Geometry – Mensuration Practice test

1.

The four walls and ceiling of a room of length 25 ft., breadth 12 ft. and height 10 ft. are to be painted. Painter A can Paint 200 sqr.ft in 5 days, painter B can paint 250 sqr.ft in 2 days. If A & B work together, in how many days do they finish the job?

Answer: Option C

Explanation:

Total area to be painted = 25*12 +2(10*12 + 10*25) = 1040 sqr.ft

A paints = 200/5 = 40 sqr.ft  per day
B paints = 250/2 = 125 sqr.ft  per day

A + B = 40 + 125 = 165 sqr.ft     

Number of days = 1040/165 = 6 10/33 

2.

There is a rectangular Garden whose length and width is 60m X 20m.There is a walkway of uniform width around garden. Area of walkway is 516m2. Find width of walkway?

Answer: Option C

Explanation:

let the width of rectangle be x so the length & breath is increased by 2x

so new total area along with walkway is (60+2x)*(20+2x)
so  (60+2x)*(20+2x)-60*20=516
=>(60+2x)*(20+2x)=1716
=>(30+x)*(10+x)=429=33*13
=>x=3
3.

In a trapezoid ABCD, the diagonals intersect at E. If area of triangle ∆ABE = 72 and area of triangle ∆CDE = 50, then area of trapezium ABCD is

Answer: Option B

Explanation:

Let h1 and h2 be the heights of the triangles CDE and ABE respectively .

The area of CDE ∆CDE = ½*CD* h1 = 50 . . . (i) ,
And area of triangle ABE, ∆ABE = ½*AB* h2 =72. . . (ii)
Also triangles ABE and CDE are similar hence CD/AB = h1/h2 = k .
Hence we get on dividing (i) by (ii)
=> CD*h1/AB*h2 = 50 / 72
=> k 2 = 25/36 (putting AB/CD = h1/h2 = k)
we get k = 5/6.
Hence area of trapezium ABCD = ½*(AB+CD)*(h1+h2) = ½*AB*h 1 (1+k)^ 2 = 72*(1+5/6)^ 2= 242
4.

The dimensions of a certain machine are 48″ X 30″ X 52″. If the size of the machine is increased proportionately until the sum of its dimensions equals 156″. What will be the increase in the shortest side?

Answer: Option A

Explanation:

Sum of present dimension 48+30+52=130.

New dimension =156.

Increase in dimension = 26.

Ratio of dimensions = 48:30:52 =>24:15:26

Therefore increase in the shortest side15*(26)/(24+15+26)=6

5.

The length of the side of a square is represented by x+2. The length of the side of an equilateral triangle is 2x. If the square and the equilateral triangle have equal perimeter, then the value of x is _______

Answer: Option D

Explanation:

Since the side of the square is x + 2, its perimeter = 4 (x + 2) = 4x + 8
Since the side of the equilateral triangle is 2x, its perimeter = 3 * 2x = 6x
Also, the perimeters of both are equal.
(i.e.) 4x + 8 = 6x
(i.e.) 2x = 8 ? x = 4.

6.

A cone of height 9 cm with diameter of its base 18 cm is carved out from a wooden solid sphere of radius 9 cm. The percentage of the wood wasted is :

Answer: Option D

Explanation:

Volume of sphere = [((4/3π)*9*9*9)]cm3.
Volume of cone = [(1/3π)*9*9*9]cm3.
Volume of wood wasted = [((4/3π)*9*9*9)-((1/3π)*9*9*9)]cm3
= (π*9*9*9)cm3.
Therefore required percentage = [((π*9*9*9)/((4/3π)*9*9*9))*100]% = ((3/4)*100)% = 75%.

7.

A parallelogram has sides 30 m and 14 m and one of its diagonals is 40 m long. Then, its area is :

Answer: Option A

Explanation:

Let ABCD be a ||gm in which AB = 30 m, BC = 14 m & AC = 40 m.
Clearly, area of || gm ABCD = 2(area of ∆ ABC).
Let a = 30, b = 14 & c = 40.
Then, s=(1/2)(a+b+c) = 42
Therefore area of ∆ ABC = √s(s-a)(s-b)(s-c)
=√42*12*28*2 = 168 m2.
Therefore area of ||gm = (2*168) m2 = 336 m2

8.

The edge of a cuboid  are in the ratio 1 : 2 : 3 and its surface area is 88 cm2. The volume of the cuboid is :

Answer: Option B

 Explanation:

Let the dimensions of the cuboid be x, 2x and 3x.
Then, 2(x*2x + 2x*3x + x*3x) = 88
= 2x2 + 6x2 + 3x2 = 44 = 11x2 = 44  = x2  = 4 = x = 2.
Therefore volume of the cuboid = (2*4*6) cm3  = 48 cm3.

9.

If the radius of a circle is diminished by 10%, then its area is diminished by :

Answer: Option B

Explanation:

Let the original radius be R cm.
New radius = (90% of R) cm =[(90/100)*R] cm = 9R/10 cm.
Original area = πR2.
Diminished area = [πR2-π(9R/10)2] cm2 = [(1-(81/100) πR2] cm2 = ((19/100 πR2) cm2.
Decrease% = [(19πR2/100)*(1/πR2)*100]% = 19%.

10.

A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man’s eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60º. What is the distance between the base of the tower and the point P?

Answer: Option D

Explanation:

 

One of AB, AD and CD must have given.

 

So, the data is inadequate

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