<p>Ok guys heres another Q...</p>
<p>2 The probability of A’s winning a hockey match is 3/2 times the
probability of B’s winning a cricket match. The probability of C’s winning a
race is twice the probability of B’s winning the cricket match. What is the
maximum probability of A’s winning the hockey match?</p>
<p>(A) ¼ (B) ½ (C) 2/3 (D)¾ (E) 1</p>
<p>ok Since C's probability is 2b, the maximum probability B can be is 1/2, making C 1. (1 is the highest probability possible for any event) Since B is 1/2, multiply b by 3/2, yielding A's probability, 3/4.</p>
<p>Yep. It's sort of a trick. I think a lot of people would answer (E), but if A was 1, B would be 2/3, and C would be 4/3, which makes no sense whatsoever.</p>
<p>to humna:</p>
<p>I thought it was C, but I forgot to take Harry's crayons into account. So If Tom had 140 and that was 3 times more than Harry, then like Hoopsplaya238
said, Harry would have 46.666666666666... crayons, which isn't possible.</p>
<p>So then the only logical answer would be the one that evenly divides into both 7 and 3.
B)120 doesn't work because 120/7 is 17.1428....
D)147 divides evenly by both 7 and 3, but 147/7=21 and Dick has to have less than 21 crayons so that doesn't work
E) 63 works but the question asks for the maximum number Tom can have, and 105 is greater than 63.</p>
<p>thnx to rebecca, dchow and hoopsplaya.
Time once again 4 another Q... Cnt make head or tail of it</p>
<p>What is the smallest positive integer that leaves a remainder of four when divided by any of the integers 6, 8 and 10?</p>
<p>6a + 4
8b + 4
10c + 4</p>
<p>least common denominator (6,8,10)=120, so then the answer would be 120+4=124.
dont know if its right or wrong. just my approach.</p>
<p>good approach.</p>
<p>Thnx tsengunn. Could you pls elaborate a little or elucidate it in another way perhaps???</p>
<p>I got another prob</p>
<p>A certain triangle has sides 1,1 and s. What are all the values of s for which this triangle is obtuse?</p>
<p>All i do is ask Qs isn't it?</p>
<p>Here is the solution to your problem : </p>
<p>Let's say AB=1 AC=1 and BC=s . It says the triangle is obtuse.
ABC obtuse => (BAC) > 90 . If (BAC) = 90 then BC=s = sqrt2 (Pythagorean Theorem )
(BAC) > 90 => s > sqrt 2
On the other hand we know that BC < AC+AB s < 1+1 = 2
So sqrt2 < s < 2 Solution for s is (sqrt2 , 2)</p>
<p>Humna: This is about the smallest positive integer question that you wanted some more elaboration on. If this were a real SAT math question, you could always just test each choice and see which one it is! But, let's say you really wanted to know how to do this problem...</p>
<p>OK, here's how to do it, and I like long, clear explanations that take things slowly, step by step, and as an unofficial tutor at my high school I learned that it works! I think that some people think that I talk to them like they're retarded, but this is the best way I know of to get you to really know how to do it.</p>
<p>Take the number 48. Divide it by 6. What do you get? You get 8. And what's the remainder? Zero.</p>
<p>You want to find a number that gives a remainder of 4. So what do you do? You add 4. So, 52/6 would be 8 and a remainder of 4.</p>
<p>OK, but what number gives a remainder of 4 when divided by 6, 8, and 10?
The correct answer will be:
4 + a multiple of 6
4 + a multiple of 8
and 4 + a multiple of 10.</p>
<p>In other words, we are looking for a number that is 4 more than a multiple of 6, 8, and 10. And we want the smallest positive integer we can get our little hands on.</p>
<p>So here's the plan: We find the smallest positive integer that is a multiple of 6, 8, and 10, and then add 4 to it. </p>
<p>That will give us the smallest positive integer that leaves a remainder of 4 when divided by 6, 8, and 10.</p>
<p>I hope you know how to find the least common multiple, which is 120. If you don't, don't hesitate to ask. It's a lot better to suffer a tiny bit of embarrassment and get the SAT question right than be shy and never end up knowing the answer.</p>
<p>Now, add 4 to our LCM (because we want a remainder of 4) and we get 124.</p>
<p>And there you have it. Hope this helps! And remember that on the SAT you can always try out the shortcut I showed you!</p>
<p>Humna: Here is my take on the question that you were confused on: "Tom has 7 times as many crayons as Dick and 3 times as many as Harry. If Dick has less than 21 crayons, what is the maximum number of crayons that Tom can have?"</p>
<p>I am convinced that there is indeed a typo to the problem, because none of the choices gives the right answer! </p>
<p>Here's how I approached it, again using my slow, step-by-step method:</p>
<p>So, Tom has 7 times as many crayons as Dick and 3 times as many as Harry.</p>
<p>OK. What if Tom had 14 crayons. Dick would have 2, and Harry would have 4.666667 crayons, which doesn't make sense.</p>
<p>So we can't just willy-nilly this one. </p>
<p>The number of crayons that Tom has must be a multiple of 21, because it has to divide evenly between 3, and it has to divide evenly between 7, and the only numbers that do that are multiples of 21.</p>
<p>So the number of crayons Tom has is a multiple of 21. </p>
<p>We also know that Dick has fewer than 21 crayons. That means that Tom has fewer than 21 x 7, or 147 crayons. </p>
<p>We want to find the maximum number of crayons for our boy Tom. Now, what is the largest multiple of 21 that is less than 147? </p>
<p>We have to go down 21: 147-21 = 126.</p>
<p>126 is the correct answer. Tom would have 126 crayons, Dick would have 18 crayons, and Harry would have 42 crayons.</p>
<p>Unfortunately, that's not one of the choices! But there's absolutely no way the answer can't be 126, because the number of crayons for Tom and Dick and Harry would be okay according to the problem.</p>
<p>So there's a typo.</p>
<p>Dchow...Thnx a lot man..sir( u are a tutor after all) I will be posting up some other Qs as well and I would really appreciate if you could take the time out to answer them. Many thnx. </p>
<p>Also baphomet, thnk you.</p>
<p>Sure, I'd love to help! I'm still in high school by the way...going to be a senior in like a month. Also, I should've said that for the problem about the crayons, you could also use the same shortcut. Take each choice (A-E) and divide by 3 to see if you get a whole number. Then take each choice again (A-E) and divide by 7 to see if you get a whole number that is less than 21. You can cross off whatever choices don't work, and then pick the biggest number of the choices left, because you want to find the maximum number of crayons.
A lot of times for the Math section, you can use shortcuts that take up much less time than it does to try to solve the problem like you would in school.</p>
<p>Another day, another Q...</p>
<p>In the square PQRS, point T is the midpoint of side QR. If the are of PQRS is 3,what is the area of quadrilateral PQTS?</p>
<p>SOS</p>
<p>If the area of PQRS is 3, then that of PQTS will be 3 * 3/4 = 9/4, I guess.</p>
<p>I think the best approach is to draw the picture in accurate scale and measure it. But that's not what I did.
All you have to do is to recognize that what part of the square is PQTS or the triangle TRS.</p>
<p>here's the solution to your problem : </p>
<p>A(PQRS) = 3 therefore each side of the square has a length of sqrt3.
A(PQTS) = A(PQRS) - A(TRS) = 3 - A(TRS)
A( TRS ) = TR* RS * 0.5 = sqrt3 / 2 * sqrt3 * 0.5 = 3/4 = 0.75
A ( PQTS ) = 3 - 0.75 = 2.25</p>
<p>For the love of God, Will sumbudy answer my Q. I posted it 2 days ago!!!!</p>
<p>Humna, two people have already answered your problem!</p>
<p>In area, perimeter, and volume problems, you are given data about geometric figures, and you are asked to supply some missing information. Here's the procedure, taken from Gruber's Complete Preparation for the New SAT:</p>
<p>Step 1. If you are not given a diagram, draw your own. Make it as accurate as possible, but don't waste time making it perfect (Draw square, label points P,Q,R,S, and T, draw a line connecting S and T).</p>
<p>Step 2. "Determine the formula that relates to the quantities involved in your problem." (Recognize that STR is 1/4 of the whole square, and that the trapezoid is 3/4 the area of the whole square, so to find the area of the trapezoid, you take 3/4 of the area of the square.)</p>
<p>Step 3: "Substitute the given information for the unknown quantities in your problem." (3/4 times 3 = 9/4)</p>
<p>"When doing volume, area, and perimeter problems, keep this hint in mind: Often the solutions to such problems can be expressed as the sum of the areas or volumes or perimeters of simpler figures. In such cases do not hesitate to break down your original figure into simpler parts."</p>
<p>Very sound advice. In this problem, you could have broken down the trapezoid into a rectangle and a triangle, and you could have found the area of each shape and added them up. That would take longer, but you would still get the correct answer.</p>
<p>Keeping these points in mind, I want you to try to avoid asking us any more problems about area, perimeter, or volume until you apply these steps on your own and try to work things out. You may be surprised at how many problems you can solve on your own! </p>
<p>If you really do get stumped (that means, if you apply the steps above and worked on the problem for ONE HOUR without getting the right answer), ask for a HINT.</p>
<p>Use that and work for about 20 to 30 minutes at most. Then ask for another HINT, until you can work the problem out on your own.</p>
<p>Here are two problems to get you going, taken from Gruber's:</p>
<ol>
<li><p>A wheel travels 33 yards in 15 revolutions. What is its diameter? (Assume pi = 22/7.)</p></li>
<li><p>If the diagonal of a square is 16" long, what is the area of the square?</p></li>
</ol>