29-07-2021 Answer

 1. Find the area of a regular hexagon inscribed in a circle of radius 18 cms.

Answer: 841.75 sq.cms approx

Sollution: In a regular hexagon the radius of the circle is equal to it's side as the center of the circle forms an equilateral triangle with the end points of a side.

Area of the desired regular hexagon = 6 * Area of equilateral triangles formed

= 6 * 1/4 * SQRT(3) * side^2

= 6/4 * SQRT(3) * 18 * 18 = 841.75 sq.cms approx

2. A rectangle of length 11 meters and width 8 meters is rolled about its length. The volume of the resulting cylinder is

Answer: 77 cu.m

Solution: L = 11 m

Height of the cylinder = h = 8m

Let the radius of cylinder be r.

2*pi*r = L, so r = L/(2*pi)

Volume = pi* r^2 * h

= (11 * 11 * 8) /(4 * 22/7) = 77 cu.m

3. The radii of top and bottom of a bucket are 25cms and 20cms respectively. If the volume of the bucket is 67100 cubic cms, find the height of the bucket.

Answer: 42 cms

Solution: The volume of a tapered bucket is given by the formula,

v= pi/3 * height * (rb*rb + rt*rb + rt*rt) where rb is bottom radius and rt is top radius.

Hence volume = 22/7 * 1/3 * h * (400+500+625) =  67100 cm.cube,

Solving, h = 42 cms.

4. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

Answer: 2:1

Solution: Volume of cone = 1/3 * pi * r^2 * h

Volume of hemisphere = 2/3 * pi * r ^ 3

As both are equal  h = 2r,

As the radius of the hemisphere is also it's height, the ratio is h:r = 2:1

5. 64 small solid spheres are made by melting a solid sphere of radius 8 cm. The ratio of the surface areas of this single large sphere to a smaller sphere will be

Answer: 16:1

Solution: Surface area of larger sphere = 4 * 22/7 * 8 * 8

Let the radius of smaller sphere be r.

Volume of one larger sphere = 4/3 * 22/7 *  8 * 8 * 8 * 1/64 = 4/3 * 22/7 * r^3.

Hence r=2cms.

Surface area of smaller spehere = 4 * 22/7 * 2 * 2

Hence required ration = 64:4 = 16:1

6. A circular swimming pool is surrounded by a concrete walk which is 4 feet wide. If the area of the walk is 11/25 th of the area of the pool, then the radius of the pool is

Answer: 20 feet

Solution: Let radius of pool be r.

Area of pool = pi*r*r

Area walk = pi * (r+4) * (r+4) - pi*r*r

It is given,

pi * (r+4) * (r+4) - pi*r*r =  11/25 * pi*r*r,

(r+4)^2 = r^2 * (11/25  + 1),

Taking square root,

(r+4) = r * 6/5,

Solving, r=20

7. Four circles each of radius 7cm are placed in a manner so that each circle touches two other adjacent circles. The area enclosed by the four circles is

Answer: 42 sq.cm

Solution: Assume one of this circle to be an incircle of the square.

The area of the square = 14*14 = 196 sq.cm

The area of the circle = 22/7 * 7 * 7 = 154 sq.cm

Difference in area = 196 -154 = 42 sq.cm

Now we are interested in finding the area of the space in the center enclosed space.

Hence the area enclosed by one circle in the center = 42/4 sq.cm

Thus area enclosed by four circles = 4 * 42/4 = 42 sq.cm

8. If the area of incircle of a square is 9*22/7 sq.cm, what is the area of the square?

Answer: 36 sq.cms

Solution: The incircle radius = Half of side of square.

As the area of circle = 22/7 * r^2,

22/7 * r^2 = 9*22/7 sq.cm,

r = 3cms.

Hence side of the square = s = 2r = 6cms.

Area of the square = s*s = 36 sq.cms

9. Find the area of a circle inscribed in a square of area 441 sq.cm.

Answer: 346.5 sq.cm

Solution: As area of square = 441 sq.cms, side of square = SQRT(441) = 21 cms.

Side of the square = Diameter of the circle. Hence radius = 21/2 cms.

Area of circle = 22/7 * 21/2 * 21/2 = 11 * 3 * 21/2 = 346.5 sq.cm

10. Two sides of a triangle are 8 cm and 24 cm. The third side is S. Which of the following must be true?

Answer: 16 < x < 32

Solution: Since the sum of any two sides is greater than the third side,

(8 + 24) > x Or x < 32

Also (x + 8) > 24 or x >16

From the above two conditions 16 < x < 32.

11. A square of side 8 cm is inscribed in a circle. Find the ratio of the area of square to circle.

Answer: 14:22

Solution: Area of square = 8^2 = 64 sq cm.

Diameter of the circle = diagonal of the square = 8√2 cm.

Area of the circle = π * [(8√2)/2]^2 = 32*π.

Required area = 64 : 32π = 2:22/7 = 14:22

12. Arun was all bent on building a new house. He carefully got the blue print of his house designed by his friend Ashwin, a civil engineer. He wanted to build a room of dimension 27 by 48 ft and lay tiles in this room. Each tile was of dimension 2 by 3 ft. How many tiles should Arun buy?

Answer: 216

Solution: Number of tiles required = Area of floor/area of one tile = 27*48/2*3

=27 * 8 = 216

13. Two roads which are of same width and of length 200 metres and 500 metres respectively intersect perpendicularly. If the overall area covered by the roads is 60000 sq.m, what is the width of the roads?

Answer: 100m

Solution: Let the width of the roads be b.

200b + 500b - b*b = 60000, (As the roads intersect, the common area should be subtracted)

b*b -700b + 60000 =0,

(b-600)(b-100)=0,

So b=600m or b=100m. From the given options 100m is the answer.

14. 20m * 34m * 20m is the dimension(breadth,length and height) of a box. How many cubes of edge 5m can be inserted into it?

Answer: None of these

Solution: Ans: None of the above (as answer is 96)

Along breadth 20m, num of cubes = 20m/5m = 4,

along length = 34m/5m = 6,

along height = 20m/5m =4.

So total num of cubes thatcan be inserted = 4*6*4 = 96

Note: You cannot just divide volume of box by volume of cube.

15. The number of revolutions of a wheel of diameter 49cms making in covering a distance of 177.10 meters is

Answer: 115

Solution: Permiter of wheel = 2*22/7 * 49/2 = 22*7 cms

Number of revolutions = 17710/(22*7) = 115

16. When a wire is given the shape of a square it's area is 121 sq.cm. If the same wire is given the shape of a circle, it's area will be

Answer: 154 sq.cm

Solution: As a square, area = 121 sq.cm. Hence side = 11cms.

Perimeter of square = 4*side = 44 cms = Perimeter of circle = 2*22/7 * r

Hence r = 7 cms.

Hence area of circle = 22/7  * 7 * 7 = 22*7 = 154 sq.cm

17. Semi circular lawns are attached to the edges of a rectangular plot measuring 42 feet * 28 feet. The total area of lawn is

Answer: 2002 sq.ft

Solution: The sides becomes the diameter.

Hence area of two semi circular lawns along length = 2 * 1/2 * pi * 21 * 21

area of two semi circular lawns along breadth = 2 * 1/2 * pi * 14 * 14

Total area of lawn = 22/7 *  (21*21 + 14*14) = 2002 sq.feet

18. Two circles of 6 cm each intersect each other as shown in the figure. The area of the overlapping region approximately is

Answer: 6.54 sq.cm

Solution: Area of the sector = 60/360  *  pi * r^2 =  1/6 * pi * 36

Area of the equilateral triangle = 1/4 * sqrt(3) * 36

Area of shaded region = Twice the area of (Sector area - Equilateral triangle)

 = 2 * 36 ( 1/6 * pi - 1/4 * sqrt(3))

 As pi = 22/7 and sqrt(3) is approximately 1.732,

 The area of the shaded region = 6.54 sq.cm approximately

19. If the diagonal and the area of a rectangle are 25 m and 168 m^2, what is the length of the rectangle?

Answer: 24m

Solution: Length = l and breadth = b

Area = lb = 168, b=168/l - eqn1

Diagonal = sqrt(l^2 + b^2) = 25

l^2 = 625 - b^2 = 625  - 168^2/l^2

l^2 = (25+168/l)(25-168/l)

Solving l=24

20. A regular solid has 30 edges and 12 faces. How many vertices does it have?

Answer: None of these

Solution: Ans: None of the above (As answer is 20)

Num of faces + Num of vertices = Num of edges + 2.

Num of vertices = 30+2-12 = 20

21. A regular solid has 12 edges and 8 vertices. How many faces does it have?

Answer: 6

Solution: Num of faces + Num of vertices = Num of edges + 2.

Num of faces = 12+2-8 = 6

22. If a new square is formed with the diagonal of a square as an edge, what is the ratio between the area of the two squares?

Answer: 1:2

Solution: Let side of square be a. 

Diagonal = a*SQRT(2).

Ratio of area = a^2:a*a^2 = 1:2

23. Two sides of a triangle are 12 cm and 9 cm. The third side is x. Which of the following must be true?

Answer: 3 < x < 21

Solution: Since the sum of any two sides is greater than the third side,

(12 + 9) > x Or x < 21

Also (x + 9) > 12 or x >3

From the above two conditions 3 < x < 21.

24. A semi circular window has a diameter of 63 cms. What is it's perimeter?

Answer: 162 cms

Solution: Perimeter = Diameter + 22/7 * radius

= 63 + 22/7 * 63/2 = 63 + 11*9 = 162 cms.

25. Raj drives along a rectangular park at 24kmph and completes one round in 4 minutes. If the ratio of length and breadth of the park is 3:2, what is the dimension of the park?

Answer: 480m * 320m

Solution: Speed = 24kmph = 24 * 5/18 meters/sec = 20/3 meters per sec = 400 meters/minute.

In 4 minutes he covers 4*400 = 1600 meters which is nothing but the perimeter of the park.

So 1600 = 2(length+breadth).

As length = 3x and breadth = 2x,

1600 = 2(3x+2x), solving x=160m. Hence the dimension is 480m*320m