Free Mixed Bag 3 Practice Test - CAT

Question 1

N is a natural number. How many values of N exist, such that $N^{2} + 24 N + 21$ has exactly three factors?

A. 0

B. 1

C. 2

D. 3

>

SOLUTION

Solution : C

Solution:Option c

For $N^{2} + 24 N + 21$ to have exactly three factors, it must be square of a prime number.

Let $N^{2} + 24 N + 21 = a^{2}$ where a is a prime number.

$(N + 12)^{2} - 123 = a^{2}$

(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 $N^{2} + 24 N + 21$ 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.

B. pq + 1

E. p*q

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?

A. 994

B. 1988

\sqrt{1988}

D. 3976

E. none of these

SOLUTION

Solution : B

Solution:

Let A = x and C =y

Then B = xy

$D = \frac{1}{x}$

$E = \frac{1}{x y}$

$F = \frac{1}{y}$

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.

A. 26

B. 34

C. 78

D. Can’t be determined

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: $12 \times \frac{13}{2} = 78$ . So, sum of jersey numbers of each of the groups $= \frac{78}{3} = 26$

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.

A. 12.75

B. 13.75

C. 10

D. 15

SOLUTION

Solution : B

Option(b)

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

$(\frac{y}{8}) = (\frac{8}{16})$ ; y=4.

$(\frac{x}{y}) = (\frac{x}{4}) = (\frac{3}{8})$ ; x=1.5.

Thus area of shaded region: $\frac{1}{2}$ ((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

$10^{6} \times 9 - (9.9.8.7.6.5.4) = 9000000 - 544320 = 8455680$

Question 7

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

A. 66

B. 26

C. 31

D. 28

E. 54

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.

A. 900

B. 901

C. 902

D. 903

E. none of these

SOLUTION

Solution : C

option c

Case 1:When none of them are zero

Number of +ve integral solutions of x + y + z = 15 is $^{14} C_{2}$ however, x y z can be -ve also total 8 ways for each set of x y z Total $^{14} C_{2} \times 8 = 728$

Case 2:When one of them is zero

No of solutions = $3 \times 4 \times^{14} C_{1} = 168$ Case 3: When two of them are zeroes

Number of solutions = $3 \times 2 = 6$

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 $7^{12}$ and $7^{13}$ , respectively, what is the value of P?

7^{10}

7^{11}

- 7^{13}

D. None of these

SOLUTION

Solution : D

The distance between R and S is $7^{13} - 7^{12} = 6 \times 7^{12}$

The distance between two consecutive points is constant.

Value of $Q = 7^{12} - 6 \times 7^{12} = - 5 \times 7^{12}$

Value of $P = - 5 \times 7^{12} - 6 \times 7^{12} = - 11 \times 7^{12}$

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.

A. 60

B. 65

C. 73

D. 48

SOLUTION

Solution : B

Let us first look at the conventional approach.
7x - 5y = 3
$\Rightarrow$ 7x = 5y+3 ----------------(1)
and
-7x + 5y = 3
$\Rightarrow$ 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 $= \frac{90}{7} \times 2 = 24 o r 25$ ; 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 x²-px+q=0, then what is the equation of the curve whose roots are ([1/r] + s) and ([1/s]+r)?

A. px²-q(1-p)x+(1+q²)=0

B. qx²+(p-q)x+(q²-1)=0

C. (p-q)x²+pqx-(p+1)(q+1)=0

D. qx²-p(q+1)x+(1+q)²=0

E. qx²-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

A. ½

B. 1

C. ¼

D. 2

E. ¾

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 $1215^{x}$ , remainder = 0. What is the maximum possible value for x?

A. 31

B. 15

C. 32

D. 33

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 $1215 = 3^{5} \times 5$ .

Required answer will be the highest power of $3^{5}$ in 315!

(No need to find to find the highest power of 5 in 315! as that will always be more than that of $3^{5}$ )

To find out highest power of $3^{5}$ , 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 $3^{5}$ 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?

\frac{1}{2}

\frac{10}{21}

\frac{13}{24}

\frac{1}{3}

\frac{13}{34}

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. = $\frac{9!}{4! \times 5!} = 126$

Number of ways of reaching C from A= HHVV= $^{4} C_{2} = 6$ . Number of ways of reaching B from C= HHHVV= $^{5} C_{3} = 10$

Required Probability = $\frac{10 \times 6}{126} = \frac{10}{21}$

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?

\frac{q}{4 \sqrt{2} + 2 \sqrt{2} + π}

\frac{q}{2 (2 + 2 \sqrt{2} + π)}

\frac{q}{2 + 2 \sqrt{2} + π}

\frac{2 q}{2 + 2 \sqrt{2} + π}

SOLUTION

Solution : B

Option(b)
Solution: Shortcut:- Using Assumption

Finding the total lengths of all the paths

Assuming radius of the circle =2

Straight paths= 2x4=8

Each side of the square = 2 $\sqrt{2}$ x4=8 $\sqrt{2}$

Circular moat perimeter= 4 $Π$

q (Total) = 4( 2+2 $\sqrt{2}$ + $Π$ )

Distance from centre to the moat = 2

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

Free Mixed Bag 3 Practice Test - CAT

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

Related ExamsMixed Bag 4Mixed Bag 5Mixed Bag 6

Related Exams
Mixed Bag 4
Mixed Bag 5
Mixed Bag 6