Practice MCQ Questions and Answer on Simplification

261.

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

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

262.

If $$\frac{a}{{b + c}}{\text{ = }}\frac{b}{{c + a}}$$   = $$\frac{c}{{a + b}}{\text{ = k,}}$$    then find the value of k is

• (A) -$$ \frac{1}{2}$$
• (B) 1
• (C) -1
• (D) $$\frac{1}{2}$$

Show Answer

 $$\eqalign{ & \frac{a}{{b + c}} = k \cr & \Rightarrow a = k\left( {b + c} \right)\,....(i) \cr & \frac{b}{{c + a}} = k \cr & \Rightarrow b = k\left( {c + a} \right)\,....(ii) \cr & \frac{c}{{a + b}} = k \cr & \Rightarrow c = k\left( {a + b} \right)\,....(iii) \cr & {\text{Adding (i), (ii) and (iii) we get,}} \cr & a + b + c \cr & = k\left( {2a + 2b + 2c} \right) \cr & \Rightarrow k = \frac{{a + b + c}}{{2(a + b + c)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr} $$

263.

The smallest fraction which should be subtracted from the sum of $${\text{1}}\frac{3}{4}$$,  $${\text{2}}\frac{1}{2}$$,  $$5\frac{7}{{12}}$$,  $${\text{3}}\frac{1}{3}$$  and  $${\text{2}}\frac{1}{4}$$  to make the result a whole number is = ?

• (A) $$\frac{5}{{12}}$$
• (B) $$\frac{7}{{12}}$$
• (C) $$\frac{1}{2}$$
• (D) 7

Show Answer

 $$\eqalign{ & {\text{Sum of given fractions}} \cr & {\text{ = }}\frac{7}{4} + \frac{5}{2} + \frac{{67}}{{12}} + \frac{{10}}{3} + \frac{9}{4} \cr & = \left( {\frac{{21 + 30 + 67 + 40 + 24}}{{12}}} \right) \cr & = \frac{{185}}{{12}} \cr & {\text{The whole number just }} \cr & {\text{less than }}\frac{{185}}{{12}}{\text{ is 15}} \cr & {\text{let }}\frac{{185}}{{12}} - x = 15 \cr & {\text{Then,}} \cr & x = \left( {\frac{{185}}{{12}} - 15} \right) \cr & \,\,\,\,\,\, = \frac{5}{{12}} \cr} $$

264.

$$1888 \div 32 \div 8 = ?$$

• (A) 7.375
• (B) 9.485
• (C) 29.5
• (D) 472

Show Answer

 $$\eqalign{ & 1888 \div 32 \div 8 \cr & = \frac{{1888}}{{32}} \div 8 \cr & = 59 \div 8 \cr & = \frac{{59}}{8} \cr & = 7.375 \cr} $$

265.

A fires 5 shots to B's 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:

• (A) 30 birds
• (B) 60 birds
• (C) 72 birds
• (D) 90 birds

Show Answer

 $$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{total}}\,{\text{number}}\,{\text{of}}\,{\text{shots}}\,{\text{be}}\,x. \cr & {\text{Then,}}\,{\text{Shots}}\,{\text{fired}}\,{\text{by}}\,{\text{A}} = \frac{5}{8}x \cr & {\text{Shots}}\,{\text{fired}}\,{\text{by}}\,{\text{B}} = \frac{3}{8}x \cr & {\text{Killing}}\,{\text{shots}}\,{\text{by}}\,{\text{A}} \cr & = \frac{1}{3}{\text{of}}\frac{5}{8}x = \frac{5}{{24}}x \cr & {\text{Shots}}\,{\text{missed}}\,{\text{by}}\,{\text{B}} \cr & = \frac{1}{2}{\text{of}}\,\frac{3}{8}x = \frac{3}{{16}}x \cr & \therefore \frac{{3x}}{{16}} = 27\,{\text{or}}\,x = {\frac{{27 \times 16}}{3}} = 144 \cr & {\text{Birds}}\,{\text{killed}}\,{\text{by}}\,{\text{A}} \cr & = \frac{{5x}}{{24}} = {\frac{5}{{24}} \times 144} = 30 \cr} $$

266.

Simplify the following $$\frac{{0.75}}{{10}} + \frac{{0.31}}{{100}} + \frac{{0.09}}{{1000}}$$

• (A) 0.03215
• (B) 0.05621
• (C) 0.07819
• (D) 0.08561

Show Answer

 $$\eqalign{ & \frac{{0.75}}{{10}} + \frac{{0.31}}{{100}} + \frac{{0.09}}{{1000}} \cr & = \frac{{75}}{{1000}} + \frac{{3.1}}{{1000}} + \frac{{0.09}}{{1000}} \cr & = \frac{{78.19}}{{10000}} \cr & = 0.07819 \cr} $$

267.

The sum of the first 35 terms of the series $$\frac{1}{2}$$ $$ + $$ $$\frac{1}{3}$$ $$ - $$ $$\frac{1}{4}$$ $$ - $$ $$\frac{1}{2}$$ $$ - $$ $$\frac{1}{3}$$ $$ + $$ $$\frac{1}{4}$$ $$ + $$ $$\frac{1}{2}$$ $$ + $$ $$\frac{1}{3}$$ $$ - $$ $$\frac{1}{4}$$ . . . . . is = ?

• (A) $$ - \frac{1}{2}$$
• (B) $$ - \frac{1}{2}$$
• (C) $$\frac{1}{4}$$
• (D) None of these

Show Answer

 Clearly , sum of first 6 terms is zero , So sum of first 30 terms = 0 ∴ Required sum $$\eqalign{ & {\text{ = }}\left( {\frac{1}{2} + \frac{1}{3} - \frac{1}{4} - \frac{1}{2} - \frac{1}{3}} \right) \cr & = - \frac{1}{4} \cr} $$

268.

The value of $$0.\overline {45} \times 1.\overline {22} $$   is:

• (A) $$0.\overline 3 $$
• (B) $$0.\overline 4 $$
• (C) $$0.\overline 6 $$
• (D) $$0.\overline 5 $$

Show Answer

 $$\eqalign{ & 0.\overline {45} \times 1.\overline {22} \cr & = \frac{{45}}{{99}} \times \frac{{122 - 1}}{{99}} \cr & = \frac{{45}}{{99}} \times \frac{{121}}{{99}} \cr & = \frac{5}{9} \cr & = 0.\overline 5 {\text{ Answer}} \cr} $$

269.

If $$\sqrt {{\text{4096}}} $$  = 64, then the value of $$\sqrt {{\text{40}}{\text{.96}}} $$   $$ + $$ $$\sqrt {{\text{0}}{\text{.4096}}} $$   $$ + $$ $$\sqrt {{\text{0}}{\text{.004096}}} $$    $$ + $$ $$\sqrt {{\text{0}}{\text{.00004096}}} $$     up to two place of decimals is = ?

• (A) 7.09
• (B) 7.10
• (C) 7.11
• (D) 7.12

Show Answer

 $$\eqalign{ & {\text{According to question,}} \cr & \sqrt {40.96} {\text{ + }}\sqrt {0.4096} {\text{ + }}\sqrt {0.004096} {\text{ + }}\sqrt {0.00004096} \cr & \Rightarrow \sqrt {\frac{{4096}}{{100}}} + \sqrt {\frac{{4096}}{{10000}}} + \sqrt {\frac{{4096}}{{1000000}}} + \sqrt {\frac{{4096}}{{100000000}}} \cr & \Rightarrow \frac{{64}}{{10}} + \frac{{64}}{{100}} + \frac{{64}}{{1000}} + \frac{{64}}{{10000}} \cr & \Rightarrow 6.4 + 0.64 + 0.064 + 0.0064 \cr & \Rightarrow 7.11 \cr} $$

270.

If $$\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1$$     and $$\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0$$     where a, b, c, p, q, r are non-zero real numbers, then $$\frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}}$$    is equal to = ?

• (A) 0
• (B) 1
• (C) 3
• (D) 9

Show Answer

 $$\eqalign{ & \frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0\,\,\,\, \cr & \Rightarrow aqr + bpr + cpq = 0....({\text{i}}) \cr & \frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1\,\, \cr & \Rightarrow {\left( {\frac{p}{a} + \frac{q}{b} + \frac{r}{c}} \right)^2} = 1 \cr & \Rightarrow {\text{ }}\frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}}\, + 2\left( {\frac{{pq}}{{ab}} + \frac{{pr}}{{ac}} + \frac{{qr}}{{bc}}} \right) = 1 \cr & \Rightarrow \frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}} + \frac{{2\left( {pqc + prb + qra} \right)}}{{abc}} = 1 \cr & \Rightarrow \frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}} = 1....\left[ {{\text{using (i)}}} \right] \cr} $$