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Question 1 of 30
1. Question
1 pointsWhat sum of money must be given as simple interest for six months at 4% per annum in order to earn RS.150 interest ?
Correct
\(Explanation\)
\(P=\frac{150\times100}{4}\times\frac{2}{1}=Rs.7500\)Incorrect
\(Explanation\)
\(P=\frac{150\times100}{4}\times\frac{2}{1}=Rs.7500\) 
Question 2 of 30
2. Question
1 pointsAt some rate of simple interest, A lent RS 6000 to B for 2 years and RS 1500 to C for 4 years and received RS 900 as interest from both of them together. The rate of interest per annum was
Correct
\(EXPLANATIONS\)
if rate of interest be \(R\%\)P.a.then.
\(SI=\frac{principle \times Time \times Rate}{100}\)
\(\therefore\frac{6000\times2\times R}{100}+\frac{1500\times4\times R}{100}\)
\(=900\)
\(\Rightarrow120R+60R=900\)
\(\Rightarrow180R=900\)
\(\Rightarrow R=\frac{900}{180}=5\%\)Incorrect
\(EXPLANATIONS\)
if rate of interest be \(R\%\)P.a.then.
\(SI=\frac{principle \times Time \times Rate}{100}\)
\(\therefore\frac{6000\times2\times R}{100}+\frac{1500\times4\times R}{100}\)
\(=900\)
\(\Rightarrow120R+60R=900\)
\(\Rightarrow180R=900\)
\(\Rightarrow R=\frac{900}{180}=5\%\) 
Question 3 of 30
3. Question
1 pointsA lent RS 5000 to B for 2 years and RS 3000 to C for 4 years on simple interest at the same rate of interest and received RS 2200 in all from both as interest. The rate of interest per annum is
Correct
Let the rate of interest per annum be \(r%\)
According to the questions,
\(\frac {5000\times2\times r}{100}+\frac{3000\times4\times r}{100}=2200\)
\(\Rightarrow100r+120r=2200\)
\(\Rightarrow220r=2200\)
\(\Rightarrow r=\frac{2200}{220}=10\%\)Incorrect
Let the rate of interest per annum be \(r%\)
According to the questions,
\(\frac {5000\times2\times r}{100}+\frac{3000\times4\times r}{100}=2200\)
\(\Rightarrow100r+120r=2200\)
\(\Rightarrow220r=2200\)
\(\Rightarrow r=\frac{2200}{220}=10\%\) 
Question 4 of 30
4. Question
1 pointsA man took a loan from a bank at the rate of 12% per annum at the simple interest . After 3 years he had to pay Rs. 5,400 as interest only for the period. the principal amount borrowed by him was:
Correct
\(Explanation\)
Let the principle be \(x\)
\(SI=\frac{Principal \times Rate \times Time}{100}\)
\(\Rightarrow 5400=\frac{x\times12\times3}{100}\)
\(\Rightarrow x=\frac{5400\times100}{12\times3}=Rs.1,500\)
Incorrect
\(Explanation\)
Let the principle be \(x\)
\(SI=\frac{Principal \times Rate \times Time}{100}\)
\(\Rightarrow 5400=\frac{x\times12\times3}{100}\)
\(\Rightarrow x=\frac{5400\times100}{12\times3}=Rs.1,500\)

Question 5 of 30
5. Question
1 pointsIf the simple interest on a certain sum of money for 15 months at \(7\frac{1}{2}\%\) per annum exceeds the sum for 8 months at \(12\frac{1}{2}\%\) per annum by Rs.32.50, then the sumof money in
Correct
Incorrect

Question 6 of 30
6. Question
1 pointsA some of money becomes \(\frac{41}{40}\) of it self in \(\frac{1}{4}\) years at a certain rate of simple interest. The rate of interest per annum is
Correct
\(Explanation:\)
Let the principal be Re. 1
\(\therefore S.I. =\frac{41}{40}1=\frac{1}{40}\)
Now, rate \(=\frac{Interest\times100}{Prinicipal\times Time}\)
\(=\frac{\frac{1}{40}\times100}{1\times\frac{1}{40}}=\frac{100\times4}{40}=10\)%
Incorrect
\(Explanation:\)
Let the principal be Re. 1
\(\therefore S.I. =\frac{41}{40}1=\frac{1}{40}\)
Now, rate \(=\frac{Interest\times100}{Prinicipal\times Time}\)
\(=\frac{\frac{1}{40}\times100}{1\times\frac{1}{40}}=\frac{100\times4}{40}=10\)%

Question 7 of 30
7. Question
1 pointsRs. 6,000 becomes Rs. 7,200 in 4 years at a certain rate of simple interest. If the rate becomes 1.5 times of it self, the amount of the same principal in 5 years will be
Correct
\(Explanation:\)
Using Rule 1,
SI\(=Rs. (72006000)\)
\(Rs.1200\)
\(\therefore\) SI\(=\frac{PTR}{100}\)
\(\Rightarrow\) \( 1200=\frac{6000\times R \times4}{100}\)
\(\Rightarrow\) \( R=\frac{1200\times100}{6000\times4}=5\)%
New rate of R \(=5\times1.5=7.5\)%
Then, SI \(=\frac{6000\times7.5\times5}{100}\)
\(=Rs. 8250\)
\(\therefore\) Amount \(=Rs. (6000+2250)\)
\(=Rs.8250\)
Incorrect
\(Explanation:\)
Using Rule 1,
SI\(=Rs. (72006000)\)
\(Rs.1200\)
\(\therefore\) SI\(=\frac{PTR}{100}\)
\(\Rightarrow\) \( 1200=\frac{6000\times R \times4}{100}\)
\(\Rightarrow\) \( R=\frac{1200\times100}{6000\times4}=5\)%
New rate of R \(=5\times1.5=7.5\)%
Then, SI \(=\frac{6000\times7.5\times5}{100}\)
\(=Rs. 8250\)
\(\therefore\) Amount \(=Rs. (6000+2250)\)
\(=Rs.8250\)

Question 8 of 30
8. Question
1 pointsA sum amounts to double in 8 years by simple interest. Then the rate of simple interest per annum is
Correct
\(Explanation:\)
Principle \(=Rs.x\)
Amount\(=Rs. 2x\)
\(\therefore\) Interest \(=2xx=Rs. x\)
\(therefore\) Rate \(=\frac{S.I. \times100}{Principle\times Time}\)
\({x\times100}{x\times8}=\frac{25}{2}\)
\(=12.5\)% per annum
Incorrect
\(Explanation:\)
Principle \(=Rs.x\)
Amount\(=Rs. 2x\)
\(\therefore\) Interest \(=2xx=Rs. x\)
\(therefore\) Rate \(=\frac{S.I. \times100}{Principle\times Time}\)
\({x\times100}{x\times8}=\frac{25}{2}\)
\(=12.5\)% per annum

Question 9 of 30
9. Question
1 pointsIn what time will be simple interest be \(\frac{2}{5}\) of the principle at 8 percent per annum ?
Correct
\(Explanation:\)
Let the principle be \(x\)
\(\therefore\) Interest \(=\frac{2}{5} x\)
Rate \(=8\)% per annum
\(\therefore Time=\frac{Interest\times100}{Principle\times Rate}\)
\(=\frac{\frac{2}{5} x\times100}{x\times8}=\frac{40}{8}=5 years\)
Incorrect
\(Explanation:\)
Let the principle be \(x\)
\(\therefore\) Interest \(=\frac{2}{5} x\)
Rate \(=8\)% per annum
\(\therefore Time=\frac{Interest\times100}{Principle\times Rate}\)
\(=\frac{\frac{2}{5} x\times100}{x\times8}=\frac{40}{8}=5 years\)

Question 10 of 30
10. Question
1 pointsThe simple interest on a sum after 4 years is \(\frac{1}{5}\) of the sum. The rate of interest per annum is
Correct
\(Explanation:\)
Let principle \(=Rs. 100\)
S.I. \(=100\frac{1}{5}=Rs.20\)
Rate \(=\frac{20\times100}{100\times4}=5\)%
Incorrect
\(Explanation:\)
Let principle \(=Rs. 100\)
S.I. \(=100\frac{1}{5}=Rs.20\)
Rate \(=\frac{20\times100}{100\times4}=5\)%

Question 11 of 30
11. Question
1 pointsEqual some of money are lent to X and Y at 7.5% per annum for a period of 4 years and 5 years respectively. If the difference in interest, paid by them was Rs. 150, the sum of lent to each was
Correct
\(Explanation:\)
Let the sum lent be \(x\)
\(\therefore \frac{x\times7.5\times5}{100}\frac{x\times7.5\times4}{100}=150\)
\(\Rightarrow \frac{x\times7.5\times1}{100}=150\)
\(\Rightarrow x=\frac{150\times100}{7.5}=Rs.2000\)
Incorrect
\(Explanation:\)
Let the sum lent be \(x\)
\(\therefore \frac{x\times7.5\times5}{100}\frac{x\times7.5\times4}{100}=150\)
\(\Rightarrow \frac{x\times7.5\times1}{100}=150\)
\(\Rightarrow x=\frac{150\times100}{7.5}=Rs.2000\)

Question 12 of 30
12. Question
1 pointsIf the simple interest for 6 years be equal to the 30% of the principal, it will be equal to the principal after
Correct
\(Explanation:\)
Let the principal be P and rate of interest be r%
According to the equation,
\(\frac{30p}{100}=\frac{P\times R\times6}{100}\)
\(\Rightarrow 30=6R\)
\(\Rightarrow R=5\)
Now let interest be equal to principal in T years.
\(\therefore P=\frac{P\times5\times T}{100}\)
\(\Rightarrow T=\frac{100}{5}=20 years.\)
Incorrect
\(Explanation:\)
Let the principal be P and rate of interest be r%
According to the equation,
\(\frac{30p}{100}=\frac{P\times R\times6}{100}\)
\(\Rightarrow 30=6R\)
\(\Rightarrow R=5\)
Now let interest be equal to principal in T years.
\(\therefore P=\frac{P\times5\times T}{100}\)
\(\Rightarrow T=\frac{100}{5}=20 years.\)

Question 13 of 30
13. Question
1 pointsThe simple interest on a certain sum for 8 months at 4% per annum is Rs. 129 less than the simple interest on the same sum for 15 months at 5% per annum. The sum is:
Correct
\(Explanation:\)
Let the sum be \(x\)
\(\frac{x\times5\times15}{100\times12}\frac{x\times4\times8}{100\times12}\)
\(\Rightarrow \frac{x}{100\times12}(7532)=129\)
\(\Rightarrow x=\frac{129\times1200}{43}=Rs. 3600\)
Incorrect
\(Explanation:\)
Let the sum be \(x\)
\(\frac{x\times5\times15}{100\times12}\frac{x\times4\times8}{100\times12}\)
\(\Rightarrow \frac{x}{100\times12}(7532)=129\)
\(\Rightarrow x=\frac{129\times1200}{43}=Rs. 3600\)

Question 14 of 30
14. Question
1 pointsA person invested a total sum of Rs. 7900 in three different schemes of simple interest at 3%, 5% and 8% per annum. At the end of one year he got same interest in all three schemes. What is the money (in Rs.) invested at 3%?
Correct
\(Explanation:\)
Ratio of the rates of S.I. \(=3:5:8\)
\(\therefore\) Ratio of investments
\(=\frac{1}{3}:\frac{1}{5}:\frac{1}{8}\)
\(40:24:15\)
Sum of terms of ratio
\(=40+24+15=79\)
\(\therefore\) Investment at the rate of 3% P.a.
\(Rs. (\frac{40}{79}\times7900)\)
\(Rs. 4000\)
Incorrect
\(Explanation:\)
Ratio of the rates of S.I. \(=3:5:8\)
\(\therefore\) Ratio of investments
\(=\frac{1}{3}:\frac{1}{5}:\frac{1}{8}\)
\(40:24:15\)
Sum of terms of ratio
\(=40+24+15=79\)
\(\therefore\) Investment at the rate of 3% P.a.
\(Rs. (\frac{40}{79}\times7900)\)
\(Rs. 4000\)

Question 15 of 30
15. Question
1 pointsAn old article is available for Rs. 12,ooo at cash payment or is available for Rs. 7,ooo cash payment and a monthly installment of Rs. 630 fro 8 months. The rate percent per annum is
Correct
\(Explanation:\)
Interest
\(=(7000+630\times8)12000\)
\(=(7000+5040)12000\)
\(1204012000=Rs. 40\)
Total principal
\(=5000+4370+3740+3110+2480+1850+1220+590\)
\(=Rs. 22360\)
Rate \(=\frac{40\times100\times12}{22360\times1}=2.1\)%
Incorrect
\(Explanation:\)
Interest
\(=(7000+630\times8)12000\)
\(=(7000+5040)12000\)
\(1204012000=Rs. 40\)
Total principal
\(=5000+4370+3740+3110+2480+1850+1220+590\)
\(=Rs. 22360\)
Rate \(=\frac{40\times100\times12}{22360\times1}=2.1\)%

Question 16 of 30
16. Question
1 pointsAt what percent per annum will Rs. 3000/ amounts to Rs. 3993/ in 3 years if the interest is compounded annually?
Correct
\(Explanation:\)
Using Rule 1.
\(P=Rs. 3000. A=Rs. 3993. n=3 years\)
\(A=P(1+\frac{r}{100})^n\)
\(\therefore(1+\frac{r}{100})^n=\frac{A}{P}\)
\((1+\frac{r}{100})^3=\frac{3993}{3000}=\frac{1331}{1000}\)
\((1+\frac{r}{100})^3=(frac{11}{10})^3\)
\(\Rightarrow 1+\frac{r}{100}=\frac{11}{10}\)
\(\Rightarrow \frac{r}{100}=\frac{11}{10}1\)
\(\Rightarrow \frac{r}{100}=\frac{1}{100} \Rightarrow r=\frac{100}{10}\)
\(\therefore r=10\)%
Incorrect
\(Explanation:\)
Using Rule 1.
\(P=Rs. 3000. A=Rs. 3993. n=3 years\)
\(A=P(1+\frac{r}{100})^n\)
\(\therefore(1+\frac{r}{100})^n=\frac{A}{P}\)
\((1+\frac{r}{100})^3=\frac{3993}{3000}=\frac{1331}{1000}\)
\((1+\frac{r}{100})^3=(frac{11}{10})^3\)
\(\Rightarrow 1+\frac{r}{100}=\frac{11}{10}\)
\(\Rightarrow \frac{r}{100}=\frac{11}{10}1\)
\(\Rightarrow \frac{r}{100}=\frac{1}{100} \Rightarrow r=\frac{100}{10}\)
\(\therefore r=10\)%

Question 17 of 30
17. Question
1 pointsThe compound interest on Rs. 2000 in 2 years if the rate of interest is 4% per annum for the first year and 3% per annum per the second year, will be
Correct
\(Explanation:\)
Using Rule 3,
Amount \(=2000(1+\frac{4}{100})(1+\frac{3}{100})\)
\(=2000\times1.04\times1.03\)
\(=Rs. 2142.40\)
\(\therefore CI =Rs. (2142.402000)\)
\(=Rs. 142.40\)
Incorrect
\(Explanation:\)
Using Rule 3,
Amount \(=2000(1+\frac{4}{100})(1+\frac{3}{100})\)
\(=2000\times1.04\times1.03\)
\(=Rs. 2142.40\)
\(\therefore CI =Rs. (2142.402000)\)
\(=Rs. 142.40\)

Question 18 of 30
18. Question
1 pointsA sum of Rs. 2000 amounts to Rs. 4000 in 2 years at compound interest. In how many years will the same amount be come Rs.8000 ?
Correct
\(Explanation:\)
\(A=P(1+\frac{R}{100})^T\)
\(\Rightarrow 4000=2000(1+\frac{R}{100})^2\)
\(\Rightarrow 2=(1+\frac{R}{100})^2\)
\(\Rightarrow 1+{R}{100}=\sqrt2\) ………(1)
\(\therefore 8000=2000(1+{R}{100})T\)
\(\Rightarrow 4= (\sqrt2)^T\)
\(\Rightarrow (\sqrt2)^4=(\sqrt2)^T\)
\(\Rightarrow T=4 years\)
Incorrect
\(Explanation:\)
\(A=P(1+\frac{R}{100})^T\)
\(\Rightarrow 4000=2000(1+\frac{R}{100})^2\)
\(\Rightarrow 2=(1+\frac{R}{100})^2\)
\(\Rightarrow 1+{R}{100}=\sqrt2\) ………(1)
\(\therefore 8000=2000(1+{R}{100})T\)
\(\Rightarrow 4= (\sqrt2)^T\)
\(\Rightarrow (\sqrt2)^4=(\sqrt2)^T\)
\(\Rightarrow T=4 years\)

Question 19 of 30
19. Question
1 pointsIf the amount is 2.25 times of the sum after 2 years at compound interest (compound annually), the rate of interest per annum is:
Correct
\(Explanation:\)
Suppose P=Rs. 100
and amount A=Rs. 225
\(A=P(1+\frac{r}{100})^t\)
or \(225=100(1+\frac{r}{100})^2\)
or \(\frac{225}{100}=[1+\frac{r}{100}]^2\)
or \(1+\frac{r}{100}=\frac{15}{10}\)
or \(\frac{100+r}{100}=\frac{15}{10}\)
or \(100+r=150\)
or \(r=50%\)
Incorrect
\(Explanation:\)
Suppose P=Rs. 100
and amount A=Rs. 225
\(A=P(1+\frac{r}{100})^t\)
or \(225=100(1+\frac{r}{100})^2\)
or \(\frac{225}{100}=[1+\frac{r}{100}]^2\)
or \(1+\frac{r}{100}=\frac{15}{10}\)
or \(\frac{100+r}{100}=\frac{15}{10}\)
or \(100+r=150\)
or \(r=50%\)

Question 20 of 30
20. Question
1 pointsA sum of money doubles it self in 4 years at compound interest. It will amount to 8 times it self at the same rate of interest in:
Correct
\(Explanation:\)
A sum of \(Rs. x\) becomes \(Rs. 2x\) in 4 years.
Similarly. \(Rs. 2x\) will becomes \(2\times2x\)
\(=Rs. 4x\) in next 4 years and \(Rs. 4x\)
will become \(2\times4x=Rs. 8x\) in at another 4 years.
So, the total time \(=4+4+4=12 years.\)
Incorrect
\(Explanation:\)
A sum of \(Rs. x\) becomes \(Rs. 2x\) in 4 years.
Similarly. \(Rs. 2x\) will becomes \(2\times2x\)
\(=Rs. 4x\) in next 4 years and \(Rs. 4x\)
will become \(2\times4x=Rs. 8x\) in at another 4 years.
So, the total time \(=4+4+4=12 years.\)

Question 21 of 30
21. Question
1 pointsA sum of Rs. 12,000, deposited at compound interest becomes double after 5 years. How much will it be after 20 years ?
Correct
\(Explanation:\)
\(A=P(1+\frac{R}{100})^T\)
\(24000=12000(1+\frac{R}{100})^5\)
\(\Rightarrow 2=(1+\frac{R}{100})^5\)
\(\Rightarrow 2^4=(1+\frac{R}{100})^20\)
i.e The sum amounts to \(Rs. 192000\) afters 20 years.
Incorrect
\(Explanation:\)
\(A=P(1+\frac{R}{100})^T\)
\(24000=12000(1+\frac{R}{100})^5\)
\(\Rightarrow 2=(1+\frac{R}{100})^5\)
\(\Rightarrow 2^4=(1+\frac{R}{100})^20\)
i.e The sum amounts to \(Rs. 192000\) afters 20 years.

Question 22 of 30
22. Question
1 pointsA certain sum of money amounts to Rs. 2,420 in 2 years and Rs. 2,662 in 3 years at some rate of compounded annually. The rate of interest per annum is:
Correct
\(Explanation:\)
Let the rate of interest be r% per annum.
According to the question.
\(4840=P(1+\frac{r}{100})^2\) …….(1)
\(5324=P(1+\frac{r}{100})^3\)…..(2)
on dividing equation (2) by equation (1), we have,
\(1+\frac{r}{100}=\frac{5324}{4840}=1+\frac{484}{4840}\)
\(\Rightarrow \frac{r}{100}=\frac{484}{4840}\)
\(\Rightarrow =10\)%
Incorrect
\(Explanation:\)
Let the rate of interest be r% per annum.
According to the question.
\(4840=P(1+\frac{r}{100})^2\) …….(1)
\(5324=P(1+\frac{r}{100})^3\)…..(2)
on dividing equation (2) by equation (1), we have,
\(1+\frac{r}{100}=\frac{5324}{4840}=1+\frac{484}{4840}\)
\(\Rightarrow \frac{r}{100}=\frac{484}{4840}\)
\(\Rightarrow =10\)%

Question 23 of 30
23. Question
1 pointsA sum of money invested at compound interest amounts to Rs. 650 at the end of first year and Rs. 676 at the end of second year. The sum of money is:
Correct
\(Explanation:\)
Interest on Rs. 650 for 1 year
\(=676650=Rs. 26\)
So, \(r=\frac{26}{650}\times100\)
\(\Rightarrow r=4\)% per annum
\(P=\frac{A}{[1+\frac{r}{100}]^t}=\frac{650}{[1+\frac{4}{100}]^t}\)
\(\frac{650}{\frac{26}{25}}=650\times\frac{25}{26}=Rs. 625\)
Incorrect
\(Explanation:\)
Interest on Rs. 650 for 1 year
\(=676650=Rs. 26\)
So, \(r=\frac{26}{650}\times100\)
\(\Rightarrow r=4\)% per annum
\(P=\frac{A}{[1+\frac{r}{100}]^t}=\frac{650}{[1+\frac{4}{100}]^t}\)
\(\frac{650}{\frac{26}{25}}=650\times\frac{25}{26}=Rs. 625\)

Question 24 of 30
24. Question
1 pointsA sum of money is invested at 20% compound interest (compounded annually.) It would fetch Rs. 723 more in 2 years if interest is compounded half yearly. The sum is
Correct
Ans: 30,000
Incorrect
Ans: 30,000

Question 25 of 30
25. Question
1 pointsOn certain sum of money the compound interest for 2 years is Rs. 282.15 and the simple interest for the same period of time is Rs. 270. The rate of interest per annum is
Correct
\(Explanation:\)
Using Rule 10,
If SI on a certain sum for two years is \(x\) and CI is \(y\), then
\(y=x(1+\frac{r}{200})\)
\(\Rightarrow 282.15=270(1+\frac{r}{100})\)
\(\Rightarrow 1+\frac{r}{200}=\frac{282.15}{270}1\)
\(\Rightarrow \frac{r}{200}=\frac{12.15}{270}\)
\(\Rightarrow r=\frac{12.15\times200}{270}=9\)%
Incorrect
\(Explanation:\)
Using Rule 10,
If SI on a certain sum for two years is \(x\) and CI is \(y\), then
\(y=x(1+\frac{r}{200})\)
\(\Rightarrow 282.15=270(1+\frac{r}{100})\)
\(\Rightarrow 1+\frac{r}{200}=\frac{282.15}{270}1\)
\(\Rightarrow \frac{r}{200}=\frac{12.15}{270}\)
\(\Rightarrow r=\frac{12.15\times200}{270}=9\)%

Question 26 of 30
26. Question
1 pointsThe compound interest on a certain sum of money 2 years at 10% per annum is Rs. 420. The simple interest on the same sum at the same rate and for the same time will be
Correct
\(Explanation:\)
If the principal be P then
\(C.I. =P[(1+\frac{R}{100})^T1]\)
\(\Rightarrow 420=P[(1+\frac{10}{100})^21]\)
\(\Rightarrow420=P(\frac{121100}{100})\)
\(\Rightarrow 420=\frac{P\times21}{100}\)
\(\Rightarrow P=\frac{420\times100}{21}=Rs.2000\)
\(\therefore S.I.=\frac{PRT}{100}\)
\(=\frac{2000\times10\times2}{100}=Rs. 400\)
Incorrect
\(Explanation:\)
If the principal be P then
\(C.I. =P[(1+\frac{R}{100})^T1]\)
\(\Rightarrow 420=P[(1+\frac{10}{100})^21]\)
\(\Rightarrow420=P(\frac{121100}{100})\)
\(\Rightarrow 420=\frac{P\times21}{100}\)
\(\Rightarrow P=\frac{420\times100}{21}=Rs.2000\)
\(\therefore S.I.=\frac{PRT}{100}\)
\(=\frac{2000\times10\times2}{100}=Rs. 400\)

Question 27 of 30
27. Question
1 pointsIf the compound interest on a certain sum for 2 years at 4% p.a. is Rs. 102. the simple interest at the same rate of interest for 2 years would be
Correct
\(Explanation:\)
If the sum be P. then
\(C.I. =P[(1+\frac{R}{100})^T1]\)
\(\Rightarrow 102=P[(1+\frac{4}{100})^21]\)
\(\Rightarrow 102=P[\frac{26}{25}^21]\)
\(\Rightarrow 102=P(\frac{676}{625}1)\)
\(\Rightarrow 102=P(\frac{676625}{625})\)
\(\Rightarrow 102=P\times\frac{51}{625}\)
\(=\frac{102\times625\times2}{51}=Rs. 1250\)
\(\therefore S.I.=\frac{1250\times2\times4}{100}=Rs. 100\)
Incorrect
\(Explanation:\)
If the sum be P. then
\(C.I. =P[(1+\frac{R}{100})^T1]\)
\(\Rightarrow 102=P[(1+\frac{4}{100})^21]\)
\(\Rightarrow 102=P[\frac{26}{25}^21]\)
\(\Rightarrow 102=P(\frac{676}{625}1)\)
\(\Rightarrow 102=P(\frac{676625}{625})\)
\(\Rightarrow 102=P\times\frac{51}{625}\)
\(=\frac{102\times625\times2}{51}=Rs. 1250\)
\(\therefore S.I.=\frac{1250\times2\times4}{100}=Rs. 100\)

Question 28 of 30
28. Question
1 pointsA sum of Rs. 210 was taken as a loan. This is to be paid back in of interest be 10% compounded annually, then the value of each installment.
Correct
Ans: 121
Incorrect
Ans: 121

Question 29 of 30
29. Question
1 pointsThe income of a company increases 20% per year. If the income is Rs. 26,64,000 in the year 2012, then its income in the year 2010 was:
Correct
\(Explanation:\)
Using Rule 1.
Let the income of company in 2010 be Rs. P
According to the equation,
\(A=P(1+\frac{R}{100})^T\)
\(\Rightarrow 2664000=P(1+\frac{20}{100})^2\)
\(\Rightarrow 2664000=P(1+\frac{1}{5})^2\)
\(\Rightarrow 2664000=P\times(\frac{6}{5})^2\)
\(\Rightarrow P=\frac{2664000\times5\times5}{6\times6}\)
\(Rs. 1850000\)
Incorrect
\(Explanation:\)
Using Rule 1.
Let the income of company in 2010 be Rs. P
According to the equation,
\(A=P(1+\frac{R}{100})^T\)
\(\Rightarrow 2664000=P(1+\frac{20}{100})^2\)
\(\Rightarrow 2664000=P(1+\frac{1}{5})^2\)
\(\Rightarrow 2664000=P\times(\frac{6}{5})^2\)
\(\Rightarrow P=\frac{2664000\times5\times5}{6\times6}\)
\(Rs. 1850000\)

Question 30 of 30
30. Question
1 pointsThe amount of Rs. 10,000 after 2 years. compounded annually with the rate of interest being 10% per annum during the first year and 12% per annum during the second year, would be ( in Rupees)
Correct
\(Explanation:\)
\(A=P(1+\frac{R1}{100})(1+\frac{R2}{100})\)
\(=10000(1+\frac{10}{100})(1+\frac{12}{100})\)
\(10000\times\frac{110}{100}\times{112}{100}\)
\(=Rs. 12320\)
Incorrect
\(Explanation:\)
\(A=P(1+\frac{R1}{100})(1+\frac{R2}{100})\)
\(=10000(1+\frac{10}{100})(1+\frac{12}{100})\)
\(10000\times\frac{110}{100}\times{112}{100}\)
\(=Rs. 12320\)