International - Quizzes (S/P) - Part 70

Rohit

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Rohit

K is the 19th from right R is the 7th from left of K.

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K is the 19th from right R is the 7th from left of K.

On the Moon, there is no air because the Moon's gravitational field is too weak to retain an atmosphere.

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On the Moon, there is no air because the Moon's gravitational field is too weak to retain an atmosphere.

Net change (-40 + 20 + $\frac{(-40)*20}{100}$)% = (-20 + $\frac{-800}{100}$)% = (-20 - 8)% = -28% = 28%

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Net change (-40 + 20 + $\frac{(-40)*20}{100}$)% = (-20 + $\frac{-800}{100}$)% = (-20 - 8)% = -28% = 28%

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

$\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}-\sqrt{\frac{16}{81}}$ = $\sqrt{\frac{15*15}{27*27}}-\sqrt{\frac{5*5}{12*12}}-\sqrt{\frac{4*4}{9*9}}$ = $\frac{15}{27}-\frac{5}{12}-\frac{4}{9}$ = $\frac{60-45-48}{108}$ = $\frac{-33}{108}$ = $\frac{-11}{36}$

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$\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}-\sqrt{\frac{16}{81}}$ = $\sqrt{\frac{15*15}{27*27}}-\sqrt{\frac{5*5}{12*12}}-\sqrt{\frac{4*4}{9*9}}$ = $\frac{15}{27}-\frac{5}{12}-\frac{4}{9}$ = $\frac{60-45-48}{108}$ = $\frac{-33}{108}$ = $\frac{-11}{36}$

1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12 = 1, 1, 1, 2, 4, 5, 5, 7, 8, 12, 12, 16, 17 Total numbers of values N = 13 Hence median value = (N + 1)/2 = (13 + 1)/2 = 7th 7th Number = 5

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1, 2, 4, 1, 5, 12, 7, 8, 5, 1, 16, 17, 12 = 1, 1, 1, 2, 4, 5, 5, 7, 8, 12, 12, 16, 17 Total numbers of values N = 13 Hence median value = (N + 1)/2 = (13 + 1)/2 = 7th 7th Number = 5

The formula for the sum of the first n terms of an arithmetic progression is $S_n\mathit{ }\mathit{=}\mathit{ }\frac{n}{2}(a+an)$ $\frac{30}{2}(12+12*30)$ = 15*(12+360) = 15 * 372 = 5580

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The formula for the sum of the first n terms of an arithmetic progression is $S_n\mathit{ }\mathit{=}\mathit{ }\frac{n}{2}(a+an)$ $\frac{30}{2}(12+12*30)$ = 15*(12+360) = 15 * 372 = 5580

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

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