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SAT Math Problems Thread
DrSteve
1266 replies24 threadsRegistered User 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
636 replies 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
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.
@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.
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
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.
If your initial guess is wrong simply adjust your guess up or down accordingly.
When you practice any question, it is generally better to know how to do it in 2 ways.
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?
4! x 3! = 144 arangements
So there are 3! ways to arrange the males, 3! ways to arrange the females, and 2 possible ways to configure the males and females. By the counting principle, we get 2*3!*3! = 72
(As an example of the error, the configuration _M_MM_ has two males next to each other)
Pckeller's method gives 6*3*2*2*1*1 = 72 as well.
Now here is a very nice, challenging probability question:
A store sells surfboards that are made so perfectly that for every 1000 surfboards, 1 is bad. An inspector comes into the store and conducts a "99% accurate" inspection to all 1000 surfboards to see if surfboards are good or bad, based on the inspection. If my surfboard is inspected "bad," then what is the probability (to the nearest 1%) that my surfboard is actually bad?