{"id":35099,"date":"2019-09-20T16:57:25","date_gmt":"2019-09-20T11:27:25","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=35099"},"modified":"2019-10-22T15:25:21","modified_gmt":"2019-10-22T09:55:21","slug":"simple-interest-and-compound-interest-questions-for-rrb-ntpc-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/simple-interest-and-compound-interest-questions-for-rrb-ntpc-set-2-pdf\/","title":{"rendered":"Simple Interest and Compound Interest Questions for RRB NTPC Set-2 PDF"},"content":{"rendered":"

Simple Interest and Compound Interest\u00a0Questions for RRB NTPC Set-2 PDF<\/em><\/strong><\/span><\/h1>\n

Download RRB NTPC Simple & Compound Interest\u00a0 Questions and Answers PDF. Top 25 RRB NTPC\u00a0 Maths\u00a0 questions based on asked questions in previous exam papers very important for the Railway NTPC exam.<\/p>\n

Download Simple and Compound Interest Questions for RRB NTPC Set-2 PDF<\/a><\/p>\n

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Question 1:\u00a0<\/b>In what time would the simple interest on a certain sum be 0.125 times the principal at 10% per annum ?<\/p>\n

a)\u00a0$1\\frac{1}{4}$ years<\/p>\n

b)\u00a0$1\\frac{3}{4}$ years<\/p>\n

c)\u00a0$2\\frac{1}{4}$ years<\/p>\n

d)\u00a0$2\\frac{3}{4}$ years<\/p>\n

e)\u00a0None of these<\/p>\n

Question 2:\u00a0<\/b>How much will a sum of 12,0007 deposited at a rate of 9% per annum (simple interest) for 13 years amount to ?<\/p>\n

a)\u00a014,040<\/p>\n

b)\u00a020,650<\/p>\n

c)\u00a013,404<\/p>\n

d)\u00a027,800<\/p>\n

e)\u00a026,040<\/p>\n

Instructions<\/b><\/p>\n

Directions for the next five questions: Study the following table carefully to answer the questions.
\nRate of interest (pcpa) offered by five companies on deposits under different schemes<\/strong><\/p>\n

\"\"<\/p>\n

Question 3:\u00a0<\/b>Mr. X deposited an amount in Scheme II with company C for two years. After that he withdrew the amount and reinvested only the principal amount in scheme IV of Company B for two years. Total amount of simple interest accrued from the two schemes is Rs. 14800. What was the principal amount?<\/p>\n

a)\u00a0Rs. 48000<\/p>\n

b)\u00a0Rs. 42000<\/p>\n

c)\u00a0Rs. 40000<\/p>\n

d)\u00a0Can\u2019t be determined<\/p>\n

e)\u00a0None of these<\/p>\n

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Question 4:\u00a0<\/b>Abishek invested an amount of Rs. 45000 for two years with company B under scheme III, Which offers compound interest, Jeevan invested an equal amount for two years with company C under Scheme IV, which offer simple Interest. Who earned more interest and how much?<\/p>\n

a)\u00a0Abhishek, Rs. 1875<\/p>\n

b)\u00a0Jeevan, Rs. 1875<\/p>\n

c)\u00a0Abhishek, Rs. 1962<\/p>\n

d)\u00a0Jeevan, Rs. 1962<\/p>\n

e)\u00a0None of these<\/p>\n

Question 5:\u00a0<\/b>Mr. Lal invested Rs. 30000 in company A under Scheme II, which offers simple interest and Rs. 48000 in Company D under Scheme II, which offers compound interest. What will be the total amount of interest earned by Mr Lal in two years?<\/p>\n

a)\u00a0Rs. 14728.80<\/p>\n

b)\u00a0Rs. 17428.50<\/p>\n

c)\u00a0Rs. 14827.70<\/p>\n

d)\u00a0Rs. 16728.20<\/p>\n

e)\u00a0None of these<\/p>\n

Question 6:\u00a0<\/b>The compound interest accured on an amount of Rs.22, 000 at the end of two years is Rs.5, 596.8. What would be the simple interest accrued on the same amount at the same rate in the same period?<\/p>\n

a)\u00a0Rs.5,420<\/p>\n

b)\u00a0Rs.5,360<\/p>\n

c)\u00a0Rs.5,280<\/p>\n

d)\u00a0Rs.5,140<\/p>\n

e)\u00a0None of these<\/p>\n

Question 7:\u00a0<\/b>The simple interest accrued on an amount of Rs. 22,500 at the end of four years is Rs.10,800. What would be the compound interest accrued on the same amount at the same rate of interest at the end of two years?<\/p>\n

a)\u00a0Rs. 16, 908<\/p>\n

b)\u00a0Rs. 5, 724<\/p>\n

c)\u00a0Rs. 28, 224<\/p>\n

d)\u00a0Rs. 8, 586<\/p>\n

e)\u00a0None of these<\/p>\n

Question 8:\u00a0<\/b>The simple interest accrued on a certain principal is Rs 35,672 in seven years at the rate of 8 pcpa. What would be the compound interest accrued on that principal at the rate of 2 pcpa in 2 years?<\/p>\n

a)\u00a0Rs 2573.48<\/p>\n

b)\u00a0Rs 2564.86<\/p>\n

c)\u00a0Rs 2753.86<\/p>\n

d)\u00a0Rs 2654.48<\/p>\n

e)\u00a0None of these<\/p>\n

Question 9:\u00a0<\/b>The simple interest accrued on an amount of Rs. 22,500 at the end of four years is Rs.10,800. What would be the compound interest accrued on the same amount at the same rate of interest at the end of two years?<\/p>\n

a)\u00a0Rs. 16, 908<\/p>\n

b)\u00a0Rs. 5, 724<\/p>\n

c)\u00a0Rs. 28, 224<\/p>\n

d)\u00a0Rs. 8, 586<\/p>\n

e)\u00a0None of these<\/p>\n

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Question 10:\u00a0<\/b>A sum of Rs16800 is divided into two parts. One part is lent at the simple interest of 6% per annum and the other at 8% per annum. After 2 years total sum received is Rs19000. The sum lent at 6% of simple interest is<\/p>\n

a)\u00a0` 12200<\/p>\n

b)\u00a0` 12000<\/p>\n

c)\u00a0` 11000<\/p>\n

d)\u00a0` 10000<\/p>\n

e)\u00a0None of these<\/p>\n

Question 11:\u00a0<\/b>Rs.X triples itself in 6 years at simple interest of R%. A person invested Rs.X at R% per annum compound interest compounded annually. If the amount received after 3 years is Rs.38394, then find the simple interest received on Rs.X at R% per annum for 4 years.<\/p>\n

a)\u00a0Rs.21540<\/p>\n

b)\u00a0Rs.22460<\/p>\n

c)\u00a0Rs.21600<\/p>\n

d)\u00a0Rs.20480<\/p>\n

e)\u00a0None of these<\/p>\n

Question 12:\u00a0<\/b>Raman deposited Rs.P in a scheme for 3 year at the rate of 10% per annum simple interest. Total amount obtained after 3 years with Rs. 1000 was\u00a0 deposited in a bank which offered 20% per annum compound interest. After 2 years compound interest obtained from the bank was Rs. 6160. Then what is the value of P(in Rs.)?<\/p>\n

a)\u00a0Rs. 15000<\/p>\n

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

c)\u00a0Rs. 10000<\/p>\n

d)\u00a0Rs. 9000<\/p>\n

e)\u00a0Rs. 18000<\/p>\n

Instructions<\/b><\/p>\n

Question 13:\u00a0<\/b>The difference between Compound interest and Simple interest for 2 years for Principle Rs 32000 is Rs 720. What is the rate of interest ?<\/p>\n

a)\u00a014%<\/p>\n

b)\u00a015%<\/p>\n

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

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

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

Question 14:\u00a0<\/b>A sum of money at compound interest yields an amount of Rs 1,25,000 in 3 years and Rs 1,95,312.5 in 5 years. What is the rate of interest per annum ?<\/p>\n

a)\u00a015%<\/p>\n

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

c)\u00a025%<\/p>\n

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

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

Question 15:\u00a0<\/b>Ajay invested a certain amount of money in simple interest for a period of 3 years at 17% simple interest and gets an interest of 2856. If he invested the total amount received in a scheme which offered compound interest at a rate which is 3% more than the earlier rate for a period of 2 years. Then find the total interest he received from the scheme at the end of 2 years (approximately)?<\/p>\n

a)\u00a03756<\/p>\n

b)\u00a03607<\/p>\n

c)\u00a03475<\/p>\n

d)\u00a04500<\/p>\n

e)\u00a03721<\/p>\n

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 <\/p>\n

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

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

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

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

Let principal amount be Rs. x.
\n(x\u00d79\u00d72)\/100 + (x\u00d79.5\u00d72)\/100 = 14800
\nor (0.18+0.19) = 14800
\nX = 14800\/0.37 = Rs. 40000<\/p>\n

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

Interest earned by Abhishek = 45000 \u00d7 (1+ 8\/100)^2- 45000
\n= 45000 \u00d7 27\/25 \u00d7 27\/25 = 45000
\n= 52488 – 45000 = 7488<\/p>\n

Interest earned by Jeevan = (45000\u00d710.5\u00d72)\/100 = 9450
\nRequired difference = 9450 – 7488 = 1962
\nHence, Jeevan earned Rs. 1962 more.<\/p>\n

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

(30000\u00d79.5\u00d72)\/100 + (48000\u00d7 109\/100\u00d7 109\/100- 48000) = 5700 + (57028.8 – 48000)
\n= 5700 + 9028.8 = Rs. 14728.80<\/p>\n

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

Let the rate of interest be R% per annum.
\nCI = P [(1 + $\\frac{R}{100})^{T}$ -1]
\n=5596.8
\n= 22000 [(1 + $\\frac{R}{100})^{2}$-1]
\n$\\frac{5596.8}{22000}$ = (1+ $\\frac{R}{100})^{2}$ -1
\n(1 + $\\frac{R}{100})^{2}$ = 1 + $\\frac{5596.8}{22000}$<\/p>\n

(1 + $\\frac{R}{100})^{2}$ = $\\frac{22000 + 5596.8}{22000}$ = $\\frac{27596.8}{22000}$
\n(1+$\\frac{R}{100})^{2}$ = $\\frac{275968}{220000}$ = $\\frac{12544}{10000}$
\n$\\frac{R}{100}$ = $\\sqrt{\\frac{12544}{10000}}$ – 1 = $\\frac{112}{100}$
\n= $\\frac{R}{100}$ = $\\frac{112}{100}$ – 1 = $\\frac{112 – 100}{100}$ = $\\frac{12}{100}$
\nR=12
\nSI = $\\frac{Principal \u00d7 Time \u00d7 Rate}{100}$ = $\\frac{22000 \u00d7 2 \u00d7 100}{100}$ = Rs.5280<\/p>\n

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

Let the rate of interest be r
\nSimple Interest = $\\frac{P t r}{100}$<\/p>\n

10800 = $\\frac{22500}{100}$ x 4 x r<\/p>\n

r = $\\frac{10800}{22500}$ x $\\frac{100}{4}$ = 12%<\/p>\n

Compound Interest = 22500(1 + $\\frac{12}{100}$)2<\/sup> –\u00a022500
\n= 22500 x $\\frac{112}{100}$ x\u00a0$\\frac{112}{100}$ – 22500
\n= 28224 – 22500
\n= 5724<\/p>\n

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

PTR\/100 = 35672 => P = 35672 * 100 \/ (7*8) = 63700
\nCompound interest in 2 years = P(1+r\/100)^2 – P = 63700(1+1\/50)^2 – 63700 = 2573.48<\/p>\n

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

Let the rate of interest be r
\nSimple Interest = $\\frac{P t r}{100}$<\/p>\n

10800 = $\\frac{22500}{100}$ x 4 x r<\/p>\n

r = $\\frac{10800}{22500}$ x $\\frac{100}{4}$ = 12%<\/p>\n

Compound Interest = 22500(1 + $\\frac{12}{100}$)2<\/sup>\u00a0–\u00a022500
\n= 22500 x $\\frac{112}{100}$ x\u00a0$\\frac{112}{100}$ – 22500
\n= 28224 – 22500
\n= 5724<\/p>\n

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

Let\u00a0x<\/em>\u00a0be the amount lent for 6%
\n16800 –\u00a0x<\/em>\u00a0is the amount lent for 8%
\n$\\frac{2×6}{100}$ + $\\frac{2(16800-x)8}{100}$ + 16800 = 19000
\nx<\/em>\u00a0= 12200<\/p>\n

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

Given, Principal = Rs.X
\nRate of interest = R%
\nTime duration = 6 years
\nAmount after 6 years = Rs.3X
\nSimple Interest after 6 years = Rs.3X – Rs.X = Rs.2X
\n$\\dfrac{X\\times R\\times6}{100} = 2X$<\/p>\n

$R = \\dfrac{100}{3}$% $= \\dfrac{1}{3}$
\nAmount received after 3 years at compound interest $= X \\times \\dfrac{4}{3}\\times\\dfrac{4}{3}\\times\\dfrac{4}{3} = 2.37X$
\nGiven, $2.37X = 38394$
\n=> $X = 16200$
\nHence, Principal = Rs.16200
\nSimple Interest at $\\dfrac{1}{3}$ in 4 years = $4\\times\\dfrac{1}{3}\\times16200 = Rs.21600$<\/p>\n

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

Raman deposited Rs.P in a scheme for 3 year at the rate of 10% per annual simple interest.
\nSo interest obtained from that scheme after 3 years = $\\text{Deposited amount}\\times \\text{Rate of interest}\\times \\text{Time}$
\n= $P\\times 10$%$\\times 3$
\n= $30P$%
\n= $\\dfrac{30P}{100}$
\n= .3P
\nTotal amount obtained after 3 years = Interest + Deposited amount
\n= .3P + P
\n= 1.3P
\nNow as per question, we have to add Rs.1000 in that amount.
\nSo it will be (1.3P + 1000). Eq.(1)
\nNow it will be deposited in bank at the rate of 20% per annual compound interest for 2 years. Then compound interest after 2 years = Rs. 6160 Eq.(2)
\nFormula for compound interest (CI) = $\\text{Principal amount}[(1 + \\dfrac{\\text{Rate of interest}}{100})^{\\text{Time}} – 1]$
\nPut the values from Eq.(1) and Eq.(2) in the formula.
\n$6160 = (1.3P + 1000)[(1 + \\dfrac{20}{100})^{2} – 1]$
\n$6160 = (1.3P + 1000)[(\\dfrac{120}{100})^{2} – 1]$
\n$6160 = (1.3P + 1000)[(\\dfrac{6}{5})^{2} – 1]$
\n$6160 = (1.3P + 1000)[(\\dfrac{36}{25}) – 1]$
\n$6160 = (1.3P + 1000)[(\\dfrac{36-25}{25})]$
\n$6160 = (1.3P + 1000)[(\\dfrac{11}{25})]$
\n$6160 = (1.3P + 1000)[(\\dfrac{11}{25})]$
\n$6160\\times\\dfrac{25}{11} = (1.3P + 1000)$
\n$560\\times25 = (1.3P + 1000)$
\n14000 = (1.3P + 1000)
\n14000 – 1000 = 1.3P
\n1.3P = 13000
\n$P = \\dfrac{13000}{1.3}$
\n$P = \\dfrac{130000}{13}$
\nP = 10000.
\nHence, option c is the correct answer.<\/p>\n

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

We have CI=$P(1+\\dfrac{R}{100})^{2}-P$<\/p>\n

CI=$P((1+\\dfrac{R}{100})^{2}-1)$<\/p>\n

CI=$32000((1+\\dfrac{R}{100})^{2}-1)$<\/p>\n

let R\/100 =r
\nCI=$32000((1+r)^{2}-1)$
\nCI=$32000*(r^{2}+2r)$
\nSI=$32000*2*R\/100$
\nSI=64000r
\n$32000*(r^{2}+2r)-64000r$=720
\n$3200(r^{2}+2r)-6400r$=72
\n$3200r^{2}+6400r-6400r$=72
\n$r^{2}$=72\/3200
\n$r^{2}$=0.0225
\nr=0.15
\nR=0.15*100
\nR=15%<\/p>\n

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

A=$P(1+\\dfrac{R}{100})^{3}$<\/p>\n

125000=$P(1+\\dfrac{R}{100})^{3}$<\/p>\n

195312.5=$P(1+\\dfrac{R}{100})^{5}$<\/p>\n

$(1+\\dfrac{R}{100})^{2}$=1.5625<\/p>\n

$1+(R\/100)$=1.25
\nR\/100 =0.25
\nR=25%<\/p>\n

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

As per the given question,
\n$\\frac{P \\times 3 \\times 17}{100} = 2856$
\n$P = 5600$
\nHe invested the total amount i.e (5600 + 2856) 8456 in a scheme for two years at 20% (17% +3%).
\nFor the first year, the interest accumulated will be 20% of 8456 i.e 1691.2
\nThe total amount will be 8456 + 1691.2 = 10147.2
\nFor the second year, the interest accumulated will be 20% of 10147.2 i.e 2029.44
\nTotal interest for two years will be 1691.2 + 2029.44 = 3720.64(approximately)
\nHence, option E is the correct answer.<\/p>\n

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We hope this Simple & Compound Interest\u00a0 questions for RRB NTPC Exam will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Simple Interest and Compound Interest\u00a0Questions for RRB NTPC Set-2 PDF Download RRB NTPC Simple & Compound Interest\u00a0 Questions and Answers PDF. Top 25 RRB NTPC\u00a0 Maths\u00a0 questions based on asked questions in previous exam papers very important for the Railway NTPC exam. Question 1:\u00a0In what time would the simple interest on a certain sum be 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