Class 8Term 2 Practice 50 questions with Answers

 

Class 8 Term 2 Practice 50 Questions with Answers


  1. In an auditorium, the number of rows is equal to the number of chairs in each row. If  the capacity of the auditorium is to accommodate 1764 chairs, find the number of  chairs in each row.

ANSWER: 

  1. Find the square of the number 82 using the property (a + b)2 = a2 + b2 + 2ab 

ANSWER: 

  1. Find the square root of a number whose prime factorisation is 2 × 2 × 3 × 3 × 3 × 3 × 5 ×  5 .

ANSWER: 

Square root = 2 x 3 x 3 x 5 = 90


  1.  Find the smallest square number that is divisible by each of the numbers 4, 9 and 10. 

ANSWER: 

LCM of 4, 9, 10= 2 × 2 × 3 × 3 × 5= 180


Here, the prime factor 5 does not have a pair. Therefore 180 is multiplied by 5 then the number obtained is a perfect square.

180 × 5 = 900

So, 900 is the smallest square that is divisible by 4, 9, and 10



  1. Find the Pythagorean triplet whose smallest member is 10.

ANSWER: 

Smallest member = 2m = 10

2m = 10

m = 10/2 = 5

m² - 1 = 5² - 1 = 25 - 1 = 24

m² + 1 = 5² + 1 = 25 + 1 = 26

The required triplet is 10 , 24 and 26.


  1. Find the square root by division method 1024. 

ANSWER: 

  1. Simplify :2x²(x + 2) – 3x (x² – 3) – 5x(x + 5)

ANSWER: 


  1.  Add : 2x²y²– 3xy + 4, 5 + 7xy – 3x²y², and 4x²y² + 10xy 

ANSWER: 

 2x²y²– 3xy + 4+ 5 + 7xy – 3x²y²+ 4x²y² + 10xy =3x²y²+14xy+9

  1.  Subtract  3xy + 5yz – 7xz + 1 from -4xy + 2yz – 2xz + 5xyz + 1 

ANSWER: 

=-4xy + 2yz – 2xz + 5xyz + 1 - [ 3xy + 5yz – 7xz + 1]

=-4xy + 2yz – 2xz + 5xyz + 1 - 3xy - 5yz + 7xz - 1

=-7xy-37z+5xz+5xyz

  1. Simplify 7x²(3x – 9) + 3 and find its values for x = 4 and x = 6 

ANSWER: 

Value of polynomial at x= 4 is 339 and 2271 at x=6

  1. Simplify:  (5x – 6) (2 x – 3) + (3 x + 5)²   

ANSWER: 

10x²-27x+18 +9x²+30x+25 =19x²+3x+43

  1. show that

ANSWER: 

ANSWER: 

(i) 1

(ii) (1+1)x2=4

  1. Evaluate 

ANSWER: 

(5²-4³)(-24)

=(25- 64)(-2)4

=-39x16=-624

  1. Simplify

ANSWER: 

(t²+3)(t²-3) =t⁴ - 9


  1. Write the following numbers in usual form:  i) 100 × 7 + 10 × 9 + 1 × 8 ii) 1000 × 3 + 100 × 1 + 10 × 5 + 1 × 9

ANSWER: 

  1. Find m so that (–3)m + 1 × (–3)5 = (–3)7 

ANSWER: 

m+6=7

m=1


  1. Simplify

ANSWER: 

  1. In a two-digit number, the unit's digit is 7 more than the ten’s digit. Sum of the digits is  half of the whole number. Find the digits and number. 

ANSWER: 


  1. The denominator of a fraction is 1 more than twice its numerator. If the numerator and the denominator are both decreased by 1 then the number obtained is 1/3 Find the fraction.

ANSWER: 

Let the numerator be x

The fraction will be = x/(2x+1)

X-1 /2x = ⅓

3x-3=2x

x=3

Thus the required fraction = 3/7


  1. In a two-digit number, unit's digit is 3 more than the ten's digit. The number formed  by interchanging the digits and the original number are in the ratio 7 : 4. Find the  number. 

ANSWER: 

Number formed is = 10x + x + 3 = 11x+3

The number =11×3+x=33+3=36.

ANSWER: 

x+0.25-3x=4.5

-2x=1.5-0.25

x=1.25/2

x=0.625

  1. Divide 64 into two parts such that three times the greater part will be equal to five  times the smaller one. 

ANSWER: 

Let x be the greater part. Then 64 - x is the smaller part.

3x= 5(64-x)

3x = 320 - 5x

8x = 320

X = 320/8

x= 40

Two parts are 40 and 24

  1.  Solve for x,

ANSWER: 

(8x-¾)/ ( 63x +4/7)=¼

7(8x-3)4(63x+4)=14

7(8x-3)4(63x+4)=14x 47

7(8x-3) = 63x +4

56x-21=63x+4

-7x=25

x=-25/7


  1.  If ‘x’ is subtracted from thrice the rational number 3/5 to obtain the value of 1/2, then find the value  of x.

ANSWER: 

  1. Find x in the adjoining figure: 


ANSWER: 

125+125+x=360

x=110

  1.  If the longer side of a trapezium is 12 cm and the distance between the parallel sides is 6  cm. Find the smallest parallel side if the area of figure is 72 cm2. 

ANSWER: 

½6(12+b)=72

3(12+b)=72

12+b=72/3=24

b=12cm


  1. Find the length of the altitude of a rhombus if lengths of its two diagonals are 12cm  and 16cm respectively.

ANSWER: 

AO = AC/2 = 16/2 = 8 cm

BO = 12/2 = 6 cm

AB2 = AO2 + BO2

AB2 = (8)2 + (6)2

AB2 = 64 + 36

AB2 = 100

AB = 10 cm

Therefore, the length of the side of the rhombus is 10 cm.

Area of Rhombus = ½ d1 x d2 = ½ x 12 x 16 = 96 cm²

Area of rhombus = base x height = 10 x height = 96cm²

Height  = 9.6 cm


  1. Find the angles of x, y in the figure given below

ANSWER: 

Sum of all interior angles of a pentagon is 540

2x+100+90+90=540

2x=260

x=130

  1. Find the value of x and y

ANSWER: 

y=90

BAD = 90

x+45=90

x=45

  1.  How many sides does a regular polygon have if the measure of an exterior angle is 450. 

ANSWER: 

=360 / 45 = 8 sides 


  1. If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours? 

ANSWER: 

=48 x 15 / 30 = 24 workers 

  1. A train is moving at a uniform speed of 75 km/hour. (i) How far will it travel in 20 minutes? (ii) Find the time required to cover a distance of 250 km. 

ANSWER: 


  1.  A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?

ANSWER: 

  1.  If 6x pens cost Rs 12x²-36x, find the cost of one pen. 

ANSWER: 

  1. The cost of 6 balls is Rs. 42. What would be the cost of 10 balls, 15 balls and 20 balls?  Write them in the form of a table. 

ANSWER: 

S

  1. Quantities u and v vary directly and when u = 12 and v = 16. Which of the following  is not a possible pair of corresponding values of u and v?  (i) 6 and 8  (ii) 15 and 20 (iii) 18 and 22. 

ANSWER: 

  1.  Factorise   (x2 + y2) (x2– y2)  

ANSWER: 

x4-y4

  1. What values of (x, y) satisfy the given equation x3 − 2x2y + 2xy2 − y2 = 0 

 A. (1, 1) B. (1, 2) C. (2, 1) D. (3, 1) 

ANSWER: 

Substitute the options in the given equation and check whether it is satisfied

1³+(-2.1.1)+2.1.1-1=1-2+2-1=0

Therefore solution(1,1)


  1. Factorise: 5y2 – 20y – 8z + 2yz  

ANSWER: 

(y-4)(5y+2z)

  1. Verify that (5x + 8)2 – 160x = (5x – 8)2  

ANSWER: 

(5x)² + 2(5x)(8) + 64 - 160x


= 25x² + 80x + 64 - 160x


= 25x² - 80x + 64


= (5x - 8)²


  1. Factorise: 25x3y – 81xy 

ANSWER: 

-xy*(5x+9)*(5x-9)

  1. Divide using factor method

ANSWER: 

=(x4-2x²+5x²-10)

=(x²-2)(x²+5) / (x²+5) 

= (x²-2)

  1. Factorise the expressions and divide them as directed. (y2 + 7y + 10) ÷ (y + 5)

ANSWER: 

(y+2)(y+5)/(y+5) = y+2


  1. The edge of a cube is 2 cm. Find the total surface area of the cuboid formed by three  such cubes joined edge to edge. 

ANSWER: 

l=6cm, b= 2cm, h= 2cm

Surface area of cuboid formed =2(lb+bh+hl) 

=2(6x2+2x2+2x6)=2x28

= 56 cm²

  1. The perimeter of the floor of a room is 50 m and its height is 2.5 m. Find the area of  four walls of the room. 

ANSWER: 

Area of  four wall=2hl+2bh = 2(l+b)x h 

= perimeter x h

=50 x 2.5 =125 m²

  1. A village has a population of 5000. It requires 120 liters of water per head per day. It  has a tank measuring 20 m by 20 m by 3 m. For how many days the water of this tank will last? 

ANSWER: 

  1.  A cubical box with lid has a length 45cm find the cost of painting inside and outside of the box at Rs.150 per sq. m. 

ANSWER: 

Area of box = 2 x 6a²

= 2 x 6 x 0.45 x 0.45 = 2.43m²

Cost = 2.43 ×150=Rs.364.5

  1. An aquarium is in the form of a cuboid whose external measures are 80 cm x 30 cm x 40 cm. The base, side faces and back face are to be covered with the coloured paper.  Find the area of paper needed.

ANSWER: 

The area of paper needed = base area + side area + back area 

= 80x30+2x30x40+80x40

= 2400+2400+3200 

=8000cm²

  1. What will be the labor charges for digging a cubical pit of 8 m at the rate of Rs.15 per m3.



ANSWER: 

Edge  = 8 m

Volume of  pit=a³=(8×8×8)=512 m³

Cost of digging per m³ volume = Rs. 15

Cost of digging 512 m³ volume = Rs. (512×15) = Rs.7680

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