Give me any SAT Reasoning Test math question (or a question from a simulated practice test) and I'll post a solution. Specify detail (just the math work or explicit reasoning or w/e) if you want. I'll post ridiculously detailed solutions by default.
ArachnotronPosts: 1,761Registered UserSenior Member
Harambee:
Image 1:
#15:
Since OD bisects <AOF, you know that m<AOD = m<DOF, and since you know that OC bisects <AOE, that m<AOB + m<BOC = m<COD + m<DOE, and since OB bisects <AOD, you know that m<BOC + m<COD = x = 40
Now since m<BOC + m<COD = 40, and since we know that OD bisects <AOF, we know that
40 + m<BOC + m<COD = 30 + m<DOE. Substituting, it is easy to see that the expression boils down to 40 + 40 = 30 + m<DOE, which gives that m<DOE is 50.
Finally, we want to find <BOE, which we can do by adding m<BOC + m<COD + m<DOE, which from the information we have derived simplifies to 40 + 50 = 90
#17:
Since we remove every 4 inches, it is useful to find out how many triangles we will remove. This can be found by: 80/4 = 20. Then, notice that for every 1 inch you remove, you gain back 2 inches. So we gain back 20(2-1) = 20 inches, which would make our total length 100 .
ArachnotronPosts: 1,761Registered UserSenior Member
Image 2:
#20
You know that the total rope length will be y + 4x, so the insight needed is an expression of y in terms of x. Luckily, we have area which gives xy = 4000. Now, rearranging, we see that x = 4000/y, which we can substitute into our expression to yield 16000/y, choice B
ArachnotronPosts: 1,761Registered UserSenior Member
Image 3:
#7
Sum of the arc measures is 2pi or 360 degrees. It's divided into 5 equal segments. ABC gets two of these segments, which leaves 3 of these segments to AEC. Thus ratio of ABC to AEC is 2:3, choice B .
Tom and Bill agreed to race across a 50-foot pool and back again. They started together, but Tom finished 10 feet ahead of Bill. If their rates were constant, and Tom finished the race in 27 seconds, how long did Bill take to finish it?
Replies to: Math help center
#5 pg. 472 of Blue Book
#6 pg. 472 of BB
#8 pg. 473 of BB
1. http://img16.imageshack.us/img16/3329/q34g.jpg
2. http://img193.imageshack.us/img193/9058/78224058.jpg
3. http://img195.imageshack.us/img195/6129/52012792.jpg
Thanks.
Image 1:
#15:
Since OD bisects <AOF, you know that m<AOD = m<DOF, and since you know that OC bisects <AOE, that m<AOB + m<BOC = m<COD + m<DOE, and since OB bisects <AOD, you know that m<BOC + m<COD = x = 40
Now since m<BOC + m<COD = 40, and since we know that OD bisects <AOF, we know that
40 + m<BOC + m<COD = 30 + m<DOE. Substituting, it is easy to see that the expression boils down to 40 + 40 = 30 + m<DOE, which gives that m<DOE is 50.
Finally, we want to find <BOE, which we can do by adding m<BOC + m<COD + m<DOE, which from the information we have derived simplifies to 40 + 50 = 90
#17:
Since we remove every 4 inches, it is useful to find out how many triangles we will remove. This can be found by: 80/4 = 20. Then, notice that for every 1 inch you remove, you gain back 2 inches. So we gain back 20(2-1) = 20 inches, which would make our total length 100 .
#20
You know that the total rope length will be y + 4x, so the insight needed is an expression of y in terms of x. Luckily, we have area which gives xy = 4000. Now, rearranging, we see that x = 4000/y, which we can substitute into our expression to yield 16000/y, choice B
#7
Sum of the arc measures is 2pi or 360 degrees. It's divided into 5 equal segments. ABC gets two of these segments, which leaves 3 of these segments to AEC. Thus ratio of ABC to AEC is 2:3, choice B .
Tom's rate: 100 ft/27 sec
Bill's rate: 90 ft/27 sec.
(27sec/90ft)(100ft) = 30 seconds.
filling entries of an initially empty 2008