The simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is?
(a) Rs. 1550
(b) Rs. 1650
(c) Rs. 1750
(d) Rs. 2000
(e) None of these
Solution: (c)
P=Rs. 4000
CI for 2 years @10%
CI\(\displaystyle \begin{array}{l}=\left[ {4000\times {{{\left( {1+\frac{{10}}{{100}}} \right)}}^{2}}-4000} \right]\\=4000\left[ {{{{\left( {\frac{{11}}{{10}}} \right)}}^{2}}-1} \right]\\=840\end{array}\)
Given, Half the CI is SI
SI=420
SI= \(\displaystyle \frac{{PRT}}{{100}}\)
\(\displaystyle \begin{array}{l}420=\frac{{P\times 8\times 3}}{{100}}\\P=\frac{{420\times 100}}{{24}}\\P=1750\end{array}\)
A sum of money placed at compound interest doubles itself in 4 years. It will amount to eight times itself at the same rate in
(a) 2 years
(b) 4 years
(c) 8 years
(d) 10 years
(e) 12 years
Solution: (e)
Rs 1 becomes Rs 2 in 4 years
Rs 2 becomes Rs 4 in another 4 years
Rs 4 becomes Rs 8 in another 4 years
\(\displaystyle \Rightarrow \)Rs 1 becomes Rs 8 in (4 + 4 + 4) i.e., 12 years.
Sachin had ₹ 23,000. He invested some amount in scheme A at SI at 20% and the remaining amount in scheme B at CI at 10%. If Sahil got the same amount from both of them at the end of one year, how much (in ₹) did he invest in scheme B ?
(a) ₹10000
(b) ₹11000
(c) ₹11500
(d) ₹12000
(e) ₹13000
Solution: (d)
Let the amount invested in scheme B is ₹ x.
Therefore, Amount invested in scheme A be ₹ (23000 – x).
According to the question,
\(\displaystyle \left( {23000-x} \right)+\frac{{\left( {23000-x} \right)\times 2\times 1}}{{100}}=x+x{{\left( {1+\frac{{10}}{{100}}} \right)}^{1}}-x\)
\(\displaystyle \Rightarrow \left( {23000-x} \right)\times \frac{6}{5}=\frac{{11x}}{{10}}\)
\(\displaystyle \Rightarrow 11x=276000-12x\)
\(\displaystyle \Rightarrow x=12000\)
Sanjay purchased a hotel worth rupees 10 lakhs and Anita purchased a car worth Rs. 16 lakh. The value of hotel every year increase by 20% of the previous value and the value of car every depreciates by 25%. What is the difference between the price of hotel and car after 3 years?
(a) 10,53,000
(b) 10,63,000
(c) 11,53,000
(d) 10,43,000
(e) None of these
Solution: (a)
Amount of the hotel after 3 years = \(\displaystyle 1000000{{\left( {1+\frac{{20}}{{100}}} \right)}^{3}}\)
\(\displaystyle 1000000{{\left( {\frac{6}{5}} \right)}^{3}}\)
=17,28,000
Amount of the car after 3 years = \(\displaystyle 1600000{{\left( {1-\frac{{20}}{{100}}} \right)}^{3}}\)
\(\displaystyle 1600000{{\left( {\frac{3}{4}} \right)}^{3}}\)
= 6,75,000
Difference = 17,28,000 – 6,75,000 = 10,53,000.
A has lent some money to B at 6% p.a. and C at 10% at the end of the year he has gain the overall interest at 8% p.a. in what ratio has he lent the money to A and B?
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 2 : 3
(e) 3 : 2
Solution (a)
Using Allegation & Mixture
So, the ratio is- 2 : 2 = 1:1.
An amount of Rs. ‘x’ at compound interest at 20% per annum for 3 year becomes ‘y’. what is y : x?
(a) 3 : 1
(b) 36 : 25
(c) 216 : 125
(d) 125 : 216
(e) Cannot be determined
Solution: (c)
Let P = ₹ x, r = 20%, t = 3 year, A = ₹ y
We know that, Amount= \(\displaystyle =P\times {{\left( {1+\frac{r}{{100}}} \right)}^{n}}\)
\(\displaystyle y=x\times {{\left( {1+\frac{{20}}{{100}}} \right)}^{3}}\)
\(\displaystyle \frac{y}{x}={{\left( {\frac{6}{5}} \right)}^{3}}=\frac{{216}}{{125}}\)
\(\displaystyle \Rightarrow y:x=216:125\)
A man saves Rs.20,000 at the beginning of each year and puts the money in a bank that pays 10% interest per year, interest being compounded annually. What would be the total savings of the man at the end of 6 years?
(a) Rs. 196840
(b) Rs. 169840
(c) Rs. 189480
(d) Rs. 199480
(e) Rs. 168840
Solution: (b)
The first Rs. 20000 would become 20000 \(\displaystyle {{\left( {1.1} \right)}^{6}}\) after 6 years, the second will become 20000 \(\displaystyle {{\left( {1.1} \right)}^{5}}\), the third will become 20000 \(\displaystyle {{\left( {1.1} \right)}^{4}}\), the fourth will become 20000 \(\displaystyle {{\left( {1.1} \right)}^{3}}\), the fifth will become 20000 \(\displaystyle {{\left( {1.1} \right)}^{2}}\) and the sixth will become 20000 (1.1).
Total amount = \(\displaystyle 20000\left[ {(1.1)+{{{\left( {1.1} \right)}}^{2}}+{{{\left( {1.1} \right)}}^{3}}+{{{\left( {1.1} \right)}}^{4}}+{{{\left( {1.1} \right)}}^{5}}+{{{\left( {1.1} \right)}}^{6}}} \right]\)
= \(\displaystyle \left( {20000} \right)\left( {1.1} \right)\left[ {1+\left( {1.1} \right)+{{{\left( {1.1} \right)}}^{2}}+{{{\left( {1.1} \right)}}^{3}}+{{{\left( {1.1} \right)}}^{4}}+{{{\left( {1.1} \right)}}^{5}}} \right]\)
= \(\displaystyle \left( {20000} \right)\frac{{{{{(1.1)}}^{6}}-1}}{{1.1-1}}\)
= \(\displaystyle 20000(7.72)\)
= \(\displaystyle Rs.169840\)
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