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655-705 (Hard)|   Algebra|   Arithmetic|      
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Bunuel
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If P = 1/(10^2 + 1) + 2/(10^2 + 2) + 3/(10^2 + 3) + ... + 10/10^2 + 10 04 Feb 2020, 03:41
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59% (02:04) correct 41% (02:19) wrong based on 426 sessions

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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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B
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Nikhilgaur29
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If P = 1/(10^2 + 1) + 2/(10^2 + 2) + 3/(10^2 + 3) + ... + 10/10^2 + 10 04 Feb 2020, 04:07
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(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 04 Feb 2020, 04:00
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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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If P = 1/(10^2 + 1) + 2/(10^2 + 2) + 3/(10^2 + 3) + ... + 10/10^2 + 10 04 Feb 2020, 03:53
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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.

IMO-B
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If P = 1/(10^2 + 1) + 2/(10^2 + 2) + 3/(10^2 + 3) + ... + 10/10^2 + 10 04 Feb 2020, 04:12
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\(P = \frac{1}{10^2 + 1} + \frac{2}{10^2 + 2} + \frac{3}{10^2 + 3} + ... + \frac{10}{10^2 + 10}\)

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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If P = 1/(10^2 + 1) + 2/(10^2 + 2) + 3/(10^2 + 3) + ... + 10/10^2 + 10 04 Feb 2020, 08:33
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Bunuel
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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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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If P = 1/(10^2 + 1) + 2/(10^2 + 2) + 3/(10^2 + 3) + ... + 10/10^2 + 10 08 Feb 2020, 20:21
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Bunuel
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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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.

Answer: B
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If P = 1/(10^2 + 1) + 2/(10^2 + 2) + 3/(10^2 + 3) + ... + 10/10^2 + 10 13 Apr 2026, 22:16
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