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12 Quant questions I cannot solve - help please!

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vstraesser Just gettin' started!
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12 Quant questions I cannot solve - help please! Post Sat Dec 31, 2011 10:24 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    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

    Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!

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    neelgandham Community Manager
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    Post 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
    Attachments

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    Post 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

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    Last edited by neelgandham on Sun Jan 01, 2012 10:57 am; edited 1 time in total

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    Post 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
    Attachments

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    Post 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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    Post Sun Jan 01, 2012 11:13 am
    Question 5

    If xyz>0 , is x >0 ?

    Quote:
    (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

    Quote:
    (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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    Post 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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    Post 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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    Post 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 =2/3 Option D

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    Post 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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    Post Mon Jan 02, 2012 12:25 pm
    Neelgandham, thanks so much!

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    Post 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 Smile

    IMO Answer E

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    Post 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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    Post 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 = ?

    Quote:
    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.

    Quote:
    From 2.
    w is less than 30.

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

    Source: http://www.beatthegmat.com/gmat-prep-sight-seeing-tour-t10477.html

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    Post 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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