Simple Interest and Compound Interest — Exercise 8(E)

The sum which amounts to ₹ 840 in 5 years at the rate of 8% per annum simple interest is

1. ₹ $\left(\dfrac{840 \times 5 \times 8}{100}\right)$

2. ₹ $\left(\dfrac{100 \times 840}{100 - 5 \times 8}\right)$

3. ₹ $\left(\dfrac{100 + 5 \times 8 \times 100}{840}\right)$

4. ₹ $\left(\dfrac{100 \times 840}{100 + 5 \times 8}\right)$

Answer

Let the sum be ₹ P.

Here,

A = ₹ 840, R = 8% p.a. and T = 5 years.

$\text{A} = \text{P} + \dfrac{\text{P} \times \text{R} \times \text{T}}{100}\\[1em] \Rightarrow 840 = \text{P} + \dfrac{\text{P} \times 8 \times 5}{100}\\[1em] \Rightarrow 840 = \dfrac{\text{P}(100 + 5 \times 8)}{100}\\[1em] \Rightarrow \text{P} = \dfrac{100 \times 840}{100 + 5 \times 8}$

Hence, Option 4 is the correct option.

If ₹ 64 amount to ₹ 83.20 in 2 years, what will ₹ 86 amount to in 4 years at the same rate per cent per annum?

1. ₹ 114.80

2. ₹ 124.70

3. ₹ 127.40

4. ₹ 137.60

Answer

Here,

P = ₹ 64, A = ₹ 83.20 and T = 2 years.

S.I. = A − P = ₹ (83.20 − 64) = ₹ 19.20

R = $\dfrac{100 \times \text{S.I.}}{\text{P} \times \text{T}}$% p.a. = $\left(\dfrac{100 \times 19.20}{64 \times 2}\right)$% p.a. = 15% p.a.

Now, P = ₹ 86, R = 15% p.a. and T = 4 years.

S.I. = $\dfrac{\text{P} \times \text{R} \times \text{T}}{100}$

$= ₹ \left(\dfrac{86 \times 15 \times 4}{100}\right)$ = ₹ 51.60

Amount = ₹ (86 + 51.60) = ₹ 137.60

Hence, Option 4 is the correct option.

The simple interest at $x$% for $x$ years will be ₹ $x$ on a sum of

1. ₹ $x$

2. ₹ $100x$

3. ₹ $\left(\dfrac{100}{x}\right)$

4. ₹ $\left(\dfrac{100}{x^2}\right)$

Answer

Let the sum be ₹ P.

Here,

S.I. = ₹ $x$, R = $x$% p.a. and T = $x$ years.

$\text{S.I.} = \dfrac{\text{P} \times \text{R} \times \text{T}}{100}\\[1em] \Rightarrow x = \dfrac{\text{P} \times x \times x}{100}\\[1em] \Rightarrow 100x = \text{P}x^2\\[1em] \Rightarrow \text{P} = \dfrac{100}{x}$

Hence, Option 3 is the correct option.

Due to a fall in the rate of interest from 13% p.a. to $12\dfrac{1}{2}$% p.a., a money-lender's yearly income diminishes by ₹ 104. His capital is

1. ₹ 20,800

2. ₹ 21,400

3. ₹ 22,300

4. ₹ 24,000

Answer

Let the capital be ₹ P.

Fall in the rate of interest = $\left(13 - 12\dfrac{1}{2}\right)$% p.a. = $\dfrac{1}{2}$% p.a.

$\Rightarrow \dfrac{\text{P} \times \dfrac{1}{2} \times 1}{100} = 104\\[1em] \Rightarrow \dfrac{\text{P}}{200} = 104\\[1em] \Rightarrow \text{P} = 104 \times 200\\[1em] \Rightarrow \text{P} = 20800$

Hence, Option 1 is the correct option.

A lends a sum of money for 10 years at 5% simple interest. B lends double that amount for 5 years at the same rate of interest. Which of the following statements is true in this regard?

1. A will get twice the amount of interest that B would get.

2. B will get twice the amount of interest that A would get.

3. A and B will get the same amount as interest.

4. B will get four times the amount of interest that A would get.

Answer

Let the sum lent by A be ₹ P. Then, the sum lent by B = ₹ 2P.

Interest received by A = ₹ $\left(\dfrac{\text{P} \times 5 \times 10}{100}\right) = ₹ \left(\dfrac{\text{P}}{2}\right)$

Interest received by B = ₹ $\left(\dfrac{2\text{P} \times 5 \times 5}{100}\right) = ₹ \left(\dfrac{\text{P}}{2}\right)$

So, A and B will get the same amount as interest.

Hence, Option 3 is the correct option.

B borrowed ₹ 720 from A at 8% simple interest for 3 years and lent the same sum to C at $10\dfrac{1}{2}$%, simple interest for 2 years. In the whole transaction B.

1. lost ₹ 21.60

2. gained ₹ 21.60

3. neither gained nor lost

4. none of these

Answer

Interest paid by B to A = ₹ $\left(\dfrac{720 \times 8 \times 3}{100}\right)$ = ₹ 172.80

Interest received by B from C = ₹ $\left(720 \times \dfrac{21}{2} \times 2 \times \dfrac{1}{100}\right)$ = ₹ 151.20

As ₹ 151.20 < ₹ 172.80, so B has a loss in the whole transaction.

Loss = ₹ (172.80 − 151.20) = ₹ 21.60

Hence, Option 1 is the correct option.

₹ 800 amounts to ₹ 920 in 3 years at simple interest. If the interest rate is increased by 3%, it would amount to

1. ₹ 992

2. ₹ 1,056

3. ₹ 1,112

4. ₹ 1,182

Answer

Here,

P = ₹ 800, A = ₹ 920 and T = 3 years.

S.I. = A − P = ₹ (920 − 800) = ₹ 120

R = $\dfrac{100 \times \text{S.I.}}{\text{P} \times \text{T}}$% p.a.

= $\left(\dfrac{100 \times 120}{800 \times 3}\right)$% p.a. = 5% p.a.

New rate = (5 + 3)% p.a. = 8% p.a.

New S.I. = ₹ $\left(\dfrac{800 \times 8 \times 3}{100}\right)$ = ₹ 192

New amount = ₹ (800 + 192) = ₹ 992

Hence, Option 1 is the correct option.

Rahul lent ₹ 6,000 to Manick for 2 years and ₹ 1,500 to Arif for 4 years and received altogether from both ₹ 900 as simple interest. The rate of interest is

1. 4% p.a.

2. 5% p.a.

3. 10% p.a.

4. 12% p.a.

Answer

Let the rate of interest be R% p.a.

Since the total simple interest received is ₹ 900,

$\Rightarrow \dfrac{6000 \times \text{R} \times 2}{100} + \dfrac{1500 \times \text{R} \times 4}{100} = 900\\[1em] \Rightarrow 120\text{R} + 60\text{R} = 900\\[1em] \Rightarrow 180\text{R} = 900\\[1em] \Rightarrow \text{R} = 5$

Hence, Option 2 is the correct option.

Consider the following statements:

If a sum of money is loaned at simple interest, then the

(i) Money gets doubled in 5 years if the rate of interest is $16\dfrac{2}{3}$%. p.a.

(ii) Money gets doubled in 5 years if the rate of interest is 20%. p.a.

(iii) Money becomes four times in 10 years if it gets doubled in 5 years.

Of these statements,

1. (i) and (iii) are correct

2. (ii) alone is correct

3. (ii) and (iii) are correct

4. (iii) alone is correct

Answer

Let the sum be ₹ P. If the money gets doubled in 5 years, then S.I. = ₹ P and T = 5 years.

R = $\dfrac{100 \times \text{S.I.}}{\text{P} \times \text{T}}$% p.a.

= $\left(\dfrac{100 \times \text{P}}{\text{P} \times 5}\right)$% p.a.

= 20% p.a.

So, Statement (i) is false and Statement (ii) is true.

If the money gets doubled in 5 years, then S.I. for 5 years = ₹ P.

∴ S.I. for 10 years = ₹ 2P

Amount after 10 years = ₹ (P + 2P) = ₹ 3P

So, the money becomes three times and not four times in 10 years. Thus, Statement (iii) is false.

Therefore, (ii) alone is correct.

Hence, Option 2 is the correct option.

In what time will a sum of money double itself at $6\dfrac{1}{4}$% p.a. simple interest?

1. 5 years

2. 8 years

3. 12 years

4. 16 years

Answer

Let the sum be ₹ P. Then, A = ₹ 2P and S.I. = ₹ (2P − P) = ₹ P.

Also, R = $6\dfrac{1}{4}$% p.a. = $\dfrac{25}{4}$% p.a.

$\text{T} = \dfrac{100 \times \text{S.I.}}{\text{P} \times \text{R}}\\[1em] = \dfrac{100 \times \text{P}}{\text{P} \times \dfrac{25}{4}}\\[1em] = 100 \times \text{P} \times \dfrac{1}{\text{P}} \times \dfrac{4}{25}\\[1em] = \dfrac{400}{25}\\[1em] = 16$

Hence, Option 4 is the correct option.

A sum of money becomes $\dfrac{8}{5}$ of itself in 5 years at a certain rate of interest. The rate per cent per annum is

1. 5%

2. 8%

3. 10%

4. 12%

Answer

Let the sum be ₹ P. Then, A = ₹ $\left(\dfrac{8}{5}\text{P}\right)$ and T = 5 years.

S.I. = A − P = ₹ $\left(\dfrac{8\text{P}}{5} - \text{P}\right) = ₹ \left(\dfrac{3\text{P}}{5}\right)$

$\text{R} = \dfrac{100 \times \text{S.I.}}{\text{P} \times \text{T}}$%

$= \left(100 \times \dfrac{3\text{P}}{5} \times \dfrac{1}{\text{P} \times 5}\right)$%

$= \dfrac{60}{5}$%

= 12%

Hence, Option 4 is the correct option.

In what time will the simple interest on ₹ 780 at 5% be equal to the simple interest on ₹ 600 at $6\dfrac{1}{2}$%?

1. 2 years

2. $3\dfrac{1}{2}$ years

3. 5 years

4. always

Answer

Let the required time be T years.

S.I. on ₹ 780 at 5% p.a. = ₹ $\left(\dfrac{780 \times 5 \times \text{T}}{100}\right)$ = ₹ 39T

S.I. on ₹ 600 at $\dfrac{13}{2}$% p.a. = ₹ $\left(600 \times \dfrac{13}{2} \times \text{T} \times \dfrac{1}{100}\right)$ = ₹ 39T

As the two interests are equal to 39T for every value of T, so the two simple interests are always equal.

Hence, Option 4 is the correct option.

Two equal sums of money are deposited in two banks, each at 15% per annum, for $3\dfrac{1}{2}$ years and 5 years. If the difference between their interests is ₹ 144, each sum is

1. ₹ 460

2. ₹ 500

3. ₹ 640

4. ₹ 720

Answer

Let each sum be ₹ P.

S.I. for 5 years = ₹ $\left(\dfrac{\text{P} \times 15 \times 5}{100}\right) = ₹ \left(\dfrac{75\text{P}}{100}\right)$

S.I. for $3\dfrac{1}{2}$ years = ₹ $\left(\text{P} \times 15 \times \dfrac{7}{2} \times \dfrac{1}{100}\right) = ₹ \left(\dfrac{105\text{P}}{200}\right)$

Since the difference between the two interests is ₹ 144,

$\Rightarrow \dfrac{75\text{P}}{100} - \dfrac{105\text{P}}{200} = 144\\[1em] \Rightarrow \dfrac{150\text{P} - 105\text{P}}{200} = 144\\[1em] \Rightarrow \dfrac{45\text{P}}{200} = 144\\[1em] \Rightarrow \text{P} = \dfrac{144 \times 200}{45}\\[1em] \Rightarrow \text{P} = 640$

Hence, Option 3 is the correct option.

A sum of ₹ 2,500 is lent out in two parts, one at 12% and another one at $12\dfrac{1}{2}$%. If the total annual income is ₹ 306, the money lent at 12% is

1. ₹ 1,200

2. ₹ 1,240

3. ₹ 1,300

4. ₹ 1,340

Answer

Let the money lent at 12% be ₹ $x$. Then, the money lent at $12\dfrac{1}{2}$% = ₹ $(2500 - x)$.

Since the total annual interest is ₹ 306,

$\Rightarrow \dfrac{x \times 12 \times 1}{100} + (2500 - x) \times \dfrac{25}{2} \times 1 \times \dfrac{1}{100} = 306\\[1em] \Rightarrow \dfrac{12x}{100} + \dfrac{25(2500 - x)}{200} = 306\\[1em] \Rightarrow 24x + 25(2500 - x) = 306 \times 200\\[1em] \Rightarrow 24x + 62500 - 25x = 61200\\[1em] \Rightarrow -x = -1300\\[1em] \Rightarrow x = 1300$

Hence, Option 3 is the correct option.

Out of a sum of ₹ 625, a part was lent at 5% and the other at 10% simple interest. If the interest on the first part after 2 years is equal to the interest on the second part after 4 years, then the second sum (in ₹) is

1. 125

2. 200

3. 250

4. 300

Answer

Let the first part be ₹ $x$. Then, the second part = ₹ $(625 - x)$.

Since the interest on the first part after 2 years equals the interest on the second part after 4 years,

$\Rightarrow \dfrac{x \times 5 \times 2}{100} = \dfrac{(625 - x) \times 10 \times 4}{100}\\[1em] \Rightarrow 10x = 40(625 - x)\\[1em] \Rightarrow 10x = 25000 - 40x\\[1em] \Rightarrow 50x = 25000\\[1em] \Rightarrow x = 500$

Second part = ₹ (625 − 500) = ₹ 125

Hence, Option 1 is the correct option.

The compound interest on ₹ 540 at $16\dfrac{2}{3}$% per annum for 2 years is

1. ₹ 180

2. ₹ 192.50

3. ₹ 195

4. ₹ 735

Answer

Here,

P = ₹ 540, R = $16\dfrac{2}{3}$% p.a. = $\dfrac{50}{3}$% p.a. and n = 2 years.

$\text{Amount after 2 years} = 540 \times \left(1 + \dfrac{50}{3 \times 100}\right)^2\\[1em] = 540 \times \dfrac{7}{6} \times \dfrac{7}{6}\\[1em] = 735$

∴ C.I. = Amount − Principal = ₹ (735 − 540) = ₹ 195

Hence, Option 3 is the correct option.

The difference between the simple interest and the compound interest on ₹ 600 for 1 year at 10% per annum, reckoned half-yearly is

1. Nil

2. ₹ 1.50

3. ₹ 4.40

4. ₹ 6.50

Answer

Here,

P = ₹ 600, R = 10% p.a. and T = 1 year.

S.I. = $\dfrac{\text{P} \times \text{R} \times \text{T}}{100}$

$= ₹ \left(\dfrac{600 \times 10 \times 1}{100}\right)$ = ₹ 60

For compound interest reckoned half-yearly, rate = 5% per half year and $n$ = 2 half years.

$\text{Amount after 1 year} = 600 \times \left(1 + \dfrac{5}{100}\right)^2\\[1em] = 600 \times \dfrac{21}{20} \times \dfrac{21}{20}\\[1em] = 661.50$

C.I. = ₹ (661.50 − 600) = ₹ 61.50

∴ Difference = C.I. − S.I. = ₹ (61.50 − 60) = ₹ 1.50

Hence, Option 2 is the correct option.

Simple interest on a sum at $12\dfrac{1}{2}$% per annum for 2 years is ₹ 256. The compound interest on the same sum at the same rate and for the same period is

1. ₹ 262.40

2. ₹ 264

3. ₹ 265.80

4. ₹ 272

Answer

Let the sum be ₹ P.

Here,

S.I. = ₹ 256, R = $\dfrac{25}{2}$% p.a. and T = 2 years.

$\text{P} = \dfrac{100 \times \text{S.I.}}{\text{R} \times \text{T}}\\[1em] = \dfrac{100 \times 256}{\dfrac{25}{2} \times 2}\\[1em] = 100 \times 256 \times \dfrac{2}{25} \times \dfrac{1}{2}\\[1em] = 1024$

Now, for compound interest,

$\text{Amount after 2 years} = 1024 \times \left(1 + \dfrac{25}{2 \times 100}\right)^2\\[1em] = 1024 \times \dfrac{9}{8} \times \dfrac{9}{8}\\[1em] = 1296$

∴ C.I. = ₹ (1,296 − 1,024) = ₹ 272

Hence, Option 4 is the correct option.

Fill in the blanks :

(i) If S.I. and C.I. denote simple interest and compound interest respectively on the same sum at the same rate of interest for the same period, then the relation between S.I. and C.I. is _________

(ii) The basic difference between simple interest and compound interest is that in simple interest, the _________ remains same throughout the loan period while in compound interest it varies.

(iii) If the compound interest (compounded annually) on a sum of money at a certain rate of interest for $n$ years is equal to the simple interest on the same sum at the same rate for the same period, then $n$ = _________.

(iv) _________ = Principal + Interest

Answer

(i) If S.I. and C.I. denote simple interest and compound interest respectively on the same sum at the same rate of interest for the same period, then the relation between S.I. and C.I. is C.I. ≥ S.I.

(ii) The basic difference between simple interest and compound interest is that in simple interest, the principal remains same throughout the loan period while in compound interest it varies.

(iii) If the compound interest (compounded annually) on a sum of money at a certain rate of interest for $n$ years is equal to the simple interest on the same sum at the same rate for the same period, then $n$ = 1.

(iv) Amount = Principal + Interest

Assertion (A) : Simple interest on a given amount is always less than or equal to the compound interest on the same amount for the same time period and at the same rate of interest per annum.

Reason (R) : S.I. = $\dfrac{PRT}{100}$

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true but Reason (R) is false.

4. Assertion (A) is false but Reason (R) is true.

Answer

In simple interest, the interest is calculated on the original principal throughout the loan period, whereas in compound interest the interest of each period is added to the principal for the next period.

For a period of 1 year, both are equal and for a period of more than 1 year, the compound interest is greater than the simple interest.

So, S.I. ≤ C.I. and hence Assertion (A) is true.

The formula S.I. = $\dfrac{\text{P} \times \text{R} \times \text{T}}{100}$ is the correct formula for simple interest, so Reason (R) is true.

However, the formula S.I. = $\dfrac{\text{P} \times \text{R} \times \text{T}}{100}$ alone does not explain why the compound interest is greater than or equal to the simple interest. So, Reason (R) is not the correct explanation of Assertion (A).

Hence, Option 2 is the correct option.

Assertion (A) : The compound interest on a sum of ₹ P for $n$ years at $r$% per annum is given by the formula $P\left(1 + \dfrac{rn}{100}\right)$.

Reason (R) : Compound Interest (CI) = Final amount – Original amount

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true but Reason (R) is false.

4. Assertion (A) is false but Reason (R) is true.

Answer

The amount on a sum of ₹ P for $n$ years at $r$% per annum, compounded annually, is given by

$\text{A} = \text{P}\left(1 + \dfrac{r}{100}\right)^n$

and the compound interest is given by

$\text{C.I.} = \text{P}\bigg\lbrace \left(1 + \dfrac{r}{100}\right)^n - 1 \bigg\rbrace$

The given formula $\text{P}\left(1 + \dfrac{rn}{100}\right)$ is the amount under simple interest and not the compound interest. So, Assertion (A) is false.

Compound Interest = Final amount − Original amount is the correct relation, so Reason (R) is true.

Hence, Option 4 is the correct option.

Simple interest on a certain amount is $\dfrac{9}{16}$ of the principal. If the numbers representing the rate of interest (in per cent) and time (in years) be equal, then the time for which the principal is lent out, is:

1. 5 years

2. 6 years

3. $6\dfrac{1}{2}$ years

4. $7\dfrac{1}{2}$ years

Answer

Let the principal be ₹ P. Then, S.I. = ₹ $\left(\dfrac{9}{16}\text{P}\right)$.

Let the rate be R% per annum and the time be R years.

$\Rightarrow \text{S.I.} = \dfrac{\text{P} \times \text{R} \times \text{T}}{100}\\[1em] \Rightarrow \dfrac{9\text{P}}{16} = \dfrac{\text{P} \times \text{R} \times \text{R}}{100}\\[1em] \Rightarrow \text{R}^2 = \dfrac{9}{16} \times 100\\[1em] \Rightarrow \text{R}^2 = \dfrac{900}{16}\\[1em] \Rightarrow \text{R} = \sqrt{\dfrac{900}{16}}\\[1em] = \dfrac{\sqrt{900}}{\sqrt{16}}\\[1em] = \dfrac{30}{4}\\[1em] = \dfrac{15}{2}$

So, time = R years = $\dfrac{15}{2}$ years = $7\dfrac{1}{2}$ years.

Hence, Option 4 is the correct option.

The compound interest earned by Deepak on a certain amount at the end of 2 years at the rate of 5% p.a, was ₹ 1,025. The amount received by Deepak at the end of 2 years (principal + interest) is:

1. ₹ 11,025

2. ₹ 12,025

3. ₹ 12,725

4. ₹ 14,225

Answer

Let the principal be ₹ P.

Here,

R = 5% p.a. and n = 2 years.

$\text{C.I.} = \text{P}\bigg\lbrace \left(1 + \dfrac{5}{100}\right)^2 - 1 \bigg\rbrace\\[1em] \Rightarrow 1025 = \text{P}\left(\dfrac{21}{20} \times \dfrac{21}{20} - 1\right)\\[1em] \Rightarrow 1025 = \text{P}\left(\dfrac{441 - 400}{400}\right)\\[1em] \Rightarrow 1025 = \dfrac{41\text{P}}{400}\\[1em] \Rightarrow \text{P} = \dfrac{1025 \times 400}{41}\\[1em] \Rightarrow \text{P} = 10000$

Amount = Principal + C.I. = ₹ (10,000 + 1,025) = ₹ 11,025

Hence, Option 1 is the correct option.

Which of the following is a better deal for you?

1. You borrowed ₹ 1,00,000 at 20% p.a. simple interest for 2 years

2. You borrowed ₹ 1,00,000 at 15% p.a, compounded annually for 2 years

3. In both (1) and (2), you will have to pay the same amount after 2 years

4. Data insufficient

Answer

Option 1 : P = ₹ 1,00,000, R = 20% p.a. and T = 2 years.

S.I. = $\dfrac{\text{P} \times \text{R} \times \text{T}}{100}$

$= ₹ \left(\dfrac{100000 \times 20 \times 2}{100}\right)$ = ₹ 40,000

Option 2 : P = ₹ 1,00,000, R = 15% p.a. and n = 2 years.

$\text{Amount after 2 years} = 100000 \times \left(1 + \dfrac{15}{100}\right)^2\\[1em] = 100000 \times \dfrac{23}{20} \times \dfrac{23}{20}\\[1em] = 132250$

C.I. = ₹ (1,32,250 − 1,00,000) = ₹ 32,250

As ₹ 32,250 < ₹ 40,000, so borrowing at 15% p.a. compounded annually costs less.

Hence, Option 2 is the correct option.

What would be the amount after 4 years when interest is compounded half-yearly, interest rate = 10% per 2 years, principal = ₹ 5,000?

1. ₹ 5,000 $\left(1 + \dfrac{10}{100}\right)^4$

2. ₹ 5,000 $\left(1 + \dfrac{5}{100}\right)^8$

3. ₹ 5,000 $\left(1 + \dfrac{2.5}{100}\right)^8$

4. ₹ 5,000 $\left(1 + \dfrac{2.5}{100}\right)^4$

Answer

Interest rate = 10% per 2 years = $\dfrac{10}{2}$% per annum = 5% per annum

As the interest is compounded half-yearly,

Rate = $\dfrac{5}{2}$% per half year = 2.5% per half year

Time = 4 years = (4 × 2) half years = 8 half years

$\text{Amount after 4 years} = 5000\left(1 + \dfrac{2.5}{100}\right)^8$

Hence, Option 3 is the correct option.

Amit wants to invest ₹ 1,00,000 in a bank. He has two options as below:

Scheme 1: Rate of interest = 3% p.a and time period = 2 years.

Scheme 2: Rate of interest = 2% p.a and time period = 3 years.

Which scheme is better for Amit?

1. Scheme 1

2. Scheme 2

3. Both the schemes will give the same amount

4. None of these

Answer

Assuming that the schemes use simple interest,

Scheme 1 : P = ₹ 1,00,000, R = 3% p.a. and T = 2 years.

S.I. = $\dfrac{\text{P} \times \text{R} \times \text{T}}{100}$

$= ₹ \left(\dfrac{100000 \times 3 \times 2}{100}\right)$ = ₹ 6,000

Scheme 2 : P = ₹ 1,00,000, R = 2% p.a. and T = 3 years.

S.I. = $\dfrac{\text{P} \times \text{R} \times \text{T}}{100}$

$= ₹ \left(\dfrac{100000 \times 2 \times 3}{100}\right)$ = ₹ 6,000

As the interest earned is the same in both the schemes, both will give the same amount.

Hence, Option 3 is the correct option.

Reena invested ₹ 5,000 for 6 months at 10% annual rate of interest, compounded annually. Vineeta invested the same amount of money for 6 months at the same annual rate of interest but compounded half-yearly. Who will get more money at the time of maturity?

1. Reena

2. Vineeta

3. Both will get the same amount

4. Data inadequate

Answer

For Reena : As the interest is compounded annually and the period is only 6 months, no compounding takes place within the period.

Interest = ₹ $\left(5000 \times 10 \times \dfrac{1}{2} \times \dfrac{1}{100}\right)$ = ₹ 250

Amount = ₹ (5,000 + 250) = ₹ 5,250

For Vineeta : Rate = 5% per half year and $n$ = 6 months = 1 half year.

Amount = ₹ $\bigg\lbrace 5000 \times \left(1 + \dfrac{5}{100}\right) \bigg\rbrace = ₹ \left(5000 \times \dfrac{21}{20}\right)$ = ₹ 5,250

As both the amounts are equal, both will get the same amount at the time of maturity.

Hence, Option 3 is the correct option.

At a certain rate of simple interest, a certain sum doubles itself in 10 years. It will treble itself in:

1. 12 years

2. 15 years

3. 20 years

4. 24 years

Answer

Let the sum be ₹ P.

When the sum doubles itself, A = ₹ 2P and S.I. = ₹ (2P − P) = ₹ P in 10 years.

So, S.I. for 10 years = ₹ P.

When the sum trebles itself, A = ₹ 3P and S.I. = ₹ (3P − P) = ₹ 2P.

$\text{Time taken} = \dfrac{2\text{P}}{\text{P}} \times 10$

= 2 x 10

= 20

Hence, Option 3 is the correct option.

If the compound interest on a certain sum of money for 3 years at 10% p.a be ₹ 993, then what would be the simple interest?

1. ₹ 800

2. ₹ 900

3. ₹ 950

4. ₹ 960

Answer

Let the sum be ₹ P.

Here,

R = 10% p.a. and $n$ = 3 years.

$\text{C.I.} = \text{P}\bigg\lbrace \left(1 + \dfrac{10}{100}\right)^3 - 1 \bigg\rbrace\\[1em] \Rightarrow 993 = \text{P}\left(\dfrac{11}{10} \times \dfrac{11}{10} \times \dfrac{11}{10} - 1\right)\\[1em] \Rightarrow 993 = \text{P}\left(\dfrac{1331 - 1000}{1000}\right)\\[1em] \Rightarrow 993 = \dfrac{331\text{P}}{1000}\\[1em] \Rightarrow \text{P} = \dfrac{993 \times 1000}{331}\\[1em] \Rightarrow \text{P} = 3000$

Now, S.I. = $\dfrac{\text{P} \times \text{R} \times \text{T}}{100}$

$= ₹ \left(\dfrac{3000 \times 10 \times 3}{100}\right)$ = ₹ 900

Hence, Option 2 is the correct option.

The differences between the compound interest and the simple interest on a certain sum at 5% p.a for 2 years is ₹ 150. The sum is:

1. ₹ 30,000

2. ₹ 40,000

3. ₹ 50,000

4. ₹ 60,000

Answer

Let the sum be ₹ P.

Here,

R = 5% p.a. and T = 2 years.

S.I. = $\dfrac{\text{P} \times 5 \times 2}{100} = ₹ \left(\dfrac{\text{P}}{10}\right)$

$\text{C.I.} = \text{P}\bigg\lbrace \left(1 + \dfrac{5}{100}\right)^2 - 1 \bigg\rbrace\\[1em] = \text{P}\left(\dfrac{21}{20} \times \dfrac{21}{20} - 1\right)\\[1em] = \text{P}\left(\dfrac{441 - 400}{400}\right)\\[1em] = \dfrac{41\text{P}}{400}$

Given,

The difference between compound interest and simple interest = ₹ 150.

$\Rightarrow \dfrac{41\text{P}}{400} - \dfrac{\text{P}}{10} = 150\\[1em] \Rightarrow \dfrac{41\text{P} - 40\text{P}}{400} = 150\\[1em] \Rightarrow \dfrac{\text{P}}{400} = 150\\[1em] \Rightarrow \text{P} = 60000$

Hence, Option 4 is the correct option.

Price of 1 kg of an Indian spice is ₹ 12,000. If its price increases at the rate of 10% every 6 months, what would be the price of 1 kg spice after 2 years?

1. ₹ 12,000 $\left(1 + \dfrac{10}{100}\right)^2$

2. ₹ 12,000 $\left(1 + \dfrac{10}{100}\right)^4$

3. ₹ 12,000 $\left(1 + \dfrac{5}{100}\right)^2$

4. ₹ 12,000 $\left(1 + \dfrac{5}{100}\right)^4$

Answer

Here, the price increases at the rate of 10% every 6 months, i.e. the rate is 10% per half year.

Time = 2 years = (2 × 2) half years = 4 half years

Price of 1 kg spice after 2 years = $12000\left(1 + \dfrac{10}{100}\right)^4$

Hence, Option 2 is the correct option.

$x$ is the simple interest on $y$ and $y$ is the simple interest on $z$, both for same period and the same rate. $x, y$ and $z$ are connected as:

1. $x^2 = yz$

2. $xy = z$

3. $y^2 = xz$

4. $z^2 = xy$

Answer

Let the rate be R% per annum and the time be T years.

As $x$ is the simple interest on $y$,

$x = \dfrac{y \times \text{R} \times \text{T}}{100} \qquad \text{.....(i)}$

As $y$ is the simple interest on $z$,

$y = \dfrac{z \times \text{R} \times \text{T}}{100} \qquad \text{.....(ii)}$

Dividing equation (i) by equation (ii),

$\dfrac{x}{y} = \dfrac{y}{z}\\[1em] \Rightarrow y^2 = xz$

Hence, Option 3 is the correct option.