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SAT Math Problems Thread

DrSteveDrSteve Registered User Posts: 1,289 Senior Member
I thought I would start a thread of math problems that students have asked me about while preparing for their SAT. After I post each question I'll give some time for you to post your attempted solutions before providing my own. I'll try to provide the Difficulty Level and Topic for each one. Here is the first one:

Level 3 Number Theory

1. If an integer n is divisible by 15, and 25, what is the next larger integer divisible by these numbers?

(A) n + 15
(B) n + 50
(C) n + 75
(D) n + 125
(E) n + 150
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Replies to: SAT Math Problems Thread

  • CandyPants16CandyPants16 Registered User Posts: 133 Junior Member
  • DrSteveDrSteve Registered User Posts: 1,289 Senior Member
    That's correct. Can you tell us how you did it?
  • Sparkkid1234Sparkkid1234 Registered User Posts: 124 Junior Member
    edited April 2014
    I have a fast yet somewhat ridiculous solution here. I just find a common number that is divisible by 15 and 25 see if any fits the letter from A to E
  • meumeumeumeu Registered User Posts: 96 Junior Member
    edited April 2014
    Since it is a level 3 problem, one should expect that it wont be that easy and any answer that attracts you instantly is probably a wrong answer.For example, n+50 seems attempting for me since it is the next larger integer that is a multiple of 5.

    How did I do it?

    GCF x LCM= 15 x 25

    GCF is for sure equal to 5.


    (25 x 15)/ 5 = 75 ,so the answer is C.

  • DrSteveDrSteve Registered User Posts: 1,289 Senior Member
    edited April 2014
    @Sparkkid1234 Not a ridiculous solution at all. You are simply taking the lcm of 15 and 25 which is 75 - a nice quick and elegant solution.

    @meumeu Your solution is similar. You didn't have to do quite so much work. Computing the lcm directly is just a bit quicker than computing both the product and the lcm.
  • DrSteveDrSteve Registered User Posts: 1,289 Senior Member
    This problem can also be done by picking numbers. If you're comfortable finding the lcm, then you can use 75 for n. If not, the product 375 will work as well.
  • DrSteveDrSteve Registered User Posts: 1,289 Senior Member
    Level 3 Number Theory

    2. Dana has pennies, nickels and dimes in her pocket. The number of dimes she has is three times the number of nickels, and the number of nickels she has is 2 more than the number of pennies. Which of the following could be the total number coins in Dana’s pocket?

    (A) 14
    (B) 15
    (C) 16
    (D) 17
    (E) 18
  • CHD2013CHD2013 Registered User Posts: 2,528 Senior Member
    d = 3n, n = 2p so the total number of coins has to be a multiple of 9 (the minimum is 1 penny, two nickels and 6 dimes). The only multiple of 9 in the answer set is e) 18.
  • MITer94MITer94 Registered User Posts: 4,747 Senior Member
    ^ The problem says the number of nickels is 2 more than the number of pennies (not 2x more).

    Letting p be the number of pennies, then the number of coins is p + (p+2) + 3(p+2) = 5p + 8. The number of coins must be 3 (mod 5) (i.e. 3 more than a multiple of 5) so (E) 18 is the answer.
  • CHD2013CHD2013 Registered User Posts: 2,528 Senior Member
    ^ Thanks. I agree with you as far as 5p + 8. At that point, wouldn't it make sense just to plug in numbers for p. Starting with p=1 or total coins = 13. Since that is not an answer option move on to p=2 and total coins = 18. 18 is answer E) so you're done.
  • DrSteveDrSteve Registered User Posts: 1,289 Senior Member
    I prefer to avoid algebra altogether and simply take a guess for the number of pennies. If we guess that there are 2 pennies, it follows that there are 4 nickels and 12 dimes. So the total is 2+4+12=18, choice (E).

    If your initial guess is wrong simply adjust your guess up or down accordingly.
  • meumeumeumeu Registered User Posts: 96 Junior Member
    ^^ That's genius.However, the answers may come as variables not numbers.
    When you practice any question, it is generally better to know how to do it in 2 ways.
  • CandyPants16CandyPants16 Registered User Posts: 133 Junior Member
    Yeah, I always just plug in numbers that meet whatever is stipulated in the question.
  • CHD2013CHD2013 Registered User Posts: 2,528 Senior Member
    ^Plugging in is a great strategy, but it doesn't always make sense. Some problems don't lend themselves to plugging in and other problems can be solved easier using other strategies.
  • SATManiac98SATManiac98 Registered User Posts: 66 Junior Member
    edited April 2014
    Here's another one:

    3 men and 3 women stand in 1 line. Two or more of the same gender cannot stand next to each other. How many different arrangements are possible?
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