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# Simple and Compound Interest Questions for RRB NTPC Set-3 PDF

Download RRB NTPC Simple & Compound Interest Questions Set-3 PDF. Top 10 RRB NTPC questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Question 1: The amount (in Rs) received at 10% per annum compound interest after 3 yrs is Rs 1,19,790. What was the principal?

a) 90000

b) 1,00,000

c) 80000

d) 75000

Question 2: What is the total earning after 2 years from ₹ 18,750 on variable compound interest if the rate of interest for the first 4 year is 4% and for the second, it is 8%?

a) ₹ 1,740

b) ₹ 1,760

c) ₹ 1,670

d) ₹ 2,310

Question 3: In how many months will Rs 8,000 yield Rs 2,648 as compound interest at 20% per annum compounded semi-annually?

a) 18

b) 24

c) 12

d) 30

Question 4: What is the rate of interest (in %) if simple interest earned on a certain sum for the 3rd year is Rs 2,000 and compound interest earned in 2 years is Rs 4,160?

a) 8

b) 10

c) 12

d) 6

Question 5: On what sum of money. the interest for one year at 12% p.a. compounded half yearly is ₹ 1,545?

a) ₹ 12,500

b) ₹ 12,875

c) ₹ 25,750

d) ₹ 24,300

Question 6: A sum of ₹800 invested on simple interest becomes ₹1200 in 8 years. What will be simple interest for 6 years on the sum at the same rate of interest?

a) ₹240

b) ₹210

c) ₹250

d) ₹300

Question 7: A sum of ₹2000 is invested on simple interest for three years at the rate of 10% per annum,then the amount will be:

a) ₹2900

b) ₹2600

c) ₹2300

d) ₹2500

Question 8: A sum of ₹ 4200 becomes ₹ 5082 in 3 years rate simple interest. What is the rate of interest per annum?

a) 8%

b) 9%

c) 5%

d) 7%

Question 9: The difference between simple and compound interests compounded annually, on a certain sum of money for 2 years at 6% pa is ₹ 3.6. What is the sum?

a) ₹800

b) ₹750

c) ₹1,000

d) ₹1,250

Question 10: The simple interest for the second year on a certain sum at a certain rate of interest is ₹ 1000. What is the sum of the interest accrued on it for the $6^{th}, 7^{th}$ and $8^{th}$ years?

a) ₹3200

b) ₹3300

c) ₹3000

d) ₹3630

$P(1+\frac{R}{100})^{3}$=119790

$P(1+\frac{10}{100})^{3}$=119790

$P(\frac{11}{10})^{3}$=119790

P=119790*1000/1331
P=Rs 90000

P=18750

1st year = 4%

2nd year = 8%

Interest for 1st year $= \frac{18750\times 4\times 1}{100} = 750$

Principal after 1st year $= 18750+750 = 19500$

Interest for 2nd year $=\frac{19500\times 8\times 1}{100} = 1560$

Total interest = 750+1560 = 2310

Option D is correct.

$P(1+\frac{R/2}{100})^{2n}-P$=2648
$(1+\frac{R/2}{100})^{2n}-1=2648/8000 4) Answer (A)$PTR/100$=2000$PR$=200000$P(1+\frac{R}{100}^{2})-P$=4160$P((200+R)(R))$=4160 Dividing both the equations we have 1/(R+200) =20/4160 R+200=208 R=8% 5) Answer (A) In compound interest, Final amount,A$=P(1+r/n)^nt$. where P is principal amount,r is rate of interest,n is number of times interest applied and t is number of times period elapsed. So,interest gain$=A-P.$So,$P(1+0.12/2)^2-P=1545.$or,$P×1.1236-P=1545.$or,$P=(1545/0.1236).$or,$P=12500.$A is correct choice. 6) Answer (D) SI= (P*R*T)/100 Given , P= 800 RS T= 8 years SI= Amount – P= 1200-800 = 400 Rs SI = (400*R*8)/100 R = (25/4) or 6.25 Now, given , T= 6 Years, P = 800 Rs, R= 6.25% SI = (800*6*6.25)/100 = ₹300 7) Answer (B) Simple Interest (SI) is given as =$\frac{PRT}{100}$, where P is Principle amount, R is Rate of Interest,T is Time period Here, P = 2000, R = 10 % So, SI =$\frac{(2000)(10)(3)}{100}$= 600 Hence the total amount will be SI + P = 2000 + 600 = 2600. 8) Answer (D) Let the rate of interest be R%. Simple Interest = Rs.5082 – Rs.4200 = Rs.882$\dfrac{4200\times R\times3}{100} = 882$=> 3R = 21 => R = 7 Therefore, Rate of interest = 7% 9) Answer (C) Let the principal be Rs.10000x Simple Interest for 1 year at 6% per annum = Rs.600x Simple Interest for 2 years at 6% per annum = Rs.600*2 = Rs.1200x Amount of Compound Interest for 2 years at 6% per annum =$10000x \times \dfrac{106}{100} \times \dfrac{106}{100} = Rs.11236x
Compound Interest = Rs.11236x – Rs.10000x = Rs.1236x
Difference between SI and CI = 1236x – 1200x = Rs.36x
Given, 36x = Rs.3.6
=> x = 0.1
Therefore, Principal = Rs.10000x = Rs.10000*0.1 = Rs.1000