# Free Algebra 03 Practice Test - CAT

### Question 1

In how many ways is it possible to choose a white square and a black square on a chess-board so that the squares must not lie in the same row or column?(CAT 2002)

#### SOLUTION

Solution :D

Option (d)

There are 32 white and 32 black squares on the chessboard

Number of ways of choosing the white square = 32

When a white square is selected, we cannot select the black square lying on the row or column of the white

Square. We have 8 such black squares for every white square selected.

Hence we have 32-8=24 black squares which can be selected for every white square selected

Total number of possibilities= 32 × 24= 768. Answer is option (d)

### Question 2

How many numbers can be made with digits 0, 7, 8 which are greater than 0 and less than a million?

#### SOLUTION

Solution :D

Option (d)

Since we are looking for numbers < a million, we can represent all the numbers in this range as “abcdef” . each

of the digits can be selected in 3 ways.

Total possible cases = 36 -1 = 728. We subtract one case where each of “abcdef” take 0.

### Question 3

A survey of 200 people in a community who watched at least one of the three channels – BBC, CNN and DD-showed that 80% of the people watched DD, 22% watched BBC, and 15% watched CNN.

If 5% of people watched DD and CNN. 10% watched DD and BBC, then what percentage of people watched BBC and CNN only?

#### SOLUTION

Solution :A

Since 80% watch DD, hence 20% do not watch DD. Let those who watch BBC and CNN only be x%, then 12 + 10 –x = 20

x = 2

### Question 4

A survey of 200 people in a community who watched at least one of the three channels – BBC, CNN and DD-showed that 80% of the people watched DD, 22% watched BBC, and 15% watched CNN.

If 5% of people watched DD and CNN. 10% watched DD and BBC, what percentage of people watched all the three channels?

#### SOLUTION

Solution :D

By zonal theorem:

I+II+III=100%

But only II is known , hence option (d)

### Question 5

A survey of 200 people in a community who watched at least one of the three channels – BBC, CNN and DD-showed that 80% of the people watched DD, 22% watched BBC, and 15% watched CNN.

If 5% of people watched DD and CNN. 10% watched DD and BBC, what is the maximum percentage of people who can watch all the three channels?

#### SOLUTION

Solution :B

Option (b)

Use the maxima of all shortcut

S=117% and X=100%

S−Xn−1=172=8.5

### Question 6

Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over phone-to-phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English and only one Englishman knows French. What is the minimum number of calls needed for the above purpose?

#### SOLUTION

Solution :(c) There have to be 2 calls from each person to the Englishman who knows French to get all the information. So, there should be 10 calls. But when the fifth guy call he would get all the information of the previous 4 guys alongwith Englishman’s information. Hence, 1 call can be saved. So, the total number of calls = 9.

### Question 7

Consider a triangle drawn on the X – Y plane with its three vertices at (41, 0), (0, 41) and (0, 0), each vertex being represented by its (X, Y) coordinates. The number of points with integer coordinates inside the triangle (excluding all the points on the boundary) is (CAT 2005)

#### SOLUTION

Solution :A

Option (a) The equation forming from the data is x + y < 41, as we cannot count the perimeter of the triangle.

Thus,total number of cases will fall in an AP with first term 39 with cd =1 and last term =1. Thus number of coordinates =

### Question 8

If a_{1} = 1 and a_{n+1} – 3a_{n} + 2 = 4n for every positive integer n, then a_{100} equals (CAT 2005)

^{99}– 200

^{99}+ 200

^{100}– 200

^{100}+ 200

#### SOLUTION

Solution :C

Option (c)

Shortcut: Double Substitution

Write each answer option in terms of n (where n=100)

3

^{n-1}– 2n3

^{n-1}+ 2n3

^{n}- 2n3

^{n}+ 2nNow at n=1, we should get 1. Put n=1 in the answer options and eliminate those options where you do not get 1

-1 ≠1

2 ≠1

1

5≠1 thus, the answer is option (c)

### Question 9

If the equation x^{3 }– ax^{2} + bx – a = 0 has three real roots then the following is true
(CAT 2000)

#### SOLUTION

Solution :D

(d) Let f(x) = x

^{3}– ax^{2}+ bx – a =0In the given equation, there are 3 sign changes, therefore there are at most 3 positive roots (Descarte’s Rule). If f(-x), there is no sign change. Thus there is no negative real root, i.e., if α, β and γ are the roots then they are all positive and we have

f(x) = (x - α)(x – β)(x - γ) = 0

→ b = αβ + βγ +γα → a = α + β + γ = αβγ

→

→ αβ, βγ, γα > 1 → b > 3.

Thus b ≠ 1.

### Question 10

The **maximum **number of non-negative real roots of 2^{x}– x – 1 = 0 equals
(CAT 2003)

#### SOLUTION

Solution :B

Option (b) It is clear that the equation 2

^{x}- x - 1 = 0 is satisfied by x = 0 and 1 only. For x > 1, f(x) = 2^{x}– x – 1 starts increasing.Alternately

Draw the graph of

y=x+1

y=2^{x for x<0 there is only one root. }

### Question 11

The fourth term of an arithmetic progression is 8. What is the sum of first 7 terms of the arithmetic progression?(CAT 1994)

#### SOLUTION

Solution :C

Option (c) Fourth term = 8 → a + 3d = 8

sum of seven terms

=

### Question 12

Two men X and Y started working for a certain company at similar jobs on January 1, 1950. X asked for an initial salary of Rs.300 with an annual increment of Rs. 30. Y asked for an initial salary of Rs. 200 with a rise of Rs. 15 every six months. Assume that the arrangements remained unaltered till December 31, 1959. Salary is paid on the last day of the month. What is the total paid to them as salary during the period ?(CAT 2001)

#### SOLUTION

Solution :A

(a) For total salary paid to X

= 12 × (300 + 330 + 360 + 390 + 420 + 450 + 480 + 510 + 540 + 570)

== 60 × 870 = Rs. 52,200

For total salary paid to Y

= 6 × [200 + 215 + 230 + 245 +………………. 20 terms]

= 6 × 10 × [2 × 200 + 19 × 15][sum of A. P.]

= 60 × [400 + 285] = Rs. 41,100

Total sum of both = Rs. 93,300

### Question 13

How many numbers can be formed from 1, 2, 3, 4, 5(without repeating), where the digit at the unit’s place must be greater than that in the ten’s place?(CAT 1998)

#### SOLUTION

Solution :B

Option (b)

Case 1- units place= 1 (0 possibilities)

Case 2- units place = 2 (tens place=1 possibility (2) and 3! Arrangements for the remaining places) = 1x3!

Case 3- units place = 3 (tens place=2 possibilities (1,2) and 3! Arrangements for the remaining places) = 2x3!

Case 4- units place = 4 (tens place=3 possibilities (1,2,3) and 3! Arrangements for the remaining places) = 3x3!

Case 5- units place = 5 (tens place=4 possibilities (1,2,3,4) and 3! Arrangements for the remaining places) = 4x3!

Total = 60

### Question 14

A man has 9 friends: 4 boys and 5 girls. In how many ways can he invite them if there have to be exactly 3 girls in the invitees? (CAT 1996)

#### SOLUTION

Solution :B

Option (b) 3 girls can be selected out of 5 girls in 5C3 ways. Since the number of boys to be invited is not to be given, he can invite or not invite them in 24 ways.

Hence required numbers of ways =5C3×24=10×16=160.

### Question 15

The sum of 3rd and 15th elements of an arithmetic expression is equal to the sum of 6th, 11th and 13th elements of the progression. Then which element of the series should necessarily be equal to zero(CAT 2003)

#### SOLUTION

Solution :C

(c) T

_{n}= a + (n - 1)d. Hence we get 3^{rd}+ 5^{th}term = (a + 2d) + (a + 4d) = 2a + 6d. Similarly, 6, 11 and 13^{th}terms = (a + 5d) + (a + 10d) + (a + 12d) = 3a + 27d. Now 2a + 6d = 3a + 27d, hence a + 11d = 10. This means that 12^{th}term is zero.