# CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

Find the interest on Rs. 500 for 9 years @ 5% p.a

- (a)
Rs. 245

- (b)
Rs. 225

- (c)
Rs. 200

- (d)
Rs. 340

Answer: Option B

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**Explanation** :

S.I. = $\frac{P\times R\times T}{100}=\frac{500\times 5\times 9}{100}$ = Rs. 225

Hence, option (b).

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

What sum lent out at 12% p.a. will amounts to Rs. 620 in 2 years

- (a)
600

- (b)
4500

- (c)
550

- (d)
500

Answer: Option D

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**Explanation** :

Let initial principal be P. Amount due is 620.

⇒ S.I. = 620 - P

⇒ 620 – P = $\frac{P\times 12\times 2}{100}$

⇒ P = 500

Hence, option (d).

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

Find the percent p.a. at which Rs. 500 amounts to Rs. 800 in 12 years?

- (a)
5%

- (b)
5.5%

- (c)
4.5%

- (d)
6%

Answer: Option A

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**Explanation** :

S.I. = 800 – 500 =$\frac{500\times R\%\times 12}{100}$

⇒ R = 5%

Hence, option (a).

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

The simple interest on a sum of money at the end of 16 years is $\frac{4}{5}$th of the sum itself. Find the rate percent p.a.

- (a)
4.5%

- (b)
5%

- (c)
5.5%

- (d)
6%

Answer: Option B

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**Explanation** :

S.I. =$\frac{4}{5}p=\frac{P\times R\times 16}{100}$

⇒ R = 5%

Hence, option (b)

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

To repay a sum of money borrowed 10 months earlier a man agreed to pay Rs. 2119. Find the amount borrowed if the rate interest charged was 2.25% p.a.

- (a)
Rs. 2080

- (b)
Rs. 2000

- (c)
Rs. 1800

- (d)
Rs. 1900

Answer: Option A

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**Explanation** :

Rate of interest for a year is given as 2.25%

⇒ Rate of interest for only per month = $\frac{2.25}{12}\%$

⇒ S.I. = 2119 – P = $\frac{\left(P\times \frac{2.25}{12}\times 10\right)}{100}$ (Considering time in months and rate of interest as %/month)

⇒ P = 2080

Hence, option (a).

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

What sum of money will amount to Rs. 8800 in 12 years at the same rate of simple interest at which Rs. 1706 amounts to Rs. 3412 in 10 years?

- (a)
4250

- (b)
4100

- (c)
4000

- (d)
None of these

Answer: Option C

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**Explanation** :

S.I. = 3412 – 1706 = $\frac{1706\times R\times 10}{100}$

⇒ R = 10%

Now, 8800 – P = $\frac{P\times 10\times 12}{100}$

⇒ P = 4000.

Hence, option (c).

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

A man deposited at a bank and withdraws Rs. 26,880 after 10 years at the rate of 6% S.I. Find the principal amount.

- (a)
Rs. 15,800

- (b)
Rs. 16000

- (c)
Rs. 16,400

- (d)
Rs. 16,800

Answer: Option D

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**Explanation** :

Amount due = P$\left(1+\frac{RT}{100}\right)$

⇒ P = $\frac{Amountdue}{\left(1+\frac{RT}{100}\right)}$

⇒ P = $\frac{26880}{\left(1+{\displaystyle \frac{10\times 6}{100}}\right)}$ = 16,800

Hence, option (d).

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

certain sum of money becomes three times in 12 years at simple interest. The rate A of interest is

- (a)
15%

- (b)
$12\frac{1}{2}$%

- (c)
$14\frac{2}{3}$%

- (d)
$16\frac{2}{3}$%

Answer: Option D

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**Explanation** :

Let principal be P, hence, amount due = 3P

⇒ S.I. – 3P – P = 2P

2P = $\frac{P\times R\times 12}{100}=R=16\frac{2}{3}$%

Hence, option (d).

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

Due to a fall in the rate of interest from 10% to $8\frac{1}{2}\%$ p.a. a money lender’s yearly income diminishes by Rs. 90. His capital is

- (a)
Rs. 6000

- (b)
Rs. 7000

- (c)
Rs. 10000

- (d)
Rs. 9000

Answer: Option A

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**Explanation** :

Interest accumulated = $\frac{\left(PRT\right)}{100}$

$P\times \frac{10}{100}-P\times \frac{8.5}{100}=90$

⇒ P = 6000

Hence, option (a).

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

An equal amount of money was lent at 5% and at 8% p.a. S.I. The total interest received after 3 years was Rs. 7800. Find how much was lent at 8%.

- (a)
Rs. 17,000

- (b)
Rs. 16,000

- (c)
Rs. 20,000

- (d)
Rs. 14,000

Answer: Option C

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**Explanation** :

Let the equal amount lent at both interest rates be Rs. P each.

∴ $\frac{P\times 5\times 3}{100}+\frac{P\times 8\times 3}{100}=7800$

⇒ P = Rs. 20,000.

Hence, option (c).

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

A sum of money doubles in 20 years at simple interest. In how many years will it be triple?

- (a)
40 yrs

- (b)
35 yrs

- (c)
30 yrs

- (d)
45 yrs

Answer: Option A

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**Explanation** :

S.I. = 2P – P = $\frac{P\times R\times 20}{100}$

⇒ R = 5%

Now, 3P - P = $\frac{P\times 5\times T}{100}$

T = 40 years

**Alternately,**

Let amount P doubles in 20 years. Hence, interest P (= 2P - P) is accumulated in 20 years.

For the same amount to triple, an interest of 2P (= 3P - P) has to be accumulated.

Since, interest accumulated has to double, the time required will also be double.

∴ Time required = 40 years.

Hence, option (a).

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

A sum of money triples in 10 years at simple interest. In how many years will it quadruple?

Answer: 15

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**Explanation** :

Let amount P triples in 20 years. Hence, interest 2P (= 3P - P) is accumulated in 10 years.

For the same amount to quadruple, an interest of 3P (= 4P - P) has to be accumulated.

Since, interest accumulated has to become 1.5 times, the time required will also be 1.5 times.

∴ Time required = 10 × 1.5 = 15 years.

Hence, 15.

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

A sum of money becomes 10 times in 18 years. In how many years will it become 100 times?

Answer: 198

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**Explanation** :

Let amount P becomes 10 times in 18 years. Hence, interest 9P (= 10P - P) is accumulated in 18 years.

For the same amount to become 100 times, an interest of 99P (= 100P - P) has to be accumulated.

Since, interest accumulated has to becomes 11 times, the time required will also be 11 times.

∴ Time required = 18 × 11 = 198 years.

Hence, 198.

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

A person invested a certain sum of money at a certain rate of interest, and twice the amount at a rate which was 4percentage points more than the percentage at which the first amount was invested. If on the average, the interest rate works out to be 19% at what rate was the first amount invested?

- (a)
$16\frac{1}{3}$%

- (b)
16%

- (c)
15%

- (d)
$14\frac{3}{2}$%

Answer: Option A

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**Explanation** :

$P\times \frac{r}{100}+2P\times \frac{\left(r+4\right)}{100}=3P\times \frac{19}{100}$

⇒ r = $16\frac{1}{3}$%

Hence, option (a).

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

If Rs. 2800 amount to be Rs. 3696 in 4 years at S.I., find what sum will amount to Rs. 6336 in 11 years at half the rate.

- (a)
Rs. 4400

- (b)
Rs. 4000

- (c)
Rs. 4700

- (d)
Rs. 3900

Answer: Option A

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**Explanation** :

3696 – 2800 = $\frac{2800\times R\times 4}{100}$

⇒ R = 8%

Now, 6336 – P = $\frac{P\times {\displaystyle \frac{8}{2}}\times 11}{100}$

⇒ P = 4400

Hence, option (a).

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?

- (a)
3.6

- (b)
6

- (c)
18

- (d)
Can't determine

- (e)
None of these

Answer: Option B

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**Explanation** :

Let rate = R% and time = R years.

Then, S.I. = $\frac{1200\times R\times R}{100}$ = 432.

⇒ $12{R}^{2}$ = 432

⇒ ${R}^{2}$ = 36

⇒ R = 6

Hence, option (b).

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

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?

- (a)
3.46%

- (b)
4.5%

- (c)
5%

- (d)
6%

- (e)
None of these

Answer: Option A

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**Explanation** :

Let the original rate be R%. Then, new rate = (2R)%.

Here, original rate is for 1 year; the new rate is for only 4 months i.e. 1/3 year.

.∴ S.I. =$\frac{725\times R\times 1}{100}+\frac{362.5\times 2R\times 1}{100\times 3}=33.50$

⇒ (2175 + 725)R = 33.50 × 100 × 3

⇒ (2175 + 725)R = 10050

⇒ (2900)R = 10050

⇒ R = 10050/2900 = 3.46

∴ Original rate = 3.46%

Hence, option (a).

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?

- (a)
Rs. 6400

- (b)
Rs. 6500

- (c)
Rs. 7200

- (d)
Rs. 7500

- (e)
None of these

Answer: Option A

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**Explanation** :

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x).

Then, $\frac{\left(x\times 14\times 2\right)}{100}+\frac{\left(13900-x\right)\times 11\times 2}{100}=3508$

⇒ 28x - 22x = 350800 - (13900 x 22)

⇒ 6x = 45000

⇒ x = 7500.

So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.

Hence, option (a).

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest?

- (a)
3%

- (b)
4%

- (c)
5%

- (d)
6%

- (e)
None of these

Answer: Option D

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**Explanation** :

S.I. = Rs. (15500 - 12500) = Rs. 3000.

Rate =$\frac{3000}{12500\times 4}\times 100=6\%$

Hence, option (d).

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**CRE 1 - Simple Interest | Arithmetic - Simple & Compound Interest**

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:

- (a)
5%

- (b)
7%

- (c)
$7\frac{1}{8}\%$

- (d)
10%

Answer: Option D

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**Explanation** :

Let the rate be R% p.a.

Then, $\frac{\left(5000\times R\times 2\right)}{100}+\frac{3000\times R\times 4}{100}=2200$

⇒ 100R + 120R = 2200

⇒ R = 2200/220 = 10.

∴ Rate = 10%.

Hence, option (d).

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