Search found 228 matches
For a+b to be even, either both a and b must be even or both a and b must be odd. Since we will be choosing 5 possible "a" numbers and 5 possible "b" numbers, the total number of possibilities will be 25. The question is: How many of those will fit the above criteria? We have 3 w...
- by elias.latour.apex
Sat Jan 13, 2018 6:06 am- Forum: Problem Solving
- Topic: If a is to be chosen at random from the integers...
- Replies: 2
- Views: 815
If f(11) = 0 then one of the two terms inside (x-p) or (x-q) must be 0. So let's arbitrarily assign p=11.
Similarly if f(20) = 0 then we can assign q=20.
So what's f(10)? It will be (10-11) (10-20) = (-1) (-10) = 10
- by elias.latour.apex
Fri Jan 12, 2018 5:35 am- Forum: Problem Solving
- Topic: Let f(x)= (x-p)(x-q). If f(11)=f(20)=0, then f(10)=?
- Replies: 4
- Views: 1054
Here's another solution that some may find easier. We have 6 empty seats and three couples. How many people can sit in the first seat? 6 people -- any of the people waiting to be seated. However, once that happens, the person sitting next to him or her must be the other half of the couple, so we hav...
- by elias.latour.apex
Fri Jan 12, 2018 5:32 am- Forum: GMAT Math
- Topic: Permutation
- Replies: 4
- Views: 3969
Well, first we must determine how many possible starting points there are for the A. Since there are 9 possibilities for x, and 6 possibilities for y, there are 54 starting points for the triangle. But don't think that (A) is the answer. That number is far too low. Once we have selected A, we can se...
- by elias.latour.apex
Mon Jan 08, 2018 7:27 pm- Forum: Problem Solving
- Topic: Right triangle ABC is to be drawn
- Replies: 2
- Views: 735
The process of elimination is pretty straightforward. Let's run down the list. A. In all moderately dim light in which people without the disorder can read large print, people with the disorder cannot read such print. It does not say that. In fact, it says that people with this disorder have normal ...
- by elias.latour.apex
Mon Jan 08, 2018 5:14 am- Forum: Critical Reasoning
- Topic: GMATPrep EP1: People with a certain eye disorder
- Replies: 1
- Views: 726
Well, there are two ways I can see to solve this problem. Both involve determining the maximum number of possibilities, and that is what I will start by doing. VERMONT has 7 letters, but we are forming 5-letter combinations. So for the first letter, we have 7 choices. For the second, we have 6 (as w...
- by elias.latour.apex
Sun Jan 07, 2018 5:55 am- Forum: Problem Solving
- Topic: What percent of 5 letter combinations
- Replies: 1
- Views: 625
Imagine that we have 6 marbles in our hand and 4 empty bowls. We select one of the marbles. Where can we place it? In any of the 4 bowls. So we have 4 choices. Now we have the second marble in our hand. Where can we place it? Again, in any of the 4 bowls. So we have 4 more choices. The same goes for...
- by elias.latour.apex
Sat Jan 06, 2018 10:38 am- Forum: Problem Solving
- Topic: How many ways are there of placing 6 marbles in 4 bowls, if
- Replies: 1
- Views: 688
To make things easy, let us assume that our odd numbers are 1 and that our even numbers are 2. AB+C is even. So what if C is 1? Then AB must also be odd. In other words, all three numbers must be 1 (odd). What if C is 2? Then AB must also be even. So AB must equal 2 or 4. So these are our options: 1...
- by elias.latour.apex
Sat Jan 06, 2018 10:35 am- Forum: Problem Solving
- Topic: If A,B,C are three integers and AB+C is even . . . .
- Replies: 3
- Views: 727
The easiest solution is to assume that at the time both pumps started working on the basement, there were 12 units of water in the basement. This is the least common multiple of 4 and 3. Accordingly, the basement had 24 units of water in it when it was full. Since the pump X evacuated 12 units of wa...
- by elias.latour.apex
Sat Jan 06, 2018 10:14 am- Forum: Problem Solving
- Topic: Working at constant rate, pump X pumped out . . . . .
- Replies: 1
- Views: 2525
To have a zero at the end, a number must be multiplied by 10. Another way to think of this is that the number must be multiplied by the prime factors 5 and 2. 135 can be factored into 5 and 27. There are no other factors of 5 or 2. 42 can be factored into 2 and 21. There are no other factors of 5 or...
- by elias.latour.apex
Sat Jan 06, 2018 10:02 am- Forum: Problem Solving
- Topic: How many zeros does 135^19 x 42^17 end with?
- Replies: 1
- Views: 577
Critical reasoning questions are best handled by reading the question first. In this case, we find that the administrators like a certain criterion for eliminating workers. Our job is to show that this criterion is bad. Upon reading the text, we discover that the criterion is to eliminate those work...
- by elias.latour.apex
Fri Jan 05, 2018 6:34 am- Forum: Critical Reasoning
- Topic: Due to recent budget constraints
- Replies: 1
- Views: 563
Dispute
Meaning no disrespect to the Economist GMAT Tutor, but any of a number of answers is possible. It depends entirely on one's understanding of the question. The question states: "... the difference among the first 3 men was 2 [inches]...," a statement that EGT has taken to mean that man 1 is...
- by elias.latour.apex
Fri Jan 05, 2018 5:54 am- Forum: Problem Solving
- Topic: Sequence and Series
- Replies: 2
- Views: 860
This is an extremely controversial question drawn from the Manhattan GMAT CAT tests. Supposedly (A) is the best answer. The reasoning expressed by MGMAT instructors is that (A) offers an alternate explanation for the decrease in profit margins. Supposedly, increased competition is what is really cau...
- by elias.latour.apex
Tue Jan 02, 2018 5:27 pm- Forum: Critical Reasoning
- Topic: Two years ago, the cost of the raw material used in a partic
- Replies: 1
- Views: 830
The argument claims that many German computer users illegally copy software. Why? Because more computers are sold than commercial programs. Is this argument convincing? Not really. Perhaps many German computer users write their own programs, and thus do not need to buy commercial programs. This is a...
- by elias.latour.apex
Tue Jan 02, 2018 6:10 am- Forum: Critical Reasoning
- Topic: A recent survey found that more computers
- Replies: 1
- Views: 574
Statement 1 tells us that the number may be 111 or 111,111 or even 111,111,111. If you know the divisibility rules for 3, then you will know that each of these numbers is divisible by 3. To determine whether a number is divisible by 3, add up the unit digits. If the result is divisible by 3, then th...
- by elias.latour.apex
Tue Jan 02, 2018 6:04 am- Forum: Data Sufficiency
- Topic: A repunit is a positive integer that contains only the digit
- Replies: 2
- Views: 783