CLASS 8 MATHEMATICS SET B ANSWER KEY

INDIAN SCHOOL MUSCAT MIDDLE SECTION SECOND PERIODIC TEST - 2018-19 ANSWER KEY - MATHEMATICS (SET-B) CLASS 8 Q.NO1

SECTION A - FILL IN THE BLANKS 3

3

2 7   /    7   2  and ( x + 5)

(b)

7 Reciprocal of    2  Factors of (x2 + 5x) are

(c)

Standard form of 0.0000579 = 5.79 ×10-5

(d) (e)

The HCF of 3x2y and –15x3y2 is (2-1 × 5-1 × 3-1)0 = 1

(a)

is x

3

3x2y

SECTION B – ‘1’ MARK QUESTIONS

Q.NO2 (a)

Factorise : x(2x – 5) – 7(2x – 5) = (2x – 5) (x – 7)

(b)

Factorise: ( p2 – 49) = (p2 – 72) =(p + 7) (p – 7)

(c)

Write the usual form of 2.12 × 10 – 4 = 0.000212

(d)

Evaluate :

(e)

Divide 44xy2z3 by 11xyz2 = 4yz

2 5  2 13 25( 13) 2 8 1 = 7 = 7 =2-1= 7 2 2 2 2

SECTION - C ( ‘2’ MARK EACH – TOTAL ( 10 MARKS ) )

Q.NO

2 m 6

3

3

m5

2 2 2 Find the value of ‘m’ for         or 7 7 7 Or 2m+9 =m+5 or 2m – m = 5 – 9 or m = - 4

2   7

2 m  63

2   7

m 5

2 or   7

2 m 9

2   7

m 5

2 6 15 6 125 2 6  36  5 6  53 5 63 = = 7 =5-3-(-7) = 54 =625 7 6 7 6 6 5 6 5  2 3 5

4

Evaluate:

5

Factorise and divide : 4xy(9m2 – 6mn + n2) ÷ 2y (3m – n) 4xy(3m – n)2÷2y (3m – n) = 2x (3m – n)

6

Factorise : p2 – 12p – 45. = p2 – 15p + 3p – 45 = p(p – 15) +3 (p – 15) = (p – 15)(p + 3)

7

Factorise completely: (y – 3)2 – (y + 3)2 ={(y – 3) +(y + 3)} {(y – 3) – (y + 3)} =(y – 3 + y + 3) (y – 3 – y – 3) = 2y (– 6) = –12y