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Comment on Avoiding Common Mistakes - Part II
At 3:25 in the example for no
In that case, no. of boys would be 0 and hence the combined statement also becomes insufficient.
Great question.
Great question.
In this instance, the number of students can be assumed to be greater than 1, since the sentence states that there is a class of "students" (plural)
is it generally safe to
In MANY real-world questions,
In MANY real-world questions, the numbers are typically positive integers. That said, it isn't ALWAYS the case. For example if we're talking about the time it takes for a pump to fill a pool, we can have non-integer values for the time it takes, the rate at which the pump pumps water and the volume of water in the pool.
Hi Brent, in part 3 resource,
https://www.beatthegmat.com/gmat-prep-donuts-mmmmmmm-t11773.html
Question link: https:/
Question link: https://gmatclub.com/forum/at-the-bakery-lew-spent-a-total-of-6-00-for-o...
Although this is an official GMAT question, the wording is a little bit ambiguous. The question does not say that exactly 1 donut at exactly 1 cupcake were purchased.
Here's my full solution: https://gmatclub.com/forum/at-the-bakery-lew-spent-a-total-of-6-00-for-o...
Cheers,
Brent
Dear Brent. Please help me to
From the articles Common Data Sufficiency Mistakes – Part II and Par III there are practice questions which confuse me. From these two questions I have one question “what is the easiest and fastest way to find values of lets x and y?”
1 question: https://www.beatthegmat.com/confused-on-the-solutions-given-in-og-13-t179210.html
Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamps did she buy?
(1) She bought $4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps and $0.29 stamps.
0.15x + 0.29y
X=?
Statement 1: 0.15x + 0.29y=4.40 or 15x+29y=440
I automatically assumed that there are several values where x and y as integers. I didn’t understand in explanations how they come up with the answer 10. Is there an easier way to tackle such statements?
2 question: https://www.beatthegmat.com/question-harder-for-ds-t39645.html?cityevent=Miami-13-648
Marta bought several pencils, if each pencil was either a 23 cent or 21 cent pencil, how many 23 cent pencils did she buy?
1) marta bought 6 pencils
2) the total value of the pencils she bought was 130 cents.
23y + 21x
Y=?
Statement 1: multiples values to get 6 pencils > Insufficient
Statement 2: 23y + 21x=130 > Sufficient
Here I immediately recognized that I can “find” values of y and x but I am not sure which way is faster and easier. Is there any uniform approach? We cannot just assume automatically that there is only one possible value for x and y, can we?
Great questions, Yulia!
Great questions, Yulia!
Question #1 (with my solution): https://gmatclub.com/forum/joanna-bought-only-0-15-stamps-and-0-29-stamp...
Unfortunately there's no easier way to solve this question. It requires us to check several possible scenarios.
The key takeaway here, however, is that producing a linear equation with two variables does not necessarily mean the statement is insufficient.
Please review my full solution (linked above) and let me know if you have any questions.
----------------------------
Question #2 (with my solution): https://gmatclub.com/forum/martha-bought-several-pencils-if-each-pencil-...
Notice that statement 2 tells us that Martha spent only $1.30.
Since pencils cost around 25 since each, we can see that there aren't many possible outcomes to consider.
Given this, it's conceivable that there may be only one valid solution.
The same applies (to a lesser extent) to question #1 above.
Consider this analogous question: If x and y integers such that 0 < x < y, what is the value of x?
Statement 1: x + y = 500
Since the sum equals a big number (500), we can immediately see that there are MANY MANY possible solutions to this equation, which means there are many possible values of x.
So statement 1 is insufficient
Now consider what happens if we change statement 1 as follows:
Statement 1: x + y = 4
Since the sum is small (and since x and y are positive integers), there aren't many possible solutions,
In fact, only one solution (x = 1 and y = 3) satisfies the given information.
So when it comes to questions of this nature, it all comes down to the number of possible solutions.
Does that help?
Thank you Brent. I wish the
Thank you for explanations. Over the night I thought of a different approach. What if in such cases we look at unit digits. Ex. 23y + 21x=130 > to get zero we need 3x2 + 1x4 = ..0
Same story with 15x+29y=440, only by multiplying with 10 we can get zero digit of 440.
I also understood your solution. Thank you.
Thanks Yulia! I'm glad you
Thanks Yulia! I'm glad you like the videos!
You're absolutely right. Examining the units digits is a great way to go.
By the way, with 15x + 29y = 440, we can get a units digit of 0 if x and y are multiples of 10, OR if y is a multiple of 5.
For example, if y = 5 and x = 3, we get: 15(3) + 29(5) = 45 + 145 = 190
Cheers,
Brent
Brilliant video Brent and you
Thanks for the kind words!
Thanks for the kind words! You made my day.