Explore the complete solution of Exercise 1.3 from the new Math 9th syllabus designed by the Punjab Curriculum and Textbook Board (PCTB), Lahore. This detailed guide simplifies the latest 9th class math concepts to help students excel in their studies
1. The sum of three consecutive integers is forty-two. Find the three integers.
Let the integers be \(x – 1\), \(x\), and \(x + 1\):
\[(x – 1) + x + (x + 1) = 42\]
\[3x = 42 \]
\[ x = 14\]
The integers are:
\[x – 1 = 13, \quad x = 14, \quad x + 1 = 15\]
Answer:
\[13, 14, 15\]
2. The diagram shows right-angled \(\triangle ABC\) in which the length of \(\overline{AC}\) is \((\sqrt{3} + \sqrt{5})\) cm. The area of \(\triangle ABC\) is \((1 + \sqrt{15})\) cm². Find the length of \(\overline{AB}\) in the form \((a\sqrt{3} + b\sqrt{5})\), where \(a\) and \(b\) are integers.
The area of \(\triangle ABC\) is:
\[\text{Area} = \frac{1}{2} \cdot \overline{AC} \cdot \overline{AB}\]
Substitute:
\[1 + \sqrt{15} = \frac{1}{2} \cdot (\sqrt{3} + \sqrt{5}) \cdot \overline{AB}\]
\[2(1 + \sqrt{15}) = (\sqrt{3} + \sqrt{5}) \cdot \overline{AB}\]
\[2 + 2\sqrt{15} = (\sqrt{3} + \sqrt{5}) \cdot \overline{AB}\]
Let \(\overline{AB} = a\sqrt{3} + b\sqrt{5}\). Substitute:
\[(\sqrt{3} + \sqrt{5}) \cdot (a\sqrt{3} + b\sqrt{5}) = 2 + 2\sqrt{15}\]
Expand:
\[a \cdot 3 + b \cdot 5 + (a + b)\sqrt{15} = 2 + 2\sqrt{15}\]
Equating coefficients:
\[3a + 5b = 2 \quad \text{(1)}\]
\[a + b = 2 \quad \text{(2)}\]
From (2):
\[b = 2 – a\]
Substitute into (1):
\[3a + 5(2 – a) = 2\]
\[3a + 10 – 5a = 2\]
\[-2a = -8 \]
\[ a = 4\]
Substitute \(a = 4\) into \(b = 2 – a\):
\[b = 2 – 4 = -2\]
\[\overline{AB} = 4\sqrt{3} – 2\sqrt{5}\]
Answer
\[\overline{AB} = 4\sqrt{3} – 2\sqrt{5}\]
3. A rectangle has sides of length \(2 + \sqrt{18}\) m and \(\left(5 – \frac{4}{\sqrt{2}}\right)\) m. Express the area of the rectangle in the form \(a + b\sqrt{2}\), where \(a\) and \(b\) are integers.
Simplify the sides:
\[2 + \sqrt{18} = 2 + 3\sqrt{2}, \quad 5 – \frac{4}{\sqrt{2}} = 5 – 2\sqrt{2}\]
The area is:
\[\text{Area} = (2 + 3\sqrt{2}) \cdot (5 – 2\sqrt{2})\]
Expand:
\[= 2 \cdot 5 + 2 \cdot (-2\sqrt{2}) + 3\sqrt{2} \cdot 5 + 3\sqrt{2} \cdot (-2\sqrt{2})\]
\[= 10 – 4\sqrt{2} + 15\sqrt{2} – 6 \cdot 2\]
\[= 10 – 12 + 11\sqrt{2}\]
\[= -2 + 11\sqrt{2}\]
Answer
\[-2 + 11\sqrt{2}\]
4. Find two numbers whose sum is 68 and difference is 22.
Let the numbers be \(x\) and \(y\), where:
\[x + y = 68 \quad \text{(1)}\]
\[x – y = 22 \quad \text{(2)}\]
Add (1) and (2):
\[2x = 90 \]
\[ x = 45\]
Subtract (2) from (1):
\[2y = 46 \]
\[ y = 23\]
The numbers are:
\[x = 45, \quad y = 23\]
Answer:
\[45, 23\]
5. The weather in Lahore was unusually warm during the summer of 2024. TV news reported the temperature as high as 48°C. By using the formula, \((°F = \frac{9}{5} °C + 32)\), find the temperature in the Fahrenheit scale.
Solutions
Convert 48°C to °F:
\[°F = \frac{9}{5} \times 48 + 32\]
\[°F = 86.4 + 32\]
\[°F = 118.4\]
Answer
The temperature is \(118.4°F\).
6. The sum of the ages of the father and son is 72 years. Six years ago, the father’s age was 2 times the age of the son. What was the son’s age six years ago?
Find the son’s age six years ago:
Let the son’s age be \(x\) and the father’s age be \(72 – x\).
\[(72 – x) – 6 = 2(x – 6)\]
\[66 – x = 2x – 12\]
\[66 + 12 = 3x\]
\[78 = 3x\]
\[x = 26\]
Son’s age six years ago:
\[26 – 6 = 20\]
Answer
The son’s age six years ago was \(20\) years.
7. Mirha sells a toy for Rs. 1520. What will the selling price be to get a 15% profit?
Find the selling price for a 15% profit:
\[\text{Selling Price} = \text{Cost Price} + \text{Profit}\]
\[\text{Profit} = \frac{15}{100} \times 1520 = 228\]
\[\text{Selling Price} = 1520 + 228 = 1748\]
Answer
The selling price is Rs. \(1748\).
8. The annual income of Tayyab is Rs. 9,60,000, while the exempted amount is Rs. 1,30,000. How much tax would he have to pay at the rate of 0.75%?
Calculate the tax for Tayyab:
Taxable income:
\[9,60,000 – 1,30,000 = 8,30,000\]
Tax:
\[\frac{0.75}{100} \times 8,30,000 = 6225\]
Answer
The tax amount is Rs. \(6225\).
9. Find the compound markup on Rs. 3,75,000 for one year at the rate of 14% compounded annually.
Find the compound markup:
Formula for compound interest:
\[A = P \left(1 + \frac{r}{100}\right)^n\]
Substitute \(P = 3,75,000\), \(r = 14\), \(n = 1\):
\[A = 3,75,000 \left(1 + \frac{14}{100}\right)\]
\[A = 3,75,000 \times 1.14 = 4,27,500\]
Compound markup:
\[4,27,500 – 3,75,000 = 52,500\]
Answer
The compound markup is Rs. \(52,500\).
