# Free Algebra 02 Practice Test - CAT

In the X – Y plane, the area of the region bounded by the graph of | x + y | + | x – y | = 4 is(2005)

A. 8
B. 12
C. 16
D. 20

#### SOLUTION

Solution : C

Let x ≥ 0, y ≥ 0 and x ≥ y

Then, | x + y | + | x – y | = 4

→ x + y + x – y = 4 → x = 2

And in case x ≥ 0, y ≥ 0, x ≤ y

x + y + y – x = 4 → y = 2

Area in the first quadrant is 4.

By symmetry, total area 4 × 4 = 16 unit.

A five-digit number is formed using the digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?(1993)

A. 6666600
B. 6666660
C. 6666666
D. None of these

#### SOLUTION

Solution : A

Keeping one digit in fixed position, another four can be arranged in 4 ! Ways = 24 ways. Thus each of 5 digit will occur in each of five places 4! Times. Hence, the sum of digits in each position is 24 (1+3+5+9) = 600. So, the sum of all numbers = 6000 (1+10+100+1000+10000) = 6666600.

SHORTCUT

Let n=number of digits

Then, The sum of all possible numbers is given by (n)!(sum of all the digits)(1111….n times)= 5!(18)(11111)= 6666600

Once I had been to the post office to buy five-rupee, two-rupee and one-rupee stamps. I paid the clerk Rs. 20, and since he had no change, he gave me three more one-rupee stamps. If the number of stamps of each type that I had ordered initially was more than one, what was the total number of stamps that I bought?

A. 10
B. 9
C. 12
D. 8

#### SOLUTION

Solution : A

Option (a)

The number of stamps that were initially bought were more than one of each type. Hence the total number of stamps

2(5 rupees) + 2(2 rupees) + 3(1 rupee) + 3(1 rupee) = 10 tickets

One year payment to the servant is Rs. 90 plus one turban. The servant leaves after 9 months and receives Rs. 65 and a turban. Then find the price of the turban (CAT 1998)

A. Rs. 10
B. Rs. 15
C. Rs. 7.5
D. Cannot be determined

#### SOLUTION

Solution : A

Option (a)

Let turban be of cost Rs. X, so, payment to the servant = 90 + x for 12 months

For 9 month = (9/12) × (90 + x) = 65 + x → x = Rs. 10

(Note:- you can also go from answer options)

You can collect as many rubies and emeralds as you can. Each ruby is worth Rs. 4 crores and each emerald is worth of Rs. 5 crore. Each ruby weights 0.3 kg and each emerald weights 0.4 kg. Your bag can carry at the most 12 kg. What you should collect to get the maximum wealth ? ( CAT 1998)

A. 20 rubies and 15 emeralds
B. 40 rubies
C. 28 rubies and 9 emeralds
D. None of these

#### SOLUTION

Solution : B

Option (b)

Number of emeralds which can be carried in total = 30. Value = 30 x 5 = 150 cr

Number of rubies which can be carried in total = 40. Value = 40 x 4 = 160 cr. Only rubies will give the maximum wealth

Find the following sum(CAT 2000)

A.
B.
C.
D.

#### SOLUTION

Solution : D

Option (d)

nth term, This equation can be rewritten as

S =

=

A, B and C are 3 cities that form a triangle and where every city is connected to every other one by at least one direct route. There are 33 routes direct and indirect from A to C and there are 23 direct routes from B to A. How many direct routes are there from A to C ?(CAT 2000)

A. 15
B. 10
C. 20
D. 25

#### SOLUTION

Solution : B

Option (b)

Let the number of direct routes from A to B be x, from A to C be z and that from C to B be y. Then the total number of routes from A to C are = xy + z = 33. Since the number of direct routes from A to B are 23, x = 23. Therefore 23y + z = 33. Then y must take value 1 and then z = 10, thus answer = (b).

m is the smallest positive integer such that for any integer n < m, the quantity n3 – 7n2 + 11n – 5 is positive. What is the value of m?(CAT 2001)

A.
B.
C.
D. None of these

#### SOLUTION

Solution : D

Option (d)

n=1 is a root of the equation

(n - 1)(n2 – 6n + 5) = (n - 1)2(n - 5)

Now, (n - 1)2 is always positive.

Now, for n < 5, the expression gives a negative quantity.

Therefore, the least value of n will be 6. Hence m = 7 .

P and Q are two positive integers such that PQ = 64. Which of the following cannot be the value of P+Q?

A. 20
B. 65
C. 16
D. 35

#### SOLUTION

Solution : D

Option (d)

PxQ = 64 = 1 × 64 = 2 × 32 = 4 × 16 = 8 × 8.

Corresponding values of P + Q are 65, 34, 20, 16.

Therefore, P + Q cannot be equal to 35.

An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, …., 9 such that the first digit of the code is nonzero. The code, handwritten on a slip, can however potentially create confusion, when read upside down-for example, the code 91 may appear as 16. How many codes are there for which no such confusion can arise?(2003)

A. 80
B.  78
C. 71
D. 69

#### SOLUTION

Solution : D

Option (d)

The available digits are 0,1,2,3……………..9. The first digit can be chosen in 9 ways ( 0 not acceptable ), the second digit can be accepted in 9 ways ( digit repetition are not allowed). Thus the code can be made in 9 × 9 =81 ways.

Now there are only four digits which can create confusion 1,6,8,9.Total number of ways can confusion arise = 4 × 3 =12

Thus required answer = 81 – 12 = 69.

Let a, b, c, d and e be integers such that a = 6b = 12c, and 2b = 9d = 12e. Then which of the following pairs contains a number that is not an integer ?(CAT 2003)

A. | |
B. | |
C. | |
D. | |

#### SOLUTION

Solution : D

Option (d)

Assumption

Now you can eliminate answer options. All the options, other than option (d) give integral values.

Consider the set Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1, 2, 3, ………., 96. How many of these sets contain 6 or any integral multiple thereof (i.e., any one of the numbers 6, 12, 18, ………..)?

A. 80
B. 81
C. 82
D. 83

#### SOLUTION

Solution : A

Option (a)

Calculating “n” which does not satisfy the requisite condition (i.e. the value of “n” which will not yield subsets with an integral multiple of 6).

The terms will fall in an AP with n =1 as the first term. The general form of the AP is 6k+1.

The last term will be 91. Thus, total number of sets which do not satisfy the condition= 906 = 15+1 = 16.

Total number of terms = 96.

Thus, number of sub sets possible = 96 - 16 = 80. Option (a).

Let S1 be a square of side a. Another square S2 is formed by joining the mid-points of the sides of S1. The same process is applied to S2 to form yet another square S3, and so on. If A1, A2, A3, ……… be the areas and P1, P2, P3, ……… be the perimeters of S1, S2, S3, ……., respectively, then the ratio equals :(CAT 2003)

A.
B.
C.
D.

#### SOLUTION

Solution : C

Option (c)

From the given condition in question

Area and perimeter of S1 = a2, 4a

Area and perimeter of S2 =

Area and perimeter of S3 =

Area and perimeter of S4

These are 2 infinite GPs , which can be solved as follows

Required ratio

=

=

=

Suppose n is an integer such that the sum of the digits of n is 2, and 1010<n<1011. The number of different values for n is:

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

#### SOLUTION

Solution : A

Option (a)

There are 11 digits in this number

When sum of digits is 2, there are 2 options

There are 2 ones, with one 1 fixed in the first position

The question in now based on the arrangement of 10 zeroes and 1 one

Number of possibilities = 10!9!1! = 10 ways

When 2 is the first digit = one possibility.

Total number of possibilities = 11

If ab+c=bc+a=ca+b=r then, r cannot take any value except :(CAT 2004)

A. 12
B. -1
C. 12 or -1
D. 12 or -1

#### SOLUTION

Solution : C

Option (c)

As ab+c=bc+a=ca+b=a+b+cb+c+c+a+a+b

=a+b+c2(a+b+c)=r=12 (Assuming a + b + c ≠ 0)

If a + b + c = 0

ab+c = aa+b+ca (by adding and subtracting a in the denominator) =a0a=aa=r=1

(similarly bc+a=ca+b= -1)

Hence r can take only 12 or -1 as values. Choice (c)