Quant Quiz On Mensuration

1.    What will be the length of the
diagonal of that square plot whose area is equal to the area of a rectangular
port of length 45 meters and breadth 40 meters?
(a) 42.5
meters
(b) 60 meters
(c) Data
inadequate
(d) 45 meters

2.    The length of a rectangle is 20%
more than its breadth. What will be the ratio of the area of a rectangle to
that of a square whose side is equal to the breadth of the rectangle?
(a)
2:1
(b) 6:5
(c) 3:4
(d) Data inadequate

3.    If the length of a certain
rectangle is decreased by 4 cm and the width is increased by 3 cm, a square
with the same area as the original rectangle would result. Find the perimeter
of the original rectangle.
(a) 40
cm
(b) 50 cm
(c) 60
cm
(d) 30 cm

4.    The base of triangular field is
three times its altitude. If the cost of cultivating the field at Rs. 24.68 per
hectare be Rs. 333.18, find its base and height.
(a) 200
m
(b) 275 m
(c) 300
m
(d) 600 m

5.    The difference between two
parallel sides of a trapezium is 4 cm. The perpendicular distance between them
is 19 cm. If the area of the trapezium is 475 cm2, find the lengths
of the parallel sides.
(a) 27 cm & 23
cm
(b) 22 cm & 25cm
(c) 20 cm & 27
cm
(d) 23 cm & 25 cm

6.    A farmer wishes to start a 100sq
m triangular vegetable garden. Since he has only 30m barbed wire, he fences
three sides of the garden letting his house compound wall act as the fourth
side fencing. The dimension of the garden is.
(a) 15 m * 6.67
m
(b) 20 m * 5 m
(c) 30 m *
3.33m
(d) 40m * 2.5 m

7.    The perimeter of 5 squares is 24
cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square
equal in area to the sum of the areas of these squares is:
(a) 31
cm
(b) 62 cm
(c) 124
cm
(d) 961 cm

8.    If the ratio of areas of two
squares is 225:256, then the ratio of their perimeters is:
(a)
256:225
(b) 225:256
(c)
16:15
(d) 15:16

9.    The sides of a triangle are 3 cm,
4 cm, and 5 cm. The area (in cm2) of the triangle formed by joining
the mid-points of the sides of this triangle is:
(a)
3/4
(b) 3/2
(c)
3
(d) 6

10.  One of sides of a right-angled triangle is
twice the other, and the hypotenuse is 10 cm. The area of the triangle is :
(a) 20 cm2
(b) 33 1/3 cm2
(c) 40 cm2

(d) 50 cm2

Answer with solution:
1.   (b)
Area= (45* 40) m=
1800 m2,
(Diagonal)2/2= 1800,
Diagonal = 60m
2.  (b)
Let breadth be x meters.
Then length= 120% of x = 120/100 x= 6x/5 m
Therefore, the required ratio=
{(6x/5)*x*x}/x*x = 6:5
3.  (b)
Let x and y be the length and breadth
of the rectangle respectively.
Then, x ­- 4 =y + 3 or x – y = 7
………..(i)
Area of rectangle = x*y,
Area of square = (x-4) * (y+3)
So, (x-4) * (y+3)= x*y,
or 3x – 4y = 12……………………………….(ii)
Solving eq (i) & (ii) x= 16 and y
= 9
Therefore perimeter of rectangle =
2(x+ y)= 2(16+9)= 50 cm
4.  (c)
Area of the field = Total cost/
Rate = 333.18/ 24.68 hectares = 13.5 hectares
= (13.5 * 1000) m=
135000 m2
Let altitude = x meters and base = 3x
meters.
Then, 1/2 * 3x * x = 135000
x2 = 90000
x= 300 m
5.   (a)
Let the two parallel sides of the
trapezium be a cm and b cm,
Then, a-b = 4……………………….(i)
And, 1/2 *(a+b)*19 = 475
(a+b) = (475*2)/19 =50…………(ii)
Solving eq(i)& (ii) a= 27 cm and b
= 23 cm
6.    (b)
According to quest, 2b + L = 30, (as
fencing is only for 3 sides)
So, L =30 – 2b
Now area of the garden = 100 sq m,
L * b =100, OR b*(30-2b) = 100
b2 -15b+ 50 =0,
now b= 5 & b=10
if b = 5 then L = 20 and if b= 10 then
L= 10
since the garden is rectangular so the
dimension would be 20m * 5 m
7.    (c)
The sides of the 5 squares are (24/4),
(32/4), (40/4), (76/4), (80/4) i.e., 6 cm, 8 cm , 10 cm , 19 cm, 20 cm
Area of the new square = [62 +
82 + (10)2 + (19)2 + (20)2]
= (36 + 64 + 100 + 361 + 400) cm2=
961 cm2
Side of the new square = √961 cm= 31
cm
Perimeter of the new square = (4*31)
cm = 124 cm
8.    (d)
A1  / A=225/256
= (15)2/(16)2
=15/16
Now 4 A1 / 4A2= 15/16
9.   (b)
A= 3 cm, b= 4 cm, and c= 5 cm
It is a right-angled triangle with
base = 3 cm and height= 4 cm
So its area= (1/2 * 3 * 4) cm2=
6 cm2
Area of required triangle= (1/4 * 6)
cm= 3/2 cm2
10.  (a)
Let the sides be a cm and 2a cm
Then, a2 + (2a)=
(10)2
5 a2= 100
a= 20 cm2

Area = (1/2 *a *2a) = a2=
20 cm
2

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11 Comments

1. sandeep says:

Gr8t adda

2. bacha party says:

1-b
2-b
3-b
4-
5-a
6-b
7-c
8-d
9-b
10-a

3. manish says:

Prevention of Corrosion
1. The rusting of iron can be prevented by painting, oiling, greasing,galvanising, chrome plating, anodising or
making alloys.

2. Galvanisation is a method of protecting steel
and iron from rusting by coating them with a thin layer of zinc.

3. Alloying is a very good method of improving the
properties of a metal.We can get the desired properties by this
method

4. When iron is mixed with nickel and chromium, we
get stainless steel, which is hard and does not
rust.
5. Gold is alloyed with either silver or copper to make it hard

6. one of the metals in Alloy is mercury, then the
alloy is known as an amalgam.

7. brass, an alloy of copper and zinc (Cu and Zn),

8. bronze, an alloy of copper and tin (Cu and Sn),

9. Solder, an alloy of lead and tin (Pb and Sn),
has a low melting point and is used for welding
electrical wires together.

4. gusto says:

Dude answer to question 4 is C and to question no 10 is B

5. Vartika says:

Refresh!!!!

6. kumar says:

Thank u manish

7. Ani@v says:

In question no 3 how got the equation 1.plz..explain me.

8. Sureshkumar Pandali says:

please do not make the typing mistakes

9. shikha says:

Thanks

10. Ashish Jadhav says:

Q no. 6?

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