# CAT 2020 QA Slot 3 | Previous Year CAT Paper

**1. CAT 2020 QA Slot 3 | Modern Math - Permutation & Combination**

How many integers in the set {100, 101, 102, ..., 999} have at least one digit repeated?

Answer: 252

**Explanation** :

Here we have to find all 3-digit numbers which have at least one digit repeated.

This can also be calculated by subtracting all 3-digit number which do not have any repeated digits from total 3-digit numbers.

Total 3-digit numbers = 9 × 10 × 10 = 900

3-digit numbers without repetition,

∴ The number 3-digit numbers without repetition = 9 × 9 × 8 = 648

Therefore, the number of 3-digit numbers which have at least one digit repeated 900 – 648 = 252.

Hence, 252.

Workspace:

**2. CAT 2020 QA Slot 3 | Arithmetic - Time, Speed & Distance**

A and B are two railway stations 90 km apart. A train leaves A at 9:00 am, heading towards B at a speed of 40 km/hr. Another train leaves B at 10:30 am, heading towards A at a speed of 20 km/hr. The trains meet each other at

- A.
11:45 a.m.

- B.
11 a.m.

- C.
10:45 a.m.

- D.
11:20 a.m.

Answer: Option B

**Explanation** :

Train A starts at 9 a.m. whereas Train B starts at 10:30 a.m.

Distance travelled by A till 10:30 a.m. = 40 × 1.5 = 60 kms.

At 10:30 a.m. remaining distance between A and B = 90 – 60 = 30 kms.

Time taken for both of them to meet now = 30/(40 + 20) = 0.5 hours i.e., 30 mins.

∴ Both trains meet half after 10:30 a.m. i.e., at 11 a.m.

Hence, option (b).

**Concept**:

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**3. CAT 2020 QA Slot 3 | Arithmetic - Average**

A batsman played n + 2 innings and got out on all occasions. His average score in these n + 2 innings was 29 runs and he scored 38 and 15 runs in the last two innings. The batsman scored less than 38 runs in each of the first n innings. In these n innings, his average score was 30 runs and lowest score was x runs. The smallest possible value of x is

- A.
3

- B.
4

- C.
1

- D.
2

Answer: Option D

**Explanation** :

Average score in n innings is 30 and in n + 2 innings is 29. Score in last 2 innings is 38 and 15

∴ 30n + 38 + 15 = 29(n + 2)

⇒ n = 5

∴ Total runs score in first n innings = 5 × 30 = 150

Also, he scored less than 38 runs in each of these 5 innings.

To calculate the lowest possible score in an innings we try to maximize the score of other 4 innings. Maximum runs score in each of the 4 innings can be 37.

∴ x + 4 × 37 = 150

⇒ x = 2

Hence, option (d).

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**4. CAT 2020 QA Slot 3 | Geometry - Quadrilaterals & Polygons**

In a trapezium ABCD, AB is parallel to DC, BC is perpendicular to DC and ∠BAD = 45°. If DC = 5 cm, BC = 4 cm, the area of the trapezium in sq.cm is

Answer: 28

**Explanation** :

The following diagram can be drawn using the information given.

Since, ∆ADE is an isosceles right triangle hence, DE = AE = 4 cm

⇒ AB = BE + AE = 5 + 4 = 9 cm

∴ Area of trapezium = $\frac{1}{2}\times (CD+AB)\times BC$ = $\frac{1}{2}\times 14\times 4$ = 28 cm^{2}.

Hence, 28.

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**5. CAT 2020 QA Slot 3 | Arithmetic - Time & Work**

A contractor agreed to construct a 6 km road in 200 days. He employed 140 persons for the work. After 60 days, he realized that only 1.5 km road has been completed. How many additional people would he need to employ in order to finish the work exactly on time?

Answer: 40

**Explanation** :

In 60 days 1.5 kms out of 6 kms is built.

∴ In 60 days 1.5/6 = 1/4^{th} work is done.

To complete the whole work, it would take 4 × 60 = 240 days, i.e., 180 more days.

Days remaining now is 200 – 60 = 140 days.

∴ To complete the remaining work in 140 days, contractor will have to hire more people. Assuming he hires x more men.

The work which 140 men would take 180 more days, now need to be completed by 140 + x men in 140 days.

∴ 180 × 140 = 140 × (140 + x)

⇒ x = 40

∴ Contractor hires 40 more people.

Hence, 40.

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**6. CAT 2020 QA Slot 3 | Algebra - Progressions**

If x₁ = - 1 and x_{m} = x_{m+1} + (m + 1) for every positive integer m, then x_{100} equals

- A.
-5151

- B.
-5150

- C.
-5050

- D.
-5051

Answer: Option C

**Explanation** :

Given, x_{m} = x_{m+1} + (m + 1)

⇒ x_{m+1} = x_{m} - (m + 1)

Put m = 1 ⇒ x_{2} = -1 – 2

Put m = 2 ⇒ x_{3} = -1 - 2 – 3

Put m = 3 ⇒ x_{4} = - 1 – 2 - 3 – 4

And so on.

∴ x_{100} = - 1 – 2 – 3 -4 … - 100

⇒ x_{100} = - (100 × 101)/2 = - 5050

Hence, option (c).

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**7. CAT 2020 QA Slot 3 | Algebra - Surds & Indices**

If a, b, c are non-zero and 14^{a} = 36^{b} = 84^{c}, then $6\mathrm{b}\left(\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{a}}\right)$ is equal to

Answer: 3

**Explanation** :

Assuming, k = 14^{a} = 36^{b} = 84^{c}

⇒ 14 = k^{1/a}

⇒ 36 = 6^{2} = k^{1/b} or 6 = k^{1/2b}

⇒ 84 = k^{1/c}

We know, 84 = 6 × 14

⇒ k^{1/c} = k^{1/2b} × k^{1/a}

⇒ k^{1/c – 1/a} = k^{1/2b}

⇒ $\frac{1}{c}-\frac{1}{a}=\frac{1}{2b}$

⇒ $2\mathrm{b}\left(\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{a}}\right)$ = 1

Hence, $6\mathrm{b}\left(\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{a}}\right)$ = 3

Hence, 3.

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**8. CAT 2020 QA Slot 3 | Algebra - Number Theory**

Let m and n be natural numbers such that n is even and 0.2 < $\frac{\mathrm{m}}{20},\frac{\mathrm{n}}{\mathrm{m}},\frac{\mathrm{n}}{11}$ < 0..5. Then m – 2n equals

- A.
4

- B.
3

- C.
2

- D.
1

Answer: Option D

**Explanation** :

We know, 0.2 < m/20 <0.5 ⇒ 4 < m < 10

Values m can take = {5, 6, 7, 8, 9}

We know, 0.2 < n/11 <0.5 ⇒ 2.2 < n < 5.5

Values n can take = {3, 4, 5}

Since n is even, the only possible value of n is 4.

Also, 0.2 < $\frac{\mathrm{m}}{20},\frac{\mathrm{n}}{\mathrm{m}},\frac{\mathrm{n}}{11}$ < 0.5 ⇒ m/5 < n < m/2

Since n = 4 ⇒ 20 > m > 8

The only possible value of m is 9.

∴ m – 2n = 9 – 2 × 4 = 1.

Hence, option (d).

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**9. CAT 2020 QA Slot 3 | Algebra - Functions & Graphs**

If f(x + y) = f(x)f(y) and f(5) = 4, then f(10) – f(-10) is equal to

- A.
0

- B.
15.9375

- C.
3

- D.
14.0625

Answer: Option B

**Explanation** :

Given, f(x + y) = f(x) × f(y)

Substitute y = 0

⇒ f(x + 0) = f(x) × f(0)

f(x) = f(x) × f(0)

[f(x) cannot be ‘0’ since it is given that f(5) = 4]

∵ f(0) = 1

f(5) = 4 (Given)

f(10) = f(5 + 5) = f(5) × f(5) = 4 × 4 = 16

f(10) = 16

We know that f(0) = 1

f(0) = f(10 - 10) = f(10 + (-10)) = f(10) × f(-10) = 1

16 × f(-10) = 1

f(-10) = 1/16 = 0.0625

f(10) - f(-10) = 16 - 0.0625 = 15.9375.

Hence, option (b).

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**10. CAT 2020 QA Slot 3 | Geometry - Coordinate Geometry**

The points (2, 1) and (-3, -4) are opposite vertices of a parallelogram. If the other two vertices lie on the line x + 9y + c = 0, then c is

- A.
15

- B.
12

- C.
13

- D.
14

Answer: Option D

**Explanation** :

One of the diagonals of the parallelogram connects the points (2, 1) and (-3, -4).

Line x + 9y + c = 0, represents the other diagonal of the parallelogram.

We know, diagonals of a parallelogram bisect each other.

∴ x + 9y + c = 0 should pass through the midpoint of the line joining (2, 1) and (-3, -4).

The midpoint of the line joining (2, 1) and (-3, -4) = $\left(\frac{2-3}{2},\frac{1-4}{2}\right)$ = (-0.5, -1.5)

⇒ Point (-0.5 , -1.5) lies on x + 9y + c = 0.

⇒ -0.5 + 9 × (-1.5) + c = 0

⇒ -0.5 - 13.5 + c = 0

⇒ c = 14.

Hence, option (d).

Workspace:

**11. CAT 2020 QA Slot 3 | Arithmetic - Time, Speed & Distance**

Vimla starts for office every day at 9 am and reaches exactly on time if she drives at her usual speed of 40 km/hr. She is late by 6 minutes if she drives at 35 km/hr. One day, she covers two-thirds of her distance to office in one-thirds of her usual time to reach office, and then stops for 8 minutes. The speed, in km/hr, at which she should drive the remaining distance to reach office exactly on time is

- A.
26

- B.
28

- C.
29

- D.
27

Answer: Option B

**Explanation** :

Let the distance from home to office be d kms.

∴ $\frac{d}{35}-\frac{d}{40}=\frac{6}{60}$

⇒ d = 28 kms.

∴ Normal time taken = $\frac{28}{40}$ = 0.7 hour = 42 minutes.

Now, she covers 2/3^{rd} of this distance in 1/3rd time i.e., in 14 minutes.

She now takes rest for 8 minutes.

∴ Time elapsed = 14 + 8 = 22 minutes.

Time remaining = 20 minutes.

In 20 minutes she needs to travel 1/3^{rd} of the distance i.e., 28/3 kms.

∴ Time taken = $\frac{28/3}{20/60}$ = 28 kmph.

Hence, option (b).

Workspace:

**12. CAT 2020 QA Slot 3 | Arithmetic - Profit & Loss**

A man buys 35 kg of sugar and sets a marked price in order to make a 20% profit. He sells 5 kg at this price, and 15 kg at a 10% discount. Accidentally, 3 kg of sugar is wasted. He sells the remaining sugar by raising the marked price by p percent so as to make an overall profit of 15%. Then p is nearest to

- A.
25

- B.
35

- C.
22

- D.
31

Answer: Option A

**Explanation** :

Let the cost price of sugar be Rs. 100/kg and marked price will be = Rs. 120/kg

∴ Total cost for the man = 35 × 100 = Rs. 3500

He earns 15% profit on this, hence total selling price = 3500 × 1.15 = Rs. 4025

Total Selling price for 5 kg sugar = 5 × 120 = 600

Total Selling price for 15 kg sugar = 15 × 120 × 0.9 = 1620

Total Selling price for 3 kg sugar = 3 × 0 = 0

Total Selling price for remaining 12 kg sugar = 12 × 120(1 + p%)

∴ 4025 = 600 + 1620 + 0 + 1440(1 + p%)

⇒ 1805 = 1440(1 + p%)

⇒ 1 + p% ≈ 1.25

⇒ p = 25%

Hence, option (a).

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**13. CAT 2020 QA Slot 3 | Algebra - Quadratic Equations**

Let m and n be positive integers, If x² + mx + 2n = 0 and x² + 2nx + m = 0 have real roots, then the smallest possible value of m + n is

- A.
8

- B.
7

- C.
5

- D.
6

Answer: Option D

**Explanation** :

First, we have the quadratic equation, x² + mx + 2n = 0

Since its roots are real, that means D ≥ 0.

⇒ m^{2} – 8n ≥ 0

⇒ m ≥ 2√(2n) …(1)

Also, we have the quadratic equation, x² + 2nx + m = 0

Since its roots are real, that means D ≥ 0.

⇒ (2n)^{2} – 4m ≥ 0

⇒ 4n^{2} – 4m ≥ 0

⇒ n^{2} ≥ m …(2)

From (1) and (2)

⇒ n^{2} ≥ m ≥ 2√(2n)

⇒ n^{3/2} ≥ 2^{3/2}

⇒ n ≥ 2

Hence, least possible value of n = 2.

⇒ m ≥ 2√(2n)

m will be least when n is least

⇒ m ≥ 2√(4)

⇒ m ≥ 4

Hence, least possible value of m = 4.

So, the smallest possible values of m and n are 4 and 2 respectively.

Hence, smallest value of m + n = 4 + 2 = 6.

Hence, option (d).

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**14. CAT 2020 QA Slot 3 | Arithmetic - Time, Speed & Distance**

Anil, Sunil, and Ravi run along a circular path of length 3 km, starting from the same point at the same time, and going in the clockwise direction. If they run at speeds of 15km/hr, 10 km/hr, and 8 km/hr, respectively, how much distance in km will Ravi have run when Anil and Sunil meet again for the first time at the starting point?

- A.
4.6

- B.
4.2

- C.
5.2

- D.
4.8

Answer: Option D

**Explanation** :

Speeds of Anil, Sunil and Ravi, which are 15 kmph, 10 kmph and 8 kmph respectively.

Length of the track is 3 km.

Time taken by Anil to complete one round = 3/15 × 60 = 12 minutes.

Time taken by Sunil to complete one round = 3/10 × 60 = 18 minutes.

Both Anil and Sunil will reach at the starting points after LCM[12, 18] = 36 minutes.

The distance covered by Ravi in 36 minutes (= 0.6 hours) = 8 × 0.6 = 4.8 km.

Hence, option (d).

**Concept**:

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**15. CAT 2020 QA Slot 3 | Algebra - Number Theory | Algebra - Inequalities & Modulus**

Let N, x and y be positive integers such that N = x + y, 2 < x < 10 and 14 < y < 23. If N > 25, then how many distinct values are possible for N?

Answer: 6

**Explanation** :

Given, 2 < x < 10

x can take any of the values from {3, 4, 5, 6, 7, 8, 9}

Also, 14 < y < 23

y can take any of the values from {15, 16, 17, 18, 19, 20, 21, 22}

The highest value N (i.e., x + y) can take = 9 + 22 = 31 (when x = 9; y = 22)

The lowest value N (i.e., x + y) can take = 3 + 15 = 18 (when x = 3; y = 15)

But, N = x + y > 25. Hence the different values of x + y are {31, 30, 29, 28, 27, 26}.

Hence, x + y, and thereby N can take 6 distinct values.

Hence, 6.

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**16. CAT 2020 QA Slot 3 | Arithmetic - Percentage**

In the final examination, Bishnu scored 52% and Asha scored 64%. The marks obtained by Bishnu is 23 less, and that by Asha is 34 more than the marks obtained by Ramesh. The marks obtained by Geeta, who scored 84%, is

- A.
399

- B.
439

- C.
417

- D.
357

Answer: Option A

**Explanation** :

Let the maximum marks be 100x.

Marks scored by Bishnu = 52x and Asha = 64x

Bishnu scored 23 less, Asha scored 34 more than the marks obtained by Ramesh

∴ Asha – Bishnu = 34 + 23 = 57

⇒ 64x – 52x = 57

⇒ 12x = 57

Now, Geeta scored 84%, her score = 84x = 7 × 12x

= 7 × 57 = 399.

Hence, option (a).

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**17. CAT 2020 QA Slot 3 | Arithmetic - Average**

Dick is thrice as old as Tom and Harry is twice as old as Dick. If Dick's age is 1 year less than the average age of all three, then Harry's age, in years, is

Answer: 18

**Explanation** :

Let Tom's age to be 'x' years.

∴ Dick's age = 3x.

∴ Harry's age = 6x.

Given, Dick's age is 1 year less than the average age of all three,

∴ 3x = (x + 3x + 6x)/3 – 1

⇒ 9x = 10x – 3

⇒ x = 3

∴ Harry’s age = 6x = 6 × 3 = 18.

Hence, 18.

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**18. CAT 2020 QA Slot 3 | Geometry - Circles | Geometry - Coordinate Geometry**

The vertices of a triangle are (0, 0), (4, 0) and (3, 9). The area of the circle passing through these three points is

- A.
123π/7

- B.
205π/9

- C.
14π/3

- D.
12π/5

Answer: Option B

**Explanation** :

The diagram can be drawn as follows:

Height of the given triangle = 9 and base = 4

∴ Area of the triangle = $\frac{1}{2}\times 9\times 4$ = 18

We know area of a triangle = abc/4R

(where a, b, and c are the sides of the triangle and R is the circumradius)

a = AB = 4 units

b = BC = $\sqrt{{9}^{2}+{1}^{2}}=\sqrt{82}$ units

c = AC = $\sqrt{{3}^{2}+{9}^{2}}=\sqrt{90}$ units

The sides of the triangle are 4, √82 and √90

⇒ 18 = $\frac{4\times \sqrt{82}\times \sqrt{90}}{4R}$

⇒ R = $\frac{\sqrt{205}}{3}$

∴ Area of the circumcircle = πR2 = $\frac{205\mathrm{\pi}}{9}$

Hence, option (b).

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**19. CAT 2020 QA Slot 3 | Algebra - Logarithms**

$\frac{2\times 4\times 8\times 16}{\left({{\mathrm{log}}_{2}4)}^{2}\right({{\mathrm{log}}_{4}8)}^{3}({{\mathrm{log}}_{8}16)}^{4}}$ equals

Answer: 24

**Explanation** :

Given, $\frac{2\times 4\times 8\times 16}{\left({{\mathrm{log}}_{2}4)}^{2}\right({{\mathrm{log}}_{4}8)}^{3}({{\mathrm{log}}_{8}16)}^{4}}$

= $\frac{2\times {2}^{2}\times {2}^{3}\times {2}^{4}}{\left({{\mathrm{log}}_{2}{2}^{2})}^{2}\right({{\mathrm{log}}_{{2}^{2}}{2}^{3})}^{3}({{\mathrm{log}}_{{2}^{3}}{2}^{4})}^{4}}$

= $\frac{{2}^{10}}{{\left(2\right)}^{2}{\left(\frac{3}{2}\right)}^{3}{\left(\frac{4}{3}\right)}^{4}}$

= $\frac{{2}^{10}\times {2}^{3}\times {3}^{4}}{{2}^{2}\times {3}^{3}\times {2}^{8}}$

= 2^{3} × 3 = 24

Hence, 24.

Workspace:

**20. CAT 2020 QA Slot 3 | Arithmetic - Simple & Compound Interest**

A person invested a certain amount of money at 10% annual interest, compounded half-yearly. After one and a half years, the interest and principal together became Rs 18522. The amount, in rupees, that the person had invested is

Answer: 16000

**Explanation** :

Rate of interest is 10% per annum. Hence, interest for half a year is 5%.

The Principal, P is invested for 1.5 years or 3 terms of half years.

Therefore, P (1 + 5%)^{3} = 18522.

⇒ P × 1.05^{3} = 18522

⇒ P × ${\left(\frac{105}{100}\right)}^{3}$= 18522

⇒ P × ${\left(\frac{21}{20}\right)}^{3}$ = 2 × 213

⇒ P = 2 × 20^{3} = 16000.

Hence, 16000.

Workspace:

**21. CAT 2020 QA Slot 3 | Algebra - Logarithms**

If log_{a}30 = A, log_{a}(5/3) = -B and log_{2}a = 1/3, then log_{3}a equals

- A.
(A + B)/2 - 3

- B.
2/(A + B) - 3

- C.
2/(A + B - 3)

- D.
(A + B - 3)/2

Answer: Option C

**Explanation** :

Since all the 4 options have A + B, lets add A and B.

∴ A + B = log_{a}30 + (-log_{a}(5/3))

= log_{a}30 – log_{a}(5/3) = log_{a}(30 × 3/5)

= log_{a}(18) = log_{a}(2 × 3^{2})

⇒ A + B = log_{a}(2) + 2log_{a}(3)

⇒ A + B = 3 + 2log_{a}(3) [∵ log_{2}a = 1/3 ∴ log_{a}2 = 3]

⇒ log_{a}(3) = (A + B - 3)/2

⇒ log_{3}(a) = 2/(A + B - 3)

Hence, option (c).

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**22. CAT 2020 QA Slot 3 | Arithmetic - Mixture, Alligation, Removal & Replacement**

Two alcohol solutions, A and B, are mixed in the proportion 1 : 3 by volume. The volume of the mixture is then doubled by adding solution A such that the resulting mixture has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol in solution B is

- A.
92%

- B.
90%

- C.
89%

- D.
94%

Answer: Option A

**Explanation** :

Given, A and B are mixed in the ratio 1 : 3

Let 10 liters of A and 30 liters of B is mixed initially i.e., total 40 liters solution.

Now volume is double by adding 40 liters of A.

∴ In the solution A = 10 + 40 = 50 liters and B = 30 liters

Concentration of alcohol in this 80-liter solution = 72%

Let the alcohol concentration in B = b%

∴ 80 × 72% = 50 × 60% + 30 × b%

⇒ 576 = 300 + 30b

⇒ b = 92%

Hence, option (a).

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**23. CAT 2020 QA Slot 3 | Algebra - Surds & Indices**

How many pairs (a, b) of positive integers are there such that a ≤ b and ab = 4^{2017}?

- A.
2019

- B.
2017

- C.
2020

- D.
2018

Answer: Option D

**Explanation** :

Given, a × b = 4^{2017} = 2^{4034}

Here, a and b and integers and will be some power of 2 since RHS is 2^{4034}.

Let’s say a = 2^{x} and b = 2^{y}. [x and y are non-negative integers]

Also, a ≤ b, Hence, x ≤ y.

Now, 2^{x} × 2^{y} = 2^{4034}

Since x ≤ y, x ≤ 4034/2 = 2017

The values x can take is 0, 1, 2, …. till 2017.

Hence, x can take a total of 2018 values.

Hence, option (d).

Workspace:

**24. CAT 2020 QA Slot 3 | Geometry - Coordinate Geometry**

The area, in sq. units, enclosed by the lines x = 2, y = |x – 2| + 4, the X-axis and the Y-axis is equal to

- A.
6

- B.
10

- C.
8

- D.
12

Answer: Option B

**Explanation** :

We can plot x = 2 and y = |x - 2| + 4

The required area is sum of the triangle (orange) + rectangle (green).

Area of triangle (orange) = 1/2 × 2 × 2 = 2

Area of rectangle = 2 × 4 = 8

∴ Total required area = 2 + 8 = 10 sq. units.

Hence, option (b).

Workspace:

**25. CAT 2020 QA Slot 3 | Algebra - Number Theory | Venn Diagram**

How many of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7?

- A.
42

- B.
41

- C.
40

- D.
43

Answer: Option B

**Explanation** :

There are total 120 numbers given.

Integers not divisible by 2, 5 and 7 = 120 – (numbers that are divisible by at least one of 2, 5 and 7)

We know, n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C).

n(x) = number of multiples of x.

∴ n(2 ∪ 5 ∪ 7) = n(2) + n(5) + n(7) - n(2 ∩ 5) - n(5 ∩ 7) - n(7 ∩ 2) + n(2 ∩ 5 ∩ 7)

⇒ n(2 ∪ 5 ∪ 7) = n(2) + n(5) + n(7) - n(10) - n(35) - n(14) + n(70)

⇒ n(2 ∪ 5 ∪ 7) = 60 + 24 + 17 - 12 - 3 - 8 + 1 = 79.

Now, Integers not divisible by 2, 5 and 7 = 120 – (numbers that are divisible by at least one of 2, 5 and 7)

= 120 – 79 = 41

∴ Out of 120 numbers given, 41 numbers are not divisible by any of 2, 5 and 7.

Hence, option (b).

Workspace:

**26. CAT 2020 QA Slot 3 | Algebra - Simple Equations**

Let k be a constant. The equations kx + y = 3 and 4x + ky = 4 have a unique solution if and only if

- A.
|k| = 2

- B.
k = 2

- C.
|k| ≠ 2

- D.
k ≠ 2

Answer: Option C

**Explanation** :

For 2 equations a_{1}x + b_{1}y = c_{1 }and a_{2}x + b_{2}y = c_{2} to have a unique solution, the condition is: $\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$.

∴ $\frac{k}{4}=\frac{1}{k}$

⇒ k^{2} = 4

⇒ k = ±2 or |k| ≠ 2.

Hence, option (c).

Workspace:

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