In a school with 10 teachers, one retires and immediately a new teacher of age 25 years joins. As a result, the average age of the teacher reduces by 3. The age of the retired teacher is [#1024]

25 years + (3*10) years = 55 years

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25 years + (3*10) years = 55 years

5² + X² = 13² => 25 + X² = 169 => X² = 196 - 25 => X² = 144 => X = 12

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5² + X² = 13² => 25 + X² = 169 => X² = 196 - 25 => X² = 144 => X = 12

A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B (10*60)s * 1m/s = 600m Let A covers a distance of X after starting of B, Then X + 600m = 2X => 2X - X = 600m => X = 600m Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km

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A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B (10*60)s * 1m/s = 600m Let A covers a distance of X after starting of B, Then X + 600m = 2X => 2X - X = 600m => X = 600m Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km

33

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33

The formula to calculate the area of a circle is A = πr², where A is the area and r is the radius of the circle.

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The formula to calculate the area of a circle is A = πr², where A is the area and r is the radius of the circle.

47

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47

25 years => A+B+C = 22*3 => A+(B+C) = 66 => A+(25*2) = 66 => A = 66-50 => A = 16 After 9 years A = 16+9 = 25 years

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25 years => A+B+C = 22*3 => A+(B+C) = 66 => A+(25*2) = 66 => A = 66-50 => A = 16 After 9 years A = 16+9 = 25 years

$\frac{45}{100}*X = \frac{30}{100} * Y$ = $\frac{X}{Y} = \frac{30*100}{100*45}$ = $\frac{X}{Y} = \frac{2}{3}$ = X:Y = 2:3

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$\frac{45}{100}*X = \frac{30}{100} * Y$ = $\frac{X}{Y} = \frac{30*100}{100*45}$ = $\frac{X}{Y} = \frac{2}{3}$ = X:Y = 2:3

X * (X-12) = 40 * 4 X = 20 X + (X-12) = 20 + 20 - 12 = 28

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X * (X-12) = 40 * 4 X = 20 X + (X-12) = 20 + 20 - 12 = 28

7 and 8

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7 and 8

7 => x * 2x = 98 => x² = $\frac{98}{2}$ => x² = 49 => x = 7

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7 => x * 2x = 98 => x² = $\frac{98}{2}$ => x² = 49 => x = 7