# Free Mixed Bag 3 Practice Test - CAT

### Question 1

N is a natural number. How many values of N exist, such that N2+24N+21 has exactly three factors?

#### SOLUTION

Solution :C

Solution:Option c

For N2+24N+21 to have exactly three factors, it must be square of a prime number.

Let N2+24N+21=a2 where a is a prime number.

(N+12)2−123=a2

(N+12+a)(N+12-a) = 123

Either N+12+a =123 and N+12-a = 1

Or N+12+a = 41 and N+12-a=3

In the first case N = 50 and a = 61.

In the second case N =10 and a =19

In either case N2+24N+21 is the square of a prime number. So two such values exist.

### Question 2

In an arithmetic progression, the p^{th} term is 1/q and the q^{th} term is 1/p. Find the sum of the first ‘pq’ terms of the progression.

#### SOLUTION

Solution :C

Solution: OPTION (c)

Using Assumption

Let p = 1 and q= 2

Then as stated 1

^{st}term is ½ and 2^{nd}term is 1. So we have an AP with common difference ½.So the AP is: ½, 1, 3/2, 2, 5/2….

Now sum of first pq terms : sum of first two terms = ½ + 1 = 3/2

Put p = 1 and q = 2 in the options:

Correct option is (c)

### Question 3

Let A, B, C,... X,Y,Z be positive numbers such that AC=B,BD=C...XZ=Y.Given that A+B=1988 find max value of Y+Z?

#### SOLUTION

Solution :B

Solution:

Let A = x and C =y

Then B = xy

D=1x

E=1xy

F=1y

G = x

H = xy

I = y

So we see that this is a sequence with a period six. So A = Y and B = Z

Given A+B = 1988, so max value of Y+Z = 1988.

### Question 4

In a football team, 12 players are divided into three groups of 4 each and are given different jersey numbers numbering from 1 to 12 in such a way that the sum of the jersey numbers of each of the three groups is same. Find the sum of jersey numbers of each group.

#### SOLUTION

Solution :A

The question is about adding all the numbers from 1 to 12 and then dividing by 3. So, sum of numbers from 1 to 12: 12×132=78. So, sum of jersey numbers of each of the groups =783=26

### Question 5

In the given figure three squares a, b and c of side 3, 5 and 8 units are placed adjacent each other. Find the area of the shaded region.

#### SOLUTION

Solution :B

Option(b)

Let x and y be the bases of smallest two triangles.

(y8)=(816); y=4.

(xy)=(x4)=(38); x=1.5.

Thus area of shaded region: 12 ((4*8)-(1.5*3))=13.75

### Question 6

A Swedish watch company is assigning ID cards for its employees. The ID is made of 7 digits. How many ID numbers can exist such that at least one of the digits repeat?

9456745

7898760

8455680

None of these

#### SOLUTION

Solution :C

Total number of combinations = No repeats + 1 number repeat + 2 number repeat +.....7 number repeat

Total number of combinations - No repeats = 1 number repeat + 2 number repeat +...... 7 number repeat

Total number of combinations - No repeats = Atleast one number repeat

106×9−(9.9.8.7.6.5.4)=9000000−544320=8455680

### Question 7

If f(x) = min (7x + 3, 8x – 6) for 0 < x < 4, then determine the maximum value of f(x).

#### SOLUTION

Solution :B

Soln:

Equate the two terms to get the point of intersection

7x+3=8x-6

X=9, which is greater than the given constraint

Minimum/Maximum of two increasing function will also be an increasing function.

Here 7x + 3 and 8x – 6 are both increasing functions, so f(x) will also be an increasing function.

Based on the constraints given, max of f(x) will occur at x = 4 i.e. f(4) = min (31, 26) = 26

Hence option (b)

### Question 8

Find the number of integral solutions to |x| + |y| + |z| = 15.

#### SOLUTION

Solution :C

option c

Case 1:When none of them are zero

Number of +ve integral solutions of x + y + z = 15 is 14C2 however, x y z can be -ve also total 8 ways for each set of x y z Total 14C2×8=728

Case 2:When one of them is zero

No of solutions = 3×4×14C1=168 Case 3: When two of them are zeroes

Number of solutions = 3×2=6

Total= 902, option (c)

### Question 9

The integers P,Q,R and S shown on the number line below are all equally spaced. If R and S are equal to 712 and 713, respectively, what is the value of P?

#### SOLUTION

Solution :D

The distance between R and S is 713−712=6×712

The distance between two consecutive points is constant.

Value of Q=712−6×712=−5×712

Value of P=−5×712−6×712=−11×712

The correct answer is option (d).

### Question 10

Considering all 2-digit natural numbers, how many values of "y" do not satisfy the equation |7x-5y|=3, given that "x" and "y" are positive integers.

#### SOLUTION

Solution :B

Let us first look at the conventional approach.

7x - 5y = 3

⇒ 7x = 5y+3 ----------------(1)

and

-7x + 5y = 3

⇒ 7x+3 = 5y ----------------(2)

Solving equation (1), we get the first integral value for y, at which x is an integer at y = 5. Values of y will increase in steps of 7 (the coefficient of x). The next few values of "y" satisfying the equation will be 5,12,19.......96.

Number of terms = 13 (considering 2 digit numbers)

Solving equation (2), we get the first integral value for y at y = 2. Values of y will increase in steps of 7. Hence the second AP will be 2, 9, 16, 23...... 93.

Number of terms = 12 (considering 2 digit numbers)

Number of values of y which satisfy this equation = 25

Therefore, number of values which do not satisfy this equation = 90 - 25 = 65. The answer is option (b).

Shortcut:

We know that there are 90 2-digit numbers.

The values which satisfy for "y" form an AP with a common difference = 7 (the coefficient of x).

Hence, the number of terms in that AP =907×2=24 or 25; since there are two APs.

The answer has to be either 90-24 = 66 or 90-25 = 65. Answer is option (b).

### Question 11

If r,s are the roots of the equation x^{2}-px+q=0, then what is the equation of the curve whose roots are ([1/r] + s) and ([1/s]+r)?

^{2}-q(1-p)x+(1+q

^{2})=0

^{2}+(p-q)x+(q

^{2}-1)=0

^{2}+pqx-(p+1)(q+1)=0

^{2}-p(q+1)x+(1+q)

^{2}=0

^{2}-p(q+1)x+(1+q)=0

#### SOLUTION

Solution :D

option d

put r=1 and s=1, then p=2 and q=1.

Roots of the curve’s equation = 2,2

Substitute values of p,q,r,s in options to check where 2,2 satisfy as roots. This happens only in option (d)

### Question 12

Find the minimum value of the expression, P = 2^{x+1} + 1/ [4^{(x/2 + 3/2)}] , given that x is real

#### SOLUTION

Solution :B

The expression can be rewritten as

2

^{x+1}+ 2^{-2 (x/2 +3/2)}= 2x+1 + 2

^{-x-3}Using the concept of AM ≥ GM

Therefore, the min value occurs when both the numbers are same. In that case 2^{(x+1) }= 2^{-(x+3)}. X = -2. Thus the number = ½.so ½ +1/2 = 1.

### Question 13

When 315! is divided by 1215x, remainder = 0. What is the maximum possible value for x?

#### SOLUTION

Solution :A

The question is based on highest power of a number in a factorial.

Here since 1215 is composite number, prime factorize 1215 i.e 1215=35×5.

Required answer will be the highest power of 35 in 315!

(No need to find to find the highest power of 5 in 315! as that will always be more than that of 35)

To find out highest power of 35, we will first find the highest power of 3 and then divide it by 5.

Highest power of 3 in 315! = 155 (105 + 35 + 11 + 3 + 1)

Highest power of 35 in 315! = 31 (highest power of 5 we will get is 77)

Required answer is 31.

### Question 14

A person has to move from A to B. He can only move upwards or towards the right. Find the probability that he passes through C?

#### SOLUTION

Solution :B

Option (b)

Totally there are 4 vertical distances and 5 horizontal distances VVVVHHHHH. Number of ways in

which a person can go from A to B via C= Arrangement of 9 things out of which 4 are identical and of one type and 5 are identical and of another type. = 9!4!×5!=126

Number of ways of reaching C from A= HHVV= 4C2=6. Number of ways of reaching B from C= HHHVV= 5C3=10

Required Probability = 10×6126=1021

### Question 15

A house is at the center of the several flat paths which surround it: 4 straight paths that travel from the house to its circular moat, where they meet up with a perfectly circular path which borders the moat; that circular path circumscribes a square path which has its corners at the ends of the 4 straight paths—see the diagram to the right. If the total length of all of the pathways is q kilometers, then which expression represents the distance from the house to the circular moat?

#### SOLUTION

Solution :B

Option(b)

Solution: Shortcut:- Using Assumption

Finding the total lengths of all the pathsAssuming radius of the circle =2

Straight paths= 2x4=8

Each side of the square = 2√2 x4=8√2

Circular moat perimeter= 4Π

q (Total) = 4( 2+2√2 + Π)

Distance from centre to the moat = 2

Look for 2 in the answer options. Answer is option (b)