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PS problem

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PS problem

by prernamalhotra » Fri Jul 18, 2014 3:58 am
Is there an easy way to solve this question:

Q.What is 1/(1)(2)+1/(2)(3)+1/(3)(4)+1/(4)(5)+1/(5)(6)+1/(6)(7)+1/(7)(8)+1/(8)(9)+1/(9)(10)?

1) 2/5
2) 3/5
3) 7/10
4) 46/55
5) 9/10

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by GMATGuruNY » Fri Jul 18, 2014 4:40 am
prernamalhotra wrote:Is there an easy way to solve this question:

Q.What is 1/(1)(2)+1/(2)(3)+1/(3)(4)+1/(4)(5)+1/(5)(6)+1/(6)(7)+1/(7)(8)+1/(8)(9)+1/(9)(10)?

1) 2/5
2) 3/5
3) 7/10
4) 46/55
5) 9/10
WRITE IT OUT and LOOK FOR A PATTERN.

Sum of the first 2 terms:
1/2 + 1/6 = 2/3.

Sum of the first 3 terms:
2/3 + 1/12 = 3/4.

Sum of the first 4 terms:
3/4 + 1/20 = 4/5.

In each case:
The NUMERATOR of the sum = the NUMBER OF TERMS.
THE DENOMINATOR of the sum = NUMERATOR + 1.

The given sum is composed of 9 terms.
Sum of the first 9 terms = 9/10.

The correct answer is E.
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by GMATinsight » Fri Jul 18, 2014 5:33 am
prernamalhotra wrote:
Is there an easy way to solve this question:

Q.What is 1/(1)(2)+1/(2)(3)+1/(3)(4)+1/(4)(5)+1/(5)(6)+1/(6)(7)+1/(7)(8)+1/(8)(9)+1/(9)(10)?

1) 2/5
2) 3/5
3) 7/10
4) 46/55
5) 9/10
Hi Prerna,

Please refer the easiest way to solve this question

Question:1/(1)(2)+1/(2)(3)+1/(3)(4)+1/(4)(5)+1/(5)(6)+1/(6)(7)+1/(7)(8)+1/(8)(9)+1/(9)(10)?

1/(1x2) = (1/1)-(1/2)
1/(2x3) = (1/2)-(1/3)
1/(3x4) = (1/3)-(1/4)
1/(4x5) = (1/4)-(1/5)
1/(5x6) = (1/5)-(1/6)
1/(6x7) = (1/6)-(1/7)
1/(7x8) = (1/7)-(1/8)
1/(8x9) = (1/8)-(1/9)
1/(9x10) = (1/9)-(1/10)

If you add them all then all terms cancel except first and the last term

Therefore Required sum of terms = (1/1)-(1/10) = 9/10

Answer: Option E
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