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Mathematics

Simplify :

(i) 715×56\dfrac{7}{15} \times \dfrac{5}{6}

(ii) −524×625\dfrac{-5}{24} \times \dfrac{6}{25}

(iii) 7−18×−914\dfrac{7}{-18} \times \dfrac{-9}{14}

(iv) −95×−103\dfrac{-9}{5} \times \dfrac{-10}{3}

(v) −28×−87-28 \times \dfrac{-8}{7}

(vi) 8−21×−143\dfrac{8}{-21} \times \dfrac{-14}{3}

Rational Numbers

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Answer

(i) 715×56\dfrac{7}{15} \times \dfrac{5}{6}

We have:

=715×56=73×16=7×13×6=718\phantom{=} \dfrac{7}{15} \times \dfrac{5}{6} \\[1em] = \dfrac{7}{3} \times \dfrac{1}{6} \\[1em] = \dfrac{7 \times 1}{3 \times 6} \\[1em] = \dfrac{7}{18}

Hence the answer is 718\dfrac{7}{18}

(ii) −524×625\dfrac{-5}{24} \times \dfrac{6}{25}

We have:

=−524×625=−124×65=−14×15=−1×14×5=−120\phantom{=} \dfrac{-5}{24} \times \dfrac{6}{25} \\[1em] = \dfrac{-1}{24} \times \dfrac{6}{5} \\[1em] = \dfrac{-1}{4} \times \dfrac{1}{5} \\[1em] = \dfrac{-1 \times 1}{4 \times 5} \\[1em] = \dfrac{-1}{20}

Hence the answer is −120\dfrac{-1}{20}

(iii) 7−18×−914\dfrac{7}{-18} \times \dfrac{-9}{14}

First, express 7−18\dfrac{7}{-18} with a positive denominator: 7×(−1)18×(−1)=−718\dfrac{7 \times (-1)}{18 \times (-1)} = \dfrac{-7}{18}.

We have:

=−718×−914=−118×−92=(−1)×(−9)18×2=(1)×(9)18×2[product of two negative integers is positive]=936=14\phantom{=} \dfrac{-7}{18} \times \dfrac{-9}{14} \\[1em] = \dfrac{-1}{18} \times \dfrac{-9}{2} \\[1em] = \dfrac{(-1) \times (-9)}{18 \times 2} \\[1em] = \dfrac{(1) \times (9)}{18 \times 2} \hspace{2cm}\text{[product of two negative integers is positive]} \\[1em] = \dfrac{9}{36} \\[1em] = \dfrac{1}{4}

Hence the answer is 14\dfrac{1}{4}

(iv) −95×−103\dfrac{-9}{5} \times \dfrac{-10}{3}

We have:

=−95×−103=−35×−101=−31×−21=(−3)×(−2)1×1=3×21×1[product of two negative integers is positive]=61=6\phantom{=} \dfrac{-9}{5} \times \dfrac{-10}{3} \\[1em] = \dfrac{-3}{5} \times \dfrac{-10}{1} \\[1em] = \dfrac{-3}{1} \times \dfrac{-2}{1} \\[1em] = \dfrac{(-3) \times (-2)}{1 \times 1} \\[1em] = \dfrac{3 \times 2}{1 \times 1} \quad \text{[product of two negative integers is positive]} \\[1em] = \dfrac{6}{1} = 6

Hence the answer is 6

(v) −28×−87-28 \times \dfrac{-8}{7}

Express -28 as −281\dfrac{-28}{1}

We have:

=−281×−87=−41×−81=(−4)×(−8)1×1=(4)×(8)1×1[product of two negative integers is positive]=321=32\phantom{=} \dfrac{-28}{1} \times \dfrac{-8}{7} \\[1em] = \dfrac{-4}{1} \times \dfrac{-8}{1} \\[1em] = \dfrac{(-4) \times (-8)}{1 \times 1} \\[1em] = \dfrac{(4) \times (8)}{1 \times 1} \quad \text{[product of two negative integers is positive]} \\[1em] = \dfrac{32}{1} = 32

Hence the answer is 32

(vi) 8−21×−143\dfrac{8}{-21} \times \dfrac{-14}{3}

First, express 8−21\dfrac{8}{-21} with a positive denominator: 8×(−1)−21×(−1)=−821\dfrac{8 \times (-1)}{-21 \times (-1)} = \dfrac{-8}{21}.

We have:

=−821×−143=−83×−23=(−8)×(−2)3×3=8×23×3[product of two negative integers is positive]=169\phantom{=} \dfrac{-8}{21} \times \dfrac{-14}{3} \\[1em] = \dfrac{-8}{3} \times \dfrac{-2}{3} \\[1em] = \dfrac{(-8) \times (-2)}{3 \times 3} \\[1em] = \dfrac{8 \times 2}{3 \times 3} \quad \text{[product of two negative integers is positive]} \\[1em] = \dfrac{16}{9}

Hence the answer is 169\dfrac{16}{9}

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