8) Simple Interest & compound interest

 1) A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:

Rs. 650
Rs. 690
Rs. 698
Rs. 700
Answer: Option
Explanation:

S.I. for 1 year = Rs. (854 - 815) = Rs. 39.

S.I. for 3 years = Rs.(39 x 3) = Rs. 117.

 Principal = Rs. (815 - 117) = Rs. 698.


2) 

What would be the annual interest accrued on a deposit of Rs. 10,000 in a bank that pays a 4 % per annum rate of simple interest?

formula: SI = P x R x T / 100
Solution : 

Here, P = 10000, R = 4, T = 1
SI = P x R x T / 100
SI = 10000 x 4 x 1 / 100
SI = 400
Thus, the annual interest would be Rs. 400

3. 

A man borrowed a certain sum of money at the rate of 6 % per annum for the first two years, 9% per annum for the next three years, and 14% per annum for the period beyond 5 years. If he pays a total interest of Rs. 22,800 at the end of 9 years, find the amount he borrowed.

Solution: 

Let the borrowed sum be P.
SI for first 2 years + SI for next 3 years + SI for next 4 years = 22800
(P x 6 x 2 / 100) + (P x 9 x 3 / 100) + (P x 14 x 4 / 100) = 22800
95 P / 100 = 22800
P = 24000

Therefore, Borrowed sum = Rs. 24,000 

4. 

At what annual rate of interest will a sum of money be thrice in 10 years?

Solution : 

Amount = Principal + SI
If the sum of money would be thrice the principal after 10 years, the SI would be twice the principal.
SI = 2 x P
(P x R x T / 100) = 2 X P
R x T / 100 = 2
R x T = 200
R x 10 = 200
R = 20 %
Thus, the required rate of interest is 20 %



5. If Rs. 5000 amounts to Rs. 5832 in two years compounded annually, find the rate of interest per annum.

We are given:

  • Principal P=5000P = 5000

  • Amount after 2 years A=5832A = 5832

  • Time t=2t = 2 years

  • Compounded annually → n=1n = 1

We use the compound interest formula:

A=P(1+r100)tA = P \left(1 + \frac{r}{100}\right)^t

Substitute the known values:

5832=5000(1+r100)25832 = 5000 \left(1 + \frac{r}{100}\right)^2

Step 1: Divide both sides by 5000:

58325000=(1+r100)2\frac{5832}{5000} = \left(1 + \frac{r}{100}\right)^2 1.1664=(1+r100)21.1664 = \left(1 + \frac{r}{100}\right)^2

Step 2: Take square root on both sides:

1.1664=1+r100\sqrt{1.1664} = 1 + \frac{r}{100} 1.08=1+r1001.08 = 1 + \frac{r}{100}

Step 3: Solve for rr:

r100=1.08−1=0.08\frac{r}{100} = 1.08 - 1 = 0.08 r=0.08×100=8%r = 0.08 \times 100 = \boxed{8\%}

✅ Answer: 8% per annum

 


6. Find the compound interest on Rs. 10,000 at 10% per annum for a time period of three and a half years.

o calculate compound interest on Rs. 10,000 at 10% per annum for 3.5 years, we’ll use the formula:

A=P(1+r100)tA = P \left(1 + \frac{r}{100}\right)^t

Given:

  • P=10,000P = 10,000

  • r=10%r = 10\%

  • t=3.5t = 3.5 years = 312=723 \frac{1}{2} = \frac{7}{2} years

  • Compounded annually

Step 1: Apply the formula:

A=10,000(1+10100)3.5=10,000(1.10)3.5A = 10,000 \left(1 + \frac{10}{100} \right)^{3.5} = 10,000 \left(1.10\right)^{3.5}

We calculate (1.10)3.5(1.10)^{3.5}:

(1.10)3.5≈1.103×1.100.5=1.331×1.0488≈1.3967(1.10)^{3.5} \approx 1.10^3 \times 1.10^{0.5} = 1.331 \times 1.0488 \approx 1.3967

Now calculate the amount:

A≈10,000×1.3967=13,967A \approx 10,000 \times 1.3967 = 13,967

Step 2: Find Compound Interest:

CI=A−P=13,967−10,000=Rs.3,967\text{CI} = A - P = 13,967 - 10,000 = \boxed{Rs. 3,967}

✅ Final Answer: Compound Interest = Rs. 3,967 (approx)

Let me know if you want it calculated with half-yearly compounding instead.

Comments

Post a Comment

Popular posts from this blog

5) Apti on mathematical ability ( Time & Work )

 1) A+B+C can do a work in 12 days. B+ C can do work in 18 days. A+ B can do work un 15 days. If b do the work alone, then how many days needed We are given: A + B + C can do the work in 12 days B + C can do the work in 18 days A + B can do the work in 15 days We are to find how many days B alone will take to complete the work. We are given: A + B + C can do the work in 12 days B + C can do the work in 18 days A + B can do the work in 15 days We are to find how many days B alone will take to complete the work. Step 1: Convert to Work Rates Let the total work be LCM of 12, 18, 15 = 180 units (assume total work is 180 units for easier calculation). Now compute each group’s work per day : (A + B + C)’s rate = 180 / 12 = 15 units/day (B + C)’s rate = 180 / 18 = 10 units/day (A + B)’s rate = 180 / 15 = 12 units/day Step 2: Subtract to find individual rates From above: A + B + C = 15 B + C = 10 So, A = 15 − 10 = 5 units/day Also...
Read more

6. Apti on Mathematical ability ( LCM , HCF )

 1.  If the HCF of two numbers is 12 and their LCM is 360, find the numbers. We know that the product of the HCF and LCM of two numbers is equal to the product of the two numbers. So, for two numbers a and b with HCF = 12 and LCM = 360: HCF × LCM = a × b 12 × 360 = a × b 4320 = a × b Now, we need to find two numbers whose product is 4320 and HCF is 12. There can be multiple pairs of numbers that satisfy this condition, and one such pair is: a = 120 and b = 36 Because 120 × 36 = 4320 and the HCF of 120 and 36 is 12. If you want valid pairs, here are a few: 12 and 360 60 and 72 36 and 120 2.  Find the HCF of 36, 48, and 72. Prime factorization of 36: 2 2 × 3 2 Prime factorization of 48: 2 4 × 3 Prime factorization of 72: 2 3 × 3 2 Common factors: 2 2 and 3 (take the minimum power) So, the HCF of 36, 48, and 72 is 2 2 × 3 = 12. 3. What is the largest three-digit number that is exactly divisible by the HCF of 24 and 36? he HCF of 24 and 36 is 12. To find the ...
Read more

9) Total Collect

1) Inspection and walkthrough is done by ? walkthrough is done by Author/Developer Inspection is done by:  Formal Review Team including: Moderator (leads the session) Author (person who wrote the document/code) Reviewers (peers, QA, devs, domain experts) Scribe (documents defects) 2) if you found a a high priority, high seberity defect in production, what you will say, its defect or bug or failure ? answer : failure
Read more