Top 30+ IIMB UG 2026 Linear Equations Questions, Download PDF

We are given that the pair of linear equations $$ax+6y=13$$ and $$7x+by=7$$ has no solution.

We know that this pair of linear equation has no solutions if:

$$\dfrac{a}{7}=\dfrac{6}{b}\neq\dfrac{13}{7}$$

So $$\dfrac{a}{7}=\dfrac{6}{b}$$

or $$ab=7*6$$

or $$ab=2*3*7$$

Therefore, the number of factors of the product $$ab$$ is$$=(1+1)(1+1)(1+1)=8$$

Therefore, $$k=8$$

Hence, Option C is correct.

The given equation is $$9x + 2y = 9-3xy$$

$$3xy+9x+2y=9$$

In this equation, if we look at the first two terms, we can take $$3x$$ common and we will get $$(y+3)$$.

So, if we can write one more term as $$(y+3)$$, then, we can take $$(y+3)$$ as common from the LHS.

Add $$6$$ on both sides.

$$3xy+9x+2y+6=9+6$$

Take $$3x$$ common from the first two terms and $$2$$ common from the last two terms.

$$3x\left(y+3\right)+2\left(y+3\right)=15$$

$$\left(3x+2\right)\times\left(y+3\right)=15$$

15 on the RHS has to be written in the form of product of two numbers (as we have a product of two numbers on LHS)

$$15=1\times15, 3\times5, -1\times-15, -3\times-5\ \ $$

Now, $$(3x+2)$$ can be either one of $$5$$, $$-5$$, $$3$$, $$-3$$, $$15$$, $$-15$$, $$1$$, $$-1$$.

So, we get the value of $$x$$ as $$1$$, $$\dfrac{-7}{3}$$, $$\dfrac{1}{3}$$, $$\dfrac{-5}{3}$$, $$\dfrac{13}{3}$$, $$\dfrac{-17}{3}$$, $$\dfrac{-1}{3}$$, $$-1$$.

Since, it is given that the solution is a non-negative integer,

We get the value of $$x$$ as $$1$$.

Hence, $$(3x+2)$$ = $$\left(3\times1\right)+2$$ = 5

So, we get the value of $$(y+3)$$ as 3.

$$y=(3-3)=0$$

Hence, $$x=1$$ and $$y=0$$.

$$(x\times y)=(1\times0)=0$$

Option B.

5a + 4b + 6c = 100 --> (1)
and 10a+6b+9c = 180 --> (2)
(2) - 1.5*(1) gives a = 12

The given equation is $$4xy+4x+y=26$$

In this equation, if we look at the first two terms, we can take $$4x$$ common and we will get $$(y+1)$$.

So, if we can write one more term as $$(y+1)$$, then, we can take $$(y+1)$$ as common.

Add $$1$$ on both sides.

$$4xy+4x+y+1=26+1$$

Take $$4x$$ common from the first two terms and 1 common from the next two terms.

$$4x\left(y+1\right)+1\left(y+1\right)=27$$

$$\left(4x+1\right)\times\left(y+1\right)=27$$

27 on the RHS has to be written in the form of product of two numbers (as we have a product of two numbers on LHS)

$$27=1\times27, 3\times9, -1\times-27, -3\times-9\ \ $$

Now, $$(4x+1)$$ can be either one of $$9$$, $$-9$$, $$3$$, $$-3$$, $$27$$, $$-27$$, $$1$$, $$-1$$.

So, we get the value of $$x$$ as $$2$$, $$\dfrac{-10}{4}$$, $$\dfrac{2}{4}$$, $$-1$$, $$\dfrac{26}{4}$$, $$-7$$, $$0$$, $$\dfrac{-2}{4}$$.

Since, it is given that the solution is a positive integer,

We get the value of $$x$$ as $$2$$.

Hence, $$(4x+1)$$ = $$\left(4\times2\right)+1$$ = 9

So, we get the value of $$(y+1)$$ as 3.

$$y=(3-1)=2$$

Hence, $$x=2$$ and $$y=2$$.

$$(x+y)=(2+2)=4$$

a+b = b+c => a=c
Similarly, b+c = a+c => a=b
So, a=b=c=k (Let's assume)
Now, $$2k = 2k^3$$ $$=> 2k^3 -2k = 0$$
=> $$k(k^2-1) = 0$$
=> $$k = \pm 1 = a = b = c$$
=> $$a^3 + b^3 + c^3 = \pm 3$$
So, both A and C are possible.
Thus, D is the correct answer.

Given initially

$$M+O+G = 145$$

After selling the fruits

$$\dfrac{4}{5}M+\dfrac{1}{2}O+\dfrac{1}{3}G = 66$$

Since we can only sell a natural number of fruits, $$G$$ must be a multiple of $$3$$ as well, and $$O$$ must be a multiple of $$2$$, and M must be a multiple of 5.

Thus, $$G$$ must be a multiple of $$75$$ and $$O$$ must be a multiple of $$50$$ ( given that they are multiples of 25 in the question)

Hence, G can be $$75, 150,..$$ while O can be $$50,100,..$$

From the original equation, we can see that $$O+G$$ must be less than $$145$$. Thus the only values of $$O$$ and $$G$$ that satisfy this are $$(50,75)$$

Thus, $$ M = 145 - (75+50) = 20$$

$$\frac{1}{a}+\frac{5}{b}=\frac{1}{5}$$

$$\ \frac{\ b+5a}{ab}=\frac{1}{5}$$

5b + 25a = ab

ab - 5b - 25a = 0

ab - 5b - 25a + 125 = 125

(a-5)(b-25) = 125

a - 5 = $$a_1$$

b - 25 = $$b_1$$

$$a_1b_1=125$$

It is given that b is an even integer less than 50. This implies $$b_1$$ is an odd integer less than 25.

Only possible integral values for $$b_1$$ are 1 and 5.

Therefore, there are two possible pairs of (a,b).

Alternate explanation:

This can be represented as $$ a = \frac{5b}{b-25}$$.

As a and b are positive, b>25. Hence the choices are b=26, 28, 30 ... 48.

Out of these 26 and 30 give integral answers for a when substituted. Hence, number of pairs = 2

Let $$u = \dfrac{1}{p-q}$$ and $$v = \dfrac{1}{p+q}$$

Now, replacing, we get:

$$15u+18v=5$$ .... Eq 1

$$10u+36v=6$$ ..... Eq 2

Multiplying Eq 1 by 2,

We get:

$$30u+36v=10$$... Eq 3

Subtracting Eq 2 from Eq 3:

$$20u = 4$$

$$u = \dfrac{4}{20} = \dfrac{1}{5}$$

Now substitute $$u = \dfrac{1}{5}$$ back into Equation 1 to find $$v$$:

$$15\left(\dfrac{1}{5}\right) + 18v = 5$$

$$3 + 18v = 5$$

$$18v = 2$$

$$v = \dfrac{2}{18} = \dfrac{1}{9}$$

Substituting u and v back into the initial equations:

$$\dfrac{1}{p-q} = \dfrac{1}{5} \implies p - q = 5 \quad \text{--- (Equation 4)}$$

$$\dfrac{1}{p+q} = \dfrac{1}{9} \implies p + q = 9 \quad \text{--- (Equation 5)}$$

Adding Eq 4 and 5 to find p:

$$(p - q) + (p + q) = 5 + 9$$

$$p = 7$$

Let the two numbers be A and B.
So, A-B = 2 and 3A+2B = 86
Multiplying the first equation by two and adding it to the second equation, we get

2A-2B + 3A + 2B = 4 + 86 = 90
So, 5A = 90 or A=18
Hence, B=16 and their sum equals 18+16 = 34

8 beedis, 11 cigars and 14 cigarettes cost 380 Rs. .... (a)
5 beedis, 7 cigars and 9 cigarettes cost 240 Rs. ......(b)
Subtracting (a) from (b), we get
3 beedis, 4 cigars and 5 cigarettes cost 140 Rs .... (c)
Subtracting (c) from (b), we get
2 beedis, 3 cigars and 4 cigarettes cost 100 Rs. ....(d)
Subtracting (d) from (c) we get
1 beedi, 1 cigar and 1 cigarette cost 40 Rs

Let the fixed enrollment fee, book insurance fee and book issuance fee be x, y and z.

When the student borrows 4 books, he pays Rs. 554.

The equation will be x + 4y + 4z = 554......     ........Eq 1

Similarly, when he borrows 8 books, he pays Rs. 1094.

The equation will be x + 8y + 8z = 1094.................Eq 2

We need to find the amount when the student issues 6 books.  i.e. x + 6y + 6z

Adding Eq 1 & 2

we get,

2x + 12y + 12z = 1648

Dividing both side with 2,

we get x + 6y + 6z = 824

Hence, the answer is 824.

Let the numbers be xy and ab

Thus, xy = 10x+y and ab = 10a+b

It is given that sum of 10x+y+10a+b = 60

10(x+a) + y+b = 60

We see that y+b has to a multiple of 10, so y+b can be 10,20 (but 20 is not possible)

Thus, y+b = 10 such that both numbers are prime.

Only possible when numbers are 3 and 7.

So, numbers are 10x+3 and 10a+7

Thus. 10(x+a)+10 = 60

x+a = 5

So, two possibilities:

= (1,4) and (2,3)

So (1,4) is not possible as the difference between numbers will be more than 10.

So, (33,27) are the possible numbers.

33*27 = 891

Option B is the correct answer.

Given that we are unknown to the ages of arjun and atul .

Let them be x and y .

Given (x +1/y)(y +1/x) = 49/6

xy +1/xy = 37/6

Solving this we get xy = 6 .

Now we know that none of them are 1 year .

Hence we can have x , y to be 2 and 3 in any order .

Let Ram current age be r and Shyam's be s

$$\frac{r+10}{s+10}=\frac{6}{7}$$

$$7r+10\ =\ 6s$$ .....(I)

Their age differece will always remain same thus s-r = 5 ....(II)

Putting eqn value of s from eqn(II) to eqn (I)

$$7r+10\ =\ 6(r+5)$$

$$7r+10\ =\ 6r+30$$

On solving we get r =20

5200 mangoes were left in the whole farm.

Had he plucked 20 more mangoes from each tree, 4400 mangoes would have left in the farm.

5200-4400=800

Number of trees = 800/20=40 trees

Total number of mangoes in the farm initilally= 240*40=9600

x mangoes from each tree

9600-40x= 5200

x= 4400/40= 110

Option D

Let the price of bat be ‘a’, ball be ‘b’ and glove be ‘c’. So we have
2a + 3b + 7c = 950
7a + 6b + 2c = 1750
Adding both the equations, we get
9a + 9b + 9c = 2700
a + b + c = 300
Hence, the price of 1 bat, 1 ball and 1 glove is 300.

Let the amount with B = x

=>Amount with A, C and D will be x + 80, x-50 and x+120+x-50 = 2x + 70 respectively.

Given x + x +80+x-50+2x+70 = 500

=> x = 80

Amount with A= 160

Amount with B= 80

Amount with C= 30

Amount with D=230

Since Amounts with C and D are not multiples of 20, they should have atleast one Rs 10 note

Least number of ten rupee notes = 2

Let the number of 5 rupee notes be x, 10 rupee notes be y and 20 rupee notes be z.
We know that,
x + y + z = 100 -----------------------(1)
5x + 10y + 20z = 1500--------------(2)
5x + 20y + 10z = 1100-------------(3)

Multiplying equation (1) by 5 and substituting it in equations (2) and (3), we get,
5y + 15z = 1000-------(4)
15y + 5z = 600 ----(5)

(4)*3 - (5) => 40z = 2400
z = 60
Therefore, the number of 20 rupee notes is 60.
Therefore, option D is the right answer.

We are given that the man has 2 types of grain in 2:3 ratio, and the total grain with the man is 100 kgs.

$$\therefore$$  quantity of grain A= $$\dfrac{\ 2}{\left(2+3\right)}\times\ 100=\ \dfrac{\ 2}{5}\times\ 100=40\ kgs$$

and quantity of grainB= $$100-40=60kgs$$

let Cost per kg of grain A be Rs k, then Cost per kg of grain B will be Rs (k+10)

Further, lets assume that Sita buys $$\ x\ kg$$ of grain A, and $$\ y\ kg$$ of grain B

while Gita buys $$\ m\ kg$$ of grain A and $$\ n\ kg$$ of grain B .

$$\ \therefore\ \ x+m=40\ ......\left(1\right)\ $$

$$\&\ \ y+n=60\ ......\left(2\right)\ $$

Given , Gita buys $$20\ kg\ $$ of grain B , $$i.e\ n=20\ $$

$$\therefore\ y=60-20=\ 40\ \ \ \left(eqn\ 2\right)$$

Now, Sita pays Rs 200 more than Gita.

Amount paid by Sita= $$x\times\ k\ +y\times\ \left(k+10\right)$$

Amount paid by Gita= $$m\times\ k\ +n\times\ \left(k+10\right)$$

$$\therefore\ x\times\ k\ +\ y\times\ \left(k+10\right)-\left\{m\times\ k\ +n\times\ \left(k+10\right)\right\}=200$$

$$or,\ \left(x+y-m-n\right).k\ +10.\left(y-n\right)=200$$

$$or,\ \ \left(x-m+40-20\right).k+10.\left(40-20\right)=200$$

$$or,\ \ \left(x-m+20\right).k+200=200$$

$$or,\ \ \left(x-m+20\right).k=0$$

Now , k= cost per Kg cannot be 0, hence $$x-m+20=0\ $$

$$or,\ x-m=-20\ .....\left(3\right)$$

Solving Eqn 1 and 3,

$$x=10\ kg$$

Hence, Amount of grain Sita bought=$$10\ kg$$

Let x, y and z be the number of coins of each denomination => x+4y+16z=100.
Max value of z=6. For z=6, max value of y=1.
Hence there are 2 cases y=0,1 for z=6. For z=5, there are 3 cases (y=5, x=0), (y=4, x=4), (y=3, x=8).
For z=4, there are 3 cases (y=9, x=0), (y=8, x=4) and (y=7, x=8).
Hence total of 8 cases