Practice MCQ Questions and Answer on Simplification

191.

The value of $$\frac{{\left(x-y\right)}^3+{\left(y-z\right)}^3+{\left(z-x\right)}^3}{{\left(x^2-y^2\right)}^3+{\left(y^2-z^2\right)}^3+{\left(z^2-x^2\right)}^3}$$       is = ?

• (A) 0
• (B) 1
• (C) $${\left[2\left(x+y+z\right)\right]}^{-1}$$
• (D) $${\left[\left(x+y\right)\left(y+z\right)\left(z+x\right)\right]}^{-1}$$    

Show Answer

 $$\eqalign{ & {\text{Since }}\left( {x - y} \right) + \left( {y - z} \right) + \left( {z - x} \right) = 0 \cr & {\text{So,}}{\left( {x - y} \right)^3} + {\left( {y - z} \right)^3} + {\left( {z - x} \right)^3} \cr & = 3\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right) \cr & {\text{Since}}\left( {{x^2} - {y^2}} \right) + \left( {{y^2} - {z^2}} \right) + \left( {{z^2} - {x^2}} \right) = 0 \cr & {\text{So,}}\left( {{x^2} - {y^2}} \right) + \left( {{y^2} - {z^2}} \right) + \left( {{z^2} - {x^2}} \right) \cr & = 3\left( {{x^2} - {y^2}} \right)\left( {{y^2} - {z^2}} \right)\left( {{z^2} - {x^2}} \right) \cr & \therefore {\text{Given expression }} \cr & {\text{ = }}\frac{{3\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)}}{{3\left( {{x^2} - {y^2}} \right)\left( {{y^2} - {z^2}} \right)\left( {{z^2} - {x^2}} \right)}} \cr & = \frac{1}{{\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)}} \cr & = {\left[ {\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)} \right]^{ - 1}} \cr} $$

192.

When $$\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)$$    is divided by $$\left(\frac{2}{5}-\frac{5}{9}+\frac{3}{5}-\frac{7}{18}\right)\text{,}$$     then the result is = ?

• (A) $$\text{5}\frac{1}{10}$$
• (B) $$\text{2}\frac{1}{18}$$
• (C) $$\text{3}\frac{1}{6}$$
• (D) $$\text{3}\frac{3}{10}$$

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 $$\eqalign{ & {\text{According to the question,}} \cr & \frac{{\frac{1}{2} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6}}}{{\frac{2}{5} - \frac{5}{9} + \frac{3}{5} - \frac{7}{{18}}}} \cr & \Rightarrow \frac{{\frac{{30 - 15 + 12 - 10}}{{60}}}}{{\frac{{36 - 50 + 54 - 35}}{{90}}}} \cr & \Rightarrow \frac{{17}}{{60}} \times \frac{{90}}{5} \cr & \Rightarrow 5\frac{1}{{10}} \cr} $$

193.

1559.95 - 7.99 × 24.96 - ? = 11542

• (A) 14
• (B) 24
• (C) 32
• (D) 18

Show Answer

 Let the number be a Given, 1559.95 - 7.99 × 24.96 - a2 = 1154 ⇒ 1560 - 8 × 25 - a2 = 1154 ⇒ 1360 - 200 - a2 = 1154 ⇒ a2 = 1360 - 1154 ⇒ a2 = 206 ⇒ a ≈ $$\sqrt {206} $$ ⇒ a ≈ 14

194.

(71 × 29 + 27 × 15 + 8 × 4) equals = ?

• (A) 3450
• (B) 3458
• (C) 2496
• (D) None of these

Show Answer

 71 × 29 + 27 × 15 + 8 × 4 = 2059 + 405 +32 = 2496

195.

If x is the square of the number when $$\left(\frac{2}{5}\text{ of }6\frac{1}{4}÷\frac{3}{7}\right)$$    of $$1\frac{2}{7}$$ is divided by $$11\frac{1}{4},$$  then the value of 81x is:

• (A) 36
• (B) 16
• (C) 9
• (D) 4

Show Answer

 $$\eqalign{ & \frac{{\left( {\frac{2}{5}{\text{ of }}6\frac{1}{4} \div \frac{3}{7}} \right) \times 1\frac{2}{7}}}{{11\frac{1}{4}}} = \sqrt x \cr & \frac{{\left( {\frac{2}{5}{\text{ of }}\frac{{25}}{4} \div \frac{3}{7}} \right) \times \frac{9}{7}}}{{\frac{{45}}{4}}} = \sqrt x \cr & \frac{{\frac{5}{2} \times \frac{7}{3} \times \frac{9}{7}}}{{\frac{{45}}{4}}} = \sqrt x \cr & \frac{{15}}{2} \times \frac{4}{{45}} = \sqrt x \cr & x = \frac{4}{9} \cr & 81x = 81 \times \frac{4}{9} \cr & 81x = 36 \cr} $$

196.

If 3.352 - (9.759 - x) - 19.64 = 7.052, then what is the value of x?

• (A) -6.181
• (B) 13.581
• (C) 33.099
• (D) 39.803

Show Answer

 3.352 - (9.759 - x) - 19.64 = 7.052 ⇒ 3.352 - 9.759 + x - 19.64 - 7.052 = 0 ⇒ x = 36.451 - 3352 ∴ x = 33.099

197.

If $$a=\frac{x}{x+y}$$   and $$b=\frac{y}{x-y}\text{,}$$   then $$\frac{ab}{a+b}$$   is equal to = ?

• (A) $$\frac{xy}{x^2+y^2}$$
• (B) $$\frac{x^2+y^2}{xy}$$
• (C) $$\frac{x}{x+y}$$
• (D) $${\left(\frac{y}{x+y}\right)}^2$$

Show Answer

 $$\eqalign{ & \frac{{ab}}{{a + b}} = \frac{{\frac{x}{{x + y}} \times \frac{y}{{x - y}}}}{{\frac{x}{{x + y}} + \frac{y}{{x - y}}}} \cr & \frac{{ab}}{{a + b}} = \frac{{\frac{{xy}}{{{x^2} - {y^2}}}}}{{\frac{{x\left( {x - y} \right) + y\left( {x + y} \right)}}{{{x^2} - {y^2}}}}} \cr & \frac{{ab}}{{a + b}} = \frac{{xy}}{{{x^2} + {y^2}}} \cr} $$

198.

Simplify : $$\text{8}\frac{1}{2}-\left[3\frac{1}{4}+\left\{1\frac{1}{4}-\frac{1}{2}\left(1\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\right\}\right]$$

• (A) $$4\frac{1}{2}$$
• (B) $$4\frac{1}{6}$$
• (C) $$9\frac{1}{2}$$
• (D) $$\frac{2}{9}$$

Show Answer

 $$\eqalign{ & {\text{According to the question,}} \cr & {\text{8}}\frac{1}{2} - \left[ {3\frac{1}{4} + \left\{ {1\frac{1}{4} - \frac{1}{2}\left( {1\frac{1}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right] \cr & \Rightarrow \frac{{17}}{2} - \left[ {\frac{{13}}{4} + \left\{ {\frac{5}{4} - \frac{1}{2}\left( {\frac{3}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right] \cr & \Rightarrow \frac{{17}}{2} - \left[ {\frac{{13}}{4} + \left\{ {\frac{5}{4} - \frac{1}{2}\left( {\frac{{9 - 2 - 1}}{6}} \right)} \right\}} \right] \cr & \Rightarrow \frac{{17}}{2} - \left[ {\frac{{13}}{4} + \left\{ {\frac{5}{4} - \frac{1}{2} \times 1} \right\}} \right] \cr & \Rightarrow \frac{{17}}{2} - \left[ {\frac{{13}}{4} + \left\{ {\frac{{5 - 2}}{4}} \right\}} \right] \cr & \Rightarrow \frac{{17}}{2} - \left[ {\frac{{13}}{4} + \frac{3}{4}} \right] \cr & \Rightarrow \frac{{17}}{2} - \frac{{16}}{4} \cr & \Rightarrow \frac{{34 - 16}}{4} \cr & \Rightarrow \frac{9}{2} \cr & \Rightarrow 4\frac{1}{2} \cr} $$

199.

By what least number should 675 be multiplied so as to obtain a perfect cube number ?

• (A) 3
• (B) 5
• (C) 24
• (D) 40

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 $$\eqalign{ & {\text{According to question,}} \cr & {\text{5}}\,\,{\text{| 675}} \cr & - - - - - - \cr & 5\,\,|\,\,135 \cr & - - - - - - \cr & 5\,\,|\,\,27 \cr & - - - - - - \cr & 3\,\,|\,\,9 \cr & - - - - - - \cr & 3\,\,|\,\,3\quad \cr & - - - - - - \cr & \,\,\,\,\,|\,\,1 \cr & {\text{Factors are}} \cr & = \,5 \times 5 \times 3 \times 3 \times 3 \cr} $$ ∴ It must be multiplied by 5 to make a perfect cube.

200.

If 1² + 2² + 3² + . . . . . + p² = $$\left[\frac{\text{p}\left(\text{p}+1\right)\left(2\text{p}+1\right)}{6}\right]\text{,}$$     then 1² + 3² + 5² + . . . . . + 17² is = ?

• (A) 969
• (B) 1785
• (C) 980
• (D) 1700

Show Answer

 12 + 32 + 52 + . . . . . + 172 = (12 + 22 + 32 + 42 + . . . . . + 172) - (22 + 42 + . . . . . + 162) = (12 + 22 + 32 + 42 + . . . . . + 172) - 22(12 + 22 + . . . . . + 82) Using given formula $$\eqalign{ & \left[ {\frac{{p\left( {p + 1} \right)\left( {2p + 1} \right)}}{6}} \right] \cr & = \frac{{17 \times 18 \times 35}}{6} - 4\left[ {\frac{{8 \times 9 \times 17}}{6}} \right] \cr & = 1785 - 4 \times 204 \cr & = 1785 - 816 \cr & = 969 \cr} $$