XAT 2024 - QA

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.

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

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

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.

The number of item combinations Rohit and his friends could have ordered veg items were $$^{16}C_2$$
And the number of ways they could have ordered non-veg items would be $$^9C_3$$

The total number of item combinations that Rohit and his friends could choose from would then be $$^{16}C_2\times\ ^9C_3$$

Let's denote the number of veg gluten-free items as $$V\sim G$$ and the number of non-veg gluten-free items as $$NV\sim G$$

The total number of ways Bela and her friends could have chosen the items from the menu would then be 
$$NV\sim G\times\ ^{N\sim G}C_2$$

We are given that Rohit and his friend's number of combinations were twelve times this value, giving us the relation

$$^{16}C_2\times\ ^9C_3=12\times\ NV\sim G\times\ ^{N\sim G}C_2$$
$$^{16}C_2\times\ ^9C_3=12\times\ NV\sim G\times\ \frac{\left(N\sim G\right)\left(N\sim G\ -1\right)}{2}$$
$$2^4\times\ 3\times\ 5\times\ 7=\ NV\sim G\times\ \left(N\sim G\right)\left(N\sim G\ -1\right)$$

At this point, we need two consecutive numbers on the left-hand side to account for the $$\left(N\sim G\right)\left(N\sim G\ -1\right)$$ on the right-hand side. 

We know that $$NV\sim G$$ must be 9 or less; trying to put this value as values from 9 to 1 and get two consecutive digits from the remaining numbers, 

The value of $$NV\sim G$$ can be 7, giving the value of $$\left(N\sim G\right)\left(N\sim G\ -1\right)$$ as 16 x 15
OR
The value of $$NV\sim G$$ can be 8, giving the value of $$\left(N\sim G\right)\left(N\sim G\ -1\right)$$ as 15 x 14

In either of these cases, the number of items on the menu not having gluten would be 7+16 = 8+15 = 23
This means that 2 of the items on the menu will not have gluten in them. 

Therefore, Option B is the correct answer. 

In order for the line two be coincident, the ratio of the coefficients of the variables and the constants from both the equations must be the same. 

That is, $$\frac{3}{q}=\frac{21}{r}=\frac{p}{-7}$$

Let's consider each option individually.
Checking through the easier options first.

Option A: p and q must have opposite signs
In order for  $$\frac{3}{q}=\frac{p}{-7}$$ to be true, p and q must be of opposite signs as pq = -21
Hence, this option does not contradict the lines coinciding. 

Option D: r and q must have the same signs

Similar to option A, for $$\frac{3}{q}=\frac{21}{r}$$ to hold true, q and r must be of the same signs. 
Hence, this option too does not contradict the lines coinciding. 

Option E: p cannot be zero
In order for q and r to have real values, the fractions $$\frac{3}{q}=\frac{21}{r}=\frac{p}{-7}$$ must not be zero. 
Putting p=0 would give the value of p/(-7) as 0, which would then not give real values of q and r
Hence, p can never be zero. This statement, too, does not contradict the two lines being coincident. 

Now that A, D, and E are not our answers, we need to consider the more complex options. 

Option B: r is the smallest amongst p, q, r
We would want $$\frac{3}{q}=\frac{21}{r}$$ to hold true.
If we take q and r to be positive, we must take p to be negative, in that case p would be negative and the smallest. So, in order to consider the possibility of r being the smallest, we must take q and r as negative values.
Now, with q and r negative, a smaller number would have a bigger magnitude or a bigger absolute value.
We can try putting in values at this point to see if this would contradict the lines coinciding.
taking r as -21 and q as -3, we can take p as 7
These values would make the lines coincident, with r being the smallest value.
Hence, this options too does not necessarily contradcit the lines coinciding.

Option C: q is the largest amongst p, q, r
Follwoing the same logic as Optoin B, we cannot take q and r to be negative, as that would make p positive, making p the largest value.
Therefore, in order to consider the possibility of q being the largest, we must take positive values of q and r
Now, comparing $$\frac{3}{q}=\frac{21}{r}$$, if q is larger than r, then the numerator of the first term would be smaller than the numerator of the second term, and the denominator of the first term would be larger than the denominator of the second term
Making the fraction $$\frac{3}{q}$$ strictly smaller than $$\frac{21}{r}$$
If this option is true, the lines can never coincide, as the ratios of the coefficient can never be equal.

Therefore, Option C would be the correct answer. 

abbb can be written as (1000*a+100*b+10*b+1*b) = 1000a+111b

Now, we need to check to the necessary and sufficient condition for which (1000a+111b) is divisible by a.

We know that 1000a is always divisible by 'a', hence, we need to check for which condition, 111b is always divisible by a.

111b can be written as (3*37*b) => (3b*37) must be divisible by a.

a can't be a factor of 37, which implies 'a' is a factor of 3b => 3b is divisible by 'a'

The correct option is E

It is given that ABC is a right-angled triangle, where $$\angle\ B\ =\ 90^{\circ\ }$$

Now, we can use Pythagoras theorem to calculate the value of AC, which is equal to $$\sqrt{\ 18^2+24^2}=\sqrt{\ 324+576}=\sqrt{\ 900}\ =\ 30$$ cm.

Here, we have drawn a common tangent MN, which is perpendicular to BC => Triangle MNC is a right-angled triangle.

It is also be concluded that traingle MNC is similar to traingle ABC

Hence, the ratio of the sides of triangle MNC is the same as the ratio of the sides of triangle ABC.

Thus, AB: BC: AC = MN: NC: MC = 18:24:30 = 3: 4: 5

Let the length of side MN be 3x => the length of NC = 4x, and the length of MC = 5x.

In triangle MNC, the circle inside it is the in circle of triangle MNC.

Thus, the in radius (r) of the circle in triangle MNC = (3x+4x-5x)/2 = x.

We know that BC = BN+NC, where BN is the diameter of the first circle, which is equal to 2r = 2x (since, r = x), and NC = 4x

=> 24 = 2x+4x 

=> 6x = 24 => x = 4 cm.

Hence, the radius of the circle (x = r) is 4 cm.

The correct option is B

Let the total number of jewels be X. 
The jewels in the first box would be X/3
The jewels in the second box would be kX/5
And the number of jewels in the third box is 66

The jewels of the second and third boxes should add up to 2X/3 jewels, giving us the relation. 
$$\frac{kX}{5}+66=\frac{2X}{3}$$
$$66=\frac{X\left(10-3k\right)}{15}$$
$$X\left(10-3k\right)=2\times\ 3^2\times\ 5\times\ 11$$

We are given that k is a positive integer. 
Taking k= 1, we get (10 - 3k) to be 7, which is not present on the right-hand side. 
Taking k = 2, we get (10 - 3k) to be 4, which is again not present on the right-hand side. 
Taking k = 3, we get (10 - 3k) to be 1, which is possible on the right-hand side and would give the value fo X to be 990. 

Taking any further value of k would give a negative value of (10 - 3k), which would not be possible. 

Therefore, the king must have had 990 jewels. 

Hence, Option A is the correct answer.  

Let the price of one kg of potato be P, one kg of cabbage be C, and one kg of onion be O. 
we are given the equations, 
$$5C+4P=80$$    (i)
$$4C+5O=93$$   (ii)

In equation (ii), we get 3 as the unit digit on the right-hand side. This is only possible with 8 + 5 on the left-hand side. 
So, the unit place value of C must be 2, which means that C can have values 2, 12, 22, 32, and so on. The value of O must be an odd number. 

Using this information in equation (i), we can see that the value of 5C will always have 0 as its unit digit. Meaning that the value of 4P must also have its unit digit as zero. 
This would tell that P is a multiple of 5. 
4P can then only be 20, 40 or 60 for C to also have a positive integer value. 

If P is 5, C would be 12, giving the value of O to be 9
If P is 10, C would be 8, and this is consistent with our criteria that the unit digit of C must be 2
If P is 15, C would be 4, which is again not consistent with the same criteria. 

Therefore, the price of one kg of potato would be Rs. 5, one of of cabbage would be Rs. 12 and one kg of onion would be Rs. 9

Aman can buy a maximum of 11 kg of onions for Rs. 99, leaving him with Rs. 1

Therefore, Option E is the correct answer. 

Let the price of one kg of potato be P, one kg of cabbage be C, and one kg of onion be O.
we are given the equations,
$$5C+4P=80$$ (i)
$$4C+5O=93$$ (ii)

In equation (ii), we get 3 as the unit digit on the right-hand side. This is only possible with 8 + 5 on the left-hand side.
So, the unit place value of C must be 2, which means that C can have values 2, 12, 22, 32, and so on. The value of O must be an odd number.

Using this information in equation (i), we can see that the value of 5C will always have 0 as its unit digit. Meaning that the value of 4P must also have its unit digit as zero.
This would tell that P is a multiple of 5.
4P can then only be 20, 40 or 60 for C to also have a positive integer value.

If P is 5, C would be 12, giving the value of O to be 9
If P is 10, C would be 8, and this is consistent with our criteria that the unit digit of C must be 2
If P is 15, C would be 4, which is again not consistent with the same criteria.

Therefore, the price of one kg of potato would be Rs. 5, one of cabbage would be Rs. 12, and one kg of onion would be Rs. 9

The question says that Aman buys only potatoes and onions. The money left after purchasing these must be less than Rs. 5 so that he cannot buy an additional kg of potatoes. 
Now, we can buy a maximum of 10 kg of onions, which would cost us Rs. 90, forcing Aman to buy 2 kg of potatoes to meet the above conditions. 

So, the number of kg of opinion bought can vary from 1 to 10, giving us a total of 10 possible scenarios. We need to check where the number of potatoes bought in kg outweighs the onions bought. 

We see that when Aman buys 7 kg of onions, he also has to buy 7 kg of potatoes. So only when he buys 10, 9 and 8 kg of onions is the weight of onions bought more than the weight of the potatoes bought. 

So, the probability that more onions are bought than potatoes would be $$\frac{3}{10}$$

Therefore, Option A is the correct answer.    

Here, we need to represent all the given data in a three sets venn-diagram that is Data Analytics (A), Database Handling (H), and Coding (C).

If we write all the data that is given in the table, we will get the following diagram:

Now we need to infer the characteristics of x, which can be concluded from the given information.

We know that $$2-x\ge\ 0$$, hence the possible values of x is 0, 1, 2

It is also known that $$x-2\ge\ 0$$, which implies the possible values of x is 2, 3, 4, ....

Now, from both equation, the only common value of x is 2.

Hence, x can take only 2 as the value.

The correct option is B

The first clue is the minimum value provided:

That is not one of the eleven values that we already have; this must mean that one of the missing values, A or B, must be 2

Since we are given that A is less than B, it must be A, which is equal to 2. 

Now, writing down the twelve known numbers, 
2, 5, 6, 7, 8, 12, 16, 19, 21, 21, 27, 29

we are also provided that the median is of the data set was 12, that would be possible if the value 12 was shifter from 6th position to 7th position.
This must mean that the other missing value would also be in the first 6 values of the data set, lying in the range $$2\le B\le12$$

Now, narrowing down our focus to the lower hinge, the value is given to be 6.5
The first six values would be 2, 5, 6, 7, 8 and one unknown value

The lower hinge would be 6.5 only if the missing value appears after 7
Narrowing down the possible values of B to be 7, 8, 9, 10, 11, 12

For the first question, of all the given options, only option C lies in this range and hence would be the answer. 

Therefore, option C is the correct answer. 

The first clue is the minimum value provided:

That is not one of the eleven values that we already have; this must mean that one of the missing values, A or B, must be 2

Since we are given that A is less than B, it must be A, which is equal to 2.

Now, writing down the twelve known numbers,
2, 5, 6, 7, 8, 12, 16, 19, 21, 21, 27, 29

we are also provided that the median of the data set was 12, which would be possible if the value 12 was shifted from the 6th position to the 7th position.
This must mean that the other missing value would also be in the first 6 values of the data set, lying in the range $$2\le B\le12$$

Now, narrowing down our focus to the lower hinge, the value is given to be 6.5
The first six values would be 2, 5, 6, 7, 8 and one unknown value

The lower hinge would be 6.5 only if the missing value appears after 7
Narrowing down the possible values of B to be 7, 8, 9, 10, 11, 12

In order to find the average work-experience of all the teachers, we must add up all the values 
adding up the known values we get
2+5+6+7+8+12+16+19+21+21+27+29 = 173

Looking at the options, the total sum of all 13 values must be 
A: 156
B: 162.5
C: 169
D:  175.5
E: 182

The sum of 13 values can only be 182 of all the given options, which would be the case when B is 9

Therefore, option E is the correct answer. 

That is not one of the eleven values that we already have; this must mean that one of the missing values, A or B, must be 2

Since we are given that A is less than B, it must be A, which is equal to 2.

Now, writing down the twelve known numbers,
2, 5, 6, 7, 8, 12, 16, 19, 21, 21, 27, 29

we are also provided that the median of the data set was 12, which would be possible if the value 12 was shifted from the 6th position to the 7th position.
This must mean that the other missing value would also be in the first 6 values of the data set, lying in the range $$2\le B\le12$$

Now, narrowing down our focus to the lower hinge, the value is given to be 6.5
The first six values would be 2, 5, 6, 7, 8 and one unknown value

The lower hinge would be 6.5 only if the missing value appears after 7
Narrowing down the possible values of B to be 7, 8, 9, 10, 11, 12

In the context of the third question, we are given that the average of the 13 values was 15, meaning that the sum of the 13 values was 195
As we calculated previously, the sum with the value we had would be 173 +B

we also have to add the mistaken value once more to this number to get 195
Let's take the unknown number to be U, giving us the equation
173 + B+ U = 195
B+U = 22

Where B must be one of 7, 8, 9, 10, 11 or 12
And U must be one of 5, 6, 7, 8, 12, 16, 19, 21, 21, 27, 29

trying to substitute the values of B into the equation, the only value which yields valid values for both B and U would be 10

Therefore, Option C is the correct answer. 

It is given that the runs in any over is an integer. So, the run rate multiplied by overs should give us an integer.

1 ≤ N - 2 < N + 6 ≤ 20

The range of N is [3, 14].

After N overs, the run rate is given as 7.43. So, 7.43*N has to be an integer. Let's try putting all the possible values N and see where we get an integer.

3*7.43=22.29

4*7.43=29.72

5*7.43=36.15

6*7.43=44.58

7*7.43=52.01

Here, we got an integer; since the run rate is reduced to 2 decimal points, We can not get an exact integer.

So, N can be 7. Let's also check the remaining possible values of N.

8*7.43=59.44

9*7.43=66.87

10*7.43=74.3

11*7.43=81.73

12*7.43= 89.16

13*7.43=96.59

14*7.43=102.02

N can also be 14.

If N=14, the (N+2)*8.11=16*8.11 should also be an integer. But it is 129.76. So, N is not 14. The only possible value is 7.

N-2 --- 8.0 = 5*8=40

N --- 7.43 = 52.01

N+2 --- 8.11 = 9*8.11=72.99

N+4--- 8.45 = 11*8.45 = 92.95

N+6 --- 8.08 = 13*8.08= 105.04

Every number is almost an integer, so N will be 7.

Now, let's calculate the runs scored in each of these overs.

After five overs, the score is 40

After seven overs, the score is 52

So, the sum of scores in 6 and 7 overs is 12. They have to score a minimum of 6 in an over. So, 6 is the sixth over, and 6 in the seventh over.

After the 9th over, the score was 73. They scored 21 in the 8th and 9th over.

After the 11th over, the score was 93, they scored 20 in the 10th and 11th.

After the 13th over, the score was 105, they scored 12 in the 12th and 13th over.

The lowest number of runs is scored in 6, 7, 12, and 13 overs. So, 7 is the correct answer.

It is given that the runs in any over is an integer. So, the run rate multiplied by overs should give us an integer.

1 ≤ N - 2 < N + 6 ≤ 20

The range of N is [3, 14].

After N overs, the run rate is given as 7.43. So, 7.43*N has to be an integer. Let's try putting all the possible values N and see where we get an integer.

3*7.43=22.29

4*7.43=29.72

5*7.43=36.15

6*7.43=44.58

7*7.43=52.01

Here, we got an integer; since the run rate is reduced to 2 decimal points, We can not get an exact integer.

So, N can be 7. Let's also check the remaining possible values of N.

8*7.43=59.44

9*7.43=66.87

10*7.43=74.3

11*7.43=81.73

12*7.43= 89.16

13*7.43=96.59

14*7.43=102.02

N can also be 14.

If N=14, the (N+2)*8.11=16*8.11 should also be an integer. But it is 129.76. So, N is not 14. The only possible value is 7.

N-2 --- 8.0 = 5*8=40

N --- 7.43 = 52.01

N+2 --- 8.11 = 9*8.11=72.99

N+4--- 8.45 = 11*8.45 = 92.95

N+6 --- 8.08 = 13*8.08= 105.04

Every number is almost an integer, so N will be 7.

Now, let's calculate the runs scored in each of these overs.

After five overs, the score is 40

After seven overs, the score is 52

So, the sum of scores in 6 and 7 overs is 12. They have to score a minimum of 6 in an over. So, 6 is the sixth over, and 6 in the seventh over.

After the 9th over, the score was 73. They scored 21 in the 8th and 9th over.

After the 11th over, the score was 93, they scored 20 in the 10th and 11th.

After the 13th over, the score was 105, they scored 12 in the 12th and 13th over.

We know the runs scored in every pair of options except Option D. So, only option D could be the answer.

It is given that the runs in any over is an integer. So, the run rate multiplied by overs should give us an integer.

1 ≤ N - 2 < N + 6 ≤ 20

The range of N is [3, 14].

After N overs, the run rate is given as 7.43. So, 7.43*N has to be an integer. Let's try putting all the possible values N and see where we get an integer.

3*7.43=22.29

4*7.43=29.72

5*7.43=36.15

6*7.43=44.58

7*7.43=52.01

Here, we got an integer; since the run rate is reduced to 2 decimal points, We can not get an exact integer.

So, N can be 7. Let's also check the remaining possible values of N.

8*7.43=59.44

9*7.43=66.87

10*7.43=74.3

11*7.43=81.73

12*7.43= 89.16

13*7.43=96.59

14*7.43=102.02

N can also be 14.

If N=14, the (N+2)*8.11=16*8.11 should also be an integer. But it is 129.76. So, N is not 14. The only possible value is 7.

N-2 --- 8.0 = 5*8=40

N --- 7.43 = 52.01

N+2 --- 8.11 = 9*8.11=72.99

N+4--- 8.45 = 11*8.45 = 92.95

N+6 --- 8.08 = 13*8.08= 105.04

Every number is almost an integer, so N will be 7.

Now, let's calculate the runs scored in each of these overs.

After five overs, the score is 40

After seven overs, the score is 52

So, the sum of scores in 6 and 7 overs is 12. They have to score a minimum of 6 in an over. So, 6 is the sixth over, and 6 in the seventh over.

After the 9th over, the score was 73. They scored 21 in the 8th and 9th over.

After the 11th over, the score was 93, they scored 20 in the 10th and 11th.

After the 13th over, the score was 105, they scored 12 in the 12th and 13th over.

The lowest number of runs is scored in 6, 7, 12, and 13 overs. So, 7 is the correct answer.