 1⁄12 of the tank

In 4 hours, Tank A and B can effectively fill 1⁄12 × 4 = 1⁄3 of the tank.

Time taken to fill the remaining 1−1/3=2/3of the tank =(2/3)/(1/6) = 4 hours

5.D

Speed of the man in still water = 12 km/hr

Speed of the stream = 3 km/hr

Speed downstream = (12+3) = 15 km/hr

Speed upstream = (12-3) = 9 km/hr

Average Speed = (Speed downstream × Speed downstream)/Speed in still water=(15×9)/12=(15×3)/4=45/4=11 1/4 km/hr

6.B

By the rule of alligation, we have

% Concentration of water

in pure water : 100 % Concentration of water in the given mixture : 10

Mean % concentration

20

20 – 10 = 10 100 – 20 = 80

Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8

Here Quantity of the mixture = 16 litres

Quantity of water : 16 = 1 : 8

Quantity of water = 16×1/8=2 litre

7.A

CP=20×8.5+35×8.75=170+306.25=476.25

Profit=40%

SP=((100+Profit%)/100)×CP=((100+40)/100)×476.25=140/100×476.25=140/4×19.05=35×19.05

Total quantity = 20 + 35 = 55 Kg

SP per Kg=(35×19.05)/55=(7×19.05)/11≈(7×19)/11≈133/11≈12

8.D

If an amount P1 is lent out at simple interest of R1% per annum and another amount P2 at simple interest rate of R2% per annum, then the rate of interest for the whole sum can be given by

R=(P1R1+P2R2)/P1+P2

P1 = Rs. 12000, R1 = 10%

P2 = ?, R2 = 20%

R = 14%

14=(12000×10+P2×20)/12000+P2

12000×14+14P2=120000+20P2

6P2=14×12000−120000=48000

P2=8000

Total amount invested = (P1 + P2) = (12000 + 8000) = Rs. 20000

9.A

Let the sum of Rs.725 is lent out at rate R% for 1 year

Then, at the end of 8 months, ad additional sum of 362.50 more is lent out at rate 2R% for remaining 4 months(1/3 year)

Total Simple Interest = 33.50

[(725×R×1)/100]+[(362.50×2R×1/3)/100]=33.50

(725×R×1)/100+(362.50×2R)/300=33.50

R=502.5/145=3.46%

10.B

Let sum = x

Time = 20 years

Simple Interest = x

R = (100×SI)/PT=(100×x)/(x×20)=100/20=5%

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

### Set C

1.There is 80% increase in an amount in 8 years at simple interest. What will be the compound interest of Rs. 14,000 after 3 years at the same rate?

A. Rs.3794

B. Rs.3714

C. Rs.4612

D. Rs.4634

2.The difference between simple interest and compound on Rs. 2400 for one year at 10% per annum reckoned half-yearly is:

A. Rs. 4

B. Rs. 6

C. Rs. 3

D. Rs. 2

3. A tree increases annually by 1⁄5 th of its height. If its height today is 50 cm, what will be the height after 2 years?

A. 64 cm

B. 72 cm

C. 66 cm

D. 84 cm

4.The sum of ages of 5 children born at the intervals of 3 years each is 50 years. Find out the age of the youngest child?

A. 6 years

B. 5 years

C. 4 years

D. 3 years

5.Ayisha’s age is 1/6th of her father’s age. Ayisha ‘s father’s age will be twice the age of Shankar’s age after 10 years. If Shankar’s eight birthdays was celebrated two years before, then what is Ayisha ‘s present age.

A. 10 years

B. 12 years

C. 8 years

D. 5 years

6.My brother is 3 years elder to me. My father was 28 years of age when my sister was born while my mother was 26 years of age when I was born. If my sister was 4 years of age when my brother was born, then what was the age my father when my brother was born?

A. 35 years

B. 34 years

C. 33 years

D. 32 years

7. Kamal was 4 times as old as his son 8 years ago. After 8 years, Kamal will be twice as old as his son. Find out the present age of Kamal.

A. 40 years

B. 38 years

C. 42 years

D. 36 years

8.A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet?

A. 126 sq. ft.

B. 64 sq. ft.

C. 100 sq. ft.

D. 102 sq. ft.

9.The length of a room is 5.5 m and width is 3.75 m. What is the cost of paying the floor by slabs at the rate of Rs. 800 per sq. metre.

A. Rs.12000

B. Rs.19500

C. Rs.18000

D. Rs.16500.

10.The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. What is the length of the rectangle?

A. 18 cm

B. 16 cm

C. 40 cm

D. 20 cm

Answers:

1.D

Let P = Rs.100

Simple Interest = Rs. 80 ( ∵ 80% increase is due to the simple interest)

Rate of interest=(100×SI)/PT=(100×80)/100×8=10% per annum

Now let’s find out the compound interest of Rs. 14,000 after 3 years at 10%

P = Rs.14000

T = 3 years

R = 10%

Amount after 3 years =P(1+R/100)^T=14000(1+10/100)^3=14000(110/100)^3=14000(11/10)^3=14×11^3=18634

Compound Interest = Rs.18634 – Rs.14000 = Rs.4634

2.B

3.B

This problem is similar to the problems we saw in compound interest.

We can use the formulas of compound interest here as well.

Rate of increase = 1/5×100=20%

Height after 2 years = P(1+R/100)^T=50(1+20/100)^2=50(1+1/5)^2=50(6/5)^2=(50×6×6)/(5×5)=2×6×6=72 cm

4.C

5.D

Consider Ayisha’s present age = x

Then her father’s age = 6x

Given that Ayisha ‘s father’s age will be twice the age of Shankar’s age after 10 years

Shankar’s age after 10 years = ½(6x + 10) = 3x + 5

Also given that Shankar’s eight birthdays was celebrated two years before

Shankar’s age after 10 years = 8 + 12 = 20

3x + 5 = 20

x = 15/3 = 5

Ayisha ‘s present age = 5 years

6.D

Let my age = x

Then

My brother’s age = x + 3

My mother’s age = x + 26

My sister’s age = (x + 3) + 4 = x + 7

My Father’s age = (x + 7) + 28 = x + 35

age my father when my brother was born = x + 35 – (x + 3) = 32

7.A

Let the age of the son before 8 years = x.

Then age of Kamal before 8 years ago = 4x

After 8 years, Kamal will be twice as old as his son

4x + 16 = 2(x + 16)

x = 8

Present age of Kamal = 4x + 8 = 4×8 +8 = 40

8.A

Let l = 9 ft.

Then l + 2b = 37

2b = 37 – l = 37 – 9 = 28

b = 28/2 = 14 ft.

Area = lb = 9 × 14 = 126 sq. ft.

9.D

Area = 5.5 × 3.75 sq. metre.

Cost for 1 sq. metre. = Rs. 800

Hence total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = Rs. 16500

10.C

Let breadth = x cm

Then length = 2x cm

Area = lb = x × 2x = 2×2

New length = (2x – 5)

New breadth = (x + 5)

New Area = lb = (2x – 5)(x + 5)

But given that new area = initial area + 75 sq.cm.

(2x – 5)(x + 5) = 2x^2 + 75

2x^2 + 10x – 5x – 25 = 2x^2 + 75

5x – 25 = 75

5x = 75 + 25 = 100

x = 100/5 = 20 cm

Length = 2x = 2 × 20 = 40cm