Basic Algebra(Exams > Cat > Quantitaitve Aptitude ) Questions and answers for exam Preparation

Exams > Cat > Quantitaitve Aptitude

BASIC ALGEBRA MCQs

Total Questions : 90 | Page 2 of 9 pages

Question 11. Find the number of ways of distributing 8 similar balls into 4 different boxes so that none of the boxes are empty.

11!8!
7!4!
7!3!
7!4!3!
None of these

Discuss Question

Answer: Option D. -> 7!4!3!
:
D
This is a type of “similar to different” questions with a lower limit of 1.
Let the 4 boxes be A,B,C,D
A+B+C+D=8
Applying a lower limit of 1
A+B+C+D=4
Applying 0’s and 1’s method:
No. of zeroes = 4 ,No. of ones = 3
Total number of ways =

\frac{7!}{(4! 3!)}

Question 12. Determine the no. of 4 letter words those can be formed from the letters of the word: CHASSIS.

Discuss Question

Answer: Option D. -> 208
:
D
We have 1 C, 1 H, 1 A, 1 I and 3 S.
The four letter word can be of the form: abcd, aabc, aaab
Case 1: abcd (none repeating): 4 letters can be selected in

^{5} C_{4}

ways. Can be arranged in

4!

Ways.
Case 2: aabc (two repeating): a can only be S. 2 can be selected in

^{4} C_{2}

ways. Can be arranged in

\frac{4!}{(2!)}

ways.
Case 3: aaab (three repeating): a can only be S. 1 can be selected in

^{4} C_{1}

ways. Can be arranged in

\frac{4!}{(3!)}

Ways.
So, total no. of words possible

= (^{5} C_{4} \times 4!) + (^{4} C_{2} \times (\frac{4!}{2!}) + (^{4} C_{1} \times (\frac{4!}{3!})))

= 120 + 72 + 16 = 208.

Question 13. If m parallel lines in a plane are intersected by a family of n parallel lines, calculate the number of parallelograms formed.

mn2
14[mn(m−1)(n−1)]
14[mn(m+1)(n+1)]
(m!n!)2

Discuss Question

Answer: Option B. -> 14[mn(m−1)(n−1)]
:
B
A parallelogram is formed when two select 2 lines from the group of m lines, and 2 from the group of n line. This can be done in =

^{m} C_{2}

and

^{n} C_{2}

ways. So, total no. of parallelograms formed

=^{m} C_{2} \times^{n} C_{2}

= (\frac{1}{4}) [\frac{(m! n!)}{((m - 2)! (n - 2)!)}]

= \frac{1}{4} [m n (m - 1) (n - 1)]

2nd method:-By substitution of any values in choices. Let

m = 2 & n = 2

then total parallelograms =1. Only option C will satisfy. Then option (c).

Question 14. In a college a committee of 7 people has to be selected from a group of 8 fourth year and 6 third year students. In how many ways can this committee be selected if in the committee, majority of fourth year students is required?

2320
1960
2216
2416

Discuss Question

Answer: Option D. -> 2416
:
D
Majority of fourth year means out of 7, minimum 4 should be from fourth year.
Considering different possibilities:

\begin{matrix} No.from 4th year & Ways of choosing 4th year & No.from 3rd year & Ways of choosing 3rd year & wys of choosing commitee 4 & 8C4=70 & 3 & 6C3=20 & 70 \times 20 =1400 5 & 8C5=56 & 2 & 6C2=15 & 56 \times 15=840 6 & 8C6=28 & 1 & 6C1=6 & 28 \times 6 =168 7 & 8C7=8 & 0 & 6C0=1 & 8 \times 1=8 Total & 2416 \end{matrix}

So, total no. of ways = 2416.

Question 15. Calculate the number of ways in which 10 different students can be divided into two groups under two different teachers containing 6 and 4 students respectively.

10!6!
10C6×10C4
10!6!4!×2!
10!4!
None of these

Discuss Question

Answer: Option C. -> 10!6!4!×2!
:
C
No. of ways =

\frac{10!}{(6! 4!)} \times 2!

Here we multiply by

2!

Because the groups are distinct.

Question 16. Department of Science and Technology wants to form a committee of three people from a panel of 7 people, out of which 3 are scientists, 3 are bureaucrat and one is both scientist as well bureaucrat. In how many ways can the committee be formed if it should have at least one scientist and one bureaucrat?

Discuss Question

Answer: Option D. -> 33
:
D

\begin{matrix} C A S E & No. of ways 2 SCIENTISTS, 1 BUREAUCRAT & ^{3} C_{2} \times^{3} C_{1} = 3 \times 3 = 9 1SCIENTIST, 2BUREAUCRAT & ^{3} C_{1} \times^{3} C_{2} = 3 \times 3 = 9 2 OUT OF 6 (3 SCIENTIST, 3 BUREAUCRAT), & ^{6} C_{2} \times 1 = 151 W H O I S B O T H \end{matrix}

So, total number of ways = 9 + 9 + 15 = 33. Hence option (e)

Question 17. Find the number of ways in which 15 players can be equally distributed into three different groups-A,B,C.

[(15!)(5!)3](13!)
[(15!)(5!)3]
[(15!)(5!)]
15!3

Discuss Question

Answer: Option B. -> [(15!)(5!)3]
:
B
In distribution order is important.
The no. of ways in which mn different things can be distributed equally into m groups containing n things each =

[\frac{(m n!)}{(n!)^{m}}]

Here, the groups are different, hence we do not divide by

3! .

So, required no. of ways =

[\frac{(15!)}{(5!)^{3}}]

Question 18. Find the no. of ways in which 15 players can be equally divided into three groups

[(15!)(5!)3](13!)
[(15!)(5!)3]
[(15!)(5!)]
15!3

Discuss Question

Answer: Option A. -> [(15!)(5!)3](13!)
:
A
The no. of ways in which mn different things can be divided equally into m groups containing n things each =

[\frac{(m n!)}{(n!)^{m}}] (\frac{1}{m!})

So, required no. of ways =

[\frac{(15!)}{(5!)^{3}}] (\frac{1}{3!})

We divide by

3!

Since the groups are of equal size and hence, indistinguishable.

Question 19. Calculate the number of ways in which 10 different students can be divided into two groups containing 6 and 4 students respectively.

10!6!
10C6×10C4
10!6!4!
10!4!
None of these

Discuss Question

Answer: Option C. -> 10!6!4!
:
C
No. of ways =

\frac{10!}{(6! 4!)}

2nd method:-Select any six then other four will be automatically selected. i.e

^{10} C_{6} = \frac{10!}{(6! 4!)}

Question 20. Vivek has 31 friends. He wants to invite some of them in a manner that he can throw maximum number of parties; also each party should have same number of guests and different set of persons. How many parties can Vivek throw?