Topic - Permutation & Combination

Solved ? - Yes

Source - 2IIM ]]>

Topic - Permutation & Combination

Solved ? - Yes

Source - 2IIM ]]>

a) 6

b) 3

c) 4

d) More than 6 ]]>

3x = 33 – 4|y|

Since x and y are integers, and since |y| is always positive regardless of the sign of y, this means that when you subtract a multiple of 4 from 33, you need to get a multiple of 3.

Since 33 is already a multiple of 3, in order to obtain another multiple of 3, you will have to subtract a multiple of 3 from it. So, y has to be a positive or a negative multiple of 3.

y = 3, -3, 6, -6, 9, -9, 12, -12...etc.

For every value of y, x will have a corresponding integer value.

So there are infinite integer values possible for x and y.

a) 4

b) 6

c) 10

d) 8 ]]>

12 can be written as 12 × 1, or 6 × 2, or 3 × 4. To start with, we can eliminate the possibilities where the two terms are negative as |y| + 4 cannot be negative.

Further, we can see that |y| + 4 cannot be less than 4. So, among the values, we can have |y| +4 take values 4, 6 or 12 only, or |y| can take values 0, 2 and 8 only.

When |y| = 0, |x| - 3 = 3, |x| = 6, x can be +6 or -6. Two pairs of values are possible: (6, 0) and (-6, 0)

When |y| = 2, |x| - 3 = 2, |x| = 5, x can be +5 or --5. There are four possible pairs here: (5, 2) , (-5, 2), (5, -2), (-5, -2)

When |y| = 8, |x| - 3 = 1, |x| = 4, x can be +4 or --4. There are four possible pairs here: (4, 8) , (-4, 8), (4, -8), (-4, -8)

Correct Answer: 10 Pairs

]]>a) 2

b) 4

c) 0

d) 1 ]]>

The first equation is a pair of lines defined by the equations

y = 8 – x ------- (i) {when y is positive}

y = x – 8 ------- (ii) {when y is negative}

With the condition that x ≤ 8 (because if x becomes more than 8, |y| will be forced to be negative, which is not allowed)

The second equation is a pair of lines defined by the equations:

y = 6 – x ------- (iii) {when x is positive}

y = 6 + x ------- (iv) {when x is negative}

with the condition that y cannot be greater than 6, because if y > 6, |x| will have to be negative.

On checking for the slopes, you will see that lines (i) and (iii) are parallel. Also (ii) and (iv) are parallel (same slope).

Lines (i) and (iv) will intersect, but only for x = 1; which is not possible as equation (iv) holds good only when x is negative.

Lines (ii) and (iii) do intersect within the given constraints. We get x = 7, y = -1. This satisfies both equations.

Only one solution is possible for this system of equations.

]]>a) 2

b) 4

c) 3

d) 1 ]]>

Case 1

x > 0. Now, |x| = x

x^2 – 7x – 18 = 0

(x – 9) (x + 2) = 0

x is either –2 or +9.

Case 2

x < 0. Now, |x| = –x

x^2 + 7x – 18 = 0

(x + 9) (x – 2) = 0

x is either –9 or +2.

However, in accordance with the initial assumption that x < 0, x can only be –9 (cannot be +2).

Hence, this equation has two roots: –9 and +9.

Alternatively, we can treat this as a quadratic in |x|, the equation can be written as |x|2 – 7 |x| – 18 = 0.

Or, (|x| – 9) (|x| + 2) = 0

|x| = 9 or –2. |x| cannot be –2.

|x| = 9, x = 9 or –9.

Correct Answer: There are 2 solutions.

]]>a) 40

b) 21

c) 20

d) 41 ]]>

If roots are real, D > 0

81 – 4|k| > 0

4|k| < 81

|k| < 20.25

Hence, –20.25 < k < 20.25

The integer values that k can take are –20, –19, –18 … 0 … 18, 19 and 20.

41 different values (Remember to include 0. Answer choice 'A' will definitely be 40

]]>a) 6

b) 4

c) 8

d) More than 8 ]]>

From the quadratic formula: (−11 ± √D)/2, we see that the numerator has to be an even number for the roots to be integers.

This implies that the discriminant should be a perfect square and be square of an odd number. (Only then we will have odd + odd = even in the numerator)

D = 121 – 4|p| = 121 – 4|p|

4|p| cannot be negative => D can take values 121, 81, 49, 25, 9, 1

|p| can be 0, 10, 18, 24, 28, 30

p can take 0, ±10, ± 18, ± 24, ± 28, ± 30

Correct Answer: More than 8 values are possible.

]]>a) 9

b) 10

c) 12

d) 20 ]]>

2x = 103 – 5y

This says that when you subtract a multiple of 5 from 103, you get an even number. You have to subtract an odd multiple of 5 from 103 in order to get an even number. There are 20 multiples of 5 till 100, ten of which are odd. (Note that you cannot subtract 105, or higher multiples as they result in a negative value for x.)

So, y can have ten integer values. x also has 10 integer values, each corresponding to a particular value of y.

y = 1, gives us a potential value for x, so do y = 3, 5, 7….19. y can take 10 values totally.

]]>a) 11.5

b) 9

c) 9.5

d) 12 ]]>

According to the question, Min + Max = 13

Median - (Min + Med + Max)/3 = Median - (12 + Med)/3 = 2.

Solving, we get Median = 9.5. ]]>

Which of the following statements if true would imply that the above system of equations does not have a unique solution?

i. a1/a2 = b1/b2 = c1/c2 ≠ d1/d2

ii. a1/a2 = a2/a3 ; b1/b2 = b2/b3

iii. a1, a2, a3 are integers; b1, b2, b3 are rational numbers, c1, c2, c3 are irrational numbers

- The equations are inconsistent or
- Two equations can be combined to give the third

Now, let us move to the statements.

i. a1/a2 = b1/a2 = c1/c2 ≠ d1/d2

This tells us that the first two equations cannot hold good at the same time.

Think about this:

x + y + z = 3;

2x + 2y + 2z = 5.

Either the first or the second can hold good. Both cannot hold good at the same time.

So, this will definitely not have any solution.

ii. a1/a2 = a2/a3 and b1/b2 = b2/b3

a1, a2, a3 are in GP, b1, b2, b3. This does not prevent the system from having a unique solution.

For instance, if we have

x + 9y + 5z = 11

2x + 3y – 6z = 17

4x + y – 3z = 15

This could very well have a unique solution.

iii. a1, a2, a3 are integers; b1, b2, b3 are rational numbers, c1, c2, c3 are irrational numbers

This gives us practically nothing. This system of equations can definitely have a unique solution.

So, only Statement i tells us that a unique solution is impossible.

]]>a) 6

b) 0

c) -16

d) 2 ]]>

Product of the roots = ab = (−7)/1 = -7.

Now, The second equation 2x^2 + px + q = 0 has roots a + 1 and b + 1.

Sum of the roots = a + 1 + b + 1 = a + b + 2 = (−p)/2 = -5 + 2 = -3 = (−p)/2 => -p = -6 => p = 6.

Product of the roots = (a + 1)(b + 1) = ab + a + b + 1 = q/2. We know the values of ab and a+b. Substituting this, we get, -7 + (-5) + 1 = q/2 => -11 = q/2 => q = - 22.

Hence, p = 6 and q = -22. => p + q = 6 + (-22) = -16.

]]>a) 6 or 3

b) 12 or 2

c) 8 or 4

d) 12 or 4 ]]>

Let r1 and r2 be the two roots

r1 = r2 + 1

Therefore,

r1 + 1 + r2 = (r1r2) ∗ r2 − 5

r2 + 1 + r2 = (r2 + 1) ∗ r2 − 5

2r2 + 6 = (r2)^2 + r2

(r2)^2 − r2 − 6 = 0

Solving, we get r2 = 3 or -2

Therefore, r1 = 4 or -1

Hence product of the two roots is either 12 or 2.

a) 3

b) 1

c) 2

d) none ]]>

x^2 – 7x - 30 = 0

(x - 10 ) (x - 3 ) = 0

x = 10, x = -3

But x = -3 is not possible as we have considered x > 0, thus 1 solution for this case.

Case 2: x < 0

x^2 + 7x - 30 = 0

(x + 10 )(x - 3)=0

x = -10 and x = 3

Only x = -10 is permissible.

Thus this equation has 2 real solutions

Alternatively, we can think of the above as a quadratic in |x|

x^2 – 7|x| - 30 = 0 can be factorized as

(|x| -10) ( |x| + 3) = 0

|x| cannot be -3, |x| can only be 10. x can take 2 real values.

a) 7

b) 9

c) 11

d) 13 ]]>