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

Let us try to calculate possible amounts of money raised by firms Alfloo and Elavalaki.

Firm Alfloo: 

This is the least possible money raised if in 2010 money raised increased by 2 crores. Here, the total sum is 23.

So this case is not possible. Therefore, in 2010 money raised must increase by 1 crore. 

(Note: Again if money raised in 2011 is ‘4’ then least possible sum is 22 for combination 1-2-4-5-4-3-2-1)

Firm Elavalaki: 

Hence, the money raised in year 2015 can be:  2 + 1 + 3 + 1 + 1 or 0 = 7 or 8 crores. 

The largest possible total amount of money (in Rs. crores) that could have been raised in 2013. 

Hence, the total money raised = 17 crores. 

If Elavalaki raised Rs. 3 crore in 2013, then these cases are considered.

Now the smallest possible amount of money that could have been raised by all the companies in
2012 is 4 + 1+ 0 + 2 + 4 = 11 crores. 

If the total amount of money raised in 2014 is Rs. 12 crores, then

Hence, statement “Bzygoo raised the same amount of money as Elavalaki in 2013” is not possible. 

Let us try to find the amount of money raised by a firm:

Bzygoo:

Czechy: 

Here, a + b + c = 9 – 2 = 7.

Only possible combination is (a, b, c) = (2, 3, 2).

(Note: Other possibility of last year of existence is not there as it will not add up to 7.)

Drjbna: 

For firms Czechy and Drjbna only, exact amounts raised by them can be concluded. 

Step 1:

Let total score of Day 1, Day 2, Day 3, Day 4 and Day 5 be D1, D2, D3, D4 and D5 respectively.

Then, D1 + D2 = 30

D2 + D3 = 31

D3 + D4 = 32

D4 + D5 = 34

As per condition 2, D3 = D4 = 16

Therefore, D1 = 15, D2 = 15, D3 = 16, D4 = 16 and D5 = 18.

As per condition 1 and 3, 

So, 2a + c = 16, where c > a. Therefore, possible combinations are (4, 8), (5, 6) but since all scores
of Chatur were multiples of 3, hence, (a, c) = (5, 6). This also implies Chatur’s unique highest score in the competition on Day 2 must be 9 and b = 3.

Step 2: 

Here 9 > x > y and q > 6 > p

and x + y = 6 and p + q = 12

Possible combinations of (x, y) = (15, 1), (4, 2)

Possible combinations of (p, q) = (5, 7), (4, 8) 

Step 1:

Let total score of Day 1, Day 2, Day 3, Day 4 and Day 5 be D1, D2, D3, D4 and D5 respectively.

Then, D1 + D2 = 30

D2 + D3 = 31

D3 + D4 = 32

D4 + D5 = 34

As per condition 2, D3 = D4 = 16

Therefore, D1 = 15, D2 = 15, D3 = 16, D4 = 16 and D5 = 18.

As per condition 1 and 3, 

So, 2a + c = 16, where c > a. Therefore, possible combinations are (4, 8), (5, 6) but since all scores
of Chatur were multiples of 3, hence, (a, c) = (5, 6). This also implies Chatur’s unique highest score in the competition on Day 2 must be 9 and b = 3.

Step 2: 

Here 9 > x > y and q > 6 > p

and x + y = 6 and p + q = 12

Possible combinations of (x, y) = (15, 1), (4, 2)

Possible combinations of (p, q) = (5, 7), (4, 8) 

Step 1:

Let total score of Day 1, Day 2, Day 3, Day 4 and Day 5 be D1, D2, D3, D4 and D5 respectively.

Then, D1 + D2 = 30

D2 + D3 = 31

D3 + D4 = 32

D4 + D5 = 34

As per condition 2, D3 = D4 = 16

Therefore, D1 = 15, D2 = 15, D3 = 16, D4 = 16 and D5 = 18.

As per condition 1 and 3, 

So, 2a + c = 16, where c > a. Therefore, possible combinations are (4, 8), (5, 6) but since all scores
of Chatur were multiples of 3, hence, (a, c) = (5, 6). This also implies Chatur’s unique highest score in the competition on Day 2 must be 9 and b = 3.

Step 2: 

Here 9 > x > y and q > 6 > p

and x + y = 6 and p + q = 12

Possible combinations of (x, y) = (15, 1), (4, 2)

Possible combinations of (p, q) = (5, 7), (4, 8) 

Step 1:

Let total score of Day 1, Day 2, Day 3, Day 4 and Day 5 be D1, D2, D3, D4 and D5 respectively.

Then, D1 + D2 = 30

D2 + D3 = 31

D3 + D4 = 32

D4 + D5 = 34

As per condition 2, D3 = D4 = 16

Therefore, D1 = 15, D2 = 15, D3 = 16, D4 = 16 and D5 = 18.

As per condition 1 and 3, 

So, 2a + c = 16, where c > a. Therefore, possible combinations are (4, 8), (5, 6) but since all scores
of Chatur were multiples of 3, hence, (a, c) = (5, 6). This also implies Chatur’s unique highest score in the competition on Day 2 must be 9 and b = 3.

Step 2: 

Here 9 > x > y and q > 6 > p

and x + y = 6 and p + q = 12

Possible combinations of (x, y) = (15, 1), (4, 2)

Possible combinations of (p, q) = (5, 7), (4, 8) 

Step 1:

Let total score of Day 1, Day 2, Day 3, Day 4 and Day 5 be D1, D2, D3, D4 and D5 respectively.

Then, D1 + D2 = 30

D2 + D3 = 31

D3 + D4 = 32

D4 + D5 = 34

As per condition 2, D3 = D4 = 16

Therefore, D1 = 15, D2 = 15, D3 = 16, D4 = 16 and D5 = 18.

As per condition 1 and 3, 

So, 2a + c = 16, where c > a. Therefore, possible combinations are (4, 8), (5, 6) but since all scores
of Chatur were multiples of 3, hence, (a, c) = (5, 6). This also implies Chatur’s unique highest score in the competition on Day 2 must be 9 and b = 3.

Step 2: 

Here 9 > x > y and q > 6 > p

and x + y = 6 and p + q = 12

Possible combinations of (x, y) = (15, 1), (4, 2)

Possible combinations of (p, q) = (5, 7), (4, 8) 

Step 1: 

It is given that average number of coins per sack in the boxes are all distinct integers, so, each box will have average of 1, 2, 3, 4, 5, 6, 7, 8 and 9 coins per sack in any order. 

Therefore, each row and column will have (9*10)/2 = 45 coins in total, since each row and column has same number of coins. 

Also, since average number of coins per sack is an integer value therefore, the sum of coins in each sack in a box should be divisible by 3. 

Average of ‘1’ is only possible for the combination of (1, 1, 1). 

This combination is possible for boxes in 3rd column - 3rd row because this satisfies 2 conditions that minimum value is 1 and median is also 1. 

Average ‘9’ is only possible for the combination of (9, 9, 9). 

This combination is possible in 2nd row-3rd column (Box in 3rd row 1st column has median 8, therefore, it cannot have an average value of 9). 

Box in 3rd row - 1st column satisfies condition (iii) only therefore, the only combination of coins that satisfies the mentioned condition is (7, 8, 9). Box in 3rd row/2nd column must satisfy conditions (i) and (iii). 

So, minimum and maximum number of coins in a bag are 1, and 9 respectively. Hence, possible combinations are (1, 5, 9) or (1, 8, 9) but only (1, 8, 9) will give the sum of 45 for Row-3 

Step 2: 

If average number of coins in a box is ‘2’ then, total sum of coins is suppose to be 6 which means each sack must have less than 5 coins and that is only possible for box in 2nd row - 2nd column. 

This box satisfy only one condition and that must be (I).

Hence, the only possible combination is (1, 2, 3). 

[We take (1, 1, 4) then two given conditions will be satisfied which contradicts.]

Step 1: 

It is given that average number of coins per sack in the boxes are all distinct integers, so, each box will have average of 1, 2, 3, 4, 5, 6, 7, 8 and 9 coins per sack in any order. 

Therefore, each row and column will have (9*10)/2 = 45 coins in total, since each row and column has same number of coins. 

Also, since average number of coins per sack is an integer value therefore, the sum of coins in each sack in a box should be divisible by 3. 

Average of ‘1’ is only possible for the combination of (1, 1, 1). 

This combination is possible for boxes in 3rd column - 3rd row because this satisfies 2 conditions that minimum value is 1 and median is also 1. 

Average ‘9’ is only possible for the combination of (9, 9, 9). 

This combination is possible in 2nd row-3rd column (Box in 3rd row 1st column has median 8, therefore, it cannot have an average value of 9). 

Box in 3rd row - 1st column satisfies condition (iii) only therefore, the only combination of coins that satisfies the mentioned condition is (7, 8, 9). Box in 3rd row/2nd column must satisfy conditions (i) and (iii). 

So, minimum and maximum number of coins in a bag are 1, and 9 respectively. Hence, possible combinations are (1, 5, 9) or (1, 8, 9) but only (1, 8, 9) will give the sum of 45 for Row-3 

Step 2: 

If average number of coins in a box is ‘2’ then, total sum of coins is suppose to be 6 which means each sack must have less than 5 coins and that is only possible for box in 2nd row - 2nd column. 

This box satisfy only one condition and that must be (I).

Hence, the only possible combination is (1, 2, 3). 

[We take (1, 1, 4) then two given conditions will be satisfied which contradicts.]

Step 1: 

It is given that average number of coins per sack in the boxes are all distinct integers, so, each box will have average of 1, 2, 3, 4, 5, 6, 7, 8 and 9 coins per sack in any order. 

Therefore, each row and column will have (9*10)/2 = 45 coins in total, since each row and column has same number of coins. 

Also, since average number of coins per sack is an integer value therefore, the sum of coins in each sack in a box should be divisible by 3. 

Average of ‘1’ is only possible for the combination of (1, 1, 1). 

This combination is possible for boxes in 3rd column - 3rd row because this satisfies 2 conditions that minimum value is 1 and median is also 1. 

Average ‘9’ is only possible for the combination of (9, 9, 9). 

This combination is possible in 2nd row-3rd column (Box in 3rd row 1st column has median 8, therefore, it cannot have an average value of 9). 

Box in 3rd row - 1st column satisfies condition (iii) only therefore, the only combination of coins that satisfies the mentioned condition is (7, 8, 9). Box in 3rd row/2nd column must satisfy conditions (i) and (iii). 

So, minimum and maximum number of coins in a bag are 1, and 9 respectively. Hence, possible combinations are (1, 5, 9) or (1, 8, 9) but only (1, 8, 9) will give the sum of 45 for Row-3 

Step 2: 

If average number of coins in a box is ‘2’ then, total sum of coins is suppose to be 6 which means each sack must have less than 5 coins and that is only possible for box in 2nd row - 2nd column. 

This box satisfy only one condition and that must be (I).

Hence, the only possible combination is (1, 2, 3). 

[We take (1, 1, 4) then two given conditions will be satisfied which contradicts.]

Step 1: 

It is given that average number of coins per sack in the boxes are all distinct integers, so, each box will have average of 1, 2, 3, 4, 5, 6, 7, 8 and 9 coins per sack in any order. 

Therefore, each row and column will have (9*10)/2 = 45 coins in total, since each row and column has same number of coins. 

Also, since average number of coins per sack is an integer value therefore, the sum of coins in each sack in a box should be divisible by 3. 

Average of ‘1’ is only possible for the combination of (1, 1, 1). 

This combination is possible for boxes in 3rd column - 3rd row because this satisfies 2 conditions that minimum value is 1 and median is also 1. 

Average ‘9’ is only possible for the combination of (9, 9, 9). 

This combination is possible in 2nd row-3rd column (Box in 3rd row 1st column has median 8, therefore, it cannot have an average value of 9). 

Box in 3rd row - 1st column satisfies condition (iii) only therefore, the only combination of coins that satisfies the mentioned condition is (7, 8, 9). Box in 3rd row/2nd column must satisfy conditions (i) and (iii). 

So, minimum and maximum number of coins in a bag are 1, and 9 respectively. Hence, possible combinations are (1, 5, 9) or (1, 8, 9) but only (1, 8, 9) will give the sum of 45 for Row-3 

Step 2: 

If average number of coins in a box is ‘2’ then, total sum of coins is suppose to be 6 which means each sack must have less than 5 coins and that is only possible for box in 2nd row - 2nd column. 

This box satisfy only one condition and that must be (I).

Hence, the only possible combination is (1, 2, 3). 

[We take (1, 1, 4) then two given conditions will be satisfied which contradicts.]

Step 1: 

It is given that average number of coins per sack in the boxes are all distinct integers, so, each box will have average of 1, 2, 3, 4, 5, 6, 7, 8 and 9 coins per sack in any order. 

Therefore, each row and column will have (9*10)/2 = 45 coins in total, since each row and column has same number of coins. 

Also, since average number of coins per sack is an integer value therefore, the sum of coins in each sack in a box should be divisible by 3. 

Average of ‘1’ is only possible for the combination of (1, 1, 1). 

This combination is possible for boxes in 3rd column - 3rd row because this satisfies 2 conditions that minimum value is 1 and median is also 1. 

Average ‘9’ is only possible for the combination of (9, 9, 9). 

This combination is possible in 2nd row-3rd column (Box in 3rd row 1st column has median 8, therefore, it cannot have an average value of 9). 

Box in 3rd row - 1st column satisfies condition (iii) only therefore, the only combination of coins that satisfies the mentioned condition is (7, 8, 9). Box in 3rd row/2nd column must satisfy conditions (i) and (iii). 

So, minimum and maximum number of coins in a bag are 1, and 9 respectively. Hence, possible combinations are (1, 5, 9) or (1, 8, 9) but only (1, 8, 9) will give the sum of 45 for Row-3 

Step 2: 

If average number of coins in a box is ‘2’ then, total sum of coins is suppose to be 6 which means each sack must have less than 5 coins and that is only possible for box in 2nd row - 2nd column. 

This box satisfy only one condition and that must be (I).

Hence, the only possible combination is (1, 2, 3). 

[We take (1, 1, 4) then two given conditions will be satisfied which contradicts.]

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.