Evaluate the following:
(i) 23×60\dfrac{2}{3} \times 6032×60
(ii) 47×280\dfrac{4}{7} \times 28074×280
(iii) 23\dfrac{2}{3}32 of 1491\dfrac{4}{9}194
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⇒2×603⇒2×20⇒40\Rightarrow \dfrac{2 \times 60}{3}\\[1em] \Rightarrow 2 \times 20\\[1em] \Rightarrow 40⇒32×60⇒2×20⇒40
Hence, 23×60=40\dfrac{2}{3} \times 60 = 4032×60=40.
⇒4×2807⇒4×40⇒160\Rightarrow \dfrac{4 \times 280}{7}\\[1em] \Rightarrow 4 \times 40\\[1em] \Rightarrow 160⇒74×280⇒4×40⇒160
Hence, 47×280=160\dfrac{4}{7} \times 280 = 16074×280=160.
⇒23×149⇒23×139⇒2×133×9⇒2627\Rightarrow \dfrac{2}{3} \times 1\dfrac{4}{9}\\[1em] \Rightarrow \dfrac{2}{3} \times \dfrac{13}{9}\\[1em] \Rightarrow \dfrac{2 \times 13}{3 \times 9}\\[1em] \Rightarrow \dfrac{26}{27}⇒32×194⇒32×913⇒3×92×13⇒2726
Hence, 23\dfrac{2}{3}32 of 149=26271\dfrac{4}{9} = \dfrac{26}{27}194=2726.
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(i) What number should be added to 512\dfrac{5}{12}125 to get 2382\dfrac{3}{8}283?
(ii) What number should be subtracted from 5 to get 15131\dfrac{5}{13}1135?
(i) 25×37\dfrac{2}{5} \times \dfrac{3}{7}52×73
(ii) 35×89\dfrac{3}{5} \times \dfrac{8}{9}53×98
(iii) 7×1237 \times 1\dfrac{2}{3}7×132
Find the reciprocal of each of the following fractions:
(i) 913\dfrac{9}{13}139
(ii) 2382\dfrac{3}{8}283
(i) 821÷4\dfrac{8}{21} \div 4218÷4
(ii) 415÷25\dfrac{4}{15} \div \dfrac{2}{5}154÷52
(iii) 8÷568 \div \dfrac{5}{6}8÷65
(iv) 514÷785\dfrac{1}{4} \div \dfrac{7}{8}541÷87
(v) 513÷1195\dfrac{1}{3} \div 1\dfrac{1}{9}531÷191