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12 Quant questions I cannot solve - help please!

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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12 Quant questions I cannot solve - help please!

by vstraesser » Sat Dec 31, 2011 10:24 am
Hi there, I just took an mba.com CAT and couldn't solve some questions. Since they only provide the right answer but no explanations, I would really appreciate if you guys could help me out here. I took screenshots and compiled those in one file (see attachment). Thanks so much, and a happy new year to all of you. Cloudy

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by neelgandham » Sun Jan 01, 2012 10:39 am
Question 1

In the rectangular coordinate system below, the area triangular region PQR is ?

Draw a line segment from point R to the x axis(at N), parallel to y axis
Draw a line segment from point R to the y axis(at M), parallel to x axis

Now, We know that area of triangular region PQR = Area of rectangle OMRN - (Area of triangle RMQ + Area of triangle RNP + Area of triangle POQ)

Area of rectangle OMRN = 7*4 (Length * Breadth)
Area of right angled triangle RMQ = 1/2 * 1 * 7 = 3.5 (!/2 * base * height)
Area of right angled triangle RNP = 1/2 * 3 * 4 = 6 (!/2 * base * height)
Area of right angled triangle POQ= 1/2 * 4 * 3 = 6 (!/2 * base * height)
area of triangular region PQR = 28 - (3.5+6+6) = 12.5 Square units
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by neelgandham » Sun Jan 01, 2012 10:44 am
Question 2

If n is a positive integer and the product of all the integers from 1 to n inclusive is a multiple of 990, what is the least possible value of n?

A. 10
B. 11
C. 12
D. 13
E. 14

990*X = 2*3^2*5*11*X = 1*2*3*..n (Where X is a positive integer and 2,3,5,11 are prime factors)
For the product to have 11 as a prime factor, the value of n > 11. Since the least possible value of n is asked, it should be 11 option B
Last edited by neelgandham on Sun Jan 01, 2012 10:57 am, edited 1 time in total.
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by neelgandham » Sun Jan 01, 2012 10:51 am
Question 3

In the figure below, a square is inscribed in a circle. If the area of the inscribed square is 16 units, what is the area of the circular region ?

1) 2pi
2) 4pi
3) 8pi
4) 12pi
5) 16pi

Area of a Square = a^2 = 16 (Where a = length of the side of the square). Implies, that the value of a = 4.
Length of the diagonal of the square = Length of the side * √2 = 4√2 = Diameter of the circle.
So, the radius of the circle = 2√2
Area of the circle = pi*r*r, where r is the radius of the circle. = pi* 2√2*2√2 = 8pi

Answer C
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by neelgandham » Sun Jan 01, 2012 11:05 am
(2^x) - (2^(x-2)) = 3*(2^13), What is the value of X?

A) 9
B) 11
C) 13
D) 15
E) 17

(2^x) - (2^(x-2)) = 3*(2^13)
(2^(x-2)) *( 2^2 -1) = 3*(2^13) (Taking the common term (2^(x-2)) out of the parenthesis)
(2^(x-2))*3 = 3*(2^13)
(2^(x-2)) = (2^13) (Divide left hand side and right hand side by 3)
Since the bases are equal(=2)and are not 0 or 1, we can equate the exponents
x-2 = 13
x = 15 Answer : Option D
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by neelgandham » Sun Jan 01, 2012 11:13 am
Question 5

If xyz>0 , is x >0 ?
(1)xy>0
If xy>0 then
x>0, y>0 and z>0 OR
x<0, y<0 and z>0
Insufficient to answer the question Is x >0
(1)xz>0
If xz>0 then
x>0, z>0 and y>0 OR
x<0, z<0 and y>0
Insufficient to answer the question Is x >0

From 1 and 2
z>0, y>0 and x>0, Hence sufficient! Option C
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by vstraesser » Mon Jan 02, 2012 5:01 am
Thanks so much for your help, much appreciated. If you guys have the time and the expertise to come up with some more solutions to the rest of the questions, I would highly appreciate it. Thanks so much.
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by neelgandham » Mon Jan 02, 2012 5:06 am
Question 6 :

A certain company employs 6 senior officers and 4 junior officers. If a committee is to be created that is made up of 3 senior officers and 1 junior officer, how many different committees are possible?

A. 8
B. 24
C. 58
D. 80
E. 210


Number of ways of picking 3 seniors from 6 : 6C3 = 20
Number of ways of picking 1 junior from 4: 4C1 = 4
Total # of committees = 6C3 * 4C1 = 20 * 4 = 80 Answer D
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by neelgandham » Mon Jan 02, 2012 5:12 am
Question 7

When a certain tree was first planted, it was 4 feet tall, and the height of the tree increased by a constant amount each year for the next 6 years. At the end of the 6th year, the tree was 1/5 taller than it was at the end of the 4th year. By how many feet did the height of the tree increase each year?

A. 3/10
B. 2/5
C. 1/2
D. 2/3
E. 6/5

Let the 'constant amount each year' be x. then by the end of Year
1, the height of the tree = 4+x
2, the height of the tree = 4+2x
3, the height of the tree = 4+3x
4, the height of the tree = 4+4x
5, the height of the tree = 4+5x
6, the height of the tree = 4+6x

From the question 4+6x = (1+(1/5))* 4+4x, Implies
4+6x = (6/5)4+4x
20+30x = 24+24x
6x = 4
x =[spoiler]2/3[/spoiler] Option D
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by neelgandham » Mon Jan 02, 2012 5:16 am
Question 8

If (1/5)^m (1/4)^18 = 1/ 2(10)^35, then m =

a.17
b.18
c.34
d.35
e.36

(1/5)^m (1/4)^18 = 1/ 2(10)^35
(1/5)^m * (1/2*2)^18 = 1/(2(2*5)^35)
(1/5)^m * (1/2)^36 = 1/(2(2^35)(5^35))
(1/5)^m * (1/2)^36 = 1/(2^36 * 5^35)
multiply left hand side and right hand side with 2^36, leaving:
1/5^m = 1/5^35
Bases are equal, so should be exponents. Hence the value of m = 35
Option D
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by vstraesser » Mon Jan 02, 2012 12:25 pm
Neelgandham, thanks so much!
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by neelgandham » Mon Jan 02, 2012 12:31 pm
Question 9

Which of the following inequalities has a solution set, when graphed on the number line, is a single line segment of finite length?

A. x^4 >= 1, implies x<-1 and x>1.A graph with two lines(not line segments) - Not the answer
B. x^3 <= 27, implies x <3. A graph with one line(not line segments) - Not the answer
C. x^2 >= 16, implies x<-4 and x>4.A graph with two lines(not line segments) - Not the answer
D. 2 <= |x| <= 5, implies -5<x<-2 and 2<x<5. A graph with two line segments(not one) - Not the answer
E. 2 <= 3x+4 <= 6 - The only choice :)

IMO Answer E
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by neelgandham » Mon Jan 02, 2012 1:18 pm
Question 10

The perimeter of a certain isosceles right triangle is 16 +16 sqrt(2). What is the length of the hypotenuse of the triangle

A. 8
B. 16
C. 4 sqrt(2)
D. 8 sqrt(2)
E. 16 sqrt(2)

Let the length of the perpendicular sides be a, Hypotenuse = a*Square root(2) = ?
16 + 16*(Square root(2)) = a+a+(a*Square root(2))
16*(1+(Square root(2)) = a*Square root(2)* (1+(Square root(2))
Multiply both sides with 1/(1+(Square root(2))
16 = a*Square root(2) = Hypotenuse !

Answer B
IMO : B
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by neelgandham » Mon Jan 02, 2012 1:23 pm
Question 11

On a certain sight-seeing tour, the ratio of the number of women to the number of children was 5 to 2. What was the number of men on the sight-seeing tour?

(1) On the sight-seeing tour, the ratio of the number of children to the number of men was 5 to 11

(2) The number of women on the sight-seeing tour was less than 30.


Solution:

women = w
men = m
children = c

given : w/c = 5/2 , to find m = ?
From 1.
c/m = 5/11 => 11c = 5m , 5c = 2w (given) => 2w/5 = 5m/11 => m= 22w/25. For m to be a integer (number of men), w has to be multiple of 25. Like 25, 50, 75 and so....
So a alone is insufficient.
From 2.
w is less than 30.

So from 1 andv2 , w = 25 => m = 22

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by neelgandham » Mon Jan 02, 2012 1:33 pm
Question 12:

For every integer 'k' from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) * (1/2^k). If 'T' is the sum of first 10 terms in the sequence, then 'T' is:-

(a) Greater than 2
(b) Between 1 & 2
(c) Between 1/4 & 1
(d) Between 1/4 & 1/2
(e) Less than 1/4


First term = 1/2
Second term = - 1/2^2
Third term = 1/2^3 and .....

Sum of terms = T = 1/2 - 1/2^2 + 1/2^3 - ... -1/2^10
1/2 - 1/2^2 = 1/4
1/2^3 -1/2^4 = 1/16 = 1/4^2
1/2^5 -1/2^6 = 1/64 = 1/4^3
1/2^7 -1/2^8 = 1/256 = 1/4^4
1/2^9 -1/2^10 = 1/1024 = 1/4^5
Sum = 1/4 + 1/4^2 + 1/4^3 + 1/4^4 + 1/4^5 > 1/4, So eliminate A,B,E
Now Sum = 1/4 + 1/4^2 + 1/4^3 + 1/4^4 + 1/4^5

Sum of terms 1/4^2 + 1/4^3 + 1/4^4 + 1/4^5 < 1/4, so, Sum = 1/4 + 1/4^2 + 1/4^3 + 1/4^4 + 1/4^5 < 1/2. Answer D
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