Help with OG Problems

oxfordbound wrote:Hi all,


pg 242:
#176. As x increases from 165 to 166, which of the following must increase? (answer is i and ii)
i. 2x-5
ii. 1- 1/x
iii. 1/ x^2 - x

(i) 2x - 5 will increase when x increases for all x

(ii) 1/x will decrease when x increases for all x >0 (for example - as x goes from 1 to 2, 1/x goes from 1 to 1/2)
-1/x will increase when 1/x increases
Thus 1-1/x will increase as x increases for all x>0

(iii) 1/x^2 is a really small fraction when x >> 1

for example if x = 100, 1/x^2 = 0.0001

So, in the expression 1/x^2 - x

1/x^2 is very small compared to x and thus can be neglected. (its like saying 0.0001 is much smaller than 100 so we can forget the 0.0001)

Thus the expression becomes -x.

Now as x increases, -x decreases for all x.

Thus (iii) decreases when x increases

Answer (i) and (ii)

hi I HAD TO LOOK UP OG , because after solving for s/t I suspected either two factors for s OR two factors for t; it turned out to be all factors for both s and t

1/s + 1/t = (s+t)/st, since s can be factored as the product of two integers from 100 and 200 inclusive and t can be factored out as the product of any integer(s) between 100 and 201 inclusive we can rewrite this,as Factor(1)*Factor(2)*(1+X)/st where Factor(1)*Factor(2) is indeed s --> (1+X)/t, where t is the product of integers from 100 to 201 inclusive; two factors are missing from X -- NO solution

corrected problem --> (s+t)/st= (1+201)/t=202/t

oxfordbound wrote:
pg 214:
# 89. If s is the product of two integers from 100 to 200 inclusive and t is the product of integers between 100 and 201 inclusive. What is 1/s + 1/t in terms of t? (answer is 202/t)

again exclude iii. 1/ (x^2 - x)

oxfordbound wrote: pg 242:
#176. As x increases from 165 to 166, which of the following must increase? (answer is i and ii)
i. 2x-5
ii. 1- 1/x
iii. 1/ x^2 - x

#181.
1 stroke 1/2 left
2 stroke 1/2-(1/2 * 1/2)=1/4 left
3 stroke 1/4-(1/4 * 1/2)=1/8 left
4 stroke 1/8-(1/8 * 1/2)=1/16 left

1/2 + 1/4 + 1/8 + 1/16 = (8+4+2+1)/16 = 15/16

#225 Cutting into fourths makes up 4 pieces each with the length 1/4. Cutting into thirds makes up 3 pieces each with the length 1/3. The following distances between 0 and 1 yard may exist:
0-1/4 --> 1/4
1/3-1/4 --> (1/3-1/4)=1/12
1/3-{1/4+1/4} --> (1/2-1/3)=1/6

oxfordbound wrote: pg 244:
#181. If 1/2 of the air in a tank is removed with each stroke of a vaccum pump, what fraction of the original amount of air has been removed after 4 strokes? (answer is 15/16)

pg 261:
#225. A straight pipe 1 yard in length was marked off in fouths and also thirds. if the pipe was then cut into seperate pieces at each of these markings, which of the following gives all the different lengths of the pieces, in fractions of a yard? (answer is 1/12, 1/6, 1/4)[/u]

If s is the product of the integers from 100 to 200 inclusive, and if t is the product of the integers from 100 to 201 inclusive, what is 1/s + 1/t in terms of t?

A. (201)^2/t
B. {(202)(201)}/t
C. 201/t
D. 202/t
E. (202)(201)/t^2

t = s*201.
We can plug in any values for s and t that satisfy this relationship.
Let s = 2.
Then t = 2*201 = 402.
1/s + 1/t = 1/2 + 1/402 = 402/804 + 2/804 = 404/804 = 101/201. This is our target.

Now we plug t=402 into the answers to see which yields our target of 101/201.

Only answer choice D works:
202/t = 202/402 = 101/201.

The correct answer is D.

As X increases from 165 to 166,which of the following must increase?

i)2x-5

ii)1-1/x

iii)1/(x^2-x)


A) i only

B)iii only

C)i and ii

d)i and iii

e) ii and iii

Ignore the values given. The question becomes:

As x increases, which of the following must also increase?

Plug in x=2 and x=3.

I. 2x-5
If x=2, then 2x-5 = 2*2 - 5 = -1.
If x=3, then 2x-5 = 2*3 - 5 = 1.
The value of 2x-5 increases.
Eliminate any answer choice that does not include i. Eliminate B and E.

ii)1-1/x
If x=2, then 1 - 1/2 = 1/2.
If x=3, then 1 - 1/3 = 2/3.
The value of 1 - 1/x increases.
Eliminate any answer choice that does not include ii. Eliminate A and D.

The correct answer is C.

Nycgrl wrote:If 1/2 of the air in a tank is removed with each stroke of a vacuum pump, what fraction of the original amount of air has been removed after 4 strokes ?

1.15/16
2. 7/8
3. 1/4
4- 1/8
5. 1/16

We can plug in a value for the tank. We should plug in a value that can be divided by 2 four times without yielding a fraction. (The answer choices lead us in the right direction: the biggest denominator is 16.)

Let tank = 16 units.
After the 1st stroke, 1/2*16 = 8 units are removed.
16-8 = 8 units are left.
After the 2nd stroke, 1/2*8 = 4 more units are removed.
8-4 = 4 units are left.
After the 3rd stroke, 1/2*4 = 2 more units are removed.
4-2 = 2 units are left.
After the last stroke, 1/2*2 = 1 more unit is removed.
2-1 = 1 unit is left.

Since only 1 unit is left, 16-1 = 15 total units were removed.
Total removed/Tank = 15/16.

The correct answer is A.

A straight pipe 1 yard in length was marked off in
fourths and also in thirds. If the pipe was then cut into
separate pieces at each of these markings, which of
the following gives all the different lengths of the
pieces, in fractions of a yard?

(A)1/6 and 1/4 only
(B)1/4 and 1/3 only
(C)1/6 ,1/4 and 1/3
(D)1/12 ,1/6, and 1/4
(E)1/12, 1/6, and 1/3

Place all the markings over a common denominator of 12:



Since 4/12 - 3/12 = 1/12, the correct answer must include 1/12. Eliminate A, B and C.
Since the pieces at each end = 1/4, the correct answer must include 1/4. Eliminate E.

The correct answer is D.