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Manager  Joined: 02 Jan 2013
Posts: 54
GMAT 1: 750 Q51 V40 GPA: 3.2
WE: Consulting (Consulting)
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Question 7:

x + y = 350
x = 25*p, y = 25*q where p and q may not have any other factors in common

Substituting into the equation we have: p+q = 14
The only solutions for pair (p,q) in which p and q do not have common factors are: (1,13), (3,11),(5,9)

There are exactly 3 pairs: (25,325), (75, 275), (125, 225)

Manager  Joined: 02 Jan 2013
Posts: 54
GMAT 1: 750 Q51 V40 GPA: 3.2
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Question 8
x.377910 = 3300k

=> x.(19.17.13.5.3^2.2) = 11.5^2.3.2^2.k
=> x. 19.17.13.3 = 11.5.2
=> x = 110.k/(19.17.13.3)

Therefore, the smallest number for x is 110 (when k = 19.17.13.3)

Manager  Joined: 02 Jan 2013
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Question 10

1. x = 1 ? Not necessary. Let x = 3, and y = 0, then x^y = 1
2. x = 1 and y = 0. Not necessary (see item 1)
3. x = 1 or y = 0. Not necessary. Let x = -1, y=2

Originally posted by caioguima on 17 Apr 2013, 16:03.
Last edited by caioguima on 18 Apr 2013, 04:48, edited 2 times in total.
Manager  Joined: 02 Jan 2013
Posts: 54
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2
Question 9:

The WONDERFUL thing is that 37*27 = 999, therefore, we can write the sum as:
1/3 + 1/9 + 1/27 + 1/37 =
333/999 + 111/999 + 37/999 + 27/999 =
508/999 =
0.508508508508508...

Note that the decimals repeat themselves in period of 3. Since 101 divided by 3 gives remainder 2, we're looking for the position #2 in the repeated set 508.

Therefore, the digit will be 0.

Manager  Joined: 04 Mar 2013
Posts: 58
Location: India
Concentration: Strategy, Operations
Schools: Booth '17 (M)
GMAT 1: 770 Q50 V44 GPA: 3.66
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Question 3.
How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

That comes to the no. of ways [1,2,3,4,5] can be grouped,

i.e 5C1+C52+5C3+5C4+5C5
=> 5+10+10+5+1

31 So D
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Originally posted by aceacharya on 17 Apr 2013, 16:59.
Last edited by aceacharya on 17 Apr 2013, 18:23, edited 1 time in total.
Manager  Joined: 04 Mar 2013
Posts: 58
Location: India
Concentration: Strategy, Operations
Schools: Booth '17 (M)
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Question 10
The answer would be none of these E

I. X=1 ; x could be any no. if y=o
II. Y=0; x could become 1 therefore negating this statement
III. x=1 or y=0; well x could very well be -1; so not necessary
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Manager  Joined: 04 Mar 2013
Posts: 58
Location: India
Concentration: Strategy, Operations
Schools: Booth '17 (M)
GMAT 1: 770 Q50 V44 GPA: 3.66
WE: Operations (Manufacturing)
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Question 5

I get that the answer is 19 that is C, but i am unable to find out how!
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Director  Joined: 25 Apr 2012
Posts: 660
Location: India
GPA: 3.21
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8. The product of a positive integer x and 377,910 is divisible by 3,300, then the least value of x is:

A. 10
B. 11
C. 55
D. 110
E. 330

Sol: Given that (x* 377910/ 3*11*100)= Integer and therefore we need to least value of x

On simplifying we get (x* 12597/11*10). Now 12597 is not divisible by 11 and 100. Therefore for the expression to be to result an integer, least value of X should be 110

Ans D
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Director  Joined: 25 Apr 2012
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4. The functions f and g are defined for all the positive integers n by the following rule: f(n) is the number of perfect squares less than n and g(n) is the number of primes numbers less than n. If f(x) + g(x) = 16, then x is in the range:

A. 30 < x < 36
B. 30 < x < 37
C. 31 < x < 37
D. 31 < x < 38
E. 32 < x < 38

Sol: Given
f(n) is the number of perfect squares less than n and g(n) is the number of primes numbers less than n.
If f(x) + g(x) = 16, then x is in the range:

Best thing will be to assume value of X and check in which of the above range it satisfies for all values of X

Let us take x - 32 so we have f(n) = 5 (1,4,9,16 and 25) and G(n) = 11 (2,3,5,7,11,13,17,19,23,29 and 31)
Clearly f(n) +g(n)= 16. Note that for any value of X in the range for option C, we get the value of F(n)+ G(n)= 16 and Hence C should be the Answer.

To cross check, lets look at option A and Assume X = 32 then f(n)= 5 and g(n) = 11. F(n)+G(n)= 16 but if take X= 31, f(n)=5 and g(n)=10 and therefore their sum is not equal to 16

Consider option E if x= 33 then f(n) =5 and g(n)= 11 and f(n)+ g(n)= 16 but if X =37 then f(n)= 6 (1,,4,9,16,25 and 36) and g(n)= 11 and f(n)+g(n)= 17

Therefore Ans should be C
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Director  Joined: 25 Apr 2012
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5. Which of the following is a factor of 18!+1?

A. 15
B. 17
C. 19
D. 33
E. 39

Sol: We have to find factor of 18!+1

We can say that 18! and 18! +1 are consecutive nos and therefore both the nos are co-prime. Therefore any number which is a factor of 18! will not be factor 18!+1

18!= 18*17*16*15*14*.....*1
Looking at option Choices we have 15,17, 33 (11*3) and 39 (3*13) as factors of 18!.

Therefore answer should be C i.e 19
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Director  Joined: 25 Apr 2012
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7. The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible?

A. 1
B. 2
C. 3
D. 4
E. 5

Sol:
Greatest Common divisors of 2 nos (n1,n2) is 25 and n1+n2= 350
Think of any 2 nos whose GCD will be 25 i.e (25,50), (25,25), (25,100), (25,125), (25,150)...
Now we are bounded by the condition that n1 +n2=350

Consider n1 =25 then n2= 325 , GCD is 25
Consider n1 =50, n2=300, GCD is 50 (Cannot be the pair)
Consider n1=75 and n2= (275), GCD is 25
Consider n1=100, n2=250, GCD is 50
n1=125, n2=225,GCD =25
n1=150, n2=200, GCD =50
n1=175, n2=175, GCD is 175

So Ans should be 3.....Option C
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Director  Joined: 25 Apr 2012
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10. If x is not equal to 0 and x^y=1, then which of the following must be true?

I. x=1
II. x=1 and y=0
III. x=1 or y=0

A. I only
B. II only
C. III only
D. I and III only
E. None

Sol: We need to find condition which will be true under all conditions

St 1 when x =1 then X^ (any value of y) will always give us 1. So A, D are possible answers
St 2 X=1 and y=0, Here and means both conditions simultaneously. But Y need not be zero as long as x= 1. Y can take any value as long as x=1 and hence B is ruled out

St 3 x=1 or y=0 here "or" means any one of the condition if true then we get x^y=1 which is true. Consider x=1 and y= 32 ----> x^y = 1, consider x= 1, y= -3, then x^y= 1
Similarly X= 2 but y=0 then 2^0 =1 Option C and D can be the possible answer

Ans Option D......

PS: I think there is a catch since St 1 covers only value of x whereas St 3 covers for all possible cases of (x,y) and I was tempted to go for option C alone as the answer. Let's see
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Director  Joined: 25 Apr 2012
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6. If the least common multiple of a positive integer x, 4^3 and 6^5 is 6^6. Then x can take how many values?

A. 1
B. 6
C. 7
D. 30
E. 36

Sol: Given LCM of +ve Integer X, 4^3 and 6^5 is 6^6

LCM of X, 2^6 and 6^5 is 6^6

Now 6^6 can be written as (2^6)*(3^6)------->Highest power of 2 and 3 is 6 only

From the given nos we already have 2^6 the highest power of 2 as 6....that means x needs to have 3^6 so possible nos with X^6 are

3^6,
2*3^6,
2^2*3^6
2^3*3^6
2^4*3^6
2^5*3^6
2^6*3^6

Ans should be 7. Note that any other combination of powers of 2 and 3 will give us value of x from any of the option combinations only.

Good Question, Tried doing it under 2 mins but could not but got it right while trying to do the same in no time pressure.
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Director  Joined: 25 Apr 2012
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9. What is the 101st digit after the decimal point in the decimal representation of 1/3 + 1/9 + 1/27 + 1/37?

A. 0
B. 1
C. 5
D. 7
E. 8

Sol: 1/3 + 1/9 + 1/27 + 1/37

1/3= 0.333333
1/9= 1/(3*3)= 0.3333333/3-----> 0.1111111
1/27= 1(3^3)= 0.037037037
1/37= 0.027027027
So if we add up
0.333
0.111
0.037
0.027
Sum is ( .508508508)...99th Digit will be 8,100th digit will be 5 and 101st digit will be 0...

I guessed it under timed condition......Would like to have faster way.

Ans should A
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6. If the least common multiple of a positive integer x, 4^3 and 6^5 is 6^6. Then x can take how many values?

A. 1
B. 6
C. 7
D. 30
E. 36

Theory for solving such questions:
1- Perform the Prime Factorization of the Number
2- Find the Highest Index among the given numbers of all the Prime No's
3- L.C.M is the product of all these prime no's with the respective highest indexes

$$4^3 = 2^6$$
$$6^5 = 2^5 * 3^5$$
Since LCM is 2^6* 3^6

For $$3^6$$ to be there, We require that to come from "X"
Since, $$2^6$$is already there,

X may contain $$2^0, 2^1, 2^2,2^3,2^4,2^5, and 2^6$$

So, X can be any of the values -
$$2^0 *3^6 2^1 *3^6 , 2^2*3^6, 2^3*3^6, 2^4*3^6, 2^5,*3^6 and 2^6*3^6$$

Hence "C"
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8. The product of a positive integer x and 377,910 is divisible by 3,300, then the least value of x is:

A. 10
B. 11
C. 55
D. 110
E. 330

Lets Break into Prime Nos First

$$3300 = 3* 2^2* 5^2 *11$$

$$377,910 = 2*3*5*7*2371$$

$$2*3*5*7*2371 * X / 3*2^2*5^2*11$$

2371 is not divisible by 2; Hence X must contain one 2(atleast), for the number to be divisible by 3300
Similarly, X should contain 01 Five and 01 eleven for the number to be divisible by 3300
Hence X must be at-least 2*5*11=110

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9. What is the 101st digit after the decimal point in the decimal representation of 1/3 + 1/9 + 1/27 + 1/37?

A. 0
B. 1
C. 5
D. 7
E. 8

Decimal Representation of :

$$1/3 = .333...$$
$$1/9 = .111...$$
$$1/27 = .037...$$
$$1/37= .027...$$

Since these digits will be repeated, and for finding 101 digits, we can ignore the part till 99th digit as it is cyclical.
Hence -
Adding .333+.111+.037+.027 = .508. Hence, 101st digit will be 0

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7. The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible?

A. 1
B. 2
C. 3
D. 4
E. 5

Since G.C.D of two integers X and Y is 25
So, X can be written as $$X= 25 a$$
and
$$Y = 25 b,$$where both a and b are co-primes.

Since sum of integers is 350, we can write as
$$25a+25b = 350 25(a+b) = 25*14$$

Now, a+b = 14 such that a and b are co-prime to each other -

Since these are numbers not digits, there is no restriction. And since ths um of two numbers is even, we can ignore the even numbers, and shal concentrate only on ODD numbers.

$$a=1;b=13 a=3;b=11 a=5;b=9$$
Hence, 3 such numbers.

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10. If x is not equal to 0 and x^y=1, then which of the following must be true?

I. x=1
II. x=1 and y=0
III. x=1 or y=0

A. I only
B. II only
C. III only
D. I and III only
E. None

All are Could be true questions, but none of them is MUST be TRUE\
1- X can be any other value than 1
2- NoT Necessarily
3- x=2 and y =0; x=1 and and y=1 can satisfy
Hence, E
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5. Which of the following is a factor of 18!+1?

A. 15
B. 17
C. 19
D. 33
E. 39

Since 18! must result in 00
And adding 1 into it = the last two digits become 01

Only multiple of 17 has 01 as last two digits.

Hence, B
Not too sure though Originally posted by imhimanshu on 18 Apr 2013, 22:52.
Last edited by imhimanshu on 18 Apr 2013, 23:39, edited 1 time in total. Re: Fresh Meat!!!   [#permalink] 18 Apr 2013, 22:52

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