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Could someone pls help me with the following problems...

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by neelgandham » Thu Dec 08, 2011 12:56 pm
Question 1

If a = 1; (a-b)/c = 1, which of the following is NOT a possible value of b ?

If (a-b)/c = 1, the value of c shouldn't be equal to 0, i.e.c !=0,
so if b = 1-c, b!= 1-0, b!=1
Answer : D

p.s. Can you please type the complete question instead of attaching screen grabs? (if possible that is)
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by neelgandham » Thu Dec 08, 2011 1:00 pm
Question 2
If x and y are positive, which of the following must be greater than 1/√(x + y) ?
1. √(x + y)/2x
2. (√x + √y)/(x + y)
3. (√x - √y)/(x + y)

Let's see what happens when each of the given expressions is greater than 1/√(x + y), i.e. let's find the conditions for which they will be great than 1/√(x + y).

I. √(x + y)/2x
.... √(x + y)/2x > 1/√(x + y)
=> (x + y) > 2x ...................... Cross-multiplication
=> y > x

y > x is a particular situation. Thus (I) is not always greater than 1/√(x + y).

II. (√x + √y)/(x + y)
.... (√x + √y)/(x + y) > 1/√(x + y)
=> (√x + √y) > (x + y)/√(x + y)
=> (√x + √y) > √(x + y)
=> (√x + √y)² > (x + y) ....................... Squaring both sides
=> (x + 2√(xy) + y) > (x + y)
=> 2√(xy) > 0
=> √(xy) > 0

√(xy) > 0 is always true as x and y are not zero. Thus (II) is always greater than 1/√(x + y).

III. (√x - √y)/(x + y)
.... (√x - √y)/(x + y) > 1/√(x + y)
=> (√x - √y) > (x + y)√(x + y)
=> (√x - √y) > √(x + y)
=> (√x - √y)² > (x + y) ....................... Squaring both sides
=> (x - 2√(xy) + y) > (x + y)
=> -2√(xy) > 0
=> √(xy) < 0

But √(xy) can't be less than zero as √(xy) > 0 as x and y are not zero. Thus (III) is never greater than 1/√(x + y).

Thus only (II) is a must be greater case.

A big thanks to Anurag for the above explanation !
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by neelgandham » Thu Dec 08, 2011 1:03 pm
Question 3
Please find the explanation with diagrams here.

https://www.beatthegmat.com/gmat-prep-qu ... 74944.html
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by neelgandham » Thu Dec 08, 2011 1:16 pm
Question 4
If k is a positive integer and the tens digit of k+5 is 4(Is this parallel? ;)), what is the tens digit of k?

1) k>35
2) The units digit of k is greater than 5

Using statement 1
k>35
Let the value of k be 36, then the value of k+5 = 41. The tens digit of integer k is 3
Let the value of k be 40, then the value of k+5 = 45. The tens digit of integer k is 4
Oops, two different answers - Insufficient!

Using statement 2
The units digit of k is greater than 5
Tens digit of k+5 is 4, implies the value of k must lie between 36 and 39 both inclusive. The tens digit of integer k is 3. Bingo!
Sufficient!

Statement 2 is sufficient, Answer B

Hello Disco22, let me know if you need any clarification !
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by Anurag@Gurome » Fri Dec 09, 2011 5:22 am
Explanation to Q1:

a = 1 and (a - b)/c = 1 implies (1 - b)/c = 1
If b = 1, then numerator will be zero and hence left hand side will be 0, which is not the case.

Hence, the correct answer is D.
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by Anurag@Gurome » Fri Dec 09, 2011 5:24 am
Explanation to Q2:

Let's see what happens when each of the given expressions is greater than 1/√(x + y), i.e. let's find the conditions for which they will be great than 1/√(x + y).

I. √(x + y)/2x
  • .... √(x + y)/2x > 1/√(x + y)
    => (x + y) > 2x ...................... Cross-multiplication
    => y > x
y > x is a particular situation. Thus (I) is not always greater than 1/√(x + y).

II. (√x + √y)/(x + y)
  • .... (√x + √y)/(x + y) > 1/√(x + y)
    => (√x + √y) > (x + y)/√(x + y)
    => (√x + √y) > √(x + y)
    => (√x + √y)² > (x + y) ....................... Squaring both sides
    => (x + 2√(xy) + y) > (x + y)
    => 2√(xy) > 0
    => √(xy) > 0
√(xy) > 0 is always true as x and y are not zero. Thus (II) is always greater than 1/√(x + y).

III. (√x - √y)/(x + y)
  • .... (√x - √y)/(x + y) > 1/√(x + y)
    => (√x - √y) > (x + y)√(x + y)
    => (√x - √y) > √(x + y)
    => (√x - √y)² > (x + y) ....................... Squaring both sides
    => (x - 2√(xy) + y) > (x + y)
    => -2√(xy) > 0
    => √(xy) < 0
But √(xy) can't be less than zero as √(xy) > 0 as x and y are not zero. Thus (III) is never greater than 1/√(x + y).

Thus only (II) is a must be greater case.
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by Anurag@Gurome » Fri Dec 09, 2011 5:27 am
Explanation to Q3:

Refer to the figure below:

Image

OC is the radius of the circle.
Hence, AB = OC implies, AB = OC = OD = OB

Hence, triangle ABO is isosceles with AB = OB.
Hence, angle BAO = angle BOA = x (say)
Hence, angle ABO = (180 - 2x)

Now on straight line AC, angle ABO = (180 - 2x)
Hence, angle CBO = 180 - (180 - 2x) = 2x

Again triangle CBO is isosceles with OB = OC
Hence, angle BCO = CBO = 2x
Hence, angle BOC = (180 - 4x)

Now on straight line AD, (angle AOB + angle BOC + angle COD)= 180
Hence, (x + (180 - 4x) + angle COD) = 180
=> angle COD = 3x

[spoiler]Statement 1:[/spoiler] angle COD = 3x = 60
Hence, angle BAO = x = 20; SUFFICIENT.

[spoiler]Statement 2:[/spoiler] angle BCO = 2x = 40
Hence, angle BAO = x = 20; SUFFICIENT.

[spoiler]The correct answer is D.[/spoiler]
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by Anurag@Gurome » Fri Dec 09, 2011 5:35 am
Explanation to Q4:

When the tens digit of k + 5 is 4, the smallest value of k is 35.

(1) When k > 35, there can be many values for which tens digit of k + 5 is 4, like 38 + 5 = 43 and 40 + 5 = 45; NOT sufficient.

(2) The units digit of k is greater than 5 implies that tens digit of k is 3 as all the possible values with tens digit 4 have units digit less than 5; SUFFICIENT.

The correct answer is B.
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