Simple Interest - 02

Explanation:-

Time = 18 + 28 + 31 + 30 + 31 + 8 = 146 days
$$\eqalign{ & {\text{Simple Interest}} \cr & {\text{ = }}\frac{{12000 \times 15 \times 146}}{{365 \times 100}} \cr & {\text{ = Rs}}{\text{. 720}} \cr & \therefore {\text{Amount will be }} \cr & {\text{ = Rs}}{\text{.}}\left( {12000 + 720} \right){\text{ }} \cr & {\text{ = Rs}}{\text{. 12720}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Sum of the 12 years age }} \cr & {\text{ = Rs}}{\text{. 100000}} \cr & {\text{Sum of the 18 years age }} \cr & = {\text{P}} + \frac{{{\text{P}} \times {\text{R}} \times {\text{T}}}}{{100}} \cr & = {\text{100000}} + \frac{{100000 \times 6 \times 6}}{{100}} \cr & = {\text{100000}} + {\text{36000}} \cr & = {\text{136000}} \cr}$$
Total expenses
= 2500 + 500 = 3000 per year
Total expenses ( 6 years )
= 3000 × 6 = Rs. 18000
Amount obtained
= 136000 - 18000
= 118000

Explanation:-

$$\eqalign{ & {\text{Let the sum be Rs}}{\text{. }}x \cr & {\text{Rate be R}}\% {\text{ p}}{\text{.a}}{\text{.}} \cr & {\text{Time be T years}}{\text{.}} \cr & {\text{Then,}} \cr & \left[ {\frac{{x \times \left( {{\text{R}} \times 2} \right) \times {\text{T}}}}{{100}}} \right] - \left( {\frac{{x \times {\text{R}} \times {\text{T}}}}{{100}}} \right) = 108 \cr & \Leftrightarrow 2x{\text{T}} = 10800\,........(i) \cr & And, \cr & \left[ {\frac{{x \times {\text{R}} \times \left( {{\text{T}} + 2} \right)}}{{100}}} \right] - \left( {\frac{{x \times {\text{R}} \times {\text{T}}}}{{100}}} \right) = 108 \cr & \Leftrightarrow 2x{\text{R}} = 18000\,.......(ii) \cr}$$
Clearly, from (i) and (ii), we cannot the find the value of x.
So, the data is inadequate.

Explanation:-

Total cost of the computer = Rs. 39000
Down payment = Rs. 17000
Balance = Rs. (39000 - 17000) = Rs. 22000.
Let the rate of interest be R% p.a.
Amount of Rs. 22000 for 5 months
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {22000 + 22000 \times \frac{5}{{12}} \times \frac{{\text{R}}}{{100}}} \right) \cr & = {\text{Rs}}{\text{.}}\left( {22000 + \frac{{275{\text{R}}}}{3}} \right) \cr}$$
The customer pays the shopkeeper Rs. 4800 after 1 month,
Rs. 4800 after 2 months, ...... and Rs. 4800 after 5 months.

Thus, the shopkeeper keeps Rs. 4800 for 4 months, Rs. 4800 for 3 months, Rs. 4800 for 2 months, Rs. 4800 for 1 months and Rs. 4800 at the end.

∴ sum of the amounts of these instalments
= (Rs. 4800 + S.I. on Rs 4800 for 4 months) + (Rs. 4800 + S.I. on Rs. 4800 for 3 months) + ...... + (Rs. 4800 + S.I. on Rs. 4800 for 1 month) + Rs. 4800

= Rs. 4800 × 5) + S.I. on Rs. 4800 for (4 + 3 + 2 + 1) months
= Rs. 24000 + S.I. on Rs. 4800 for 10 months
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {24000 + {\text{4800}} \times {\text{R}} \times \frac{{10}}{{12}} \times \frac{1}{{100}}} \right) = {\text{Rs}}{\text{.}}\left( {24000 + 40{\text{R}}} \right). \cr & \therefore 22000 + \frac{{275{\text{R}}}}{3} = 24000 + 40{\text{R}} \cr & \Rightarrow \frac{{155}}{3} = 2000 \cr & \Rightarrow {\text{R}} = \frac{{2000 \times 3}}{{155}} \cr & = 38.71\% {\text{p}}{\text{.a}}{\text{.}} \cr}$$

Explanation:-

⇒ Rs. 10 + S.I. on Rs. 10 for 11 months
= Rs. 11 + S.I. on Rs. 1 for (1 + 2 + 3 + 4 + ........... + 10) months
⇒ Rs. 10 + S.I. on Rs. 1 for 110 months
= Rs. 11 + S.I. on Rs. 1 for 55 months
S.I. on Rs. 1 for 55 months = Rs. 1
$$\eqalign{ & \therefore {\text{Rate}} = \left( {\frac{{100 \times 12}}{{1 \times 55}}} \right)\% \cr & = 21\frac{9}{{11}}\% . \cr}$$

Explanation:-

Total price of TV = Rs. 16000
Initial payment = Rs. 4000
Remaining amount = Rs. 12000
Simple interest in 15 months for Rs. 12000
$$\eqalign{ & \Rightarrow {\text{S}}{\text{.I}}{\text{. = }}\frac{{{\text{P}} \times {\text{R}} \times {\text{T}}}}{{100}} \cr & \Rightarrow {\text{S}}{\text{.I}}{\text{. = }}\frac{{12000 \times 12 \times 15}}{{100 \times 12}} \cr & \Rightarrow {\text{S}}{\text{.I}}{\text{. = Rs}}{\text{. 1800}} \cr}$$
⇒ With S.I. total amount to be paid for principal amount Rs. 12000
= Rs. (12000 + 1800)
= Rs. 13800
= Therefore, total amount he pays for the TV is
= 4000 + 13800
= Rs. 17800

Explanation:-

Let the annual instalment be Rs. x.
Then,
$$\eqalign{ & \Rightarrow \left[ {x + \left( {\frac{{x \times 3 \times 4}}{{100}}} \right)} \right] + \left[ {x + \left( {\frac{{x \times 2 \times 4}}{{100}}} \right)} \right] + \left[ {x + \left( {\frac{{x \times 1 \times 4}}{{100}}} \right)} \right] + x = 848 \cr & \Leftrightarrow \frac{{28x}}{{25}} + \frac{{27x}}{{25}} + \frac{{26x}}{{25}} + x = 848 \cr & \Leftrightarrow 106x = 848 \times 25 = 21200 \cr & \Leftrightarrow x = 200 \cr}$$
Short Cut Method : The annual payment that will discharge a debt of Rs. A due in t years at the rate of interest r % p.a. is.
$$\eqalign{ & \frac{{100{\text{A}}}}{{100t + \frac{{rt\left( {t - 1} \right)}}{2}}}. \cr & \therefore {\text{Annual instalment}} \cr & = {\text{Rs}}{\text{.}}\left[ {\frac{{100 \times 848}}{{100 \times 4 + \frac{{4 \times 4 \times 3}}{2}}}} \right] \cr & = {\text{Rs}}{\text{.}}\left( {\frac{{100 \times 848}}{{424}}} \right) \cr & = {\text{Rs}}{\text{. }}200. \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let the sum be Rs}}{\text{. }}x{\text{ and}} \cr & {\text{original rate be R}}\% \cr & {\text{Then,}} \cr & \Rightarrow \frac{{x \times \left( {{\text{R}} + 1} \right) \times 3}}{{100}} - \frac{{x \times {\text{R}} \times 3}}{{100}} = 5100 \cr & \Rightarrow 3{\text{R}}x + 3x - 3{\text{R}}x = 510000 \cr & \Rightarrow 3x = 510000 \cr & \Rightarrow x = 170000. \cr & {\text{Hence,}} \cr & {\text{Sum}} = {\text{Rs}}.170000. \cr}$$

Explanation:-

Let the rates of interest in the former and latter cases be R% and (R + 1) % p.a.
Then,
$$\eqalign{ & 5000 \times {\text{R}} \times 4 = 4000 \times \left( {{\text{R}} + 1} \right) \times 4 \cr & \Rightarrow \frac{{{\text{R}} + 1}}{{\text{R}}} = \frac{{5000 \times 4}}{{4000 \times 4}} \cr & \Rightarrow 1 + \frac{1}{4} \cr & \Rightarrow {\text{R}} = 4. \cr & {\text{Hence,}} \cr & {\text{Required rate}} = 4\% \,{\text{p}}{\text{.a}}{\text{.}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{Interest = Rs}}{\text{. 1 per day}} \cr & \therefore {\text{Interest in one year}} \cr & {\text{ = 1}} \times {\text{365 = Rs}}{\text{. 365}} \cr & \therefore {\text{S}}{\text{.I}}{\text{. = }}\frac{{{\text{P}} \times {\text{T}} \times {\text{R}}}}{{100}} \cr & \Rightarrow 365 = \frac{{{\text{P}} \times 5 \times 1}}{{100}} \cr & \Rightarrow {\text{P}} = \frac{{365 \times 100}}{5} \cr & \Rightarrow {\text{P}} = {\text{Rs}}{\text{. 7300}} \cr}$$