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198 700+ question doubts

by gdk800 » Mon Dec 06, 2010 4:10 am
Q1. Two machine types, Type R and Type S, operate at a constant rate. R does a job in 36 hours. S does same job in 18. If we use the same number of each type to do the job in 2 hours, how many Type r machines do we need?
a. 3
b. 4
c. 6
d. 9
e. 12

OA is C


Q2. If x ≠ 0, then √x2/x =
a. -1
b. 0
c. 1
d. x
e. |x|/x

In Q2. OA is mentioned as E but i think C should be fine because

Isn't Root (X^2) = +X. Example take X^2 = 9, the value of Root (X^2) = 3 and not -3.
Thus the answer should be C in this case.

Q3. What is m+n?
a. 3
b. 4
c. 5
d. 6
e. 7

Image attached.

Its a 4x4 matrix.
Image
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by goyalsau » Mon Dec 06, 2010 4:24 am
gdk800 wrote:
Q1. Two machine types, Type R and Type S, operate at a constant rate. R does a job in 36 hours. S does same job in 18. If we use the same number of each type to do the job in 2 hours, how many Type r machines do we need?
a. 3
b. 4
c. 6
d. 9
e. 12

OA is C
Machine R does the job in 36 hours. it will do 1/36 work in one hour.
Machine S does the job in 18 hours. it will do 1/18 work in one hour.

In two hours Both the machines will do x ( 2/36 + 2/18 ) = 1

x ( 2/36 + 2/18 ) = 1
x ( 1/18 + 1/9 ) = 1
x ( 1+ 2 / 18 ) = 1
x = 18/3
x = 6
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by goyalsau » Mon Dec 06, 2010 4:33 am
gdk800 wrote:
Q3. What is m+n?
a. 3
b. 4
c. 5
d. 6
e. 7

Image attached.

Its a 4x4 matrix.
Image
x + 4 = 1 , x = -3
y + 4 = -5 , y = -9
z + 4 = m ,

x + e = 7 ------ -3 + e = 7 , e = 10

y + e = n -------- -9 + 10 = n , n = 1

z + e = 10 , ------ z + 10 = 10, z = 0

z + 4 = m , -------- 0 + 4 = m , m = 4

m + n ----- 4 + 1 = 5

Answer is 5
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by Rahul@gurome » Mon Dec 06, 2010 4:35 am
gdk800 wrote:Q2. If x ≠ 0, then √x2/x =
a. -1
b. 0
c. 1
d. x
e. |x|/x

In Q2. OA is mentioned as E but i think C should be fine because

Isn't Root (X^2) = +X. Example take X^2 = 9, the value of Root (X^2) = 3 and not -3.
Thus the answer should be C in this case.
No.
√x² = |x|

In fact the representation +x has no significant meaning! What if x = -3, then +x = -3. That's why we use the absolute value representation.

Thus in this case, √x²/x = |x|/x

The correct answer is E.
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by GMATGuruNY » Mon Dec 06, 2010 4:38 am
gdk800 wrote:Q1. Two machine types, Type R and Type S, operate at a constant rate. R does a job in 36 hours. S does same job in 18. If we use the same number of each type to do the job in 2 hours, how many Type r machines do we need?
a. 3
b. 4
c. 6
d. 9
e. 12
Plug in a value for the job. Let job = 36.
Rate for R = work/time = 36/36 = 1 per hour
Rate of S = work/time = 36/18 = 2 per hour

Now let's plug in the answer choices, which represent the number needed of each type of machine:

Answer choice C: 6 of each machine
Rate for 6 of machine R = 6*1 = 6 per hour
Rate for 6 of machine S = 6*2 = 12 per hour
Combined rate for all machines = 6+12 = 18 per hour
Time = work/rate = 36/18 = 2 hours. Success!

The correct answer is C.
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by GMATGuruNY » Mon Dec 06, 2010 4:44 am
gdk800 wrote: Q2. If x ≠ 0, then √x2/x =
a. -1
b. 0
c. 1
d. x
e. |x|/x
The correct answer must work for any value of x other than 0.

Let x = -2.
√(x^2)/x = √4/(-2) = -1.
Eliminate B, C and D. (E works because |-2|/-2 = -1.)

Let x = 2.
√(2^2)/2 = √4/(2) = 1. Eliminate A.

The correct answer is E.
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by goyalsau » Mon Dec 06, 2010 4:49 am
Rahul@gurome wrote:
No.
√x² = |x|

In fact the representation +x has no significant meaning! What if x = -3, then +x = -3. That's why we use the absolute value representation.

Thus in this case, √x²/x = |x|/x
I don't understand this one, √x²= |x|

Is √ only over x^2 or is it on √ 1/x

Rahul Please explain
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by Rahul@gurome » Mon Dec 06, 2010 4:57 am
goyalsau wrote:I don't understand this one, √x²= |x|

Is √ only over x^2 or is it on √ 1/x

Rahul Please explain
The '√' is only over x².

√x² is always equals to |x| because the value of √x² should be always positive whether x is positive or not. For example whether x = 3 or x = -3, √x² = √9 = 3. Now for x = 3, √x² = 3 = x is OK. But if x = -3, then √x² = 3 = -x which is not same as x. Thus √x² = |x|, now we can represent both the positive and negative cases in the same expression.
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by goyalsau » Mon Dec 06, 2010 8:54 am
Rahul@gurome wrote: The '√' is only over x².

√x² is always equals to |x| because the value of √x² should be always positive whether x is positive or not. For example whether x = 3 or x = -3, √x² = √9 = 3. Now for x = 3, √x² = 3 = x is OK. But if x = -3, then √x² = 3 = -x which is not same as x. Thus √x² = |x|, now we can represent both the positive and negative cases in the same expression.
Thanks Rahul, But it will be really nice of you , If you can explain one more thing.


Rahul I am able to understand that Mode signs are used only to represent the both negative and Positive value of x in same expression.

I don't understand one thing.

X^ 2 = 4
X = + 2 or -2

Substituting x = 2
√x² = √(2)^2 = √4 = +2 or -2

I m just not able to understand why the value of √x² will be Positive.
As i know whenever we replace square root we always have to consider two values. But
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by GMATGuruNY » Mon Dec 06, 2010 8:58 am
goyalsau wrote:
Rahul@gurome wrote: The '√' is only over x².

√x² is always equals to |x| because the value of √x² should be always positive whether x is positive or not. For example whether x = 3 or x = -3, √x² = √9 = 3. Now for x = 3, √x² = 3 = x is OK. But if x = -3, then √x² = 3 = -x which is not same as x. Thus √x² = |x|, now we can represent both the positive and negative cases in the same expression.
Thanks Rahul, But it will be really nice of you , If you can explain one more thing.


Rahul I am able to understand that Mode signs are used only to represent the both negative and Positive value of x in same expression.

I don't understand one thing.

X^ 2 = 4
X = + 2 or -2

Substituting x = 2
√x² = √(2)^2 = √4 = +2 or -2

I m just not able to understand why the value of √x² will be Positive.
As i know whenever we replace square root we always have to consider two values. But
√ means the positive root only.

Thus, while x^2 = 4 means that x=2 or x = -2, x = √4 means that x=2.
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by goyalsau » Mon Dec 06, 2010 9:16 am
GMATGuruNY wrote: √ means the positive root only.

Thus, while x^2 = 4 means that x=2 or x = -2, x = √4 means that x=2.
Thanks Guru, Got it at last............
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by gdk800 » Mon Dec 06, 2010 2:03 pm
Much thanks to everyone for their reverts.

Here is one more question. Is there a faster way doing this question if we are given a 2 digit number?


Q. If M = √4 + 3√4 + 4√4, the value of M is
a. < 3
b. = 3
c. 3 < M < 4
d. = 4
e. > 4

OA is E

Clarification: M is equal to summation of 2nd root, 3rd root & fourth root of number 4.
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by goyalsau » Mon Dec 06, 2010 7:09 pm
gdk800 wrote:
Q. If M = √4 + 3√4 + 4√4, the value of M is
a. < 3
b. = 3
c. 3 < M < 4
d. = 4
e. > 4

Clarification: M is equal to summation of 2nd root, 3rd root & fourth root of number 4.
I got it wrong, ON the First attempt But I think there is a way to do this quickly.

M = √4 + 3√4 + 4√4

√4 = 2

4√4 = ( 4 )^1/4 = ( 2 ) ^ 2/4 = ( 2 ) ^ 1/2 = √2

M = 2 + 3√4 + √2

2 + √2 = 3.414 ( its good to know √2 = 1.414 )

M = 3.414 + 3√4

We know that any power of 1 is 1, Whether it is 1 ^1/8 or 1^ 1.11 or 1 ^ 1/3

So any root of any positive integer greater than 1, Will always be Greater than one. { Any Root of any integer greater than one will always be greater than 1 } I realized this when i marked your question wrong.... :wink:

3√4 > 1

M = 3.414 + ( value greater than 1 )

M = 4.414 + ( some value )

Now we can surely say that M > 4

option E :)
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by GMATGuruNY » Mon Dec 06, 2010 7:28 pm
gdk800 wrote:Much thanks to everyone for their reverts.

Here is one more question. Is there a faster way doing this question if we are given a 2 digit number?


Image
Since the problem asks only for an estimation, we can ballpark:

√4 = 2.
4^(1/3) > 1.
4*(1/4) = √2 ≈ 1.4.

So M > 4.4.

The correct answer is E.
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