Jimmy2588

Would someone help explain two questions from the 10 reals book for me? I tried studyhall, but i dont understand of the explanations and the second explanation was blank.

10 Reals 3rd edition:

#24 (page 420)

What is the least positive integer n for which 12n is the cube of an integer?

#25 (page 420)

[please refer to diagram if you have the book handy]

A wheel has an outer diameter of 11 inches, as shown above. The rim is 1 inch wide and the diameter of the hub is 2 inches. If each spoke extends 1/2 inch into the hub and 1/2 inch into the rim, what is the sum of the lengths of the five spokes, in inches?

Thanks so much!

Flipsta_G

24. Basically, you have to list the cubes of integers in order until you find one that is divisible by 12.

2^3 = 8 (not divisible by 12)
3^3 = 27 (not divisible by 12)
4^3 = 64 (not divisible by 12)
5^3 = 125 (not divisible by 12)
6^3 = 216 DING DING DING...divisible by 12

Now, since we are looking for what n is, so we divide 216 by 12 = 18. So the answer is 18.

25. If the spokes extend 1/2 inch into the rim and the hub, then the length of one spoke is the radius of the wheel - the hub radius - the rim radius + 1 (because of the sticking into the hub and rim). So one spoke = 11/2 -1-1+1 = 5.5 - 1 = 4.5
Since there are 5 spokes, the total length of the spokes is 4.5*5 = 22.5

Jimmy2588

why did you subtract 1 from 5.5?

xiggi

For the first one, a simple way is to list a few cubes starting from with the smallest ones.

1 = 1
2 = 8
3 = 27
4 = 64
5 = 125
6 = 216 .... Hmmm, this is interesting because 6.6.6 is also 6 x 3 x 2 x 6

2 x 6 is obviously 12, so we are onto something => 12 X 18 = 216 or 12n = 216
Thus n = 18.

A quick check may be good to make sure none of the integers 1-5 could work. Bingo, none could work, so our answer sticks.

For the second one, you need to draw a diagram. Decide if you will use the diameter or the radius. I think using the radius is easier.

Start from the center of the wheel and start drawing the problem.

A = 0 = center
B = 1 = end of hub (hub is 2 inch so that gives us 1" from the center)
C = 4.5 = start of rim (is at 4,5 because 1 inch from outer part of wheel)
D = 5.5 = end of rim or outer part of wheel (that is 5.5 from center because 1/2 of 11)

A<--1-->B<------?--------->C<--1-->D

A<--<---------spoke ------------>-->D

I have to use letters here because this is not exactly a great drawing board. On your drawing you should put the values immediately.

The distance from B to C is (5.5 - 1 - 1) or 3.5 inches. However, the spoke extends inside the rim and hub for one inch (two times 1/2 inch). This means that the each spoke is 4.5 inches long.

Now, be careful to pay close attention to the EXACT question. Your need the length of 5 spokes. On the SAT, it is easy to oberlook a small detail, rush into solving, and miss the detail.

xiggi

took too long to draw that diagram

Flipsta_G

The total radius of the wheel is 5.5. But some of that is taken away from lenth of the spokes due to the hub and the rim. The rim is 1 inch, and the hub has a radius of 1 inch. So if the spoke did not extend into the rim and hub 1/2 inch each, then the length of a spoke would be 5.5 - 2 = 3.5. But since the spoke does extend into the hub 1/2 inch and rim 1/2 inch, for a total of 1 unseen inch, the actual spoke length is 3.5 + 1.
In my initial explanation, I just subtracted 1 for the rim, 1 for the hub, and then added 1 for the double 1/2 extension. This in turn simplified to -1.

Jimmy2588

but one question, i just realized that.. wouldn't be simpler just divide 11 by 2 to get the radius of the entire wheel, which = 5.5, then subtract 1, since 1/2 inch is in the hub and 1/2 inch is in rim (1/2 + 1/2 = 1) to get 4.5.. then multiply 4.5 by 5 to get 22.5

Would that be an ok way to do the quesiton?

btw: Thanks soooooo much for the explanations! I really appreciate the time and effort!

optimizerdad

A quicker way for #24 is:

12 = (2)(2)(3) . Ask yourself: how many more 2's and 3's do you need to multiply this by, in order to get an integer^3? You need to multiply by n = (2)(3)(3), to get
12n = (2)(2)(2) (3)(3)(3) = ((2)(3))^3

So n = (2)(3)(3) = 18.

(This logic saves you from having to evaluate 1^3, 2^3, 3^3....)

xiggi

"Would that be an ok way to do the quesiton?"

It works in this case because the rim and 1/2 of the hub are exactly ONE. If the hub would have been 3 inches you could not follow your approach, because the spoke extend inside the hub by 1/2 inch, but not necessarily 1/2 inch from the center of the wheel.

Again, drawing a quick diagram is the way to approach this and leaves little chance for errors. Visualizing the SAT problems that involve distances is typically the best and fastest way. A lot of times, the answer just JUMPS at you, well before having to make any calculations. This case was no different!