11.
Find the value of the following expression.
(A) $$ - \frac{4}{3}$$
(B) $$\frac{4}{3}$$
(C) $$\frac{1}{7}$$
(D) $$ - \frac{1}{7}$$
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Solution:
$$\eqalign{ & \frac{{1\frac{2}{3} \div \frac{5}{6} \times 6 + \frac{4}{5} \times \frac{1}{2} + \frac{2}{3}}}{{2 - \left[ {1\frac{1}{3} \times \left( { - \frac{3}{5}} \right) - 6\left\{ {\frac{3}{5} - \left( {3 - \frac{3}{{10}}} \right)} \right\}} \right]}} \cr & = \frac{{\frac{5}{3} \times \frac{6}{5} \times 6 + \frac{4}{{10}} + \frac{2}{3}}}{{2 - \left[ {\frac{4}{3} \times \frac{{ - 3}}{5} - 6\left\{ {\frac{3}{5} - \frac{{27}}{{10}}} \right\}} \right]}} \cr & = \frac{{12 + \frac{{6 + 10}}{{15}}}}{{2 - \left( { - \frac{4}{5} + \frac{{21 \times 6}}{{10}}} \right)}} \cr & = \frac{{12 + \frac{{16}}{{15}}}}{{2 - \left( { - \frac{4}{5} + \frac{{63}}{5}} \right)}} \cr & = \frac{{\frac{{196}}{{15}}}}{{2 - \frac{{59}}{5}}} \cr & = \frac{{196}}{{15}} \times \frac{{ - 5}}{{49}} \cr & = \boxed{ - \frac{4}{3}} \cr} $$
12.
$$4848 \div 24 \times 11 - 222 = ?$$
(A) 200
(B) 2444
(C) 2000
(D) $$115\frac{3}{8}$$
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Solution:
$$\eqalign{ & \frac{{4848}}{{24}} \times 11 - 222 \cr & = 202 \times 11 - 222 \cr & = 2222 - 222 \cr & = 2000 \cr} $$
13.
The value of $$2.\overline {13} \div 3.\overline {12} $$ ÃÂÃÂ is:
(A) $$\frac{{239}}{{309}}$$
(B) $$\frac{{211}}{{309}}$$
(C) $$\frac{{249}}{{309}}$$
(D) $$\frac{{301}}{{309}}$$
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Solution:
$$\eqalign{ & 2.\overline {13} \div 3.\overline {12} \cr & = \frac{{213 - 2}}{{99}} \div \frac{{312 - 3}}{{99}} \cr & = \frac{{211}}{{99}} \times \frac{{99}}{{309}} \cr & = \frac{{211}}{{309}}{\text{ Answer}} \cr} $$
14.
$${\text{If }}\left[ {4 - \frac{5}{{1 + \frac{1}{{3 + \frac{1}{{2 + \frac{1}{4}}}}}}}} \right]$$ ÃÂÃÂ ÃÂÃÂ
part of a journey takes ten minutes, then to complete $$\frac{3}{5}$$th of that journey, it will take = ?
(A) 40 minutes
(B) 45 minutes
(C) 48 minutes
(D) 36 minutes
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Solution:
$$\eqalign{ & {\text{According to question,}} \cr & \,\,\,\left[ {4 - \frac{5}{{1 + \frac{1}{{3 + \frac{1}{{2 + \frac{1}{4}}}}}}}} \right] \cr & = 4 - \frac{5}{{1 + \frac{1}{{3 + \frac{1}{{\frac{9}{4}}}}}}} \cr & = 4 - \frac{5}{{1 + \frac{1}{{3 + \frac{4}{9}}}}} \cr & = 4 - \frac{5}{{1 + \frac{1}{{\frac{{31}}{9}}}}} \cr & = 4 - \frac{5}{{1 + \frac{9}{{31}}}} \cr & = 4 - \frac{5}{{\frac{{40}}{{31}}}} \cr & = 4 - \frac{{5 \times 31}}{{40}} \cr & = 4 - \frac{{31}}{8} \cr & = \frac{1}{8} \cr & {\text{According to question}} \cr & \frac{1}{8}{\text{part = 10 minutes}} \cr & {\text{1 part = 10 minutes}} \cr & \frac{3}{5}{\text{ part = 80}} \times \frac{3}{5} \cr & {\text{ = 48 minutes}} \cr} $$
15.
Evaluate : $${{ - {{\left( {4 - 6} \right)}^2} - 3\left( { - 2} \right) + \left| { - 6} \right|} \over {18 - 9 \div 3 \times 5}}$$
(A) $$\frac{3}{8}$$
(B) $$\frac{4}{7}$$
(C) $$\frac{8}{3}$$
(D) $$\frac{7}{4}$$
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Solution:
$$\eqalign{ & \frac{{ - {{\left( {4 - 6} \right)}^2} - 3\left( { - 2} \right) + \left| { - 6} \right|}}{{18 - 9 \div 3 \times 5}} \cr & = \frac{{ - {{\left( { - 2} \right)}^2} - \left( { - 6} \right) + 6}}{{18 - 3 \times 5}} \cr & = \frac{{ - 4 + 6 + 6}}{{18 - 15}} \cr & = \frac{8}{3} \cr} $$
16.
Evaluate: $$\frac{1}{{15}} + \frac{1}{{35}} + \frac{1}{{63}} + \frac{1}{{99}} + \frac{1}{{143}}$$
(A) $$\frac{4}{{39}}$$
(B) $$\frac{5}{{39}}$$
(C) $$\frac{{10}}{{39}}$$
(D) $$\frac{7}{{39}}$$
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Solution:
$$\eqalign{ & \frac{1}{{15}} + \frac{1}{{35}} + \frac{1}{{63}} + \frac{1}{{99}} + \frac{1}{{143}} \cr & = \frac{1}{2}\left[ {\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7}\,.\,.\,.\,.\,.\,\frac{1}{{11}} - \frac{1}{{13}}} \right] \cr & = \frac{1}{2}\left[ {\frac{1}{3} - \frac{1}{{13}}} \right] \cr & = \frac{1}{2} \times \frac{{10}}{{39}} \cr & = \frac{5}{{39}} \cr} $$
17.
Given that ( 12 + 22 + 32 + .......... + 102 ) = 385, then the value of ( 22 + 42 + 62 + .......... + 202 ) is equal to = ?
(A) 770
(B) 1155
(C) 1540
(D) (385)2
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Solution:
( 22 + 42 + 62 + .......... + 202 ) = 22 ( 12 + 22 + 32 + .......... + 102 ) = 4 × 385 = 1540
18.
$$\frac{{225}}{{836}} \times \frac{{152}}{{245}} \div 1\frac{{43}}{{77}} = ?$$
(A) $$\frac{{6}}{{49}}$$
(B) $$\frac{{6}}{{11}}$$
(C) $$\frac{{3}}{{28}}$$
(D) $$\frac{{1}}{{7}}$$
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Solution:
$$\eqalign{ & \frac{{225}}{{836}} \times \frac{{152}}{{245}} \div 1\frac{{43}}{{77}} \cr & = \frac{{225}}{{836}} \times \frac{{152}}{{245}} \div \frac{{120}}{{77}} \cr & = \frac{{225}}{{836}} \times \frac{{152}}{{245}} \times \frac{{77}}{{120}} \cr & = \frac{3}{{28}} \cr} $$
19.
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
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Solution:
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
20.
$$\frac{6}{{5 - \frac{5}{3}}} \div \frac{{4 - \frac{2}{{4 - \frac{1}{2}}}}}{{5 - \frac{3}{2}}} - \frac{2}{5}\,$$ ÃÂÃÂ ÃÂÃÂ $$\text{of}$$ $$\left\{ {\frac{6}{9} + \frac{2}{3}{\text{ of }}\frac{1}{2}} \right\}$$ ÃÂÃÂ $$ = ?$$
(A) $$1\frac{1}{3}$$
(B) $$2\frac{{13}}{{49}}$$
(C) $$1\frac{7}{{16}}$$
(D) $$2\frac{3}{5}$$
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Solution:
$$\eqalign{ & {\text{Given expression,}} \cr & \frac{6}{{5 - \frac{5}{3}}} \div \frac{{4 - \frac{2}{{4 - \frac{1}{2}}}}}{{5 - \frac{3}{2}}} - \frac{2}{5}\, \times \,\left\{ {\frac{6}{9} + \frac{2}{3} \times \frac{1}{2}} \right\}{\text{ }} \cr & = \frac{6}{{\left( {\frac{{10}}{3}} \right)}} \div \frac{{4 - \frac{2}{{\left( {\frac{7}{2}} \right)}}}}{{\left( {\frac{7}{2}} \right)}} - \frac{2}{5}\, \times \,\left\{ {\frac{6}{9} + \frac{1}{3}} \right\} \cr & = \frac{{6 \times 3}}{{10}} \div \frac{{4 - \frac{{2 \times 2}}{7}}}{{\left( {\frac{7}{2}} \right)}} - \frac{2}{5} \times 1 \cr & = \frac{9}{5} \div \frac{{\left( {4 - \frac{4}{7}} \right)}}{{\left( {\frac{7}{2}} \right)}} - \frac{2}{5} \cr & = \frac{9}{5} \div \left( {\frac{{24}}{7} \times \frac{2}{7}} \right) - \frac{2}{5} \cr & = \frac{9}{5} \times \frac{{49}}{{48}} - \frac{2}{5} \cr & = \frac{{147}}{{80}} - \frac{2}{5} \cr & = \frac{{147 - 32}}{{80}} \cr & = \frac{{115}}{{80}} \cr & = \frac{{23}}{{16}} \cr & = 1\frac{7}{{16}} \cr} $$