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CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - CAT MCQ


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10 Questions MCQ Test Daily Test for CAT Preparation - CAT Previous Year Questions: Linear and Quadratic Equations (July 16)

CAT Previous Year Questions: Linear and Quadratic Equations (July 16) for CAT 2024 is part of Daily Test for CAT Preparation preparation. The CAT Previous Year Questions: Linear and Quadratic Equations (July 16) questions and answers have been prepared according to the CAT exam syllabus.The CAT Previous Year Questions: Linear and Quadratic Equations (July 16) MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for CAT Previous Year Questions: Linear and Quadratic Equations (July 16) below.
Solutions of CAT Previous Year Questions: Linear and Quadratic Equations (July 16) questions in English are available as part of our Daily Test for CAT Preparation for CAT & CAT Previous Year Questions: Linear and Quadratic Equations (July 16) solutions in Hindi for Daily Test for CAT Preparation course. Download more important topics, notes, lectures and mock test series for CAT Exam by signing up for free. Attempt CAT Previous Year Questions: Linear and Quadratic Equations (July 16) | 10 questions in 20 minutes | Mock test for CAT preparation | Free important questions MCQ to study Daily Test for CAT Preparation for CAT Exam | Download free PDF with solutions
CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 1
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If x and y are real numbers such that x2 + (x − 2y − 1)2 = −4y(x + y), then the value x−2y is

[2023]

Detailed Solution for CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 1


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.

CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 2
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If then is equal to

[2023]

Detailed Solution for CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 2

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CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 3
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The number of integer solutions of equation 2 |x| (x2 + 1)= 5x2 is

[2023]


Detailed Solution for CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 3

2 |x| (x2 + 1)= 5x2 
Let |x| = k
2k (k2 + 1) = 5k2 
Either k = 0; or 2(k2 + 1) = 5k
2k2 – 5k + 2 = 0
2k2 – 4k – k + 2 = 0
2k(k – 2) –1(k – 2) = 0
(2k – 1)(k – 2) = 0
k = 0.5 or k = 2
Therefore, k which is |x|, can take the values 0, 0.5 or 2
So, x can take the values 0, -0.5, 0.5, -2, 2
Since we are looking for integral solutions, x can only take the values 0, 0.5 or 2
Therefore, there are only 3 integral solutions to 2 |x| (x2 + 1) = 5x2.

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CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 4
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Let & and β be the two distinct roots of the equation 2x2 - 6x + k = 0, such that (α + β) and αβ are the distinct roots of the equation x2 + px + p = 0. Then, the value of 8(k - p) is

[2023]

Detailed Solution for CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 4

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CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 5
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The equation x3 + (2r + 1)x2 + (4r - 1)x + 2 = 0 has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of r is

[2023]

Detailed Solution for CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 5

CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 6
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The sum of all possible values of x satisfying the equation 24x2 - 22x2 + x + 16 + 22x + 30 = 0, is

[2023]
Detailed Solution for CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 6

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CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 7
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Amal purchases some pens at ₹ 8 each. To sell these, he hires an employee at a fixed wage. He sells 100 of these pens at ₹ 12 each. If the remaining pens are sold at ₹ 11 each, then he makes a net profit of ₹ 300, while he makes a net loss of ₹ 300 if the remaining pens are sold at ₹ 9 each. The wage of the employee, in INR, is

[2021]

Detailed Solution for CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 7

Let the number of pens purchased be n. Then the cost price is 8n. The total expenses incurred would be 8n + W, where W refers to the wage.
Then SP in the first case =12 × 100 + 11 × (n−100)
Given profit is 300 in this case: 1200 + 11n - 1100 - 8n - W = 300 ⇒ 3n - W = 200
In second case: 1200 + 9n - 900 - 8n - W = -300 (Loss). ⇒ W-n = 600.
Adding the two equations: 2n = 800
n = 400.
Thus W = 600 + 400 = 1000

CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 8
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A basket of 2 apples, 4 oranges and 6 mangoes costs the same as a basket of 1 apple, 4 oranges and 8 mangoes, or a basket of 8 oranges and 7 mangoes. Then the number of mangoes in a basket of mangoes that has the same cost as the other baskets is

[2021]
Detailed Solution for CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 8

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.

CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 9
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Onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg, respectively. A family spends a fixed amount of money on onion for each of the first three months, and then spends half that amount on onion for each of the next two months. The average expense for onion, in rupees per kg, for the family over these 5 months is closest to

[2021]
Detailed Solution for CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 9

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

CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 10
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The amount Neeta and Geeta together earn in a day equals what Sita alone earns in 6 days. The amount Sita and Neeta together earn in a day equals what Geeta alone earns in 2 days. The ratio of the daily earnings of the one who earns the most to that of the one who earns the least is

[2021]
Detailed Solution for CAT Previous Year Questions: Linear and Quadratic Equations (July 16) - Question 10

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

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