The case where Amit gets the least amount of marks looks like…

The maximum mark that Amit can get is 31, He can get it if he can grab 20 marks from the people who score 48, 49 and 50.
In fact, it is possible for Amit to do that…

The difference between the highest and lowest marks if Amit = 31 - 11 = 20.

The rules of the examination are…
● A correct answer gets +3 marks.
● A wrong answer gets -1 mark.
● An unanswered question gets +1 mark.
So, the scenario is, every question is already awarded 1 mark before attempting, if the attempt is right you get +2 marks and if it is wrong you get -2 marks.
So after re-imagining the examination…
● Every Question is awarded + 1 before attempting itself.
(This means you enter the examination with 75 marks in your pocket)
● A right answer fetches +2 marks.
● A wrong answer fetches -2 marks.
Now it is up to the student to increase or decrease his total marks from 75.
If he answers more questions right than wrong he gets additional marks.
If he answers more questions wrong than right his marks decrease.
So, when Rayan scored a total of 97 marks in the examination, 75 were given to him on a platter. The remaining 22 is what he put effort to score.
This means the difference between the number of questions he got right and the number of questions he got wrong is 11.
In the most extreme case, he might have got 11 questions right and did not attempt the remaining.

He might have got a few questions wrong, but the number of right questions should always be more than the number of wrong questions by 11.
The case where the number of wrong questions is ‘x’ looks like…

The number of unattempted questions was higher than the number of attempted questions…
This means,
 64 - 2x > 11 + x + x
53 > 4x
x≤13x≤13.
The maximum value of x is 13.
The maximum number of the right answers = 11 + 13 = 24

The number of 6-digit integers that can be formed is 600. 

The number of 5-digit integers that can be formed is 600. 

The number of 4-digit integers that can be formed is 240. 
The total number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5,
using each digit at most once, is 600 + 600 + 240 = 1440 

One candidate got 30% of the polled votes, the remaining three got in the ratio of 1 : 2 : 3
The polled votes were split in 3 : 7 ratio.
The 70% of them were again split in the ratio 1 : 2 : 3
The votes were polled in the ratio of 6(3) : 7(1 : 2 : 3)
18 : 7 : 14 : 21
Let’s assume that the actual polled votes are 18x, 7x, 14x, 21x
The winner of the election received 2512 votes more than the candidate with the second highest votes.
21x - 18x = 2512
3x = 2512
Total polled votes = 18x + 7x + 14x + 21x = 60x = 20(3x) = 20(2512) = 50,240
The polled votes represent 80% of the total registered votes.
Total registered votes = 50,240 + 12,560 = 62,800.

a₁ + a₂ + a₃ + … + a_n = 300(n)
6a₁ + a₂ + a₃ + … + a_n = 400(n)
5a₁ = 100(n)
a₁ = 20(n)
The constraints of the problem are:
● Sum of n terms = 300(n)
● The first term is 20(n)
● All other terms (terms other than the first term) should be greater than or equal to the first term. (Because the sequence is a non-decreasing sequence.)
Image a scenario where all the other terms are equal to the first term. That is a case where all the terms are equal. Since the average is 300. All of them should be 300. The first term is 300.
20(n) = 300
n = 15
Now imagine that the sequence has 16 terms the first term will be 320, and all the other terms will be greater than or equal to 320. So the average can’t be 300.
Or basically, the maximum number of terms in the sequence is 15.
What is the minimum number of terms in the sequence?
Can the sequence have one term?
n = 1
a₁ = 20
But the sum of terms is not 300.
The sequence should have more than one term.
n = 2
a₁ = 2(20) = 40
The sum of terms can be 300(2). This happens by having the second term as 560.
The minimum number of terms in the sequence is 2.
The maximum number of terms in the sequence is 15.
The number of terms in the sequence can have 14 different values.
And in each case the value of a₁ is distinct.
Hence, a₁ can have 14 different values.

Let the half volume of each container be 80 cc.
Why 80cc though? Why not some other value?
We know that the process of shifting half volumes is happening three times overall.
(12)3=18(12)3=18.
So it would be wise to assume the initial volume to be some multiple of 8 so that we don’t have to deal with fractions later on.


The ratio of Sugar Syrup to Water in the second container is 5 : 6.

On average, Manu targets saving ₹550 per month.
In the first 9 months, Manu earns ₹4000 per month and spends ₹3500.
So, he ends up saving only ₹450 per month. In other words, he misses his target by ₹50 per month.
So he saves ₹50 ×× 9 = ₹450 less than his target.
In the next 3 months of the year, his expenses are ₹3700 per month. To save ₹550 per month in these three months, his income should be ₹3700 + ₹550 = ₹4250 per month.
But also, he should earn some more to compensate for the ₹450 he was short of in the first 9 months. This ₹450 is earned over a span of 3 months. Or he should earn ₹150 extra each month.
So Manu’s income in the last 3 months = ₹4250 + ₹150 = ₹4400

a + 2b = 6 
a + b = 6 - b 
Clearly, the maximum/minimum value of (a + b) depends on the value of b. 
Since a and b are positive real numbers, 
The minimum value that b can take is 0. 
The maximum value that b can take is when a is 0. 
0 + 2b = 6 
b = 3. 
When b = 0; a + b = 6 - b = 6 
When b = 3; a + b = 6 - 3 = 3 
Therefore, the minimum and maximum values of (a + b) are 3 and 6 respectively. 
The average of these extreme values is ( 3 + 6)/2 

Time taken by Anu, Tanu and Manu to complete a job is in the ratio 5 : 8 : 10 
This means that their efficiencies are in the ratio 15 : 18 : 110 
Or, their efficiencies are in the ratios 8 : 5 : 4 
This means, for instance, if Anu can wash 8 plates in an hour, Tanu and Manu can wash 5 & 4
plates respectively in one hour. 
So, let us remodel the entire question over this imaginary scenario of washing plates… 
All three of them finish the job in 4 days working 8 hours per day. 
Anu, Tanu and Manu can wash 8, 5 and 4 plates respectively in an hour. 
So the total number of plates = (8+5+4)×4×8=17×4×8 
Anu and Tanu work together for 6 days, working 6 hours 40 minutes per day. 
The total number of plates washed by them = (8+5)×6×623=13×40 
So the remaining plates to be washed =17×4×8−13×40 
=4(17×8−130) 
=4(80+56−130) 
=4×6 
So, the question is, in how many hours can Manu wash 4×6 plates? 
We know that Manu can wash 4 plates in one hour. 
Therefore, he requires 6 hours to wash 4×6 plates. 

The answer is 'Company B'

Choice A is the correct answer.

It is given,
Company A:
Revenue = 240 and cost incurred = 180
Profit = 240 - 180 = 60
Company B:
Revenue = 220 and cost incurred = 145
Profit = 220 - 145 = 75
Company C:
Revenue = 195 and cost incurred = 110
Profit = 195 - 110 = 85
Company D: Revenue = 140 and cost incurred = 160
No profit.
Company C had the highest annual profit.

The question is " Considering all three years, which company had the highest annual profit? "

Hence, the answer is 'Company C'

Choice A is the correct answer.

For all the companies in all three years, cost incurred is less than Revenue except for D in 2020.
Revenue is 20 and cost incurred is 50
Company D experienced the highest annual loss in 2020

The question is " Which of the four companies experienced the highest annual loss in any of the years? "

Hence, the answer is 'Company D'

Choice A is the correct answer.

The question is " The ratio of a company's annual profit to its annual costs is a measure of its performance. Which of the four companies had the lowest value of this ratio in 2019? "

Hence, the answer is 'Company A'

Choice A is the correct answer.

Company A:
The number of employees in the beginning of 2019 = 150
The number of employees hired in 2019 = 20
The number of employees should be at the beginning of 2020 is 150+20, i.e. 170 but there are 140 only. This
implies 30 left company A in 2019.
The number of employees in the beginning of 2020 = 140
The number of employees hired in 2020 = 35
The number of employees should be at the beginning of 2021 is 140+35, i.e. 175 but there are 150 only. This
implies 25 left company A in 2020.
The number of employees left company A in 2019 and 2020 = 30 + 25 = 55
Company B:
Similarly, the number of employees left company B in 2019 = 210 + 35 - 240 = 5
The number of employees left company B in 2020 = 240 + 45 - 250 = 35
The number of employees left company B in 2019 and 2020 = 5 + 35 = 40
Company C:
Similarly, the number of employees left company C in 2019 = 320 + 45 - 320 = 45
The number of employees left company C in 2020 = 320 + 40 - 320 = 40
The number of employees left company C in 2019 and 2020 = 45 + 40 = 85
Company D:
Similarly, the number of employees left company D in 2019 = 400 + 30 - 410 = 20
The number of employees left company D in 2020 = 410 + 35 - 400 = 45
The number of employees left company D in 2019 and 2020 = 20 + 45 = 65
The total number of employees lost in 2019 and 2020 is least for company B.

The question is " If the last location visited is Ahmednagar, then what is the total distance covered in the route (in km)? "

Hence, the answer is '35'

The question is " If the total number of widgets delivered in a day is 250 units, then what is the total distance covered in the route (in km)? "

Hence, the answer is '38'

The question is " What is the chance that the total number of widgets delivered in a day is 260 units and the route ends at Bikrampore? "

Hence, the answer is '7.56%'

Choice C is the correct answer.

The question is " If the first location visited from the warehouse is Ahmednagar, then what is the chance that the total distance covered in the route is 40 km? "

Hence, the answer is '18%'

Choice D is the correct answer.

The question is " If Ahmednagar is not the first location to be visited in a route and the total route distance is 29 km, then which of the following is a possible number of widgets delivered on that day? "

Hence, the answer is '210'

Choice A is the correct answer.

1. Over the two days, all three of them met the same total number of households, and each of them sold a total of 100 items.





2. On both days, Lahur met the same number of households and sold the same number of items.



3. Hokli could not sell any item on the second day because the first household he met on that day complained against him.



4. Tohri met 30 more households on the second day than on the first day.



5. Tohri's success rate was twice that of Lahur's on the first day, and it was 75% of Lahur's on the second day.



yx−15yx−15 = 50x50x*2
100−yx+15100−yx+15 = 50x50x*3434
xy = 100(x–15).
4x(100–y) = 150(x+15)
400x – 4xy = 150(x+15)
400x – 4*100(x–15) = 150(x+15) 400 * 15 = 150 (x+15). x = 25



y10y10 = 50255025*2
y = 40

1. Over the two days, all three of them met the same total number of households, and each of them sold a total of 100 items.





2. On both days, Lahur met the same number of households and sold the same number of items.



3. Hokli could not sell any item on the second day because the first household he met on that day complained against him.



4. Tohri met 30 more households on the second day than on the first day.



5. Tohri's success rate was twice that of Lahur's on the first day, and it was 75% of Lahur's on the second day.



yx−15yx−15 = 50x50x*2
100−yx+15100−yx+15 = 50x50x*3434
xy = 100(x–15).
4x(100–y) = 150(x+15)
400x – 4xy = 150(x+15)
400x – 4*100(x–15) = 150(x+15) 400 * 15 = 150 (x+15). x = 25



y10y10 = 50255025*2
y = 40

1. Over the two days, all three of them met the same total number of households, and each of them sold a total of 100 items.





2. On both days, Lahur met the same number of households and sold the same number of items.



3. Hokli could not sell any item on the second day because the first household he met on that day complained against him.



4. Tohri met 30 more households on the second day than on the first day.



5. Tohri's success rate was twice that of Lahur's on the first day, and it was 75% of Lahur's on the second day.



yx−15yx−15 = 50x50x*2
100−yx+15100−yx+15 = 50x50x*3434
xy = 100(x–15).
4x(100–y) = 150(x+15)
400x – 4xy = 150(x+15)
400x – 4*100(x–15) = 150(x+15) 400 * 15 = 150 (x+15). x = 25



y10y10 = 50255025*2
y = 40

1. Over the two days, all three of them met the same total number of households, and each of them sold a total of 100 items.





2. On both days, Lahur met the same number of households and sold the same number of items.



3. Hokli could not sell any item on the second day because the first household he met on that day complained against him.



4. Tohri met 30 more households on the second day than on the first day.



5. Tohri's success rate was twice that of Lahur's on the first day, and it was 75% of Lahur's on the second day.



yx−15yx−15 = 50x50x*2
100−yx+15100−yx+15 = 50x50x*3434
xy = 100(x–15).
4x(100–y) = 150(x+15)
400x – 4xy = 150(x+15)
400x – 4*100(x–15) = 150(x+15) 400 * 15 = 150 (x+15). x = 25



y10y10 = 50255025*2
y = 40

1. Over the two days, all three of them met the same total number of households, and each of them sold a total of 100 items.





2. On both days, Lahur met the same number of households and sold the same number of items.



3. Hokli could not sell any item on the second day because the first household he met on that day complained against him.



4. Tohri met 30 more households on the second day than on the first day.



5. Tohri's success rate was twice that of Lahur's on the first day, and it was 75% of Lahur's on the second day.



yx−15yx−15 = 50x50x*2
100−yx+15100−yx+15 = 50x50x*3434
xy = 100(x–15).
4x(100–y) = 150(x+15)
400x – 4xy = 150(x+15)
400x – 4*100(x–15) = 150(x+15) 400 * 15 = 150 (x+15). x = 25



y10y10 = 50255025*2
y = 40

The speeds of Ravi and Ashok converted to meters per second are 40×5/18=100/9 m/s and 50×5/18=1259 m/s50×518=125/9 m/s respectively.

The relative speed between Ravi and Ashok = 100/9+125/9=225/9 m/s

So they meet exactly after 225/225/9=9 seconds.The distance covered by R in 9 seconds = 100/9×9=100 m.

So, for all three of them to meet at the same time Vijay has to cover 54 + 100 = 154m in 9

Seconds.

∴ The speed of Vijay = 154/9 m/s=154/9×185=61.6kmph 

Let the time taken by A to fill the tank alone be x hr, which implies the time taken by B to empty the tank alone is (x-1)hr B is the drainage pipe, and the time taken by C to fill the tank is y hours

It is given that when pipe A,B and C are turned on together, the empty tank is filled in two hours. Hence, 1/x- 1/x-1 +1/y= 1/2

It is given that if pipes B and C are turned on together when the tank is empty and pipe B is turned off after one hour, then pipe C takes another one hour and 15 min to fill the remaining tank

Hence, B worked for 1hr, and C worked for 2hr 15min, which is equal to 9/4hours

In 1 hour, B worked -1/x-1 units, and in 9/4 hours, C worked 9/4y units. Hence, 9/4y-1/x-1=1

Solving both the equations we get y=3/2 and x=3. Hence, the time taken by C is 3/2 hr which is equal to 90 Min 

“Minu purchases a pair of sunglasses at Rs.1000 and sells to Kanu at 20% profit.”

This means, Minu purchased the glasses at Rs.1000 and sold them to Kanu at Rs.1200.

Minu made a profit of Rs. 200 so far.

“Then, Kanu sells it back to Minu at 20% loss.”

This means, Kanu sold the glasses back to Minu at 80% of 1200 = Rs. 960

“Finally, Minu sells the same pair of sunglasses to Tanu the total profit made by Minu from all her transactions is Rs.500”

This means, Minu has to make a further profit of Rs. 300. She achieves this by selling the glasses back to Tanu at 960 + 300 = Rs.1260

So the profit % in the transaction is 300/960×100=31.25%

In a company, 20% of the employees work in the manufacturing department.” The ratio of number of manufacturing to non-manufacturing employees = 1: 4 

So, let the number of manufacturing to non-manufacturing employees be x and 4x respectively. “the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company,” Ratio of total salaries of manufacturing to non-manufacturing employees = 1: 5 

So, let the total salaries of manufacturing to non-manufacturing employees be y and 5y respectively.

So, the ratio of average salaries of manufacturing to non-manufacturing employees will be y/x:5y/ 4x=4:5 

A positive integer less than 50, having exactly two distinct factors other than 1 and itself, is either a perfect cube below 50 or an integer that is a product of exactly two distinct primes.
Case i)
Perfect cubes below 50 are 2³ and 3³. So, two numbers here
Case ii)
For the product of two primes to be below 50, the individual primes should be below 25.
(Because, the smallest prime is 2 and multiplying 2 with anything greater than or equal to 25 yields a number greater than or equal to 50.)
2, 3, 5, 7, 11, 13, 17, 19, 23 are prime numbers less than 25.

2, 3, 5, 7 are the primes less than √50, any product of two numbers among them yields a product less than 50.

11, 13, 17, 19, 23 are the primes greater than √50, any product of two numbers among them yields a product greater than 50.
So, there are 0 pairs here.
Between the two lists 11 and 13 can pair with 2 and 3, while 17, 19, and 23 can only pair with 2.
So, there are 7 pairs here.
So, totally, there are 2 + 6 + 0 + 7 = 15 such numbers.