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Comment on Testing the Answer Choices
Great video!
Seems like that would eat up
There are instances when
There are instances when testing the answer choice can actually be faster than more conventional approaches.
Also, there may be times when you have no idea how to solve a math question using conventional techniques. In these instances, you can either skip/guess the question and move on, or you can start testing the answer choices.
In some cases, it may make more sense to skip/guess the question and use your time on other questions. In other case, it may make more sense to take the time to test the answer choices.
The "right" approach will depend on several factors, including your target scores and how much time you have remaining.
Nice way to arrange the
The most important thing is
Good point!
Good point!
Can you solve it by the
Which of the two questions
Which of the two questions (in the above video lesson) would you like me to answer?
Cheers,
Brent
2nd question
Let C = # of cars at
Let C = # of cars at dealership
Let T = # of trucks at dealership
So, equation #1: C + T = 100
Now let's deal with the info about the used vehicles.
(1/2)C = # of used cars
(1/3)T = # of used trucks
So, equation #2: (1/2)C + (1/3)T = 42
Multiply both sides of equation #2 by 6 to get: 3C + 2T = 252
Multiply both sides of equation #1 by 3 to get: 3C + 3T = 300
Subtract bottom equation from top equation to get: -T = -48
So, T = 48
Answer: D
Cheers,
Brent
Hi dear
Can you help me to solve this using table
Cars and trucks over used and unused cards
I'm not sure I understand
I'm not sure I understand your question.
I solve the question in the video using a table.
https://gre.myprepclub.com
Hey Brent is this appreacoh right?
B = J + 8
Gives 2 cars to J + 8 - 2 he will have twice cars
so B = 2(J + 6)
J + 8 = 2J + 12
J = 4
But to Get B we should remember B - 2 = 2J
B = 2(4) + 2 = 10
Question link: https://gre
Question link: https://gre.myprepclub.com/forum/bobby-has-eight-more-cars-than-jackie-i...
You're close but....
Let J = the number of cars Jackie has INITIALLY
Let B = the number of cars Bobby has INITIALLY
So, if Bobby has eight more cars than Jackie, we can write: B = J + 8
Given: If Bobby gives two of his cars to Jackie, Bobby will have twice the cars that Jackie has.
If Bobby gives 2 cars to Jackie, then....
J + 2 = the number of cars Jackie has NOW.
And B - 2 = the number of cars Bobby has NOW.
So, if Bobby NOW has twice the cars that Jackie has, we can write....
B - 2 = 2(J + 2)
I'll let you solve the system of equations from here.
Does that help?
Hey Brent, thanks for this.
You bet!
You bet!
Our two equations are:
(1) B = J + 8
(2) B - 2 = 2(J + 2)
Take equation (2) and replace B with J + 8, from equation (1)
When we do this we get: (J + 8) - 2 = 2(J + 2)
Simplify both sides of the equation: J + 6 = 2J + 4
Subtract 4 from both sides: J + 2 = 2J
Subtract J from both sides: 2 = J
Since we now know that J = 2, we can plug this value into equation (1) to get: B = 2 + 8, which means B = 10 (answer: E)
Hey Brent, quick question.
Time - 4.02.
Not sure how you arrived at 2/3*10 out of 1/3*32? Can you please elaborate on this?
The mixed number 10 2/3 means
The mixed number 10 2/3 means 10 + 2/3 (not 10 x 2/3)
First, 1/3 x 32 = 32/3
Now we need to convert the fraction 32/3 into a mixed number.
3 divides into 32 ten times with remainder 2.
So, we can say that 32/3 = 10 2/3