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# Choose the correct answer:

$(a)\; 2\large\frac{1}{2}$$\; \times\; 8 \large\frac{3}{4} \;$$\;\times\; 9\large\frac{1}{2}\;$$\times;X = 0\times \;\large\frac{1}{2}\;$$\;\times\;\large\frac{125}{1000}\;$$\;\times\; 4\large\frac{3}{4}\; (b)\; 8.9+4.7 = 7.56+1.25+Y (c) Z = \large\frac{1}{5}\;$$\times\;\large\frac{1}{2}$$\;-\;\large\frac{1}{2}\;$$\times\;\large\frac{1}{10}$

(A) X = 1, Y = 4.79, Z = 0.05 (B) X = 0, Y = 8.81, Z = 1.05 (C) X = 0, Y = 4.79, Z = 0.05 (D) X = 1, Y = 8.81, Z = 1.25

$(a)\; 2\large\frac{1}{2}$$\; \times\; 8 \large\frac{3}{4} \;$$\;\times\; 9\large\frac{1}{2}\;$$\times\;X = 0\times \;\large\frac{1}{2}\;$$\;\times\;\large\frac{125}{1000}\;$$\;\times\; 4\large\frac{3}{4}\; The right hand size is a multiple of 0, therefore X must be equal to 0; we don't have to expand the mixed fractions, multiply and divide them to calculate X. (b)\; 8.9+4.7 = 7.56+1.25+Y Y = 8.9 + 4.7 - 7.56 - 1.25 = 13.6 - 8.81 = 4.79 (c) \;Z = \large\frac{1}{5}\;$$\times\;\large\frac{1}{2}$$\;-\;\large\frac{1}{2}\;$$\times\;\large\frac{1}{10}$
$Z = \large\frac{1}{2}$$\times \large($$\large\frac{1}{5}$$- \large\frac{1}{10}) = \large\frac{1}{2}$$ \times \large\frac{2-1}{10}$$= \large\frac{1}{20}$$ = 0.05$