|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
24 Jul 2017, 08:51
Kudos
Bookmarks
bindiyajoisher
Four consecutive odd numbers = \((2x-3), (2x-1), (2x+1), (2x+3)\).
Three consecutive even numbers = \((2y-2), 2y, (2y+2)\).
\((2x-3) + (2x-1) + (2x+1) + (2x+3) = (2y-2) + (2y) + (2y+2)\)
\(8x = 6y\)
\(4x = 3y\), so, \(\frac{x}{y} = \frac{3}{4}\).
Also, given: \(101 < 2y < 200\). So, \(2y\) must be divisible of \(8\) between the numbers \(101\) and \(200\), possible values are \(104, 112, 120, ... 192\): \(\frac{192-104}{8} + 1 = 12\). Ans - A.
21 Apr 2023, 03:23
Kudos
Bookmarks
bindiyajoisher
Every time one looks at a question with fresh eyes, one finds a new method of solving.
Since we have a starting point for the middle term of the even numbers, I would say it goes from 102 to 198. Say this is M
Sum of 3 even integers = 3M = Sum of 4 odd integers
Average of 4 odd integers must be an even integer so 3M/4 = Even integer.
This means 3M/8 is an integer which means M is a multiple of 8.
First multiple of 8 in given range is 104 (8 * 13) and last is 192 (8 * 24). Number of multiples = 24 - 13 + 1 = 12
Answer (A)
26 Sep 2023, 06:39
Kudos
Bookmarks
bindiyajoisher
Let the 4 consecutive odd numbers be a, a+2, a+4, a+6 => Sum = 4a+12
Let the 3 consecutive even numbers be b, b+2, b+4 => Sum = 3b+6
=> 4a+12 = 3b+6
=> 3b = 4a+6
=> b = 4a/3 + 2 ... this implies that a must be an even multiple of 3, i.e. a multiple of 6 (since b is even)
=> The middle term = b+2 = 4a/3 + 4
=> 101 < 4a/3 + 4 < 200
=> 97 < 4a/3 < 196
=> 72.75 < a < 147
Thus, we need multiples of 6 between 73 and 147
The first is 78 = 6*13 and the last is 144 = 6*24
Number of numbers = 24 - 13 + 1 = 12
Answer A
