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f(x + f(x)) = 4*f(x) and f(1) = 41. A function on real numbers is defined as:
- (A) 8
(B) 12
(C) 16
(D) 20
(E) 24
A simple question with a promising storyline!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
px² + qx + p = 0 => x² + (q/p)x + 1 = 03. 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.
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/12.4. 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
Say, the number is n and the divisor is d.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
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 }Rahul@gurome wrote:Question Number 4: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/12.4. 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
The correct answer is B.
Thanks goyalsau!goyalsau wrote: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 }Rahul@gurome wrote:Question Number 4: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/12.4. 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
The correct answer is B.
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.
Rahul@gurome wrote:Question Number 1:f(x + f(x)) = 4*f(x) and f(1) = 41. A function on real numbers is defined as:
- (A) 8
(B) 12
(C) 16
(D) 20
(E) 24
f(5) = f(1 + 4) = f(1 + f(1)) = 4*f(1) = 4*4 = 16
The correct answer is C.
The function is defined as f(x + f(x)) = 4*f(x).ankur.agrawal wrote:Hey , How do u write f(1 + f(1)) = 4*f(1)
