Simplification — Part 1 MCQs with Explanations

Let number of notes of each denomination be x . Then x + 5 x + 10 x = 480 ⇒ 16 x = 480 ∴ x = 30. Hence, total number of notes = 3 x = 90.

Let the number of students in rooms A and B be x and y respectively. Then, x - 10 = y + 10 ⇒ x - y = 20 . . . . . (i) and x + 20 = 2(y - 20) ⇒ x - 2y = -60 . . . . . (ii) Solving (i) and (ii) we get: x = 100, y = 80 ∴ The required answer A = 100

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

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

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{price}}\,{\text{of a}}\,{\text{saree}}\,{\text{and}}\,{\text{a}}\,{\text{shirt}} \cr & \,{\text{be}}\,Rs.\,x\,{\text{and}}\,Rs.\,y\,{\text{respectively}}. \cr & {\text{Then,}}\,2x + 4y = 1600\,....(i) \cr & \,\,\,\,\,\,\,\,\,\,\,{\text{and}}\,x + 6y = 1600\,....(ii) \cr & {\text{Divide}}\,{\text{equation}}\,(i)\,by\,2, \cr & {\text{we}}\,{\text{get}}\,{\text{the}}\,{\text{below}}\,{\text{equation}}. \cr & = > \,x + 2y = 800\,\, - - - (iii) \cr & {\text{Now}}\,{\text{subtract}}\,(iii)\,{\text{from}}\,(ii) \cr & x\,\, + \,\,\,\,6y\,\, = \,\,1600\,\,\,( - ) \cr & x\,\, + \,\,\,\,2y\,\, = \,\,\,\,\,\,\,800 \cr & - - - - - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4y\,\, = \,\,\,\,\,\,800 \cr & - - - - - - - - - - \cr & {\text{Therefore}},\,y = 200 \cr & {\text{Nowapply}}\,{\text{value}}\,{\text{of}}\,y\,{\text{in}}\,(iii) \cr & = > x + 2 \times 200 = 800 \cr & = > x + 400 = 800 \cr & {\text{Therefore}}\,x = 400 \cr & {\text{Solving}}\,(i)\,{\text{and}}\,(ii)\,{\text{we}}\,{\text{get}}\, \cr & x = 400,\,y = 200 \cr & \therefore {\text{Cost}}\,{\text{of}}\,{\text{12}}\,{\text{shirts}} \cr & = Rs.\,\left( {12 \times 200} \right) \cr & = Rs.\,2400 \cr} $$

$$\eqalign{ & {\text{Let}}\,C's\,{\text{share}} = Rs.\,x \cr & \text{Then,}\,B's\,{\text{share}} = \,Rs.\,\frac{x}{4}, \cr & A's\,{\text{share}} = Rs.\,\left( {\frac{2}{3} \times \frac{x}{4}} \right) = Rs.\,\frac{x}{6} \cr & \therefore \frac{x}{6} + \frac{x}{4} + x = 1360 \cr & \Rightarrow \frac{{17x}}{{12}} = 1360 \cr & \Rightarrow x = \frac{{1360 \times 12}}{{17}} = Rs.\,960 \cr & {\text{Hence}},\,B's\,{\text{share}} = Rs.\,\left( {\frac{{960}}{4}} \right) \cr & = Rs.\,240 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{saving}}\,{\text{in}}\,{\text{N}}{\text{.S}}{\text{.C}}\,{\text{and}}\,{\text{P}}{\text{.P}}{\text{.F}}{\text{.}}\,{\text{be}}\, \cr & Rs.\,x\,{\text{and}}\,Rs.\,\left( {150000 - x} \right){\text{respectively}}. \cr & {\text{Then}},\,\frac{1}{3}x = \frac{1}{2}\left( {150000 - x} \right) \cr & \Rightarrow \frac{x}{3} + \frac{x}{2} = 75000 \cr & \Rightarrow \frac{{5x}}{6} = 75000 \cr & \Rightarrow x = \frac{{75000 \times 6}}{5} = 90000 \cr & \therefore {\text{Savings}}\,{\text{in}}\,{\text{Public}}\,{\text{Provident}}\,{\text{Fund}} \cr & = Rs.\,\left( {150000 - 90000} \right) \cr & = Rs.\,60000 \cr} $$

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

$$\eqalign{ & {\text{Original}}\,{\text{share}}\,{\text{of}}\,{\text{1}}\,{\text{person}} = \frac{1}{8} \cr & {\text{New}}\,{\text{share}}\,{\text{of}}\,{\text{1}}\,{\text{person}} = \frac{1}{7} \cr & {\text{Increase}} = {\frac{1}{7} \times \frac{1}{8}} = \frac{1}{{56}} \cr & \therefore {\text{Required}}\,{\text{fraction}} \cr & = \frac{{ {1/56} }}{{ {1/8} }} = {\frac{1}{{56}} \times \frac{8}{1}} = \frac{1}{7} \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{capacity}}\,{\text{of}}\,{\text{1}}\,{\text{bucket}} = x \cr & {\text{Then,}}\,{\text{the}}\,{\text{capacity}}\,{\text{of}}\,{\text{tank}} = 25x \cr & {\text{New}}\,{\text{capacity}}\,{\text{of}}\,{\text{bucket}} = \frac{2}{5}x \cr & \therefore {\text{Required}}\,{\text{number}}\,{\text{of}}\,{\text{buckets}} \cr & = \frac{{25x}}{{\left( {2x/5} \right)}} \cr & = {25x \times \frac{5}{{2x}}} \cr & = \frac{{125}}{2} \cr & = 62.5 \cr} $$

Suppose the man works overtime for x hours. Now, working hours in 4 weeks = (5 * 8 * 4) = 160. ∴ 160 * 2.40 + x * 3.20 = 432 ⇒ 3.20 x = 432 - 384 = 48 ⇒ x = 15. Hence, total hours of work = (160 + 15) = 175.

Explanation 1 $$\eqalign{ & {\text{Let number of children}} = n \cr & {\text{Then, number of books each child will get }} = \frac{n}{8} \cr & {\text{Total books distributed}} = n \times \frac{n}{8} = \frac{{{n^2}}}{8} \cr & {\text{If the number children}} = \frac{n}{2}, \cr & {\text{number of books each child will get}} = 16 \cr & {\text{Total books distributed }} = \frac{n}{2} \times 16 = 8n{\text{ }} \cr & \therefore \frac{{{n^2}}}{8} = 8n \cr & \Rightarrow \frac{n}{8} = 8 \cr & \Rightarrow n = 64 \cr & {\text{Total number of books distributed}} \cr & = 8n = 8 \times 64 = 512 \cr} $$ Explanation 2 If number of children was half, each child would have got 16 books. Therefore, actually each child got $$\frac{{16}}{2}$$ = 8 Books And the number of children is 8 × 8 = 64 Hence, total number of books distributed = 64 × 8 = 512 Explanation 3 Let number of children $$ = n$$ Then, number of books each child will get $$ = \frac{n}{8}$$ If the number children $$ = \frac{n}{2}$$, number of books each child will get $$ = 16$$ More children, less books (indirect proportion). Therefore, $$\eqalign{ & n : \frac{n}{2} = 16 : \frac{n}{8} \cr & \Rightarrow \frac{{{n^2}}}{8} = 8n \cr & \Rightarrow \frac{n}{8} = 8 \cr & \Rightarrow n = 64 \cr} $$ Therefore, total number of books distributed = 8n = 8 × 64 = 512

Let the number of hens be x and the number of cows be y. Then, x + y = 48 . . . . . (i) and 2x + 4y = 140 ⇒ x + 2y = 70 . . . . . (ii) Solving (i) and (ii) we get: x = 26, y = 22 ∴ The required answer = 26

$$\eqalign{ & \frac{{{{\left( {469 + 174} \right)}^2} - {{\left( {469 - 174} \right)}^2}}}{{\left( {469 \times 174} \right)}} = ? \cr & {\text{We}}\,{\text{ know that}}, \cr & 4ab = {\left( {a + b} \right)^2} - {\left( {a - b} \right)^2} \cr & \therefore \frac{{{{\left( {469 + 174} \right)}^2} - {{\left( {469 - 174} \right)}^2}}}{{\left( {469 \times 174} \right)}} \cr & = \frac{{4 \times 469 \times 174}}{{469 \times 174}} \cr & = 4 \cr} $$

Suppose their paths cross after x minutes Then, 11 + 57x = 51 - 63x ⇒ 57x + 63x = 51 - 11 ⇒ 120x = 40 ⇒ x = $$\frac{1}{3}$$ Number of floors covered by David in $$\frac{1}{3}$$ min. $$\eqalign{ & = {\frac{1}{3} \times 57} \cr & = 19 \cr} $$ So, their paths cross at (11 + 19) i.e., 30 th floor

$$\eqalign{ & 1 + \frac{1}{{1 + \frac{2}{{2 + \frac{3}{{1 + \frac{4}{5}}}}}}} \cr & = 1 + \frac{1}{{1 + \frac{2}{{2 + \frac{3}{{\frac{9}{5}}}}}}} \cr & = 1 + \frac{1}{{1 + \frac{2}{{2 + \frac{5}{3}}}}} \cr & = 1 + \frac{1}{{1 + \frac{2}{{\frac{{11}}{3}}}}} \cr & = 1 + \frac{1}{{1 + \frac{6}{{11}}}} \cr & = 1 + \frac{1}{{\frac{{17}}{{11}}}} \cr & = 1 + \frac{{11}}{{17}} \cr & = 1\frac{{11}}{{17}} \cr} $$

$$\eqalign{ & 1 + \frac{2}{{1 + \frac{3}{{1 + \frac{4}{5}}}}} \cr & = 1 + \frac{2}{{1 + \frac{3}{{\frac{9}{5}}}}} \cr & = 1 + \frac{2}{{1 + \frac{5}{3}}} \cr & = 1 + \frac{2}{{\frac{8}{3}}} \cr & = 1 + \frac{3}{4} \cr & = \frac{7}{4} \cr} $$

$$\eqalign{ & {\text{According to question}} \cr & \frac{{9|3 - 5| - 5|4| \div 10}}{{ - 3\left( 5 \right) - 2 \times 4 \div 2}} \cr & \Rightarrow \frac{{9 \times 2 - 20 \div 10}}{{ - 15 - 2 \times 2}} \cr & \Rightarrow \frac{{18 - 2}}{{ - 15 - 4}} \cr & \Rightarrow - \frac{{16}}{{19}} \cr} $$

Given, 5 - [4 - {3 - (3 - 3 - 6)}] = 5 - [4 - {3 - (-6)}] = 5 - [4 - {3 +6}] = 5 - [4 - {9}] = 5 - [4 - 9] = 5 - [-5] = 5 + 5 = 10

$$\eqalign{ & \frac{{ - {{\left( {4 - 6} \right)}^2} - 3\left( { - 2} \right) + \left| { - 6} \right|}}{{18 - 9 \div 3 \times 5}} \cr & = \frac{{ - {{\left( { - 2} \right)}^2} - \left( { - 6} \right) + 6}}{{18 - 3 \times 5}} \cr & = \frac{{ - 4 + 6 + 6}}{{18 - 15}} \cr & = \frac{8}{3} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \frac{{\frac{5}{3} \times \frac{7}{{51}}{\text{ of }}\frac{17}{5} - \frac{1}{3}}}{{\frac{2}{9} \times \frac{5}{7}{\text{ of }}\frac{{28}}{5} - \frac{2}{3}}}\, \cr & \Rightarrow \frac{{\frac{5}{3} \times \frac{7}{{51}} \times \frac{17}{5} - \frac{1}{3}}}{{\frac{2}{9} \times \frac{5}{7} \times \frac{{28}}{5} - \frac{2}{3}}}\, \cr & \Rightarrow \frac{{\frac{5}{3} \times \frac{7}{{15}} - \frac{1}{3}}}{{\frac{2}{9} \times 4 - \frac{2}{3}}} \cr & \, \Rightarrow \frac{{\frac{7}{9} - \frac{1}{3}}}{{\frac{8}{9} - \frac{2}{3}}} \cr & \, \Rightarrow \frac{{\frac{{7 - 3}}{9}}}{{\frac{{8 - 6}}{9}}} \cr & \Rightarrow \frac{4}{2} \cr & \Rightarrow 2 \cr} $$

1 - [5 - {2 + (- 5 + 6 - 2) 2}] = 1 - [5 - {2 + (- 1) 2}] = 1 - [5 - {2 - 2}] = 1 - [5 - 0] = 1 - 5 = -4

According to the question, $$\eqalign{ & \sqrt {1.3} + \sqrt {1300} + \sqrt {0.013} \cr & \Rightarrow \sqrt {\frac{{130}}{{100}} }+ \sqrt {1300} + \sqrt {\frac{{130}}{{10000}}} \cr & \Rightarrow \frac{{11.4}}{{10}} + 36.05 + \frac{{11.4}}{{100}} \cr & \Rightarrow 1.14 + 36.05 + 0.114 \cr & \Rightarrow 37.304 \cr} $$

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

$$\eqalign{ & \frac{{4848}}{{24}} \times 11 - 222 \cr & = 202 \times 11 - 222 \cr & = 2222 - 222 \cr & = 2000 \cr} $$

Let the missing number is x. then, $$\frac{{354750}}{{4096 + x}}$$ = 55 ⇒ 55( 4096 + x ) = 354750 ⇒ 55x = 354750 - 225280 ⇒ 55x = 129470 ⇒ x = 2354

$$\eqalign{ & {\text{According to question}} \cr & \frac{{\frac{5}{3} \times \frac{7}{{51}}{\text{of}}\frac{17}{5} - \frac{1}{3}}}{{\frac{2}{9} \times \frac{5}{7}{\text{of}}\frac{{28}}{5} - \frac{2}{3}}}\, \cr & \Rightarrow \frac{{\frac{5}{3} \times \frac{7}{{51}} \times \frac{17}{5} - \frac{1}{3}}}{{\frac{2}{9} \times \frac{5}{7} \times \frac{{28}}{5} - \frac{2}{3}}}\, \cr & \Rightarrow \frac{{\frac{5}{3} \times \frac{7}{{15}} - \frac{1}{3}}}{{\frac{2}{9} \times 4 - \frac{2}{3}}} \cr & \, \Rightarrow \frac{{\frac{7}{9} - \frac{1}{3}}}{{\frac{8}{9} - \frac{2}{3}}} \cr & \, \Rightarrow \frac{{\frac{{7 - 3}}{9}}}{{\frac{{8 - 6}}{9}}} \cr & \Rightarrow \frac{4}{2} \cr & \Rightarrow 2 \cr} $$

$$\eqalign{ & {\text{Let,}} \cr & {\text{853 + }}x \div 17 = 1000 \cr & {\text{Then,}} \cr & {\text{853 + }}\frac{x}{{17}} = 1000 \cr & \Leftrightarrow \frac{x}{{17}} = 1000 - 853 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 147 \cr & \Leftrightarrow x = 147 \times 17 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 2499 \cr} $$

$$\eqalign{ & {\text{Let,}} \cr & \left( {x - 968} \right) \div 79 \times 4 = 512 \cr & {\text{Then,}} \cr & \frac{{x - 968}}{{79}} \times 4 = 512 \cr & \Rightarrow x - 968 = \frac{{512 \times 79}}{4} \cr & \Rightarrow x - 968 = 10112 \cr & \Rightarrow x = 10112 + 968 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 11080 \cr} $$

Given Expression = 1 - [ 1 - { 1 - ( 1 - 0 ) } ] = 1 - [ 1 - ( 1 - 1 ) ] = 1 - [ 1 - 0 ] = 0

$$\eqalign{ & {\text{According to the question}} \cr & {\text{ }}\frac{{{{\left( {3.4567} \right)}^2} - {{\left( {3.4533} \right)}^2}}}{{0.0034}} \cr & \left[ {\because {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right] \cr & \Rightarrow \frac{{\left( {3.4567 + 3.4533} \right)\left( {3.4567 - 3.4533} \right)}}{{0.0034}} \cr & \Rightarrow \frac{{6.91 \times 0.0034}}{{0.0034}} \cr & \Rightarrow 6.91 \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & \frac{{{{\left( {0.75} \right)}^3}}}{{1 - 0.75}} + \left[ {0.75 + {{\left( {0.75} \right)}^2} + 1} \right] \cr & \Rightarrow \frac{{{{\left( {0.75} \right)}^3}}}{{1 - 0.75}} + \frac{{{1^3} - {{\left( {0.75} \right)}^3}}}{{1 - 0.75}}\, \cr & \left[ {\because {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)} \right] \cr & \Rightarrow \frac{{{{\left( {0.75} \right)}^3} + 1 - {{\left( {0.75} \right)}^3}}}{{0.25}} \cr & \Rightarrow \frac{1}{{0.25}} \cr & \Rightarrow 4 \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & \sqrt {4 + \sqrt {44 + \sqrt {10000} } } \cr & \Rightarrow \sqrt {4 + \sqrt {44 + 100} } \cr & \Rightarrow \sqrt {4 + \sqrt {144} } \cr & \Rightarrow \sqrt {4 + 12} \cr & \Rightarrow \sqrt {16} \cr & \Rightarrow 4 \cr} $$

$$\eqalign{ & {\text{Let,}} \cr & 45 - \left[ {28 - \left\{ {37 - \left( {15 - x} \right)} \right\}} \right] = 58 \cr & {\text{then,}} \cr & {\text{45}} - \left[ {28 - \left\{ {37 - 15 + x} \right\}} \right] = 58 \cr & \Rightarrow 45 - \left[ {28 - \left\{ {22 + x} \right\}} \right] = 58 \cr & \Rightarrow 45 - \left[ {28 - 22 - x} \right] = 58 \cr & \Rightarrow 45 - \left[ {6 - x} \right] = 58 \cr & \Rightarrow 45 - 6 + x = 58 \cr & \Rightarrow 39 + x = 58 \cr & \Rightarrow x = 58 - 39 \cr & \Rightarrow x = 19 \cr} $$

$$\eqalign{ & \frac{{113 \times 4 - x \times 2}}{{13 \times 9 - 5 \times 7}} = 5 \cr & \Rightarrow \frac{{452 - 2x}}{{117 - 35}} = 5 \cr & \Rightarrow \frac{{452 - 2x}}{{82}} = 5 \cr & \Rightarrow 452 - 2x = 410 \cr & \Rightarrow 2x = 452 - 410 = 42 \cr & \Rightarrow x = 21 \cr} $$

( 2 2 + 4 2 + 6 2 + .......... + 20 2 ) = 2 2 ( 1 2 + 2 2 + 3 2 + .......... + 10 2 ) = 4 × 385 = 1540

$$\eqalign{ & {\text{According to the question,}} \cr & \frac{{\root 3 \of 8 }}{{\sqrt {16} }} \div \sqrt {\frac{{100}}{{49}}} \times \root 3 \of {125} \, \cr & \Rightarrow \frac{2}{4} \div \frac{{10}}{7} \times 5 \cr & \Rightarrow \frac{2}{4} \times \frac{7}{{10}} \times 5 \cr & \Rightarrow \frac{7}{4} \cr & \Rightarrow 1\frac{3}{4} \cr} $$

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

$$\eqalign{ & {\text{According to the question,}} \cr & \frac{{\sqrt {72} - \sqrt {18} }}{{\sqrt {12} }} \cr & \Rightarrow \sqrt {\frac{{72}}{{12}}} - \sqrt {\frac{{18}}{{12}}} \cr & \Rightarrow \sqrt 6 - \sqrt {\frac{6}{4}} \cr & \Rightarrow \frac{{\sqrt 6 }}{2} \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & \frac{7}{4} - \frac{6}{5} + \frac{{13}}{8} \cr & = \frac{{70 - 48 + 65}}{{40}} \cr & = \frac{{87}}{{40}} \cr & = 2\frac{7}{{40}} \cr} $$