The product of two consecutive odd numbers is 19043. Which is the smaller number? [#869]

137

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137

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%

36

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36

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

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

120

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120

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.

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

2

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2

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

48

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48

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

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