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)

A. 30240

B. 10500

C. 45600

D. 69760

E. 62660

SOLUTION

Solution : D

Total number of 5 digit telephone numbers which can be formed using digits 0,1,2…9 is $10^{5}$

Number of 5 digit numbers having not even a single digit repeated is $^{10} P_{5}$ = 30240

Required number of telephone numbers = $10^{5}$ -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?

A. 736

B. 768

C. 792

D. 840

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 $\frac{7!}{3!}$ 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?

A. 561

B. 664

C. 652

D. 284

E. 585

SOLUTION

Solution : B

To get a two factor product out of 50 numbers, we can choose the two numbers in $^{50} C_{2}$ ways, out of

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

Non multiples of 3 till $50 = 50 - \frac{50}{3} = 34$ .

$^{50} C_{2} -^{34} C_{2}$

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.

A. 1215

B. 1306

C. 210

D. 330

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. $^{11} C_{4} = 330$

Question 5

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

A. 40

B. 35

C. 41

D. 42

E. None of these

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= $\frac{4!}{2!}$ = 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?

A. 1435

B. 2541

C. 2381

D. 3031

E. 2600

SOLUTION

Solution : B

Case 1- No girls

Possible selections = $^{11} C_{10} = 11$

Case 2- One girl

Possible selections= $4 \times^{11} C_{9} = 220$

Case 3- Two girls

Possible selections = $^{4} C_{2} \times^{11} C_{8} = 990$

Case 4- Three girls

Possible selections = $^{4} C_{3} \times^{11} C_{7} = 1320$

Total= 2541

Question 7

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

^{5} C_{2}

^{6} C_{2}

^{12} C_{2}

^{8} C_{2}

E. None of these

SOLUTION

Solution : B

This is a question based on $S ‘ a D$ 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 = $^{6} C_{2}$

Question 8

Find the number of natural number solutions of $x_{1} + x_{2} + x_{3} + 3 x_{4} = 15$ .

A. 78

B. 94

C. 106

D. 103

E. 87

SOLUTION

Solution : B

$\begin{matrix} X_{1} & X_{2} & X_{3} & 3 X_{4} & = 15 Case 1: & X_{1} & X_{2} & X_{3} & 3 & = 15 Case 2: & X_{1} & X_{2} & X_{3} & 6 & = 15 Case 3: & X_{1} & X_{2} & X_{3} & 9 & = 15 Case 4: & X_{1} & X_{2} & X_{3} & 12 & = 15 \end{matrix}$

Case 1:
$X_{1} + X_{2} + X_{3} = 12;$ Total number of natural number solution = $^{11} C_{2} = 55$
Case 2:
$X_{1} + X_{2} + X_{3} = 9;$ Total number of natural number solution = $^{8} C_{2} = 28$
Case 3:
$X_{1} + X_{2} + X_{3} = 6;$ Total number of natural number solution = $^{5} C_{2} = 10$
Case 4:
$X_{1} + X_{2} + X_{3} = 3;$ 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= $^{7} C_{3} = 35$

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

Total No of ways of arrangement = $9 \times 35$ = 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 = $3 \times 2 = 6$ ways

The remaining two digits can be chosen and arranged among the remaining 4 digits in $^{4} P_{2}$ ways = 12 ways

Total number of ways = $6 \times 12 = 72$ 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?

A. 8

B. 2

C. 1

D. 4

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= $2 \times 2 = 4$

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?

A. 55

B. 66

C. 78

D. 27

E. 52

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 = $^{11} C_{2} = 55$

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

A. 620

B. 150

C. 120

D. 180

E. 200

SOLUTION

Solution : B

To form a rectangle we need four intersecting lines.

The first two parallel lines can be selected in $^{5} C_{2}$ ways and the other two parallel lines can be selected in $^{6} C_{2}$ ways.

Total number of parallelograms which can be formed= $10 \times 15 = 150$

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?

A. 12

B. 10

C. 18

D. 12

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 = $^{5} C_{2}$ = 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?

A. 15

B. 21

C. 19

D. 10

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 = $^{7} C_{2} = 21$

Free Modern Maths 01 Practice Test - CAT

Question 1

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Question 2

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Question 3

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Question 4

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Question 5

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Question 6

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Question 7

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Question 8

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Question 9

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Question 10

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Question 11

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Question 12

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Question 13

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Question 14

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Question 15

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