class 8 math solution wbbse

Arithmetic: Simplifying Fractions

Problem: Simplify the fraction $\frac{36}{60}$.

Solution:

1. Find the greatest common divisor (GCD) of 36 and 60.
   • Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
   • Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
   • GCD: 12
2. Divide the numerator and the denominator by the GCD: $$\frac{36 \div 12}{60 \div 12} = \frac{3}{5}$$ So, $\frac{36}{60}$ simplified is $\frac{3}{5}$.

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Algebra: Solving Linear Equations

Problem: Solve for $x$: $2x + 3 = 11$.

Solution:

1. Subtract 3 from both sides: $$2x + 3 - 3 = 11 - 3 \implies 2x = 8$$
2. Divide both sides by 2: $$\frac{2x}{2} = \frac{8}{2} \implies x = 4$$ So, the solution is $x = 4$.

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Geometry: Finding the Area of a Triangle

Problem: Find the area of a triangle with a base of 10 cm and a height of 5 cm.

Solution:

1. Use the formula for the area of a triangle: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$
2. Substitute the given values: $$\text{Area} = \frac{1}{2} \times 10 \times 5 = 25 \, \text{cm}^2$$ So, the area of the triangle is $25 \, \text{cm}^2$.

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Data Handling: Mean Calculation

Problem: Calculate the mean of the following data set: 2, 4, 6, 8, 10.

Solution:

1. Add all the numbers: $$2 + 4 + 6 + 8 + 10 = 30$$
2. Divide by the number of data points: $$\text{Mean} = \frac{30}{5} = 6$$ So, the mean is 6.

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Arithmetic: Ratio and Proportion

Problem: If $a:b = 2:3$ and $b:c = 4:5$, find the ratio $a:c$.

Solution:

1. Given $a:b = 2:3$ and $b:c = 4:5$.

2. To find $a:c$, make $b$ the same in both ratios. The least common multiple (LCM) of 3 and 4 is 12.

3. Adjust the ratios:

   $$a:b = 2 \times 4 : 3 \times 4 = 8:12$$ $$b:c = 4 \times 3 : 5 \times 3 = 12:15$$

4. Now $a:b = 8:12$ and $b:c = 12:15$. So, $a:c = 8:15$.

   Thus, $a:c = 8:15$.

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Algebra: Solving Equations with Variables on Both Sides

Problem: Solve for $x$: $3x + 5 = 2x + 15$.

Solution:

1. Subtract $2x$ from both sides: $$3x + 5 - 2x = 2x + 15 - 2x \implies x + 5 = 15$$
2. Subtract 5 from both sides: $$x + 5 - 5 = 15 - 5 \implies x = 10$$ So, the solution is $x = 10$.

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Geometry: Perimeter of a Rectangle

Problem: Find the perimeter of a rectangle with a length of 8 cm and a width of 5 cm.

Solution:

1. Use the formula for the perimeter of a rectangle: $$\text{Perimeter} = 2 \times (\text{length} + \text{width})$$
2. Substitute the given values: $$\text{Perimeter} = 2 \times (8 + 5) = 2 \times 13 = 26 \, \text{cm}$$ So, the perimeter of the rectangle is 26 cm.

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Data Handling: Median Calculation

Problem: Find the median of the following data set: 7, 9, 12, 3, 5.

Solution:

1. Arrange the numbers in ascending order: $$3, 5, 7, 9, 12$$
2. Find the middle number (median). There are 5 numbers, so the middle number is the third one: $$\text{Median} = 7$$ So, the median is 7.

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Arithmetic: Percentage Calculation

Problem: Calculate 25% of 200.

Solution:

1. Convert the percentage to a fraction: $$25\% = \frac{25}{100} = \frac{1}{4}$$
2. Multiply by 200: $$\frac{1}{4} \times 200 = 50$$ So, 25% of 200 is 50.

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Algebra: Simplifying Expressions

Problem: Simplify $3(x + 4) - 2(x - 3)$.

Solution:

1. Distribute the numbers inside the parentheses: $$3(x + 4) = 3x + 12$$ $$-2(x - 3) = -2x + 6$$
2. Combine the simplified parts: $$3x + 12 - 2x + 6 = 3x - 2x + 12 + 6 = x + 18$$ So, the simplified expression is $x + 18$.

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Geometry: Circumference of a Circle

Problem: Find the circumference of a circle with a radius of 7 cm.

Solution:

1. Use the formula for the circumference of a circle: $$\text{Circumference} = 2\pi r$$
2. Substitute the given radius: $$\text{Circumference} = 2 \times \pi \times 7 = 14\pi \approx 14 \times 3.14 = 43.96 \, \text{cm}$$ So, the circumference of the circle is approximately 43.96 cm.

Arithmetic: Compound Interest

Problem: Calculate the compound interest for a principal amount of ₹2000, an annual interest rate of 5%, and a time period of 2 years.

Solution:

1. Use the compound interest formula: $$A = P \left(1 + \frac{r}{100}\right)^t$$ where $A$ is the amount, $P$ is the principal, $r$ is the rate of interest, and $t$ is the time period.
2. Substitute the given values: $$A = 2000 \left(1 + \frac{5}{100}\right)^2 = 2000 \left(1.05\right)^2$$
3. Calculate the amount: $$A = 2000 \times 1.1025 = 2205$$
4. Find the compound interest: $$\text{Compound Interest} = A - P = 2205 - 2000 = 205$$ So, the compound interest is ₹205.

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Algebra: Quadratic Equations

Problem: Solve the quadratic equation $x^2 - 5x + 6 = 0$.

Solution:

1. Factorize the quadratic equation: $$x^2 - 5x + 6 = (x - 2)(x - 3) = 0$$
2. Set each factor to zero: $$x - 2 = 0 \implies x = 2$$ $$x - 3 = 0 \implies x = 3$$ So, the solutions are $x = 2$ and $x = 3$.

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Geometry: Volume of a Cylinder

Problem: Find the volume of a cylinder with a radius of 3 cm and a height of 5 cm.

Solution:

1. Use the formula for the volume of a cylinder: $$\text{Volume} = \pi r^2 h$$
2. Substitute the given values: $$\text{Volume} = \pi \times 3^2 \times 5 = \pi \times 9 \times 5 = 45\pi \approx 45 \times 3.14 = 141.3 \, \text{cm}^3$$ So, the volume of the cylinder is approximately $141.3 \, \text{cm}^3$.

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Data Handling: Mode Calculation

Problem: Find the mode of the following data set: 4, 1, 2, 4, 3, 4, 2, 1.

Solution:

1. Arrange the numbers and count the frequency of each number:
   • 1 appears 2 times
   • 2 appears 2 times
   • 3 appears 1 time
   • 4 appears 3 times
2. Identify the number with the highest frequency:
   • 4 appears most frequently (3 times)
   $$\text{Mode} = 4$$ So, the mode is 4.

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Arithmetic: Simple Interest

Problem: Calculate the simple interest for a principal amount of ₹1500, an annual interest rate of 6%, and a time period of 3 years.

Solution:

1. Use the simple interest formula: $$\text{Simple Interest} = \frac{P \times R \times T}{100}$$
2. Substitute the given values: $$\text{Simple Interest} = \frac{1500 \times 6 \times 3}{100} = \frac{27000}{100} = 270$$ So, the simple interest is ₹270.

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Algebra: Solving Systems of Equations

Problem: Solve the system of equations: $2x + 3y = 12$ $x - y = 2$

Solution:

1. Solve the second equation for $x$: $$x = y + 2$$
2. Substitute $x = y + 2$ into the first equation: $$2(y + 2) + 3y = 12 \implies 2y + 4 + 3y = 12 \implies 5y + 4 = 12 \implies 5y = 8 \implies y = \frac{8}{5} = 1.6$$
3. Substitute $y = 1.6$ into $x = y + 2$: $$x = 1.6 + 2 = 3.6$$ So, the solution is $x = 3.6$ and $y = 1.6$.

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Geometry: Surface Area of a Cube

Problem: Find the surface area of a cube with a side length of 4 cm.

Solution:

1. Use the formula for the surface area of a cube: $$\text{Surface Area} = 6s^2$$
2. Substitute the given side length: $$\text{Surface Area} = 6 \times 4^2 = 6 \times 16 = 96 \, \text{cm}^2$$ So, the surface area of the cube is $96 \, \text{cm}^2$.

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Data Handling: Range Calculation

Problem: Find the range of the following data set: 14, 7, 22, 9, 18.

Solution:

1. Identify the maximum and minimum values in the data set:
   • Maximum value = 22
   • Minimum value = 7
2. Calculate the range: $$\text{Range} = \text{Maximum value} - \text{Minimum value} = 22 - 7 = 15$$ So, the range is 15.

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Arithmetic: LCM Calculation

Problem: Find the least common multiple (LCM) of 12 and 15.

Solution:

1. List the multiples of each number:
   • Multiples of 12: 12, 24, 36, 48, 60, ...
   • Multiples of 15: 15, 30, 45, 60, ...
2. Identify the smallest common multiple:
   • The smallest common multiple is 60.
   $$\text{LCM} = 60$$ So, the LCM of 12 and 15 is 60.

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Algebra: Evaluating Expressions

Problem: Evaluate the expression $3a^2 - 2b + 5$ for $a = 2$ and $b = 3$.

Solution:

1. Substitute $a = 2$ and $b = 3$ into the expression: $$3(2)^2 - 2(3) + 5$$
2. Calculate the value: $$3 \times 4 - 2 \times 3 + 5 = 12 - 6 + 5 = 11$$ So, the value of the expression is 11.