**2. At what percentage per annum will Rs. 6000 amount to Rs. 7986 in 3 years if the interest is compounded yearly?**

**3. A customer invested Rs. 12,000 in two types of share at the simple rate of interest of 10% per annum and 20% per annum. If total interest at the end of the year is 14% of total amount. Find the amount invested at 10% per annum?**

**4. What annual payment will discharge a debt of Rs. 4770 dues in 3 year at the rate of S.I 6% per annum?**

**5. Sangeeta borrowed some money from a bank at the rate of 9 % per annum for first 5 years, 6 % per annum for next 3 years and 4 % per annum for beyond 8 years. If the total simple interest paid by him at the end of 10 years was Rs. 7668, how much money did he borrowed?**

**Answers and Explanations**

**1. Answer –** Option B

**Explanation –**

let the principle amount be 100

Now using slab method,

\(\frac {1}{10}\) = 10%

1. 10 (Year 1)

2. 10 1 (Year 2)

3. 10 1 1 0.1 (Year 3)

Now,

Total C.I = 33.1

Let Assume

S.I for three years if rate of interest 10% and principle amount is 100

Total S.I = 30

Now,

Difference between C.I & S.I = 33.1 â€“ 30

= 3.1 – – – – – – (i)

According to question, Difference = 310

If principle amount is 100 then difference is = 3.1

Now, difference is 310

So, Principle amount = \(\frac {310}{3.1} \times 100\) = 10000

**2. Answer –** Option B

**Explanation –**

As given in the question, the principal amount (P) = Rs 6000 and

Amount (A) = Rs 7986

Time (n) = 3 years

We need to find out the rate percent.

By using formula:-

\(p{(1 + \frac {R}{100})}^{n}\)

i.e, \(\frac {7986}{6000} = {( \frac {100 + r}{100})}^{3}\)

\(\frac {1331}{1000} = {( \frac {100 + r}{100})}^{3}\)

\({(\frac {11}{10})}^{3} = {( \frac {100 + r}{100})}^{3}\)

\(\frac {11}{10} = ( \frac {100 + r}{100})\)

i.e, r = 10%

**3. Answer –** Option D

**Explanation –**

Let Rs. P was invested at 10% and the rest are at 20%.

\(\frac {P \times 10 \times 1}{100} + (12000 – P) \times \frac {20 \times 1}{100} = \frac {12000 \times 14 \times 1}{100}\)

P + 24000 – 2P = 16800

P = Rs.7200

**4. Answer –** Option B

**Explanation –**

Annual payment = \(\frac {100x}{100t + \frac {rt(t-1)}{2}} = \frac {100 \times 4770}{100 \times 32 + (\frac {6 \times 3 (3-1)}{2})} = \frac{477000}{300 + 18} = \frac{477000}{318} = 1500\)

**5. Answer –** Option C

**Explanation –**

Let Rs. P was invested at 10% and the rest are at 20%.

7668 = \(\frac {P}{100} (9 \times 5 + 6 \times 3 + 4 \times 2)\)

7668 = \(\frac {P}{100} (45 + 18 + 8)\)

7668 = \(\frac {P}{100} \times 71\)

P = \(\frac {766800}{71}\) = Rs. 10800

**2. SBI offers 10% interest compounded half yearly. A customer deposits Rs. 1200 each on 1st January and 1st July of a year. At the end of the year, what amount would he gained as an interest?**

**3. The difference between CI and SI for 3 years @20% p.a . is Rs. 152 . What is the principal lent in each case?**

**4. What annual payment will discharge a debt of Rs.1696 due in 2 year at 12% simple interest?**

**5. The compound interest on Rs. 6250 at 12% per annum for 1 year, compounded half-yearly is**

**Answers and Explanations**

**1. Answer –** Option C

**Explanation –**

Hint: Half – yearly compounding always increases the value of the amount more than annual compounding.

**2. Answer –** Option B

**Explanation –**

Amount = \(1200 {(1 + \frac {10}{100})}^{2} + 1200 (1 + \frac {10}{100}) = 1200[( \frac {121}{100} + \frac {11}{10}) = 1200 \times \frac {231}{100} = 2772\)

CI = 2772 – (1200 + 1200) = Rs.372

**3. Answer –** Option C

**Explanation –**

Difference between CI and SI for 3 years = Rs. 152

\(P{(\frac {r}{100})}^{2} ( \frac { r}{100} + 3)\) = 152

\(P(\frac {1}{25}) (\frac { 16}{5})\) = 152

\(P = \frac {152 \times 25 \times 5}{16}\)

P = 9.5x 25x 5 = 1187.5

**4. Answer –** Option A

**Explanation –**

According to the formula,

Annual payment = \(\frac {100p}{100T + \frac {RT(T-1)}{2}}\)

= \(\frac {100 \times 1696}{100 \times 2 + \frac {24(2-1)}{2}} = \frac {1696 \times 100}{212} = 800\)

**5. Answer –** Option A

**Explanation –**

T = 2 half years Rate = 6%

i.e, CI = P\([{( 1 + \frac {r}{100})}^{2} – 1 ]\)

6250 = P\([{( 1 + \frac {6}{100})}^{2} – 1 ]\)

= 6250 \(\times\) 0.1236 = 772.5

**2. The simple and compound interest that can be earned in two years at the same rate is Rs 1,000 and Rs 1,040 respectively. What is the rate (percent per annum) of interest?**

**3. The compound interest on Rs. 100000 compounded quarterly for 9 months at the rate of 4% per annum is**

**4. The amount received at 10% per annum compound interest after 3 yrs is Rs 10,648. What was the principal (in Rs)?**

**5. The amount of Rs. 25,000 after 2 years, compounded annually with the rate of interest being 10% per annum during the first year and 15% per annum during the second year, would be (in rupees)**

**Answers and Explanations**

**1. Answer –** Option D

**Explanation –**

Amount = P\({(1 + \frac {r}{100})}^{t}\)

Here,

interest rate =r=10% ;

P1= principle=2000 ; t1 = 1 year

A1 = 2000 \({(1 + \frac {10}{100})}^{1}\)

= 2000 \(\frac {11}{10}\)

P2= 2000 + 2200= 4200 Rs

Final amount A2 = 4200\({(1 + \frac {10}{100})}^{1}\) , here t2 = 1year

= 4200\(\frac {11}{10}\)

= 4620

Total CI = 4620 â€“ (2000 + 2000) = 620 Rs

**2. Answer –** Option C

**Explanation –**

Simple interest of two year=1000

Simple interest per year=\(\frac {1000}{2} = 500 \)

Difference between simple interest and compound interest=1040-1000=40

40 rupees is the interest at the first year simple interest

Then required rate of interest=\(\frac {40}{500} \times 100 = 8%\)

**3. Answer –** Option C

**Explanation –**

P=Rs. 100000

Time, T=9 months=3 quarter yr

R = 4% per annum =1% quarterly

i.e, A = \(100000{(1 + \frac {1}{100})}^{3} = 100000 {( \frac {101}{100})}^{3} \)

= \(10000 \times \frac {101}{100} \times \frac {101}{100}\times \frac {101}{100} = \) Rs. 103030.10

Also, compound interest = A – P = Rs. 103030.10 – Rs. 100000

Compound interest = Rs. 3030.10

**4. Answer –** Option A

**Explanation –**

10% = \(\frac {1}{10}\)

Let principal be = \({10}^{3}\) = 1000

Total amount = 1000 + 3 * 100 + 3 * 10 + 1 = 1331 unit

1331 unit â‡’ 10648

1 unit â‡’ 8

1000 unit â‡’ 1000 Ã— 8 = Rs. 8000

Thus, principal = 8000.

**5. Answer –** Option D

**Explanation –**

\(25000 \times (1 + \frac {10}{100}) \times (1 + \frac {15}{100})\)

A = \(25000 \times \frac {11}{10} \times \frac {23}{210} = 125 \times 11 \times 23\) = Rs. 31625