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Comment on Marge’s Chocolates
Another way to solve this:
You can also solve this problem with a little bit of number sense. If she ate 100 chocolates over 5 days, that means she averaged 20 chocolates per day. Since the increment from one day to the next remained the same (3), then we know that on the middle day (day 3) she ate the average number of chocolates (20). Therefore, on day 4 she ate 23, and day 5 she ate 26. The extra chocolates eaten on day 4 and 5 is balanced out by subtracting 3 chocolates on day 2 and 3 less chocolates on day 1.
14,17,20,23,26
Perfect reasoning!!
Perfect reasoning!!
NOTE TO OTHERS: This solution uses a nice property that's covered in the Statistics module.
Here's the lesson that covers the property: https://www.greenlighttestprep.com/module/gre-statistics/video/804
This is good! You applied
Me in real life
:-)
:-)
Hello Brent,
I assumed chocolate eaten each day as
c-6, c-3, c, c+3 and c+6 for each consecutive day.
I think, this minimizes some time for calculation, and also is effective for higher number problems.
Is this method applicable always for questions like this?
Regards,
Anil
That's a great idea. When you
That's a great idea. When you add all 5 days, the constant terms (-6, -3, 3 and 6) add to zero, so we get: 5c = 100
This approach probably saves only a few seconds so, on test day, I wouldn't spend too long trying to force it upon your solution.
I'd say the strategy only works for a very specific handful of word problems.
Cheers,
Brent
Brent,
How do you identify problems where it would be faster to solve by testing solutions?
Great question, Jared!
Great question, Jared!
As soon as I see a word problem, I immediately consider testing answer choices as Plan A.
Then, I give myself another 15 to 20 seconds to identify a different (possibly faster) approach.
If I'm able to identify another approach, I need to quickly determine whether it will be faster than testing the answer choices.
To do so, I consider the number of steps it will take to test each answer choice as well as the number of steps with the approach.
All of this needs to take place in a very short period of time. In some cases, it will be easy to determine the faster approach, and in other cases, it won't be so easy to do so.
As you can imagine, we don't want to spend a lot of time figuring out which approach is faster. I mean, it makes no sense to waste 60 seconds determining the faster approach if that faster approach is only 10 or 15 seconds faster than the other approach.
Why can't be solved such that
C + (C+3) + (C+3) + (C+3) + (C+3) = 100?
I got different solution?
In your solution, Marge age:
In your solution, Marge age:
- C chocolates on day 1.
- C+3 chocolates on day 2.
- C+3 chocolates on day 3.
- C+3 chocolates on day 4.
- C+3 chocolates on day 5.
This does not comply with the condition that says, "Each day she ate three chocolates more than on the PREVIOUS day."
So, for example, on day 3, Marge must have eaten 3 chocolates more than on day 2. However, in your solution, she ate same number of chocolates (C+3) on both days
Likewise, on day 4, Marge must have eaten 3 chocolates more than on day 3. However, in your solution, she ate same number of chocolates (C+3) on both days
Etc.
Does that help?
I worked at Lindt chocolates