Linear Equations CAT Previous Year Questions with Answer PDF

Since x & y are real numbers, (x + 2y) & (x − 2y − 1) are both real.
A square of a real number is always non-negative.
For this reason, for the equation: (x + 2y)² + (x − 2y − 1)² = 0 to be true, both (x + 2y) & (x − 2y − 1) must be equal to 0.
x − 2y − 1 = 0
x − 2y = 1
We solve the linear equations (x + 2y = 0) & (x − 2y − 1 = 0) to get the exact values of x & y, but that is not required to solve the question at hand.

Let the cost of an apple, an orange and a mango be a, o, and m respectively.
Then it is given that:
2a + 4o + 6m = a + 4o + 8m
or a = 2m.
Also, a + 4o + 8m = 8o + 7m
10m - 7m = 4o
3m =  4o.
We can now express the cost of a basket in terms of mangoes only:
2a + 4o + 6m = 4m + 3m + 6m = 13m.

Let us assume the family spends Rs. 100 each month for the first 3 months and then spends Rs. 50 in each of the next two months.
Then amount of onions bought = 10, 5, 4, 2, 1, for months 1-5 respectively.
Total amount bought = 22kg.
Total amount spent = 100 + 100 + 100 + 50 + 50 = 400.
Average expense = 400/22 = Rs.18.18 ≈ 18

Let the amounts Neeta, Geeta, and Sita earn in a day be n, g, and s respectively. 
Then, it has been given that:
n + g = 6s - i
s + n = 2g - ii
ii-i, we get: s - g = 2g - 6s
7s = 3g.
Let g be 7a. Then s earns 3a.
Then n earns 6s - g = 18a - 7a = 11a.
Thus, the ratio is 11a:3a = 11:3

The quadratic equation of the form ∣x² − 4x − 13∣ = r has its minimum value at x = -b/2a, and hence does not vary irrespective of the value of x.
Hence at x = 2 the quadratic equation has its minimum.
Considering the quadratic part : ∣x² − 4x − 13∣ . as per the given condition, this must-have 3 real roots.
The curve ABCDE represents the function ∣x² − 4x − 13∣. Because of the modulus function, the representation of the quadratic equation becomes :
ABC'DE. 
There must exist a value, r such that there must exactly be 3 roots for the function. If r = 0 there will only be 2 roots, similarly for other values there will either be 2 or 4 roots unless at the point C'.
The point C' is a reflection of C about the x-axis. r is the y coordinate of the point C' :
The point C which is the value of the function at x = 2, = 2² − 8 − 13 = -17,
the reflection about the x-axis is 17.

x₀ = 1, x₁ = 2

Hence, the series begins to repeat itself after every 5 terms. Terms whose number is of the form 5n are 1, 5_n+1 are 2... and so on, where n=0,1,2,3,....
2021 is of the form 5_n+1. Hence, its value will be 2.

We have four inflection points -5, -2, 3, and 9.
For x < -5, all four terms (x + 5), (x - 3), (x - 9), (x + 2) will be negative. Hence, the overall expression will be positive. Similarly, when x > 9, all four terms will be positive.
When x belongs to (-2, 3), two terms are negative and two are positive. Hence, the overall expression is positive again.
We are left with the range (-5, -2) and (3, 9) where the expression will be negative.

Given f (5 + x ) = f (5 - x )
Put x = x - 5
f ( x) = f (10 - x)
∴ Let a, b be two roots of f(x) = 0, then 10 - a,10 - b are also roots of f ( x ) = 0
∴ Hence sum of the roots = a + b + 10 - a + 10 - b = 20

We know if a^b = 1
⇒ a = 1 and b is any number
or a = - 1 and b is even
a > 0 and  b is 0
case 1: x² - 13x + 42 = 0 ⇒ x = 6, 7
case 2 : x² - 7x +11 = 1 ⇒ x2 - 7x +10 = 0 ⇒ x = 2 or 5
case 3: x² - 7x +11 = -1⇒ x2 - 7x +12 = 0
⇒ x = 4 or 3
Hence number of solutions are 6 

f ( x) = x² + ax + b

∴ f (x) = x² - 4x + b f (x) = (x - 2)² + b - 4 when b ≥ 4 f ( x) ≥ 0 for all x
∴ The minimum value of b is 4 

Simultaneous equation have a unique solution only if

From the given equations, a unique solution would exist only if

⇒ k 2 ≠ 4 ⇒| k | ≠ 2

Given,


since 
m - 2n = 9 - 2 x 4 =1

Let the total marks be T and scores of Bishnu, Asha and Ramesh be a, b and c respectively.
Given, a = 52% of T = c - 23 and b = 64% of T = c + 34
Hence, (64 - 52)% of T = (c + 34) - (c - 23) = 57
i.e. 12% of T = 57
Hence, score of Geeta = 84% of T = 7 ´ 57 = 399