BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

what is the effective way of solving this?

Expert replies
by g.shankaran » Sun Jun 12, 2011 2:28 am
if x is equal to the sum of the even integers from 40 to 60, inclusive and y is the number of even integers from 40 to 60, inclusive what is the value of x + y?

1. 550
2. 551
3. 560
4. 561
5. 571
Join the discussion
Source: — Problem Solving |

by Calvin123 » Sun Jun 12, 2011 2:49 am
g.shankaran wrote:if x is equal to the sum of the even integers from 40 to 60, inclusive and y is the number of even integers from 40 to 60, inclusive what is the value of x + y?

1. 550
2. 551
3. 560
4. 561
5. 571
no of even integer between 40 to 60 is> 60-40/2+1 = 11 =y
sum of even integer from 40 to 60 is> 11/2(40+60) = 550 =x

Now x+y= 550+11= 561
ans is 4
Join the discussion

by GMATGuruNY » Sun Jun 12, 2011 3:31 am
g.shankaran wrote:if x is equal to the sum of the even integers from 40 to 60, inclusive and y is the number of even integers from 40 to 60, inclusive what is the value of x + y?

1. 550
2. 551
3. 560
4. 561
5. 571
To determine the sum of evenly spaced integers, use the following formula:

Sum = (number of integers) * (average of biggest and smallest)

To count the number of evenly spaced integers in a set:

Number of integers = (Biggest - Smallest)/(distance between each successive pair) + 1

With even integers, the distance between each successive pair is 2.
Thus, given the even integers from 40 to 60, inclusive:

Number of integers = (60-40)/2 + 1 = 11.
Average of biggest and smallest = (60+40)/2 = 50.
Sum = number * average = 11*50 = 550.

Sum + number = 550+11 = 561.

The correct answer is D.
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 vinayreguri » Sun Jun 12, 2011 3:46 am
Raw Solution:
x=(40+42+44+46+48)+(50+52+54+56+58)+60
y=(60-40)/2(inclusive of 40 & 60)=22/2=11

Now x+y=550+11=561

Ofcourse, This is time consuming.
Join the discussion

by cans » Tue Jun 14, 2011 3:50 am
x=40+42+....+60
y=11
x=(11/2)(100)=550
x+y=561
If my post helped you- let me know by pushing the thanks button ;)

Contact me about long distance tutoring!
[email protected]

Cans!!
Join the discussion

by Tega1984 » Tue Jun 14, 2011 5:18 am
o.k here's my solution. factor out 2 to be left with 2(20+21+22+23...all the way to 30) and divide by 2. 2's will cancel and you're left with 20+21+22.... Now multiply 30 by 31 to get 930. than you should multiply 19 by 20 to get 380 and subtract 380 from 930 to get 550. This 550 is the sum of all even integers from 40 to 60. Than just count the number of even integers from 40 to 60(it's 11) and sum those two numbers. 550+11=561 So the answer is 4.
g.shankaran wrote:if x is equal to the sum of the even integers from 40 to 60, inclusive and y is the number of even integers from 40 to 60, inclusive what is the value of x + y?

1. 550
2. 551
3. 560
4. 561
5. 571
Join the discussion