Question 1 :
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the length of broken part.
(A) 13 m (B) 15 m (C) 18 m
Solution :
Here,
AC2 = AB2 + BC2
AC2 = 122 + 52
AC2 = 144 + 25 ==> 169
AC = √169 = √13 ⋅ 13
AC = 13 m
So, the length of the broken part is 13 m.
Question 2 :
Find the ratio of 3 km to 300 m.
(A) 10:1 (B) 1:10 (C) 2:5
Solution :
We can compare two ratios of the same kind only.So,let us convert km to meters.
1000 m = 1 km
3 km = 3 x 1000 = 3000 m
= 3000 : 300
= 10 : 1
So, the required ratio is 10 : 1.
Question 3 :
Find BC, if the area of the triangle ABC is 36 cm2 and the height AD is 3 cm.
(A) 18 cm (B) 24 cm (C) 10 cm
Solution :
BC = Base of the triangle
AD = Height of the triangle = 3 cm
Area of triangle = (1/2) x Base x Height
= (1/2) x BC x AD
36 = (1/2) x BC x 3
(36 ⋅ 2)/3 = BC
BC = 24 cm
So, the base of the triangle is 24 cm.
Question 4 :
How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 22/7)
(A) 300 times (B) 200 times (C) 400 times
Solution :
Radius of wheel = 28 cm
Distance to be covered = 352 m = 35200 cm
Distance covered in one revolution = 2πr
= 2 ⋅ (22/7) ⋅ 28
= 176 cm
Number of revolutions
= Distance to be covered/Distance covered in one revolution
= 35200/176
= 200 times
So, the wheel has to be rotated 200 times to go 352 m.
Question 5 :
A rectangular park is 45 m long and 30 m wide. The path 2.5 m wide is constructed outside the park. Find the area of that path.
(A) 400 m2 (B) 200 m2 (C) 150 m2
Solution :
Area of path
= Area of rectangle ABCD - Area of rectangle abcd
In abcd : Length = 45 m Width = 30 m |
In ABCD : Length AB = 45 + 2.5 + 2.5 = 50 m Width BC = 30 + 2.5 + 2.5 = 35 m |
Area of path = 50 ⋅ 35 - 45 ⋅ 30
= 1750 - 1350
= 400 m2
So, the area of the path is 400 m2 .
Question 6 :
Find the value of the expression a2 + 2 ab + b2
if a = 3, b = 2
(A) 27 (B) 35 (C) 25
Solution :
a2 + 2 ab + b2 = 32 + 2 (3)(2) + 22
= 9 + 12 + 4
= 25
So, the value of the given expression is 25.
Question 7 :
In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?
(A) 27 (B) 35 (C) 25
Solution :
There are 3 computers for every 6 students
Number of computers for 1 student = 3/6 = 1/2
Number of computers for 24 students = (1/2) 24
= 12 computers
So, every 24 students need 12 computers.
Question 8 :
The side of an equilateral triangle is 3.5 cm. Find its perimeter.
(A) 5.9 cm (B) 2.7 cm (C) 10.5 cm
Solution
Side length of equilateral triangle = 3.5 cm
Perimeter of equilateral triangle = 3a
= 3 (3.5)
= 10.5 cm
So, the perimeter of the equilateral triangle is 10.5 cm.
Question 9 :
Find the value of 2.5 + 5.1
(A) 7.6 (B) 2.5 (C) 8.2
Solution
oooooooooooooooooooooo2.5 (+) ooooooooooooooooooooooo
oooooooooooooooooooooo5.1ooooooooooooooooooooooooooo
oooooooooooooooooooo_____ooooooooooooooooooooooooo
oooooooooooooooooooo07.60000oooooooooooooooooooooo
oooooooooooooooooooo_____ooooooooooooooooooooooooo
Question 10 :
The wood cutter took 12 minutes to make 3 pieces of a block of wood. How long would be needed to make 5 such pieces?
(A) 50 minutes (B) 20 minutes (C) 30 minutes
Solution :
Time taken for wood cutter to cut block of wood into 3 pieces = 12 minutes
Time taken for wood cutter to cut a block of wood into 1 piece
= 12/3 = 4 minutes
Time taken for wood cutter to cut a block of wood as 5 pieces
= 4 (5) = 20 minutes
So the time taken is 20 minutes.
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