In any polygon, the largest side length should be less than the sum of the lengths of the other sides.
Let the fourth side be ‘x’.
Case i) x is the largest side.
x < 1 + 2 + 6
x < 7
x can take the following values, 4, 5, 6
Case ii)
4 is the largest side and x is less than 4.
4 < x + 1 +2
x > 1
x can take the values 2 or 3
So, x can take the values 2, 3, 4, 5, 6.
x can take 5 values.

Since the total donation amount is 15,250 rupees, there should be at least one 250 rupee note. The remaining 15000 can exist as thirty 500 rupee notes.

Since the total donation amount is 15,250 rupees, there should be at least one 250 rupee note. The remaining 15000 can exist as thirty 500 rupee notes.

But the total number of notes will only be 31. We have 100 notes. So to keep the number of 500 notes as high as possible let’s convert the 500 rupee notes to 100 rupee notes. Every time we do this conversion we add 4 new notes.

From here, we can convert one 500 rupee note to two 250 rupee notes.

So, the maximum number of 500 rupee notes is 12

Let the number of students in the school be N.
N < 5000
N leaves a remainder of 4 when divided by 9, 10, 12, or 25.
Since 4 is less than 9, 10, 12 and 25
N leaves a remainder of 4 when divided by LCM(9, 10, 12, 25).
N leaves a remainder of 4 when divided by 900.
N = 900(x) + 4
Since N < 5000
x can range from 0 to 5
But 900(x) + 4 is a multiple of 11 only when x = 2
N = 900(2) + 4 = 1804
When we divide these 1804 students into groups of 12, we get, 150 groups.
Because, 1804 = 12(150) + 4

N people finish 35% of the project by working 7 hours a day for 10 days.
N people finish 35% of the project by working 70 hours.
N people finish 5% of the project by working 10 hours.
N people finish 65% of the project by working 130 hours.
65% of the project is done in N * 130 man hours.
The remaining 65% was actually done by (N-10) people in 14 days by working 10 hours a day.
(N-10) people finish 65% of the project by working 140 hours.
65% of the project is done in (N - 10) * 140 man hours.
N * 130 = (N - 10) * 140
13N = 14N - 140
N = 140

We know the sum of the first pair of numbers is 14 * 2 = 28
Sum of the last pair of numbers is 28 * 2 = 56
To maximize the average of all the 6 numbers, we must try to maximize the two numbers in between. This is possible when the last pair of numbers are 27 and 29.
The maximum average case is
a, b, 25, 26, 27, 29
Where a + b = 28
The average of these 6 numbers is 22.5

Let the distance between the 2 cars be D km.
Let the speeds of the two cars be ‘a’ and ‘b’ respectively and a > b.
Case I)
Cars are moving in the opposite direction (towards each other)
Relative speed = a + b
Time taken = 1.5 hrs
Case II)
Cars are moving in the same direction (Car A chasing Car B)
Relative speed = a - b
Time taken = 10.5 hrs
In both the cases the distance between the cars is the same, ‘D’.
But the time taken is in the ratio 1.5 : 10.5 or 1 : 7
Therefore, the speeds will be in the ratio 7 : 1
a + b = 7(a - b)
8b = 6a
3a = 4b
Substituting b = 60 kmph,
We get, a = 80 kmph.
D = (a + b) * 1.5 = 210km
When they move towards each other the distance covered by them is in the ratio 4 : 3 and the total distance covered by them together is 210 km.
The slower car, B, covers 3/7 th of this 210 km which is 90 km.

If Neetu intended to get a 5% annual interest, ideally all the banks should have maintained a 5% interest rate.
But Bank A returns 0.5% extra interest on 8000 rupees, which is 40 rupees.
But Bank B returns 0.6% extra interest on 5000 rupees, which is 30 rupees.
A & B combines are paying 70 rupees extra than 5%.
So Bank C should maintain such an interest rate that, the interest generated on the remaining 7000 rupees is 70 less than 5% interest.
Since 70 is 1% of 7000. The interest rate at Bank C should be 5% - 1% = 4%
If all the 20,000 rupees were invested in Bank C, the interest generated is 4% of 20,000 = 800 rupees.

O is the centroid of the triangle ABC,that means,
BO : OD = 2 : 1
CO : OE = 2 : 1
Ar(BOC) : Ar(ODC) = 2 : 1
Ar(COB) : Ar(OEB) = 2 : 1

Since BD is the median, Ar(BDA) = Ar(BDC)
This means Ar(AEOD) = 2x
2x + x + x + 2x = 108
6x = 108
Ar(AEOD) = 2x = 36

Since ED is the line joining the midpoints of AB and AC, Ar(AED) = ¼ Ar(ABC)
Ar(AED) = ¼ Ar(108) = 27
Ar(EOD) = Ar(AEOD) - Ar(AED) = 36 - 27 = 9 sq cm

Imagine 2 beakers of milk and water each of volume 500 cc

After the transfer of liquids from one beaker to another, both the beakers end up having the same volume of 500 cc again.

If there is x cc of Milk in the first beaker, there will be 500 - x cc of water in the second beaker, the remaining x cc of water should be in the second beaker!
Therefore, the amount of water in the glass and the amount of milk in the cup are equal and hence in the ratio 1 : 1.
It does not matter how many times the liquids are transferred, if the final volumes are 500 cc, then, the amount of water in the glass and the amount of milk in the cup will be in the ratio 1 : 1.

Since Alex is twice as fast as Bob and thrice as fast as Cole, let us assume that the work done by Alex in a day as 6x.
This means that the work done by Bob and Cole in a day is 3x and 2x respectively.

Since Bob can finish the job in 40 days working alone, the quantum of work to finish the job = 40 ×× 3x = 120x
The work done on the first day, the second day, and the third day is as follows.

The total work done in one cycle of 3 days = 9x + 5x + 8x = 22x
120x = 22x(5) + 9x + 1x
This means, it takes 5 complete cycles of 3 days, a full Day 1 and Day 2 to finish the job.
In one cycle Alex works twice, on Day 1 and Day 3.
So the total number of days that Alex works = 2(5) = 1 = 11 days

an=13+6(n−1)=7+6n

bm=15+7(m−1)=8+7m

For some m, n, let 7+6n=8+7m

6n−7m=1

n=6,m=5

a6=43 and b5=43

Since, L.C.M(6,7)=42, the common terms of the series are of the form 43+42k

Let’s find the smallest 4-digit number of this form…

1000/42≅23.80100042≅23.80

43+42(23)=100943+42(23)=1009

∴43+42(22)=967∴43+42(22)=967 is the largest 3-digit such value.

We need to find the area of the quadrilateral ABDE = area of rectangle ABCD + area of triangle
CDE => Area of ABCD = (7-3)*5 = 20 units, and the area of triangle CDE = (1/2)*10*5 = 25
units. 

Hence, the area of the quadrilateral ABDE = (20+25) = 45 units 

Since Ar(ABP),Ar(APQ)&Ar(AQCD) are in GP and Ar(AQCD)=4×Ar(ABP)

Ar(ABP):Ar(APQ):Ar(AQCD)=1:2:4

9x/2:9y/2:27+9z/2=1:2:4

x:y:6+z=1:2:4

y=2x

6+z=4x

x + y + z = 6

x + 2x + 4x – 6 = 6

x = 127127

z = 4x – 6 = 48−427=6748−427=67

∴ x = 2z

Therefore, x:y:z=2:4:1

BP : PQ : QC = 2 : 4 : 1 

Let the total amount be equal to T.

T=n×352

“However, if two persons receive Rs 506 each and the remaining amount is divided equally among the other persons, each of them receive less than or equal to Rs 330”

T≤2×506+(n−2)×330

n×352≤2×506+(n−2)×330

n×352≤352+n×330

n×22≤352

n≤352/22

n≤16

So, the maximum value that n can take is 16. 

Let the number of white and black shirts bought by Jayant be w and b respectively.

Then the total Cost Price (CP) =1000×w+1125×b=1000(w+b)+125×b

Since the goods are marked up by 25% and then offered at a discount of 10%, the total Selling

Price (SP) =CP×1.25×0.9=1.125

This implies that there was a 12.5% of Profit, which is given to be 51, 000

12.5%(CP) = 51,000 

CP = 4,08,000

1000(w+b)+125×b=4,08,000

w and b are positive integers (since at least one shirt of each color needs to be purchased.)

To purchase maximum number of shirts, you need to purchase minimum number of the costlier shirts, which are the blue ones

Observe that the total CP is a multiple of 1000. And for that to happen b should be a multiple of 8 in 1000(w+b)+125×b=4,08,000

So the minimum value of b = 8, in which case, w = 399. Hence, the maximum number of shirts
that can be purchased = 399 + 8 = 407 

Let the speeds of the boats be x kmph and x + 6 kmph.
After 2 hours, the distance travelled by the boats is 2x km and 2x + 12 kms respectively.
The distance between them is 60 kms.
60² = (2x)² + (2x + 12)²
3600 = 4 x² + 4 x² + 144 + 48x
900 = 2 x² + 36 + 12 x
450 = x² + 18 + 6 x
x² + 6x - 432 = 0
x² + 24x - 18x - 432 = 0
x = 18 or x = -24
The speed of the slower boat is 18 kmph.

Let the capital invested by Mr.Pinto be ₹300.
₹300 because we deal with one-third, further in the question.

The interest generated is ₹15 per year.
Let’s say the capital is invested for n years.
When return equals capital…
15 n = 300
n = 20.

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