**Q1. Rohan, Sohan and Mohan are partners in a business. Rohan, whose money has been used for 4 months, claims 1/8 of the profit. Sohan, whose money has been used for 6 months, claims 1/3 of the profit. Mohan had invested Rs. 1560 for 8 months. How much money did Rohan contribute?**

(a) Rs. 720

(b) Rs. 1220

(c) Rs. 840

(d) Rs. 1620

**Q2. Anup and Deepak started a joint firm. Anup’s investment was thrice the investment of Deepak and the period of his investment was two times the period of investment of Deepak. Deepak got Rs. 7000 as profit for his investment. Their total profit if the distribution of profit is directly proportional to the period and amount, is**

(a) 42000

(b) 56000

(c) 54000

(d) 49000

**Q3. A takes twice as much time as B and C takes thrice as much time as B to finish a work. Working together, they can finish the work in 12 days. Find the number of days needed for A to do the work alone.**

(a) 20

(b) 22

(c) 33

(d) 44

**Q4. A can give B a 200 m startup and C a 300 m startup in a race of 1 km. How many metres startup can B gives to C in a 1 km race**

(a) 130 m.

(b) 125 m.

(c) 100 m.

(d) 80 m.

**Q5. In a one km race A gives B a start of 100 m and in a one km race B gives a start of 80 m to C. In a 1 km race who will win and by how much distance from the worst performer between two losers?**

(a) A, 172 m.

(b) A, 182 m.

(c) A, 144 m.

(d) A, 192m.

**Q6. The perimeters of two squares are 24 cm and 32 cm. The perimeter (in cm) of a third square which is equal in area to the sum of the areas of the first and second square is:**

(a) 45

(b) 40

(c) 32

(d) 48

**Q7. The length of a rectangular garden is 12 m and its breadth is 5 m. Find the length of the diagonal of a square garden having the same area as that of the rectangular garden.**

(a) 2√30 m

(b) √13 m

(c) 13 m

(d) 8√15 m

**Q8. The base of a triangle is 15 cm and height is 12 cm. The height of another triangle of double the area having base 20 cm is:**

(a) 9 cm

(b) 18 cm

(c) 8 cm

(d) 12.5 cm

**Q9. A circular wire of radius 21 cm is cut and bent in the form of a rectangle whose sides are in the ratio of 6 : 5. Assuming**

**π=22/7,**

**the area enclosed by the rectangle is:**

(a) 540 cm2

(b) 1080 cm2

(c) 2160 cm2

(d) 4320 cm2

**Q10. The area (in m2) of the square which has the same perimeter as of a rectangle whose length is 48 m and is 3 times its breadth, is:**

(a) 1000

(b) 1024

(c) 1600

(d) 1042

**Solution**

S1. Ans.(a)

Sol. Let the profit 24P

Ratio of their share in profit = 24P × 1/8 : 24P × 1/3 : remaining = 3 : 8 : 13

Rohan’s profit : Mohan,s profit

= (Rohan × 4)/(1560 × 8)=3/13

Rohan = (3/13)×(1560 × 8)/4 = Rs. 720

S2. Ans.(d)

Sol. Investment of Anup : Investment of Deepak = 3 : 1

Ratio of their time period = 2 : 1

Ratio of their profit = 3 × 2 : 1 × 1 = 6 : 1

Deepak profit = 7000

Ratio of their profit = 6 : 1 and Deepak profit = 7000, So for finding total profit, we divide 7000 by Deepak’s profit proportion (1) and multiply by total profit proportion (6 + 1) = 7

Their total profit = (7/1) × 7000 = 49000

S3. Ans.(d)

Sol. Let time taken by B = x

Then time taken by A = 2x

and time taken by C = 3x

According to the question,

(1/x)+(1/2x)+(1/3x) = (1/12)

⇒1+(1/2)+(1/3) = (x/12)

⇒(6+3+2)/6 = (x/12)

∴ x = 22

Required number of days = 2x = 2 × 22

= 44 days

S4. Ans.(b)

Sol. A covers = 1000 m

B covers = 800 m

C covers = 700 m

Ratio of speeds of B : C = 800 : 700 = 8 : 7

Since, when B moves 8m, C moves 7 metre.

Therefore, when B moves 1000m, C moves 875 m. (by unitary method).

Thus, B can Give C a start of 1000 – 875 = 125 m.

S5. Ans.(a)

Sol. Ratio of speeds of A : B = 1000 : 900 = 100 : 90

Ratio of speeds of B : C = 1000 : 920 = 100 : 92

Therefore, when A moves 1000 m,

B moves 900 m

When B moves 1000 m, C moves 920 m

∴ B moves 900 m, C moves = 920/1000 × 900

= 828 m.

Since, C moves 8% less than B in the same time.

Thus, C is the worst performer and A will by him by 172 m.

S6. Ans.(b)

Sol. Required perimeter =√(24/4)^2+(32/4)^2 )×4

= 40.

S7. Ans.(a)

Sol. Length of the diagonal = √(12×5)×√2

=2√30 m.

S8. Ans.(b)

Sol. Required height =(2 × 15 × 12)/20

= 18 cm.

S9. Ans.(b)

Sol. Let, the sides of the rectangle be 6K and 5K, respectively.

∴ 2[6K+5K]=2×(22/7)×21=132⇒K=6

∴ Area of the rectangle =30K^2=1080 cm^2

S10. Ans.(b)

Sol. Breadth of the rectangle =48/3=16 m

Perimeter of the rectangle =2(l+b)=2×64=128 m

Side of the square =48/3=32

Area of the square =(32)^2=1024 m^2

Sol. Let the profit 24P

Ratio of their share in profit = 24P × 1/8 : 24P × 1/3 : remaining = 3 : 8 : 13

Rohan’s profit : Mohan,s profit

= (Rohan × 4)/(1560 × 8)=3/13

Rohan = (3/13)×(1560 × 8)/4 = Rs. 720

S2. Ans.(d)

Sol. Investment of Anup : Investment of Deepak = 3 : 1

Ratio of their time period = 2 : 1

Ratio of their profit = 3 × 2 : 1 × 1 = 6 : 1

Deepak profit = 7000

Ratio of their profit = 6 : 1 and Deepak profit = 7000, So for finding total profit, we divide 7000 by Deepak’s profit proportion (1) and multiply by total profit proportion (6 + 1) = 7

Their total profit = (7/1) × 7000 = 49000

S3. Ans.(d)

Sol. Let time taken by B = x

Then time taken by A = 2x

and time taken by C = 3x

According to the question,

(1/x)+(1/2x)+(1/3x) = (1/12)

⇒1+(1/2)+(1/3) = (x/12)

⇒(6+3+2)/6 = (x/12)

∴ x = 22

Required number of days = 2x = 2 × 22

= 44 days

S4. Ans.(b)

Sol. A covers = 1000 m

B covers = 800 m

C covers = 700 m

Ratio of speeds of B : C = 800 : 700 = 8 : 7

Since, when B moves 8m, C moves 7 metre.

Therefore, when B moves 1000m, C moves 875 m. (by unitary method).

Thus, B can Give C a start of 1000 – 875 = 125 m.

S5. Ans.(a)

Sol. Ratio of speeds of A : B = 1000 : 900 = 100 : 90

Ratio of speeds of B : C = 1000 : 920 = 100 : 92

Therefore, when A moves 1000 m,

B moves 900 m

When B moves 1000 m, C moves 920 m

∴ B moves 900 m, C moves = 920/1000 × 900

= 828 m.

Since, C moves 8% less than B in the same time.

Thus, C is the worst performer and A will by him by 172 m.

S6. Ans.(b)

Sol. Required perimeter =√(24/4)^2+(32/4)^2 )×4

= 40.

S7. Ans.(a)

Sol. Length of the diagonal = √(12×5)×√2

=2√30 m.

S8. Ans.(b)

Sol. Required height =(2 × 15 × 12)/20

= 18 cm.

S9. Ans.(b)

Sol. Let, the sides of the rectangle be 6K and 5K, respectively.

∴ 2[6K+5K]=2×(22/7)×21=132⇒K=6

∴ Area of the rectangle =30K^2=1080 cm^2

S10. Ans.(b)

Sol. Breadth of the rectangle =48/3=16 m

Perimeter of the rectangle =2(l+b)=2×64=128 m

Side of the square =48/3=32

Area of the square =(32)^2=1024 m^2