Let the price of iPhone be 100x => the final price is 103.5x.

Let the price of iPod be 100y.

The final price is 93.5y. => 103.5x = 93.5y => x/y = 935/1035 = 187/207.

Let the initial amount with Lohit, Mohit and Nohit be L, M and N respectively.

After Lohit gives money to Mohit the amount with each of them will be,
Lohit = L-M , Mohit = 2M , Nohit = N.

After Mohit gives money to Nohit the amount with each of them will be,
Lohit = L-M , Mohit = 2M-N , Nohit = 2N.

After Nohit gives money to Lohit the amount with each of them will be,
Lohit = 2(L-M) , Mohit = 2M-N , Nohit = 2N-(L-M).

Now all of them have equal amount with each other,

2(L-M) = 2M-N = 2N-(L-M)

Eqating first and second term,
2N-L+M = 2(L-M)
=> N = 3(L-M)/2....eqn1

Equating second and third term, 
2M-N = 2N-(L-M)
L=3N-M....eqn2

Equating first and second term,
2M-N = 2(L-M)
4M=2L+N....eqn3

Substituting eqn 1 in eqn3
=> 4M = 2L + 3(L-M)/2 => 8M = 4L +3L - 3M
=> L/M = 11/7

Substituting eqn 2 in eqn3
4M=2(3N-M)+N  =>  4M = 6N-2M+N => N/M = 6/7

=> L: M: N = 11: 7: 6
=> L = 55, M = 35, N = 30.

Hence the amount which Lohit originally had is Rs. 55


Alternatively,
Working backwards 
Stage                                      Lohit      Mohit       Nohit
After 3rd transaction               40          40            40
After 2nd transaction              20          40             60
After 1st transaction               20          70             30
Initially                                    55          35            30

Let the current ages of Messi and Maradona be x and y.
(x+n) + (y+n) = 80 = > x+y+2n = 80
$$ 3\times(x-n)=(y-n)$$
$$ 4\times(x-2n)=(y-2n)$$

Solving the 3 equations we get,
n = 5, x = 20 and y = 50.

The difference between their ages will always be the same and is 30 years.

CP of rice = 20*5x + 30*6x + 40*7x = 560x.
SP of rice = 560x*110/100 = 616x.
Therefore, price of the mixture = 616x/(5x+6x+7x) = $$34 \frac{2}{9}$$

Total age of group = 143.

Total age of the three groups together = 33+39+42 = 114.

Therefore, youngest + oldest = 29 years.

The age of the oldest can be 19.

So, the youngest is 10 years old.

Hence, the difference in ages is 9 years.

Let the initial average weight of 150 students be x.
$$\frac{Sum\ of\ the\ weights\ of\ 150\ students}{150} = x$$
Sum of the weights of 150 students = 150x
Let the new student, who is added, have weight m. The average weight of the new class becomes x+0.05
Therefore, [150x+m]/151 = x+0.05
150x+m = 151x+7.55
m = x+7.55
Let the weight of the student who is removed be k. Then the new average becomes x+0.05-0.02 = x+0.03
[150x+m-k]/150 = x+0.03
m-k = 150*0.03 = 4.5
k = m-4.5 = x+7.55-4.5 = x+3.05
Now sum of the weights = m+k = (x+7.55)+(x+3.05) = 2x+10.6
Hence b = 10.6

Alternative Solution:

The initial average is x,

The weight of 1st students can be calculated by subtracting the sum of initial weights from the final weights =  151(x+0.05) - 150x = x+7.55
Similarly, weight of 2nd students =  151{x+0.05} - 150{x+0.03} = x+3.05
Weight of 1st student+ weight of 2nd student = x+7.55+x+3.05 = 2x + 10.6 = 2x + b,
=> b=10.6.

Let the price of basmati and kolam be x and y respectively. Assuming that the quantity is 5 kg, the shopkeeper will mix 3 kg of basmati with 2 kg of surti kolam. So, the selling price will be 5x since he sells all quantity as basmati. But the actual cost price is less which is equal to 3kg of basmati and 2kg of surti kolam = 3x+2y.  Profit  = 5x- (3x+2y) = 2x-2y Hence, profit %= 33.33% = (2x-2y)/(3x+2y) => x=8y/3.

From, mixture 2, profit per kg = Rs 25. Hence profit for 5kg =  25*5 = (5x-4x-y)=x-y. Hence x=200 and y=75.

Let's assume for 100% profit, he is mixing them in the ratio of k:1.

In 100% profit case, the selling price will be double of the cost price.

S.P= 200(k+1) and C.P= (200 X k) + (75 X 1)

=> 200k+200= 2 X [200k + 75]

=> k=1/4

So the ratio is 1:4

Let the number of students be k.

Total weight = 24k.

5 students left and 2 students joined. So, new total number of students = k-3

New total weight = 24k - 140 + 68 = 24(k-3).

New average = 24(k-3)/(k-3) = 24.

Let the incomes be 5k and 7k and expenditures be 3t and 4t. Ratio of savings is (5k-3t)/(7k-4t). 

Let us assume that (5k-3t)/(7k-4t) = 4/5 (Option A).
25k - 15t = 28k -16t
=> t = 3k. 
Now, substituting this relationship in the equation (5k-3t)/(7k-4t), we get the numerator as -4k/-5k. Though the overall ratio is 4/5, the numerator is -4k. => 3t>5k in this case. However, expenditure cannot be greater than income and therefore, we can eliminate this case. Checking similarly, we can eliminate options B and C. 

Option D:
 (5k-3t)/(7k-4t) = 12/17
85k-51t = 84k - 48t
k = 3t.

Substituting this relationship in (5k-3t)/(7k-4t), we get the ratio as 12t/17t. Both the numerator and denominator are positive and hence, option D is a possible solution. 

Let the volume of the tank be 300L.
Initially, A and B are opened. Since the flow of A is double that of B, the volume of milk would be double that of water when the tank is 150L full.
Volume of milk = 100L
Volume of water = 50L
Now the tap C is opened.
Final volume of milk = 80% of 300 = 240L
Remaining volume of milk to be filled = 240-100 = 140L
Water to be filled = 150-140 = 10L
$$\frac{140}{4+x} = \frac{10}{2}$$
4+x = 28
x = 24

Let the initial number of candies be x.

After A woke up, number of candies left = 2/3(x-1)

After B woke up, number of candies left = 2/3(2/3(x-1)-1)

Similarly for C it is = 2/3[2/3(2/3(x-1)-1)-1]

2/3[2/3(2/3(x-1)-1)-1] = 3k+1

[2/3(2/3(x-1)-1)-1] = 9k/2+3/2

2/3(2/3(x-1)-1) = 9k/2+5/2

(2/3(x-1)-1) = 27k/4+15/4

2/3(x-1) = 27k/4+19/4

x-1 = 81k/8 + 57/8

x = 81k/8 + 65/8

To minimize x, we need to find the minimum value of k such that 81k+65 is a multiple of 8.

81k+65 can be expressed as (80k+64) + (k+1). The first part is divisible by 8. For the second part to be divisible by 8, minimum value of k is 7.

For k = 7, x = 79.

Given
Each of three varieties contain starch, hops and water in varying proportions
A contains  x% starch and y% hops. So 100-(x+y) % of water.
B contains y% starch and y% water. So 100 - (2y) % of hops.
C contains x% hops and y% water. So 100- (x+y) % of starch

Concentration of starch in final mixture is 38%,then

(2x/10) + (3y/10) + (5(100 -(x+y))/10) = 38

3x + 2y = 120 - (1)

Also, given concentration of hops in final mixture is 28%, then

(2y/10) + (3(100-2y)/10) + (5x/10) = 28

-5x + 4y = 20 - (2)

Solving (1) and (2) we get x = 20 and y = 30

∴ x + y = 50

Going from the options, we can see that if the ratio is 1:1:3, the resultant mixture will be of concentration: (1*15+1*20+3*30)/(1+1+3) = 125/5 = 25%. Option b) is the correct answer.

Assume that the teacher divided the groups in such a way that the conditions given are not satisfied. In such a case let us find the maximum number of students in group B.
In the worst possible scenario we can assume that in group B there are 10 students with 7 marks each, 8 students with 5 marks each and 3 students with 3 marks each. There will be 21 students in group B but the condition given is still not satisfied.
So, if there are 21 students, in group B, there is a possibility that the condition given is not satisfied. But, if there are 22 students in group B, then there will be at least five students with n marks each.
So, the minimum number of students to be in group B such that the criteria is fulfilled is 22.
Hence, option (2) is the correct choice.

Let the quantities of grapes, milk, and sugar he bought be 8k, 5k, and 3k respectively.
In 8k grapes- 60% is water and 40% is dry grapes i.e. 0.6 X 8k water and 0.4 X 8k dry grapes
In 5k milk, 80% is water i.e. 4k is water.
Total water in the mixture= 4.8k+4k=8.8k
Total dry grapes= 3.2k
Now after cooking only 75% of water remained.
=> water remaining= 3/4 X 8.8k= 6.6k
So, the required ratio = 3.2k/6.6k=16/33.

Let the ratio of alloy A : alloy B = 1:k.
Copper to Iron in the final alloy = [(7/20)*1+(5/17)*k]/[(13/20)*1+(12/17)*k] = 3:7

Solving this, we get k = 17/2 => the required ratio = 2:17.

Let SP = 100 => MP = 140.
New SP = 70.
Profit = 20% => CP = 70*100/120.
If the discount is 25%, the new SP is 105.
Therefore, profit % = $$\frac{105 - 70*100/120}{70*100/120} * 100$$% = 80%

In 5 tonnes of Ironite, zinc present = 0.5 tonnes and iron present = 4.5 tonnes.

Since the ratio given is 3:2, the weight of iron =3x and the weight of zinc =2x

Zinc is lost but iron will remain same in both, hence, 4.5=3x  => x=1.5

The total weight = 3x+2x = 5x = 5*1.5  =7.5

Let the initial volume of juice and water be a ml and b ml respectively.

Initially
Juice = a, Water = b

After first mixing,
Juice = a, Water = x

As the ratio gets inverted,
a/x = b/a
$$x = a^2/b$$
Juice = a , Water =$$a^2/b$$

After second mixing,
Juice = y, Water =$$a^2/b$$
As the ratio comes back to its original value,
y/[$$a^2/b$$] = a/b
y =$$a^3/b^2$$
Juice = $$a^3/b^2$$, Water = $$a^2/b$$

Ratio of juice in the original mixture to that in final mixture = a/($$a^3/b^2$$) = $$b^2/a^2$$
As a and b are natural numbers, the only plausible value is in option d where both numerator and denominator are perfect squares.

Alternate Solution:

Let the initial volume of the solution be $$O_1$$. After Ram mixes water, let the volume of the solution be $$O_2$$ and let the volume of the final solution (after Shyam mixes juice) be $$O_3$$

The amount of juice is the same in $$O$$ and $$O_1$$ as Ram only added water. Hence, $$\frac{a}{a+b} \times O = \frac{b}{a+b} \times O_1$$
This would imply that $$O_1 = \frac{a}{b} \times O$$
Similarly, The amount of water is the same in $$O_1$$ and $$O_2$$ as Shyam only added juice. Hence, $$\frac{a}{a+b} \times O_1 = \frac{b}{a+b} \times O_2$$
This would imply that $$O_2 = \frac{a}{b} \times O_1$$
Therefore, $$O_2 = \frac{a}{b} \times \frac{a}{b} \times O = \frac{a^2}{b^2} \times O$$
Therefore, $$\frac{O}{O_2} = \frac{b^2}{a^2}$$

Hence, the required ratio is the ratio of the squares of two natural numbers. Of the given options, only 4:9 is of that form and hence, that is the answer.

Let the price of basmati and kolam be x and y respectively. Assuming that the quantity is 5 kg, the shopkeeper will mix 3 kg of basmati with 2 kg of surti kolam. So, the selling price will be 5x since he sells all quantity as basmati. But the actual cost price is less which is equal to 3kg of basmati and 2kg of surti kolam = 3x+2y.  Profit  = 5x- (3x+2y) = 2x-2y Hence, profit %= 33.33% = (2x-2y)/(3x+2y) => x=8y/3.

From, mixture 2, profit per kg = Rs 25. Hence profit for 5kg =  25*5 = (5x-4x-y)=x-y. Hence x=200 and y=75.

Let's assume for 100% profit, he is mixing them in the ratio of k:1.

In 100% profit case, the selling price will be double of the cost price.

S.P= 200(k+1) and C.P= (200 X k) + (75 X 1)

=> 200k+200= 2 X [200k + 75]

=> k=1/4

So the ratio is 1:4

Each of the 15 numbers is a sum of 2 numbers taken at a time among A, B, C, D, E and F. Therefore in all possible cases A will be paired with B,C,D,E and F once hence A will be counted 5 times. This is true for each number. Therefore, 

5A + 5B + 5C+ 5D + 5E + 5F = 10+15+17+13+19+18+16+20+15+13+17+11+24+22+20

A + B + C + D + E + F = $$\frac{10+15+17+13+19+18+16+20+15+13+17+11+24+22+20}{5}$$ = 50
If the 15 numbers are written in the order of their magnitude, then the sequence is 10, 11, 13, 13, 15, 15, 16, 17, 17, 18, 19, 20, 20, 22, 24.
The least number is obtained by adding the least 2 numbers among A, B, C, D, E ad F.
Hence A + B = 10 => B must be at least 6 => C must be at least 7.
=> C + D + E + F = 40
Highest number is obtained by adding E and F. Second highest number is obtained by adding D and F. 
=> D + F = 22 and E + F = 24
=> C + D = 16 => D must be at least 9 and C can be at most 7.

=> The value of C is 7.