In a range of consecutive numbers starting with 1, all the even numbers are removed. From the remaining, consider the first 7 numbers. The sum of these 7 numbers is [#1699]

1+3+5+7+9+11+13 = 49

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1+3+5+7+9+11+13 = 49

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

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

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

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

24

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24

33

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33

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.

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

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

40

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40

11

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11

42 - (4*3) = 16 - 12 = 4

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4^{2 - (4*3) = 16 - 12 = 4}