hi, this is quincy again, i came across more unfamiliar questions while practicing....
1. approxmating#1: 39x15x17(i read all of the approxmation problems on your website...but this one..?)
approxmating#2:20x16x18=?
2. how many 2-member committees can be formed from a group of 7 people? (i thought you do 7x6=42, but the answer key had something different)
3. the simplified coefficient of the third term in the expansion of (2x+y)^6 is? (.......no idea)
4. the area of a triangle whose base is 6 times its height h is k(h)^2 and k=? (i was able to find the answer after spending 15 min afterwards...any easier way to do this?)
5. how many distinct 5 letter words, real or imaginary, can be made using the letters s,c,o,t,t,? (answer key says 60, i thought it was 6x5x4x3x2x1)
6. how many integers between 3 and 30 are relatively prime to 30?(i know how to find relatively prime, in this case, it's 1x2x4=8, but i don't know the answer when it asks "between x and y" because i don't know what should i subtract from 8, concept problem...)
7. the vertex of the parabola y=3x^2+12x+4 is (h,k) and k=? (again...i could find it after spending 15 min on it....but...i don't know if there is a trick to it)
8. the smallest root of 2x^2 +13x+20=0 is..(all i can see is quadratic formula...but it takes too long...)
9. 49=x^2+3y^2, x and y are positive integers, then y=?
lol, sorry to ask lots of questions at once...i am being a nerd here and practicing UIL numbersense 24/7 during spring break.....
Wow...I can see you have been working. I will just tell you the ones I know right off the bat and only one at a time. The others I will look up and get back to you. It has been 8 years since I have done some of these so it will take a little bit of research. And some are new. I have never seen some of them.
Well, let's tackle the first problem. The trick is you have to recognize what to do pretty fast. When I saw 39 * 15 * 17, the first thing I looked for was a way to multiply 2 numbers so I could get a multiple of 100. The easiest way to do that is to change the problem to 40 * 15 * 16 (notice I added one and subtracted one from 39 and 17 respectively.) It is really easy now: 40*15*16 = 600*16 = 9600. This should be enough to get you within 5%.
The next one is 20*16*18. Definitely, the easiest way to do this is to know that 18^3 = 5832. If you don't know that then it takes a little more time. Not much though. You can leave the 20 because it is easy to multiply. So let's look at an easy way to multiply 16*18. Change this to be 14*20 (subtract 2, add 2). Now we have 20*14*20 = 5600. That's easily within 5%.
thanks...oh man!! i should've recognized the 2nd estimation problem right away(18 cubed)!!! that should be a slap for me...16x18x20...i guess i'll have to practice my analyzing skills some more.....
#2 - This involves a background in probability and statistics. So I will do my best.
When we are dealing with problems like this try to think through it logically first. There is only one question that needs answered: Does the order matter? Let's look at the problem:
We have 7 people: A, B, C, D, E, F, and G
We are making groups of 2: AB, AC, FG, etc.
Does the order matter? Can I have AB and BA and count this as 2 groups or is this just one group. In this problem it is only one group because they are the same people, so order does matter. That means we have to use this formula C(7,2) which is calculated by:7!/[(7-2)!*2!] or 21. If order does not matter we use P(7,2) which is calculated by 7!/(7-2)!. Does that make any sense?
In general the formula for C(n,r) is given by n!/[(n-r)!*r!] and P(n,r) is given by n!/(n-r)!. You will use one of these two formulas any time you are dealing with combinations or permuatations. Let me know if I need to be more clear.
#3 - This also involves probablility and is much more complex and it takes lots of practice to be able to do this quickly.
In general if you have (ax + by)^n and you want the rth term the formula is : C(n,r-1)*a^(n-(r-1))*b^(r-1). This looks scary but on number sense tests this is usually fairly easy. Also if it is (ax - by)^n, the formula is the same except every even number term will be negative.
Let's look at some examples:
Ex [1] (2x + y)^6. The coefficient of the 3rd term is:
C(6,2) = 15.
2^(6-2) = 16.
1^2 = 1.
15*16*1 = 240. The answer is 240.
Ex [2] (x - 3y)^4. The coefficient of the x^3y term is:
The first thing to do is find out what term this is. It is the 2nd since the 1st term will be x^4, 2nd is x^3, 3rd is x^2, etc.
Remember the term is going to be negative since we are subtracting and the term we are looking for is even.
C(4,1) = 4.
1^(4-1) = 1.
3^1 = 3.
The answer is -[4*1*3] or -12.
I hope this helps. Practice a lot on these. Notice if a or b is 1, you can ignore that step. In Ex[1] ignore step #3, in Ex[2] ignore step #4.
I can only answer one more question right now because I have to go. I will try to get the rest as soon as I can. I will look up question #4. Let's look at question #5.
How many 5-letter words real or imaginary can be made using S, C, O, T, T. I think you had the right idea, except it is not 6! but 5!. So it is 5*4*3*2*1. But that is not all. If you look, you have 2 T's. This changes everything. If the letters were all distinct, the answer would be 5!. Since we have 2 T's we must divide by 2! or 2. So the answer is 5!/2! or 120/2 = 60.
If the problem said S, S, O, T, T then the answer is 5!/[2!2!] or 30. Does this help any?
Let me know if you need me to explain in greater detail.
I just got more time so I can answer at least one more. Let's look at question #4.
#4 - I have never seen this one before but it is not hard at all. It uses basic geometry.
The area of a triangle is given by A = 1/2*b*h where b is base, and h is height. If we know that the base is 6 times the height then we have another formula: b = 6h. Substituting above we get A = 1/2*6h*h = 3h^2. The answer is 3.
#6 - Think about what relatively prime means. It means that any two numbers that have a GCD of 1. Obviously 1 is always relatively prime.
In this example, you are right by saying that there are 8 numbers less than 30 that are relatively prime to 30. But we don't know how many between 3 and 30. You have to see if there are any numbers less than 3 that are relatively prime to 30 by counting. We know 1 is. Is 2? No! The GCD of 2 and 30 is 2 not 1. Is 3? No! The GCD of 3 and 30 is 3 not 1. Therefore we only need to subtract 1. The answer is 7. This didn't take up very much time at all, so this is how I would approach it.
#7 - You should know from algebra the vertex of a parabola is given by (h,k) where h = -b/(2A) in a quadratic equation ax^2 + bx + c = 0. To find k simply plug h back into the equation.
In your example we are given the formula 3x^2 + 12x + 4 = 0. We can find h very quickly by using -b/(2a) or -12/(2*3) or -2. Plugging -2 into the equation (to find k) we get 3(4) + 12(-2) + 4 = 12-24+4 = -8. So k = -8.
In number sense you have to be able to look at quadratic equations and be able to figure out quickly how to factor them. If you have this skill, problems will be MUCH easier. When I saw 2x^2 + 13x + 20 = 0, it didn't take me but a couple of seconds to see this factored to (2x + 5)(x + 4)=0. You really need to be able to do this quickly. Now all we do is solve for the roots. We get -5/2 and -4. -4 is smaller so that is the smallest root.
In this problem, there is no short-cut per say but it is still fast. Just go through all the numbers starting with x =1, then x = 2, etc until something works. You should be able to see this very quickly.
49 = x^2 + 3y^2.
With x = 1 we get 48 = 3y^2 which works with y = 4. So that's all. Usually it won't take long. Usually, less than 10 seconds max.
thanks sooooooo much (to the 10th power) for spending time answering my question, your explanation was very clear and i understood mostly everything you wrote. however, i tried to use your explanation on probability to solve some similar questions, but unfortunately i got the wrong answer...
Question: how many ways can 5 people be seated 3 at a time in 3 chairs in a row?
question: how many ways can four people sit in a row of six chairs?(the bottom number was bigger than the top number...)
and... i have one more question to ask....
a triangle has integral sides of x, 22 and 2x. the smallest value of x is.....
thank you so much again (to the tenth power), i greatly appreicate your help
I must apologize sincerely. I was working too fast and made a big mistake when I was talking about Permutations. I didn't even notice until you replied. Let me correct something:
P(n,r) = n!/(n-r)! which is not what I had above so I will change that one too. Thanks for pointing that out to me even if you didn't mean to.
The first problem is a permutation because order does not matter. We could have ABC, ACB, BCA, BAC, CAB, or CBA and they all count. So we will be using P(5,3) for your first problem which equals 60. Sometimes they might say 5 taken 3 at a time, which is the same thing. Let me know if I got it wrong but I am pretty sure the answer is 60.
Sorry about the confusion.
Your second question deals with permutations as well. I think the answer might be P(6,4) since we are taking them 4 at a time and the order does not matter for the same reasons as above. Let me know if 360 is the wrong answer.
Your last question deals with a particular concept in geometry. We know that in a triangle every side must be less than the sum of the other two sides. So if we had segments of 12, 24, and 11, this cannot form a triangle since 11+12 < 24. So for this problem we know that x + 2x > 22. Or we can think of this as 3x > 22. The smallest integral value of x that makes this possible is x = 8.
If possible try to write down the answers to your questions and also make a new thread for each set of questions. This will make things a little easier for me. Thanks.
