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ok great thanks! I was confused since i saw other posts with Screen shots directly from the Gmat prep.

got a couple of quant questions for you math experts! would appreciate some help with them on how they landed on the answers. they are taken from the GMAT prep.

1. the value (10^8-10^2)/ (10^7 - 10^3) is closest to which of the following?

a. 1 b. 10 c. 10^2 d. 10^3 e. 10^4

answer is B and i can't solve why!


2. are x and y both positive ( answer C)
(1)2x-2y=1
(2) X/Y >1

3. if x is a positive integer, what is the last common multiple of x,6, 9

1. the LCM of x and 6 is 30
2. The lcm of x and 9 is 45

why is the answer D and not C?????

4. for which of the following functions is (fx)=f(1-x) for all x?

answer is F(X) = x^2(1-x)^2

why and whats the general strategy to solve these?

FINALLY:

i can solve this but what is the quicker way w/o plug and chug?:

mark bought a set of 6 flower pots of dif sizes total cost is $8.25. Each pot cost .25cets more than the one below it in size. What was the cost in ollars of the larges pot?

a. 1.75
b.1.85
c.2
d.2.15
e.2.30


thanks so much everyone!

best,
c

1. the value (10^8-10^2)/ (10^7 - 10^3) is closest to which of the following?

a. 1 b. 10 c. 10^2 d. 10^3 e. 10^4

(10^8-10^2)/ (10^7 - 10^3)
= 10^2(10^6-1)/10^3(10^4-1)=(10^6-1)/10(10^4-1)
approximate it
i.e. 10^6/10*10^4=10

Option B.

2. are x and y both positive
(1)2x-2y=1
(2) X/Y >1

-->
1) x-y=1/2 i.e. x>y
Insufficient

2) x/y>1

for y>0; x>y
for y<0; x<y

Insufficient

from 1) &2)

x>y hence y>0
i.e. x>0

Sufficient
Hence C.

3. if x is a positive integer, what is the last common multiple of x,6, 9

1. the LCM of x and 6 is 30
2. The lcm of x and 9 is 45

-->
1) x has a 5 i.e.2^a*3^b*5^1

where a=0,1
b=0,1

now x=2^a*3^b*5^1, where;a=0,1; b=0,1
6=2*3
9=3*3
LCM= 2*3*3*5

sufficient

2) here also x has a 5 i.e. x= 3^c*5^1; c=0,1,2

now
x= 3^c*5^1; c=0,1,2
6=2*3
9=3*3

hence LCM= 2*3*3*5

Sufficient.

hence D

4. for which of the following functions is (fx)=f(1-x) for all x?


--> Better strategy is to move through options.

put x=1-x and check the left and right hand side of f(x)=f(1-x)
such as for

F(X) = x^2*(1-x)^2
f(1-x)= (1-x)^2*(1-1+x)^2= x^2*(1-x)^2

5. mark bought a set of 6 flower pots of dif sizes total cost is $8.25. Each pot cost .25cets more than the one below it in size. What was the cost in ollars of the larges pot?

a. 1.75
b.1.85
c.2
d.2.15
e.2.30

-->. its basically an AP.
Take the prices to be
x-0.5, x-0.25, x, x+0.25, x+0.5, x+0.75

hence adding above
6x+0.75=8.25
=> x=1.25

hence x+0.75=1.25+0.75=2

i.e. C


Hope this helps..!!

The value (10^8-10^2)/ (10^7 - 10^3) is closest to which of the following?

a. 1 b. 10 c. 10^2 d. 10^3 e. 10^4

The question asks for an APPROXIMATION.
Relative to 10^8 and 10^7, the values being subtracted -- 10² and 10³ -- are so small that they can safely be ignored.

Thus:
(10^8 - 10³)/(10^7 - 10²)
≈ 10^8/10^7
= 10.

The correct answer is B.

c210 wrote:Are x and y both positive?
(1) 2x-2y=1
(2) X/Y > 1

Check here:

https://www.beatthegmat.com/are-x-and-y- ... 89906.html

c210 wrote:For which of the following functions is f(x) =f(1-x) for all x?

Check here:

https://www.beatthegmat.com/gmatprep-ps- ... 91553.html

Mark bought a set of 6 flower pots of different sizes at a total cost of $8.25. Each pot cost 0.25 more than the next one below it in size. What was the cost, in dollars, of the largest pot?

A. $1.75
B. $1.85
C. $2.00
D. $2.15
E. $2.30

The prices are evenly spaced integers, with a difference of 25 cents from one price to the next.
With evenly spaced integers, average = median = sum/number.
Thus, the median of the 6 prices = 825/6 ≈ 137.
Since the total is a multiple of 25, and the difference from one price to the next is 25, each price must be a multiple of 25.
Since the median is 137, the 3 most expensive prices are 150, 175, and 200.

The correct answer is C.

thank you so very much!!

. the value (10^8-10^2)/ (10^7 - 10^3) is closest to which of the following?

a. 1 b. 10 c. 10^2 d. 10^3 e. 10^4


For this question, very easily got the answer as 10.0001

which is 10 itself...

Just went with the flow of the calculations and got it in less than 2 mins...

are x and y both positive
(1)2x-2y=1
(2) X/Y >1


Question Stem: Are X and Y both greater than 0?

Statement 1: x - y = 1/2

when x - y = 1/2, x>y but in that case x and y both > 0 or X and Y both < 0 as well. Can't say for sure...

Eg: let x = 2 and y = -2.5.

In the above case, the answer will be a positive half...

Hence insufficient... but we can take is x > y


Statement 2: x/y > 1

= x > y if x and y both are greater than 0.

OR

x < y if x and y both are less than 0.




Combined: we know from statement 1 that x > y and from statement 2 we know that x > y then x and y > 0...

Hence both statements are sufficed...


Hope this helps...

if x is a positive integer, what is the last common multiple of x,6, 9

1. the LCM of x and 6 is 30
2. The lcm of x and 9 is 45

This question can be done on the basis of trial and error very easily...

Statement 1: When LCM of x and 6 is 30.

There are a lot many combinations: (5,6) (10,6) (15,6).

In all the three options, lets take 5, 10 or 15. the only overall LCM comes out to be 90. Means take any number the LCM comes out to be 90.

Hence sufficient....


Statement 2: When LCM of x and 9 is 45.

There are again a lot many options...


lets take them as (5,9); (15,9).

take both the options i.e take both the numbers 5 and 15 both the Overall LCM of the three numbers comes out to be 90.

(i) 5,6,9 LCM is 90

(ii) 15,6,9 LCM is 90 as well...

Hence statement 2 is sufficient...


Hence the answer comes out to be a D.... and not C...

Initially even I thought as C but then i moved 1 stepped ahead and calculated the LCM is all the options that I got and got the answer as D...

Hope this helps...

for which of the following functions is (fx)=f(1-x) for all x?


--> Better strategy is to move through options.

put x=1-x and check the left and right hand side of f(x)=f(1-x)
such as for

F(X) = x^2*(1-x)^2
f(1-x)= (1-x)^2*(1-1+x)^2= x^2*(1-x)^2


Thanx killer 1987 for a wonderful explanation... I did not get this question... i got it wrong...

Thanx once again.

mark bought a set of 6 flower pots of dif sizes total cost is $8.25. Each pot cost .25cets more than the one below it in size. What was the cost in ollars of the larges pot?

a. 1.75
b.1.85
c.2
d.2.15
e.2.30

I think this is the simplest algebra question i might have ever seen... This question is nothing but simple algebra just like the ages question...


I got this question and the answer is C. i.e 2



[x + (x + .25) + (x + .50) + (x + .75) + (x + 1) (x + 1.25)] = 8.25

Therefore, x = .75

if the answer is .75 then the largest pot is .75 + 1.25 = 2

NO need for AP formula and stuff just take the smallest pot as x and form the above equation...


Hope this really helps you guyzzzzzzz.........