# Free Mixed Bag 2 Practice Test - CAT

### Question 1

Find the unit digit of the LCM of 73001–1 and 73001+1.

#### SOLUTION

Solution :D

option d

73001−1 and 73001+1 are consecutive even natural numbers

Hence, their HCF=2

LCM=product of numbers HCF=(78001)2−12=76002−12

76002 ends with 9 (use the concept of power cycle)

So 76002−1 ends in 8. We also know that as 76002 is a perfect square of an odd number, its tens digit will be even. So the required unit digit is 4.

### Question 2

If an integer N is equal to the sum of all even multiples of 15 between 295 and 615, what is the greatest prime factor of N?

#### SOLUTION

Solution :C

option (c)

Finding the sum of even multiples of 15 is equivalent to finding the sum of multiples of 30.

By observation, the first multiple of 30 greater than 295 will be equal to 300 and the last multiple of

30 smaller than 615 will be equal to 600.

The sum of a set = (the mean of the set) × (the number of terms in the set)

There are 11 terms in the set

The mean of the set = (the first term + the last term) divided by 2: (300+600)2=450

If k= the sum of this set

k=450×11

Therefore, the largest prime factor of k is 11. The correct answer is C.

### Question 3

Find the inverse of the quadratic function f.
f(x) = x^{2} - 2x , x ≥ 1

^{2}-4

#### SOLUTION

Solution :D

Using the technique for inverse function at x=1, f(x) = -1

Replace x by f(x) in the answer options and

Look for an answer options where x= -1, & f(x)=1

Answer is option (d)

2nd method: - Let y = x

^{2}- 2xx

^{2}- 2x - y = 0

### Question 4

A number sequence has 100 elements. Any of its elements (except for the first and last element) is equal to the product of its neighbors. The product of the first 50 elements, just as the product of all the elements is 27. What is the sum of the first and the second element?

#### SOLUTION

Solution :C

option c

Solution:

Let first two numbers be x and y.. so the sequence will be

x,y,yx,1x,1y,xy,x,y,.. the sequence repeats after 6 numbers. Also the product of first six numbers is 1.

So now, product of first 50 numbers will be xy=27 while product of first 100 numbers will be y2x=27. On solving we get y=9 and x=3.

So their sum = x+y=12.

Alternatively

Look at the product, it is 27 which is nothing but 33and the series is all about products. So the numbers in the series will be full of threes. Look at the answer options for numbers which can be expressed as sum of 2 powers of 3. Only 12=3+32is the possibility.

Series is 3,32,33,13,132,13,3,32.............

### Question 5

Two varieties of Coffee worth Rs. 126 per kg and Rs. 130 per kg are mixed with a third variety in the ratio 1: 1: 2. If the resulting mixture is worth Rs. 172 per kg, then the price (Rs. /kg) of the third variety of the coffee is

#### SOLUTION

Solution :Two varieties @ Rs. 126 and Rs. 130 are mixed in equal ratio of 1: 1. So, their average price can be taken as (126+130)2=Rs. 128 per kg. Now we have two varieties: one at the rate Rs. 128 per kg and other at the rate Rs. x per kg which are mixed in the ratio 2: 2 or 1: 1 to give a resulting mixture of Rs. 172 per kg.

Applying Alligation:

X−17244=11

x = 216.

So, the price of the third variant is Rs. 216 per kg.

### Question 6

A ship develops a leak 12 km from the shore. Despite the leak, the ship is able to move towards the shore at a speed of 8 km/hr. However, the ship can stay afloat only for 20 minutes. If a rescue vessel were to leave from the shore towards the ship, and it takes 4 minutes to evacuate the crew and passengers of the ship, the minimum speed of the rescue vessel in order to be able to successfully rescue the people aboard the ship is (in kmph)

#### SOLUTION

Solution :As it takes 4 mins to rescue the passengers, in 16 mins the rescue vessel should reach the ship. Now the ship too is traveling towards the shore with 8kmph.

Let the speed of the rescue vessel be X.. Then X×1660+8×1660=12

⇒ X = 37.

### Question 7

In the given isosceles right angled triangle UVW, a square PQRS is inscribed as shown in the figure. If PV:VS=2:1, what is the ratio of areas of the square to the outer triangle UVW?

#### SOLUTION

Solution :D

Ans: d. 2:5

Using variables

In the given figure, we draw QT || VW.

Δ PTQ and Δ PVS are congruent (A,A,A and side)

Hence PT=Y, QT=X=UT (also since ΔUTQ and ΔUVW are similar).

Thus UV=UW=2X+Y. Area of square: X2+Y2; Area of ΔUVW= 12 *(2X+Y)2.

Given X=2Y, X2+Y2: ( 12 *(2X+Y)2) = 5Y2: ( 25Y2* 12 ); = 25.

Using numbers,You can solve the problem faster as follows:

PV=2, SV=1 => PS= √5 ⇒ Area of square =5

PTQ congruent to PVS

PT=1 and QT=2

UT=TQ (45-45-90) ⇒ UT= 2

Now UV= 5 ⇒ WV=5 (45-45-90)

UW=5√2

(Area of square / Area of triangle )= (5(25/2)) = 2:5

### Question 8

Find f(x) + g(x)

#### SOLUTION

Solution :D

option (d)

f(x) = [(x + 1)/x] [(x + 2)/(x +1)] … [(x +n + 1)/(x + n)] = [(x + n + 1)/x]

g(x) = [(x – 1)/x] [(x -2)/(x – 1)] … [(x-n-1)/(x – n)] = [(x – n -1)/x]

f(x) + g(x) = (x + n + 1 + x - n – 1)/x = 2

Hence option (d)

Shortcut:- Single Substitution

Put x=1, n=0

f(x)= 2

g(x)=0 => f(x) +g(x) =2

put x=1 and n=0 and look for 2 in the answer options. Answer is option (d)

### Question 9

Ram can complete a work in 10 hours, Ajit in 12 hours and Suresh in 15 hours. All of them began the work together, but Ram had to leave the work after 2 hours of the start and Ajit left 3 hours before the completion of the work. How long did the work last?

#### SOLUTION

Solution :In 1 hour Ram can do 10% of the work, Ajit can do 10012% of the work and Suresh can do 10015% of the work.

So, total number of hours = 2 + 2 + 3 = 7 hours.

### Question 10

Cosider a, b, c : real numbers. X=a+ba−b,Y=b+cb−c,Z=c+ac−a

What is tha value of XY + YZ + XZ ?

#### SOLUTION

Solution :E

Option (e)

Assume a=1, b=2, c=3

X=-3 Y=-5 Z=2

XY+YZ+XZ=15-10-6=-1

Assume a=2, b=3 and c=4

X=-5 Y=-7 Z=3

XY+YZ+XZ= 35-21-15= -1

Hence answer =e

### Question 11

The total number of ways in which a 5 digit number divisible by 3 can be formed from the digits 0,1,2,3,4,5 without repetition is?

#### SOLUTION

Solution :C

option (c )

Divisibility test for 3- sum of digits should be divisible by 3.

Case 1:- When 0 is not included → sum = 15 divisible by 3

Number of combinations = 5! = 120

Case 2:- When 3 is not used. → Sum = 12 divisible by 3

Number of combinations - 4.4..3.2 =96

Total = 120+96= 216

### Question 12

Let A and B be digits (that is, A and B are integers between 0 and 9 inclusive). If the product of the three-digit integers 2A5 and 13B is divisible by 36, determine the number of possible ordered pairs (A,B).

#### SOLUTION

Solution :D

option d

For the product (2A5)(13B) to be divisible by 36, we need it to be divisible by both 4 and 9. Since 2A5 is odd, it does not contain a factor of 2.Therefore, 13B must be divisible by 4.

For a positive integer to be divisible by 4, the number formed by its last two digits must

be divisible by 4, i.e. 3B is divisible by 4, i.e. B= 2 or B = 6.

Case 1: B = 2

In this case, 132 is divisible by 3, but not by 9. Therefore, for the original product to be

divisible by 9, we need 2A5 to be divisible by 3.

For a positive integer to be divisible by 3, the sum of its digits is divisible by 3, i.e.

2 + A + 5 = A + 7 is divisible by 3.

Therefore, A = 2 or 5 or 8.

Case 2: B = 6

In this case, 136 contains no factors of 3, so for the original product to be divisible by 9,

we need 2A5 to be divisible by 9.

For a positive integer to be divisible by 9, the sum of its digits is divisible by 9, i.e.

2 +A+5 is divisible by 9. Therefore, A is 2.

Therefore, the four possible ordered pairs are ⇒A, B = 2,2, 8,2, 5,2, 2,6.

### Question 13

f_{n}= f_{n – 1}, if n is odd= 2f_{n-1}, if n is even.If f_{0} = 1, then find the value of f_{20} + f_{21}.

#### SOLUTION

Solution :B

Soln:

Option (b)

f

_{4}+ f_{5}= 2f_{3}+ f_{4}= 2f_{2}+ 2f_{3}= 4f_{1}+ 2f_{2}= 4f_{0}+ 4f_{1}= 4f_{0}+ 4f_{0}= 8f_{0}= 8Hence option (e)

OR starting with

f(0) = 1

F(1) = 1

F(2) = 2

F( 3) = 2

F(4) = 4

F (5) = 4

F(6) = 8

F(7) = 8

……………..

…………….

F(N) =2

^{N/2}F(N+1) = 2

^{N/2}

Therefore f( 20) = 2

^{10}and f(21) = 2^{10}f(20) + f(21) =1024+1024 = 2048

### Question 14

Find the maximum number of people who liked all three actors.

#### SOLUTION

Solution :A

As we know, S = I + 2 II + 3 III

and X = I + II + III

S - X = II + 2 III

If we want to maximize exactly III. We need to put exactly II = 0

(S−X)2⇒(224−120)2=52

Short Cut: - Maximum of all can be directly calculated as

S−Xn−1, Where n is number of things.

Answer is (224−120)2=52. Option (a)

### Question 15

Find the minimum number of people who liked all three actors.

#### SOLUTION

Solution :C

1st method: -

80 people like Shahrukh, it means at least 20 people like Amir or Akshay or Amir and Akshay both.

Similarly atleast 36 people like Shahrukh or Akshay or Shahrukh and Akshay both.

Again atleast 20 people like people like Shahrukh or Amir or Shahrukh and Amir both.

40 + 56 + 40 = 136

It means that if there is no interection in these three courses, 136 would be maximum number of people who like Shahrukh, Amir, Akshay alone or 2 at a time.

Thus it gives atleast 0% people like all three.

2nd method: - Use minimum of all technique taught in class

120-80= 40

120-64=56

120-80=40

Sum of all differences = 136

120−136<0. Hence, the minimum is 0. Option (e)