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.
SOLUTION
Solution : B
Solution: The fig can be considered as:
Since the triangles are similar:
r1+r2r2+r3=r1−r2r2−r3Substituting r1= 20 and r3= 5 , r2 = 10. Option (b).
Question 2
Find the height of the right cylinder whose volume is 511 cm3 and the area of the base is 36.5 cm2.
SOLUTION
Solution : C
Ans: c.
Soln: Volume of a cylinder = πr2h
Area of the base = π r2
Thus, height = (VolSurface Area)=51136.5= 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 cm2. What is the volume of the cylinder?
SOLUTION
Solution : A
Solution:- Total surface Area: 2πr2+2πrh2πrh=32⇒1+rh=32⇒rh=12⇒h=2r
TSA:(2πr2+2πrh) = 924 ⇒(2πr2+2πr×2r) = 924 ⇒ r = 7 and h = 14
Volume = πr2h=227×7×7×14 = 2156Option(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.
SOLUTION
Solution : D
Solution:- Volume of original rod =πr2h
Changed volume of changed rod = π(r4)2h1
But, Volume remains constant.
πr2h = π(r4)2h1⇒h1 = 16hOption(d).
Question 5
If the diagonals of a rhombus are 18 cm and 24 cm respectively, then find its perimeter.
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).
Question 6
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:
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).
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:
SOLUTION
Solution : C
Area=πr2=6.25π
Rate of flow= (πr2x1000) mm3/min=6250π mm2/minVolume of cone=13 πr2h = 13 π (200)2x 240 mm3=80πx40000=3200000π mm3Thus, total time taken is:(Volume of cone in mm3)(Rate of water flow in mm3/min) = 51.2minThus answer option (c) is correct.
Question 8
If the inner rectangle is 8cm x 6 cm. Find the area of the shaded region.
SOLUTION
Solution : D
Diagonal of rectangle: √(82+62), =10.
Half of diagonal = radius of circle = 5.
Area of shaded region: π52 - (8x6) = 30.57
Question 9
In the figure find the area outside the trapezium, given square is of side 16 cm.
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
Question 10
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.
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
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?
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
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?
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
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?
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?
SOLUTION
Solution : D
Circumfrence of the base of cone = 2πr = side of triangle = a
radius (r) = a2π
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?
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.)