Free Basic Geometry - 01 Practice Test - CAT

Question 1

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

A. 7

B. 10

C. 12.5

D. 15

SOLUTION

Solution : 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 2

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

A. 3.5 cm

B. 10.5 cm

C. 14 cm

D. None of these

SOLUTION

Solution : 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 3

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?

A. 2156 cc

B. 2183 cc

C. 2492 cc

D. None of these

SOLUTION

Solution : 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 4

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.

A. 2 times

B. 12 times

C. 8 times

D. 16 times

SOLUTION

Solution : 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 5

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

A. 15 cm

B. 42 cm

C. 60 cm

D. 68 cm

SOLUTION

Solution : 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 6

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:

A. 1:2:3

B. 1:4:6

C. 1:6:9

D. None of these

SOLUTION

Solution : 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 7

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:

A. < 30 mins

B. 30 < t < 50 mins

C. 50 < t < 75 mins

D. >75 mins

SOLUTION

Solution : C

Area= $π$ r $^{2}$ =6.25 $π$

Rate of flow= ( $π$ r $^{2}$ x1000) mm $^{3}$ /min

=6250 $π$ mm $^{2}$ /min

Volume of cone= $\frac{1}{3}$ $π$ r $^{2}$ h = $\frac{1}{3}$ $π$ (200) $^{2}$ x 240 mm $^{3}$

=80 $π$ x40000

=3200000 $π$ mm $^{3}$

Thus, total time taken is:

$\frac{(V o l u m e o f c o n e i n m m^{3})}{(R a t e o f w a t e r f l o w i n m m^{3} / m i n)}$ = 51.2min

Thus answer option (c) is correct.

Question 8

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

A. 44 cm

^{2}

B. 34.25 cm

^{2}

C. 32.50 cm

^{2}

D. None of these

SOLUTION

Solution : D

Diagonal of rectangle: $\sqrt{(8^{2} + 6^{2})}$ , =10.

Half of diagonal = radius of circle = 5.

Area of shaded region: $π$ 5 $^{2}$ - (8x6) = 30.57

Question 9

In the figure find the area outside the trapezium, given square is of side 16 cm.

A. 128 cm

^{2}

B. 154 cm

^{2}

C. 168 cm

^{2}

D. 28 cm

^{2}

SOLUTION

Solution : A

Required Area = Area of triangle AFG + Area of triangle DEF + Area of triangle BCD =

$(\frac{1}{2} \times 6 \times 8) + (\frac{1}{2} \times 8 \times 6) + (\frac{1}{2} \times 10 \times 16)$

= 128 cm $^{2}$

Question 10

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.

A. 42 min

B. 68 min

C. 54 min

D. 82 min

SOLUTION

Solution : A

Ans:a 2 $π$ r=308 2 $\times$ $(\frac{22}{7}) \times r$ = 308 Find 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 11

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?

\sqrt{5}

B. 1

\frac{1}{4}

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

SOLUTION

Solution : 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 12

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?

A. 1 cm

B. 2.5 cm

C. 3.5 cm

D. 4 cm

SOLUTION

Solution : B

Solution: - Volume of earth taken out: $π$ r $^{2}$ 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 13

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?

\frac{a}{2 π}

B. a

\frac{a}{2}

D. 2

π

SOLUTION

Solution : B

The cone will have slant height same as the triangle side.

Question 14

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 radius of the cone formed?

A. a

\frac{a}{4}

\frac{a}{π}

D. None of these

SOLUTION

Solution : D

Circumfrence of the base of cone = 2 $π$ r = side of triangle = a

radius (r) = $\frac{a}{2 π}$

Question 15

Sixteen cylindrical sprite cans are placed in a square carton such that each can either touches another can or a wall of the carton. Each can is of unit radius. What is the bottom area of the carton?

A. 16 sq units

B. 64 sq units

C. 32 sq units

D. None

SOLUTION

Solution : B

Ans: b (Add the radii of cans in 1 row. We get 8 as length. Same for breadth. So, carton dimensions: 8 by 8. Bottom area: 8x8=64 sq units.)

Free Basic Geometry - 01 Practice Test - CAT

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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