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21
A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to 8 times?
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{Let}}, \cr & {\text{Principal}} = Rs.\,100 \cr & {\text{Amount}} = Rs.\,200 \cr & {\text{Rate}} = r\% \cr & {\text{Time}} = 4\,{\text{years}} \cr & {\text{Now}}, \cr & A = P \times {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^n} \cr & 200 = 100 \times {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^4} \cr & 2 = {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^4} - - - - - - \left( i \right) \cr & {\text{If}}\,{\text{sum}}\,{\text{become}}\,{\text{8}}\,{\text{times}}\,{\text{in}}\,{\text{the}}\,{\text{time}}\,n\,{\text{years}} \cr & {\text{then,}} \cr & 8 = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} \cr & {2^3} = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} - - - - - - \left( {ii} \right) \cr & {\text{Using}}\,{\text{eqn}}\,\left( i \right)in\left( {ii} \right),\,{\text{we}}\,{\text{get}} \cr & {\left( {{{\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]}^4}} \right)^3} = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} \cr & {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^{12}} = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} \cr & {\text{Thus}},\,n = 12\,{\text{years}}. \cr} $$
22
Divide Rs. 6000 into two parts so that simple interest on the first part for 2 years at 6% p.a. may be equal to the simple interest on the second part for 3 years at 8% p.a.
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{Let}}\,{1^{st}}\,{\text{part}}\,{\text{is}}\,x\,{\text{and}}\,{2^{nd}}\,{\text{part}}\,{\text{is}}\,\left( {6000 - x} \right) \cr & {\text{According}}\,{\text{to}}\,{\text{question}}, \cr & {\frac{{x \times 2 \times 6}}{{100}}} = \frac{{ {\left( {6000 - x} \right) \times 3 \times 8} }}{{100}} \cr & 12x = 144000 - 24x \cr & Or,\,36x = 144000 \cr & Or,\,x = \frac{{144000}}{{36}} = Rs.\,4000 \cr & {1^{st}}\,{\text{part}} = Rs.\,4000 \cr & {2^{nd}}\,{\text{part}} = Rs.\,2000 \cr} $$
23
A sum of money becomes $$\frac{7}{4}$$ of itself in 6 years at a certain rate of simple interest. Find the rate of interest.
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{sum}}\,is\,{\text{Rs}}.\,100, \cr & {\text{Then}}\,{\text{it}}\,{\text{become}}\,\frac{7}{4}\,{\text{times}} \cr & i.e.\,{\text{Rs}}.\,\frac{{700}}{4}\,{\text{in}}\,6\,{\text{years}}. \cr & {\text{Interest}} = {\frac{{700}}{4}} - 100 = Rs.\,\frac{{300}}{4} \cr & {\text{Hence}}, \cr & {\text{Rate}} = \frac{{{\text{Total}}\,{\text{interest}}}}{{{\text{Given}}\,{\text{time}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{300}}{4} \times 6 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 12 {\frac{1}{2}} \% \cr} $$
24
If a certain sum of money becomes doubles at simple interest in 12 years, what would be the rate of interest per annum?
Discuss
Answer & Solution
Answer: Option A
Solution:
Let,
Principal, P = Rs. 100;
Amount, A = Rs. 200;
Time = 12 years;
Interest = Rs. 100;
Rate of interest
$$\eqalign{ & = \frac{{{\text{Total Interest}}}}{{{\text{Given Time}}}} \cr & = \frac{{100}}{{12}} \cr & = 8\frac{1}{3}\% \cr} $$
25
If a sum of Rs. 13040 is to be repaid in two equal installments at $$3\frac{3}{4}$$ % per annum, what is the amount of each installment?
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{each}}\,{\text{installment}}\,{\text{be}}\,P \cr & {\text{Hence}}, \cr & {\frac{x}{{ {\left( {\frac{{100}}{{100 + r}}} \right) + {{\left( {\frac{{100}}{{100 + r}}} \right)}^2}} }}} \cr & Or,\,\frac{x}{{ {1 + {\frac{{15}}{{400}}} } }} + \frac{x}{{ {1 + {{\left( {\frac{{15}}{{400}}} \right)}^2}} }} = Rs.\,13040 \cr & {\text{On}}\,{\text{solving,}}\,{\text{it}}\,{\text{gives}}\,x = Rs.\,6889 \cr} $$
26
What is the amount of equal installment, if a sum of Rs. 1428 due 2 years hence has to be completely repaid in 2 equal annual installments starting next year?.
Discuss
Answer & Solution
Answer: Option D
Solution:
As short - cut of installment for 2 installments is given by
Installment = $$ {\frac{{\text{P}}}{{ {\left( {\frac{{100}}{{100}} + {\text{R}}} \right) + {{\left( {\frac{{100}}{{100}} + {\text{R}}} \right)}^2}} }}} $$
There is the need of rate (R) which is unavailable in the question.
so, we cannot determine the answer
27
A milkman sells cow milk at the rate of Rs. 55 litre including a profit 12 per cent. He also sells buffalo milk at the rate of Rs. 36 per litre including a profit of 20%. How much profit will he earn in five days if he sells 8 litres of cow milk and 10 litres of buffalo milk per day?
Discuss
Answer & Solution
Answer: Option E
Solution:
$$\eqalign{ & {\text{Total}}\,{\text{cow}}\,{\text{milksold}}\,{\text{in}}\,{\text{five}}\,{\text{days}} \cr & = 5 \times 8 = 40\,{\text{litres}} \cr & {\text{Total}}\,{\text{buffalo}}\,{\text{milk}}\,{\text{sold}}\,{\text{in}}\,{\text{five}}\,{\text{days}} \cr & = 5 \times 10 = 50\,{\text{litres}} \cr & {\text{Hence}}, \cr & {\text{SP}}\,{\text{of}}\,{\text{40}}\,{\text{litres}}\,{\text{cow}}\,{\text{milk}} \cr & = 40 \times 55 = Rs.\,2200 \cr & {\text{Hence}}, \cr & {\text{Profit}}\,{\text{on}}\,{\text{cow}}\,{\text{milk}} \cr & = \frac{{ {2200 \times 12} }}{{112}} = Rs.\,235.71 \cr & {\text{Profit}}\,{\text{on}}\,{\text{buffalo}}\,{\text{milk}} \cr & = \frac{{ {50 \times 36 \times 20} }}{{120}} = Rs.\,300 \cr & {\text{Thus}}, \cr & {\text{Total}}\,{\text{Profit}} \cr & = 235.71 + 300 = Rs.\,535.71 \cr} $$
28
The population of vultures in a particular locality is decreases by certain rate of interest (compounded annually). If the current population of vultures be 29160 and the ratio of decrease in population for second year and 3rd year be 10 : 9. What was the population of vultures 3 years ago?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & \frac{{{\text{Decrease in second year}}}}{{{\text{Decrease in third year}}}} \cr & = \frac{{100}}{{\left( {100 - r} \right)}} \cr & = \frac{{10}}{9} \to r = 10\% \cr} $$
Let the population of vultures 3 years ago be P, then
$$\eqalign{ & P \times {\left[ {1 - \left( {\frac{{10}}{{100}}} \right)} \right]^3} = 29160 \cr & \to P = 40000 \cr} $$
29
A man had 1000 hens at the beginning of year 2001 and the number of hens each year increases by 10% by giving birth. At the end of each year we double the no. of hens by purchasing the same no. of hens as there is the no. of hens with us at the time. What is the no. of hens at the beginning of 2004?
Discuss
Answer & Solution
Answer: Option B
Solution:
1000 == 10%↑ ⇒ 1100 == Double ⇒ 2200 == 10%↑ ⇒ 2420 == Double ⇒ 4840 == 10%↑ ⇒ 5324 == Double ⇒ 10648
30
The difference between simple and compound interest for the fourth year is Rs. 7280 at 20% p.a. What is the principal sum?
Discuss
Answer & Solution
Answer: Option B
Solution:
Difference between Compound interest and Simple interest for the fourth year is Rs. 7280
$$\eqalign{ & P{\left( {\frac{6}{5}} \right)^3}\left( {\frac{1}{5}} \right) - \frac{P}{5} = 7280 \cr & \frac{P}{5}\left[ {{{\left( {\frac{6}{5}} \right)}^3} - 1} \right] = 7280 \cr & \frac{P}{5}\left[ {\frac{{216}}{{125}} - 1} \right] = 7280 \cr & \frac{P}{5} \times \frac{{91}}{{125}} = 7280 \cr & \therefore P = 50000 \cr} $$