The least number by which 2450 must be multiplied to make it a perfect square, is [#1028]

2

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2

=> 2x * 3x = 96 => 6x² = 96 => x² = 96/6 => x² = 16 => x = 4 2x + 3x = 5x = 5 * 4 = 20

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=> 2x * 3x = 96 => 6x² = 96 => x² = 96/6 => x² = 16 => x = 4 2x + 3x = 5x = 5 * 4 = 20

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

48

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48

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

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

20% = $\frac{25}{100+25} * 100$% $\frac{25}{125} * 100$% = 20%

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20% = $\frac{25}{100+25} * 100$% $\frac{25}{125} * 100$% = 20%

6²-1 : 6 = 35 : 6 11²-1 : 11 = 120 : 11

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6²-1 : 6 = 35 : 6 11²-1 : 11 = 120 : 11

X + (X+2) + (X+4) + (X+6) + (X+8) = 95 => 5X + 20 = 95 => 5X = 95-20 => 5X = 75 => X = 15 Hence (X+8) = 15+8 = 23

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X + (X+2) + (X+4) + (X+6) + (X+8) = 95 => 5X + 20 = 95 => 5X = 95-20 => 5X = 75 => X = 15 Hence (X+8) = 15+8 = 23