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word problems :

Expert replies
by \'manpreet singh » Sat Jul 07, 2012 10:23 pm
Q.1)For any positive integer n, the sum of the first
n positive integers equalsn(n+1)/2 . What is the sum
of all the even integers between 99 and 301 ?
(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150

Q.2)This year Henry will save a certain amount of his
income, and he will spend the rest. Next year Henry
will have no income, but for each dollar that he saves
this year, he will have 1 + r dollars available to spend.
In terms of r, what fraction of his income should Henry
save this year so that next year the amount he has
available to spend will be equal to half the amount that
he spends this year?
Join the discussion
Source: — Problem Solving |

by eagleeye » Sat Jul 07, 2012 10:28 pm
'manpreet singh wrote:Q.1)For any positive integer n, the sum of the first
n positive integers equalsn(n+1)/2 . What is the sum
of all the even integers between 99 and 301 ?
For an evenly spaced consecutive set,

Sum = number of terms * (average of first and last term)

We have first term as 100, last as 300, and even integers means the difference is 2.
Number of terms = (last - first)/(difference) + 1 = (300-100)/2 + 1 = 101.
Average of first and last = (100+300)/2 = 200.
Hence Sum = 101*200 = 20200.
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by eagleeye » Sat Jul 07, 2012 10:39 pm
'manpreet singh wrote:
Q.2)This year Henry will save a certain amount of his
income, and he will spend the rest. Next year Henry
will have no income, but for each dollar that he saves
this year, he will have 1 + r dollars available to spend.
In terms of r, what fraction of his income should Henry
save this year so that next year the amount he has
available to spend will be equal to half the amount that
he spends this year?
Let the fraction saved this year be x. Therefore the fraction spent = 1-x.
Next year the money he has to spend = (1+r) for each dollar saved = x*(1+r)
This should be equal to half the money spent this year => x*(1+r) = (1-x)/2
=> 2x+2rx = 1-x => x(2+1+2r) = 1 => x = 1/(3+2r).
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by GMATGuruNY » Sun Jul 08, 2012 2:54 am
'manpreet singh wrote:Q.1)For any positive integer n, the sum of the first
n positive integers equalsn(n+1)/2 . What is the sum
of all the even integers between 99 and 301 ?
(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150
I posted an explanation here:

https://www.beatthegmat.com/for-any-posi ... tml#361604
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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by GMATGuruNY » Sun Jul 08, 2012 2:56 am
'manpreet singh wrote: 2)This year Henry will save a certain amount of his
income, and he will spend the rest. Next year Henry
will have no income, but for each dollar that he saves
this year, he will have 1 + r dollars available to spend.
In terms of r, what fraction of his income should Henry
save this year so that next year the amount he has
available to spend will be equal to half the amount that
he spends this year?
I posted a solution here:

https://www.beatthegmat.com/official-gui ... 61258.html
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion

by \'manpreet singh » Tue Jul 10, 2012 8:04 am
Thanks Eagle and Mitch .
But I feel it is tough to immediately take the algebraic route.
So I think the plug and chuck method is the best for this case.
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