Let the principal be \( P \).

Since the money doubles in 10 years, the total amount becomes \( 2P \).
So, Simple Interest (SI) = \( 2P - P = P \)

Use the formula:
$$\text{SI} = \frac{P \cdot R \cdot T}{100}$$ Substituting values: $$P = \frac{P \cdot R \cdot 10}{100}$$ Cancel \( P \) on both sides: $$1 = \frac{R \cdot 10}{100} \Rightarrow R = \frac{100}{10} = \boxed{10\%}$$
Final Answer:
$$\boxed{10\% \text{ per annum}}$$

Let the number of correct answers be \(x\) and wrong answers be \(y\).

Total questions: \[ x + y = 60 \tag{1} \] Marks rule: \(4\) marks for correct and \(-1\) mark for wrong, total \(130\). \[ 4x - y = 130 \tag{2} \] From (1): \(y = 60 - x\). Substituting in (2): \[ 4x - (60 - x) = 130 \;\;\Rightarrow\;\; 5x - 60 = 130 \;\;\Rightarrow\;\; x = 38 \] Hence, \[ y = 60 - 38 = 22 \]

Solution

Given: 4/5 of the students are boys, and the rest are girls. 2/5 of boys and 1/4 of girls are absent.

Step-by-Step Solution

• Total students = N
• Boys: (4/5) × N = 4N/5
• Girls: (1/5) × N = N/5
• Boys absent: (2/5) × (4N/5) = 8N/25
• Girls absent: (1/4) × (N/5) = N/20
• Boys present: (4N/5 - 8N/25 = 12N/25)
• Girls present: (N/5 - N/20 = 3N/20)
• Total present = (12N/25 + 3N/20)

Final Calculation

LCM of 25 and 20 is 100:
(12N/25 = 48N/100), (3N/20 = 15N/100)
Total present = (48N/100 + 15N/100 = 63N/100)

Answer: 63/100

$\frac{M_1 H_1 D_1}{W_1}$=$\frac{M_2 H_2 D_2}{W_2}$ ----(i)

Here 

$M_1=2, H_1=2, W_1=2$

$M_2=?, H_2=18, W_2=18$ 

Using (i)

$\frac{2 \times 2 \times 1}{2}$=$\frac{M_2 \times 18 }{18}$

$M_2=2$

Hence, 2 cats are needed to catch 18 mice.