CAT - Inequalities

Question:

Solve the inequality x³ – 5x² + 8x – 4 > 0.

A.  (2, )

B.  (1, 2) U (2, )

C.  (-, 1) U (2, )

D.  (-, 1)

Correct Answer :  (B)

Explanation:

Let a, b, c be the roots of this cubic equation

a + b + c = 5

ab + bc + ca = 8

abc = 4

This happens when a = 1, b = 2 and c = 2 {This is another approach to solving cubic equations}

The other approach is to use polynomial remainder theorem

If you notice, sum of the coefficients = 0

=>  P(1) = 0

=>  (x - 1) is a factor of the equation. Once we find one factor, we can find the other two by dividing the polynomial by (x - 1) and then factorizing the resulting quadratic equation.

(x - 1) (x - 2) (x - 2) > 0

Let us call the product (x - 1)(x - 2)(x - 2) as a black box.

If x is less than 1, the black box is a –ve number

If x is between 1 and 2, the black box is a +ve number

If x is greater than 2, the black box is a +ve number

Since we are searching for the regions where black box is a +ve number, the solution is as follows:

1 < x < 2 OR x > 2

Answer Choice (B)

Difficulty level 2

CAT Functions

Question

Qn. f(x + y) = f(x)f(y) for all x, y, f(4) = +3 what is f(-8)?

A.      1/3

B.      1/9

C.      9

D.      6

Answer Choice: B

Explanatory Answer

f(x + 0) = f(x) f(0)

f(0) = 1

f(4 + -4) = f(0)

f(4 + -4) = f(4) f(-4)

1 = +3 x f(-4)

f(-4) = 1/3

f(-8) = f(-4 + (-4)) = f(-4) f(-4)

f(-8) = 1/3 x 1/3 = 1/9 

Answer Choice (B)

CAT Progressions

Question

Second term of a GP is 1000 and the common ratio is   where n is a natural number. Pn is the product of n terms of this GP. P6 > P5 and P6 > P7, what is the sum of all possible values of n?

A.            5

B.            9

C.            8

D.            11

Correct Answer

Choice B

Explanatory Answer

Common ratio is positive, and one of the terms is positive => All terms are positive

P₆ = P₅ * t₆ => If P₆ > P₅, t₆ > 1

P₇ = P₆ * t₇ => If P₆ > P₇, t₇ < 1

T₆ = t₂ * r⁴ = 1000r⁴;

T₇ = t₂ * r⁵ = 1000r⁵

1000r⁴ > 1 and 1000r⁵ < 1

1/r^4 < 1000 and 1/r^5 > 1000

1/r = n

n⁴ < 1000 and n⁵ > 1000, where n is a natural number

n⁴ < 1000 => n < 6

n⁵ > 1000 => n > 4

n could be 4 or 5. Sum of possible values = 9

Answer Choice (B)

CAT Counting: Question with digits

This is an interesting question from Counting. Simple framework, but one needs to be very careful with the enumeration. One can get wrong answers in a number of ways.

Question

If we listed all numbers from 100 to 10,000, how many times would the digit 3 be printed?

A.      3980

B.      3700

C.      3840

D.      3780

Correct Answer

Choice A

Explanatory Answer

We need to consider all three digit and all 4-digit numbers.

Three-digit numbers: A B C. 3 can be printed in the 100’s place or10’s place or units place.

Ø  100’s place:  3 B C. B can take values 0 to 9, C can take values 0 to 9. So, 3 gets printed in the 100’s place 100 times

Ø  10’s place:  A 3 C. A can take values 1 to 9, C can take values 0 to 9. So, 3 gets printed in the 10’s place 90 times

Ø  Unit’s place:  A B 3. A can take values 1 to 9, B can take values 0 to 9. So, 3 gets printed in the unit’s place 90 times

So, 3 gets printed 280 times in 3-digit numbers

Four-digit numbers: A B C D. 3 can be printed in the 1000’s place, 100’s place or10’s place or units place.

Ø  1000’s place:  3 B C D. B can take values 0 to 9, C can take values 0 to 9, D can take values 0 to 9. So, 3 gets printed in the 100’s place 1000 times.

Ø  100’s place:  A 3 C D. A can take values 1 to 9, C & D can take values 0 to 9. So, 3 gets printed in the 100’s place 900 times.

Ø  10’s place:  A B 3 D. A can take values 1 to 9, B & D can take values 0 to 9. So, 3 gets printed in the 10’s place 900 times.

Ø  Unit’s place:  A B C 3. A can take values 1 to 9, B & C can take values 0 to 9. So, 3 gets printed in the unit’s place 900 times.

3 gets printed 3700 times in 4-digit numbers.

So, there are totally 3700 + 280 = 3980 numbers

Answer Choice (A)

Questions and solutions on Permutation and Combination

Permutation and Combination.

Questions
1. Sum of three natural numbers a, b and c is 10. How many ordered triplets (a, b, c) exist?

2. In how many ways 11 identical toys be placed in 3 distinct boxes such that no box is empty?

Correct Answers:
1. 9C2
2. 10C2


Explanatory Answers

1. Sum of three natural numbers a, b and c is 10. How many ordered triplets (a, b, c) exist?

a + b + c = 10. Now, let us place ten sticks in a row

|     |     |     |    |     |     |     |     |     |

This question now becomes the equivalent of placing two '+' symbols somewhere between these sticks. For instance

|     |     |     | + |     |     |     |     |  + |, This would be the equivalent of 4 + 5 + 1. or, a = 4, b = 5, c = 1.

There are 9 slots between the sticks, out of which one has to select 2 for placing the '+'s. 

The number of ways of doing this would be 9C2. Bear in mind that this kind of calculation counts ordered triplets. (4,5,1) and (1, 4, 5) will both be counted as distinct possibilities.

We can also do a + b + c = n where a, b, c have to be whole numbers (instead of natural numbers as in this question) with a small change to the above approach. Give it some thought.


2. In how many ways 11 identical toys be placed in 3 distinct boxes such that no box is empty?
This is nothing but number of ways of having a,b,c such that a + b + c = 11, where a, b, c are natural numbers.

10C2. 

The two questions stated above are almost identical ones.

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CAT Geometry Solutions

Have given below the solutions to the questions on basic geometry

1. Perimeter of a triangle with integer sides is equal to 15. How many such triangles are possible?

This is just a counting question, with the caveat that sum of two sides should be greater than the third. Let us assume a < b < c

a = 1, Possible triangle 1, 7, 7
a = 2, possible triangle 2, 6, 7
a = 3, possible triangles 3, 6, 6 and 3, 5, 7
a = 4, possible triangles 4, 4, 7 and 4, 5, 6

Again, from comments,
a = 5, possible triangle is 5,5,5,

There are totally 7 triangles possible

2. Triangle ABC has integer sides x, y, z such that xz = 12. How many such triangles are possible?

xz = 12

x,z can be 1, 12 or 2, 6 or 3, 4

Possible triangles
1-12-12
2-6-5; 2-6-6; 2-6-7
3-4-2; 3-4-3; 3-4-5; 3-4-6.

As pointed out in the comments section, I have missed the triangle 3-4-4.

There are totally 9 triangles.

3. Triangle has sides a^2, b^2 and c^2. Then the triangle with sides a, b, c has to be - a) Right angled b) Acute-angled c) Obtuse angled d) can be any of these three

Assuming a < b < c, we have a^2 + b^2 > c^2. This implies the triangle with sides a, b, c has to be acute-angled.

P.S: Big thanks to 'maniac' for pointing out the errors