# Free Modern Maths 01 Practice Test - CAT

### Question 1

The number of 5 digit telephone numbers having at least one of their digits repeated is? (0 can be the starting digit of the number)

#### SOLUTION

Solution :D

Total number of 5 digit telephone numbers which can be formed using digits 0,1,2…9 is 105

Number of 5 digit numbers having not even a single digit repeated is 10P5 = 30240Required number of telephone numbers = 105 -30240 = 69760

### Question 2

In how many ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order?

#### SOLUTION

Solution :D

Since the order of vowels will always remain the same despite these occupying different positions -> if we assume each vowel as X then our question is same as asking "arrange JPTRXXX" => in all 7!3! ways. => Choice (d) is the right answer

### Question 3

In how many ways can you select 2 numbers till 50, which will have their product as a multiple of 3?

#### SOLUTION

Solution :B

To get a two factor product out of 50 numbers, we can choose the two numbers in 50C2 ways, out of

which we need to negate pairs wherein both numbers are not multiples of 3.

Non multiples of 3 till 50=50−503=34.

50C2−34C2

### Question 4

To set up a farm, a farmer goes to buy some animals. He selects from cows, sheep, goats, chickens and pigs. In how many ways can he select 15 animals such that he selects atleast 1 cow, 2 sheep, 2 goats and 3 chicken.

#### SOLUTION

Solution :D

C+S+G+Ch+P = 15

Removing 1+2+2+3 =8

C+S+G+Ch+P=7

This is the arrangement of 7 zeroes and 4 ones. 11C4=330

### Question 5

If all the letters of the word RASAM can be arranged as in a dictionary, what is its rank?

#### SOLUTION

Solution :C

First we fix A as the first letter

No. of words possible with A as the first letter = 4! = 24

No. of words possible with M as the first letter= 4!2! = 12

No. of words possible with RAA as the first 3 letters = 2

No. of words possible with RAM as the first 3 letters = 2

Next word will be RASAM

Rank= 41

### Question 6

A team of 10 is to be formed from 4 girls and 11 boys. How many ways this can be done by taking at most 3 girls in the team?

#### SOLUTION

Solution :B

Case 1- No girls

Possible selections = 11C10=11

Case 2- One girl

Possible selections= 4×11C9=220

Case 3- Two girlsPossible selections =4C2×11C8=990

Case 4- Three girls

Possible selections = 4C3×11C7=1320

Total= 2541

### Question 7

How many integral solutions are there for the equation A+B+C = 10 where A≥1,B≥2,C≥3?

#### SOLUTION

Solution :B

This is a question based on S‘aD with a varying lower limit.

Take out 1+2+3=6 from the RHS. Equation changes to

A+B+C = 4. Answer will the arrangement of 4 zeroes and 2 ones = 6C2

### Question 8

Find the number of natural number solutions of x1+x2+x3+3x4=15.

#### SOLUTION

Solution :B

X1X2X33X4=15Case 1:X1X2X33=15Case 2:X1X2X36=15Case 3:X1X2X39=15Case 4:X1X2X312=15

Case 1:

X1+X2+X3=12; Total number of natural number solution = 11C2=55

Case 2:

X1+X2+X3=9; Total number of natural number solution = 8C2=28

Case 3:

X1+X2+X3=6; Total number of natural number solution = 5C2=10

Case 4:

X1+X2+X3=3; Total number of natural number solution = 1

Total number of natural number solution = 55 + 28 + 10 + 1 = 94

### Question 9

In how many ways can you put 7 letters into their respective envelopes such that exactly 3 go into the right envelope?

#### SOLUTION

Solution :No of ways in which the 3 correct envelopes can be selected= 7C3=35

Derangement of 4 envelopes & letters = 9 (derangement value for 4 is 9)

Total No of ways of arrangement = 9×35 = 315

### Question 10

The number of four digit numbers that can be formed from the digits 2,3,5,6,7,8 so that the digits do not repeat and the terminal digits are even is

#### SOLUTION

Solution :If the first digit and last digit of the four digit number needs to be even, then we have to choose numbers out of 2, 6 and 8

The first digit can be chosen in 3 ways and the last digit in 2 ways = 3×2=6 ways

The remaining two digits can be chosen and arranged among the remaining 4 digits in 4P2 ways = 12 ways

Total number of ways = 6×12=72 ways

### Question 11

There are 6 friends A,B,C,D,E and F. B does not want to sit next to D,E or F and E does not want to sit next to A,B or C. In how many ways can you arrange these 6 people around a circular table?

#### SOLUTION

Solution :D

If B, does not want any neighbour among D,E,F, B needs to be seated between only A and C. A and C can be arranged in 2! Ways

Similarly, E who does not want to be seated next to A,B or C will be seated between the other two( D,F). Those two can be arranged in 2! Ways.

Total number of arrangements= 2×2=4

### Question 12

How many numbers (N) can we have, such that 100<N<1000 with the sum of the digits of these numbers as 10?

#### SOLUTION

Solution :A

Use the concept of grouping, for a three digit number ABC

A+B+C = 10, A >1

It changes to A + B + C = 9

Number of possible solutions = 11C2=55

### Question 13

The maximum number of rectangles that can be formed from a set of 5 parallel lines intersecting another set of 6 parallel lines is

#### SOLUTION

Solution :B

To form a rectangle we need four intersecting lines.

The first two parallel lines can be selected in 5C2 ways and the other two parallel lines can be selected in 6C2 ways.

Total number of parallelograms which can be formed= 10×15=150

### Question 14

What are the numbers of ways in which 5 similar balls will be put in 4 distinct baskets when the second basket has exactly 2 balls?

#### SOLUTION

Solution :B

Let’s number the baskets as A,B,C and D

With the first constraint, 2 of the balls are going to basket 2(B), hence we have 5-2=3 balls left to distribute among the remaining baskets.

A+C+D = 3

Number of distributions possible = 5C2 = 10

### Question 15

What are the numbers of ways in which 5 similar balls will be put in 4 distinct baskets when the first basket has 0 balls?

#### SOLUTION

Solution :B

0 balls are going to basket A. hence we will consider the distribution only in the remaining 3 baskets

B+C+D = 5

Total = 7C2=21