GRADE 9 MATH PRACTICE

Problem 1 :

Evaluate the following

Solution :

Let x  =  0.7575.......  -----(1)

Multiply by 100 on both sides.

100x  =  75.7575.........  -----(2)

(2) - (1)

100x - x  =  75.7575......... - 0.7575.........

99x  =  75

x  =  75/99

Problem 2 :

Put π and 22/7 in order relation.

 (A) π > 22/7         (B) π < 22/7          (C) π = 22/7

Solution :

π  =  22/7

Problem 3 :

The value of √20-√225+√80

Solution :

√20-√225+√80

  =  √(2⋅2⋅5) - √(5⋅5⋅33)+√(2⋅2⋅2⋅2⋅5)

  =  2√5 -15 + 4√5

  =  6√5-15

Problem 4 :

A wall in the form of a rectangle has base 15 m and height 10 m. If the cost of painting the wall is $16 per square meter, find the cost for painting the entire wall.

Solution :

Area of rectangle  =  length ⋅ height

length  =  15 m and height  =  10 m

Area of rectangle  =  15 ⋅ 10

  =  150

Cost of painting the wall  =  $16

Required cost  =  150(16)

  =  $2400

Problem 6 :

Find the base of a parallelogram if its area is 40 cm² and altitude is 15 cm.

Solution :

Area of a parallelogram  =  base x height

40  = base x 15

base  =  40/15

base  =  8/3 cm

Problem 7 :

If the lengths of the sides of a triangle are 11 cm, 60 cm and 61 cm, find the area and perimeter of the  triangle.

Solution :

Since all the sides are having different measures, it is a scalene triangle.

a  =  11 cm, b  =   60 cm and c  =  61  cm

Perimeter   =  a+b+c

  =  11+60+61

  =  132

s  =  (a+b+c)/2

s  =  132/2

s  =  66

Area  =  √s(s-a)(s-b)(s-c)

  =  √6655⋅6⋅5

  =  11⋅5⋅6

  =  330 cm2

Problem 8 :

A wire of length 264 cm is cut into two equal portions. One portion is bent in the form of a circle and the other in the form of an equilateral triangle. Find the ratio of radius and side length of triangle.

Solution :

Perimeter of circle  =  Perimeter of equilateral triangle 

  =  132 cm

Circumference of circle :

2πr  =  132

2⋅(22/7)⋅r  =  132

r  =  132⋅(7/44)

r  =  21

Perimeter of triangle :

3a  =  132

a  =  44

Ratio of radius and side length of triangle :

  21 : 44

Problem 9 :

Pablo buys popsicles for his friends. The store sells single popsicles for $1 each, 3 popsicle boxes for $2 each, and 5 popsicle boxes for $3 . What is the greatest number of popsicles that Pablo can buy with $8 ?

Solution :

To get the greatest number of popsicles, we can choose two 5 popsicle box that costs $3.

5 + 5 + 2  =  13 (maximum popsicles)

So, he can buy 13 popsicles for $8.

Problem 10 :

Find the perimeter of the sector whose area is 924 square cm and the central angle is 240.

Solution :

(θ/360)πr2  =  924

(240/360) ⋅ (22/7) ⋅ r2  =  924

r2  =  21 ⋅ 21

r  =  21

Perimeter of sector  =  (θ/360)2πr

  =  (240/360) ⋅ ⋅ (22/7) 21

  =  88

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

1.  25/33


2.  π < 22/7

3.  4√5

4.  2400

5.  7500

6.  8/3 cm

7.  330, 132

8.  21 : 44

9.  181 cm²

10. 130 cm

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