A man is 3 years older than his wife and four times as old as his son. If the son becomes 15 years old after 3 years, the present age of his wife is : [#1127]

The son becomes 15 years old after 3 years. Hence son's present age = 15 years - 3 years = 12 years Man is four times as old as his son. Hence Man's present age = 12 years * 4 = 48 years Man is 3 years older than his wife. Hence present age of his wife = 48 years - 3 years = 45 years.

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The son becomes 15 years old after 3 years. Hence son's present age = 15 years - 3 years = 12 years Man is four times as old as his son. Hence Man's present age = 12 years * 4 = 48 years Man is 3 years older than his wife. Hence present age of his wife = 48 years - 3 years = 45 years.

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

1st Month -> (105 * 2) - 120 = 210 - 120 = 90 2nd Month -> (90 * 2) - 120 = 180 - 120 = 60 3rd Month -> (60 * 2) - 120 = 120 - 120 = 0

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1st Month -> (105 * 2) - 120 = 210 - 120 = 90 2nd Month -> (90 * 2) - 120 = 180 - 120 = 60 3rd Month -> (60 * 2) - 120 = 120 - 120 = 0

Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.

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Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.

LCM will be r. px= q and qy = r, hence r = pxy.

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LCM will be r. px= q and qy = r, hence r = pxy.

$\sqrt[3]{8^2}$ = $\sqrt[3]{64}$ = $\sqrt[3]{4^3}$ = 4

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$\sqrt[3]{8^2}$ = $\sqrt[3]{64}$ = $\sqrt[3]{4^3}$ = 4

A : B = 3 => $\frac{A}{B}$ = 3 $\frac{144A}{36B}$ = $\frac{144}{36}\times \frac{A}{B}$ = $\frac{144}{36}$ x 3 = 4 x 3 = 12

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A : B = 3 => $\frac{A}{B}$ = 3 $\frac{144A}{36B}$ = $\frac{144}{36}\times \frac{A}{B}$ = $\frac{144}{36}$ x 3 = 4 x 3 = 12

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1*20 + 0.5*30 = 20+15 = 35

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1*20 + 0.5*30 = 20+15 = 35

= $\sqrt{0.04} + \sqrt{1.44} + \sqrt{1.69} + \sqrt{0.0009}$ = $\sqrt{\mathrm{0.02*0.02}} + \sqrt{\mathrm{1.2*1.2}} + \sqrt{\mathrm{1.3*1.3}} + \sqrt{\mathrm{0.03*0.03}}$ = 0.2 + 1.2 + 1.3 + 0.03 = 2.73

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= $\sqrt{0.04} + \sqrt{1.44} + \sqrt{1.69} + \sqrt{0.0009}$ = $\sqrt{\mathrm{0.02*0.02}} + \sqrt{\mathrm{1.2*1.2}} + \sqrt{\mathrm{1.3*1.3}} + \sqrt{\mathrm{0.03*0.03}}$ = 0.2 + 1.2 + 1.3 + 0.03 = 2.73

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