What's YOUR favorite GMAT Quantitative shortcut?

[mixpanda]

Hello everyone!

Please post your favorite shortcut for use on the Quantitative section of the GMAT!

Mine?

Rates: If Nina can finish the project in 12 hours and Jacob can finish it in 7 hours, how long would it take them to complete the project together?

1/12 + 1/7 = 1/n

Fancy!

EDIT: I forgot to explain! So the formula is as it is because in one hour, Nina finishes a twelfth of the project, and within that same hour, Jacob finishes a seventh of the project. Those two amounts added together is the portion of the project they would complete if working together. Whatever you figure n to be, that's how many hours it would take!

Some problems go even further and are more tricky. For example:

If a blue robot finishes the work in 15 hours, and an orange robot finishes the work in 10 hours, how many of each robot would it take to finish the project in a total of 3 hours?

For this problem, you would do the following:

Figure out how much of the work would be done in one hour by one pair of the blue and orange robots together:

1/15 + 1/10 = 1/n

Find a common denominator:

2/30 + 3/30 = 1/n

5/30 = 1/n

1/6 = 1/n

n=6

It takes six hours for one pair of robots to finish the project. In order to complete that project in half that time, three hours, it would take two of each robot.

Come on guys, post!

[VP_Jim]

Great idea for a thread!

Easily, my favorite math trick is plugging in values for variables in questions such as:

"If Alex is twice as old as Bob, and if Bob is x years old, how old will Alex be five years from now, in terms of x?"

Those questions come up all the time and are super easy to solve once you get the hang of picking numbers. This approach works on pretty much every question that has variables in the answer choices.

[alexhard]

Hello everyone!

Please post your favorite shortcut for use on the Quantitative section of the GMAT!

Mine?

Rates: If Nina can finish the project in 12 hours and Jacob can finish it in 7 hours, how long would it take them to complete the project together?

1/12 + 1/7 = 1/n

Fancy!

EDIT: I forgot to explain! So the formula is as it is because in one hour, Nina finishes a twelfth of the project, and within that same hour, Jacob finishes a seventh of the project. Those two amounts added together is the portion of the project they would complete if working together. Whatever you figure n to be, that's how many hours it would take!

Some problems go even further and are more tricky. For example:

If a blue robot finishes the work in 15 hours, and an orange robot finishes the work in 10 hours, how many of each robot would it take to finish the project in a total of 3 hours?

For this problem, you would do the following:

Figure out how much of the work would be done in one hour by one pair of the blue and orange robots together:

1/15 + 1/10 = 1/n

Find a common denominator:

2/30 + 3/30 = 1/n

5/30 = 1/n

1/6 = 1/n

n=6

It takes six hours for one pair of robots to finish the project. In order to complete that project in half that time, three hours, it would take two of each robot.

Come on guys, post!

But that's the long way of doing it

Here's a shortcut: set A and B so that if you were to do it the long way, it would be 1/A + 1/B = 1/n
Then, n=AB/(A+B)
So in your example n=150/25=6

[Bara]

What a great thread!

While basic, I like dealing with percents the way I deal with sales:

If an item is 40% off, then I'm paying 60% - - so going directly for the 60% instead of the 40% then subtracting from the whole.

Developing a kind of math literacy (see: The Kingdom of Infinite Number: A Field Guide by Bryan Bunch) is the way to go.

[ladistar]

Bara, I've been looking for an outside source that can help with number properties (aside from the MGMAT book and OG). Is the "Kingdom" a good text for understanding number properties? Seems so from the description on Amazon.

[Bara]

The book does a good job. I'm also going through some Ian Stewart books now (not the Ian Stewart on here): The Magical Maze and Nature's Numbers. I don't have feedback to give yet.

I think developing math literacy is super important for the GMAT. It will allow you to move through the Quant with a sexy stealth. I also encourage you cover the basics that you'll be tested on not everything that is known to mankind with respect to number properties.

If you feel like you've covered it, and mastered it, been there and done that, but still want assurance, you might want an hour with a tutor to seal the deal and give you a hecher.

(A hecher is that symbol that denotes something is kosher, or ritually approved - - a Jewish stamp of approval)

[beatthegmat]

Moving this thread to the GMAT Math section.

@mixpanda: Thanks for starting this.

[zeal_huyi]

You should check out the GMAT Math Sheet. It got some neat concepts and techniques.

[cfarrera]

Great Idea!!!!
Here is one of my favorite shortcuts:

Different Divisors contained in 2100?

1. You breakdown the number into prime factors always starting with the smallest: 2100/2, 1050/2, 525/3,175/5,35/5,7/7
2. Now you have the following prime factors: 2,2,3,5,5,7
3. 2^2,3^1,5^2,7^1
4. Now you take the exponents and add 1 and multiply them
(2+1)(1+1)(2+1)(1+1)= 36
5. There are 36 different divisors in 2100

Voilá

Also you can use this formula when asked for the # of unique factors

[doclkk]

Great idea for a thread!

Easily, my favorite math trick is plugging in values for variables in questions such as:

"If Alex is twice as old as Bob, and if Bob is x years old, how old will Alex be five years from now, in terms of x?"

Those questions come up all the time and are super easy to solve once you get the hang of picking numbers. This approach works on pretty much every question that has variables in the answer choices.

This is great and a technique that I studied this week for my MGMAT class! Thanks!

[Svedankae]

Great Idea!!!!
Here is one of my favorite shortcuts:

Different Divisors contained in 2100?

1. You breakdown the number into prime factors always starting with the smallest: 2100/2, 1050/2, 525/3,175/5,35/5,7/7
2. Now you have the following prime factors: 2,2,3,5,5,7
3. 2^2,3^1,5^2,7^1
4. Now you take the exponents and add 1 and multiply them
(2+1)(1+1)(2+1)(1+1)= 36
5. There are 36 different divisors in 2100

Voilá

Also you can use this formula when asked for the # of unique factors

that is sick!! where did u pick that up?

[deirdre]

One good one is that xC1 = xC(x-1) = x

This can save time and scribblings in combinations questions.

[cholloway]

What a great thread!

While basic, I like dealing with percents the way I deal with sales:

If an item is 40% off, then I'm paying 60% - - so going directly for the 60% instead of the 40% then subtracting from the whole.

Developing a kind of math literacy (see: The Kingdom of Infinite Number: A Field Guide by Bryan Bunch) is the way to go.

Thanks for the tip. Just ordered it off Amazon. Had to purchase from another seller though b/c Amazon does not have any in inventory.

[Zohebh786]

Great Idea!!!!
Here is one of my favorite shortcuts:

Different Divisors contained in 2100?

1. You breakdown the number into prime factors always starting with the smallest: 2100/2, 1050/2, 525/3,175/5,35/5,7/7
2. Now you have the following prime factors: 2,2,3,5,5,7
3. 2^2,3^1,5^2,7^1
4. Now you take the exponents and add 1 and multiply them
(2+1)(1+1)(2+1)(1+1)= 36
5. There are 36 different divisors in 2100

Voilá

Also you can use this formula when asked for the # of unique factors

So for 2200, according to the above rule would there be only 16 factors?

I got the prime factors of 41, 5, 2, 2, 2. Which would make the exponents 1, 1, 3. Thus adding 1, and mutlplying gives you 4*2*2=16?

[deanm85]

I got 24 divisors by that shortcut.

2200/2=1100
1100/2=550
550/2=275
275/5=55
55/5=11
11/11=1

2^3 * 5^2 * 11

(3+1)(2+1)(1+1)=4*3*2=24