71. There is a decrease of 10% yearly on an article. If this article was bought 3 years ago and present cost is Rs. 5,832 then what was the cost of article at buying time?
(a) Rs. 7,200
(b) Rs. 7,862
(c) Rs. 8,000
(d) Rs. 8,500
(e) None of these
Solution: (c)
\(\displaystyle A=P{{\left( {1-\frac{R}{{100}}} \right)}^{n}}\)
Where A = Value of goods after n years
P = Initial Price
R = Rate of depreciation
Therefore,
\(\displaystyle \begin{array}{l}A=P{{\left( {1-\frac{R}{{100}}} \right)}^{n}}\\P=\frac{A}{{{{{\left( {1-\frac{R}{{100}}} \right)}}^{n}}}}\\\Rightarrow P=\frac{{5832}}{{{{{\left( {1-\frac{{10}}{{100}}} \right)}}^{3}}}}\\\Rightarrow P=\frac{{5832}}{{{{{\left( {1-\frac{1}{{10}}} \right)}}^{3}}}}\\\Rightarrow P=\frac{{5832}}{{{{{\left( {\frac{9}{{10}}} \right)}}^{3}}}}\\\Rightarrow P=5832\times \frac{9}{{10}}\times \frac{9}{{10}}\times \frac{9}{{10}}\\P=8000\end{array}\)
72. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
(a) Rs. 6400
(b) Rs. 6500
(c) Rs. 7200
(d) Rs. 7500
(e) None of these
Solution: (a)
Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 – x)
Then,
\(\displaystyle \begin{array}{l}\frac{{x\times 14\times 2}}{{100}}+\frac{{(13900-x)\times 11\times 2}}{{100}}=3508\\\Rightarrow 28x-22x=350800-(13900\times 22)\\\Rightarrow 6x=45000\\\Rightarrow x=7500\end{array}\)
So, sum invested in Scheme B
=Rs(13900-7500)
=Rs.6400
73. Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?
(a) 3.6
(b) 6
(c) 18
(d) Cannot be determined
(e) None of these
Solution: (b)
Let rate=R%, and Time = R years
Then,
\(\displaystyle \begin{array}{l}\frac{{1200\times R\times R}}{{100}}=432\\12{{R}^{2}}=432\\{{R}^{2}}=36\\R=6\end{array}\)
74. An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes:
(a) 10%
(b) 10.25%
(c) 10.5%
(d) 11%
(e) None of these
Solution: (b)
Let the sum be Rs.100.
Then, S.I. for first 6 months
\(\displaystyle =\frac{{100\times 10\times 1}}{{100\times 2}}\)
=Rs. 5
S.I. for last 6 months
\(\displaystyle =\frac{{105\times 10\times 1}}{{100\times 2}}\)
=Rs. 5.25
So, amount at the end of 1 year
= Rs. (100 + 5 + 5.25)
= Rs. 110.25
Therefore, Effective rate = (110.25 – 100) = 10.25%
75. A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?
(a) 3.46%
(b) 4.55%
(c) 5.36%
(d) 6%
(e) None of these
Solution: (a)
Let the original rate be R%. Then, new rate = (2R)%.
Note:
Here, original rate is for 1 year(s); the new rate is for only 4 months i.e. 1/3 year(s).
\(\displaystyle \begin{array}{l}\frac{{725\times R\times 1}}{{100}}+\frac{{362.50\times 2R\times 1}}{{100\times 3}}=33.50\\\Rightarrow (2175+725)R=33.50\times 100\times 3\\\Rightarrow (2175+725)R=10050\\\Rightarrow R=\frac{{10050}}{{2900}}=3.46\end{array}\)
76. A person borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at \(\displaystyle 6\frac{1}{4}\) p.a for 2 years. Find his gain in the transaction per year.
(a) Rs. 112.50
(b) Rs. 125
(c) Rs. 150
(d) Rs. 167.50
(e) Can’t be determined
Solution: (a)
Gain in 2 years
\(\displaystyle \begin{array}{l}=Rs.\left( {5000\times \frac{{25}}{4}\times \frac{2}{{100}}} \right)-\left( {\frac{{5000\times 4\times 2}}{{100}}} \right)\\=Rs(625-400)\\=Rs.225\end{array}\)
Therefore gain in 1 year
\(\displaystyle =\frac{{225}}{2}=Rs.112.50\)
77. A sum of Rs. 1750 is divided into two parts such that the interests on the first part at 8% simple interest per annum and that on the other part at 6% simple interest per annum are equal. The interest on each part (in Rupees) is?
(a) Rs. 60
(b) Rs. 65
(c) Rs. 70
(d) Rs. 40
(e) Rs. 30
Solution: (a)
Let, Principal = 100 units in both cases
\(\displaystyle \begin{array}{*{20}{c}} {–} & {{{1}^{{st}}}part} & {{{2}^{{nd}}}part} & {Total} \\ {\Pr incipal} & {{{{100}}_{{\times 3}}}} & {{{{100}}_{{\times 4}}}} & {+\to 700units} \\ {Interest} & {{{8}_{{\times 3}}}} & {{{6}_{{\times 4}}}} & . \\ . & . & . & . \end{array}\)
Note : Interest is same in both cases
According to the question,
700 units = Rs. 1750
\(\displaystyle \begin{array}{l}1unit=\frac{{1750}}{{700}}\\24unit=\frac{{1750}}{{700}}\times 24\\=Rs.60\end{array}\)
Alternate method
Principal = Rs. 1750
Let the first part = x
Hence second part = (1750−x)
According to the question,
\(\displaystyle \begin{array}{l}\Rightarrow x\times \frac{8}{{100}}\times 1=(1750-x)\times \frac{6}{{100}}\times 1\\\Rightarrow 4x=5250-3x\\\Rightarrow x=750\end{array}\)
Therefore, Second part = Rs. (1750−750)
= Rs. 1000
Required interest is,
\(\displaystyle \begin{array}{l}=750\times \frac{8}{{100}}\\=Rs.60\end{array}\)
78. Ramakant invested amounts in two different schemes A and B for five years in the ratio of 5 : 4 respectively. Scheme A offers 8% simple interest and bonus equal to 20% of the amount of interest earned in 5 years on maturity. Scheme B offers 9% simple interest. If the amount invested in scheme A was Rs. 20000, what was the total amount received on maturity from both the schemes?
(a) Rs. 50800
(b) Rs. 51200
(c) Rs. 52800
(d) Rs. 58200
(e) Rs. 60300
Solution: (c)
Let the amounts invested in schemes A and B be Rs. 5x and 4x respectively.
then,
5x = 20000
\(\displaystyle \Rightarrow \)x = 4000
Therefore, Amount invested in scheme B = Rs. 16000
Total interest received on maturity
\(\displaystyle \begin{array}{l}=Rs\left[ {120\%of\left( {\frac{{20000\times 8\times 5}}{{100}}} \right)+\left( {\frac{{16000\times 9\times 5}}{{100}}} \right)} \right]\\=Rs.\left( {120\%of8000+7200} \right)\\=Rs.\left( {9600+7200} \right)\\=Rs.52800\end{array}\)
79. Mohan lent some amount of money at 9% simple interest and an equal amount of money at 10% simple interest each for two years. If his total interest was Rs. 760, what amount was lent in each case?
(a) Rs. 1700
(b) Rs. 1800
(c) Rs. 1900
(d) Rs. 2000
(e) Rs. 2500
Solution: (d)
Total interest percent
= (9×2)%+(10×2)%
\(\displaystyle \Rightarrow \)38%=760
\(\displaystyle \Rightarrow \)100%=2000
Hence required principal = Rs. 2000
Alternate method
Let the amount invested = Rs. P
As per the questions,
\(\displaystyle \begin{array}{l}\Rightarrow \frac{{P\times 9\times 2}}{{100}}+\frac{{P\times 10\times 2}}{{100}}=760\\\Rightarrow \frac{{18P}}{{100}}+\frac{{20P}}{{100}}=760\\\Rightarrow 38P=76000\\\Rightarrow P=2000\end{array}\)
80. A person invests money in three different schemes for 6 years, 10 years and 12 years at 10%, 12% and 15% simple interest respectively. At the completion of each scheme, he gets the same interest. The ratio of his investment is?
(a) 6 : 3 : 2
(b) 2 : 3 : 4
(c) 3 : 4 : 6
(d) 3 : 4 : 2
(e) None of these
Solution: (a)
TLet the required ratio be x:y:z. Then,
S.I. on Rs. x for 6 years at 10% p.a. = S.I. on Rs. y for 10 years at 12% p.a.
\(\displaystyle \begin{array}{l}\Rightarrow \frac{{x\times 6\times 10}}{{100}}=\frac{{y\times 10\times 12}}{{100}}\\\Rightarrow \frac{x}{y}=\frac{2}{1}\end{array}\)
S.I. on Rs. y for 10 years at 12% p.a. = S.I. on Rs. z for 12 years at 15% p.a.
\(\displaystyle \begin{array}{l}\Rightarrow \frac{{y\times 10\times 12}}{{100}}=\frac{{z\times 12\times 15}}{{100}}\\\Rightarrow \frac{y}{z}=\frac{3}{2}\end{array}\)
\(\displaystyle \Rightarrow \)x:y=2:1 and y:z=3:2
\(\displaystyle \Rightarrow \)x:y=6:3 and y:z=3:2
\(\displaystyle \Rightarrow \)x:y:z=6:3:2.
Alternate method
Let the principal in each case = 100 units as per the question,
\(\displaystyle \begin{array}{*{20}{c}} . & {1part} & {2part} & {3part} \\ {\Pr incipal} & {{{{100}}_{{x6}}}} & {{{{100}}_{{x3}}}} & {{{{100}}_{{x2}}}} \\ {Rate\%} & {10} & {12} & {15} \\ {Time} & 6 & {10} & {12} \end{array}\)
We can see that interest earned in each case is same, i:e \(\displaystyle {{60}_{{x6}}}\), \(\displaystyle {{120}_{{x3}}}\), and \(\displaystyle {{180}_{{x2}}}\)
Hence required ratio
= 600 : 300 : 200 (of sum)
= 6 : 3 : 2
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