Solve the following CAT 2022 Questions using “x” maro G Strategy
1. For any real number x, let [x] be the largest Integer less than or equal to x. CAT 2022 Question. TITA _______
Answer Explanation CAT 2022 Question
Given 1/5 + n/25 = 1
x ki maro…. put, n = 1 to 20
So, for all values of n = 1 to 19, [1/5 + n/25] = 0
Similarly When, 1/5 +n/25 = 2, n = 45
So, for n = 20 to 44 i.e. 25 values [1/5 + n/25] = 1
So, N = 44
2. Let a, b, c be non-zero real numbers such that b2 < 4ac, and f(x) = ax2 + bx + c. If the set S consists of all integers m such that f(m) < 0, then the set S must necessarily be
1.the empty set
2.the set of all positive integers
3.the set of all integers
4.either the empty set or the set of all integers
Answer Explanation CAT 2022 Question
Answer D. Given f(x) = ax2+ bx + c
Given Given, b2< 4ac
Put a>0, we get empty set
Put a<0, we get set of all integers
For any natural number t, suppose the sum of the first t terms of an arithmetic progression is (n + 2n2). If the nth term of the progression is divisible by 9, then the smallest possible value of n is
a) 4 b) 7 c) 9 d) 8
Answer Explanation CAT 2022 Question
Sum of n terms = n + 2n2
Sum of n-1 terms = (n-1) + 2(n-1)2
So, nth term = n + 2n2 – ((n-1) + 2(n-1)2) = n + 2n2 – n + 1 -2n2 +4n -2 = 4n -1
For this to be divisible by 9, smallest value of n (from options) = 7
The largest real value of a for which the equation |x + a| + |x – 1| = 2 has an infinite number of solutions for x is
1. 0 2. 2 3. 1 4. -1
Answer Explanation CAT 2022 Question
putting x = 1
|1+a| + |1-1| = 2
So, |1+a| = 2 => a = 1 or -3
Largest value of a = 1
All the vertices of a rectangle lie on a circle of radius R. If the perimeter of the rectangle is P, then the area of the rectangle is
Answer Explanation CAT 2022 Question
l2 + b2 = (2R)2
And, 2(l+b) = P
Squaring
4(l2 + b2 + 2lb) = P2
4R2 + 2lb = P2/4
lb = P2/8 – 2R2
Let A be the largest positive integer that divides all the numbers of the form 3k + 4k + 5k and B be the largest positive integer that divides all the numbers of the form 4k + 3(4k) + 4k+2, where k is any positive integer. Then (A + B) equals
TITA ________
Answer Explanation CAT 2022 Question
3k + 4k + 5k = (4-1)k + 4k + (4+1)k = 4x when x is odd and 4x+2 when x is even
So the largest value that always divided both is 2. So, A = 2
4k + 3(4k) + 4k+2 = 4k + 3(4k) + 42*4k = (1+3+16)(4k) = 20(4k)
Minimum value of B = 20*41 = 80
So, A+B = 82
Then the maximum value of f(x) becomes 100 when a is equal to _______ TITA
Answer Explanation CAT 2022 Question
Answer 50
Case 1:
a+50 < 100
in this case the min value will happen at a+50 and max value will happen at a or 100
at x = a
f(x) = |x-a| + |x-100| + |x-a-50|
f(a) = 0 + 100 – a + 50 = 150 at a = 0
So, this case is not possible since max value is 100
Answer Explanation CAT 2022 Question
Case 1:
a+50 < 100
in this case the min value will happen at a+50 and max value will happen at a or 100
at x = a
f(x) = |x-a| + |x-100| + |x-a-50|
f(a) = 0 + 100 – a + 50 = 150 at a = 0
So, this case is not possible since max value is 100
Case 2:
a+50 = 100
a = 50
Here, max will happen at x = a
at x = a = 50
f(x) = |x-a| + |x-100| + |x-a-50|
f(a) = 0 + 50 + 50 = 100
This is possible
Case 3:
a + 50 > 100
Here the max will again happen at x = a and min will happen at x =100
at x = a
f(x) = |x-a| + |x-100| + |x-a-50|
f(a) = 0 + 100 – a + 50 = 150 at a = 0
which is not possible
So, a = 50
Consider the arithmetic progression 3, 7, 11, … and let An denote the sum of the first n terms of this progression. Then the value of
1. 404 2. 415 3. 455 4. 442
Answer Explanation CAT 2022 Question
A1 = 3
A2 = 3 + 7
A3 = 3+7+11
And so on, A25 = 3+7+11 +….+ 99 (since general term of the series is 4n-1)
Let N = A1 + A2 + …..A25 = 25(3) + 24(7) + 23(11) + …. 1(99)
The general term of the series = (26-n)(4n-1) = 105n – 4n2 – 26
Thus, ∑N = ∑(105n – 4n2 – 26) = 105(n(n+1)/2 – 4(n(n+1)(2n+1)/6 – 26n
Thus required sum = 1/25 * ( 105*25*26)/2 – 4*25*26*51/6 – 26*25) = 105*13 – 2*26*17 – 26 = 455
In an examination, there were 75 questions. 3 marks were awarded for each correct answer, 1 mark was deducted for each wrong answer and 1 mark was awarded for each unattempted question. Rayan scored a total of 97 marks in the examination. If the number of unattempted questions was higher than the number of attempted questions, then the maximum number of correct answers that Rayan could have given in the examination is
Answer Explanation CAT 2022 Question
Let number of questions correct be x, wrong be y and unattempted be z
So, x+y+z = 75 ——–I
3x – y + z = 97 ——–II
II + I will give
4x + 2z = 172
2x + z = 86
Now, z > x+y so min z = 38
So, max x = (86-38)/2 = 24
Answer Explanation CAT 2022 Question
Answer Explanation CAT 2022 Question
If a and b are non-negative real numbers such that a + 2b = 6, then the average of the maximum and minimum possible values of (a + b) is
1. 3.5 2. 4.5 3. 4 4. 3
Answer Explanation CAT 2022 Question
a+2b = 6
Max a + b when a is max and b is min so a = 6 and b = 0 => a+b = 6
Min a+b when a is min and b is max so a = 0 and b = 3 => a+b = 3
Required average = (3+6)/2 = 4.5
On day one, there are 100 particles in a laboratory experiment. On day n, where n≥2, one out of every n particles produces another particle. If the total number of particles in the laboratory experiment increases to 1000 on day m, then m equals
a)16 b) 19 c)17 d)18
Answer Explanation CAT 2022 Question
On day 2, 1 out of every 2 will produce 1 more particle, so increase of 50 and total becomes 150.
On day 3, 1 out of every 3, so 50 out of 150 will produce 1 more article, so increase of 50 and total becomes 200
This process will keep repeating and increase of 50 will happen every day.
Total increase = 1000 – 100 = 900
So number of days required after day 1 = 900/50 = 18
Thus, m = 18+1 = 19
