Test: Linear Equations- 1 - Bank Exams MCQ

Step 1: Equation

The given equation is:
2x + y = 40
Where x and y are positive integers and x ≤ y.

Step 2: Express y

From 2x + y = 40, solve for y:
y = 40 - 2x

Step 3: Apply conditions

For y to be positive:
40 - 2x > 0, so x < 20
Thus, x can take values from 1 to 19.

Now, apply the condition x ≤ y:
x ≤ 40 - 2x
Solving for x:
3x ≤ 40
x ≤ 40/3 ≈ 13.33
So, x can take integer values from 1 to 13.

Step 4: Count solutions

The possible values of x are 1, 2, 3, ..., 13. For each of these, y = 40 - 2x is valid, and the condition x ≤ y holds.

Thus, the total number of solutions is 13.

Quick Verification

x can range from 1 to 13. For each x, y = 40 - 2x is valid, and x ≤ y is true. Therefore, the number of solutions is 13.

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.

Given Inequality:

|n - 60| < |n - 100| < |n - 20|

This inequality represents the relative distances of n from 60, 100, and 20 on the number line. The key is to find the values of n for which these inequalities hold.

Step 1: Breaking down the conditions

• The absolute difference between n and 60 must be less than the absolute difference between n and 100, and

• The absolute difference between n and 100 must be less than the absolute difference between n and 20.

We will now consider four possible ranges of n and check the inequalities for each.

Step 2: Case-by-case analysis

Case 1: n < 20

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

The inequality becomes:

• 100 - n < 20 - n, which simplifies to 100 < 20, which is false.

Thus, no solutions exist for n < 20.

Case 2: 20 < n < 60

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

The inequality becomes:

• 60 - n < 100 - n, which simplifies to 60 < 100, which is always true.

• 100 - n < n - 20, which simplifies to 120 < 2n, or n > 60.

Since n > 60 contradicts the assumption 20 < n < 60, there are no solutions for 20 < n < 60.

Case 3: 60 < n < 100

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

The inequality becomes:

• n - 60 < 100 - n, which simplifies to 2n < 160, or n < 80.

• 100 - n < n - 20, which simplifies to 120 < 2n, or n > 60.

Thus, n must satisfy 60 < n < 80, so the possible values of n are 61, 62, ..., 79 (19 values).

Case 4: n > 100

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

The inequality becomes:

• n - 60 < n - 100, which simplifies to 60 < 100, which is true.

• n - 100 < n - 20, which simplifies to 100 < 20, which is false.

Thus, no solutions exist for n > 100.

Step 3: Conclusion

• From Case 1: no solutions for n < 20.

• From Case 2: no solutions for 20 < n < 60.

• From Case 3: Solutions exist for 60 < n < 80, with 19 values: 61, 62, ..., 79.

• From Case 4: no solutions for n > 100.

Thus, the number of solutions is 19.

Quick Verification:

Let's check the interval 60 < n < 100:

• The condition for this interval is n > 60 and n < 80.

• Values of n: 61, 62, ..., 79.

This gives exactly 19 values for n satisfying the inequality.

We have, p² + 5 < 5p + 14

=> p² – 5p – 9 < 0

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

Let's consider x-5 as 'p'

Case 1: p ≥ 0

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

p² +5p - 24 = 0

p² + 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

y² + 3y - 18 ≥ 0

⇒ y² + 6y - 3y - 180

⇒ y(y + 6) -3(y + 6) ≥ 0

⇒ (y - 3)(y + 6) ≥ 0

⇒ y ≥ 3 and y ≤ - 6

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.

Given:

We are given the inequality:
2 ≤ |x – 1| × |y + 3| ≤ 5

We know that both x and y are negative integers.

Step 1: Understand the behavior of the absolute values

Since both x and y are negative, we need to look at the absolute values:

• |x – 1| represents the distance of x from 1.

• |y + 3| represents the distance of y from -3.

Step 2: Possible values of |x – 1| and |y + 3|

• |x – 1| can take values from 2 to 5, as it must satisfy the inequality 2 ≤ |x – 1| ≤ 5.

• |y + 3| must be such that the product |x – 1| × |y + 3| lies between 2 and 5.

Step 3: Case-by-case analysis

Case 1: |x – 1| = 2

• 2 ≤ |x – 1| × |y + 3| ≤ 5 implies |y + 3| can be 1 or 2.

  • x = -1 (since |x – 1| = 2)

  • Possible values of y: y = -4, -2, -5, -1.

  • Number of pairs: 4.

Case 2: |x – 1| = 3

• |y + 3| = 1 (since 3 × 1 = 3 lies within the range).

  • x = -2 (since |x – 1| = 3)

  • Possible values of y: y = -4, -2.

  • Number of pairs: 2.

Case 3: |x – 1| = 4

• |y + 3| = 1 (since 4 × 1 = 4 lies within the range).

  • x = -3 (since |x – 1| = 4)

  • Possible values of y: y = -4, -2.

  • Number of pairs: 2.

Case 4: |x – 1| = 5

• |y + 3| = 1 (since 5 × 1 = 5 lies within the range).

  • x = -4 (since |x – 1| = 5)

  • Possible values of y: y = -4, -2.

  • Number of pairs: 2.

Step 4: Total number of pairs

Total pairs = 4 + 2 + 2 + 2 = 10.

Quick Verification:

To check quickly, let’s list the possible values for x and y:

• For x = -1: y = -4, -2, -5, -1 (4 combinations)

• For x = -2: y = -4, -2 (2 combinations)

• For x = -3: y = -4, -2 (2 combinations)

• For x = -4: y = -4, -2 (2 combinations)

Summing up, we get 10 valid pairs of (x, y).

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• To maximize abc – (a + b + c) with |a|, |b|, |c| all different and each in [–10, 10], we want a big positive product abc and a negative sum (so subtracting it adds).

• Take two large negatives and one large positive with the three biggest distinct magnitudes: –10, –9, 8.

• Compute:

  • abc = (–10) × (–9) × 8 = 720.

  • a + b + c = –10 – 9 + 8 = –11.

  • abc – (a + b + c) = 720 – (–11) = 731.

Trying all-positive 10, 9, 8 gives 720 – 27 = 693 < 731. Other sign choices give less. Hence the maximum is 731.