Page 19 - pdf N.C.E.R.T INTEGERS CH 1

INTEGERS 19

Or, we can do it this way,

(–25) × 37 × 4 = (–25) × 4 × 37 = [(–25) × 4] × 37 = (–100) × 37 = –3700
Which is the easier way?
Obviously the second way is easier because multiplication of (–25) and 4 gives
–100 which is easier to multiply with 37. Note that the second way involves
commutativity and associativity of integers.
So, we find that the commutativity, associativity and distributivity of integers help to
make our calculations simpler. Let us further see how calculations can be made
easier using these properties.
(ii) Find 16 × 12

16 × 12 can be written as 16 × (10 + 2).
16 × 12 = 16 × (10 + 2) = 16 × 10 + 16 × 2 = 160 + 32 = 192
(iii) (–23) × 48 = (–23) × [50 – 2] = (–23) × 50 – (–23) × 2 = (–1150) – (– 46)

= –1104
(iv) (–35) × (–98) = (–35) × [(–100) + 2] = (–35) × (–100) + (–35) × 2
= 3500 + (–70) = 3430

(v) 52 × (– 8) + (–52) × 2
(–52) × 2 can also be written as 52 × (–2).
Therefore, 52 × (– 8) + (–52) × 2 = 52 × (– 8) + 52 × (–2)

= 52 × [(– 8) + (–2)] = 52 × [(–10)] = –520

TRY THESE

Find (ñ 49) × 18; (ñ25) × (ñ31); 70 × (–19) + (–1) × 70 using distributive property.

EXAMPLE 2 Find each of the following products:
(i) (–18) × (–10) × 9 (ii) (–20) × (–2) × (–5) × 7
(iii) (–1) × (–5) × (– 4) × (– 6)

SOLUTION

(i) (–18) × (–10) × 9 = [(–18) × (–10)] × 9 = 180 × 9 = 1620
(ii) (–20) × (–2) × (–5) × 7 = – 20 × (–2 × –5) × 7 = [–20 × 10] × 7 = – 1400
(iii) (–1) × (–5) × (– 4) × (– 6) = [(–1) × (–5)] × [(– 4) × (– 6)] = 5 × 24 = 120

EXAMPLE 3 Verify (–30) × [13 + (–3)] = [(–30) × 13] + [(–30) × (–3)]

SOLUTION (–30) × [13 + (–3)] = (–30) × 10 = –300