**Linear Equations** constitute an important portion of the CAT Quantitative Ability (QA) section. 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; practice **all the important Formulas** on CAT Linear Equations. You can check out these Linear Equation questions from **CAT Previous year’s papers**. Practice a good number of sums on CAT **Linear Equation** questions so that you can answer these questions with ease in the exam. In this article, we will look into some important Linear Equation Questions for CAT QA. These are a good source for practice; If you want to practice these questions, you can download this Important CAT Linear Equation Questions PDF below, which is completely Free.

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**Question 1: **Suppose you have a currency, named Miso, in three denominations: 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos?

a) 17

b) 16

c) 18

d) 15

e) 19

**1) Answer (C)**

**Solution:**

If two 50 Misos are used, the 107 can be paid in only 1 way.

If one 50 Miso is used, the number of ways of paying 107 is 6 – zero 10 Miso, one 10 Miso and so on till five 10 Misos.

If no 50 Miso is used, the number of ways of paying 107 is 11 – zero 10 Miso, one 10 Miso and so on till ten 10 Misos.

So, the total number of ways is 18

**Question 2: **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

**2) 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 3: **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

**3) 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 4: **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

**4) 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)

**Question 5: **At a certain fast food restaurant, Brian can buy 3 burgers, 7 shakes, and one order of fries for Rs. 120 exactly. At the same place it would cost Rs. 164.5 for 4 burgers, 10 shakes, and one order of fries. How much would it cost for an ordinary meal of one burger, one shake, and one order of fries?

a) Rs. 31

b) Rs. 41

c) Rs. 21

d) Cannot be determined

**5) Answer (A)**

**Solution:**

Let the price of 1 burger be x and the price of 1 shake be y and the prize of 1 french fries be z

3x + 7y + z = 120

4x + 10y + z = 164.5

=> x + 3y = 44.5

=> x = 44.5 – 3y

=> 3(44.5 – 3y) + 7y + z = 120 => z = 120 – 133.5 + 2y

So, x+y+z = 44.5 – 3y + y -13.5 + 2y = 31

So, the cost of a meal consisting of 1 burger, 1 shake and 1 french fries = Rs 31

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

**Question 6: **The number of solutions $(x, y, z)$ to the equation $x – y – z = 25$, where x, y, and z are positive integers such that $x\leq40,y\leq12$, and $z\leq12$ is

a) 101

b) 99

c) 87

d) 105

**6) Answer (B)**

**Solution:**

x – y – z = 25 and $x\leq40,y\leq12$, $z\leq12$

If x = 40 then y + z = 15. Now since both y and z are natural numbers less than 12, so y can range from 3 to 12 giving us a total of 10 solutions.Similarly, if x = 39, then y + z = 14. Now y can range from 2 to 12 giving us a total of 11 solutions.

If x = 38, then y + z = 13. Now y can range from 1 to 12 giving us a total of 12 solutions.

If x = 37 then y + z = 12 which will give 11 solutions.

Similarly on proceeding in the same manner the number of solutions will be 10, 9, 8, 7 and so on till 1.

Hence, required number of solutions will be (1 + 2 + 3 + 4 . . . . + 12) + 10 + 11

= 12*13/2 + 21

78 + 21 = 99

**Question 7: **How many different pairs(a,b) of positive integers are there such that $a\geq b$ and $\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$?

**7) Answer: 3**

**Solution:**

$\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$

=> $ab = 9(a + b)$

=> $ab – 9(a+b) = 0$

=> $ ab – 9(a+b) + 81 = 81$

=> $(a – 9)(b – 9) = 81, a > b$

Hence we have the following cases,

$ a – 9 = 81, b – 9 = 1$ => $(a,b) = (90,10)$

$ a – 9 = 27, b – 9 = 3$ => $(a,b) = (36,12)$

$ a – 9 = 9, b – 9 = 9$ => $(a,b) = (18,18)$

Hence there are three possible positive integral values of (a,b)

**Question 8: **If $5^x – 3^y = 13438$ and $5^{x – 1} + 3^{y + 1} = 9686$, then x + y equals

**8) Answer: 13**

**Solution:**

$5^x – 3^y = 13438$ and $5^{x – 1} + 3^{y + 1} = 9686$

$5^{x} + 3^{y}*15 = 9686*5$

$5^{x} + 3^{y}*15 = 48430$

16*$3^y$=34992

$3^y$ = 2187

y = 7

$5^x$=13438+2187=15625

x=6

x+y = 13

**Question 9: **In 2010, a library contained a total of 11500 books in two categories – fiction and nonfiction. In 2015, the library contained a total of 12760 books in these two categories. During this period, there was 10% increase in the fiction category while there was 12% increase in the non-fiction category. How many fiction books were in the library in 2015?

a) 6160

b) 6600

c) 6000

d) 5500

**9) Answer (B)**

**Solution:**

Let the number of fiction and non-fiction books in 2010 = 100a, 100b respectively

It is given that the total number of books in 2010 = 11500

100a+100b = 11500 ——-Eq 1

The number of fiction and non-fiction books in 2015 = 110a, 112b respectively

110a+112b = 12760 ——-Eq 2

On solving both the equations we get, b=55, a= 60

The number of fiction books in 2015 = 110*60=6600

**Question 10: **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

**10) 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.