# Free Modern Maths 02 Practice Test - CAT

At a board meeting, each person shook hands with everyone present. Totally there were 45 handshakes. Mid- way through the meeting, the 3 foreign delegates left as they had a flight to catch. The number of women is now 1 more than the number of men. When the meeting concluded, each person shook hands with only the person of the same gender. What were the total number of handshakes in the meeting?

A. 50
B. 54
C. 52
D. 36

#### SOLUTION

Solution : B

Answer = b

Initially nC2 = 45 =>  n=10

3 people leave, implies that 7 people are left.

number of women = one more than number of men

no. of women = 4. men =3

Handshakes at the end = 4C2+3C2=9
Total handshakes = 45+9 =54.

Find the value of x such that: 7Cx1+7Cx=8Cx+2

A. 2
B. 3
C. 4
D. 5
E. 6

#### SOLUTION

Solution : B

Soln:  We have nCr1+nCr=n+1Cr
7Cx1+7Cx=8Cx

8Cx=8Cx+2=8C(8x2)

Either x = x + 2 or  x + 2 = (8 - x - 2)

But xx+2

2x + 2 = 8

x = 3

Hence option (b)

Alternatively, go from answer options.

Find the value of x for which the value of 16Cx is maximum for any natural number x.

A.
B.
C.
D.

#### SOLUTION

Solution : D

Soln:  Maximum value of 16Cx occurs when x = 162 = 8.

Hence option (d)

If nCr=5 and nPr=120, then determine the value of n, r.

A. 5, 3
B. 4, 2
C. 7, 4
D. 6, 4
E. 5, 4

#### SOLUTION

Solution : E

Soln:  nPrnCr = 1205 = 2

nCr=(nPr)(r!)

nPr[nPrr!]

r! = 24

r = 4

Given nPr = 120

nP4 = 120

n!(n4)!=120

n(n-1)(n-2)(n-3) = 120

n = 5

Hence option (e)

Determine the no. of terms in the expansion of (a+b+c+d)5 and (1+a+a2+a3)5

A. 56, 16
B. 32, 24
C. 165, 32
D. 32, 32

#### SOLUTION

Solution : A

No. of terms in expansion of (a+b+c+d)5 = Number of solution of (a + b + c + d) = 5, 8C3 = 56

No. of terms in expansion of (1+a+a2+a3)5=(5×3)+1=16

Hence option (a)

How many different three digit numbers can be formed with the digits 1, 2, 3, 4, 5 and 6 so that none of the digits are repeated?

A. 120
B. 130
C. 150
D. 100
E. 80

#### SOLUTION

Solution : A

Suppose we start with the unit’s place. Unit place can be filled by any of the 6 digits given, so it can be filled in 6 ways. Now coming the tens place, it can be filled by any of the six digits except the one which has been used at unit’s place. So, it can be filled in 5 ways. Similarly the hundred’s place can be filled by any of the six digits except those two which have been used at unit’s and tens’ place. So, it can be filled in 4 ways.

4 ways   5 ways   6 ways

So, total numbers of numbers possible = 6×5×4=120. Hence option (a)

The number of natural numbers which lie between 108 and 109 which have products of their digits as 6 is___.

#### SOLUTION

Solution :

108<n<109

n is a 9 digit number

Product of 6 is possible in the following cases

Exactly one digit is 6 and the remaining digits are 1

Number of possible numbers= 9

One digit is 2 and the other is 3, remaining are 1 = 9P2 = 72

Total possible numbers = 72+9 = 81

Find the value of ‘n’ if nP6=12nP4.

A. 8
B. 11
C. 9
D. 7
E. 10

#### SOLUTION

Solution : A

Soln:
n!(n6)!=12×n!(n4)!

(n4)!=12×(n6)!

(n4)(n5)=4×3

n = 8

Hint: go from answer options. Only option(a) satisfies given equation.

5 Nephrologists decide to hold daily meetings such that (i) At least one Nephrologist attend each day. (ii) A different set of Nephrologists must attend on different days. (iii) On day N for each 1d<N, at least one Nephrologist must attend who was present on day d. How many maximum days can meetings be held?

A. 14
B. 16
C. 20
D. None of these

#### SOLUTION

Solution : B

We need to find the largest possible number of subsets of {1, 2, 3, 4, 5} such that no 2 subsets are disjoint. Fix one element from the set to be presented in each subset and we can have 24 such possibilities.

Because one element is common. So, remaining 4 can be selected 4C0 + 4C1 + 4C2 + 4C3 + 4C4 = 24 = 16 ways. There are total 16 days.

Ravi has five friends. In how many ways can he invite one or more for a dinner?

A. 32
B. 31
C. 25
D. 120

#### SOLUTION

Solution : B

The question is about selecting 1 or more things from 5 different things.

He can select one in 5C1 ways

He can select two in 5C2 ways

....

He can select all five in 5C5 ways.

Hence total no. of ways = 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 25 – 1 = 31.

Determine the number of ways in which 5 prizes can be distributed among 4 students.

A. 45
B. 54
C. 20
D. 5!4
E. 9

#### SOLUTION

Solution : A

This is a type of “Different to Different” question.

5 prizes can be distributed among 4 students in 45 ways.

Hence option (a)

Find out the number of ways in which 18 flowers can be arranged in a necklace/garland.

A.

18!

B.

17!

C.

16!

D.

12(17!)

#### SOLUTION

Solution : D

This is a case of “Circular Permutation” where you cannot differentiate between clockwise and anticlockwise directions.

No. of arrangements = 12(181)!=12(17!)

Hence option (d)

5 people dine together at a round table. They dine together till each of them dines with different neighbours. Determine the number of days they dine together.

A. 12
B. 24
C. 30
D. 7
E. 120

#### SOLUTION

Solution : A

5 people can sit together around a round table in (51)!=4! Ways. But in anticlockwise and clockwise arrangements each of them will have the same neighbour. So, no. of arrangements possible = 12(4!)=12.

So, they dine together for 12 days.

Hence option (a)

A bag has 3 red balls, 2 yellow balls and 3 black balls. They are drawn one by one and placed in a row. Find the number of ways they can be arranged.

A. 280
B. 410
C. 560
D. 712
E. 478

#### SOLUTION

Solution : C

Soln:

There are a total of 8 balls with 3 of one kind, 2 of other kind and rest 3 of the same kind. These can be arranged in 8!(3!×2!×3!)=560

Consider a Master Set S= {1,2,3,4….12} How many subsets can be formed which will contain one or more elements of S (including all S) such that the elements of the sets are integral multiples of the smallest subset of the set.

A. 2246
B. 2824
C. 3452
D. 2102
E. 1857

#### SOLUTION

Solution : D

If 1 is the smallest element of the set, all or none of the other elements can be selected in 211 ways, as for each number from 2 to 12, there are two options, of getting selected or not getting selected. There are 11 such numbers 2-12 including both, therefore 2.2.2…..11 times = 211

Similarly, If 2 is the smallest element in the set, 4,6,8,10 and 12 can be selected in 25 ways

If 3 is the smallest element in the set 6,9,12 23 different sets

Required Solution = 211 + 25 + 23 + 22 + 21 + 21 + 6 = 2102