Let the price of required variety = Rs. P/kg

Then, respective amounts were m kg, m kg and 2m kg

= 126m + 135m + 2pm = 153 x 4m

=> 2p = 351

p = 175.5 / kg

The ratio of the ages of A and B is 3 : 5.
The ratio of the ages of B and C is 3 : 5.

B's age is the common link to both these ratio. Therefore, if we make the numerical value of the ratio of B's age in both the ratios same, then we can compare the ages of all 3 in a single ratio.

The can be done by getting the value of B in both ratios to be the LCM of 3 and 5 i.e., 15.

The first ratio between A and B will therefore be 9 : 15 and
the second ratio between B and C will be 15 : 25.

Now combining the two ratios, we get A : B : C = 9 : 15 : 25.

Let their ages be 9x, 15x and 25x.
Then, the sum of their ages will be 9x + 15x + 25x = 49x

The question states that the sum of their ages is 147.
i.e., 49x = 147 or x = 3.

Therefore, B's age = 15x = 15*3 = 45

Let ratio of the incomes of Pavan and Amar be 4x and 3x

and Ratio of their expenditures be 3y and 2y

4x - 3y = 1889 ......... I

and

3x - 2y = 1889 ...........II

I and II

y = 1889

and x = 1889

Pavan's income = 7556

Frequency of step of A:B:C = 5 : 6 : 7

But in terms of size of step, 6A = 7B = 8C

Therefore, Ratio of speeds of A, B and C = 5/6 : 6/7 : 7/8 = 140 : 144 : 147

Minimum number of chocolates are possible when he purchases maximum number of costliest chocolates.

Thus,          2 x 5 + 5 x 2 =Rs.20

Now Rs.100 must be spend on 10 chocolates as 100 = 10 x 10.

Thus minumum number of chocolates = 5 + 2 + 10 = 17

Given ratio of pens and pencils = 3 :2

Number of Pens = 3x

Number of Pencils = 2x

Average number of pencils & Pens =  3 x + 2 x 2 = 180

5x = 360

=> x = 72

Hence, the number of pencils = 2x = 72 x 2 = 144.

Let the two numbers be x and y

Given x : y = 3 : 7 .....(1)

Now, x+6 : y+6 = 5 : 9 .....(2)

From (1), x = 3y/7

From (2), 5y - 9x = 24

=> 5y - 9(3y/7) = 24

=> y = 21

=> From(1), x = 9

Hence, the two numbers be 9 and 21

Given total rupees = 20 Rs
No. of one rupee stamps = 3
Now, remaining money = Rs. 17
With that he buys only 2 and 5 rupee stamps
Let number of Rs. 5 stamps = K
Let number of Rs. 2 stamps = L
5K + 2L = 17
K = 3, L = 1 (possible)
L = 6, K = 1 (possible)
=> But given that they are different in number so, K is not equal to 3
one rupee stamps = 3
2 two stamps = 6
5 rupee stamps = 1
Total number of stamps = 10.

Let the incomes of A and B be 3P and 4P.

If each saves Rs. 100 per month, then their expenditures = Income - savings = (3P - 100) and (4P - 100).

The ratio of their expenditures is given as 1 : 2.

Therefor, (3P - 100) : (4P - 100) = 1 : 2

Solving, We get P = 50. Substitute this value of P in 3P and 4P.

Thus, their incomes are : Rs.150 and Rs.200

Let the number of failed students be x

=> Number of passed students = 4x 

So total number of students was 5x

From the given data,

If total number of students be 5x ? 35

=> 4x-35-9/x+9 = 2/1

=> 4x ? 44 = 2(x + 9)

=> 4x ? 2x = 18 + 44

=> x = 31

Total number = 31×5

= 155