2What principal will amount to ₹ 16000 in 6 years at 10% simple interest?
A
11000
B
12000
C
10000
D
13000
Correct Ans:C
Explanation:
Let the principal be ₹ p, given rate of interest is 10% and time = 6 years. Amount received at the end of 6 years = 16000 ₹ Amount = Principal + S.I = p + (p×r×t)/100
⇒ 16000 = p + (p×10×6)/100
= p + 6p/10
= 16p/10
⇒ P = 16000×(10/16)
= 1000×10
= 10000 ₹ Principal should be ₹ 10000
C
3A sum of money doubles it'self in 8 years. What is the rate of interest?
A
11.50%
B
12.50%
C
13.50%
D
14.50%
Correct Ans:B
Explanation:
Let the principal be ₹ p, rate of interest be r and time given, t = 8. Amount = 2p (since the money gets doubled), interest earned = p Which means S.I = p, we know that S.I = (p×t×r) / 100 p = (p×8×r) / 100
⇒ 8r/100 = 1
⇒ r=100/8
⇒ r = 12.5 The rate of interest is 12.5%
B
4What will the ratio of simple interest earned by certain amount at the same rate of interest for 6 years and that for 9 years:
A
1 : 3
B
4 : 3
C
2 : 3
D
5 : 3
Correct Ans:C
Explanation:
We know that,
SI = Pnr/100
Let the principal be P and rate of interest be R
SI = [P×6×R/100] / [P×9×R/100]
SI = 6/9 = $$\frac{2}{3}$$
SI = 2 : 3
C
5A certain amount of money amounts to ₹ 720 in 2 years and to ₹ 870 in 4($$\frac{1}{2}$$) years. Find the rate of interest if S.l. is reckoned:
A
10%
B
12%
C
14%
D
16%
Correct Ans:A
Explanation:
Formula to find sum without rate of interest, P = (A1×n2- A2×n1) / n2-n1 P = (720×9/2 - 870×2) / (9/2) -2 P = (3240 - 1740)/(5/2) = (1500×2)/5 P = 600 we know that, SI = A - P A = 720; P = 600 SI = 720 - 600 = 120 To find rate of interest SI = Pnr/100 120 = (600×2×r)/100 Rate of interest, r = 10%
A
6Nalini borrowed ₹ 1075 from her friend at 7% per annum. She returned the amount after 7 years. How much amount did she pay?
A
1501.75
B
1601.75
C
1701.75
D
1801.75
Correct Ans:B
Explanation:
Given principal, p = 1075 ₹, rate of interest r= 7%, time, t=7 years Simple interest,
SI = (p×r×t)/100
⇒ SI = (1075×7×7)/100
⇒ SI = 526.75. Amount = Principal + S.I
Amount paid by Nalini to her friend is 1601.75 ₹
B
7If the difference between Cl and Sl earned on a certain amount at 15% pa at the end of 3 years is ₹3200, find out the principal:
A
₹45050
B
₹45150
C
₹45350
D
₹45450
Correct Ans:B
Explanation:
The difference between Cl and Sl for 3 years = Pr2/1002[3 + (r/100)] 3200 = (P×15×15)/(100×100)[3 + (15/100)] 3200 = (P×3×3)/(20×20)[63/20] P = 3200[(20×20×20)/(9×63)] P = 45,149.91 P = 45,150
B
8A sum of ₹ 800 is lent at the rate of 4% per annum. Find the interest at the end of 6 years:
A
192
B
172
C
182
D
202
Correct Ans:A
Explanation:
Given principal, p = 800 ₹, rate of interest r= 4%, time, t=6 years Simple interest, SI = (p×r×t)/100 ⇒ SI = (800×4×6)/100 ⇒ SI = 192 At the end of 6 years, interest earned is 192
A
9A sum at simple interest of 13($$\frac{1}{2}$$)% per annum amounts to ₹ 3080 in 4 years. Find the sum:
A
₹ 2200
B
₹ 2200
C
₹ 2100
D
₹ 2000
Correct Ans:D
Explanation:
We know Amount - ₹ 3080; R = 27/2 %; N = 4 yrs Let the sum be x. SI = Amount - Principal SI = 3080 - x SI = PNR/100 3080 -×= (x × 27 × 4)/(100×2) 3080 -×= 27x/50 154000 - 50x = 27x 154000 = 27x + 50x 154000 = 77x x = 154000/77 x = ₹ 2000 ∴ sum is ₹ 2000.
D
10The simple interest on a sum of money is 1/9 of the principal and the number of years is equal to the rate percent per annum. The rate percent per annum is:
A
10/3%
B
11/3%
C
13/3%
D
14/3%
Correct Ans:A
Explanation:
Let the principal be ₹x and rate of interest be r. Given, interest = 1/9 of principal Simple interest = (1/9)x Also given, Number of years, N = Rate of interest, r SI = PNR/100 x/9 = (x×r×r)/100 r2 = 100/9 r = 10/3 ∴ rate of interest = 10/3%.