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By Edwin Mason,2014-07-15 22:54
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    3 If p is a positive odd integer, what is the remainder when p is divided by 4?

    (1) When p is divided by 8, the remainder is 5.

(2) p is the sum of the squares of two positive integers.

(1) 足够回答问题?因为 p = 8q + 5。;q 是商。?

所以 p = 4(2q + 1) + 1。那么 p 4 除时余数是 1

    (2) 这个条件难很多。设 p = m^2 + n^2。因为 p 是奇数?所以 m n 必须一奇一偶。设 m 是偶数?那么 m^2 就必须是 4 的倍数?所以可以不考虑。现在只要看 n^2。因为 n 是奇数?所以 n 可以写

    Originally Posted by rits700

    the positive integer k has exactly 2 positive prime factors 3 and 7. If k has a total of

    6 positive factors including 1 & k, what is the value of k?

1) 3^2 is a factor of k

    2) 7^2 is not a factor of k

If p is the smallest positive integer such that p^3/3920 is also an integer, what is the

    sum of the digits of p?

(A) 5

    (B) 7

    (C) 9

    (D) 11

    (E) 13

Answer: A

m = p*p*p/ 3920

    m = p*p*p/ 2*2*2*2*5*7*7

    m = p*p*p/ [(4^2)*5*(7^2)]

Hence p must be equal to 4*5*7

    thus p = 140

    Hence sum of digits = 1+4+0 = 5

    Posted by Prachi Pareekh at 3/19/2008 05:11:00 AM 0 comments Links to this post

    Labels: Integers, Problem Solving Tuesday, March 18, 2008

    Problem Solving - 44

    Samar tried to type his new 7-digit phone number on a form, but what appeared on

    the form was 39269, since the '4' key on his computer no longer works. His

    secretary has decided to make a list of all of the numbers that could be Samar's new

    number. How many numbers will there be on the list?

(A) 21

    (B) 24

    (C) 25

    (D) 30

    (E) 36

Answer: A

    It is clear from the question that there are two 4's missing as it is a seven digit number.

    Total number of ways of choosing places for these two missing 4's in these 7 digits is 7C2 = 21 Once you fix the place for these two 4's rest all numbers will occupy remaining places.

Posted by Prachi Pareekh at 3/18/2008 08:51:00 AM 1 comments Links to this post

    Labels: Permutations and Combinations, Problem Solving

    Tuesday, February 05, 2008

    Data Sufficiency - 44

    In triangle ABC, AB has a length of 10 and D is the midpoint of AB. What is the length of line segment DC?

(1) Angle C= 90

    (2) Angle B= 45

Answer: A

    From statement (1): it is given that angle C = 90 degrees ...this implies that ABC is a right angle triangle with AB as the hypotenuse and DC as the median. We know that --- In all right triangles, the median on the hypotenuse is the half of the hypotenuse. Hence DC=5

    Posted by Prachi Pareekh at 2/05/2008 12:41:00 AM 0 comments Links to this post

    Labels: Data Sufficiency, Geometry

    Sunday, February 03, 2008

    Data Sufficiency - 43

    In the figure shown, what is the value of x?

(1) The length of line segment of QR is equal to the length of line segment RS

    (2) The length of line segment of ST is equal to the length of line segment TU

Answer: C

From statement (1): Length of line segment of QR is equal to the length of line

    segment RS ..this implies angle RQS = angle RSQ = p(say)

    From statement (2): Length of line segment of ST is equal to the length of line

    segment TU .. this implies angle TUS = angle TSU = q(say)

Hence p+p+angle QRS = 180 --- eq(1) and q+q+angle UTS = 180 --- eq(2)

    Thus, p+q+x = 180

Now because angle RPT = 90, QRS+UTS= 90

    adding eq(1) and eq(2) we get:

    2p+2q+QRS+UTS = 360

    2p+2q+90=360

    p+q = 270/2 = 135

Now x = 180-p-q..hence the answer C

    x = 180 - (p+q) = 180 - 135 = 45

Posted by Prachi Pareekh at 2/03/2008 07:33:00 AM 1 comments Links to this post

    Labels: Data Sufficiency, Geometry, GMAT Prep

    Thursday, January 31, 2008

    Data Sufficiency - 42

    Is a-3b an even number?

1). b=3a+3

    2). b-a is an odd number

Answer: C

From statement (1): Given that b=3a+3

    Thus a-3b=a-3(3a+3) = -8a-9 which may be even, odd, integer, non-integer, rational etc ... Hence insufficient

    From statement (2): Given that b-a is an odd number implies b is of the form b=(2k+1)+a where k is an integer

    Thus a-3b= a-3[(2k+1)+a] = -2a -6k-3 which may be even, odd, integer, non-integer, rational etc ..Hence insufficient

Taking statement (1) and (2) together: -8a-9=-2a-6k-3 for some integer k

    or -6a=-6k+6=-6(k+1) implies a=k+1

    Thus a is an integer, either odd or even

    Now statement (2) tells us that b is also an integer and that exactly one of {a,b} is even

    If a is even and b is odd, a-3b is odd

    If b is even and a is odd a-3b is odd

Thus (1) and (2) combined tell us that a-3b is an odd number...hence sufficient

    Posted by Prachi Pareekh at 1/31/2008 11:18:00 PM 2 comments Links to this post

    Labels: Data Sufficiency, Integers, Numbers

    Wednesday, January 30, 2008

    Problem Solving - 43

    If the sum of four consecutive positive integers a three digit multiple of 50, the mean of the these integers must be one of x possible values, where x=

(A) 7

    (B) 8

    (C) 9

    (D) 10

(E) more than 10

Answer: C

Suppose four integers are a, a+1, a+2 and a+3

    Hence a + (a+1) + (a+2) + (a+3) = 4a+6

    Now 4a+6 will be a multiple of 50 i.e 4a+6=50m where m can take any value from {2,3...,19}

But a = (50m-6)/4 = (25m-3)/2 must be an integer, so m must be odd.

Thus m can be any odd integer from 3 to 19

    3=1+1*2

    19=1+9*2

So there are 9 different values for m, a and 4a+6, as well as (4a+6)/4

    Posted by Prachi Pareekh at 1/30/2008 06:25:00 PM 0 comments Links to this post

    Labels: Integers, Problem Solving

    Data Sufficiency - 41

    Sania has a circular garden in her backyard. She puts poles A,B and C on the circumference of her garden. Then she ties ropes between these poles. Is length of one of the ropes is equal to the diameter of her garden?

    1. Slope of line joining pole A and B is 3/4 and slope of line joining poles B and C is -4/3

    2. Length of line joining pole A and B is 12 and length of line joining B and C is 5

Answer: A

    From statement (1): Given that the slope of line AB is 3/4 and slope of line BC is -4/3. This implies that the product of slopes = -1. Hence AB perpendicular BC and B is a right angle. Thus AC is a diameter which implies ABC form a semi-circle.

    Hence sufficient

    From statement (2): Given that length of line AB is 12 and length of BC is 5. However this does not imply that ABC is a right angled triangle. We can draw number of different triangles with the same given two sides but with different third side.

    Hence insufficient

Posted by Prachi Pareekh at 1/30/2008 06:07:00 PM 0 comments Links to this post

    Labels: Data Sufficiency, Geometry

    Problem Solving - 42

    E is a collection of four odd integers and the greatest difference between any two integers in E is 4. The standard deviation of E must be one of how many numbers?

(A) 3

    (B) 4

    (C) 5

    (D) 6

    (E) 7

Answer: B

    Suppose the integers are 1, 3 and 5. Therefore the four integers can be:

1, 5, 5, 5

    1, 3, 5, 5

    1, 3, 3, 5

    1, 1, 5, 5

    1, 1, 1, 5

    1, 1, 3, 5

    Here two pairs have the same standard deviation. thus in all we have four different standard deviations

    Posted by Prachi Pareekh at 1/30/2008 05:58:00 PM 0 comments Links to this post

    Labels: Problem Solving, Statistics

    Wednesday, January 23, 2008

Data Sufficiency - 40

    In XY plane, does the line with equation y=3x+2 contain point (r,s)?

1) (3r + 2 - s)(4s + 9 - s) = 0

    2) (4r - 6 - s)(3r + 2 - s) = 0

Answer: C

    Given that y = 3x+2 implies that does 3x+2-y = 0 contains the point (r,s) implies is (3r+2-s) = 0 ?

From statement (1): (3r+2-s)(4s+9-s) = 0 implies either (3r+2-s) = 0 or (4s+9-s) = 0.

    Now when (3r+2-s)...the line passes through (r,s)

    When (4r+9-s) = 0 ...we cannot determine that whether the line passes through (r,s) or not.

    Hence insufficient

From statement (2): (4r-6-s)(3r+2-s) = 0 implies either (4r-6-s) = 0 or (3r+2-s) = 0

    Now when (4r-6-s) = 0 ... we cannot determine that whether the line passes through (r,s) or not

    When (3r+2-s) = 0..the line passes through (r,s)

    Hence insufficient

    Taking statement (1) and (2) together: (3r+2-s)(4s+9-s) = 0 and (4r-6-s)(3r+2-s) =0... We cannot have both 4r+9-s=0 and 4r-6-s=0 so it is (3r+2-s) = 0 ... only this equation makes both the equations to be 0

    Hence sufficient

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