{"id":27093,"date":"2019-04-05T16:22:52","date_gmt":"2019-04-05T10:52:52","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=27093"},"modified":"2019-04-05T16:22:52","modified_gmt":"2019-04-05T10:52:52","slug":"compound-interest-questions-for-ssc-chsl-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/compound-interest-questions-for-ssc-chsl-pdf\/","title":{"rendered":"Compound Interest Questions for SSC CHSL PDF"},"content":{"rendered":"

Compound Interest Questions for SSC CHSL PDF:<\/strong><\/span><\/h2>\n

SSC CHSL Physics Previous Year Questions download PDF based on previous year question paper of SSC exams. 25 Very important Physics Previous Year questions for SSC CHSL Exam.<\/p>\n

Download Compound Interest Questions for SSC CHSL PDF<\/a><\/p>\n\n

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Question 1:\u00a0<\/b>On a certain sum of money, the simple interest for 2 years is Rs. 350 at the rate of 4% per annum. It was invested at compound interest at the same rate for the same duration as before, how much more interest would be earned ?<\/p>\n

a)\u00a0Rs. 3.50<\/p>\n

b)\u00a0Rs. 7<\/p>\n

c)\u00a0Rs. 14<\/p>\n

d)\u00a0Rs. 35<\/p>\n

Question 2:\u00a0<\/b>The compound interest on a sum of Rs. 5000 at 8% per annum for 9 months when interest is compound quarterly is:<\/p>\n

a)\u00a0Rs. 300<\/p>\n

b)\u00a0Rs. 300.12<\/p>\n

c)\u00a0Rs. 306.04<\/p>\n

d)\u00a0Rs. 308<\/p>\n

Question 3:\u00a0<\/b>A certain amount grows at an annual interest rate of 12%, compounded monthly. Which of the following equations can be solved to find the number of years, y, that it would take for the investment to increase by a factor of 64 ?<\/p>\n

a)\u00a0$64=(1.01)^{12y}$<\/p>\n

b)\u00a0$\\frac{1}{64}= (1.04)12y$<\/p>\n

c)\u00a0$64=(1.04)^{12y}$<\/p>\n

d)\u00a0$8=(1.01)^{6y}$<\/p>\n

Question 4:\u00a0<\/b>What Is the compound interest (in Rs.) on a principal sum of Rs. 2800 for 2 years at the rate of 12% per annum?<\/p>\n

a)\u00a0687.18<\/p>\n

b)\u00a0634.46<\/p>\n

c)\u00a0712.32<\/p>\n

d)\u00a0568.68<\/p>\n

Question 5:\u00a0<\/b>If interest being compound half yearly then what sum (in Rs.) will amount to Rs. 38416 in 2 years at the rate of 80% per annum at compound interest ?<\/p>\n

a)\u00a014000<\/p>\n

b)\u00a015000<\/p>\n

c)\u00a010000<\/p>\n

d)\u00a012000<\/p>\n

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Question 6:\u00a0<\/b>A sum of Rs. 12,000, deposited at compoundd interest becomes double after 5 years. How much will it be after 20 years ?<\/p>\n

a)\u00a0Rs. 1,44,000<\/p>\n

b)\u00a0Rs. 1,20,000<\/p>\n

c)\u00a0Rs. 1, 50,000<\/p>\n

d)\u00a0Rs. 1,92,000<\/p>\n

Question 7:\u00a0<\/b>The difference between simple interest and compound interest of a certain sum of money at 20% per annum for 2 years is Rs. 48. Then the sum is<\/p>\n

a)\u00a0Rs. 1,000<\/p>\n

b)\u00a0Rs. 1,200<\/p>\n

c)\u00a0Rs. 1,500<\/p>\n

d)\u00a0Rs. 2,000<\/p>\n

Question 8:\u00a0<\/b>A sum of Rs. 12,000 deposited at compound interest becomes double after 5 years. After 20 years, it will become<\/p>\n

a)\u00a0Rs. 48,000<\/p>\n

b)\u00a0Rs. 96,000<\/p>\n

c)\u00a0Rs. 1,90,000<\/p>\n

d)\u00a0Rs. 1,92,000<\/p>\n

Question 9:\u00a0<\/b>The difference between the compound interest and simple interest for the amount Rs. 5,000 in 2 years is Rs.32. The rate of interest is<\/p>\n

a)\u00a05%<\/p>\n

b)\u00a08%<\/p>\n

c)\u00a010%<\/p>\n

d)\u00a012%<\/p>\n

Question 10:\u00a0<\/b>There is 100% increase to an amount in 8 years, at simple interest. Find the compound interest of Rs. 8000 after 2 years at the same rate of interest.<\/p>\n

a)\u00a0Rs. 2500<\/p>\n

b)\u00a0Rs. 2000<\/p>\n

c)\u00a0Rs. 2250<\/p>\n

d)\u00a0Rs. 2125<\/p>\n

SSC CHSL FREE MOCK TEST<\/a><\/p>\n

Question 11:\u00a0<\/b>The compound interest generated on a sum of Rs 5000, in two years at 5% per annum, the interest is being compounded half yearly is.<\/p>\n

a)\u00a0Rs 557<\/p>\n

b)\u00a0Rs 489<\/p>\n

c)\u00a0Rs 519<\/p>\n

d)\u00a0Rs 362<\/p>\n

Question 12:\u00a0<\/b>The difference between simple interest and compound interest (at the same rate of interest) for two years on a sum of Rs. 25000 is Rs. 1000. What is the common rate of interest?<\/p>\n

a)\u00a010%<\/p>\n

b)\u00a020%<\/p>\n

c)\u00a030%<\/p>\n

d)\u00a040%<\/p>\n

Question 13:\u00a0<\/b>Sohan deposited Rs 1 lac in a bank at a certain simple interest rate such that the amount got doubled after 5 years. If Sohan had deposited the same amount at the same interest rate compounded annually, the amount after 2 years would have been<\/p>\n

a)\u00a0Rs 1.44 lacs<\/p>\n

b)\u00a0Rs 1.21 lacs<\/p>\n

c)\u00a0Rs 1.25 lacs<\/p>\n

d)\u00a0Rs 1.69 lacs<\/p>\n

Question 14:\u00a0<\/b>A certain sum of money will amount to 24200 in 2 years at 10% per annum compounded annually. What will this sum amount to if it is invested at 8% per annum simple interest for 3 years?<\/p>\n

a)\u00a025000<\/p>\n

b)\u00a024800<\/p>\n

c)\u00a024200<\/p>\n

d)\u00a024000<\/p>\n

Question 15:\u00a0<\/b>What is the difference between simple interest and compound interest (at the same rate of interest) for two years on a sum of Rs. 20000? The common rate of interest is 10%.<\/p>\n

a)\u00a0Rs. 200<\/p>\n

b)\u00a0Rs. 400<\/p>\n

c)\u00a0Rs. 600<\/p>\n

d)\u00a0Rs. 800<\/p>\n

FREE SSC MATERIAL – 18000 FREE QUESTIONS<\/a><\/p>\n

Question 16:\u00a0<\/b>If the difference between simple interest and compound interest for two years on a sum of Rs. 50000 is Rs. 12500 when the rate of interest is the same, what is the common rate of interest?<\/p>\n

a)\u00a040%<\/p>\n

b)\u00a050%<\/p>\n

c)\u00a045%<\/p>\n

d)\u00a035%<\/p>\n

Question 17:\u00a0<\/b>Ganesh invests a sum of money in a compound interest which pays 20% annually. How many years will it take for Ganesh to get back double the amount he invested?<\/p>\n

a)\u00a03<\/p>\n

b)\u00a04<\/p>\n

c)\u00a05<\/p>\n

d)\u00a0Cannot be determined<\/p>\n

Question 18:\u00a0<\/b>A bank offers 12% compound interest on a sum of Rs. 10,000 compounded every quarter. What will be the approximate interest earned at the end of 1 year?<\/p>\n

a)\u00a0Rs 1055<\/p>\n

b)\u00a0Rs 1155<\/p>\n

c)\u00a0Rs 1255<\/p>\n

d)\u00a0Rs 1355<\/p>\n

Question 19:\u00a0<\/b>The compound interest on a certain sum of money for 2 years at 5% per annum is 410. The simple interest on the same sum at the same rate and for the same time is<\/p>\n

a)\u00a0400<\/p>\n

b)\u00a0300<\/p>\n

c)\u00a0350<\/p>\n

d)\u00a0405<\/p>\n

Question 20:\u00a0<\/b>The compound interest on =1,800 at 10% per annum for a certain period of time is 378. Find the time in years.<\/p>\n

a)\u00a02.0 years<\/p>\n

b)\u00a02.8 years<\/p>\n

c)\u00a03.0 years<\/p>\n

d)\u00a02.5 year<\/p>\n

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Question 21:\u00a0<\/b>If the compound interest on a certain sum for two years at 12% per annum is 2,544, the simple interest on it at the same rate for 2 years will be<\/p>\n

a)\u00a02,400<\/p>\n

b)\u00a02,500<\/p>\n

c)\u00a02,480<\/p>\n

d)\u00a02,440<\/p>\n

Question 22:\u00a0<\/b>A man borrows Rs. 21000 at 10% compound interest. How much he has to pay equally at the end of each year, to settle his loan in two years ?<\/p>\n

a)\u00a0Rs. 12000<\/p>\n

b)\u00a0Rs. 12100<\/p>\n

c)\u00a0Rs. 12200<\/p>\n

d)\u00a0Rs. 12300<\/p>\n

Question 23:\u00a0<\/b>The time in which 80,000 ambunts to 92,610 at 10% p.a. at compound interest, interest being compounded semi annually is :<\/p>\n

a)\u00a01.5 years<\/p>\n

b)\u00a02 years<\/p>\n

c)\u00a02.5 years<\/p>\n

d)\u00a03 years<\/p>\n

Question 24:\u00a0<\/b>A sum of money at compound interest will amount to 650 at the end of the first year and 676 at the end of the second year. The amount of money is<\/p>\n

a)\u00a01,300<\/p>\n

b)\u00a0650<\/p>\n

c)\u00a01,250<\/p>\n

d)\u00a0625<\/p>\n

Question 25:\u00a0<\/b>A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to four times itself ?<\/p>\n

a)\u00a012 years<\/p>\n

b)\u00a013 years<\/p>\n

c)\u00a08 years<\/p>\n

d)\u00a016 years<\/p>\n

FREE MOCK TEST FOR SSC STENOGRAPHER<\/a><\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Rate of interest = 4%\u00a0and time period = 2 years<\/p>\n

Let principal sum = Rs.\u00a0$x$<\/p>\n

Simple interest = $\\frac{P\\times R\\times T}{100}$<\/p>\n

=> $\\frac{x\\times4\\times2}{100}=350$<\/p>\n

=> $x=\\frac{35000}{8}=4375$<\/p>\n

Now, interest earned under compound interest = $P[(1+\\frac{R}{100})^T-1]$<\/p>\n

=\u00a0$4375[(1+\\frac{4}{100})^2-1]$<\/p>\n

=\u00a0$4375[(\\frac{26}{25})^2-1]$<\/p>\n

= $4375\\times\\frac{676-625}{625}$<\/p>\n

= $7\\times51=Rs.$ $357$<\/p>\n

$\\therefore$ Difference in interest = $357-350=Rs.$ $7$<\/p>\n

=> Ans – (B)<\/p>\n

2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Principal sum = Rs. 5000<\/p>\n

Rate of interest = 8% and time period = $\\frac{9}{12}=\\frac{3}{4}$ years<\/p>\n

Compound interest\u00a0when interest is compound quarterly\u00a0= $P[(1+\\frac{R}{400})^{4T}-1]$<\/p>\n

= $5000[(1+\\frac{8}{400})^{\\frac{3}{4}\\times4}-1]$<\/p>\n

= $5000[(1+\\frac{1}{50})^3-1]$<\/p>\n

= $5000[(\\frac{51}{50})^3-1]$<\/p>\n

= $5000\\times(\\frac{132651-125000}{125000})$<\/p>\n

= $\\frac{7651}{25}=Rs.$ $306.04$<\/p>\n

=> Ans – (C)<\/p>\n

3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Rate of interest = 12% p.a. = 1% per month<\/p>\n

Time = $12y$ months<\/p>\n

Let principal = Re 1 and thus amount = Rs. 64<\/p>\n

$\\therefore$ $A=P(1+\\frac{R}{100})^T$<\/p>\n

=> $64=1(1+\\frac{1}{100})^{12y}$<\/p>\n

=> $64=(1.01)^{12y}$<\/p>\n

=> Ans – (A)<\/p>\n

4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Principal sum = Rs. 2800<\/p>\n

Rate of interest = 12% and time = 2years<\/p>\n

Compound interest = $P[(1+\\frac{R}{100})^T-1]$<\/p>\n

=\u00a0$2800[(1+\\frac{12}{100})^2-1]$<\/p>\n

= $2800[(\\frac{28}{25})^2-1]$<\/p>\n

= $2800\\times(\\frac{784-625}{625})$<\/p>\n

= $2800\\times\\frac{159}{625}=Rs.$ $712.32$<\/p>\n

=> Ans – (C)<\/p>\n

5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let principal sum = Rs. $P$ and amount = Rs. 38,416<\/p>\n

Rate of interest = 80% and time = 2 years<\/p>\n

Amount if interest being compound half yearly\u00a0= $P(1+\\frac{R}{200})^{2T}$<\/p>\n

=>\u00a0$P(1+\\frac{80}{200})^{2\\times2}=38,416$<\/p>\n

=>\u00a0$P\\times(\\frac{7}{5})^4=38,416$<\/p>\n

=> $P=38,416\\times\\frac{625}{343\\times7}$<\/p>\n

=> $P=16\\times625=Rs.$ $10,000$<\/p>\n

=> Ans – (C)<\/p>\n

6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

For compound interest A = $p(1+\\frac{r}{100})^{n}$ where p is principal amount, r is rate and t is time
\nafter 5 years it gets doubled
\nhence putting the values we will get $(1+\\frac{r}{100})^{5}$ = 2
\nnow after 20 years total amount will be $p((1+\\frac{r}{100})^{5})^{4}$ = 16p = $16 \\times 12000$ = 192000<\/p>\n

7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Difference between simple interest and compound interest for two years will be
\n48 = $\\frac{2pr}{100} – ( \\frac{2pr}{100} + \\frac{pr^2}{10^4})$ \u00a0(where p is principal amount and r is rate per annum)
\nPutting r=20% and solving above equation for p, we will get p = 1200 rs.<\/p>\n

8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

As we know $P(1+ \\frac{r}{100})^t$ is amount of compound interest where r is rate, P is principal amount and t is time.
\nSo $12000(1+ \\frac{r}{100})^5 = 2 \\times 12000$
\nor $(1+ \\frac{r}{100}) = 2^(\\frac{1}{5})$ \u00a0 \u00a0 \u00a0eq(1)
\nNow after 20 years compound interest will be = $12000(1+ \\frac{r}{100})^20$
\nor \u00a0 \u00a0 $12000 (2^{(\\frac{1}{5})})^{20}$ \u00a0 \u00a0 \u00a0 \u00a0 (from eq.(1))
\nor \u00a0 \u00a0 $ 12000 \\times 16 = 192000$<\/p>\n

9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Difference between compound interest and simple interest for 2 years will be
\n= $(P((1+\\frac{r}{100})^2) – P) – 2P \\frac{r}{100} = 32$ (where P is principal amount 5000 and r is rate )
\nafter solving above equation we will get r = 8%<\/p>\n

10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

I = PTR\/100<\/p>\n

there is 100% increase in amount means, interest = principle.<\/p>\n

given, T = 8yrs.<\/p>\n

I = P<\/p>\n

P = P*8*R\/100<\/p>\n

R = 12.5%<\/p>\n

compound interest of 8000\/- at 12.5% for 2 years is<\/p>\n

CI = total amount – 8000\/-<\/p>\n

= P$(1+R\/100)^{n}$ – 8000\/-<\/p>\n

= 8000$(1+12.5\/100)^{2}$ – 8000\/-<\/p>\n

= 10125 – 8000<\/p>\n

= 2125\/-<\/p>\n

so the answer is option D.<\/p>\n

GK Questions And Answers PDF<\/a><\/p>\n

11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

The formula to calculate compound interest is,<\/p>\n

$A=P(1+\\frac{r}{n})^{nt}$ where,<\/p>\n

A = Amount<\/p>\n

P = Principal<\/p>\n

r = rate of interest<\/p>\n

n = number of times compounded in a year<\/p>\n

t = time period in years<\/p>\n

Applying the above formula we get,<\/p>\n

Compound interest generated after two years is<\/p>\n

= Amount – Principal<\/p>\n

= $5000 (1 + \\frac{0.05}{2})^{(2*2)}$ – $5000$= $1.0509*5000 – 5000$ = $5519 – 5000$ = Rs $519$<\/p>\n

Hence Option C.<\/p>\n

12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Simple interest for 2 years on Rs. 25000 at rate \u2018r\u2019 = $\\frac{25000*2*r}{100}=500r$
\nSimple interest for each year = $250r$
\nCompound interest for 2 years on Rs. 25000 at rate \u2018r\u2019 =
\nSimple interest for 1st year + Simple interest for 2nd year
\nSimple interest for 1st year = $250r$
\nSimple interest for 2nd year = $\\frac{(25000+250r)*r}{100}$ = $250r + 2.5r^2$
\nThe difference between interest is given as Rs. 1000
\nSo we get,
\n$250 + 2.5r^2 + 250r – 500r = 1000$
\n$2.5r^2 = 1000$
\n$r^2 = 400$
\n$ r = 20$%
\nThus, Option B.<\/p>\n

13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

If the amount gets doubled after 5 years, the simple interest accrued will be equal to P (principal amount).
\n=> $P = \\frac{P \\times r \\times 5}{100}$ => r = 20%
\nSo, the amount after 2 years under compound interest will be
\n$1(1+\\frac{20}{100})^2 = 1(1.2)^2 = 1.44$ lacs.
\nThus, A is the correct answer.<\/p>\n

14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

We know that the formula for compound interest is given by
\nA = $P( 1 + \\frac{r}{100})^n$
\nHere A = 24200, n = 2 and r = 10, so on putting these values in the equation, we get
\n24200 = $P( 1 + \\frac{10}{100})^2$
\n=> P = 24200\/1.21 = 20,000
\nIf it is invested at 8% per annum simple interest for 3 years, the interest earned would be
\nI = (P*R*T)\/100 = 20000*3*8\/100 = 4800
\nHence, the amount will be 20,000 + 4800 = 24800<\/p>\n

15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Simple interest for 2 years on Rs. 20000 at rate 10% = $\\frac{20000*2*10}{100}=4000$ rupees
\nCompound interest for 2 years on Rs. 20000 at rate 10% =
\nSimple interest for 1st year + Simple interest for 2nd year
\nSimple interest for 1st year = $2000$ rupees
\nPrincipal for the second year will be 20000+2000 = Rs. 22000
\nSimple interest for 2nd year = $\\frac{(22000)*10}{100}=2200$ rupees
\nTotal simple interest = Rs. 4000
\nTotal compound interest = Rs. 2000 + Rs. 2200 = Rs. 4200
\nThe difference is Rs. 200
\nThus, Option A.<\/p>\n

16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Simple interest for 2 years on Rs. 50000 at rate \u2018r\u2019 = $\\frac{50000*2*r}{100}=1000r$
\nSimple interest for each year = $500r$
\nCompound interest for 2 years on Rs. 50000 at rate \u2018r\u2019 = Simple interest for 1st year + Simple interest for 2nd year
\nSimple interest for 2nd year = Simple interest for 1st year +simple interest on interest of first year
\n=$(500r)+(500r)+500r(\\frac{r}{100})$ =$1000r+5r^{2}$
\nDifference between the simple and the compound interest = $1000r+5r^{2} – 1000r$ = $5r^{2}$
\n$5r^{2} = 12500$
\n$r = 50$%
\nHence, Option B.<\/p>\n

17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Le the amount he invested be $P$.
\nAfter 3 years the amount will be,
\n$A=P(1+r)^t$
\n$=>A=P(1+0.2)^3$
\n$=>A=P(1.2)^3$
\n$=>A=1.728P$
\nAfter 4 years the amount will be,
\n$A=P(1.2)^4$
\n$=>A=2.0736P$
\nThus the amount will get doubled after 4 years.
\nHence Option B.<\/p>\n

18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

If the rate of interest is 12% per annum, the corresponding rate for 3 months in 12%\/4 = 3% per quarter.
\nSo, the amount at the end of the year = $10,000 * (1+\\frac{3}{100})^4$ = Rs 11255.09
\nHence the interest earned = Amount – Principal = 11255.09 – 10000 = Rs 1255.09<\/p>\n

19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

we know that :<\/p>\n

1. For first year , compound interest and simple interest is same if the principal amount and rate of interest is same in both cases.<\/p>\n

2. From 2nd year onwards ,the compound interest is normal interest plus the interest on accumulated amount due to interest untill last cycle.<\/p>\n

3. Every year simple interest remains same if Rate of Interest and principal amount remains same .<\/p>\n

Let the compound interest for 1st year be be Rs y<\/p>\n

For two years ,CI = Rs 410<\/p>\n

y + y + $\\frac{5}{100}$y = 410<\/p>\n

$\\frac{41y}{20}$ = 410<\/p>\n

y = 200<\/p>\n

So for two years , Simple Interest = 200 + 200 = Rs 400<\/p>\n

 <\/p>\n

20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Principal Amount (P) = Rs 1800<\/p>\n

Rate of Interest = 10%<\/p>\n

Compound Interest = Rs 378<\/p>\n

Let the time period be T<\/p>\n

So<\/p>\n

C.I = P$(1+ \\frac{R}{100})^T$ – P<\/p>\n

378 = 1800$(1+ \\frac{10}{100})^T$ – 1800<\/p>\n

T = 2 years<\/p>\n

21)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let the Rate of Interest be R and Principal Amount be Rs P<\/p>\n

So, R = 12%<\/p>\n

Time Period (T) = 2 years<\/p>\n

Compound Interest = P$(1 + \\frac{R}{100})^T$ – P<\/p>\n

2544 = P$(1 + \\frac{12}{100})^2$ – P<\/p>\n

P = Rs 10000<\/p>\n

Now SI with same rate of interest = $\\frac{P \\times R \\times T}{100}$ = $\\frac{10000 \\times 12 \\times 2}{100}$ = Rs 2400<\/p>\n

22)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

We now that if Rs z is the amount to be paid by a person after n years then the present value of that Rs z is given as = $\\frac{z}{(1 + \\frac{R}{100})^n}$<\/p>\n

where R is the Rate of Interest (Compounded Interest Rate)<\/p>\n

So let assume that the amount of equal installment be Rs y<\/p>\n

and hence we can say that if R = 10% per annum<\/p>\n

and Principal Amount = Rs 21000<\/p>\n

then present value of 1st installment + present value of 2nd installment = Rs 21000<\/p>\n

$\\frac{z}{(1 + \\frac{10}{100})^1}$ +$\\frac{z}{(1 + \\frac{10}{100})^2}$ = 21000<\/p>\n

$\\frac{2.1z}{(1.21)}$ = 21000<\/p>\n

z = Rs 12100<\/p>\n

 <\/p>\n

23)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Given that Rs 80000 becomes Rs 92610 at 10% per annum .<\/p>\n

semi-annual Rate become = $\\frac{10}{2}$ = 5%<\/p>\n

Let number of semi-annual cycles required = T<\/p>\n

So Using ,<\/p>\n

Compounded Amount = P$(1+\\frac{R}{100})^T$<\/p>\n

92610 = 80000$(1+\\frac{5}{100})^T$<\/p>\n

T = 3 semi annual cycles = 1.5 years<\/p>\n

24)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

let the sum of money be Rs P<\/p>\n

and Rate of Interest be R% per annum<\/p>\n

Compounded Amount = P$(1 + \\frac{R}{100})^n$<\/p>\n

n – number of time periods<\/p>\n

Compounded amount after 1 year = P$(1 + \\frac{R}{100})^1$ = 650………….(1)<\/p>\n

Compounded amount after 2 year = P$(1 + \\frac{R}{100})^2$ = 676………….(2)<\/span><\/p>\n

Dividing both equation 1 and equation 2<\/span><\/p>\n

$(1 + \\frac{R}{100})$ = 1.04
\n<\/span><\/p>\n

and from here it can be concluded that R = 4% . Now using equation 1 and value of R <\/span><\/p>\n

P$(1 + \\frac{4}{100})^1$ = 650
\n<\/span><\/span><\/p>\n

P = Rs 625<\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

25)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

it is given that A sum of money placed at compound interest doubles itself in 4 years<\/p>\n

here we need to make the money 4 times<\/p>\n

imagine that we invested Rs P<\/p>\n

and hence ,<\/p>\n

P becomes 2P in 4 years and so this 2P will become 4P in another 4 years and hence total 8 years are required to make Rs P –> Rs 4P<\/p>\n\n

FREE SSC CHSL MOCK TEST SERIES<\/a><\/p>\n

FREE SSC STUDY MATERIAL<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Compound Interest Questions for SSC CHSL PDF: SSC CHSL Physics Previous Year Questions download PDF based on previous year question paper of SSC exams. 25 Very important Physics Previous Year questions for SSC CHSL Exam. Take a free mock test for SSC CHSL Download SSC CHSL Previous Papers Question 1:\u00a0On a certain sum of money, […]<\/p>\n","protected":false},"author":21,"featured_media":27199,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,378],"tags":[358,1525,1751,1697],"class_list":{"0":"post-27093","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-ssc-chsl","10":"tag-ssc-chsl","11":"tag-ssc-chsl-2019","12":"tag-ssc-chsl-free-mock","13":"tag-ssc-chsl-previous-papers"},"better_featured_image":{"id":27199,"alt_text":"Compound Interest Questions for SSC CHSL PDF","caption":"Compound Interest Questions for SSC CHSL PDF","description":"Compound Interest Questions for SSC CHSL 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