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

Exams > Cat > Quantitaitve Aptitude

BASIC GEOMETRY MCQs

Total Questions : 93 | Page 7 of 10 pages

Question 61. A fan has 3 equally spaced blades. The central dial has an area of 3

π

^{2}

. Length of each blade is (20-

\sqrt{3}

)cm. If a triangle is drawn from the tips of the blades, find its area.

900 sq cm
300 √3 sq cm
900 sq cm
400√3 sq units

Discuss Question

Answer: Option B. -> 300 √3 sq cm
:
B
Ans: b) 300

\sqrt{2}

sq cm
Soln: Radius of the dial at the centre:

\sqrt{\frac{3 π}{π}}

total length of the blade from the centre: (20-

\sqrt{3}

\sqrt{3}

; = 20.
Half of each side: 30 cos 30. = 10

\sqrt{3}

. Total length of each side: 20

\sqrt{3}

.
Thus , area:

(\frac{\sqrt{3}}{4}) (20 \sqrt{3})^{2}

. =300

\sqrt{2}

sq cm

Question 62. A square hole of cross sectional area 4 cm

^{2}

is drilled across a cube with its length parallel to a side of the cube. If an edge of the cube measures 5 units, what is the total surface area of the structure so formed?

158 cm2
190 cm2
166 cm2
182 cm2

Discuss Question

Answer: Option D. -> 182 cm2
:
D
Ans: d) 182 cm

^{2}

New SA=(original SA)+(SA of square channel formed through the cube)-(squares on 2 faces)
= (25x6)+(2x5x4)-(4+4) ;=182 sq units

Question 63. If (9p

^{2}

+4), (2p+7) and (8p

^{2}

-1) are three sides of a triangle. Which of the following CANNOT be a value for p?

2
√2
2√2
None of these

Discuss Question

Answer: Option C. -> 2√2
:
C
Ans: c 2

\sqrt{2}

.
Soln: Reverse Gear Approach: Use answer options
The largest side should be smaller than sum of other 2 sides.
The sides are 76, 4

\sqrt{2}

+7, 63; Here the sum of last two sides is 75.6 which is lesser than 76

Question 64.

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 65.

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 66.

Find the height of the right cylinder whose volume is 511 cm $^{3}$ and the area of the base is 36.5 cm $^{2}$ .

3.5 cm
10.5 cm
14 cm
None of these

Discuss Question

Answer: Option C. -> 14 cm
:
C

Ans: c.

Soln: Volume of a cylinder = πr $^{2}$ h

Area of the base = π r $^{2}$

Thus, height = $(\frac{V o l}{S u r f a c e A r e a}) = \frac{511}{36.5}$ = 14.

Question 67.

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.

2 times
12 times
8 times
16 times

Discuss Question

Answer: Option D. -> 16 times
:
D

Solution:- Volume of original rod = $π r^{2}$ h
Changed volume of changed rod = $π {(\frac{r}{4})}^{2} h_{1}$
But, Volume remains constant.
$π r^{2}$ h = $π {(\frac{r}{4})}^{2} h_{1} \Rightarrow h_{1}$ = 16h

Option(d).

Question 68.

If the diagonals of a rhombus are 18 cm and 24 cm respectively, then find its perimeter.

15 cm
42 cm
60 cm
68 cm

Discuss Question

Answer: Option C. -> 60 cm
:
C

Ans: c

Find length of sides by considering the half of diagonals

(pythagoras theorem): $\sqrt{9^{2} + 12^{2}}$ = 15. Perimeter: 4 $\times$ 15. Option(c).

Question 69.

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: -

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 70.

Water flows at the rate of 10 m per mins from a cylindrical pipe of radius 2.5 mm. A conical vessel of diameter 40 cm and depth 24 cm is filled with water flowing from this pipe. The time taken to fill the conical vessel is: