Free Basic Algebra - 03 Practice Test - CAT

Question 1

The value of $\sqrt{6 + \sqrt{6 + \sqrt{6 + . . . . . \infty}}}$ is:

A. 2

B. -2

C. 3

D. -2 or 3

SOLUTION

Solution : C

Let y be the given expression, then

$y = \sqrt{(6 + y)}$

$y^{2} = 6 + y$

$y^{2} - y - 6 = 0$

(y – 3)(y + 2) = 0

y = 3, y = -2

So, the value of the given expression is 3.

Question 2

If $x^{3} - k^{2} x^{2} + (6 k + 5) x - 5 k = 0$ has two roots 3 and 5, then which of the following is the third root?

A. 1

B. 2

C. 3

D. 4

SOLUTION

Solution : A

Assume $α$ be the 3rd root.

Sum of the roots, i.e., $3 + 5 + α = k^{2}$

Product of the roots, i.e. $3 \times 5 \times α = 5 k$

So, $8 + α = 9 α^{2}$

Then, $α = 1 o r - \frac{8}{9}$

Hence the 3rd root is 1.

Question 3

A basket has 5 oranges and 4 apples. In how many ways can you make a selection if you have to take at least 1 orange and 1 apple?

345

365

465

SOLUTION

Solution : D

At least 1 orange can be selected in $2^{5} - 1 = 31$ ways.

At least 1 apple can be selected in $2^{4} - 1 = 15$ ways.

Total no. of selections possible $= 31 \times 15 = 465$

Question 4

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?

A. 2320

B. 1960

C. 2216

D. 2416

SOLUTION

Solution : 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 5

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?

SOLUTION

Solution : D

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

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

Question 6

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?

^{31} C_{15}

31!

^{31} C_{17}

15!

SOLUTION

Solution : A

Soln:

As n is odd, $^{n} C_{r}$ is maximum when r = $\frac{(n - 1)}{2}$ =15 or $\frac{(n - 1)}{2}$ = 16.

So, number of parties possible $=^{31} C_{16} =^{31} C_{15} .$

Hence option (a)

Question 7

How many different straight lines can be formed by joining 14 different points of which 6 are collinear and rest are non-collinear?

A. 76

B. 63

C. 64

D. 77

SOLUTION

Solution : D

No. of lines formed by joining 14 points = $^{14} C_{2}$

No. of lines formed by joining 6 points = $^{6} C_{2}$

Out of the 14 points, when you join the 6 collinear points, you don’t get $^{6} C_{2}$ lines, but you get only one line.

So, no. of lines $=^{14} C_{2} -^{6} C_{2} + 1 = 91 - 15 + 1 = 77$ .

2nd method:- By selection, $^{8} C_{2} +^{8} C_{1} \times^{6} C_{1} + 1 = 77$ .

Question 8

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

\frac{m n}{2}

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

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

\frac{(m! n!)}{2}

SOLUTION

Solution : 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 9

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

196

216

217

208

SOLUTION

Solution : 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 10

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

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

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

$[\frac{(15!)}{(5!)}]$

$\frac{15!}{3}$

SOLUTION

Solution : 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 11

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

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

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

$[\frac{(15!)}{(5!)}]$

$\frac{15!}{3}$

SOLUTION

Solution : 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 12

In how many ways can a pack of 52 cards be divided equally among 4 players?

\frac{(52!)}{[(4!)^{13}]}

\frac{\frac{(52!)}{[(13!)^{4}]}}{4!}

\frac{(52!)}{[(13!)^{4}]}

\frac{(52!)}{(4!)}

E. None of these

SOLUTION

Solution : C

Here as it has to be divided among 4 different players, the order is important.

No. of ways = $\frac{(52!)}{[(13!)^{4}]}$ .

Question 13

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

\frac{10!}{6!}

^{10} C_{6} \times^{10} C_{4}

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

\frac{10!}{4!}

SOLUTION

Solution : 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 14

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.

$\frac{10!}{6!}$

$^{10} C_{6} \times^{10} C_{4}$

$\frac{10!}{6! 4!} \times 2!$

$\frac{10!}{4!}$

SOLUTION

Solution : C

No. of ways = $\frac{10!}{(6! 4!)} \times 2!$

Here we multiply by $2!$ Because the groups are distinct.

Question 15

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

\frac{11!}{8!}

\frac{7!}{4!}

\frac{7!}{3!}

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

SOLUTION

Solution : 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 = $\frac{7!}{(4! 3!)}$ .

Free Basic Algebra - 03 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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