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Mathematics Form 3 Chapter 1 Indices Mathematics Form 3 Chapter 1 Indices
1 – 2 n 5 3
–1 2 2
3 –3
2
(g) (12 × 27) ÷ 8 3 (b) 5 × 5 = 5 × 5 PT3 Standard Practice 1 6. A (3a b ) × 7a b
1 – 2 5 n + 1 = 5 5 + 3 = 9a b × 7a b
–2 4
3 –3
3 2
= (2 × 3 × 3 ) ÷ (2 ) 3 n + 1 = 8 = (9 × 7)a –2 + 3 4 – 3
3
2
b
1 2 (1 + 3) × 1 Section A
2 × – 3 × –
= 2 2 3 × 3 2 n = 7 = 63ab

1 4

= 2 × 3 2 1. m  = m 5 0 3 2 3 6

3
4
5
= 8 × 9 (c) 3 = 9 n 1 B 7c × 7 ÷ 49c × 343c
2
3
1
= 72 3 = 3 2n = m 5 × 4 = 7c × 1 ÷ 7 c × 7c 2
2 2
2
2
3
2 = 2n = m 4 5 = 7 1 – 2 + 1 2 1 – + 2
c
2
12. (a) 2x y × 5xy 2 n = 1 = c
3 2
= 2 × 5x 3 + 1 2 + 2 Answer: C 5 1
y
1
4 4
= 10x y (d) (b ) = b 9 C 1 d 2 – 3 = d 3
n 3
5
5

2
–1 2
–1 2
3n
b = b 9 2. 7 k × 7k 1 3 = 7 k × 7 k 3 1 3
3 2
d
3
(b) 81m n ÷ 27m n 3n = 9 3 2 (7 ) k 3 2 = 
4 2
3 2
5
2 2
4
= 3 m n ÷ 3 m n n = 3 (49k) –1 + 3 – 3 2 + – 3 1
3 2
3
4 2
3
n
= 3 4 − 3 m 3 − 4 2 − 2 = 7 k 2 2 ≠ 5d
3
3 = 7 k
–1 2
= (e) 27(5 ) = 125(3 ) D e × e × e × e × e × e = e 6
n
n
m = k 2
1
  n = 125 7 Answer: C
5
3
27
a
4 –1
(c) (3a b ) ×       3 Answer: A Section B
3 –1 2
n
5
5
b
b
= 9a 3 × 2 – 4 –1 × 2+1 3 = 3 3. 3q × 5q = (3 × 5)q 4 + 5
4
5
= 9a b n = 3 = 15q 9 1. a × a 3 a 6
2 –1
5
a 2
= 9 Answer: A
b
(a ) a 8
3 2
–3k
(hk ) × h k 4. 2 k + 1 × 2 = 32
2 2
3 –3
(d) k + 1 –3k 5
h k 2 = 2 2
4 –4
k
= h 1 × 2 + 3 – 4 2 × 2 – 3 – (–4) 1 – 2k = 5 a –3 a 3
= h 2 +3 – 4 4 – 3 + 4 2k = –4
k
1
= hk 5 k = –2 3  2 —
a
Answer: D a 3
5
2 –3 3
2 4
(2r s ) × r s 3 4
(e) y 6 3r s 
1 5. A = y 6 – 5 4 3
8r s y 5 2. × (16r s )
2 2
2 –6 2
= y 5r s
–3 7
2 3 2 × 3 + – 1
5
7
1
4
= × r 2 2 s –3 × 3 + 4 – 2 B y × y = y 7 + 4 — × 4
2 3 = y 11 3 r 3 × 4 s 2 3 3 3
4 4
5
= r 6 + – 1 2 s –9 + 4 – 2 C y + y 8 = 5r s × 4 2 × 2 × r 2 × 2 × s –6 × 2
–3 7
2
3
8 –7
= r s 3 2
D (y ) = 1 = 81r s 3 r s
11 0
–3 7 × 4
3 –9
5r s
12
13. (a) 2 ÷ 8 = 2 n Answer: B 5 184
2 ÷ 2 = 2 n = 5 r 3 + 3 – (–3) × S 2 – 9 – 7
12
3
2 12 – 3 = 2 n
2 = 2 n 5 184 9 –14
9
n = 9 = 5 r s
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A3
BOOKLET ANS MATH F3.indd 3 03/01/2020 10:20 AM