Help with math!

Test Preparation SAT Preparation
1

Hey guys. I am taking the SAT on jan 23( after two weeks) and I need help with some questions.
1,The city library donated some children’s books to Mr.
Clark’s first-grade class. If each student takes 4 books,
there will be 20 books left. If 3 students do not take a
book and the rest of the students take 5 books each,
there will be no books left. How many books were
donated to the class?
(A) 120
(B) 140
© 160
(D) 175
(E) 185

2.Bernardo drives to work at an average speed of 50
miles per hour and returns along the same route at
an average speed of 25 miles per hour. If his total
travel time is 3 hours, what is the total number of
miles in the round-trip?
(A) 225
(B) 112.5
© 100
(D) 62.5
(E) 50
3.There are 6 red, 6 brown, 6 yellow, and 6 gray scarves
packaged in 24 identical, unmarked boxes, 1 scarf per
box. What is the least number of boxes that must be
selected in order to be sure that among the boxes
selected 3 or more contain scarves of the same color?
(A) 3
(B) 6
© 7
(D) 8
(E) 9
4.A carton contains b boxes of paper plates, and each
box contains n plates. If the carton costs c dollars,
what is the cost per paper plate, in dollars, when the
plates are bought by the carton?
(A) cbn
(B) c/bn
© bn/c
(D) cn/b
(E) n/cb

 Also, if you have college board's blue book, 2nd edition, please check out practice 4, section 6 and question # 17,and #15  on page 981. Thank you very much.
2

@ethiololita6 I’ll get you started, but ultimately I want you to solve the questions.

  1. Let S be the number of students and B be the number of books. We have

B = 4S + 20 (each student takes 4 books, 20 books left over)

If 3 students do not take a book, then S-3 students take books. The S-3 students take 5 books each with no remainder, so

B = 5(S-3)

Aha! Two equations, two variables. You can solve for S and B by substituting B.

  1. You can let d be the distance between home and school, in miles. He drives d miles at 50 mph and d miles at 25 mph. So his total time taken is d/50 + d/25 = 3. Solve for d. Remember that the answer is 2d, since that is the total number of miles.

You can also guess at the answers pretty easily.

3. Figure out what the "worst case" is. That is, imagine a case where we pick many boxes but cannot come up with 3 of the same color.

  1. One carton costs c dollars, which buys bn paper plates. How much does it cost for one paper plate?
3

I thought @pckeller would have playfully cracked #1 with his omnipotent numeric proclivity by now, but since that’s not the case, here goes nothing.

If b is the number of books, then the number of students on the one hand is
(b-20)/4, but on the other hand it’s b/5+3.

We can solve the equation
(b-20)/4 = b/5+3
algebraically or in many different ways with a sheer power of the TI family.

Or - we can simply start plugging the given answers into (b-20)/4 and b/5+3 until these two expressions become equal to each other. (Usually this kind of thing done starting with choice C.)
(160-20)/4=35, 160/5+3=35. Bingo!
(Incidentally, we just found the number of students.)
The answer is C indeed.

4

ETA.
I realize that from a number-crunching purist’s point of view this solution is contaminated with algebra, but one could avoid dealing with dirty unknowns by using plain reasoning based on the same idea, substituting ‘b’ with the word ‘books’, and so on.

5

@gcf101 It’s funny that you should mention this! This is one problem that I admit that I used algebra for. But then the students in my SAT class teased me for it – they had solved it by trial and error with the answer choices, even less formally than you did. So it seems that I have been a bad influence.

6

@pckeller - it was gnawing on me that my solution lacked a numeric harmony. And then the revelation came upon me: What would @pckeller do? So I sneak peeked into your SAT mental kitchen, and that was it! (I guess, your students have free passes to that sacred space.) Again:

Let’s pick a number for the number of students and see if it works.
Let it be 30.
30x4 + 20 = 140. (30-3)x5 = 135.

(Since all the answers are multiple of five, and in order to get the number of books we multiply the number of students by 4 and then add 20, 5 should divide - a wink to @MITer94 - the number of students.)

So, we bravely pick the next multiple of 5 to catch up with the increase of 140:
35x4 + 20 = 160. (35-3)x5 = 160.
Done!

7

Just can’t let sleeping thread lie…

@pckeller Trying to recreate your students solving path for #1:
Going through the answer choices,
A: (120-20)/4=25; (25-3)x5=110. 120/=110.
B: (140-20)/4=30; (30-3)x5=135. 140/=135.
C: (160-20)/4=35; (35-3)x5=160. 160=160.
The end.

Your students are so darn smart.

8

@ MITer94
Thanks!I got the correct answer for every question.But, I used xiggi’s formula to find the average of the two speeds on question #2.Then using the total time, I found the total distance.Is my way reliable or risky?

9

@pckeller and @gcf 101 ,STOP TALKING ABOUT MY QUESTIONS !!!

10

@ethiololita6 Getting a bit possessive, eh? :slight_smile:

It’s actually very helpful to learn different methods of solving the same question: you never know when one of them may come in handy.

PS
It’s all my fault; @pckeller just made a small comment!

11

Non-algebraically:
Under the given scenario (total distances driven at 25 mi/hr and at 50 mi/hr are equal) the ratio of the total time on all 25 mi/hr segments to the total time on all 50 mi/her segments is an inverse ratio of the corresponding speeds, i.e. 50:25, or 2:1
Breaking the total travel time of 3 hrs at the ratio of 2:1, we see that Bernardo drove 2 hours at 25 mi/hr and 1 hour at 50 mi/hr; the total distance then is 25(2)+50(1)=100.

12

@ethiololita6 isn’t the average speed just the harmonic mean of 50 and 25?

If you know the average speed round-trip, and the total time, then you know total distance.

13

Well ok, but…

You could also use trial and error. No kidding, I started with c. @DrSteve always says to start with c but I just picked it because it looked easiest. Then:

If the round trip is 100 then each way is 50. Going at 50 mph that takes one hour. Coming back at 25 mph takes 2 hours. Total time: 3 hours.

14

Just to clarify my stance on “Starting with choice ©,” my advice is the following:

In many SAT math problems you can get the answer simply by trying each of the answer choices until you find the one that works. Unless you have some intuition as to what the correct answer might be, then you should always start in the middle with choice © as your first guess. The reason for this is simple. Answers are usually given in increasing or decreasing order. So very often if choice © fails you can eliminate one or two of the other choices as well.

Exceptions: If the word least appears in the problem, then start with the smallest number as your first guess. Similarly, if the word greatest appears in the problem, then start with the largest number as your first guess.

Note: On the revised SAT there are only 4 choices, so I have changed the name of this strategy to “Starting with choice (B) or ©.”

15

@DrSteve ^Yay for binary search. O(log n) running time.

Starting with © definitely works for problem 2. since you can tell whether your guess is too high or too low, but it is also easy to get into the pitfall of doing 100/50 = 2, 100/25 = 4, 2+4 = 6, and erroneously picking (E) 50. I think as long as you are careful, whichever solution works for you is fine.