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Some PS Problems

by oddball » Wed Nov 17, 2010 12:35 pm
1. A function on real numbers is defined as:
Image
  • (A) 8
    (B) 12
    (C) 16
    (D) 20
    (E) 24
2. In the interior of a forest there are more than thirty apes. A certain number of apes equal to the square of one-eighth of the total number are playing and having great fun. The remaining twelve apes are on a hill and the echo of their shrieks by the adjoining hills frightens them. They came and join the apes in the forest and play with enthusiasm. What is the total number of apes?
  • (A) 16
    (B) 32
    (C) 40
    (D) 48
    (E) 64
3. What can you say about the roots of px² + qx + p = 0?
  • (A) The roots are same.
    (B) One root is double the other.
    (C) One root is the square of the other.
    (D) One root is the reciprocal of the other.
    (E) The roots are equal in magnitudes but differ in signs.
4. When an integer z is divided by 6, the remainder is 1. Which of the following cannot be an integer?
  • (A) z/15
    (B) z/12
    (C) z/9
    (D) z/4
    (E) z/3
5. A number when divided by a certain divisor leaves a remainder of 19; if twice the number is divided by the same divisor, then the remainder is 7. Then which of the following is true?
  • (A) The divisor is 31
    (B) The number can only be 50.
    (C) The number lies between 10 and 20.
    (D) Both (A) and (B).
    (E) Can't Determine
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by Rahul@gurome » Wed Nov 17, 2010 12:42 pm
Question Number 1:
1. A function on real numbers is defined as:
Image
  • (A) 8
    (B) 12
    (C) 16
    (D) 20
    (E) 24
f(x + f(x)) = 4*f(x) and f(1) = 4

f(5) = f(1 + 4) = f(1 + f(1)) = 4*f(1) = 4*4 = 16

The correct answer is C.
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by Rahul@gurome » Wed Nov 17, 2010 12:44 pm
Question Number 2:
2. In the interior of a forest there are more than thirty apes. A certain number of apes equal to the square of one-eighth of the total number are playing and having great fun. The remaining twelve apes are on a hill and the echo of their shrieks by the adjoining hills frightens them. They came and join the apes in the forest and play with enthusiasm. What is the total number of apes?
  • (A) 16
    (B) 32
    (C) 40
    (D) 48
    (E) 64
A simple question with a promising storyline! :)

Say, total number of apes = n and n > 30
Therefore, number of "having fun" apes = (n/8)²
Remaining number of apes = 12

Thus, (n/8)² + 12 = n
=> n² - 64n + 12*64 = 0
=> n² - 48n - 16n + 16*48 = 0
=> (n - 16)(n - 48) = 0
=> n = 16 or 48

But n > 30 => n = 48

The correct answer is D.
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by Rahul@gurome » Wed Nov 17, 2010 12:47 pm
Question Number 3:
3. What can you say about the roots of px² + qx + p = 0?
  • (A) The roots are same.
    (B) One root is double the other.
    (C) One root is the square of the other.
    (D) One root is the reciprocal of the other.
    (E) The roots are equal in magnitudes but differ in signs.
px² + qx + p = 0 => x² + (q/p)x + 1 = 0

Any quadratic expression with roots a and b can be written as (x - a)(x - b) = x² - (a + b)x + ab. Thus, the constant term is the product of the roots. In the given expression the constant term is 1. Thus the roots must be reciprocal.

The correct answer is D.
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by Rahul@gurome » Wed Nov 17, 2010 12:51 pm
Question Number 4:
4. When an integer z is divided by 6, the remainder is 1. Which of the following cannot be an integer?
  • (A) z/15
    (B) z/12
    (C) z/9
    (D) z/4
    (E) z/3
When z is divided by 6, the remainder is 1. It implies that z is not divisible by 6 or its multiples. Thus z/(Multiple of 6) can never be an integer. Only such option is z/12.

The correct answer is B.
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by Rahul@gurome » Wed Nov 17, 2010 1:02 pm
Question Number 5:
5. A number when divided by a certain divisor leaves a remainder of 19; if twice the number is divided by the same divisor, then the remainder is 7. Then which of the following is true?
  • (A) The divisor is 31
    (B) The number can only be 50.
    (C) The number lies between 10 and 20.
    (D) Both (A) and (B).
    (E) Can't Determine
Say, the number is n and the divisor is d.
Therefore, (n = ad + 19) and (2n = bd + 7), where a and b are constant integers.
We can easily conclude that d is greater than 19.

If we multiply the first expression by 2, we get (2n = 2ad + 38)
Which must be equal to (2n = bd + 7). Therefore, d = 38 - 7 = 31

(This is a bit tricky. Think about it! As we multiply n by 2, the remainder becomes 7. Whereas multiplication results in 38. Therefore 38 must be greater than than d and of the form (md + 7) => md = 31 => Only possible integer value of m is 1. Thus d = 31)

One possible value of n is 50. But n may be any integer of the given forms, i.e (31a + 1)

The correct answer is A.
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by goyalsau » Wed Nov 17, 2010 10:27 pm
Rahul@gurome wrote:Question Number 4:
4. When an integer z is divided by 6, the remainder is 1. Which of the following cannot be an integer?
  • (A) z/15
    (B) z/12
    (C) z/9
    (D) z/4
    (E) z/3
When z is divided by 6, the remainder is 1. It implies that z is not divisible by 6 or its multiples. Thus z/(Multiple of 6) can never be an integer. Only such option is z/12.

The correct answer is B.
Rahul is there any way by which we can calculate the first multiple of 15 , 9 , 4 , 3 . Considering the fact that Number must be in the form of { 6(x) + 1 }
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by Rahul@gurome » Wed Nov 17, 2010 11:08 pm
goyalsau wrote:
Rahul@gurome wrote:Question Number 4:
4. When an integer z is divided by 6, the remainder is 1. Which of the following cannot be an integer?
  • (A) z/15
    (B) z/12
    (C) z/9
    (D) z/4
    (E) z/3
When z is divided by 6, the remainder is 1. It implies that z is not divisible by 6 or its multiples. Thus z/(Multiple of 6) can never be an integer. Only such option is z/12.

The correct answer is B.
Rahul is there any way by which we can calculate the first multiple of 15 , 9 , 4 , 3 . Considering the fact that Number must be in the form of { 6(x) + 1 }
Thanks goyalsau!
Your comment made me realized that the options has got some problems and my solution too. If a number is z is of the form (6n + 1). It is not divisible by multiples of 6 or multiples of 2 or multiples of 3. This is because as z is always odd and one more than multiple of 3. Thus, none of the options is an integer.
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by goyalsau » Wed Nov 17, 2010 11:59 pm
Rahul@gurome wrote:
Thanks goyalsau!
Your comment made me realized that the options has got some problems and my solution too. If a number is z is of the form (6n + 1). It is not divisible by multiples of 6 or multiples of 2 or multiples of 3. This is because as z is always odd and one more than multiple of 3. Thus, none of the options is an integer.

:wink::wink::wink::wink::wink::wink::wink::wink::wink::wink:
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by ankur.agrawal » Thu Nov 18, 2010 12:44 am
Hey , How do u write f(1 + f(1)) = 4*f(1)
Rahul@gurome wrote:Question Number 1:
1. A function on real numbers is defined as:
Image
  • (A) 8
    (B) 12
    (C) 16
    (D) 20
    (E) 24
f(x + f(x)) = 4*f(x) and f(1) = 4

f(5) = f(1 + 4) = f(1 + f(1)) = 4*f(1) = 4*4 = 16

The correct answer is C.
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by Rahul@gurome » Thu Nov 18, 2010 1:45 am
ankur.agrawal wrote:Hey , How do u write f(1 + f(1)) = 4*f(1)
The function is defined as f(x + f(x)) = 4*f(x).
Thus, for x = 1, f(1 + f(1)) = 4*f(1)
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