{"id":39478,"date":"2019-12-18T17:44:42","date_gmt":"2019-12-18T12:14:42","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=39478"},"modified":"2019-12-18T17:54:45","modified_gmt":"2019-12-18T12:24:45","slug":"simple-and-compound-interest-questions-for-rrb-ntpc-set-3-pdf-2","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/simple-and-compound-interest-questions-for-rrb-ntpc-set-3-pdf-2\/","title":{"rendered":"Simple and Compound Interest Questions for RRB-NTPC Set-3 PDF"},"content":{"rendered":"

Simple and Compound Interest Questions PDF<\/span><\/span><\/strong><\/h1>\n

Download RRB NTPC\u00a0 SI & CI Questions Set-3 PDF. Top 10 RRB NTPC questions based on asked questions in previous exam papers very important for the Railway NTPC exam.<\/p>\n

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Question 1:\u00a0<\/b>What is the difference (in Rs) between compound interest (compounded annually) and simple interest for 3 years on a principal of Rs 3000 at the annual rate of 20% ?<\/p>\n

a)\u00a0464<\/p>\n

b)\u00a0384<\/p>\n

c)\u00a0356<\/p>\n

d)\u00a0424<\/p>\n

Question 2:\u00a0<\/b>The difference between simple interest on a certain sum at the rate of 8% per annum for 18 months and 24 months is Rs 452. What is the sum (in Rs)?<\/p>\n

a)\u00a010500<\/p>\n

b)\u00a014600<\/p>\n

c)\u00a011300<\/p>\n

d)\u00a013600<\/p>\n

Question 3:\u00a0<\/b>Simple interest on a sum for 10 years is equal to 5% of the principal. In how many years interest will be equal to the principal?<\/p>\n

a)\u00a0100<\/p>\n

b)\u00a0150<\/p>\n

c)\u00a0200<\/p>\n

d)\u00a0250<\/p>\n

Question 4:\u00a0<\/b>A sum of money invested at simple interest becomes $\\frac{13}{10}$ of itself in 2 years and 6 months. What is the rate (in percentage) of interest per annum?<\/p>\n

a)\u00a010<\/p>\n

b)\u00a015<\/p>\n

c)\u00a012<\/p>\n

d)\u00a018<\/p>\n

Question 5:\u00a0<\/b>The simple interest on a sum of money is $\\frac{16}{25}$ of the principal. The number of years is equal to the rate of interest per annum. What is the rate (in percentage) of interest per annum?<\/p>\n

a)\u00a04<\/p>\n

b)\u00a016<\/p>\n

c)\u00a08<\/p>\n

d)\u00a012<\/p>\n

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Question 6:\u00a0<\/b>A sum of Rs 5000 becomes Rs 8000 in 3 years, when invested in a scheme of simple interest. If the same sum is invested in a scheme of compound interest with same yearly interest rate (compounding of interest is yearly), then what will be the amount (in Rs) after 3 years?<\/p>\n

a)\u00a08640<\/p>\n

b)\u00a08260<\/p>\n

c)\u00a08880<\/p>\n

d)\u00a09220<\/p>\n

Question 7:\u00a0<\/b>The simple interest on a sum is $\\frac{5}{9}$ of the principal for 25 years. What is the rate (in percentage) of interest per annum?<\/p>\n

a)\u00a0$\\frac{25}{9}$<\/p>\n

b)\u00a0$\\frac{9}{2}$<\/p>\n

c)\u00a0$\\frac{20}{9}$<\/p>\n

d)\u00a0$\\frac{15}{4}$<\/p>\n

Question 8:\u00a0<\/b>Simple interest on a sum of Rs 67400 for 14 months is Rs 4718. What is the rate (in percentage) of interest per annum?<\/p>\n

a)\u00a05.5<\/p>\n

b)\u00a06<\/p>\n

c)\u00a08<\/p>\n

d)\u00a07<\/p>\n

Question 9:\u00a0<\/b>Some part of Rs 17500 was lent at the rate of 24% per annum simple interest and the remaining part at the rate of 10% per annum simple interest. The total interest received after 5 years is Rs 13300. What is the ratio of money lent at the rate of 24% and 10%?<\/p>\n

a)\u00a012 : 13<\/p>\n

b)\u00a03 : 4<\/p>\n

c)\u00a03 : 2<\/p>\n

d)\u00a013 : 22<\/p>\n

Question 10:\u00a0<\/b>A sum of Rs 4000 becomes Rs 5800 in 3 years, when invested in a scheme of simple interest. If the same sum is invested in a scheme of compound interest with same yearly interest rate (compounding of interest is done yearly), then what will be the amount (in Rs) after 2 years?<\/p>\n

a)\u00a04430<\/p>\n

b)\u00a05450<\/p>\n

c)\u00a05290<\/p>\n

d)\u00a04970<\/p>\n

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Question 11:\u00a0<\/b>The difference between the simple interest and compound interest (compounded annually) on Rs. 40,000 for 3 years at 8% per annum is<\/p>\n

a)\u00a0Rs.684.32<\/p>\n

b)\u00a0Rs.788.48<\/p>\n

c)\u00a0Rs.784.58<\/p>\n

d)\u00a0Rs.4000<\/p>\n

Question 12:\u00a0<\/b>The difference between compound interest and simple interest on a certain sum of money for 2 years at 5% per annum is Rs. 41. What is the sum of money?<\/p>\n

a)\u00a0Rs.7200<\/p>\n

b)\u00a0Rs.9600<\/p>\n

c)\u00a0Rs.16400<\/p>\n

d)\u00a0Rs.8400<\/p>\n

Question 13:\u00a0<\/b>The difference between simple and compound interest (compounded annually) on a sum of money for 3 years at 10% per annum is Rs. 93. The sum (in Rs.) is:<\/p>\n

a)\u00a030000<\/p>\n

b)\u00a030300<\/p>\n

c)\u00a03000<\/p>\n

d)\u00a03030<\/p>\n

Question 14:\u00a0<\/b>The compound interest on a certain sum for 2 years at 10% per annum is Rs. 525 . The simple interest on the same sum for double the time at half the rate percent per annum is :<\/p>\n

a)\u00a0Rs.520<\/p>\n

b)\u00a0Rs.550<\/p>\n

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

d)\u00a0Rs. 515<\/p>\n

Question 15:\u00a0<\/b>The compound interest on Rs. 64,000 for 3 years, compounded annually at 7.5% p.a. is<\/p>\n

a)\u00a0Rs. 14,400<\/p>\n

b)\u00a0Rs. 15,705<\/p>\n

c)\u00a0Rs. 15,507<\/p>\n

d)\u00a0Rs. 15,075<\/p>\n

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Question 16:\u00a0<\/b>The difference between compound interest and simple interest on an amount of Rs. 15,000 for 2 years is Rs. 96. The rate of interest per annum is?<\/p>\n

a)\u00a06%<\/p>\n

b)\u00a07%<\/p>\n

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

d)\u00a09%<\/p>\n

Question 17:\u00a0<\/b>The 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)<\/p>\n

a)\u00a011,320<\/p>\n

b)\u00a012,000<\/p>\n

c)\u00a012,320<\/p>\n

d)\u00a012,500<\/p>\n

Question 18:\u00a0<\/b>The difference between the compound interest and the simple interest on Rs. 6250 at 8% per annum in 2 years is<\/p>\n

a)\u00a0Rs.30<\/p>\n

b)\u00a0Rs.40<\/p>\n

c)\u00a0Rs.50<\/p>\n

d)\u00a0Rs.60<\/p>\n

Question 19:\u00a0<\/b>The difference between compound interest and simple interest on Rs. 5000 for 2 years at 8% per annum payable yearly is<\/p>\n

a)\u00a0Rs.30<\/p>\n

b)\u00a0Rs.31<\/p>\n

c)\u00a0Rs.33<\/p>\n

d)\u00a0Rs.32<\/p>\n

Question 20:\u00a0<\/b>The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Rs. 1. The sum (in Rs.) is:<\/p>\n

a)\u00a0620<\/p>\n

b)\u00a0630<\/p>\n

c)\u00a0640<\/p>\n

d)\u00a0625<\/p>\n

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

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

Principal sum = Rs. 3,000<\/p>\n

Rate of interest = 20% and time period = 3 years<\/p>\n

=> Difference between compound and simple interest = $[P(1+\\frac{R}{100})^T-P]-(\\frac{P \\times R\\times T}{100})$<\/p>\n

= $[3000(1+\\frac{20}{100})^3-3000]-(\\frac{3000 \\times 20\\times 3}{100})$<\/p>\n

= $[3000(\\frac{6}{5})^3-3000]-(1800)$<\/p>\n

= $3000\\times(\\frac{216}{125})-4800$<\/p>\n

= $(24\\times216)-4800$<\/p>\n

= $5184-4800=Rs.$ $384$<\/p>\n

=> Ans – (B)<\/p>\n

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

Let the sum be = Rs. $100x$<\/p>\n

Rate of interest = 8%<\/p>\n

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

According to ques,<\/p>\n

=> $(\\frac{100x\\times8\\times2}{100})-(\\frac{100x\\times8\\times1.5}{100})=452$<\/p>\n

=> $16x-12x=452$<\/p>\n

=> $x=\\frac{452}{4}=113$<\/p>\n

$\\therefore$ Sum = $100\\times113=Rs.$ $11,300$<\/p>\n

=> Ans – (C)<\/p>\n

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

Let principal amount = Rs.\u00a0$100x$<\/p>\n

=> Simple interest = $\\frac{5}{100}\\times100x=Rs.$ $5x$<\/p>\n

Let rate of interest = $r\\%$ and time period = 10 years<\/p>\n

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

=> $\\frac{100x\\times r\\times10}{100}=5x$<\/p>\n

=> $10r=5$<\/p>\n

=> $r=\\frac{5}{10}=0.5\\%$<\/p>\n

Let the interest will be equal to the principal in = $t$ years at 0.5% rate<\/p>\n

=> $\\frac{100x\\times 0.5\\times t}{100}=100x$<\/p>\n

=> $t=\\frac{100}{0.5}=200$ years<\/p>\n

=> Ans – (C)<\/p>\n

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

Let sum of money invested = Rs.\u00a0$100x$<\/p>\n

=> Amount under simple interest = $\\frac{13}{10}\\times100x=Rs.$ $130x$<\/p>\n

Thus, simple interest = $130x-100x=Rs.$ $30x$<\/p>\n

Let rate of interest = $r\\%$ and time period = 2.5 years<\/p>\n

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

=> $\\frac{100x\\times r\\times2.5}{100}=30x$<\/p>\n

=> $2.5r=30$<\/p>\n

=> $r=\\frac{30}{2.5}=12\\%$<\/p>\n

=> Ans – (C)<\/p>\n

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

Let sum of money invested = Rs.\u00a0$100x$<\/p>\n

=> Simple interest = $\\frac{16}{25}\\times100x=Rs.$ $64x$<\/p>\n

Let rate of interest = time period = $x$<\/p>\n

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

=> $\\frac{100x\\times x\\times x}{100}=64x$<\/p>\n

=> $x^2=64$<\/p>\n

=> $x=\\sqrt{64}=8\\%$<\/p>\n

=> Ans – (C)<\/p>\n

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

Principal sum = Rs. 5000 and time period = 3 years<\/p>\n

=> Amount after simple interest = Rs. 8000<\/p>\n

Thus, simple interest = Rs. (8000-5000) = Rs. 3000<\/p>\n

Let rate of interest = $r\\%$<\/p>\n

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

=> $\\frac{5000\\times r\\times3}{100}=3000$<\/p>\n

=> $150r=3000$<\/p>\n

=> $r=\\frac{3000}{150}=20\\%$<\/p>\n

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

= $5000(1+\\frac{20}{100})^3$<\/p>\n

= $5000(1+\\frac{1}{5})^3=5000(\\frac{6}{5})^3$<\/p>\n

=\u00a0$5000\\times\\frac{216}{125}$<\/p>\n

= $40\\times216=Rs.$ $8640$<\/p>\n

=> Ans – (A)<\/p>\n

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

Let principal amount = Rs. $9x$<\/p>\n

=> Simple interest = $\\frac{5}{9}\\times9x=Rs.$ $5x$<\/p>\n

Let rate of interest = $r\\%$ and time period = 25 years<\/p>\n

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

=> $\\frac{9x\\times r\\times25}{100}=5x$<\/p>\n

=> $\\frac{9r}{4}=5$<\/p>\n

=> $x=\\frac{5\\times4}{9}=\\frac{20}{9}\\%$<\/p>\n

=> Ans – (C)<\/p>\n

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

Principal sum = Rs. 67,400 and simple interest = Rs. 4718<\/p>\n

Let rate of interest = $r\\%$ and time period = $\\frac{14}{12}=\\frac{7}{6}$ years<\/p>\n

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

=> $\\frac{67,400\\times r\\times7}{6\\times100}=4718$<\/p>\n

=> $\\frac{4718r}{6}=4718$<\/p>\n

=> $r=6\\%$<\/p>\n

=> Ans – (B)<\/p>\n

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

Let sum lent for 24% = Rs.\u00a0$100x$ and sum lent for 10% = Rs. $(17500-100x)$<\/p>\n

Time period = 5 years<\/p>\n

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

According to ques,<\/p>\n

=> $\\frac{100x\\times24\\times5}{100}+\\frac{(17500-100x)\\times10\\times5}{100}=13300$<\/p>\n

=> $120x+(175-x)\\times50=13300$<\/p>\n

=> $120x+(175\\times50)-50x=(175\\times76)$<\/p>\n

=> $70x=175(76-50)$<\/p>\n

=> $x=\\frac{175\\times26}{70}=65$<\/p>\n

$\\therefore$ Required ratio = $\\frac{100\\times65}{17500-(100\\times65)}$<\/p>\n

= $\\frac{6500}{11000}=\\frac{65}{110}=13:22$<\/p>\n

=> Ans – (D)<\/p>\n

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

Principal sum = Rs. 4000 and time period = 3 years<\/p>\n

=> Amount after simple interest = Rs. 5800<\/p>\n

Thus, simple interest = Rs. (5800-4000) = Rs. 1800<\/p>\n

Let rate of interest = $r\\%$<\/p>\n

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

=> $\\frac{4000\\times r\\times3}{100}=1800$<\/p>\n

=> $120r=1800$<\/p>\n

=> $r=\\frac{1800}{120}=15\\%$<\/p>\n

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

= $4000(1+\\frac{15}{100})^2$<\/p>\n

= $4000(1+\\frac{3}{20})^2=4000(\\frac{23}{20})^2$<\/p>\n

=\u00a0$4000\\times\\frac{529}{400}$<\/p>\n

= $10\\times529=Rs.$ $5290$<\/p>\n

=> Ans – (C)<\/p>\n

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

Principal (P) = Rs. 40,000<\/p>\n

Rate of interest (r) = 8% and time period (t) = 3 years<\/p>\n

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

= $\\frac{40,000 \\times 8 \\times 3}{100}$<\/p>\n

= $400 \\times 24=Rs.$ $9600$<\/p>\n

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

= $40,000[(1+\\frac{8}{100})^3-1]$<\/p>\n

=\u00a0$40,000[(\\frac{27}{25})^3-1]$<\/p>\n

= $40,000 (\\frac{19683-15625}{15625})=40,000 \\times \\frac{4058}{15625}$<\/p>\n

= $Rs.$ $10388.48$<\/p>\n

$\\therefore$ C.I. – S.I. = $10388.48-9600=Rs.$ $788.48$<\/p>\n

=> Ans – (B)<\/p>\n

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

Let the given sum = Rs. $100x$<\/p>\n

Rate of interest = 5% and time period = 2 years<\/p>\n

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

= $100x [(1 + \\frac{5}{100})^2 – 1]$<\/p>\n

= $100x [(\\frac{21}{20})^2 – 1] = 100x (\\frac{441 – 400}{400})$<\/p>\n

= $100x \\times \\frac{41}{400} = 10.25x$<\/p>\n

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

= $\\frac{100x \\times 5 \\times 2}{100} = 10x$<\/p>\n

=> Difference\u00a0between simple and compound interests = $10.25-10x = 41$<\/p>\n

=> $0.25x = 41$<\/p>\n

=> $x = \\frac{41}{0.25} = 164$<\/p>\n

$\\therefore$ Value of given sum = $100 \\times 164 = Rs. 16,400$<\/p>\n

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

Let the given sum = Rs. $1000x$<\/p>\n

Rate of interest = 10% and time period = 3 years<\/p>\n

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

= $1000x [(1 + \\frac{10}{100})^3 – 1]$<\/p>\n

= $1000x [(\\frac{11}{10})^3 – 1] = 1000x (\\frac{1331 – 1000}{1000})$<\/p>\n

= $1000x \\times \\frac{331}{1000} = 331x$<\/p>\n

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

= $\\frac{1000x \\times 10 \\times 3}{100} = 300x$<\/p>\n

=> Difference\u00a0between simple and compound interests = $331-300x = 93$<\/p>\n

=> $31x = 93$<\/p>\n

=> $x = \\frac{93}{31} = 3$<\/p>\n

$\\therefore$ Value of given sum = $1000 \\times 3 = Rs. 3,000$<\/p>\n

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

Let the sum = Rs.\u00a0$100x$<\/p>\n

For compound interest, rate of interest (r) = 10% and time (t) = 2 years<\/p>\n

=> $C.I. = P[(1+\\frac{r}{100})^t-1]$<\/p>\n

=> $100x[(1+\\frac{10}{100})^2-1]=525$<\/p>\n

=> $100x[(\\frac{11}{10})^2-1]=525$<\/p>\n

=> $100x \\times (\\frac{121-100}{100})=525$<\/p>\n

=> $21x=525$<\/p>\n

=> $x=\\frac{525}{21}=25$<\/p>\n

$\\therefore$ Sum invested = Rs. 2500<\/p>\n

For simple\u00a0interest, rate of interest (r) = 5% and time (t) = 4 years<\/p>\n

=> $S.I.= \\frac{P \\times r \\times t}{100}$<\/p>\n

= $\\frac{2500 \\times 5 \\times 4}{100}$<\/p>\n

= $25 \\times 20=Rs.$ $500$<\/p>\n

=> Ans – (C)<\/p>\n

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

Principal (P) = Rs. 64,000<\/p>\n

Time\u00a0(t) = 3 years and rate of interest under compound interest (r) = 7.5%<\/p>\n

=> $C.I. = P[(1+\\frac{r}{100})^t-1]$<\/p>\n

= $64000[(1+\\frac{7.5}{100})^3-1]$<\/p>\n

= $64000[(1+\\frac{15}{200})^3-1]$<\/p>\n

= $64000[(\\frac{43}{40})^3-1]$<\/p>\n

= $64000 \\times \\frac{(43)^3-(40)^3}{64000}$<\/p>\n

= $79507-64000=Rs.$ $15,507$<\/p>\n

=> Ans – (C)<\/p>\n

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

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

Principal amount = Rs. 10,000<\/p>\n

Rate of interest = 10% for first year and 12% for second year<\/p>\n

=> Amount after interest = $10,000 \\times \\frac{110}{100} \\times \\frac{112}{100}$<\/p>\n

= $110 \\times 112=Rs.$ $12,320$<\/p>\n

=> Ans – (C)<\/p>\n

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

Principal (P) = Rs. 6250<\/p>\n

Rate of interest (r) = 8% and time period (t) = 2 years<\/p>\n

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

= $\\frac{6250 \\times 8 \\times 2}{100}$<\/p>\n

= $62.5 \\times 16=Rs.$ $1000$<\/p>\n

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

= $6250[(1+\\frac{8}{100})^2-1]$<\/p>\n

=\u00a0$6250[(\\frac{27}{25})^2-1]$<\/p>\n

= $6250 \\times (\\frac{104}{625})$<\/p>\n

= $Rs.$ $1040$<\/p>\n

$\\therefore$ C.I. – S.I. = $1040-1000=Rs.$ $40$<\/p>\n

=> Ans – (B)<\/p>\n

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

Principal (P) = Rs. 5000<\/p>\n

Rate of interest (r) = 8% and time period (t) = 2 years<\/p>\n

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

= $\\frac{5000 \\times 8 \\times 2}{100}$<\/p>\n

= $50 \\times 16=Rs.$ $800$<\/p>\n

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

= $5000[(1+\\frac{8}{100})^2-1]$<\/p>\n

=\u00a0$5000[(\\frac{27}{25})^2-1]$<\/p>\n

= $5000 \\times (\\frac{104}{625})$<\/p>\n

= $Rs.$ $832$<\/p>\n

$\\therefore$ C.I. – S.I. = $832-800=Rs.$ $32$<\/p>\n

=> Ans – (D)<\/p>\n

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

Rate of interest\u00a0(r) = 4 % and time (t) = 2 years<\/p>\n

Difference\u00a0between simple and compound interests (d) = Rs. 1<\/p>\n

Then, the sum = $\\frac{d \\times (100)^t}{(r)^t}$<\/p>\n

= $\\frac{1 \\times (100)^2}{(4)^2} = \\frac{10000}{16}$<\/p>\n

= Rs. 625<\/p>\n

=> Ans – (D)<\/p>\n\n

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We hope this Time, Speed & Distance Questions set-3 pdf for RRB NTPC exam will be highly useful for your Preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Simple and Compound Interest Questions PDF Download RRB NTPC\u00a0 SI & CI Questions Set-3 PDF. Top 10 RRB NTPC questions based on asked questions in previous exam papers very important for the Railway NTPC exam. Download RRB NTPC Previous Papers PDF Question 1:\u00a0What is the difference (in Rs) between compound interest (compounded annually) and simple […]<\/p>\n","protected":false},"author":21,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,1127,2071,2067,31,3171,1679,1897,1605,1603,1601,1673],"tags":[2641,1593,3332,3336,2148,2092,3334],"class_list":{"0":"post-39478","1":"post","2":"type-post","3":"status-publish","4":"format-standard","6":"category-downloads","7":"category-featured","8":"category-ibps-rrb","9":"category-ibps-rrb-clerk-en","10":"category-ibps-rrb-clerk","11":"category-railways","12":"category-railways-en","13":"category-rrb-group-d","14":"category-rrb-group-d-2","15":"category-rrb-je","16":"category-rrb-ntpc","17":"category-rrb-ntpc-je","18":"category-rrb-para-medical-categories","19":"tag-how-to-prepare-rrb-ntpc","20":"tag-rrb-ntpc","21":"tag-rrb-ntpc-free-mock","22":"tag-rrb-ntpc-free-tests","23":"tag-rrb-ntpc-gk","24":"tag-rrb-ntpc-gk-questions","25":"tag-rrb-ntpc-notification"},"better_featured_image":null,"yoast_head":"\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n