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Test: Linear Equations- 1 - CAT MCQ


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10 Questions MCQ Test Quantitative Aptitude (Quant) - Test: Linear Equations- 1

Test: Linear Equations- 1 for CAT 2024 is part of Quantitative Aptitude (Quant) preparation. The Test: Linear Equations- 1 questions and answers have been prepared according to the CAT exam syllabus.The Test: Linear Equations- 1 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Linear Equations- 1 below.
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Test: Linear Equations- 1 - Question 1

The number of solutions of the equation 2x + y = 40 where both x and y are positive integers and x <= y is:

Detailed Solution for Test: Linear Equations- 1 - Question 1

y = 38 => x = 1

y = 36 => x = 2

y = 14 => x = 13

y = 12 => x = 14 => Cases from here are not valid as x > y.

Hence, there are 13 solutions.

Test: Linear Equations- 1 - Question 2

For a real number x the condition |3x - 20| + |3x - 40| = 20 necessarily holds if

Detailed Solution for Test: Linear Equations- 1 - Question 2

Case 1: x ≥ 40/3

we get 3x-20 +3x-40 = 20

6x=80


Case 2
we get 3x - 20 + 40 - 3x = 20
we get 20 = 20
So we get x 
Case 3x  < 20/3
we get 20-3x+40-3x =20
40=6x
x = 20/3
but this is not possible
so we get from case 1,2 and 3

Now looking at options
we can say only option C satisfies for all x .
Hence 7<x<12.

Test: Linear Equations- 1 - Question 3

The number of integers n that satisfy the inequalities | n - 60| < n - 100| < |n - 20| is 

Detailed Solution for Test: Linear Equations- 1 - Question 3

We have |n - 60| < |n - 100| < |n - 20|
Now, the difference inside the modulus signified the distance of n from 60, 100, and 20 on the number line.

This means that when the absolute difference from a number is larger, n would be further away from that number.

The absolute difference of n and 100 is less than that of the absolute difference between n and 20.

Hence, n cannot be ≤ 60, as then it would be closer to 20 than 100. Thus we have the condition that n>60.

The absolute difference of n and 60 is less than that of the absolute difference between n and 100.

Hence, n cannot be ≥ 80, as then it would be closer to 100 than 60.

Thus we have the condition that n<80.

The number which satisfies the conditions are 61, 62, 63, 64……79. Thus, a total of 19 numbers.

Alternatively

as per the given condition: |n - 60| < |n - 100| < |n - 20|

Dividing the range of n into 4 segments. (n < 20, 20<n<60, 60<n<100, n > 100)

1) For n < 20.

|n-20| = 20-n, |n-60| = 60- n, |n-100| = 100-n

considering the inequality part: |n - 100| < n - 20|

100 -n < 20 -n,

No value of n satisfies this condition.

2) For 20 < n < 60.

|n-20| = n-20, |n-60| = 60- n, |n-100| = 100-n.

60- n < 100 – n and 100 – n < n – 20

For 100 -n < n – 20.

120 < 2n and n > 60. But for the considered range n is less than 60.

3) For 60 < n < 100

|n-20| = n-20, |n-60| = n-60, |n-100| = 100-n

n-60 < 100-n and 100-n < n-20.

For the first part 2n < 160 and for the second part 120 < 2n.

n takes values from 61 …………….79.

A total of 19 values

4) For n > 100

|n-20| = n-20, |n-60| = n-60, |n-100| = n-100

n-60 < n – 100.

No value of n in the given range satisfies the given inequality.

Hence a total of 19 values satisfy the inequality.

Test: Linear Equations- 1 - Question 4

The inequality of p2 + 5 < 5p + 14 can be satisfied if:

Detailed Solution for Test: Linear Equations- 1 - Question 4

We have, p+ 5 < 5p + 14

=> p2 – 5p – 9 < 0

=> p<6.4 or p>-1.4
Hence, p ≤ 6, p > −1 will satisfy the inequalities

Test: Linear Equations- 1 - Question 5

Consider the equation:

|x-5|2 + 5 |x - 5| - 24 = 0

The sum of all the real roots of the above equationis:

Detailed Solution for Test: Linear Equations- 1 - Question 5

Let's consider x-5 as 'p'

Case 1: p ≥ 0

|x - 5| |2 + 5|x - 5| - 24 = 0

p2 +5p - 24 = 0

p+ 8p - 3p - 24 = 0

p(p + 8) -3 (p + 8) = 0

(p + 8) (p - 3) = 0

p = -8 and p = 3

x - 5 = 3,x = 8 This is a real root since x is greater than 5.

x - 5 = -8, x = -3. This root can be negated because x is not greater than 5.

Case 2: p < 0

p2 - 5p - 24 = 0

p2 - 8p + 3p - 24 = 0

p=8, -3

x - 5 = 8, x = 13. This root can be negated because x is not less than 5

x - 5 = -3, x = 2. This is a real root because x is less than 5.

The sum of the real roots = 8 + 2 = 10

Test: Linear Equations- 1 - Question 6

If y2 + 3y – 18 ≥ 0, which of the following is true?

Detailed Solution for Test: Linear Equations- 1 - Question 6

y2 + 3y - 18 ≥ 0

⇒ y2 + 6y - 3y - 180

⇒ y(y + 6) -3(y + 6) ≥ 0

⇒ (y - 3)(y + 6) ≥ 0

⇒ y ≥ 3andy ≤ - 6

Test: Linear Equations- 1 - Question 7

Consider the function f(x) = (x + 4)(x + 6)(x + 8) ⋯ (x + 98). The number of integers x for which f(x) < 0 is:

Detailed Solution for Test: Linear Equations- 1 - Question 7

The critical points of the function are -4, -6, -8, … , -98 ( 48 points).

For all integers less than -98  and greater than -4 f(x) > 0 always .

for x= -5, f(x) < 0

Similarly, for x= -9, -13, …., -97 (This is an AP with common difference -4)

Hence, in total there are 24 such integers satisfying f(x)< 0.

Test: Linear Equations- 1 - Question 8

If 2 ≤ |x – 1| × |y + 3| ≤ 5 and both x and y are negative integers, find the number of possible combinations of x and y.

Detailed Solution for Test: Linear Equations- 1 - Question 8

2 ≤ |x – 1| × |y + 3| ≤ 5
The product of two positive number lies between 2 and 5.
As x is a negative integer, the minimum value of |x – 1| will be 2 and the maximum value of |x – 1| will be 5 as per the question.

When, |x – 1| = 2, |y + 3| can be either 1 or 2
So, for x =  -1, y can be – 4 or – 2 or – 5 or -1.
Thus, we get 4 pairs of (x, y)

When |x – 1| = 3, |y + 3| can be 1 only
So, for x = – 2, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

When |x – 1| = 4, |y + 3| can be 1 only
So, for x = – 3, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

When |x – 1| = 5, |y + 3| can be 1 only
So, for x = – 4, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

Therefore, we get a total of 10 pairs of the values of (x, y)
Hence, option A is the correct answer.

Test: Linear Equations- 1 - Question 9

The minimum possible value of the sum of the squares of the roots of the equation x2 + (a + 3) x - (a + 5) = 0 is

Detailed Solution for Test: Linear Equations- 1 - Question 9

Let the roots of the equation x2 + (a + 3) x- (a + 5) = 0 be equal to p, q

Hence, p + q = -(a + 3) and p x q = -(a + 5)

Therefore, p2 + q= a+ 6a + 9 + 2a +10 = a+ 8a + 19 = (a+4)2 + 3

As (a + 4)2 is always non negative, the least value of the sum of squares is 3

Test: Linear Equations- 1 - Question 10

a, b, c are integers, |a| ≠ |b| ≠|c| and -10 ≤ a, b, c ≤ 10. What will be the maximum possible value of [abc – (a + b + c)]?

Detailed Solution for Test: Linear Equations- 1 - Question 10

|a| ≠ |b| ≠|c| and -10 ≤ a, b, c ≤ 10

Expression: [abc - (a + b + c)]

For maximum value, two of a,b and c should be negative, as all three negative will make abc negative.

Thus, max value will occur if a= -10, b = -9, c = 8

⇒ Max value = [(-10 × 9 × 8)-(-10-9+8)]

= 720 + 11 = 731

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