XAT 2024 Question Paper

Let's assume that the number of students who initially took Arts is x and the number of students who initially took Science is y.

After 20 students shifted from Arts to MatEco, the number of Science students was twice that of Arts students.

That is y=2*(x-20) => 2x-y-40=0-->1

Now, 45 of the students who originally enrolled in Science have switched to MatEco. This makes the number of Arts students twice the number of Science students.

So, x-20=2*(y-45) => x-2y+70=0-->2

Now that we have two linear equations. Let's solve them.

Multiply Equation 1 with -2 and add it to Equation 2.

-4x+2y+80=0

x-2y+70=0

-3x+150=0

3x=150=>x=50

So, the original number of students enrolled in Arts is 50.

Substitute the value of x in equation 1 to get the y value.

y=60

There are 65 (20 from arts, 45 from Science who shifted to the MatEco) in MatEco. All these study Economics; apart from them, the people currently in Arts also study Economics. There are 50-20=30 people currently in Arts. So, the total number of people who study Economics is 65+30=95.

The number of students from 1 to 10 is in AP. Let's assume the series as follows:

a, a+d, a+2d, a+3d, a+4d, a+5d, a+6d, a+7d, a+8d, a+9d.

Given that the sum of the number of students in the first four classes is 462.

So, a +a+d+ a+2d+ a+3d=4a+6d=462 => 2a +3d=231--->1

It is also given that the total number of students from Class I to V is twice the total number of students from Class VI to X.

So, 5a+10d = 2*(5a+35d).

5a+10d=10a+70d

5a+60d=0 --> 2

Solving equations 1 and 2.

Multiply equation 1 with 20.

We get 40a+60d=4620--->3. Subtract equation 2 from equation 3.

We get 35a=4620

=> a=132.

Substitute a value in equation 2. We get 60d=-660=>d=-11.

Now, we have to calculate the number of students in class VI.

That is a+5d=132-5*11=77.

Let's assume the initial speed of the flight is F. And the time taken is t.

Given the distance to be travelled, 11200.

We know distance = speed *time.

So, 11200=Ft-->1

Due to bad weather, the flight started with a three-hour delay. Given that the flight speed increased by 100kmph, the delay would be reduced to one hour. So, it is nothing but the travel time is reduced by 2 hours.

In that case, the equation will become 11200=(F+100)*(t-2)-->2

Equation 1 can be re-written as t=$$\frac{11200}{F}$$-->3

Now Equation 2: $$\frac{11200}{F+100}$$=t-2

$$\Rightarrow$$ = $$\frac{11200}{F+100}$$+2=t

$$\frac{11200+2\left(F+100\right)}{F+100}=t$$

$$t=\frac{11400+2F}{F+100}$$-->4

Divide equation 3 by equation 4:'

$$\frac{t}{t}=\frac{\frac{11200}{F}}{\frac{11400+2F}{F+100}}$$

$$\Rightarrow$$ $$\frac{\left(11400+2F\right)}{F+100}=\frac{11200}{F}$$

$$\left(11400+2F\right)\cdot F=\left(F+100\right)\cdot11200$$

$$\left(11400F+2F^2\right)=\left(11200F+1120000\right)$$

$$\ 2F^2+200F-1120000=0\ \Rightarrow\ F^2+100F-560000=0$$

So, F=700 or -800. Speed can't be negative. So, the speed is 700km/hr.

Now, if the speed is increased by 350, the new speed will be 1050.

Hence, the time taken will be $$\frac{11200}{1050}=\frac{32}{3}=10\ \frac{2}{3\ }hrs=10hrs\ 40\min$$

The flight has to start at 6:30 am, but it is delayed by 3 hrs. So, the actual time is 9:30 am.

10hrs: 40mins from that will be 8:10 pm

Let's reduce the RHS first :-$$2 \log_{25}(2x - 4)$$ = $$\frac{2}{2}\log_5(2x-4)$$ = $$ \log_{5}(2x - 4)$$

Here we used the formulae $$\log_{a^n}b^m=\frac{m}{n}\log_ab$$

Now LHS=RHS

$$\log_5(x - 2) =  \log_{5}(2x - 4)$$

The bases are equal.

So, x-2=2x-4

x=2, but 2 is not in the domain as it will lead to log 0. Hence, there are no solutions for this.

The cost of running all three shows a day is 10,000+3*5,000=25,000.

The has to be 80% occupancy of total seats. In total there are 200*3 shows=600 seats.

Out of that 80% is $$\frac{80}{100}\cdot600=420$$

Since we need to maximize the profit. We shall take the vacant 120 seat in show 3 which cost less.

The income from first show 200 * 250= 50,000

The income from second show 200 * 250= 50,000

The income form third show 80*200= 16,000

The total income is 1,16,000-25,000=91,000

Given that the plate costs 120, they are selling it for 160, and the number of buyers is 300; the current profit is 40 per customer.

It is crucial to note that any increase of price by 10 will decrease the number of customers by 10, and conversely, any decrease of price will increase the number of buyers by 10.

Let's say the maximum profit is gained by increasing the profit by x times 10. In that case, the number of customers will decrease by x times 10.

The profit will be (40+10x)(300-10x); this has to be maximum.

Let's expand this expression.

(40+10x)(300-10x)=12000+3000x-400x-$$100x^2$$

12000+2600x-$$100x^2$$

This has to be maximum.

If we multiply it by -1, it has to be minimum.

$$100x^2$$-12000-2600x=$$100x^2$$-2*130*10x+$$130^2$$= $$(10x-130)^2$$-12000-$$130^2$$

$$(10x-130)^2$$-28900

Before, we multiplied the expression by -1 to revert that again, multiply by -1.

The new expression is 28900 - $$(10x-130)^2$$. The maximum value of this expression is 28900.

The line AC is tangent to the circle. So, the line DE is perpendicular to AC and is radius to the circle.

Look at the $$\triangle\ $$ADC and  $$\triangle\ $$DEC. $$\angle\ $$DAC = $$\angle\ $$EDC, $$\angle\ $$C=$$\angle\ $$C. And both D and E are right angle triangles.

So, both the triangles are similar.

$$\triangle\ $$ ADC is a right-angled triangle.

$$AC^2$$=$$AD^2$$+$$DC^2$$

$$AC^2$$=$$200^2+150^2=250^2$$

AC=250, AD=200, DE=r, DC=150 

$$\frac{AD}{DE}=\frac{AC}{DC}$$

$$\frac{200}{r}=\frac{250}{150}$$

$$\Rightarrow$$ r=120.

Given half pond is inside the triangle, so he has to only buy half the area of the circle.

Cost for buying area of semi circle =$$\pi\ \frac{r^2}{2}$$*1400=$$\pi\ \frac{120^2}{2}$$*1400= 3,16,80,000

The question states that the path forms a right-angle triangle at Rivu. So, let's consider the distance from Raju to Rivu as x and that from Rivu to Ratan as y.

i)

Now, $$\triangle\ ABC$$ form a right-angled triangle. Using Pythagoras theorem

$$AB^2+BC^2=AC^2$$

=>$$x^2+y^2=80^2=6400$$ --->1

The area of $$\triangle\ ABC$$ is $$\frac{1}{2}\cdot b\cdot h$$=$$\frac{1}{2}\cdot x\cdot y$$ ---->2

If we consider the base of the triangle as AC of length 80(side of rectangle) and height as altitude from B of length 25 (half the breadth of rectangle), we can get the area as $$\frac{1}{2}\cdot b\cdot h$$=$$\frac{1}{2}\cdot 80\cdot 25=1000$$.

Now, use this in equation 2.

$$\frac{1}{2}\cdot x\cdot y$$=1000

xy=2000--->3

$$\left(x+y\right)^2=x^2+y^2+2xy$$=6400+4000=10400

x+y=20$$\sqrt{\ 26}$$

This is the distance the ball travels at 10m/s. The time taken is 2$$\sqrt{\ 26}$$+5 sec for holding.

So, 1 alone is sufficient.

ii) Given the area of the triangle is 1000

$$\frac{1}{2}\cdot x\cdot y$$=1000

=> xy=2000

With this alone, we can't find the travel time.

Let's say the unknown number is x—the LCM of x, 990= 6930.

LCM is the greatest power of a prime number in those two numbers.

Now let's prime factorize 990 and 6930.

990=2*5*11*$$3^2$$

6930=2*$$3^2$$*5*7*11

The number 990 has all the powers of prime numbers, which are also present in 6930, except 7. So, x definitely has to have 7.

Then, it is given that the GCD of x and 550 is 110. So, x has to be a factor of 110. 110 is not a multiple of 7. So, we have to multiply the 110 by 7. That is 770, which is the minimum value of x.

The sum of digits of x is 7+7+0=14

First, we have to find the roots of the cubical equation.

One root has to be figured out using the trial and error method.

Let's try 1.

$$P(x) = 2x^3 - 11x^2 + 17x - 6$$

$$P(1) = 2*1^3 - 11*1^2 + 17 - 6=1$$

So, 1 is not a root.

Let's try 2,

$$P(2) = 2*2^3 - 11*2^2 + 17*2 - 6=0$$

Hence, two is one of the roots of this equation.

Now, using the synthetic division method cubical equation can be written as (x-2)($$2x^2-7x+3$$)=0

The roots of $$2x^2-7x+3$$ are 1/2, 3.

The area of circles with radii 3, 2, 1/2 are $$9\pi $$, $$4\pi$$, $$\frac{\pi}{4}$$.

Their ratio od areas is 9:4:$$\frac{1}{4}$$

$$\Rightarrow$$ 36:16:1 is the ratio.