‘A’ starts his journey at 1:00 p.m. from a location P with a speed of 1 m/sec. ‘B’ starts his journey from the same location P and along the same direction at 1:10 p.m. with a speed of 2 m/sec. If ‘B’ meets ‘A’ at the location Q, then the distance PQ is : [#1122]

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

(1000+100)m / (20m/s) = 55 seconds

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(1000+100)m / (20m/s) = 55 seconds

2⁶ = 64 3⁵ = 243 4⁴ = 256 5³ = 125

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2⁶ = 64 3⁵ = 243 4⁴ = 256 5³ = 125

In ascending order = -2, -1, 1, 2, 3, 5, 6 median = $\frac{7+1}{2}$th term = 4th term = 2

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In ascending order = -2, -1, 1, 2, 3, 5, 6 median = $\frac{7+1}{2}$th term = 4th term = 2

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

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

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

Area = πr² π1² : π3² = π1 : π9 = 1:9

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Area = πr² π1² : π3² = π1 : π9 = 1:9

An even number is a whole number that is divisible by 2, such as 4, 6, or 8.

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An even number is a whole number that is divisible by 2, such as 4, 6, or 8.

100 - 98 = 2

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100 - 98 = 2

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

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

The difference of speed = 22m/s - 20m/s = 2m/s As the train with speed 22m/s will cross the other, hence this train have to go the smallest length of the trains distance ahead of the other to cross completely. Hence This train will be ahead of 100m with a speed of 2m/s. Hence The time to cross each other will take = 100m / (2m/s) = 50 sec

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The difference of speed = 22m/s - 20m/s = 2m/s As the train with speed 22m/s will cross the other, hence this train have to go the smallest length of the trains distance ahead of the other to cross completely. Hence This train will be ahead of 100m with a speed of 2m/s. Hence The time to cross each other will take = 100m / (2m/s) = 50 sec