Question

प्रश्न 35. कोई धन 4 वर्ष 500 रु. तथा छः वर्ष 550 रु. हो जाता है, तो
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Answer 1

${ \tt{ \red{ \huge{ \underline{ \underline{ QUESTION- }}}}}}$




▪ A sum of money amounts to Rs. 500 in 4 years and Rs. 550 in 6 years. Then, find the money??




${ \tt{ \red{ \underline{ \underline{ \huge{SOLUTION- }}}}}}$




${ \sf{ \blue{let \: the \: principal \: be \: Rs . \: P}}} \\ \\ { \sf{ \blue{and \: the \: rate \: be \: R \: percent}}}$




${ \pink{ \sf{ \underline{ \underline{Simple \: Interest \: (S .I.)}}}}}$




${ \underline{ \boxed{ \sf{ \green{S.I . = \frac{P \times R \times T}{100}}}}}}$




${ \sf{ \pink{ \underline{ \underline{amount = P + S.I .}}}}}$




where,




${ \star{ \blue{ \sf{ \: \: P = principal}}}} \\ \\ { \star{ \blue{ \sf{ \: \: R = rate}}}} \\ \\ { \star{ \blue{ \sf{ \: \: T = time}}}}$




${ \purple{ \underline{ \bold{CASE\: 1}}}}$




${ \orange{ \sf{principal \: = P }}} \\ \\ { \orange{ \sf{rate = R}}} \\ \\ { \orange{ \sf{ time = 4 \: years}}} \\ \\ { \orange{ \sf{ amount = Rs . \: 500}}}$




${ \red{ \sf{ S .I . = \frac{P \times R \times 4}{100}}}}$




${ \implies{ \sf{ \red{S .I. = \frac{4PR}{100}}}}}$




${ \sf{ \blue{amount = \frac{4PR}{100} + P = \: Rs. \: 500}}}$




${ \implies{ \blue{ \sf{ \frac{2PR }{50} + P = 500}}}}$




${ \implies{ \blue{ \sf{ \frac{2PR + 50P}{50} = 500}}}}$




${ \implies{ \blue{ \sf{2PR + 50P = 500 \times 50}}}}$




${ \implies{ \blue{ \sf{2PR + 50P = 25000}}}} - - - { \sf{eqn(1)}}$




$\bold{ \purple{ \underline{CASE \: 2}}}$




${ \orange{ \sf{ principal = P}}} \\ \\ { \orange{ \sf{rate = R}}} \\ \\ { \orange{ \sf{time = 6}}} \\ \\ { \orange{ \sf{ amount = Rs . \: 550}}}$




${ \sf{ \red{S .I. = \frac{P \times R \times 6}{100}}}}$




${ \implies{ \sf{ \red{S.I. = \frac{3PR}{50}}}}}$




${ \blue{ \sf{amount = \frac{3PR }{50} + P = Rs . \: 550}}}$




${ \implies{ \blue{ \sf{ \frac{3PR + 50P}{50} = 550}}}}$




${ \implies{ \sf{ \blue{3PR + 50P = 550 \times 50}}}}$




${ \implies{ \blue{ \sf{3PR + 50P = 27500}}}} - - - { \sf{eqn(2)}}$




▪ subtracting eqn (1) from eqn (2)....




${ \pink{ \sf{3PR + 50P = 27500}}} - - - { \sf{eqn(2)}} \\ { \pink{ \sf{2PR + 50P = 25000}}} - - - { \sf{eqn (1)}}\\ { \sf{ - \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: \: \: - }} \\ { \green{ \sf{PR \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2500}}}$




▪ putting the value of PR in eqn (1)...




${ \red{ \sf{2PR + 50P = 25000}}} - - - { \sf{eqn(1)}}$




${ \implies{ \sf{ \red{2(2500) + 50P = 25000}}}}$




${ \implies{ \red{ \sf{5000 + 50P = 25000}}}}$




${ \implies{ \red{ \sf{50P = 25000 - 5000}}}}$




${ \implies{ \red{ \sf{50P = 20000}}}}$




${ \implies{ \red{ \sf{P = \frac{20000}{50}}}}}$




${ \implies{ \boxed{ \boxed{ \red{ \sf{ \: \: \: P = Rs. \: 400 \: \: \: }}}}}}$