1. At what rate of simple interest per annum the sum of money get doubled in 10 years?
a. 5%
b. 10%
c. 15%
d. 20%
Answer: b. 10%
Explanation: Let the required number be \( x \).
According to the question:
Explanation:
Let the Principal be \( P \).
Since the amount doubles itself, the Amount (\( A \)) = \( 2P \).
Simple Interest (\( SI \)) = \( A – P = 2P – P = P \).
Given: Time (\( T \)) = 10 years.
Step 1: Apply the Simple Interest Formula
\[ SI = \frac{P \times R \times T}{100} \]
Substituting the values:
\[ P = \frac{P \times R \times 10}{100} \]
Step 2: Solve for Rate (\( R \))
\[ 1 = \frac{R \times 10}{100} \]
\[ R = \frac{100}{10} = 10 \]
The rate of interest is 10% per annum.
2. A borrows 800 at the rate of 12% per annum simple interest and B borrows 910 at the rate of 10% per annum, simple interest. In how many years will their amounts of debt be equal ?
a. 20 years
b. 21 years
c. 22 years
d. 23 years
Answer: c. 22 years
Explanation:
Let the time after which the amounts will be equal be \( t \) years.
We know that: \( \text{Amount} = \text{Principal} + \text{Simple Interest} \)
Formula for \( SI = \frac{P \times R \times T}{100} \)
Step 1: Set up the equation for both cases
Case 1 (800 at 12%): \( 800 + \frac{800 \times 12 \times t}{100} \)
Case 2 (910 at 10%): \( 910 + \frac{910 \times 10 \times t}{100} \)
Step 3: Solve for \( t \)
\[ t = \frac{110}{5} = 22 \]
The amounts will be equal after 22 years.
3. 7. A person deposited 400 for 4 years and 600 for 3 years at the same rate of simple interest in a bank. Altogether he received 340 as interest. The rate of simple interest per annum was?
a. 5%
b. 7.5%
c. 10%
d. 12.5%
Answer: c. 10%
Explanation:
Let the common rate of interest be \( r \).
Total Interest = (Interest on ₹400 for 4 years) + (Interest on ₹600 for 3 years)
Step 1: Apply the SI Formula
\[ 340 = \left( \frac{400 \times 4 \times r}{100} \right) + \left( \frac{600 \times 3 \times r}{100} \right) \]
Step 3: Solve for \( r \)
\[ r = \frac{340}{34} = 10 \]
The rate of interest is 10% per annum.
4. B invested a certain sum of money at 7% per annum simple interest. If he receives total sum of of Rs. 1,365 after one year, the sum he invested was?
a. 18,000
b. 18,500
c. 19,000
d. 19,500
Answer: d. 19,500
Explanation:
Let the principal amount be \( P \).
Given: Simple Interest (\( SI \)) = 1365, Rate (\( R \)) = 7%, and Time (\( T \)) = 1 year.
Step 1: Use the Simple Interest Formula
\[ SI = \frac{P \times R \times T}{100} \]
Substituting the values:
\[ 1365 = \frac{P \times 7 \times 1}{100} \]
Step 2: Solve for \( P \)
\[ 1365 = \frac{7P}{100} \]
\[ 7P = 1365 \times 100 \]
\[ P = \frac{136500}{7} \]
\[ P = 19500 \]
The principal amount is ₹19,500.
5. . If the principal amount is Rs. 14,000, then the total amount after simple interest for 3 years at a rate of 5% per annum will be?
a. Rs. 2,100
b. Rs. 16,100
c. Rs. 17,500
d. Rs. 18,000
Answer: b. Rs. 16,100
Explanation:
To find the total amount, we first calculate the Simple Interest and then add it to the Principal.
Step 1: Calculate Simple Interest (SI)
Using the formula: \[ SI = \frac{P \times R \times T}{100} \]
Given: \( P = 14000 \), \( R = 5\% \), and \( T = 3 \) years.
\[ SI = \frac{14000 \times 5 \times 3}{100} \]
\[ SI = 140 \times 15 = 2100 \]
Step 2: Calculate the Total Amount
\[ \text{Amount} = \text{Principal} + SI \]
\[ \text{Amount} = 14000 + 2100 = 16100 \]
The total amount after 3 years will be ₹16,100.
6. If a sum of Rs. 2,000 amounts to Rs. 2,420 in 3 years at a certain rate of simple interest per annum, then what will the same sum amount to in 5 years?
a. Rs. 2500
b. Rs. 2600
c. Rs. 2700
d. Rs. 2800
Answer: c. Rs. 2700
Explanation:
To find what the sum will amount to in 5 years, we first need to find the Rate of Interest (\( R \)) from the first condition and then apply it to the second condition.
Step 1: Find the Rate of Interest (\( R \))
Principal (\( P \)) = 2000, Amount (\( A \)) = 2420, Time (\( T \)) = 3 years.
Simple Interest (\( SI \)) = \( 2420 – 2000 = 420 \).
\[ SI = \frac{P \times R \times T}{100} \]
\[ 420 = \frac{2000 \times R \times 3}{100} \]
\[ 420 = 60R \Rightarrow R = 7\% \]
Step 2: Calculate Interest for 5 years
Now, find the SI for the same principal at 7% for 5 years:
\[ SI = \frac{2000 \times 7 \times 5}{100} \]
\[ SI = 20 \times 35 = 700 \]
Step 3: Calculate the New Total Amount
\[ \text{Amount} = \text{Principal} + \text{New SI} \]
\[ \text{Amount} = 2000 + 700 = 2700 \]
The sum will amount to ₹2,700 in 5 years.
7. The simple interest on 4,000 in 3 years at the rate of x% per annum equals the simple interest on 5,000 at the rate of 12% per annum in 2 years. The value of x is?
a. 7
b. 8
c. 9
d. 10
Answer: d. 10
Explanation:
According to the question, the Simple Interest earned in both cases is equal.
Formula: \[ SI = \frac{P \times R \times T}{100} \]
Step 1: Set up the equation
Let the missing rate of interest be \( x \).
Interest on 4000 = Interest on 5000
\[ \frac{4000 \times x \times 3}{100} = \frac{5000 \times 12 \times 2}{100} \]
Step 2: Simplify and solve for \( x \)
Cancel 100 from both denominators:
\[ 4000 \times 3 \times x = 5000 \times 12 \times 2 \]
\[ 12000x = 120000 \]
\[ x = \frac{120000}{12000} \]
\[ x = 10 \]
The rate of interest is 10% per annum.
8. Azim borrowed a certain sum which amounted to Rs. 23,200 in Rs. 36 months at 15% per annum simple interest. What is the borrowed sum ?
a. Rs. 15,000
b. Rs. 16,000
c. Rs. 17,000
d. Rs. 18,000
Answer: b. Rs. 16,000
Explanation:
Let the principal amount be \( P \).
Given: Amount (\( A \)) = 23,200, Rate (\( R \)) = 15%, and Time (\( T \)) = 3 years.
Simple Interest (\( SI \)) = \( \text{Amount} – \text{Principal} = 23200 – P \).
Step 1: Apply the Simple Interest Formula
\[ SI = \frac{P \times R \times T}{100} \]
\[ 23200 – P = \frac{P \times 15 \times 3}{100} \]
Step 2: Solve for \( P \)
\[ 23200 – P = \frac{45P}{100} \]
Multiply the entire equation by 100 to remove the fraction:
\[ 2320000 – 100P = 45P \]
\[ 2320000 = 145P \]
\[ P = \frac{2320000}{145} = 16000 \]
The principal amount is ₹16,000.
9. Ram deposited a certain sum of money in a company at 12% per annum simple interest for 4 years and deposited equal amount in fixed deposit in a bank for 5 years at 15% per annum simple interest. If the difference in the interest from two sources is 1350, then the sum deposited in each company will be?
a. Rs. 4,000
b. Rs. 5,000
c. Rs. 6,000
d. Rs. 7,000
Answer: b. Rs. 5,000
Explanation:
Let the principal amount deposited be \( P \).
According to the question, the difference between the interest earned from two different periods and rates is ₹1,350.
Step 1: Set up the equation
Interest 1 (15% for 5 yrs) – Interest 2 (12% for 4 yrs) = 1350
\[ \left( \frac{P \times 15 \times 5}{100} \right) – \left( \frac{P \times 12 \times 4}{100} \right) = 1350 \]
Step 2: Simplify and solve for \( P \)
\[ \frac{75P}{100} – \frac{48P}{100} = 1350 \]
\[ \frac{27P}{100} = 1350 \]
\[ 27P = 135000 \]
\[ P = \frac{135000}{27} \]
\[ P = 5000 \]
The amount deposited was ₹5,000.
10. Ramesh lends Rs. 700 for 3 years to a person and lends Rs. 900 for 4 years to the other person at the same rate of simple interest. If altogether he receives Rs. 285 as interest, what is the rate of interest per annum ?
a. 2%
b. 3%
c. 4%
d. 5%
Answer: d. 5%
Explanation:
Let the common rate of interest be \( r\% \).
According to the question, the total interest earned from two different sums is ₹285.
Step 1: Set up the equation
Total Interest = (Interest on ₹700) + (Interest on ₹900)
\[ \left( \frac{700 \times 3 \times r}{100} \right) + \left( \frac{900 \times 4 \times r}{100} \right) = 285 \]
Step 2: Simplify and solve for \( r \)
\[ 21r + 36r = 285 \]
\[ 57r = 285 \]
\[ r = \frac{285}{57} \]
\[ r = 5 \]
The rate of interest is 5% per annum.
11. A man buys a TV priced at Rs. 16000. He pays Rs. 4000 at once and the rest after 15 months on which he is charged a simple interest at the rate of 12% per year.The total amount he pays for the TV is ?
a. Rs. 17,800
b. Rs. 18,200
c. Rs. 19,800
d. Rs. 19,200
Answer: a. Rs. 17,800
Explanation:
Initial amount = 16,000.
After a payment of 4,000, the remaining Principal (\( P \)) = \( 16000 – 4000 = 12000 \).
Step 1: Convert Time to Years
Time (\( T \)) = 15 months = \( \frac{15}{12} \) years = \( \frac{5}{4} \) years.
Step 2: Calculate Simple Interest (SI)
Given: Rate (\( R \)) = 12% per annum.
\[ SI = \frac{P \times R \times T}{100} \]
\[ SI = \frac{12000 \times 12 \times 5}{100 \times 4} \]
\[ SI = 120 \times 3 \times 5 = 1800 \]
Step 3: Calculate the Final Amount
Total amount to be paid = Original Principal + Interest
\[ \text{Total} = 16000 + 1800 = 17800 \]
The total amount after interest will be ₹17,800.
12. A certain sum under simple interest at a certain rate of interest per annum amounts to ₹1,200 in 2 years and to ₹1,500 in 3 years. The rate of interest per annum is?
a. 20%
b. 30%
c. 40%
d. 50%
Answer: d. 50%
Explanation:
In Simple Interest, the interest earned remains the same every year.
Amount after 3 years = ₹1,500
Amount after 2 years = ₹1,200
Step 1: Find the Interest for 1 year
Interest for 1 year = Amount (3 yrs) – Amount (2 yrs)
Interest (1 yr) = \( 1500 – 1200 = 300 \)
Step 2: Find the Principal Amount
Interest for 2 years = \( 300 \times 2 = 600 \)
Principal (\( P \)) = Amount (2 yrs) – Interest (2 yrs)
\[ P = 1200 – 600 = 600 \]
Step 3: Calculate the Rate of Interest (\( r \))
Using the formula: \[ SI = \frac{P \times r \times t}{100} \]
\[ 600 = \frac{600 \times r \times 2}{100} \]
\[ 600 = 12r \]
\[ r = \frac{600}{12} = 50\% \]
The rate of interest is 50% per annum.
13. A lends Rs. 800 to B and a certain sum to C at the same time at a simple interest of 7% per annum. If in 4 years, he altogether receives Rs. 476 as interest from the two, then the sum of money he lent to C was?
a. Rs. 700
b. Rs. 800
c. Rs. 900
d. Rs. 1000
Answer: c. Rs. 900
Explanation:
Let the amount given to C be \( x \).
The total interest earned from B and C combined is ₹476. Both were given loans at a 7% rate for 4 years.
Step 1: Set up the Total Interest equation
\[ \text{Interest from B} + \text{Interest from C} = 476 \]
\[ \left( \frac{800 \times 7 \times 4}{100} \right) + \left( \frac{x \times 7 \times 4}{100} \right) = 476 \]
Step 3: Solve for \( x \)
\[ 28x = 25200 \]
\[ x = \frac{25200}{28} = 900 \]
The amount given to C was ₹900.
14. Two equal sums were lent out at 7% and 5% S.I. respectively. The interest earned on the two loans add up to Rs. 1440 for 4 years. The amount given was?
a. Rs. 1,000
b. Rs. 2,000
c. Rs. 3,000
d. Rs. 4,000
Answer: c. Rs. 3,000
Explanation:
Let the principal amount be \( x \).
The sum of interest from two different rates over 4 years is ₹1,440.
Step 1: Set up the Total Interest equation
\[ \text{Interest}_1 + \text{Interest}_2 = 1440 \]
\[ \left( \frac{x \times 7 \times 4}{100} \right) + \left( \frac{x \times 5 \times 4}{100} \right) = 1440 \]
Step 3: Solve for \( x \)
\[ 48x = 1440 \times 100 \]
\[ x = \frac{144000}{48} \]
\[ x = 3000 \]
The principal amount is ₹3,000.
15.B lends Rs. 30,000 to two of his friends. He gives Rs. 18,000 to the first at 8% p.a. simple interest. B wants to make a profit of 12% on the whole. The simple interest rate at which he should lend the remaining sum of money to the second friend is?
a. 6.2%
b. 6.8%
c. 12%
d. 18%
Answer: d. 18%
Explanation:
Total investment = ₹30,000. This is split into two parts: ₹18,000 and the remaining ₹12,000.
Target: 12% profit on the entire ₹30,000.
Step 2: Calculate Interest from the first friend
Interest on ₹18,000 at 8%:
\[ \text{Interest}_1 = 18,000 \times \frac{8}{100} = 1,440 \]
Step 3: Calculate Required Interest from the second friend
Remaining Amount = \( 30,000 – 18,000 = 12,000 \).
Required Interest from second friend = \( 3,600 – 1,440 = 2,160 \).
Step 4: Find the Rate (\( r \)) for the remaining ₹12,000
\[ \frac{12,000 \times r \times 1}{100} = 2,160 \]
\[ 120r = 2,160 \]
\[ r = \frac{2,160}{120} = 18 \]
The rate of interest on the remaining amount should be 18%.
16. At certain rate of simple interest per annum a sum of money amounts to of itself in 10 years. What is the rate 13/8 of simple interest per annum?
a. 7.5%
b. 10%
c. 12.5%
d. 15%
Answer: c. 12.5%
Explanation:
Let the Principal amount be \( P \).
According to the question, the Amount (\( A \)) becomes \( \frac{13}{8} \) of the Principal.
Step 1: Find the Simple Interest (SI)
\[ SI = A – P \]
\[ SI = \frac{13P}{8} – P = \frac{5P}{8} \]
Step 2: Use the Simple Interest Formula
Given: Rate (\( R \)) = 5% per annum. Let the time be \( T \).
\[ SI = \frac{P \times R \times T}{100} \]
\[ \frac{5P}{8} = \frac{P \times 5 \times T}{100} \]
Step 3: Solve for \( T \)
Cancel \( P \) and \( 5 \) from both sides:
\[ \frac{1}{8} = \frac{T}{100} \]
\[ T = \frac{100}{8} = 12.5% \]
17. John invested a sum of money at an annual simple interest rate of 5%. At the end of four years the amount invested plus interest earned was Rs. 720. The amount invested was?
a. Rs. 500
b. Rs. 550
c. Rs. 600
d. Rs. 800
Answer: c. Rs. 600
Explanation:
Let the principal amount be \( x \).
Given: Amount (\( A \)) = 720, Rate (\( R \)) = 5%, and Time (\( T \)) = 4 years.
Simple Interest (\( SI \)) = \( \text{Amount} – \text{Principal} = 720 – x \).
Step 1: Use the Simple Interest Formula
\[ SI = \frac{P \times R \times T}{100} \]
Substituting the values:
\[ 720 – x = \frac{x \times 5 \times 4}{100} \]
Step 2: Simplify the equation
\[ 720 – x = \frac{20x}{100} \]
\[ 720 – x = \frac{x}{5} \]
Step 3: Solve for \( x \)
Multiply the entire equation by 5:
\[ 5(720 – x) = x \]
\[ 3600 – 5x = x \]
\[ 6x = 3600 \]
\[ x = \frac{3600}{6} = 600 \]
The principal amount is ₹600.
18. A sum of Rs. 1000 is lent out partly at 8% and the remaining at 10% per annum. If the yearly income on the average is 8.8%, the two parts respectively are?
a. Rs. 500, Rs. 500
b. Rs. 600, Rs. 400
c. Rs. 450, Rs. 550
d. Rs. 700, Rs. 300
Answer: b. Rs. 600, Rs. 400
Explanation:
Total amount = ₹1,000.
Let the amount lent at 8% be \( x \).
Then, the amount lent at 10% will be \( 1000 – x \).
The overall interest rate is 8.8%, so the total interest earned is \( 1000 \times 8.8\% = 88 \).
Step 1: Set up the Interest Equation
\[ \left( \frac{x \times 8}{100} \right) + \left( \frac{(1000 – x) \times 10}{100} \right) = 88 \]
Step 2: Simplify and solve for \( x \)
Multiply the entire equation by 100 to remove denominators:
\[ 8x + 10(1000 – x) = 8800 \]
\[ 8x + 10000 – 10x = 8800 \]
\[ -2x = 8800 – 10000 \]
\[ -2x = -1200 \]
\[ x = 600 \]
Step 3: Find both parts
First part (at 8%) = ₹600.
Second part (at 10%) = \( 1000 – 600 = \) ₹400.
19. A sum of Rs. 700 becomes Rs. 910 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what amount will the same sum becomein 3 years ?
a. Rs. 956
b. Rs. 974
c. Rs. 994
d. Rs. 1004
Answer: c. Rs. 994
Explanation:
First, we find the initial rate of interest, then increase it by 4% to find the new total amount.
Step 1: Find the Original Rate of Interest (\( r \))
Given: Principal (\( P \)) = 700, Simple Interest (\( SI \)) = 210, and Time (\( T \)) = 3 years.
\[ SI = \frac{P \times r \times T}{100} \]
\[ 210 = \frac{700 \times r \times 3}{100} \]
\[ 210 = 21r \Rightarrow r = 10\% \]
Step 2: Calculate the New Rate and Interest
New Rate = \( 10\% + 4\% = 14\% \).
New Simple Interest:
\[ SI_{new} = \frac{700 \times 14 \times 3}{100} \]
\[ SI_{new} = 7 \times 42 = 294 \]
Step 3: Find the Total Amount
\[ \text{Total Amount} = \text{Principal} + SI_{new} \]
\[ \text{Total Amount} = 700 + 294 = 994 \]
The new amount will be ₹994.
40. The amount 2,100 became 2,352 in 2 years at simple interest. If the interest rate is decreased by 1%, what is the new interest ?
a. Rs. 150
b. Rs. 180
c. Rs. 200
d. Rs. 210
Answer: c. Rs. 994
Explanation:
First, we find the original rate of interest, then decrease it by 1% to find the new Simple Interest.
Step 1: Find the Original Rate (\( r \))
Principal (\( P \)) = 2100, Amount (\( A \)) = 2352, Time (\( T \)) = 2 years.
Simple Interest (\( SI \)) = \( 2352 – 2100 = 252 \).
Using the formula: \[ SI = \frac{P \times r \times T}{100} \]
\[ 252 = \frac{2100 \times r \times 2}{100} \]
\[ 252 = 42r \Rightarrow r = 6\% \]
Step 2: Calculate New Interest with Decreased Rate
New Rate = \( 6\% – 1\% = 5\% \).
\[ \text{New } SI = \frac{2100 \times 5 \times 2}{100} \]
\[ \text{New } SI = 21 \times 10 = 210 \]
The new interest will be ₹210.
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