9TH GRADE MATH WORKSHEET WITH SOLUTION

Problem 1 :

Find the value of k, if x - 1 is a factor of

4x3 + 3x2 - 4x + k

(A)  -3     (B)  -5     (C)  0

Solution :

Let P(x)  =  4x+ 3x- 4x + k -----(1)

Given, x - 1 is a factor of P(x).

Then, x  =  1

By applying x  =  1 in equation (1), we get

 4(1)+ 3(1)- 4(1) + k  =  0

 4 + 3 - 4 + k  =  0

3 + k  =  0

k  =  -3

So, the value of k is -3.

Problem 2 :

What is the value of the expression given below ?

∛4 × ∛16

(A)  3   (B)  4   (C)  2

Solution  :

Finding the value of ∛4 and ∛16.

By using the calculator, we get

∛4  =  1.59

∛16  =  2.52

Then,

∛4 × ∛16  =  (1.59 × 2.52)

=  4

So, the answer is 4.

Problem 3  :

Find the area of the quadrilateral whose diagonal measures the length of 50m and the perpendicular distance measures 10m and 20m respectively.

(A)  7.50 sq.m   (B)  75 sq.m   (C)  750 sq.m

Solution  :

Area of quadrilateral  =  1/2 × d × (h1 + h2)

Here d  =  50m, h=  10m and h=  20m

Then,

Area  =  1/2 × 50 × (10 + 20)

=  1/2 × 50 × 30

=  750 sq.m

So, the area of the quadrilateral is 750 sq.m

Problem 4  :

Cost of leveling a land is $12 per square meter. A land is in the form of a trapezium whose parallel sides are of lengths 18m and 12m. If its other two sides are each of length 5m, find the total cost incurred in leveling the land.

(A)  230   (B)  540   (C)  900 

Solution  :

Area of trapezium  =  1/2 × h × (a + b)

Here h  =  5m, a  =  18m and b  =  12m

Area  =  1/2 × 5 × (18 + 12)

=  1/2 × 5 × 30

Area  =  75 sq.m

To find the total cost of leveling land,

=  cost of leveling land  × Area of land

=  12 × 75

=  900

So, the answer is 900.

Problem 5 :

The perimeter of a rhombus is 20cm. One of the diagonals is of length 8cm. Find the length of the other diagonal and the area of the rhombus.

(A)  50 sq.m   (B)  32 sq.m   (C)  24 sq.m 

Solution :

Given, perimeter of a rhombus is 20cm.

Length of the one diagonal d1 is 8cm.

Let the other diagonal be d2.

Perimeter of a rhombus  =  4 sides

So, side  =  20/4  =  5cm

We know that the diagonals of a rhombus bisect each other at right angles.

Since the diagonals bisect each other, so half the length of the diagonal d1.

That is, d1/2  =  8/2  =  4cm

To find the other diagonal,

Using the Pythagorean theorem :

Here d1/2  =  4cm, side  =  5cm and d2/2  =  ?

(d1/2)2 + (d2/2)=  side2

42 + (d2/2) =  52

16 + (d2/2) =  25

 (d2/2)=  9

 d2/2  =  3

 d=  6cm

So, length of the other diagonal d2 is 6cm.

Now,

Area of the rhombus using diagonals  =  1/2 × d1 × d2

=  1/2 × 8 × 6

Area  =  24cm2.

So, area of the rhombus is 24cm2.

Problem 6  :

Evaluate the following log64 + log4

(A)  4   (B)  8   (C)  6 

Solution  :

By using the rule loga (mn)  =  logm + loga n, we get

log64 + log4  =  log2 (64 . 4)

=  log(26 . 22)

=  log2 2(6 + 2)

=  log2 28

=  8log2

=  8(1)

=  8

So, the answer is 8.

Problem 7  :

Two sets are called ---------- if they have same elements.

(A)  Equivalent sets   (B)  Null sets   (C)  Equal sets 

Solution  :

So, the answer is equal sets.

Problem 8 :

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

(A)  100cm   (B)  115cm   (C)  130cm 

Solution :

Area of the sector  =  πr× θ/360

Here A  =  924 and θ  =  240

924  =  πr× 240/360

924  =  22/7 × r× 2/3

(924 × 21)/44  =  r2

r=  441

r  =  21cm

Perimeter of the sector  =  2r + θ/360 × 2πr

=  2(21) + (240/360) × 2 × 22/7 × 21

=  42 + (2/3) × 132

=  42 + 88

=  130cm

So, the perimeter of the sector is 130cm.

Problem 9 :

Factorise 3x2 - 18xy

(A)  3x(x - 6y)   (B)  3x(x - 18y)   (C)  3(x2 - 18y)

Solution  :

3x- 18xy  =  3x(x - 6y)

So, the answer is 3x(x - 6y)

Problem 10 :

Solve 3(x + 4) + 2(y - 3)  =  17 and 7(x - 3) - 3(y + 4)  =  8

(A)  (8, 1)   (B)  (5, -2)   (C)  (0, 6)

Solution :

3(x + 4) + 2(y - 3)  =  17 -----(1)

7(x - 3) - 3(y + 4)  =  8 -----(2)

From (1), we get

3x + 12 + 2y - 6  =  17

3x + 2y  =  17 - 6

2y  =  11 - 3x

y  =  (11 - 3x)/2 -----(3)

From (2), we get

7x - 21 - 3y - 12  =  8

7x - 3y  =  41

-3y  =  41 - 7x

y  =  (7x - 41)/3 -----(4)

Equating (3) and (4), we get

(11 - 3x)/2  =  (7x - 41)/3

3(11 - 3x)  =  2(7x - 41)

33 - 9x  =  14x - 82

33 + 82  =  14x + 9x

115  =  23x

x  =  5

By applying x  =  5 in equation (3), we get

y  =  (11 - 3(5))/2

y  =  (-4)/2

y  =  -2

So, the solution is (5, -2)

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

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Unit Rates

    Dec 02, 22 04:18 PM

    Unit Rates - Concept - Examples with step by step explanation

    Read More

  2. Adding and Subtracting Fractions Worksheet

    Dec 02, 22 07:27 AM

    Adding and Subtracting Fractions Worksheet

    Read More

  3. Relation between Radian and Degree

    Dec 02, 22 07:16 AM

    Relation between Radian and Degree

    Read More