Practice MCQ Questions and Answer on Simplification

271.

$$\frac{3}{2}\times \frac{11}{5}÷\left(\frac{25}{44}\times \frac{11}{5}\right)÷\frac{33}{15}=?$$

• (A) $$\frac{1}{2}$$
• (B) $$\frac{2}{3}$$
• (C) $$\frac{126}{125}$$
• (D) $$5\frac{101}{125}$$

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Answer & Solution Answer: Option E Solution: $$\eqalign{ & \frac{3}{2} \times \frac{{11}}{5} \div \left( {\frac{{25}}{{44}} \times \frac{{11}}{5}} \right) \div \frac{{33}}{{15}} \cr & = \frac{3}{2} \times \frac{{11}}{5} \div \frac{5}{4} \div \frac{{33}}{{15}} \cr & = \frac{3}{2} \times \frac{{11}}{5} \times \frac{4}{5} \times \frac{{15}}{{33}} \cr & = \frac{6}{5} \cr & = 1\frac{1}{5} \cr} $$

272.

$$\left[2\sqrt{54}-6\sqrt{\frac{2}{3}}-\sqrt{96}\right]$$     is equal to = ?

• (A) 0
• (B) 1
• (C) 2
• (D) $$\sqrt{6}$$

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 $$\eqalign{ & {\text{According to question,}} \cr & 2\sqrt {54} - 6\sqrt {\frac{2}{3}} - \sqrt {96} \cr & = \,6\sqrt 6 - 2\sqrt {\frac{2}{3} \times 9} - 4\sqrt 6 \cr & = 6\sqrt 6 - 2\sqrt 6 - 4\sqrt 6 \cr & = 0 \cr} $$

273.

Number of digits in the square root of 62478078 is = ?

• (A) 4
• (B) 5
• (C) 6
• (D) 3

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 Number of digits in 62478078 = 8 ∴ Number of digits in its square root = 4

274.

The cost of 5 pendants and 8 chains is Rs. 145785. What would be the cost of 15 pendants and 24 chains ?

• (A) Rs. 325285
• (B) Rs. 439355
• (C) Rs. 550000
• (D) Cannot be determined

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Answer & Solution Answer: Option E Solution: $$\eqalign{ & {\text{5P + 8C = 145785}} \cr & \Rightarrow 3(5{\text{P + 8C) = 3}} \times {\text{145785}} \cr & \Rightarrow 15{\text{P + 24C = 437355}} \cr} $$

275.

If a - b = 3 and a² + b² = 29, find the value of ab.

• (A) 10
• (B) 12
• (C) 15
• (D) 18

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 2ab = (a2 + b2) - (a - b)2 2ab = 29 - 9 2ab = 20 ∴ ab = 10

276.

$$\sqrt{0.00060516}$$     is equal to = ?

• (A) 0.0246
• (B) 0.00246
• (C) 0.246
• (D) 0.000246

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 $$\eqalign{ & {\text{According to the question,}} \cr & \sqrt {0.00060516} {\text{ }} \cr & \Rightarrow \sqrt {\frac{{60516}}{{100000000}}} \cr & \Rightarrow \frac{{246}}{{10000}} \cr & \Rightarrow 0.0246 \cr} $$

277.

Simplify : $$\left[\left(1+\frac{1}{10+\frac{1}{10}}\right)\times \left(1+\frac{1}{10+\frac{1}{10}}\right)-\left(1-\frac{1}{10+\frac{1}{10}}\right)\times \left(1-\frac{1}{10+\frac{1}{10}}\right)\right]÷\left[\left(1+\frac{1}{10+\frac{1}{10}}\right)+\left(1-\frac{1}{10+\frac{1}{10}}\right)\right]=?$$

• (A) $$\frac{100}{101}$$
• (B) $$\frac{90}{101}$$
• (C) $$\frac{20}{101}$$
• (D) $$\frac{101}{100}$$

Show Answer

 $$\eqalign{ & {\text{According to question,}} \cr & \frac{{\,\left[ {\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) \times \left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) - \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right) \times \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)} \right]}}{{\,\left[ {\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) + \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)} \right]}} \cr & {\text{let ,}}1 + \frac{1}{{10 + \frac{1}{{10}}}} = \frac{{111}}{{101}} = a \cr & \,\,\,\,\,\,\,\,1 - \frac{1}{{10 + \frac{1}{{10}}}} = \frac{{91}}{{101}} = b \cr & \Rightarrow \frac{{{a^2} - {b^2}}}{{a + b}}\left[ {\because {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right] \cr & \Rightarrow \frac{{\left( {a + b} \right)\left( {a - b} \right)}}{{a + b}} \cr & \Rightarrow \left( {a - b} \right) \cr & \Rightarrow \frac{{111}}{{101}} - \frac{{91}}{{101}} \cr & \Rightarrow \frac{{20}}{{101}} \cr} $$

278.

If $$\left(x+\frac{1}{x}\right)\text{ = 2,}$$    then $$\left(x-\frac{1}{x}\right)$$   is equal to = ?

• (A) 0
• (B) 1
• (C) 2
• (D) 5

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 $$\eqalign{ & \left( {x + \frac{1}{x}} \right){\text{ = 2}} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2}{\text{ = }}{{\text{2}}^2} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 4 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2.x.\frac{1}{x} \cr & \,\,\,\,\,\,\,\, = 2 - 2 = 0 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2}{\text{ = 0}} \cr & \Rightarrow x - \frac{1}{x} = 0 \cr} $$

279.

The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Rs. 4000. The total price of 12 chairs and 3 tables is:

• (A) Rs. 3500
• (B) Rs. 3750
• (C) Rs. 3840
• (D) Rs. 3900

Show Answer

 $$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{cost}}\,{\text{of}}\,{\text{a}}\,{\text{chair}}\,{\text{and}}\,{\text{that}}\,{\text{of}}\,\,{\text{table}} \cr & \,{\text{be}}\,Rs.\,x\,{\text{and}}\,Rs.\,y\,{\text{respectively}}. \cr & {\text{Then}},\,10x = 4y\,\,\,or\,\,\,y = \frac{5}{2}x \cr & \therefore 15x + 2y = 4000 \cr & \Rightarrow 15x + 2 \times \frac{5}{2}x = 4000 \cr & \Rightarrow 20x = 4000 \cr & \therefore x = 200 \cr & {\text{So}},\,y = {\frac{5}{2} \times 200} = 500 \cr & {\text{Hence,}}\,{\text{the}}\,{\text{cost}}\,{\text{of}}\,{\text{12}}\,{\text{chairs}}\,{\text{and}}\,{\text{3}}\,{\text{tables}} \cr & = 12x + 3y \cr & = Rs.\,\left( {2400 + 1500} \right) \cr & = Rs.\,3900 \cr} $$