Mathematics — Sample Question Bank

Given,

(x + 3), 1, (3x − 7) and −5 are in proportion.

$$\begin{aligned}\therefore \frac{(x + 3)}{1} = \frac{(3x - 7)}{-5} \\[1em] \Rightarrow (x + 3)= \frac{(3x - 7)}{-5} \\[1em] \Rightarrow -5(x + 3) = (3x - 7) \\[1em] \Rightarrow -5x - 15 = 3x - 7 \\[1em] \Rightarrow -5x - 3x = 15 - 7 \\[1em] \Rightarrow -8x = 8 \\[1em] \Rightarrow x = -\frac{8}{8} \\[1em] \Rightarrow x = -1.\end{aligned}$$

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{1}{\cos θ} - \cos θ\Big) \Big(\frac{1}{\sin θ} - \sin θ\Big) \\[1em] \Rightarrow \Big(\frac{1 - \cos^2 θ}{\cos θ}\Big) \Big(\frac{1 - \sin^2 θ}{\sin θ}\Big) \\[1em] \Rightarrow \Big(\frac{\sin^2 θ}{\cos θ}\Big) \Big(\frac{\cos^2 θ}{\sin θ}\Big) \\[1em] \Rightarrow \sin θ \cos θ.\end{aligned}$$

Hence, proved (sec θ − cos θ)(cosec θ − sin θ) = sin θ cos θ.

Comparing the given quadratic equation with ax² + bx + c = 0, we get :

a = 3, b = $\sqrt{7}$ and c = 2

Discriminant = b² - 4ac

= $(\sqrt{7})^2$ - 4 x 3 x 2

= 7 - 24

= -17.

Since, D < 0, there are no real roots.

Hence, option 4 is the correct option.

Solving L.H.S of the equation $2x - \frac{5}{3} \lt \frac{3x}{5} + 10 \le \frac{4x}{5} + 11 ;\ x \in R$ we get,

$$\begin{aligned}\Rightarrow 2x - \frac{5}{3} \lt \frac{3x}{5} + 10 \\[1em] \Rightarrow \frac{6x - 5}{3} \lt \frac{3x + 50}{5} \\[1em] \text{Multiplying both side by 15, we get:} \\[1em] \Rightarrow 15\Big(\frac{6x - 5}{3}\Big) \lt 15\Big(\frac{3x + 50}{5}\Big) \\[1em] \Rightarrow 5(6x - 5) \lt 3(3x + 50) \\[1em] \Rightarrow 30x - 25 \lt 9x + 150 \\[1em] \Rightarrow 30x - 9x \lt 25 + 150 \\[1em] \Rightarrow 21x \lt 175 \\[1em] \Rightarrow x \lt \frac{175}{21} \\[1em] \Rightarrow x \lt \frac{25}{3} \text{ ........(1)}\end{aligned}$$

Solving R.H.S of the equation $2x - \frac{5}{3} \lt \frac{3x}{5} + 10 \le \frac{4x}{5} + 11 ;\ x \in R$ we get,

$$\begin{aligned}\Rightarrow \frac{3x}{5} + 10 \le \frac{4x}{5} + 11 \\[1em] \Rightarrow \frac{3x + 50}{5} \le \frac{4x + 55}{5} \\[1em] \text{Multiplying both side by 5, we get:} \\[1em] \Rightarrow 5\Big(\frac{3x + 50}{5}\Big) \le 5\Big(\frac{4x + 55}{5}\Big) \\[1em] \Rightarrow 3x + 50 \le 4x + 55 \\[1em] \Rightarrow 50 - 55 \le 4x - 3x \\[1em] \Rightarrow -5 \le x \\[1em] \Rightarrow x \ge -5 \text{ ........(2)}\end{aligned}$$

From equation (1) and (2), we get :

-5 ≤ x < $\frac{25}{3}$.

Hence, solution set equals to -5 ≤ x < $\frac{25}{3}$, x ∈ R.

CGST% = $\frac{CGST}{M.P.} \times 100$

Substituting values we get :

CGST% = $\frac{120}{2000} \times 100$ = 6%

GST% = 2 × CGST% = 2 × 6% = 12%.

Hence, Option 3 is the correct option.

Let time be n months.

By formula,

I = $P \times \frac{n(n + 1)}{2\times 12} \times \frac{r}{100}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 1040 = 2000 \times \frac{n(n + 1)}{24} \times \frac{8}{100} \\[1em] \Rightarrow 1040 = 2000 \times \frac{n(n + 1)}{300} \\[1em] \Rightarrow n(n + 1) = \frac{1040 \times 300}{2000} \\[1em] \Rightarrow n(n + 1) = 156 \\[1em] \Rightarrow n^2 + n - 156 = 0 \\[1em] \Rightarrow n^2 + 13n - 12n - 156 = 0 \\[1em] \Rightarrow n(n + 13) - 12(n + 13) = 0 \\[1em] \Rightarrow (n - 12)(n + 13) = 0 \\[1em] \Rightarrow n - 12 = 0 \text{ or } n + 13 = 0 \\[1em] \Rightarrow n = 12 \text{ or } n = -13.\end{aligned}$$

Since, no. of months cannot be negative.

∴ n = 12.

Hence, total time for which the account was held = 12 months.

Solving,

$$\Rightarrow \begin{bmatrix*}[r] 2 & 0 \\ 0 & 4 \end{bmatrix*}\begin{bmatrix*}[r] x \\ y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2.x + 0.y \\ 0.x + 4.y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x + 0 \\ 0 + 4y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 2x \\ 4y \end{bmatrix*} = \begin{bmatrix*}[r] 2 \\ -8 \end{bmatrix*} \\[1em] \Rightarrow 2x = 2 \text{ and } 4y = -8 \\[1em] \Rightarrow x = \frac{2}{2} \text{ and } y = -\frac{8}{4} \\[1em] \Rightarrow x = 1 \text{ and } y = -2.$$

Hence, Option 1 is the correct option.

B + C = $$\begin{bmatrix*}[r] 1 & 2 \\ 2 & 4 \end{bmatrix*} + \begin{bmatrix*}[r] 4 & 1 \\ 1 & 5 \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 3 \\ 3 & 9 \end{bmatrix*}$$

Substituting values we get :

$$A(B + C) - 14I = \begin{bmatrix*}[r] 1 & 3 \\ 2 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 5 & 3 \\ 3 & 9 \end{bmatrix*} - 14\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 1 \times 5 + 3 \times 3 & 1 \times 3 + 3 \times 9 \\ 2 \times 5 + 4 \times 3 & 2 \times 3 + 4 \times 9 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 5 + 9 & 3 + 27 \\ 10 + 12 & 6 + 36 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 & 30 \\ 22 & 42 \end{bmatrix*} - \begin{bmatrix*}[r] 14 & 0 \\ 0 & 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 14 - 14 & 30 - 0 \\ 22 - 0 & 42 - 14 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 30 \\ 22 & 28 \end{bmatrix*}.$$

Hence, A(B + C) - 14I = $$\begin{bmatrix*}[r] 0 & 30 \\ 22 & 28 \end{bmatrix*}.$$

Given,

Total GST paid = ₹ 600

Rate of GST = 12%

Let the price of the article be ₹ $x$.

By formula,

GST = Rate of GST × Price of the article

$$\begin{aligned}\Rightarrow 600 = \frac{12}{100} \times x \\[1em] \Rightarrow x = \frac{600 \times 100}{12} \\[1em] \Rightarrow x = \frac{60000}{12} \\[1em] \Rightarrow x = 5000.\end{aligned}$$

Thus, the price of the article = ₹ 5,000.

Hence, option 3 is the correct option.

Given,

Deposit per month (P) = ₹ 1000

Time (n) = 2 years = 24 months

Maturity value = 26 × ₹ 1000 = ₹ 26,000

(a) Total amount deposited = P × n

= 1000 × 24

= ₹ 24,000

By formula,

Interest = Maturity value − Total amount deposited

= ₹ 26,000 − ₹ 24,000

= ₹ 2,000.

Hence, interest paid by the bank = ₹ 2,000.

(b) Let the rate of interest be r% per annum.

By formula,

$I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$

Substituting values, we get :

$$\begin{aligned}\Rightarrow 2000 = 1000 \times \frac{24(24 + 1)}{2 \times 12} \times \frac{r}{100} \\[1em] \Rightarrow 2000 = 1000 \times \frac{24 \times 25}{24} \times \frac{r}{100} \\[1em] \Rightarrow 2000 = 1000 \times 25 \times \frac{r}{100} \\[1em] \Rightarrow 2000 = 250r \\[1em] \Rightarrow r = \frac{2000}{250} \\[1em] \Rightarrow r = 8.\end{aligned}$$

Hence, rate of interest (r%) = 8% per annum.