Free Algebra 02 Practice Test - CAT
Question 1
In the X – Y plane, the area of the region bounded by the graph of | x + y | + | x – y | = 4 is(2005)
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.
Question 2
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)
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
Question 3
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?
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
Question 4
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)
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)
Question 5
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)
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
Question 6
Find the following sum
(CAT 2000)
SOLUTION
Solution : D
Option (d)
nth term,
This equation can be rewritten as
S =
=
Question 7
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)
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).
Question 8
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)
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 .
Question 9
P and Q are two positive integers such that PQ = 64. Which of the following cannot be the value of P+Q?
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.
Question 10
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)
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.
Question 11
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)
|
|
|
|
SOLUTION
Solution : D
Option (d)
Assumption
Start with e. if e=12, then c=36, d=16, b=72 and a=432
Now you can eliminate answer options. All the options, other than option (d) give integral values.
Question 12
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, ………..)?
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).
Question 13
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)
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
=
=
=
→
→
Question 14
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:
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
Question 15
If ab+c=bc+a=ca+b=r then, r cannot take any value except :(CAT 2004)
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+c−a (by adding and subtracting a in the denominator) =a0−a=a−a=r=−1
(similarly bc+a=ca+b= -1)
Hence r can take only 12 or -1 as values. Choice (c)