Quantitative Reasoning Question...HELP!!

[frogger33]

Ok, so I am trying to learn this math stuff for the DAT since I got a 14 on QR and 19 and aboves on everything else my first time. Here is my question if someone can PLEASE help me!

I can figure out the problem if I add all the different possible ways out in my head, but I literally spend 5 minutes doing it. Is there a quicker way to do this problem??

A store sells candies for $.03 and another brand for $.04. How many combinations of these two brands can a boy buy and recieve no change from a quarter?
A. 0
B. 1
C. 2
D. 3
E. 4



Thanks guys!!

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[denty]

yuck!
math sucks! i hated that section that most. dont do like I did and delay until you take the exam to study. I had the same problem and was sooo rusty with math, seriously my crutch the calculator..*sigh* just do as many problems as you can. sorry i cant help you with this problem. but also search the forums there were looots of people that asked quant questions onver the summer.
best of luck

 

[frogger33]

Yeah, thanks. I am taking the DAT at the end of january so I am TRYING to figure this stuff out!!

 

[dent_girl]

is that right? 2 combo?

 

[psiyung]

Ok, so I am trying to learn this math stuff for the DAT since I got a 14 on QR and 19 and aboves on everything else my first time. Here is my question if someone can PLEASE help me!

I can figure out the problem if I add all the different possible ways out in my head, but I literally spend 5 minutes doing it. Is there a quicker way to do this problem??

A store sells candies for $.03 and another brand for $.04. How many combinations of these two brands can a boy buy and recieve no change from a quarter?
A. 0
B. 1
C. 2
D. 3
E. 4



Thanks guys!!

I'm sure there is a mathematical formula for your problem, but if you take it one step at a time and write down everything, this problem wont take more than a minute. Again you have to think quickly

1) .03(1) = .03 -- .25-.03 = .22 (not divisible by 4 -- wont work)
2) .03(2) = .06 -- .25-06 = .19 (not divisible by 4 -- wont work)
3) .03(3) = .09 -- .25-.09 = .16 -- divisible by 4 --WORKS!
4) .03(4) = .12 quickly think 25-12= 13 which wont work
5) .03(5) = 15 think 25-15 = 10 wont work
6) .03(6) = 18

Now do you see the pattern that formed. It is just adding by 3 cents
so the next will be 21 and after that 24

25-21 = 4 so this will work

I believe 2 is the answer

 

[psiyung]

You can see the answer reducing by 3 cents as well from 22, 19, 16, 13, 10, 7, 4, 1, etc...When you're trying to solve DAT problems, the formula wont always hit you right there. It's times like this you have to accomadate and use separate ways.

 

[frogger33]

Oh.... Thanks guys

I Never thought of the problem that way! That really helps. Thanks sooo much. I guess I still need lots of work!!

 

[gabrielmeraz]

What is the big deal with the 14 on QR?
From what I've heard in the forum and even from my PPH advisor is that if there is a place to screw up on the DAT is the QR section. Somehow, RC takes more prescedence over math. I tell you this because I too got a 14 on the math section and 19 and above on everything else (well, except PAT 17), and I have been encourage to apply and not go through the pain. What is your GPA? because, dude! If you can avoid taking the test again it will be to you benefit.

Ok, so I am trying to learn this math stuff for the DAT since I got a 14 on QR and 19 and aboves on everything else my first time. Here is my question if someone can PLEASE help me!

I can figure out the problem if I add all the different possible ways out in my head, but I literally spend 5 minutes doing it. Is there a quicker way to do this problem??

A store sells candies for $.03 and another brand for $.04. How many combinations of these two brands can a boy buy and recieve no change from a quarter?
A. 0
B. 1
C. 2
D. 3
E. 4



Thanks guys!!

 

[frogger33]

Yeah, the problem is that some schools have a minimum score for all sections. I would like to go to my state school (UF) bcuase it is much cheaper, but they reqiure at least a 15 on all sections.... therefore, here I am, dreading this test like no other one again. grrrr

 

[gabrielmeraz]

Yeah, the problem is that some schools have a minimum score for all sections. I would like to go to my state school (UF) bcuase it is much cheaper, but they reqiure at least a 15 on all sections.... therefore, here I am, dreading this test like no other one again. grrrr

Can your GPA make up for it?

 

[frogger33]

I wish!!

I have a 3.8 GPA. I had an interview and all, but I was told that I would have to retake the DAT and get a 15 and above on ALL sections.... so sad.

Some schools do not care (nor have a minimum), and others have a minimum of 13 or 15 on all sections. Unfortunetly for me, UF has a minimum requirement of 15 to be accepted. I only applied to two schools, so I figure I really have no choice now. I am just worried that I will do worse this time around b/c it is hard to motivate myself again.

 

[gabrielmeraz]

Well, good luck dude!

I wish!!

I have a 3.8 GPA. I had an interview and all, but I was told that I would have to retake the DAT and get a 15 and above on ALL sections.... so sad.

Some schools do not care (nor have a minimum), and others have a minimum of 13 or 15 on all sections. Unfortunetly for me, UF has a minimum requirement of 15 to be accepted. I only applied to two schools, so I figure I really have no choice now. I am just worried that I will do worse this time around b/c it is hard to motivate myself again.

 

[luder98]

psiyung's way will always work.

For this particular problem there's a "faster" way to solve it if you notice the intention of the test creator. 25 = 5^2 = 3^2 + 4^2. So, the problem can be re-written as follows:

3x + 4y = 3*3 + 4*4 (x and y are the number of the coresponding candies)

or 3(x-3) = 4(4-y)

Normally, if you have n unknowns, you need n equations to solve the problem. In this particular case, you only have one equation with two unknowns. So, there must be another piece of information. Since x and y are the numbers of candies, they must be positive integers. So, the only x and y that satisfy the equation are the ones that make:

1. Both sides of the equation equal to 0: x-3=4-y=0 or x=3 and y=4

Or

2. 4-y divisilbe by 3 and positive (1 and 2 will not work for x). Therefore, the only y is 1, resulting x=7.

The key is to break the equation into products of primes. IMO, this is one of those problems that you want to mark and come back later when time is allowed.

 

[wimmcs]

Best advice for math is to buy acethedat.com and study math and PAT... believe me, you will thank me after you study this section and you take the DAT. I studied this section (the science sections sucks!!) the night before the DAT and I can tell you that I had at least 6 or 7 ?s that were almost word for word!! It was unbelievable!!! Needless to say, I was thankful someone gave me this hint prior to taking the DAT. This is just a "pay it forward" kind of thing. Hope it works as well for you as it did for me.

 

[aar44]

For this question, start with the maximum number of 3¢ candy the boy can buy. Then subtract a piece of 3¢ and see if the remaining change will divide by 4. If it does, then the purchase is exactly equal to $0.25. The boy can buy a maximum of eight 3¢ candies (with a penny left over) for a quarter. Remember, you want no remainder, so the combination of eight 3¢ and zero 4¢ candies does not work. Seven 3¢ candies cost $0.21 and leave a remainder of four cents, which will buy one 4¢ candy, so seven 3¢ candies and one 4¢ candy cost a total of exactly $0.25. That’s one combination that works. If you try six 3¢ candies, you have 7 cents left, which leaves a remainder when divided by 4 cents. Five 3¢ candies leaves 10 cents remaining, and 10 cents also has a remainder when divided by 4 cents. Four 3¢ candies doesn’t work, as it leaves thirteen cents remaining. Three 3¢ candies leaves 16 cents remaining, and 16 is divisible by 4. The combination of three 3¢ candies and four 4¢ candies is the second combination to work. If you try the other possible combinations in the same way, you’ll see that they don’t work. Thus, there are two possible combinations and choice (C) is correct.

This is the way you have to think about this problem.