19 + Interview Questions in GENERAL QA IN AREA Page 1 InterviewSolution

1.	Two concentric circles form a ring. The inner and outer circumferences of ring are (352/7) m and (518/7) m respectively. Find the width of the ring.
Answer» Let the inner and outer radii be r and R metres. Then 2πr = (352/7) r =((352/7) X (7/22) X (1/2)) =8 m. 2πR=(528/7) R=((528/7) x (7/22) x (1/2)) = 12 m. Width of the ring = (R - r) = (12 - 8) m = 4 m

1.

Two concentric circles form a ring. The inner and outer circumferences of ring are (352/7) m and (518/7) m respectively. Find the width of the ring.

Answer»

Let the inner and outer radii be r and R metres.

Then 2πr = (352/7)

r =((352/7) X (7/22) X (1/2)) =8 m.

2πR=(528/7)

R=((528/7) x (7/22) x (1/2)) = 12 m.

Width of the ring = (R - r) = (12 - 8) m = 4 m

Discussion

2.	A wheel makes 1000 revolutions in covering a distance of 88 km. Find the radius of the wheel.
Answer» Distance covered in one revolution =((88 x 1000)/1000) = 88 m. 2πR = 88 2 x (22/7) x R = 88 R = 88 x (7/44) = 14 m.

2.

A wheel makes 1000 revolutions in covering a distance of 88 km. Find the radius of the wheel.

Answer»

Distance covered in one revolution =((88 x 1000)/1000) = 88 m.

2πR = 88

2 x (22/7) x R = 88

R = 88 x (7/44) = 14 m.

Discussion

3.	A room 5 m 55 cm long and 3 m 74 cm broad is to be paved with square tiles. Find the least number of square tiles required to cover the floor.
Answer» Area of the room = (544 x 374) cm² . Size of largest square tile = H.C.F. of 544 cm and 374 cm = 34 cm. Area of 1 tile = (34 x 34) cm² . Number of tiles required =(544374)/(3434)=176

3.

A room 5 m 55 cm long and 3 m 74 cm broad is to be paved with square tiles. Find the least number of square tiles required to cover the floor.

Answer»

Area of the room = (544 x 374) cm² .

Size of largest square tile = H.C.F. of 544 cm and 374 cm = 34 cm.

Area of 1 tile = (34 x 34) cm² .

Number of tiles required =(544*374)/(34*34)=176

Discussion

4.	Find the area of a square, one of whose diagonals is 3.8 m long.
Answer» Area of the square = (1/2)* (diagonal)² = [(1/2)3.83.8 ]m² = 7.22 m² .

Discussion

5.	If each side of a square is increased by 25%, find the percentage change in its area.
Answer» Let each side of the square be a. Then, area = a². New side =(125a/100) =(5a/4). New area = (5a/4)² =(25a² )/16. Increase in area = ((25 a² )/16)-a² =(9a²)/16. Increase% = [((9a² )/16)(1/a² )100] % = 56.25%.

5.

If each side of a square is increased by 25%, find the percentage change in its area.

Answer»

Let each side of the square be a. Then, area = a².

New side =(125a/100) =(5a/4). New area = (5a/4)² =(25a² )/16.

Increase in area = ((25 a² )/16)-a² =(9a²)/16.

Increase% = [((9a² )/16)*(1/a² )*100] % = 56.25%.

Discussion

6.	The diagonals of two squares are in the ratio of 2 : 5. Find the ratio of their areas.
Answer» Let the diagonals of the squares be 2x and 5x respectively. Ratio of their areas = (1/2)(2x)² :(1/2)(5x)² = 4x² : 25x² = 4 : 25.

6.

The diagonals of two squares are in the ratio of 2 : 5. Find the ratio of their areas.

Answer»

Let the diagonals of the squares be 2x and 5x respectively.

Ratio of their areas = (1/2)*(2x)² :(1/2)*(5x)² = 4x² : 25x² = 4 : 25.

Discussion

7.	The difference between two parallel sides of a trapezium is 4 cm. perpendicular distance between them is 19 cm. If the area of the trapezium is 475 find the lengths of the parallel sides.
Answer» Let the two parallel sides of the trapezium be a cm and b cm. Then, a - b = 4 And, (1/2) x (a + b) x 19 = 475 (a + b) =((475 x 2)/19) a + b = 50 Solving (i) and (ii), we get: a = 27, b = 23. So, the two parallel sides are 27 cm and 23 cm.

7.

The difference between two parallel sides of a trapezium is 4 cm. perpendicular distance between them is 19 cm. If the area of the trapezium is 475 find the lengths of the parallel sides.

Answer»

Let the two parallel sides of the trapezium be a cm and b cm.

Then, a - b = 4 And, (1/2) x (a + b) x 19 = 475

(a + b) =((475 x 2)/19)

a + b = 50

Solving (i) and (ii), we get: a = 27, b = 23.

So, the two parallel sides are 27 cm and 23 cm.

Discussion

8.	Find the area of a rhombus one side of which measures 20 cm and 01 diagonal 24 cm.
Answer» Let other diagonal = 2x cm. Since diagonals of a rhombus bisect each other at right angles, we have: (20)² = (12)² + (x)² = (20)² – (12)²= 256= 16 cm. So, other diagonal = 32 cm. Area of rhombus = (1/2) x (Product of diagonals) = ((1/2)x 24 x 32) cm² = 384 cm²

8.

Find the area of a rhombus one side of which measures 20 cm and 01 diagonal 24 cm.

Answer»

Let other diagonal = 2x cm.

Since diagonals of a rhombus bisect each other at right angles, we have:

(20)² = (12)² + (x)² = (20)² – (12)²= 256= 16 cm.

So, other diagonal = 32 cm.

Area of rhombus = (1/2) x (Product of diagonals) = ((1/2)x 24 x 32) cm² = 384 cm²

Discussion

9.	Find the area of a triangle whose sides measure 13 cm, 14 cm and 15 cm.
Answer» Let a = 13, b = 14 and c = 15. Then, S = (1/2)(a + b + c) = 21. (s- a) = 8, (s - b) = 7 and (s - c) = 6. Area = (s(s- a) (s - b)(s - c))^(1/2) = (21 8 7*6)^(1/2) = 84 cm² .

9.

Find the area of a triangle whose sides measure 13 cm, 14 cm and 15 cm.

Answer»

Let a = 13, b = 14 and c = 15. Then, S = (1/2)(a + b + c) = 21.

(s- a) = 8, (s - b) = 7 and (s - c) = 6.

Area = (s(s- a) (s - b)(s - c))^(1/2)

= (21 *8 * 7*6)^(1/2)

= 84 cm² .

Discussion

10.	The base of a parallelogram is twice its height. If the area of the parallelogram is 72 sq. cm, find its height.
Answer» Let the height of the parallelogram be x. cm. Then, base = (2x) cm. 2x X x =72 2x² = 72 X²=36 x = 6 Hence, height of the parallelogram = 6 cm.

10.

The base of a parallelogram is twice its height. If the area of the parallelogram is 72 sq. cm, find its height.

Answer»

Let the height of the parallelogram be x. cm. Then, base = (2x) cm.

2x X x =72

2x² = 72

X²=36

x = 6

Hence, height of the parallelogram = 6 cm.

Discussion

11.	If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.
Answer» Let x and y be the length and breadth of the rectangle respectively. Then, x - 4 = y + 3 or x - y = 7 ----(i) Area of the rectangle =xy; Area of the square = (x - 4) (y + 3) (x - 4) (y + 3) =xy <=> 3x - 4y = 12 ----(ii) Solving (i) and (ii), we get x = 16 and y = 9. Perimeter of the rectangle = 2 (x + y) = [2 (16 + 9)] cm = 50 cm.

11.

If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.

Answer»

Let x and y be the length and breadth of the rectangle respectively.

Then, x - 4 = y + 3 or x - y = 7 ----(i)

Area of the rectangle =xy; Area of the square = (x - 4) (y + 3)

(x - 4) (y + 3) =xy <=> 3x - 4y = 12 ----(ii)

Solving (i) and (ii), we get x = 16 and y = 9.

Perimeter of the rectangle = 2 (x + y) = [2 (16 + 9)] cm = 50 cm.

Discussion

12.	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. Find the length of the rectangle.
Answer» Let breadth = x. Then, length = 2x. Then, (2x - 5) (x + 5) - 2x * x = 75 5x - 25 = 75 x = 20. :. Length of the rectangle = 20 cm

12.

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. Find the length of the rectangle.

Answer»

Let breadth = x. Then, length = 2x. Then,

(2x - 5) (x + 5) - 2x * x = 75

5x - 25 = 75

x = 20.

:. Length of the rectangle = 20 cm

Discussion

13.	In two triangles, the ratio of the areas is 4 : 3 and the ratio of their heights is 3 : 4. Find the ratio of their bases.
Answer» Let the bases of the two triangles be x and y and their heights be 3h and 4h respectively. Then, ((1/2) X x X 3h)/(1/2) X y X 4h) =4/3 x/y =(4/3 X 4/3)=16/9 Required ratio = 16 : 9.

13.

In two triangles, the ratio of the areas is 4 : 3 and the ratio of their heights is 3 : 4. Find the ratio of their bases.

Answer»

Let the bases of the two triangles be x and y and their heights be 3h and 4h respectively.

Then, ((1/2) X x X 3h)/(1/2) X y X 4h) =4/3

x/y =(4/3 X 4/3)=16/9

Required ratio = 16 : 9.

Discussion

14.	Find the area of a right-angled triangle whose base is 12 cm and hypotenuse is 13 cm.
Answer» Height of the triangle = [(13)² - (12)² ]^(1/2) cm = (25)^(1/2) cm = 5 cm. Its area = (1/2)* Base * Height = ((1/2)12 5) cm² = 30 cm².

14.

Find the area of a right-angled triangle whose base is 12 cm and hypotenuse is 13 cm.

Answer»

Height of the triangle = [(13)² - (12)² ]^(1/2) cm = (25)^(1/2) cm = 5 cm.

Its area = (1/2)* Base * Height = ((1/2)*12 * 5) cm² = 30 cm².

Discussion

15.	The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.
Answer» Area of the field = Total cost/rate = (333.18/25.6)hectares = 13.5 hectares (13.5 x 10000) m² = 135000 m² . Let altitude = x metres and base = 3x metres. Then, (1/2)* 3x * x = 135000 <=>x² = 90000 <=>x = 300. Base = 900 m and Altitude = 300 m.

15.

The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.

Answer»

Area of the field = Total cost/rate = (333.18/25.6)hectares = 13.5 hectares

(13.5 x 10000) m² = 135000 m² .

Let altitude = x metres and base = 3x metres.

Then, (1/2)* 3x * x = 135000 <=>x² = 90000 <=>x = 300.

Base = 900 m and Altitude = 300 m.

Discussion

16.	A rectangular grassy plot 110 m. by 65 m has a gravel path 2.5 m wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq. metre.
Answer» Area of the plot = (110 x 65) m² = 7150 m² Area of the plot excluding the path = [(110 - 5) * (65 - 5)] m² = 6300 m² . Area of the path = (7150 - 6300) m² = 850 m² . Cost of gravelling the path = Rs.850 * (80/100)= Rs. 680

16.

A rectangular grassy plot 110 m. by 65 m has a gravel path 2.5 m wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq. metre.

Answer»

Area of the plot = (110 x 65) m² = 7150 m²

Area of the plot excluding the path = [(110 - 5) * (65 - 5)] m² = 6300 m² .

Area of the path = (7150 - 6300) m² = 850 m² .

Cost of gravelling the path = Rs.850 * (80/100)= Rs. 680

Discussion

17.	A lawn is in the form of a rectangle having its sides in the ratio 2: 3. The area of the lawn is (1/6) hectares. Find the length and breadth of the lawn.
Answer» Let length = 2x metres and breadth = 3x metre. Now, area = (1/6 )x 1000 m² = 5000/3m² So, 2x * 3x = 5000/3 x² = 2500/9 x = 50/3 therefore Length = 2x = (100/3) m = 33(1/3) m and Breadth = 3x = 3(50/3) m = 50m.

17.

A lawn is in the form of a rectangle having its sides in the ratio 2: 3. The area of the lawn is (1/6) hectares. Find the length and breadth of the lawn.

Answer»

Let length = 2x metres and breadth = 3x metre.

Now, area = (1/6 )x 1000 m² = 5000/3m²

So, 2x * 3x = 5000/3

x² = 2500/9

x = 50/3

therefore Length = 2x = (100/3) m = 33(1/3) m and Breadth = 3x = 3(50/3) m = 50m.

Discussion

18.	Find the cost of carpeting a room 13 m long and 9 m broad with a carpet 75 cm wide at the rate of Rs. 12.40 per square metre.
Answer» Area of the carpet = Area of the room = (13 * 9) m² = 117 m² . Length of the carpet = (area/width) = 117 (4/3) m = 156 m. Therefore Cost of carpeting = Rs. (156 12.40) = Rs. 1934.40.

18.

Find the cost of carpeting a room 13 m long and 9 m broad with a carpet 75 cm wide at the rate of Rs. 12.40 per square metre.

Answer»

Area of the carpet = Area of the room = (13 * 9) m² = 117 m² .

Length of the carpet = (area/width) = 117 *(4/3) m = 156 m.

Therefore Cost of carpeting = Rs. (156 * 12.40) = Rs. 1934.40.

Discussion

19.	The area of a circular field is 13.86 hectares. Find the cost of fencing it at the rate of Rs. 4.40 per metre.
Answer» Area = (13.86 x 10000) m² = 138600 m² . πR² = 138600 (R)² = (138600 x (7/22)) R = 210 m. Circumference = 2πR = (2 x (22/7) x 210) m = 1320 m. Cost of fencing = Rs. (1320 x 4.40) = Rs. 5808

19.

The area of a circular field is 13.86 hectares. Find the cost of fencing it at the rate of Rs. 4.40 per metre.

Answer»

Area = (13.86 x 10000) m² = 138600 m² .

πR² = 138600

(R)² = (138600 x (7/22))

R = 210 m.

Circumference = 2πR = (2 x (22/7) x 210) m = 1320 m.

Cost of fencing = Rs. (1320 x 4.40) = Rs. 5808

Discussion

Explore topic-wise InterviewSolutions in .

Two concentric circles form a ring. The inner and outer circumferences of ring are (352/7) m and (518/7) m respectively. Find the width of the ring.

A wheel makes 1000 revolutions in covering a distance of 88 km. Find the radius of the wheel.

A room 5 m 55 cm long and 3 m 74 cm broad is to be paved with square tiles. Find the least number of square tiles required to cover the floor.

Find the area of a square, one of whose diagonals is 3.8 m long.

If each side of a square is increased by 25%, find the percentage change in its area.

The diagonals of two squares are in the ratio of 2 : 5. Find the ratio of their areas.

The difference between two parallel sides of a trapezium is 4 cm. perpendicular distance between them is 19 cm. If the area of the trapezium is 475 find the lengths of the parallel sides.

Find the area of a rhombus one side of which measures 20 cm and 01 diagonal 24 cm.

Find the area of a triangle whose sides measure 13 cm, 14 cm and 15 cm.

The base of a parallelogram is twice its height. If the area of the parallelogram is 72 sq. cm, find its height.

If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.

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. Find the length of the rectangle.

In two triangles, the ratio of the areas is 4 : 3 and the ratio of their heights is 3 : 4. Find the ratio of their bases.

Find the area of a right-angled triangle whose base is 12 cm and hypotenuse is 13 cm.

The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.

A rectangular grassy plot 110 m. by 65 m has a gravel path 2.5 m wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq. metre.

A lawn is in the form of a rectangle having its sides in the ratio 2: 3. The area of the lawn is (1/6) hectares. Find the length and breadth of the lawn.

Find the cost of carpeting a room 13 m long and 9 m broad with a carpet 75 cm wide at the rate of Rs. 12.40 per square metre.

The area of a circular field is 13.86 hectares. Find the cost of fencing it at the rate of Rs. 4.40 per metre.