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# Q.5 Find: (i) 2.5 ×0.3 (ii) 0.1 × 51.7 (iii) 0.2 × 316.8 (iv) 1.3 × 3.1 (v) 0.5 × 0.05 (vi) 11.2 × 0.15 (vii) 1.07 × 0.02 (viii) 10.05 × 1.05 (ix) 100.01 × 1.1 (x) 100.01 × 1.1

(i) 0 2.5 × 0.3
$$\therefore$$25 × 3 = 75 and there are 2 digits (1 + 1) right to the decimal points in 2.5 and 0.3.
Thus 2.5 × 0.3 = 0.75
(ii) 0.1 × 51.7
$$\therefore$$ 1 × 517 = 517 and there are two digits (1 + 1) right to the decimal places in 0.1 and 51.7.
(iii) 0.2 × 316.8
$$\therefore$$ 2 × 3168 = 6336 and there are 2 digits (1 + 1) right to the decimal points in 0.2 and 316.8.
Thus 0.2 × 316.8 = 63.36.

(iv) 1.3 × 3.1
$$\therefore$$ 13 × 31 = 403 and there are 2 digits (1 + 1) right to the decimal points in 1.3 and 3.1.
Thus 1.3 × 3.1 – 4.03
(v) 0.5 × 0.05
$$\therefore$$ 5 × 5 = 25 and there are 3 digits (1 + 2) right to the decimal points in 0.5 and 0.05.
Thus 0.5 × 0.05 = 0.025
(vi) 11.2 × 0.15
$$\therefore$$ 112 × 15 = 1680 and there are 3 digits (1 + 2) right to the decimal points in 11.2 and 0.15.
Thus 11.2 × 0.15 = 1.680

(vii) 1.07 × 0.02
$$\therefore$$ 107 × 2 = 214 and there are 4-digits (2 + 2) right to the decimal places is 1.07 × 0.02.
Thus 1.07 × 0.02 = 0.0214
(viii) 10.05 × 1.05
$$\therefore$$ 1005 × 105 = 105525 and there are 4 digits (2 + 2) right to the decimal places in 10.05 × 1.05.
Thus 10.05 × 1.05 = 10.5525
(ix) 101.01 × 0.01
$$\therefore$$ 10101 × 1 = 10101 and there are 4 digits (2 + 2) right to the decimal places in 101.01 and 0.01.
Thus 101.01 × 0.01 = 1.0101
(x) 100.01 × 1.1
$$\therefore$$ 10001 × 11 = 110011 and there are 3 digits (2 + 1) right to the decimal points in 100.01 and 1.1.
Thus 100.01 × 1.1 = 110.011.