Answer key SESSION ENDING EXAMINATION Practice Exam -1

 Answer key SESSION ENDING EXAMINATION   Practice Exam -1

Grade: VIII                 Subject: Mathematics      Max. Marks: 80    Duration: 180 +15 Minutes

SECTION-A (1 x 20 = 20)

1.     (ab)10

2. (b)   3−8

3. b) 236 cm2

4.  (c) 96 cm2          

5.(d) one third

6.(b) m4n3p2

7.  (c) n²

8.(b) OK

9.  (c) 8x, 8(x +1), 8(x + 2) 

10. (a) x = 0

11.41cm²

12.Yes, 225 is a perfect square number since the square root of 225 is 15, which is a whole number.

13. 12mn²p

14.1/16

15. 66

16. Rs84

17. 750 sheets

18. 5/2 [q(p-q)]

19. (x-13)(x-1)

20. 4x

SECTION-B (2x8=16 marks)

21.  Subtract 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q 

18−3p+11q+5pq−2pq²+5p2q−(4p²q−3pq+5pq²−8p−7q−10)

=18−3p+11q+5pq−2pq²+5p²q−4p²q+3pq−5pq²+8p+7q+10

=28+5p+18q+8pq−7pq²+p²q

22.  Find the number of sides of a regular polygon whose each exterior angle has a measure of 45°.

Exterior angle = 360 / number of sides

45 = 360/n

Number of sides = 360/45 = 8

23. One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number? 

Let the digit in units place =xThe digit in tens place =3x

The number is 10(3x)+x---(1)

Digits are interchangedDigit in units place =3x

Digit in tens place =x

The new number =10(x)+3x ---(2)

sum of these two numbers =88

10(3x)+x+10(x)+3x=88

30x+x+10x+3x=88

44x=88⇒x=88/44⇒x=2

from (1)The number =10(3x)+x=10(3×2)+2=60+2=62

Therefore, the required numbers are 62 or 26

24.  Write a Pythagorean triplets using when the smallest member is18 . 

2m=18

M=9

Second number = m²-1 = 81-1=80

Third number =m² +1 = 81+1=82

Pythagorean Triplet = 18,80,82

25.  The length of the fence of a trapezium shaped field ABCD  is 130 m and side AB is Perpendicular to each of the parallel sides AD and BC. If  BC = 54 m, CD =19 m and AD = 42 m , find the area of the field. 

The perimeter of field is 130⟹AB+BC+CD+AD=130

AB=130−54−19−42

AB=15m

Area is given as 1/2AB(AD+BC)

=1/2×15(54+42)

=15×48=720m²

26. (a) Express Mass of Uranus = 86,800,000,000,000,000,000,000,000 kg in standard form. 

8.68 x 1025

(b) Find the value of m for which 5m÷5– 3 = 125. 

m+3=3 

m=0

27. 6 pipes are required to fill a tank in 1 hour 20 minutes. How long will it take if only 5 pipes of the same type are used? 

inverse proportion.

80×6=x×5⇒(80×6) / 5=x⇒x=96

The time taken to fill the tank by 5 pipes is 96 minutes or 1 hour 36 minutes.

28. Factorise: x4 – (y + z)4 

x4−(y+z)4=(x²)²−{(y+z)²}²

=(x²+(y+z)²)(x²−(y+z)²)

=(x²+(y+z)²)(x+y+z)(x−y−z)

=(x²+y²+z²+2yz)(x+y+z)(x−y-z)

=1×(x²+y²+z²+2yz)×(x+y+z)×(x−y−z)

SECTION-C (3x8=24 marks)

29. The diagonals of a quadrilateral are of lengths 6 cm and 8 cm. If the diagonals bisect each  other at right angles, what is the length of each side of the quadrilateral?

OA = 3 cm

OB = 4 cm

In a right triangle AOB , AB² =  OA² + OB² = 3² + 4² = 25

Each side AB = 5 cm

30. Solve:

3(7y+4)=−4(y+2)

⇒21y+12=−4y−8

⇒21y+4y=−8−12

⇒25y=−20

⇒y=−20/25

⇒y=−4/5

31. There are 2401 students in a school. P.T. teacher wants them to stand in rows & columns such that the   no. of rows to equal to the no. of columns. Find the number of rows. 

Let the number of rows be x

No of columns = x

Number of students = x² = 2401

X = 49

The no.of rows = 49

32. Simplify: (a) Add: p ( p – q), q ( q – r) and r ( r – p) 

p(p-q) + q(q-r)+r(r-p)

=(p²-pq)+(q²-qr)+(r²-pr)

=p²+q²+r²-pq-qr-pr

                      (b) Subtract 5x2 – 4y2 + 6y – 3 from 7x2 – 4xy + 8y2 + 5x –3y.

= 7x2 – 4xy + 8y2 + 5x –3y - (5x2 – 4y2 + 6y – 3)

= 7x2 – 4xy + 8y2 + 5x –3y - 5x2 + 4y2 - 6y + 3

=  2x2 – 4xy + 12y2 + 5x –9y +3

33.In a building there are 24 cylindrical pillars. The radius of each pillar is 28 cm and height is 4 m. Find the total cost of painting the curved surface area of all pillars at the rate of Rs 8 per m2. 

CSA of cylinder = 2rh = 2x22/7x0.28x4=7.04 m²

24 such pillars = 24 x 7.04 = 168.96 m²

Total cost of painting = 168.96 x 8 = Rs. 1351.68

34. Evaluate :  

2115-2115  .= 0

  (ii)

(1+8+27)-5/2 = (36)-5/2 =( 6²)-5/2 =65

35. Factorise: (a) m4– 256  = (m-4)(m+4)(m²+16)     (b) x2+ xy + 8x + 8y = (x+y)(x+8)

36. An electric pole, 14 metres high, casts a shadow of 10 metres. Find the height of a tree that casts a shadow of 15 metres under similar conditions. 

It is in direct proportion 

x=14x15/10=21

The height of tree is 21m

SECTION-D  (5x4=20 marks)

37. (i)  Find the least number which must be subtracted from 18265 to make it a perfect square. Also,  find the square root of the resulting number.

Hence the least number which must be subtracted from 18265 to make it a perfect square is 40.

OR

Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.

8=2x2x2

15=3x5

20=2x2x5

LCM = 2x2x2x3x5=120

Hence the required smallest square number is = 120x30=3600    (2 marks)

 (ii)Simplify: (i) (5x – 6) (2 x – 3) + (3 x + 5)2 

((5x - 6) • (2x - 3)) +  2 • (3x + 5)

10x^2-21x+28

          (ii ) (2x + 5y) (2 x + 3y)

\(4x^{2}+16xy+15y^{2}\)

 (3 marks)

38. (i)Find area of the below figure:(2 marks)

Area of the triangle ABC =½ x 7x5  17.5sq.m

Area of the triangle FED =½ x 7x5  17.5sq.m

Rectangle ACDF = 7x8=56 sq.m

Area of the figure = 17.5+17.5+56 = 91 sq.m

(ii)  .A well with 10m inside diameter is dug 14m deep. Earth taken out of it is spread all around to a width of 5m to form embankment. Find height embankment.

Well with 10m diameter is dug 14m deep.⇒ Cylindrical radius =10/2=5m

Volume =πr²h=22/7×5×5×14=1100 ⇒R1=5+5=10m

r=5m

⇒V=π(R2−r2)h

1100=22/7(10²−5²)h

350=(100−25)h

h=350/75=14/3

⇒h=4.66m

Hence,  height embankment is 4.66m

 (3 marks)

39. (i) Solve:

12x² +30x +14x+35 = 12x² + 8x + 39x+26

12x² +44x+35 = 12x² + 47x+26

47x-44x=35-26

3x=9

x=3

(2 marks)

(ii)  (a) If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4? 

No. of students =24

no. of sweets each gets =5

therefore, total no. of sweets 24×5=120

the no. of children reduced by 4 =24−4=20

Hence,each child will get 120/20=6 sweets.

(b) 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? 

20x6 = 30 x

120=30x

x=120/30=4

The food will last for 4 days.

(3 marks)

40.(i)  Factorise then divide 156(36y²-64)y³104(6y+8)y²

(3 marks)

(ii) Find the value of x : ( 2 marks)