1. A sum of Rs. 14,000 amounts to Rs. 22,400 in 12 years at the rate of simple interest. What is the rate of interest?
1. 7%
2. 6%
3. 5%
4. 4%
Answer & Explanation Answer:- 3 Explanation :- P= Rs. 14000 2. A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest? Answer & Explanation Answer:- 1 Explanation :- Let the original rate be R%.(for one Year) 3. An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest after one year becomes: Answer & Explanation Answer:- 2 Explanation :- Let the sum be Rs. 100. Then, 4. A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is: Answer & Explanation Answer:- 2 Explanation :- Let the Rate of Interest be R% 5. What annual payment will discharge a debt of Rs. 6450 due in 4 years at 5% per annum? Answer & Explanation Answer:- 1 Explanation :- Amount to be repaid in 4 years = Rs.6450 OR (Shortcut Formula) Let Dept be P
SI=Rs. 22400
T=12 years
R=(SI*100)/(P*T)
R=(22400*100)/(14000*12)
Therefore , R = 5%
1. 3.46%
2. None of these
3. 4.5%
4. 5%
Then, new rate = (2R)%.(for1/3 year)
Therefore
((725*R*1)/100) +((362.50*R*(1/3))/100)=33.50
(2175 + 725) R = 10050
(2900)R = 10050
R = 3.46%
1. None of these
2. 10.25%
3. 10.5%
4. 10%
S.I.(first 6 months) = Rs.((100 x (10/2) x 1)/(100) ]= Rs.5
S.I.(last 6 months) =Rs.((105 x (10/2) x 1)/(100) ] = Rs.5.25
Therefore, Amount after 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25
Effective rate = (110.25 – 100) = 10.25%
1. 5%
2. 10%
3. 7%
4. 8%
Total SI=2200
((5000*R*2)/100)+((3000*R*4)/100)=2200
100R+120R=2200
R = 10%
1. Rs.1500
2. Rs.1400
3. Rs.1800
4. Rs.1600
Let Dept to repay in one year be Rs.X
SI.(for 3 years)=(X*5*3)/100=15X/100
SI. (for 2 years)=(X*5*2)/100=10X/100
SI. (for 1 year)=(X*5*1)/100=5X/100
Total Amount to be repaid in 4 years => (X+(15X/100))+ (X+(10X/100))+ (X+(5X/100))+X=6450
=>4X+(30X/10)=6450
40X+3X=64500
X=64500/43
X= Rs. 1500
Annual Payment=x
R=Rate of Interest
Time=T
Therefore,
X=(6450*100)/430
X= Rs.1500