Complete Model Solutions Mathematics Paper

Question1.

Given,
x2?2026=?(x+2026)
To Find: x =?
When LHS=RHS
Soln. :-
a) Rule of Square root
?(x+2026) ? 0
? x+2026 ? 0
? x ? -2026

b) It should also greater than equal to 0 in left side.
x2?2026 ? 0
? x2 ? 2026
? x ? ? (2026)
? |x| = 45
Therefore,
x ? -45 and x ? 45

Hence, negative values of are not solutions.
So,
`
At x = 45 and x = 46
x = 45
x2?2026=?(x+2026) - (Given Equation)
LHS,
x2?2026 = (45) 2 2026
= -1
RHS,
?(x+2026) =?(45+2026)
= 45.50
So, LHS ? RHS, (LHS < RHS>

x = 46
LHS,
X2?2026 = (46)2- 2026
= 90
RHS,
?(x+2026) = ?(46+2026)
= 45

So, LHS ? RHS, (LHS > RHS)

So, both functions are continuous, and they are intersect between 45 and 46.
By approximation value, x = 45.51
LHS,
x2?2026 = (45.51)2- 2026
= 45.16

RHS,
?(x+2026) = ?(45.51+2026)
= 45.51

LHS ? RHS (Exactsolutioncannotbeexpressedinclosedform)
Final Answer,
Therefore, the value of x is 45.51.

Question 2.

There is a grid in which is infinite on both side (up and down direction).
At t = 0, only 1 ant is in left bottom corner.
After t = 1 second,
Any ant can copy itself to
Right cell
Up cell

Condition where copy is going, that cell must be vacant.
In image, yellow region is showing that area is fixed.
Therefore, is there a time t such that the ants have completely vacated the yellow region?

Proof:
Ant Movement,
Ants can copy only up or right.
Ants never move left or down direction
It does not leave its original cells; it only creates copies.
Nature of yellow region,
Yellow region is finite
Grid infinite but yellow area is small
Initially,

At t = 0,
An ant is in yellow region or bottom left corner
At t = 1, t = 2,
Ants are creating copy and moves up or right direction
The copy that goes out does not replace the original ant.
Original ants will stay inside the yellow region.
No matters how many ants its spread, it will always stay at least one ant in yellow region. It is because ants are duplicated, not moved.
So, there is no time t will come when yellow region will be empty.
Final answer,
No, there is no any time t which is completely vacant in yellow region.

Question 3.

Given,
There is a 10x10 grid,
Total 100 small square
There are 2x2 cards,
Every card covers 4 small square
It can also use as any number of cards
Overlapping is allowed
It is not important of colour
How many minimum 2x2 cards that cover all 10x10 grid?
Proof:
Step 1. Overlap not allowed
If overlap is not allowed then grid = 100 square
1 card = 4 square
So, number of cards required,

Thus, 25 cards can perfectly cover in all grid without overlap
Step 2. Overlap allowed
Total grid = 100 square
Every card can give 4 unique square.
Let assume, 24 cards will use,
Maximum distinct squares covered
24x4 = 96 < 100>So, if we will overlap or not, 24 cards cannot cover all 100 squares of the grid.
Therefore, at least 25 cards are necessary to cover the all grids.

Final answer,

Minimum number is 2x2 card required = 25.
Question 4.
Given,
There are 5 coloured balls.
V (Voilet), O (Orange), R (Red), B (Blue), G (Green)
These balls,
are put in one track.
And track further ahead forms a circular section that can fit exactly 3 balls

It showing starts position in diagram
Now, we have to check from starting position, it can make possible arrangement (configuration) on follow the only given new rules.
Move 1. Rotation in circular section
When 3 balls are in circle

It can rotate

VOR ORV
It is changes the order in circle.
Move 2. Track Move
Left side ball goes forward
A balls comes out from the circle
A ball is enter in circle on right side.

Now,
Use PERMUTATION arrangement
Each arrangement requires swape to create
Swaps can even or odd.

Even Permutation

When even no. of swaps are use.
Eq. 2 Swaps, 4 Swaps
Odd Permutation
When odd no. of swaps are use.
Eq. 1 swaps, 3 swaps
Nature of allowed moves
Circular rotation
Rotation of 3 balls
This is even permutation
Track move
It is combination of two swaps
This is also even permutation
Rule = Even + Even = Even
So, no matter how many moves we will do.
Result will come always even.
Odd permutation cannot come.
So,
No, all configuration is not possible.

Final Answer,
From starting position, it can obtain even permutation because both allowed moves are even permutation. So, whoever configuration of odd permutation, this cannot possible in this system.

Question 5.

Given,
p is prime number, p >2
Machine States = Color, Number
Where,
Colour = Red, Black
Number = 1,2,3,p
So, total states = 2p
Toggle Switch
Change the color
Same number
Crank
Red number +1
Black number 1
Show the machine that all 2p states are come in one cycle.
Soln.
Step1 States list,
Red list
(Red,1), (Red,2), , (Red, p)

Black List
(Black,1), (Black,2), , (Black, p)

Step 2 Effect on crank
When colour red
Keep cranking it
(Red,1) (Red, 2) (Red, p) (Red, 1)
This makes a complete cycle.
When colour black
Keep cranking it
(Black, p) (Black, p-1) (Black, 1) (Black, p)
This is also complete cycle.
Means,
Red states and Black states are connected with each other

Step 3 Role of Toggle
Toggle is only flip of colour.
Means,
(Red, x) ? (Black, x)
Due to this, Red cycle and Black cycle are connected each other.
Step 4 Now combine Crank and Toggle
From crank, it can move inside the red states and black states.
From toggle, it can go red to black and black to red.
So, from any states,
it can go any number before the use of crank.
Then by use of toggle, it can use change the colour.
Final Result,
P has exactly 2p states. From using the crank operation, all red states and black states makes one cycle. Toggle switches are connected with red and black states. So, thats why, all 2p states are connected and also in one cycle. So, starting point of states, it can reach the p machine from other states.

Question 7.

Use of combination method,
nCr =
20C4=
=
= 4845
Final answer = 4845 ways.

Question 8.
Given equation,
n3<3n>Inequality for small positive integers
Check value
n = 1,
LHS, n3 = 13 = 1
RHS, 3n = 31 = 3
LHS

n = 2,
LHS, n3 = 23 = 8
RHS, 3n = 32 = 9
LHS

n = 3,
LHS, n3 = 33 = 27
RHS, 3n = 33 = 27
LHS=RHS, (False)

n = 4,
LHS, n3 = 43 = 64
RHS, 3n = 34 = 81
LHS

n = 5,
LHS, n3 = 53 = 125
RHS, 3n = 35 = 243
LHS

n = 6,
LHS, n3 = 63 = 216
RHS, 3n = 3 = 729
LHS

For all integers , the inequality continues to hold.
Hence, the inequality is true for all integers .
Final answer: Smallest Integer z = 3
Question 9.
Given,
DOMOSOGY
Total no. words = 8
Repeated letter,
O = 3 Times
D,M,S,G,Y = 1 Time
Soln.
Use Permutation Formula,
Permutations =
So,

=
= 6720
Final Answer = 6720