Question 1: A sum of ₹12,000 is invested at a rate of 10% per annum for 2 years. Calculate the difference between the compound interest (compounded annually) and the simple interest for the same period.
- 1. ₹120
- 2. ₹150
- 3. ₹180
- 4. ₹200
Solution:
Simple Interest (SI):
\[ SI = \frac{P \times R \times T}{100} = \frac{12000 \times 10 \times 2}{100} = ₹2400 \]
Compound Interest (CI):
\[ CI = P \left(1 + \frac{R}{100}\right)^T – P \]
\[ CI = 12000 \left(1 + \frac{10}{100}\right)^2 – 12000 = 12000 \times 1.1^2 – 12000 \]
\[ CI = 12000 \times 1.21 – 12000 = ₹2520 \]
Difference:
\[ \text{Difference} = CI – SI = 2520 – 2400 = ₹120 \]
Question 2: Find the next number in the series: 2, 6, 12, 20, 30, …
- 1. 40
- 2. 42
- 3. 44
- 4. 46
Solution:
The differences between consecutive terms are:
\[ 6 – 2 = 4,\ 12 – 6 = 6,\ 20 – 12 = 8,\ 30 – 20 = 10 \]
The differences increase by 2 each time. The next difference will be \(10 + 2 = 12\).
\[ \text{Next term} = 30 + 12 = 42 \]
Question 3: If \(\sin \theta = \frac{3}{5}\), find \(\cos \theta\).
- 1. \(\frac{4}{5}\)
- 2. \(\frac{3}{5}\)
- 3. \(\frac{\sqrt{16}}{5}\)
- 4. \(\frac{1}{5}\)
Solution:
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ \cos^2 \theta = 1 – \sin^2 \theta = 1 – \left(\frac{3}{5}\right)^2 = 1 – \frac{9}{25} = \frac{16}{25} \]
\[ \cos \theta = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
Question 4: A shopkeeper marks an item at ₹500 and offers a discount of 20%. If the cost price is ₹350, find the profit percentage.
- 1. 20%
- 2. 25%
- 3. 15%
- 4. 10%
Solution:
Selling Price (SP):
\[ SP = \text{Marked Price} – \text{Discount} = 500 – (20\% \times 500) = ₹400 \]
Profit:
\[ \text{Profit} = SP – CP = 400 – 350 = ₹50 \]
Profit Percentage:
\[ \text{Profit \%} = \frac{\text{Profit}}{\text{Cost Price}} \times 100 = \frac{50}{350} \times 100 = 14.29\% \approx 15\% \]
Question 5: If the ratio of the ages of two people is 5:7, and the sum of their ages is 48, find their ages.
- 1. 20 and 28
- 2. 30 and 18
- 3. 24 and 16
- 4. 21 and 27
Solution:
Let their ages be \(5x\) and \(7x\).
\[ 5x + 7x = 48 \implies 12x = 48 \implies x = 4 \]
Thus, their ages are:
\[ 5x = 20,\ 7x = 28 \]
Question 6: A can complete a task in 12 days, and B can complete the same task in 8 days. How long will it take for A and B together to complete the task?
- 1. 4.8 days
- 2. 5 days
- 3. 6 days
- 4. 7 days
Solution:
The work rates are:
\[ \text{A’s rate} = \frac{1}{12},\ \text{B’s rate} = \frac{1}{8} \]
Combined rate:
\[ \text{Combined rate} = \frac{1}{12} + \frac{1}{8} = \frac{2}{24} + \frac{3}{24} = \frac{5}{24} \]
Total time:
\[ \text{Time} = \frac{1}{\text{Combined Rate}} = \frac{1}{\frac{5}{24}} = \frac{24}{5} = 4.8\ \text{days} \]
Question 7: A tower is 50 meters tall. From a point on the ground, the angle of elevation to the top of the tower is \(60^\circ\). Find the distance of the point from the base of the tower.
- 1. \(25\sqrt{3}\) m
- 2. \(30\sqrt{3}\) m
- 3. \(50\sqrt{3}\) m
- 4. \(100\) m
Solution:
Using \(\tan \theta = \frac{\text{Height}}{\text{Base}}\):
\[ \tan 60^\circ = \sqrt{3} = \frac{50}{\text{Base}} \]
\[ \text{Base} = \frac{50}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{50\sqrt{3}}{3} \]
Question 8: Find the missing number in the following matrix:
2 4 6
3 6 ?
4 8 12
- 1. 8
- 2. 9
- 3. 12
- 4. 10
Solution:
For each element in the row, multiply row and column:
\[ \text{Element at (2,3)} = 2 \times 3 \times 2 = 12 \]
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