Three fourth of a number is more than two third of the number by 5. Then the number is [#875]

60

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60

12

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12

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

2 Because, Reminder must be positive and it should be less then the divisor

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2 Because, Reminder must be positive and it should be less then the divisor

2:3 = 8:12 N1*N2 = HCF*LCM 8*12 = 96 N1+N2 = 8+12 = 20

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2:3 = 8:12 N1*N2 = HCF*LCM 8*12 = 96 N1+N2 = 8+12 = 20

Zero

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Zero

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%

2 * $\frac{22}{7}$ * 7 = 2 * 22 = 44

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2 * $\frac{22}{7}$ * 7 = 2 * 22 = 44

$\frac{3}{2}$ >$\frac{3}{2}\times \frac{4}{9}$ = $\frac{12}{18}$ = $\frac{2}{3}$

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$\frac{3}{2}$ >$\frac{3}{2}\times \frac{4}{9}$ = $\frac{12}{18}$ = $\frac{2}{3}$

39

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39

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