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# If P = 1/(10^2 + 1) + 2/(10^2 + 2) + 3/(10^2 + 3) + ... + 10/10^2 + 10

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If P = 1/(10^2 + 1) + 2/(10^2 + 2) + 3/(10^2 + 3) + ... + 10/10^2 + 10  [#permalink]

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04 Feb 2020, 02:41
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If $$P = \frac{1}{10^2 + 1} + \frac{2}{10^2 + 2} + \frac{3}{10^2 + 3} + ... + \frac{10}{10^2 + 10}$$ then which of the following is the best approximate value of P.

A. 0.42
B. 0.52
C. 0.57
D. 0.62
E. 0.72

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Re: If P = 1/(10^2 + 1) + 2/(10^2 + 2) + 3/(10^2 + 3) + ... + 10/10^2 + 10  [#permalink]

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04 Feb 2020, 02:53
2
Explanation:

Approximating the denominator of each fraction, one gets P as follows:
= (1+2+3+4+5)/100 + (6+7+8+9+10)/110
= 15/100 + 40/110
= 0.15 + 0.36 = 0.51 Aprox.

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Re: If P = 1/(10^2 + 1) + 2/(10^2 + 2) + 3/(10^2 + 3) + ... + 10/10^2 + 10  [#permalink]

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04 Feb 2020, 03:00
1
1st term -> $$\frac{1}{(10^2+1)}$$ =$$\frac{1}{101}$$ -> slightly < 0.01
2nd term -> $$\frac{2}{(10^2+2)}$$ =$$\frac{2}{102}$$ -> slightly < 0.02
.
.
.
10th term -> $$\frac{10}{(10^2+10)}$$ =$$\frac{1}{11}$$ -> slightly < 0.1

Therefore -> Total slightly < (0.01 + 0.02 + 0.03 + .... + 0.1) -> slightly less than 0.55 -> Answer B -> 0.52
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Re: If P = 1/(10^2 + 1) + 2/(10^2 + 2) + 3/(10^2 + 3) + ... + 10/10^2 + 10  [#permalink]

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04 Feb 2020, 03:07
2
(1+2+3...+10)/(10^2+10) < P < (1+2+3...+10)/(10^2)

This implies 55/110 < P < 55/100

0.5 < P < 0.55
Option B should be the Answer

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If P = 1/(10^2 + 1) + 2/(10^2 + 2) + 3/(10^2 + 3) + ... + 10/10^2 + 10  [#permalink]

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04 Feb 2020, 03:12
1
$$P = \frac{1}{10^2 + 1} + \frac{2}{10^2 + 2} + \frac{3}{10^2 + 3} + ... + \frac{10}{10^2 + 10}$$

Analyzing the question, there is addition of 1, 2,3...10 to $$10^2$$ in the denominator.
Adding a small value to $$10^2$$ will slightly decrease value of the fraction. Lets understand:
$$\frac{1}{100}$$ = 0.01 and $$\frac{1}{101}$$=0.009
$$\frac{2}{100}$$ = 0.02 and $$\frac{2}{102}$$ = 0.196

hence, lets take common denominator as $$10^2$$ for approximation
Adding the numerator, we get 1+2+3+....+10 = 55
55/$$10^2$$ = 0.55

As denominator will be greater than $$10^2$$, hence the value of the fraction will be slightly less that 0.55

Therefore, IMO B.
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Re: If P = 1/(10^2 + 1) + 2/(10^2 + 2) + 3/(10^2 + 3) + ... + 10/10^2 + 10  [#permalink]

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04 Feb 2020, 07:33
1
Bunuel wrote:
If $$P = \frac{1}{10^2 + 1} + \frac{2}{10^2 + 2} + \frac{3}{10^2 + 3} + ... + \frac{10}{10^2 + 10}$$ then which of the following is the best approximate value of P.

A. 0.42
B. 0.52
C. 0.57
D. 0.62
E. 0.72

1/101˜0.01
2/102˜0.02
3/103˜0.03

10/110˜0.0909
sum first 10 positive integers: n(n+1)/2=10(11)/2=55
since difference 10/100=0.1 and 10/110=0.09 is 0.01
sum sequence must be less than 0.54, which closest to 0.52

Ans (B)
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Re: If P = 1/(10^2 + 1) + 2/(10^2 + 2) + 3/(10^2 + 3) + ... + 10/10^2 + 10  [#permalink]

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08 Feb 2020, 19:21
1
Bunuel wrote:
If $$P = \frac{1}{10^2 + 1} + \frac{2}{10^2 + 2} + \frac{3}{10^2 + 3} + ... + \frac{10}{10^2 + 10}$$ then which of the following is the best approximate value of P.

A. 0.42
B. 0.52
C. 0.57
D. 0.62
E. 0.72

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Project PS Butler

We see that each denominator is actually very close to 100 (notice that the smallest denominator is 10^2 + 1 = 101 and the largest is 10^2 + 10 = 110). Therefore, 1/(10^2 + 1) is essentially 1/100, 2/(10^2 + 2) is essentially 2/100 and so on. Approximating all the fractions using the denominator 100, we have

P ≈ 1/100 + 2/100 + 3/100 + … + 10/100 = (1 + 2 + 3 + … + 10)/100 = 55/100 = 0.55

However, since 1/(10^2 + 1) is slightly less than 1/100, 2/(10^2 + 2) is slightly less than 2/100 and so on, P should be slightly less than 0.55. Therefore, the best approximate value of P is 0.52.

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Re: If P = 1/(10^2 + 1) + 2/(10^2 + 2) + 3/(10^2 + 3) + ... + 10/10^2 + 10   [#permalink] 08 Feb 2020, 19:21
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