Assertion (A): The amount on ₹ 4000 at 10% p.a compounded annually for $2\dfrac{1}{2}$ years = $4000\Big(1+\dfrac{10}{100}\Big)^\dfrac{5}{2}$.

Reason (R): When the total time is not complete number of years, for example time = m years and n months, and rate is r % p.a (compounded annually), then

A = sum $\Big(1+\dfrac{r}{100}\Big)^m \times \Big(1+\dfrac{\dfrac{r}{12}}{100}\Big)^n$

1. A is true, R is false.
2. A is false, R is true.
3. Both A and R are true.
4. Both A and R are false.

Assertion (A): A certain sum of money is borrowed at 10% compound interest for two years, it amounts to ₹ 5,000. The sum borrowed = ₹ 5,000 $\Big(1 -\dfrac{10}{100}\Big)\Big(1 -\dfrac{10}{100}\Big)$.

Reason (R): Since, amount = sum borrowed x $\Big(1+\dfrac{10}{100}\Big)\Big(1+\dfrac{10}{100}\Big)$

∴ Sum borrowed = Amount x $\Big(1 -\dfrac{10}{100}\Big)\Big(1 -\dfrac{10}{100}\Big)$.

1. A is true, R is false.
2. A is false, R is true.
3. Both A and R are true.
4. Both A and R are false.

Assertion (A): If a² - 8a - 1 = 0.

⇒ $\dfrac{a^2}{a} - \dfrac{8a}{a} - \dfrac{1}{a} = 0$

⇒ $a-\dfrac{1}{a}$ = 8 is not true.

Reason (R): The division of each term of the equation a² - 8a - 1 = 0 is defined only when a is not equal to 0 (zero).

1. A is true, R is false.
2. A is false, R is true.
3. Both A and R are true.
4. Both A and R are false.

Assertion (A): If x ≥ 5 and $x +\dfrac{1}{x}$ = m, then $x - \dfrac{1}{x}$ = m - 2.

Reason (R): $(x -\dfrac{1}{x})^2 = (x +\dfrac{1}{x})^2 - 4$

⇒ $x - \dfrac{1}{x} = \sqrt{(x+\dfrac{1}{x})^2-4}$

= $\sqrt{m^2-4}= m - 2$

1. A is true, R is false.
2. A is false, R is true.
3. Both A and R are true.
4. Both A and R are false.