Basic Geometry(Exams > Cat > Quantitaitve Aptitude ) Questions and answers for exam Preparation

Exams > Cat > Quantitaitve Aptitude

BASIC GEOMETRY MCQs

Total Questions : 93 | Page 5 of 10 pages

Question 41. If the inner rectangle is 8cm x 6 cm. Find the area of the shaded region.

44 cm2
34.25 cm2
32.50 cm2
None of these

Discuss Question

Answer: Option D. -> None of these
:
D
Diagonal of rectangle:

\sqrt{(8^{2} + 6^{2})}

, =10.

Half of diagonal = radius of circle = 5.
Area of shaded region:

π

^{2}

- (8x6) = 30.57

Question 42. There is an equilateral triangle of side ‘a’ units. Now we join any two sides of the triangle to form a cone.What is the slant height of the cone formed?

a2π
a
a2
2πa

Discuss Question

Answer: Option B. -> a
:
B
The cone will have slant height same as the triangle side.

Question 43. A rectangular field is of dimension 15.4m x12.1 m. A circular well of 0.7 m radius and 3 m depth is dug in the field. The mud, dug out from the well, is spread in the field. By how much would the level of the field rise?

1 cm
2.5 cm
3.5 cm
4 cm

Discuss Question

Answer: Option B. -> 2.5 cm
:
B
Solution: - Volume of earth taken out:

π

^{2}

h. =

π

x (0.7)

^{2}

x 3 = V.
Area of field without circle: (15.4x12.1)-(

π

(0.7

^{2}

)) = A (say).
Axh=V
h=2.5 cm

Question 44. If the right circular cone is cut into three solids of volumes

V_{1}, V_{2} a n d V_{3}

by two cuts which are parallel to the base and trisects the altitude, then

V_{1} : V_{2} : V_{3}

is:

1:2:3
1:4:6
1:6:9
None of these

Discuss Question

Answer: Option D. -> None of these
:
D
Solution: -

If The Right Circular Cone Is Cut Into Three Solids Of Vol...

The resultant figure is made of three similar triangles. The height and radius will be in ratio

\frac{1}{3} : \frac{2}{3} : 1

= 1:2:3.
r

_{1}

= 1, h

_{1}

= 1
r

_{2}

= 2, h

_{2}

= 2
r

_{3}

= 3, h

_{3}

= 3
Volume will be in the ratio of r\(^2\)h for the three circular cones.
Required Volumes are
V

_{1}

= r

_{1}^{2}

x h

_{1}

= 1
V

_{2}

= r

_{2}^{2}

x h

_{2}

- V

_{1}

= 8 - 1 = 7
V

_{3}

= r

_{3}^{2}

x h

_{3}

- (V

_{1}

+ V

_{2}

) = 27 - (7 + 1) = 19
Volumes will be in given ratio= 1:7:19.Option (d).

Question 45. The ratio between the curved surface area and the total surface area of right cylinder is 2:3 and the total surface area is 924 cm

^{2}

. What is the volume of the cylinder?

2156 cc
2183 cc
2492 cc
None of these

Discuss Question

Answer: Option A. -> 2156 cc
:
A
Solution:- Total surface Area:

\frac{2 π r^{2} + 2 π r h}{2 π r h} = \frac{3}{2} \Rightarrow 1 + \frac{r}{h} = \frac{3}{2} \Rightarrow \frac{r}{h} = \frac{1}{2} \Rightarrow h = 2 r

TSA:

(2 π r^{2} + 2 π r h)

= 924

\Rightarrow (2 π r^{2} + 2 π r \times 2 r)

= 924

\Rightarrow

r = 7 and h = 14
Volume =

π r^{2} h = \frac{22}{7} \times 7 \times 7 \times 14

= 2156
Option(a).

Question 46. In the following figure, the radii of the largest and smallest circles are 20 and 5. Find the radius of the intermediate circle.

7
10
12.5
15

Discuss Question

Answer: Option B. -> 10
:
B
Solution: The fig can be considered as:

Since the triangles are similar:

\frac{r_{1} + r_{2}}{r_{2} + r_{3}} = \frac{r_{1} - r_{2}}{r_{2} - r_{3}}

Substituting r

_{1}

= 20 and r

_{3}

= 5 , r

_{2}

= 10. Option (b).

Question 47. The top of a conical container has a circumference of 308 m. Water flows in at a rate of 12 m

^{3}

every 2 secs. When will the container be half filled, if its depth is 12m.

42 min
68 min
54 min
82 min

Discuss Question

Answer: Option A. -> 42 min
:
A
Ans:a 2

π

r=3082

\times

(\frac{22}{7}) \times r

= 308Find half volume of conical container.
Time reqd:

\frac{v o l o f \frac{1}{2} c o n t a i n e r}{f l o w r a t e}

\frac{15078.28}{6}

= 2520;

\frac{2520}{60}

=42min

Question 48. A well is dug 20 ft deep and the mud which came out is used to build a wall of width 1 ft around the well on the ground. If the height of the wall around the well is 5 ft, then what is the radius of the well?

√5
1
14
(√5+1)4

Discuss Question

Answer: Option D. -> (√5+1)4
:
D
Solution: - Let radius = r feet.
Volume of mud from well: πr

^{2}

(20) cubic feet.
Volume of wall around well: 5π((r+1)

^{2}

-r

^{2}

).
πr

^{2}

(20) =5π((r+1)

^{2}

-r

^{2}

) => 4r

^{2}

- 2r - 1=0

\Rightarrow

r =

\frac{(\sqrt{5} + 1)}{4}

Question 49. The radius of an iron rod is decreased to one fourth of its original radius. If its volume remains constant, then the length will become.