41. Geeta borrowed some money at the rate of 6% p.a for the first two years, at the rate of 9% p.a for the next three years, and at the rate of 14% p.a for the period beyond five years. If she pays a total interest of Rs.11400 at the end of nine years, how much did she boorrow?
(a) Rs 10,000
(b) Rs 11,000
(c) Rs 12,000
(d) Rs 14,000
(e) Rs 15,000
Solution: (c)
Let the sum borrowed be ‘x’, then
\(\displaystyle \begin{array}{l}\left( {\frac{{x\times 6\times 2}}{{100}}+\frac{{x\times 9\times 3}}{{100}}+\frac{{x\times 14\times 4}}{{100}}} \right)=11400\\\Rightarrow \left( {\frac{{3x}}{{25}}+\frac{{27x}}{{100}}+\frac{{14x}}{{25}}} \right)=11400\\\Rightarrow \left( {\frac{{95x}}{{100}}} \right)=11400\\\Rightarrow x=\left( {\frac{{11400\times 100}}{{95}}} \right)=12000\end{array}\)
Hence, sum borrowed = Rs. 12,000.
42. In a certain time, a sum becomes 3 times at the rate of 5% per annum. At what rate of interest, the same sum becomes 6 times in same duration?
(a) 7%
(b) 9%
(c) 12.5%
(d) 15%
(e) 17%
Solution: (c)
Net percent rate
\(\displaystyle \left( {{{r}_{2}}} \right)=\left( {\frac{{y-1}}{{x-1}}} \right)\times {{r}_{1}}\)
x is the no. of times the sum becomes of itself in the 1st scenario = 3
y is the no. of times the sum becomes of itself in the 1st scenario = 6
\(\displaystyle {{r}_{1}}\) is the rate of interest in the 1st scenario = 5%
\(\displaystyle {{r}_{2}}\) is the rate of interest in the 2nd scenario = ?
By the short trick approach, we get
Required rate percent,
\(\displaystyle \left( {{{r}_{2}}} \right)=\left( {\frac{{6-1}}{{3-1}}} \right)\times 5=\frac{{25}}{2}=12.5\%\)
Alternate method,
3P = P + S.I; S.I = 2P
i.e. if Amount = 6P then S.I = 5P
\(\displaystyle \begin{array}{l}SI=\frac{{PRT}}{{100}}\\2P=\frac{{PRT}}{{100}}\\\Rightarrow T=\frac{{200}}{5}\end{array}\)
if Amount = 6P then S.I = 5P
\(\displaystyle \begin{array}{l}SI=\frac{{PRT}}{{100}}\\5P=\frac{{P\times R\times 200}}{{100\times 5}}\\\Rightarrow R=\frac{{25}}{2}=12.5\%\end{array}\)
43. A certain sum of money amounts to 756 in 2 years and to 873 in \(\displaystyle 3\frac{1}{2}\) years at a certain rate of simple interest. The rate of interest per annum is
(a) 10%
(b) 11%
(c) 12%
(d) 13%
(e) 14%
Solution: (d)
Net percent rate
S.I .for \(\displaystyle 1\frac{1}{2}\) years
= ₹ (873 – 756) = ₹ 117
S.I. for 2 years
= \(\displaystyle (117\times \frac{2}{3}\times 2)=156\)
Therefore, Principal = 756 – 156 = 600
Now, P = 600, T = 2, S.I. = 156
\(\displaystyle R=\frac{{100\times S.I.}}{{P\times T}}\)
\(\displaystyle \frac{{100\times 156}}{{600\times 2}}=13\%\)
Alternate method:
Rate of interest = \(\displaystyle (\frac{{{{A}_{1}}-{{A}_{2}}}}{{{{A}_{2}}{{T}_{1}}-{{A}_{1}}{{T}_{2}}}})\times 100\)
\(\displaystyle (\frac{{756-873}}{{873\times 2-756\times \frac{7}{2}}})\times 100\)
\(\displaystyle (\frac{{-117}}{{1746-2646}})\times 100\)
\(\displaystyle (\frac{{-117}}{{-900}})\times 100=13\%\)
44. What sum will amount to ₹ 7000 in 5 years at \(\displaystyle 3\frac{1}{2}\) simple interest?
45. A man took a loan from a bank at the rate of 12% per annum at simple interest. After 3 years he had to pay ₹ 5,400 as interest only for the period. The principal amount borrowed by him was:
(a) ₹ 2,000
(b) ₹ 10,000
(c) ₹ 20,000
(d) ₹ 15,000
(e) ₹ 22,000
Solution: (d)
Let the principal be x.
S.I. = \(\displaystyle \frac{{principal\times rate\times time}}{{100}}\)
\(\displaystyle \Rightarrow \)\(\displaystyle 5400=\frac{{x\times 12\times 3}}{{100}}\)
\(\displaystyle \Rightarrow \)\(\displaystyle x=\frac{{5400\times 100}}{{12\times 3}}\)=15000
46. A sum of money at simple interest amounts to ₹ 1,012 in \(\displaystyle 2\frac{1}{2}\) and to ₹ 1,067.20 in 4 years. The rate of interest per annum is:
(a) 2.5%
(b) 3%
(c) 4%
(d) 5%
(e) 6%
Solution: (c)
Principal + S.I. for \(\displaystyle \frac{5}{2}\) years = 1012 …(i)
Principal + S.I. for 4 years = 1067.20 …(ii)
Subtracting equation (i) from (ii)
S.I. for \(\displaystyle \frac{5}{2}\) years = ₹ 55.20
S.I. for \(\displaystyle \frac{5}{2}\) years = \(\displaystyle 55.20\times \frac{2}{3}\times \frac{5}{2}=92\)
Principal = (1012 – 92) = 920
\(\displaystyle \Rightarrow \)Rate = \(\displaystyle \frac{{92\times 100}}{{920\times \frac{5}{2}}}\)
\(\displaystyle \frac{{2\times 92\times 100}}{{920\times 5}}=4\%\)
Alternate method:
R = \(\displaystyle (\frac{{{{A}_{1}}-{{A}_{2}}}}{{{{A}_{2}}{{T}_{1}}-{{A}_{1}}{{T}_{2}}}})\times 100\)
\(\displaystyle (\frac{{1012-1067.20}}{{1067.20\times \frac{5}{2}-1012\times 4}})\times 100\)
\(\displaystyle \frac{{-55.2}}{{(2668-4048)}}\times 100\)
\(\displaystyle \frac{{-55.2}}{{-1380}}\times 100=4\%\)
47. A sum of money lent out at simple interest amounts to 720 after 2 years and to 1020 after a further period of 5 years. The sum is:
(a) 500
(b) 600
(c) 700
(d) 710
(e) 750
Solution: (b)
Principal + SI for 2 years = ₹ 720 …. (i)
Principal + SI for 7 years = ₹ 1020 …..(ii)
Subtracting equation (i) from (ii) get,
SI for 5 years = ₹(1020 – 720) = ₹ 300
\(\displaystyle \Rightarrow \)SI for 2 years = ₹ \(\displaystyle 300\times \frac{2}{5}=120\)
\(\displaystyle \Rightarrow \)Principal = ₹ (720 – 120) = ₹ 600
Alternate method :
P = \(\displaystyle \frac{{{{A}_{2}}{{T}_{1}}-{{A}_{1}}{{T}_{2}}}}{{{{T}_{1}}-{{T}_{2}}}}\)
\(\displaystyle \frac{{1020\times 2-720\times 7}}{{2-7}}\)
\(\displaystyle \frac{{2040-5040}}{{-5}}\)
\(\displaystyle \frac{{-300}}{{-5}}=600\)
48. The sum of money, that will give ₹ 1 as interest per day at the rate of 5% per annum simple interest is
(a) ₹ 3650
(b) ₹ 36500
(c) ₹ 730
(d) ₹ 7300
(e) ₹ 7500
Solution: (d)
The sum of money will give ₹ 365 as simple interest in a year.
S.I. = \(\displaystyle \frac{{PRT}}{{100}}\)
\(\displaystyle 365=\frac{{P\times 5\times 1}}{{100}}\)
P = \(\displaystyle \frac{{365\times 100}}{5}=7300\)
49. If the simple interest on a certain sum of money for 15 months at \(\displaystyle 7\frac{1}{2}\)% per annum exceeds the simple interest on the same sum for 8 months at \(\displaystyle 12\frac{1}{2}\) per annum by ₹ 32.50, then the sum of money (in ₹ ) is:
(a) 312
(b) 312.50
(c) 3120
(d) 3120.50
(e) 3120.05
Solution: (c)
Let the sum be x.
Using formula, I = \(\displaystyle \frac{{PRT}}{{100}}\) we have
\(\displaystyle \frac{{x\times \frac{{15}}{{12}}\times \frac{{15}}{2}}}{{100}}-\frac{{x\times \frac{8}{{12}}\times \frac{{25}}{2}}}{{100}}\) = 32.50
\(\displaystyle \frac{{25x}}{{2400}}=32.50\)
\(\displaystyle x=\frac{{32.50\times 2400}}{{25}}=3120\)
\(\displaystyle \Rightarrow \)Required sum=₹3120
50. What annual instalment will discharge a debt of ₹ 6450 due in 4 years at 5% simple interest ?
(a) ₹ 1500
(b) ₹ 1835
(c) ₹ 1935
(d) ₹ 1950
(e) ₹ 1980
Solution: (a)
Let each instalment be x Then,
\(\displaystyle (x+\frac{{x\times 5\times 1}}{{100}})+(x+\frac{{x\times 5\times 2}}{{100}})+(x+\frac{{x\times 5\times 3}}{{100}})+x=6450\)
\(\displaystyle (x+\frac{x}{{20}})+(x+\frac{x}{{10}})+(x+\frac{{3x}}{{20}})+x=6450\)
\(\displaystyle \frac{{21x}}{{20}}+\frac{{11x}}{{10}}+\frac{{23x}}{{20}}+x=6450\)
\(\displaystyle \frac{{21x+22x+23x+20x}}{{20}}=6450\)
\(\displaystyle \frac{{86x}}{{20}}=6450\)
X = \(\displaystyle \frac{{6450\times 20}}{{86}}=1500\)
Alternate method,
Equal instalment = \(\displaystyle \frac{{6450\times 200}}{{4[200+(4-1)\times 5]}}\)
\(\displaystyle \frac{{6450\times 200}}{{4(215)}}\)
\(\displaystyle \frac{{6450\times 50}}{{215}}=1500\)
