Practice MCQ Questions and Answer on Bankers Discount

Solution:

 T = 4 years R = 5% Banker's Gain, BG = Rs. 200 $$\eqalign{ & TD = \frac{{BG \times 100}}{{TR}} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{200 \times 100}}{{4 \times 5}} \cr & \,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{.}}\,1000 \cr & \cr & TD = \sqrt {PW \times BG} \cr & 1000 = \sqrt {PW \times 200} \cr & 1000000 = PW \times 200 \cr & PW = \frac{{1000000}}{{200}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{. }}5000 \cr} $$

Solution:

 $$\eqalign{ & {\text{S}}{\text{.I}}{\text{.}}\,{\text{on}}\,{\text{Rs}}{\text{.}}\,{\text{1600 = T}}{\text{.D}}{\text{.}}\,{\text{on}}\,{\text{Rs}}{\text{.}}\,{\text{1680}}{\text{.}} \cr & \therefore {\text{Rs}}{\text{.}}\,{\text{1600}}\,{\text{is}}\,{\text{the}}\,{\text{P}}{\text{.W}}{\text{.}}\,{\text{of}}\,{\text{Rs}}{\text{.}}\,{\text{1680,}}\,{\text{i}}{\text{.e}}{\text{.,}} \cr & {\text{Rs}}{\text{.}}\,{\text{80}}\,{\text{is}}\,{\text{on}}\,{\text{Rs}}{\text{.}}\,{\text{1600}}\,{\text{at}}\,{\text{15}}\% . \cr & \therefore {\text{Time}} = \left( {\frac{{100 \times 80}}{{1600 \times 15}}} \right){\text{year}} \cr & = \frac{1}{3}{\text{year}} = 4\,{\text{months}}{\text{.}} \cr} $$

Solution:

 $$\eqalign{ & TD = \frac{{BG{\text{ }} \times 100}}{{TR}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{540 \times 100}}{{6 \times 12}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{90 \times 100}}{{12}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{15 \times 100}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{.}}\,750 \cr & BG = BD - TD \cr & \Rightarrow 540 = BD - 750 \cr & \Rightarrow BD = 540 + 750 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1290 \cr} $$

Solution:

 $$\eqalign{ & T.D. = \sqrt {P.W.} \times B.G. \cr & \therefore B.G. = \frac{{{{\left( {T.D.} \right)}^2}}}{{P.W.}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{.}}\,\frac{{110 \times 110}}{{1100}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{.}}\,11 \cr & \therefore B.D. = T.D. + B.G. \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{.}}\,110 + 11 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{.}}\,121 \cr} $$

Solution:

 $$\eqalign{ & BD{\text{ for }}1\frac{1}{2}{\text{ years}} = {\text{Rs}}{\text{.}}\,\,120 \cr & BD{\text{ for }}2{\text{ years}} = 120 \times \frac{2}{3} \times 2 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{.}}\,160 \cr & TD\,{\text{for}}\,2\,{\text{year}} = Rs.\,150 \cr & F = \frac{{BD \times TD}}{{\left( {BD - TD} \right)}} \cr & \,\,\,\,\,\,\,\, = \frac{{160 \times 150}}{{160 - 150}} \cr & \,\,\,\,\,\,\,\, = \frac{{160 \times 150}}{{10}} \cr & \,\,\,\,\,\,\,\, = {\text{Rs}}{\text{.}}\,2400 \cr} $$ ⇒ Rs.160 is the simple interest on Rs. 2400 for 2 years $$\eqalign{ & \Rightarrow 160 = \frac{{2400 \times 2 \times R}}{{100}} \cr & \Rightarrow R = \frac{{160 \times 100}}{{2400 \times 2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{160}}{{48}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20}}{6} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{10}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3\frac{1}{3}\,\% \cr} $$

Solution:

 Present Worth (P.W.) = 840 - 105 = Rs. 735 Therefore, S.I. on Rs. 735 = Rs. 105 S.I. on Rs. 8400 = $$\frac{{105 \times 840}}{{735}}$$ = Rs. 120

Solution:

 $$\eqalign{ & T = 6{\text{ months}} = \frac{1}{2}{\text{year}} \cr & R = 6\% \cr & TD = \frac{{BD \times 100}}{{100 + TR}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{18.54 \times 100}}{{100 + \left( {\frac{1}{2} \times 6} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{18.54 \times 100}}{{103}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{1854}}{{103}} \cr & \,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{.}}\,18 \cr} $$

Solution:

 $$\eqalign{ & {\text{Let,}}\,{\text{B}}{\text{.D}}{\text{.}} = \operatorname{Rs} .\,{\kern 1pt} 1 \cr & {\text{Then,}}{\kern 1pt} \,{\text{B}}{\text{.G}}{\text{.}} = \operatorname{Re} .\,{\kern 1pt} \frac{3}{{25}} \cr & \therefore {\text{T}}{\text{.D}}{\text{. = }}\left( {{\text{B}}{\text{.D}}{\text{. - B}}{\text{.G}}{\text{.}}} \right) \cr & = \operatorname{Rs} .\,{\kern 1pt} \left( {1 - \frac{3}{{25}}} \right) \cr & = \operatorname{Rs} .{\kern 1pt} \,\frac{{22}}{{25}} \cr & {\text{Sum}} = {\frac{{1 \times {\frac{{22}}{{25}}} }}{{1 - {\frac{{22}}{{25}}} }}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{.}}{\kern 1pt} \,\frac{{22}}{3} \cr & {\text{S}}{\text{.I}}{\text{.}}{\kern 1pt} \,{\text{on}}{\kern 1pt} {\text{Rs}}{\text{.}}\,{\kern 1pt} \frac{{22}}{3}{\kern 1pt} {\text{for}}\,{\kern 1pt} 1\frac{1}{2}\,{\text{years}}\,{\kern 1pt} {\text{is}}\,\operatorname{Rs} .\,{\kern 1pt} 1 \cr & \therefore {\text{Rate}} = \left( {\frac{{100 \times 1}}{{\frac{{22}}{3} \times \frac{3}{2}}}} \right)\% {\kern 1pt} \cr & {\kern 1pt} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 9\frac{1}{9}\% \cr} $$

Solution:

 $$\eqalign{ & TD = {\text{Rs}}{\text{.}}\,240 \cr & T = 6\,{\text{months}}\, = \frac{1}{2}\,{\text{year}} \cr & R = 15\% \cr & TD = \frac{{BG \times 100}}{{TR}} \cr & \Rightarrow 240 = \frac{{BG \times 100}}{{\left( {\frac{1}{2} \times 15} \right)}} \cr & BG = \frac{{240 \times 15}}{{100 \times 2}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{120 \times 15}}{{100}} \cr & \,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{.}}\,18 \cr & BG = BD - TD \cr & \Rightarrow 18 = BD - 240 \cr & \Rightarrow BD = 18 + 240 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{.}}\,258 \cr} $$

Solution:

 $$\eqalign{ & {\text{P}}{\text{.W}}{\text{.}} = Rs.\,\left( {540 - 90} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = Rs.\,450 \cr & \therefore {\text{S}}{\text{.I}}{\text{.}}\,{\text{on}}\,Rs.\,450 = Rs.\,90 \cr & {\text{S}}{\text{.I}}{\text{.}}\,{\text{on}}\,Rs.\,540 \cr & = Rs.\,\left( {\frac{{90}}{{450}} \times 540} \right) \cr & = Rs.\,108 \cr & \therefore {\text{B}}{\text{.D}}. = Rs.\,108 \cr} $$