# CAT Linear Equations Questions PDF [Most Important]

**Linear Equations** is a key topic in the **CAT** **Quants **section. Make sure you are aware of all the **Important Concepts in CAT Quant (Linear Equation)**. One must not miss out on the questions on Linear Equations in the QA section. Linear Equations fall under the category of **Algebra** in the CAT Quants; You can also check out these CAT Linear Equation questions from the **CAT Previous year papers**. This post will look into some important Linear Equation questions for CAT. These are good sources of practice for CAT 2022 preparation. If you want to practice these questions, you can download these Important** Linear Equation Questions for CAT** (with detailed answers) **PDF** and the video solutions below, which are completely Free.

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**Question 1: **The price of Darjeeling tea (in rupees per kilogram) is 100 + 0.10n, on the nth day of 2007 (n=1, 2, …, 100), and then remains constant. On the other hand, the price of Ooty tea (in rupees per kilogram) is 89 + 0.15n, on the nth day of 2007 (n = 1, 2, …, 365). On which date in 2007 will the prices of these two varieties of tea be equal?

a) May 21

b) April 11

c) May 20

d) April 10

e) June 30

**1) Answer (C)**

**Solution:**

Price of Darjeeling tea on 100th day= 100+(0.1*100)=110

Price of Ooty tea on nth day= 89+0.15n

Let us assume that the price of both varieties of tea would become equal on nth day where n<=100

So

89+0.15n=100+0.1n

n=220 which does not satisfy the condition of n<=100

So the price of two varieties would become equal after 100th day.

89+0.15n=110

n=140

140th day of 2007 is May 20 (Jan=31,Feb=28,March=31,April=30,May=20)

**Question 2: **When you reverse the digits of the number 13, the number increases by 18. How many other two-digit numbers increase by 18 when their digits are reversed?

a) 5

b) 6

c) 7

d) 8

e) 10

**2) Answer (B)**

**Solution:**

Let the number be xy

10y + x = 10x + y + 18

=> 9y – 9x = 18

=> y – x = 2

So, y can take values from 9 to 4 (since 3 is already counted in 13)

Number of possible values = 6

**Question 3: **A change-making machine contains one-rupee, two-rupee and five-rupee coins. The total number of coins is 300. The amount is Rs. 960. If the numbers of one-rupee coins and two-rupee coins are interchanged, the value comes down by Rs. 40. The total number of five-rupee coins is

a) 100

b) 140

c) 60

d) 150

**3) Answer (B)**

**Solution:**

Let the number of coins of the three denominations be x, y and z respectively.

x+y+z = 300

x+2y+5z = 960

2x+y+5z = 920

=> 3(x+y) + 10z = 1880

=> 3(300 – z) + 10z = 1880

=> 900 + 7z = 1880 => z = 980/7 = 140

So, the number of 5 rupee coins is 140

**Question 4: **A piece of string is 40 cm long. It is cut into three pieces. The longest piece is three times as long as the middle-sized and the shortest piece is 23 cm shorter than the longest piece. Find the length of the shortest piece.

a) 27

b) 5

c) 4

d) 9

**4) Answer (C)**

**Solution:**

Let the longest piece be x

Shortest piece = x – 23

Middle-sized piece = x/3

So, x + x – 23 + x/3 = 40 => 7x/3 = 63 => x = 27

Shortest piece = 27 – 23 = 4

**Question 5: **Out of two-thirds of the total number of basketball matches, a team has won 17 matches and lost 3 of them. What is the maximum number of matches that the team can lose and still win more than three- fourths of the total number of matches, if it is true that no match can end in a tie?

a) 4

b) 6

c) 5

d) 3

**5) Answer (A)**

**Solution:**

Total matches played = 17+3 = 20

Total matches = 20*$\frac{3}{2}$ = 30

Number of wins required = 75% of 30 = 22.5 = 23 wins

23-17 = 6 more wins are required out of 10 matches to maintain 75% win recor which means there would be 4 losses.

**Question 6: **Three friends, returning from a movie, stopped to eat at a restaurant. After dinner, they paid their bill and noticed a bowl of mints at the front counter. Sita took one-third of the mints, but returned four because she had a momentary pang of guilt. Fatima then took one-fourth of what was left but returned three for similar reason. Eswari then took half of the remainder but threw two back into the bowl. The bowl had only 17 mints left when the raid was over. How many mints were originally in the bowl?

a) 38

b) 31

c) 41

d) None of these

**6) Answer (D)**

**Solution:**

Let the total number of mints in the bowl be n

Sita took n/3 – 4. Remaining = 2n/3 + 4

Fatim took 1/4(2n/3 + 4) – 3. Remaining = 3/4(2n/3 + 4) + 3

Eswari took 1/2(3/4(2n/3+4)+3) – 2

Remaining = 1/2(3/4(2n/3+4)+3) + 2 = 17

=> 3/4(2n/3+4)+3 = 30 => (2n/3+4) = 36 => n = 48

So, the answer is option d)

Checkout: **CAT Free Practice Questions and Videos**

**Question 7: **Let a, b, x, y be real numbers such that $a^2 + b^2 = 25, x^2 + y^2 = 169$, and $ax + by = 65$. If $k = ay – bx$, then

a) $0 < k \leq \frac{5}{13}$

b) $k > \frac{5}{13}$

c) $k = \frac{5}{13}$

d) k = 0

**7) Answer (D)**

**Solution:**

$\left(ax+by\right)^2=65^2$

$a^2x^2\ +\ b^2y^2+\ 2abxy\ =\ 65^2$

$k = ay – bx$

$k^2\ =\ a^2y^2+b^2x^2-2abxy$

$(a^2 + b^2)(x^2 + y^2 )= 25* 169$

$a^2x^2+a^2y^2+b^2x^2+b^2y^2=\ 25\times\ 169$

$k^2=\ 65^2\ -\ \left(25\times\ 169\right)$

k = 0

D is the correct answer.

**Question 8: **Let k be a constant. The equations $kx + y = 3$ and $4x + ky = 4$ have a unique solution if and only if

a) $\mid k\mid\neq2$

b) $\mid k\mid=2$

c) $k\neq2$

d) $k=2$

**8) Answer (A)**

**Solution:**

Two linear equations ax+by= c and dx+ ey = f have a unique solution if $\frac{a}{d}\ne\ \frac{b}{e}$

Therefore, $\frac{k}{4}\ne\ \frac{1}{k}$ => $k^2\ne\ 4$

=> k $\ne\ $ |2|

**Question 9: **If $3x+2\mid y\mid+y=7$ and $x+\mid x \mid+3y=1$ then $x+2y$ is:

a) $-\frac{4}{3}$

b) $\frac{8}{3}$

c) $0$

d) $1$

**9) Answer (C)**

**Solution:**

We need to check for all regions:

x >= 0, y >= 0

x >= 0, y < 0

x < 0, y >= 0

x < 0, y < 0

However, once we find out the answer for any one of the regions, we do not need to calculate for other regions since the options suggest that there will be a single answer.

Let us start with **x >= 0, y >= 0,**

3x + 3y = 7

2x + 3y = 1

Hence, x = 6 and y = -11/3

Since y > = 0, this is not satisfying the set of rules.

Next, let us test **x >= 0, y < 0,**

3x – y = 7

2x + 3y = 1

Hence, y = -1

x = 2.

This satisfies both the conditions. Hence, this is the correct point.

WE need the value of x + 2y

x + 2y = 2 + 2(-1) = 2 – 2 = 0.

**Question 10: **In an examination, there were 75 questions. 3 marks were awarded for each correct answer, 1 mark was deducted for each wrong answer and 1 mark was awarded for each unattempted question. Rayan scored a total of 97 marks in the examination. If the number of unattempted questions was higher than the number of attempted questions, then the maximum number of correct answers that Rayan could have given in the examination is

**10) Answer: 24**

**Solution:**

Let the number of questions attempted be x+y out of which x are correct and y are incorrect and the number of questions unattempted be z.

It is given,

x + y + z = 75 …… (1)

3x – y + z = 97 …… (2)

(2)-(1) -> x – y = 11

(1)+(2) -> 2x + z = 86

z > x + y

z > 75 – z

z > 37.5

Minimum possible value of z is 38

2x + 38 = 86

2x = 48

x = 24

The maximum number of correct answers is 24.