GMAT Problem Solving Discussions

[vicky.verma]

Originally Posted by rd880

if n and y are positive integers such that 450y = n^3
which of the following must be an integer

1 y/(3 *2*2*5)
2 y/(3*3*2*5)
3 y/(3*2*5*5)

a none
b 1 only
c 2 only
d 3 only
e 1 2 and 3

450y = 3*3*5*5*2*y = n^3

For the above to be true y has to be : 2*2*5*3, which makes y/2*2*5*3 an integer.

Answer for me : B (1 only)

[vicky.verma]

Originally Posted by bibinvarghese

450y=n^3
450y=(y/x)^3
450=y^2/x^3
3^2*5^2*2*x^3=y^2
that means 2*x^3 must be perfect square
None satisfy these condition so answer is none

" 3^2*5^2*2*x^3=y^2
that means 2*x^3 must be perfect square
None satisfy these condition so answer is none"

If I take x=2*2*2, then 3^2*5^2*2*x^3=y^2 stands valid.
so y here becomes 3*5*2*2

P.S. This is a public forum. Even if you feel the answer someone has posted is wrong, no need to groan I hope eric.segal1 agrees

[bibinvarghese]

Originally Posted by rd880

if n and y are positive integers such that 450y = n^3
which of the following must be an integer

1 y/(3 *2*2*5)
2 y/(3*3*2*5)
3 y/(3*2*5*5)

a none
b 1 only
c 2 only
d 3 only
e 1 2 and 3

if you doesn't understand what i told i will explain in words
let n be equal to y/x
now
450y=y^3/x^3
canceling y
450x^3=y^2
now 450x^3 has to perfect square
9^2 and 5^2 are factors in 450 that are perfect square
now 2*x^3 has to perfect square That means there has to an odd power of 2 in x and all others has to be even powers
1 has both conditions wrong
2 has 3^2,2 but 5 has even power
3 has 3 having odd power and 2 with even power
So none of them can be correct.

You can groan if you feel this is useful and thank if you feel this is useful as you told this is a public forum.

[rd880]

thats the rite answer, NONE
thanks for the elaborate explanation

[eric.segal1]

Originally Posted by bibinvarghese

if you doesn't understand what i told i will explain in words
let n be equal to y/x
now
450y=y^3/x^3
canceling y
450x^3=y^2
now 450x^3 has to perfect square
9^2 and 5^2 are factors in 450 that are perfect square
now 2*x^3 has to perfect square That means there has to an odd power of 2 in x and all others has to be even powers
1 has both conditions wrong
2 has 3^2,2 but 5 has even power
3 has 3 having odd power and 2 with even power
So none of them can be correct.

You can groan if you feel this is useful and thank if you feel this is useful as you told this is a public forum.

i guess vicky verma has pointed at yr doubt precisly and i think u understand where was te flaw in yr answer afterall.
now guys throw up the nxt one and lets make this a bit more challenging.

and i think groans r not a problem, its wat its meant to be. Now throw the ext one in. we r all waiting.

[sats123]

The answer should be E

[vicky.verma]

Originally Posted by sats123

The answer should be E

dude!!!

what question are you answering to and what's the explanation for your answer choice.

[jackgreat]

hey guys,

stuck on the following ques.. explanation would be nice.

the ques invloves solving numbers with underroots so please check image attached.

choices are
(1) 8 + 3underoot7
(2) 4 + 3underoot7
(3) 8
(4) 4
(5) underroot7

thanks....

[bibinvarghese]

Originally Posted by jackgreat

hey guys,

stuck on the following ques.. explanation would be nice.

the ques invloves solving numbers with underroots so please check image attached.

choices are
(1) 8 + 3underoot7
(2) 4 + 3underoot7
(3) 8
(4) 4
(5) underroot7

thanks....

Lets approximate the whole thing ~= or less than root of 18.25
now =4+
Now from choices we can find the answer.

[getneonow]

Originally Posted by jackgreat

hey guys,

stuck on the following ques.. explanation would be nice.

the ques invloves solving numbers with underroots so please check image attached.

choices are
(1) 8 + 3underoot7
(2) 4 + 3underoot7
(3) 8
(4) 4
(5) underroot7

thanks....

Given number = sqrt[ 9sq.rt(7)+ 0.25 ]

For rough estimation eliminate 0.25 = approx 3 * 7^1/4 = approx 3* sq.rt(2.6) = approx 3 * (1.6) = approx 4.8

so (4) 4 should be the closest answer. Others are way beyond the number.

Neo