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# CAT Linear Equation Questions PDF [Most Important]

Linear Equations is one of the important topics in the CAT Quants (Algebra) section. These questions are usually not very tough, and hence you should not miss out on the questions from CAT Linear Equations. Firstly, understand the important concepts of Linear Equations such as equations of a line, slope, solving simultaneous linear equations, etc. Also, learn all the Important Formulas from Linear equations. You can check out these CAT Linear Equations-based questions from the CAT Previous year’s papers. In this post, we will look into some important Linear Equations Questions for CAT. These are a good source of practice for CAT preparation; If you want to practice these questions, you can download this CAT Linear Equation Questions PDF [Most Important] along with the detailed solutions (and video solutions) below, which is completely Free.

Question 1:Â 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)Â 17

b)Â 16

c)Â 18

d)Â 15

e)Â 19

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:Â 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)Â May 21

b)Â April 11

c)Â May 20

d)Â April 10

e)Â June 30

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:Â 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)Â 5

b)Â 6

c)Â 7

d)Â 8

e)Â 10

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:Â 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)Â 38

b)Â 31

c)Â 41

d)Â None of these

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
=>Â 3/4(2n/3+4)+3 = 30 =>Â (2n/3+4) = 36 => n = 48
So, the answer is option d)

Question 5:Â 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)Â Rs. 31

b)Â Rs. 41

c)Â Rs. 21

d)Â Cannot be determined

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

Question 6:Â 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)Â 101

b)Â 99

c)Â 87

d)Â 105

Solution:

x – y – z = 25 andÂ $x\leq40,y\leq12$,Â $z\leq12$
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 +Â z = 14. Now y can range from 2 to 12 giving us a total of 11 solutions.
If x = 38, then y +Â 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 + 2Â + 3 + 4 . . . . + 12) + 10 + 11
= 12*13/2 + 21
78 + 21 = 99

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

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

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:Â 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)Â 6160

b)Â 6600

c)Â 6000

d)Â 5500

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 = 11500Â  Â  Â  Â  Â  Â  ——-Eq 1

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

110a+112b = 12760Â  Â  Â  Â  Â  Â ——-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:Â 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)Â $0 < k \leq \frac{5}{13}$

b)Â $k > \frac{5}{13}$

c)Â $k = \frac{5}{13}$

d)Â k = 0

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

Question 11:Â A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of 5 children. How many toffees were there in his stock initially?

Solution:

Let the initial number of chocolates be 64x.

First child gets 32x+1 and 32x-1 are left.

2nd child gets 16x+1/2 and 16x-3/2 are left

3rd child gets 8x+1/4 and 8x-7/4 are left

4th child gets 4x+1/8 and 4x-15/8 are left

5th child gets 2x+1/16 and 2x-31/16 are left.

Given, 2x-31/16=0=> 2x=31/16 => x=31/32.

.’. Initially the Gentleman has 64x i.e. 64*31/32 =62 chocolates.

Question 12:Â Let A, B and C be three positive integers such that the sum of A and the mean of B and C is 5. In addition, the sum of B and the mean of A and C is 7. Then the sum of A and B is

a)Â 5

b)Â 4

c)Â 6

d)Â 7

Solution:

Given

A+(B+C)/2=5 => 2A+B+C=10….(i)

(A+C)/2 +B=7 => A+2B+C=14…..(ii)

(i)-(ii)=> B-A=4 => B=4+A.

Given, A, B, C are positive integers

If A=1=>B=5 => C=3

If A=2=>B=6 => C=0 but this is invalid as C is positive.

Similarly if A>2 C will be negative and cases are not valid.

Hence, A+B=6.

Question 13:Â Dick is thrice as old as Tom and Harry is twice as old as Dick. If Dick’s age is 1 yearÂ less than the average age of all three, then Harry’s age, in years, is

Solution:

Let tom’s age = x

=>Â Dick=3x

=>harry = 6x

Given,

3x+1 = (x+3x+6x)/3

=> x= 3

Hence, Harry’s age = 18 years

Question 14:Â Let k be a constant. The equations $kx + y = 3$ and $4x + ky = 4$ have a unique solution ifÂ and only if

a)Â $\mid k\mid\neq2$

b)Â $\mid k\mid=2$

c)Â $k\neq2$

d)Â $k=2$

Solution:

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

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

=> kÂ $\ne\$ |2|

Question 15:Â In May, John bought the same amount of rice and the same amount of wheat as he hadÂ bought in April, but spent â‚¹ 150 more due to price increase of rice and wheat by 20%Â and 12%, respectively. If John had spent â‚¹ 450 on rice in April, then how much did heÂ spend on wheat in May?

a)Â Rs.560

b)Â Rs.570

c)Â Rs.590

d)Â Rs.580

Solution:

Let John buy “m” kg of rice and “p” kg of wheat.

Now let the price of rice be “r” in April. Price in May will be “1.2(r)”

Now let the price of wheat be “w” in April . Price in April will be “1.12(w)”.

Now he spent â‚¹150 more in May , so 0.2(rm)+0.12(wp)=150

Its also given that he had spent â‚¹450 on rice in April. So (rm)=450

So 0.2(rm)= (0.2)(450)=90 Substituting we get (wp)=60/0.12 or (wp)=500

Amount spent on wheat in May will be 1.12(500)=â‚¹560