Exercise 3.2
1. Fill in the blanks:
(i) [ 18m4
(n8) ] / [ 2m(3)n3
] = 9 mn5 ]
(ii) [l4m5n(7)]
/ [2lm(3)n6] = [l3m2n]
/ 2 ]
(iii) [ 42a4b5 (c2) ] / [ 6(a)4(b)2 ] = (7) b3 c2 ]
2. Say True or False
(i) 8x 3 y ÷ 4x2 =
2xy
(ii) 7ab 3 ÷
14ab = 2b2
3. Divide
(i) 27 y3 by 3y
(ii) x3 y2
by x2y
(iii) 45x3 y2z4 by (−15xyz)
(iv) (3xy)2 by 9xy
Solution:
(i) 27y3 / 3y = (27/3) y3−1 = 9y2
(ii) x3y2 / x2y = x3−2 y2−1 = x1y1
= xy
(iii) [ 45x3 y2z4 ] / [−15xyz] = (45/−15) x3−1
y2−1 y4−1z4−1 = −3x2
yz3
(iv) (3xy)2 / [ 3 × (3xy) ] = [ (3xy)2 ] / [3 × (3xy)] = (1/3) (3xy)2−1
= (1/3) 3xy = xy
4. Simplify
Solution:
(i) [3m2
/ m ] + [2 m4 / m3 ] = 3m2−1 + 2 m4−3 = 3m
+ 2m = (3 + 2) m = 5m
(ii) [ 14p5q3
/ 2p2q ] – [ 12p3q4
/ 3q2 ] = [ (14/2) p5−2q3−1
] – [ 12/3 p3q4−3 ] = 7p3q2
− 4p3q2 = 3p2q2
5. Divide
(i) (32 y2 − 8 yz) by
2y
(ii) (4m2n3
+ 16m4 n2 − mn)
by 2mn
(iii) 5xy2 - 18x2
y3 + 6xy by 6xy
(iv) 81( p4 q2r3 + 2 p3q3 r2
− 5p2 q2r2
) by ( 3pqr)2
Solution:
(i) (32y2 – 8yz) by 2y
[ 32y2 – 8yz ]
/ 2y = [32y2 / 2y ] – [ 8yz / 2y ]
= (32/2) y2−1 – (8/2) y1−1z =
16y – 4z
(ii) (4m2 n3 + 16 m4
n2 − mn) by 2 mn
[ 4m2n3 + 16m4n2
– mn ] / 2mn = [4m2n3 / 2mn
] + [ 16m4n2 / 2mn ] – [ mn /
2mn]
= (4/2)m2−1 n3−1 + (16/2)m4−1 n2−1
– (1/2)m1−1n1−1
= 2m1n2 + 8 m3n1
– (1/2)m0n0
= 2mn2
+ 8m3n – 1/2
(iii) 5xy2 − 18x2y3
+ 6xy by 6xy
[ 5xy2 − 18x2y3
+ 6xy ] / 6xy = [ xy(5y − 18xy2
+ 6) ] / 6xy
= [ 5y − 18xy2
+ 6 ] / 6 = (5/6) y − 3xy2 + 1
(iv) 81 (p4q2r3
+ 2p3q3r2 − 5p2q2r2)
by (3pqr)2
[ 81 (p4q2r3
+ 2p3q3r2 − 5p2q2r2)
] / (3pqr)2
= [ 81 (p2q2r2)
(p2r + 2pq− 5) ] / [ 9(p2q2r2)
]
= 81/9 (p2q2r2)1−1
(p2r + 2pq− 5)
= 9(p2r + 2pq− 5) = 9p2r
+ 18pq − 45
6. Identify the errors and correct them
(i) 7 y 2 − y2
+3y 2 = 10
y2
(ii) 6xy + 3xy = 9x 2 y2
(iii) m(4m − 3) = 4m2 – 3
(iv) (4n)2− 2n + 3 = 4n 2 − 2n + 3
(v) (x − 2)(x + 3) = x2
– 6
(vi) -3p2 + 4p - 7 =
- (3p2 + 4p - 7)
Solution:
(i) 7y2 − y2
+ 3y2 = (7 – 1 +
3) y2 = (6 + 3) y2
= 9y2
(ii) 6xy + 3xy = (6 + 3) xy = 9xy
(iii) m (4m − 3) = m
(4m) + m (−3) = 4m2 − 3m
(iv) (4n)2 − 2n
+ 3 = 16n2 − 2n + 3
(v) (x − 2) (x + 3) = x
(x + 3) − 2 (x + 3)
= x(x) + (x)
× 3 + (−2) (x) + (−2) (3)
= x2 +
3x − 2x − 6 = x2 + x − 6
(vi) − 3p2 + 4p
− 7 = − (3p2 − 4p +7)
7. Statement A: If 24p2q is divided by 3pq, then
the quotient is 8p.
Statement B: Simplification of (5 x + 5) / 5 is 5x
(i) Both
A and B are true
(ii) A is
true but B is false
(iii) A is
false but B is true
(iv) Both
A and B are false
[Answer: (ii) A is true but B is false]
Solution:
[24p2q] / [3pq]
= 8p2 / p = 8p2−1 = 8p
( [5x + 5] / 5 ) = [ 5(x + 1) ] / 5 = x + 1
8. Statement A: 4x 2 + 3x -2 = 2(2x2 +
3x/2 -1)
Statement B: (2m–5)–(5–2m) = (2m–5) + (2m–5)
(i) Both
A and B are true
(ii) A is
true but B is false
(iii) A is
false but B is true
(iv) Both
A and B are false
[Answer: (i) Both A and B are true]
Solution:
(2m − 5) − (5 − 2m) = 2m − 5 − 5 + 2m
= 4m − 10
(2m − 5) + (2m − 5) = 4m − 10
Answer:
Exercise 3.2
1.
2. (i) True (ii) False
3. (i ) 9 y2 (ii) xy (iii ) − 3x2yz3 (iv) x y
4. (i) 5m (ii) 3p3q2
5. (i ) 16 y − 4z (ii) 2 mn2 +
8m3 n – 1/2 (iii ) (5/6)y - 3 xy2 + 1 (iv) = 9 p2r +18pq – 45
6. (i ) 9 y2 (ii) 9 xy (iii ) 4 m2
−3m (iv) 16 n2 − 2n + 3 (v) x2 + 1x − 6 (vi) –(3p2
–4p + 7)
7. (ii) A is true but B
is false
8. (i) both A and B are true
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