**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*