# Free Basic Geometry - 01 Practice Test - CAT

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:

r1+r2r2+r3=r1r2r2r3Substituting r1= 20 and r3= 5 , r2 = 10. Option (b).

Find the height of the right cylinder whose volume is 511 cm3 and the area of the base is 36.5 cm2.

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 = πr2h

Area of the base = π r2

Thus, height = (VolSurface Area)=51136.5= 14.

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 cm2. 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: 2πr2+2πrh2πrh=321+rh=32rh=12h=2r
TSA:(2πr2+2πrh) = 924 (2πr2+2πr×2r) = 924 r = 7 and h = 14
Volume = πr2h=227×7×7×14 = 2156

Option(a).

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 =πr2h
Changed volume of changed rod = π(r4)2h1
But, Volume remains constant.
πr2h = π(r4)2h1h1 = 16h

Option(d).

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):92+122 = 15. Perimeter: 4 × 15.  Option(c).

If the right circular cone is cut into three solids of volumes V1,V2 and V3 by two cuts which are parallel to the base and trisects the altitude, then V1:V2:V3 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 13:23:1 = 1:2:3.

r1 = 1, h1 = 1

r2 = 2, h2 = 2

r3 = 3, h3 = 3

Volume will be in the ratio of r$$^2$$h for the three circular cones.

Required Volumes are

V1 = r21 x h1 = 1

V2= r22 x h2 - V1 = 8 - 1 = 7

V3= r23 x h3 - (V1 + V2) = 27 - (7 + 1) = 19

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

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=πr2=6.25π

Rate of flow= (πr2x1000) mm3/min
=6250π mm2/min
Volume of cone=13 πr2h = 13 π (200)2x 240 mm3
=80πx40000
=3200000π mm3
Thus, total time taken is:
(Volume of cone in mm3)(Rate of water flow in mm3/min) = 51.2min
Thus answer option (c) is correct.

If the inner rectangle is 8cm x 6 cm. Find the area of the shaded region. A. 44 cm2
B. 34.25 cm2
C. 32.50 cm2
D. None of these

#### SOLUTION

Solution : D

Diagonal of rectangle: (82+62), =10. Half of diagonal = radius of circle = 5.

Area of shaded region: π52 - (8x6) = 30.57

In the figure find the area outside the trapezium, given square is of side 16 cm. A. 128 cm2
B. 154 cm2
C. 168 cm2
D. 28 cm2

#### SOLUTION

Solution : A Required Area = Area of triangle AFG + Area of triangle DEF + Area of triangle BCD =

(12×6×8)+(12×8×6)+(12×10×16)

= 128 cm2

The top of a conical container has a circumference of 308 m. Water flows in at a rate of 12 m3 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 2πr=308 2 × (227)×r = 308 Find half volume of conical container.

Time reqd:vol of 12 containerflow rate=15078.286 = 2520; 252060 =42min

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?

A. 5
B. 1
C. 14
D.  (5+1)4

#### SOLUTION

Solution : D

Solution: -  Let radius = r feet.

Volume of mud from well: πr2(20) cubic feet.

Volume of wall around well: 5π((r+1)2-r2).

πr2(20) = 5π((r+1)2-r2) => 4r2 - 2r - 1=0 r =  (5+1)4

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: πr2h. = π x (0.7)2 x 3 = V.

Area of field without circle: (15.4x12.1)-(π(0.72)) = A (say).

Axh=V

h=2.5 cm

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? A. a2π
B. a
C. a2
D. 2πa

#### SOLUTION

Solution : B

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

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
B. a4
C. aπ
D. None of these

#### SOLUTION

Solution : D

Circumfrence of the base of cone = 2πr = side of triangle = a

radius (r) = a2π

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.)