Model Question Paper 2

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{1}{3}\right)^9 \times (3^4)^2 \\[1em] \Rightarrow \frac{1}{3^9} \times 3^8\\[1em] \Rightarrow 3^{-9} \times 3^8 \\[1em] \Rightarrow 3^{-9+8}\\[1em] \Rightarrow 3^{-1} \\[1em] \Rightarrow \frac{1}{3}.\end{aligned}$$

Hence, option 4 is the correct option.

Here, S.P. = ₹ 329 and loss = 6%.

Let the cost price of the article be ₹ x.

loss = 6% of ₹ x = ₹ $\frac{6 \times x}{100} = ₹\frac{3x}{50}$

S.P. of the article = C.P. - loss

$$\begin{aligned}= x - \frac{3x}{50}\\[1em] = \Big(1 - \frac{3}{50}\Big)x \\[1em] = \Big(\frac{50 - 3}{50}\Big)x \\[1em] = \frac{47x}{50}.\end{aligned}$$

Given,

S.P. = ₹ 329

$$\begin{aligned}\therefore \frac{47x}{50} = 329 \\[1em] \Rightarrow x = \frac{329 \times 50}{47}\\[1em] \Rightarrow x = 350\end{aligned}$$

So, C.P. = ₹350

Hence, option 4 is the correct option.

Amount donated by Ravi = ₹ 3⁶ = ₹ (3 × 3 × 3 × 3 × 3 × 3) = ₹ 729.

Amount donated by Rashmi = ₹ 6³ = ₹ (6 × 6 × 6) = ₹ 216.

Since 729 > 216, Ravi donated more money.

Hence, Ravi donated more money.

The two digit numbers whose digits add up to 11 are:

29, 38, 47, 56, 65, 74, 83, 92

So, A = {29, 38, 47, 56, 65, 74, 83, 92}.

The set A has 8 elements, so n(A) = 8.

Hence, the cardinal number of set A is 8.

Let the fourth proportional be x. Then 8 : 24 :: 25 : x.

$$\begin{aligned}\Rightarrow \frac{8}{24} = \frac{25}{x}\\[1em] \Rightarrow 8 \times x = 24 \times 25\\[1em] \Rightarrow x = \frac{24 \times 25}{8}\\[1em] \Rightarrow x = \frac{600}{8} = 75\end{aligned}$$

Hence, the fourth proportional is 75.

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{3}{5}\right)^3 \times \left(\left(\frac{5}{3}\right)^{-2}\right)^4 = \left(\frac{3}{5}\right)^{2m-1}\\[1em] \Rightarrow \left(\frac{3}{5}\right)^3 \times \left(\frac{5}{3}\right)^{-8} = \left(\frac{3}{5}\right)^{2m-1}\\[1em] \Rightarrow \left(\frac{3}{5}\right)^3 \times \left(\frac{3}{5}\right)^{8} = \left(\frac{3}{5}\right)^{2m-1}\\[1em] \Rightarrow \left(\frac{3}{5}\right)^{3+8} = \left(\frac{3}{5}\right)^{2m-1}\\[1em] \Rightarrow \left(\frac{3}{5}\right)^{11} = \left(\frac{3}{5}\right)^{2m-1}\end{aligned}$$

Since the bases are equal, the powers must be equal:

$$\begin{aligned}\Rightarrow 2m - 1 = 11\\[1em] \Rightarrow 2m = 12\\[1em] \Rightarrow m = 6\end{aligned}$$

Hence, m = 6.

Since 18% of the apples go bad, the percentage of good apples = (100 − 18)% = 82%.

Let the total number of apples be x. Then 82% of x = 123.

$$\begin{aligned}\Rightarrow \frac{82}{100} \times x = 123\\[1em] \Rightarrow x = \frac{123 \times 100}{82}\\[1em] \Rightarrow x = 150\end{aligned}$$

Total number of apples = 150.

Number of bad apples = 18% of 150 = $\frac{18}{100} \times 150 = 27$.

Hence, the number of bad apples = 27.

Let the two parts be 8x and 5x.

Difference of the two parts = 24.

$$\begin{aligned}\Rightarrow 8x - 5x = 24\\[1em] \Rightarrow 3x = 24\\[1em] \Rightarrow x = 8\end{aligned}$$

First part = 8x = 8 × 8 = 64.

Second part = 5x = 5 × 8 = 40.

The number = 64 + 40 = 104.

Hence, the number is 104 and its two parts are 64 and 40.

Cost of 2.5 kg of apples = ₹ 187.5.

Cost of 1 kg of apples = $\frac{187.5}{2.5}$ = ₹ 75.

Quantity of apples bought for ₹ 330 = $\frac{330}{75} = 4.4$ kg.

Hence, 4.4 kg of apples can be bought for ₹ 330.

Let the sum (principal) be ₹ P. Here R = 5% per annum, T = 3 years and Amount = ₹ 2,875.

Amount = Principal + Interest

$$\begin{aligned}\Rightarrow \text{Amount} = P + \frac{P \times R \times T}{100} \\[1em] \Rightarrow \text{Amount} = P\left(1 + \frac{RT}{100}\right)\\[1em] \Rightarrow 2875 = P\left(1 + \frac{5 \times 3}{100}\right)\\[1em] \Rightarrow 2875 = P\left(1 + \frac{15}{100}\right)\\[1em] \Rightarrow 2875 = P \times \frac{115}{100}\\[1em] \Rightarrow P = \frac{2875 \times 100}{115}\\[1em] \Rightarrow P = 2500.\end{aligned}$$

Hence, the required sum of money is ₹ 2,500.