# PE 1 - SICI | Arithmetic - Simple & Compound Interest

**PE 1 - SICI | Arithmetic - Simple & Compound Interest**

The cost of a computer depreciates by Rs 720 during the second year and Rs 648 during the third year. What is the rate of depreciation per annum and the original cost of computer?

- A.
100/9%, Rs. 800

- B.
100/9%, Rs. 888.8

- C.
10%, Rs. 8000

- D.
10%, Rs. 800

Answer: Option C

**Explanation** :

Let the rate of depreciation be r% p.a. and cost of computer at the beginning of 2nd year be C.

⇒ C × r% = 720 … (1)

∴ Cost of the computer at the end of 2nd year (or at the beginning of 3rd year) = C(1 – r%)

⇒ C(1 – r%) × r% = 648 …(2)

⇒ 720(1 – r%) = 648

⇒ 1 – r% = 648/720 = 9/10

⇒ r = 10%

From (1) we have

C × 10% = 720

⇒ C = 7200

∴ Cost of the computer at the end of 1st year (or beginning of 2nd year) = 7200

∴ Cost of the computer at the beginning of 1st year) = 7200/0.9 = 8000

Hence, option (c).

Workspace:

**PE 1 - SICI | Arithmetic - Simple & Compound Interest**

The value of a fridge which costs Rs 6,000 depreciates by 10% each year. After how many years will its depreciated value be Rs 4374?

- A.
3

- B.
1

- C.
4

- D.
2

Answer: Option A

**Explanation** :

After 1^{st} year, the value of fridge will become 6000 – 600 = Rs 5400.

After 2^{nd} year, the value of fridge will become 5400 – 540 = Rs 4860.

After 3^{rd} year, the value of fridge will become 4860 – 486 = Rs 4374.

Thus after 3 years the value depreciates to Rs 4374.

Hence, option (a).

Workspace:

**PE 1 - SICI | Arithmetic - Simple & Compound Interest**

A sum of Rs 6,000 was taken as a loan. This is to be repaid in two equal annual installments. If the rate of interest is 20% compounded annually then find the value of each installment.

- A.
Rs. 4400

- B.
Rs. 2220

- C.
Rs. 4320

- D.
Rs. 4420

Answer: Option C

**Explanation** :

P = Rs. 6000

r = 20%

t = 2 years

Let the equal annual installment be x

∴ $P{\left(1+\frac{r}{100}\right)}^{2}$ = $x{\left(1+\frac{r}{100}\right)}^{1}$ + $x{\left(1+\frac{r}{100}\right)}^{2}$

⇒ $6000{\left(1+\frac{20}{100}\right)}^{2}$ = $x{\left(1+\frac{20}{100}\right)}^{1}$ + $x{\left(1+\frac{20}{100}\right)}^{2}$

Solving this we get

x = 4320

Hence, option (c).

Workspace:

**PE 1 - SICI | Arithmetic - Simple & Compound Interest**

A man borrowed a certain sum of money and paid back in two equal installments. If the rate of compound interest is 4% and if he paid back Rs 1014 annually. What sum did he borrow?

- A.
Rs. 1921.5

- B.
Rs. 1912.5

- C.
Rs. 1192.5

- D.
None of these

Answer: Option B

**Explanation** :

Let P be the sum.

Total amount due at the end of 1^{st} year = $x\left(1+\frac{4}{100}\right)$

Installment for 1^{st} year = 1014

Amount due after payment of installment = $x\left(1+\frac{4}{100}\right)$ - 1014.

⇒ $x\left(\frac{26}{25}\right)$ - 1014

Similarly,

Amount due at the of 2^{nd} year after payment of installment = $\left[x\left(\frac{26}{25}\right)-1014\right]\left(\frac{26}{25}\right)$ - 1014 = 0

Solving the above equation, we get

x = Rs. 1912.5

Hence, option (b).

Workspace:

**PE 1 - SICI | Arithmetic - Simple & Compound Interest**

What equal installment will discharge a debt of Rs. 1007 due in 5 years, rate of interest being 3% p.a.?

- A.
RS. 200

- B.
Rs. 190

- C.
Rs. 210

- D.
None of these

Answer: Option B

**Explanation** :

A = 1007,

t = 5 years,

r = 3%

∴ A = $x\left(1+\frac{R\times (T-1)}{100}\right)$ + $x\left(1+\frac{R\times (T-2)}{100}\right)$ + ... + x

⇒ 1007 = $x\left(1+\frac{3\times 4}{100}\right)$ + $x\left(1+\frac{3\times 3}{100}\right)$ + $x\left(1+\frac{3\times 2}{100}\right)$ + $x\left(1+\frac{3\times 1}{100}\right)$ + x

⇒ 1007= $x\left(5+\frac{3\times 10}{100}\right)$

⇒ x = $\frac{10070}{53}$ = 190

Hence, option (b).

Workspace:

**PE 1 - SICI | Arithmetic - Simple & Compound Interest**

Krishna borrows Rs. 45K from a bank at 10% compound interest. He repays it in three annual installments that are in arithmetic progression. He ends up paying 54K totally. How much did he pay in year 1?

- A.
Rs. 16,500

- B.
Rs. 19,500

- C.
Rs. 21,000

- D.
Rs. 18,000

Answer: Option B

**Explanation** :

Let the repayments be Rs "a – d", Rs "a" and Rs. "a + d"

∴ (a – d) + a + (a + d) = 54000

⇒ 3a = 54000

⇒ a = 18000

P = 45,000

r = 10% pa

t = 3 years

∴ 45000${\left(1+\frac{10}{100}\right)}^{3}$ = (18000 - d)${\left(1+\frac{10}{100}\right)}^{2}$ + 18000$\left(1+\frac{10}{100}\right)$ + (18000 + d)

⇒ 45000 × 1.331 = (18000 – d) × 1.21 + 18000 × 1.1 + (18000 + d)

⇒ 45000 × 1.331 = 18000 × (1.21 + 1.1 + 1) + d × (-1.21 + 1)

⇒ 0.21d = 18000 × 3.31 – 45000 × 1.331

⇒ 0.21d = 9000(2 × 3.31 – 5 × 1.331)

⇒ 21d = 900000(6.62 – 6.655)

⇒ d = -1500

∴ The payments are Rs. 19500, Rs. 18000 and Rs. 16,500.

Hence, option (b).

Workspace:

**PE 1 - SICI | Arithmetic - Simple & Compound Interest**

A bicycle is sold by a shopkeeper for Rs. 19,200 cash or for Rs. 4,800 cash down payment together with five equal monthly instalments. If the rate of interest charged by the shopkeeper is 12% per annum find each instalment. [Enter the nearest possible integer as your answer in Rs.]

Answer: 2965

**Explanation** :

Using the formula for installments for SI:

P$\left(1+\frac{R\times T}{10}\right)$ = x$\left(1+\frac{R\times (T-1)}{10}\right)$ + x$\left(1+\frac{R\times (T-2)}{10}\right)$ + ⋯ + x

Where, P = amount lent

x = equal installment,

R = interest rate and

T = total duration of loan.

⇒14,400$\left(1+\frac{1\times 5}{10}\right)$ = x$\left(1+\frac{1\times 4}{10}\right)$ + x$\left(1+\frac{1\times 3}{10}\right)$ + x$\left(1+\frac{1\times 2}{10}\right)$ + $\left(1+\frac{1\times 1}{10}\right)$ + x

⇒ 14,400$\left(1+\frac{5}{10}\right)$ = x$\left(5+\frac{1\times (4+3+2+1)}{100}\right)$

⇒ 14,400$\left(\frac{21}{20}\right)$ = x$\left(5+\frac{1}{10}\right)$

⇒ x ≈ 2965

Hence, 2965.

Workspace:

**PE 1 - SICI | Arithmetic - Simple & Compound Interest**

George borrows Rs 2,100 at the rate of 10% per annum compounded annually and has to repay it in two equal instalments in 2 years. What is the value of each instalment?

- A.
Rs. 1100

- B.
Rs. 1080

- C.
Rs. 1070

- D.
None of these

Answer: Option D

**Explanation** :

We know,

P${\left(1+\frac{R}{100}\right)}^{2}$ = x${\left(1+\frac{R}{100}\right)}^{1}$ + x

⇒ 2100${\left(1+\frac{10}{100}\right)}^{2}$ = x${\left(1+\frac{R}{100}\right)}^{1}$ + x

⇒ 2100 × 1.21 = 1.1x + x

⇒ 2100 × 1.21 = 2.1x

⇒ x = 1210

Hence, option (d).

Workspace:

**PE 1 - SICI | Arithmetic - Simple & Compound Interest**

John purchased a new bike worth Rs 96,000 on instalment at 5% p.a. interest compounded annually. At the end of one year, he repaid a certain amount and he repaid the balance amount of Rs 63000 at the end of the second year. What amount did he repay at the end of the first year?

- A.
Rs. 40,000

- B.
Rs. 40,800

- C.
Rs. 41,400

- D.
Rs. 42,000

Answer: Option B

**Explanation** :

We know,

P${\left(1+\frac{R}{100}\right)}^{2}$ = (1^{st} installment)${\left(1+\frac{R}{100}\right)}^{1}$ + (2^{nd} installment)

⇒ 96,000${\left(1+\frac{5}{100}\right)}^{2}$= x${\left(1+\frac{5}{100}\right)}^{1}$ + 63,000

⇒ 96000 × 1.05 = x + 60000

⇒ x = 40,800

Hence, option (b).

Workspace:

**PE 1 - SICI | Arithmetic - Simple & Compound Interest**

Find the effective rate of interest if rate of interest is 40% p.a. and interest being compounded quarterly?

- A.
45%

- B.
46.41%

- C.
48%

- D.
47.5%

Answer: Option B

**Explanation** :

We know,

P${\left(1+\frac{R/n}{100}\right)}^{(n\times t)}$= P${\left(1+\frac{{E}_{r}}{100}\right)}^{t}$

Since interest is compounded quarterly, n = 4 and t = 1, R = 40% and Er = effective rate of interest.

⇒ P${\left(1+\frac{40/4}{100}\right)}^{4}$= P${\left(1+\frac{{E}_{r}}{100}\right)}^{1}$

⇒ (1.1)^{4} = (1 + E_{r}/100)

⇒ 1.4641 = 1 + E_{r}/100

⇒ E_{r} = 46.41%

Hence, option (b).

Workspace:

**PE 1 - SICI | Arithmetic - Simple & Compound Interest**

Find the CI realized in the year at a rate of 40% p.a., with compounding being done quarterly, on a sum of Rs 1000?

- A.
420

- B.
400

- C.
424.20

- D.
464.10

Answer: Option D

**Explanation** :

We know,

A = P${\left(1+\frac{R/n}{100}\right)}^{(n\times t)}$

⇒ A = 1000${\left(1+\frac{10}{100}\right)}^{4}$ = 1464.1

⇒ Compound Interest = 1464.1 – 1000 = 464.1

Hence, option (d).

Workspace:

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