Solution:

Let the capacity of bucket be X liters

Originally in the bucket
X liters milk and 0 liters water

After first draw,
X-E liters milk and E liters water

After 2nd draw
X- [2E-(E^2/X)] and 2E - (E^2)/X liters water

Now as per the details given in the question,
X - [2E-(E^2/X)] / [2E - (E^2)/x] = 16/9

After solving the equation, we get
X = 5E

Thus the capacity of bucket is 5E liters

Solution:

We are given a fact that we have to mix equal quantities of b and c
Thus we may as well take average of them both and consider them as a single type

Let us call that type as d
The price will be 170/2 = Rs.85/Kg

Now we have to mix only a which is Rs.95/Kg and d which is Rs.85/Kg

If you compare with the target price, a is Rs.5 more than it and d is Rs.5 less than it

Thus we may mix 50% of each into the final mixture

Required a = 50 Kg
Required d = 50 Kg
D is an equal mixture of b and c thus the final mixture will have 25 kg each of b and c.

So in order to obtain the required results, we will have to mix 50 Kg of a and 25 Kg of each b and c.

Solution:

What we already know is that Sn = n (n+1)/2
What we are provided in the question is that
S2n = 3Sn
=> 2n (2n+1)/2 = 3n (n+1)/2
=> 2 (2n+1) = 3 (n+1)
=> 4n+2 = 3n+3
=> n = 1
Thus, S3n/Sn = [3n (3n+1)/2]/[n (n+1)/2] = 3*4/1*2 = 6

Which means that option c is correct.

Solution:

Let us assume that the required number is x

Now according to the question the numbers must be in proportion
(5 + x) : (11 + x) :: (19 + x) : (37 + x)
185 + 42x + x^2 = 209 + 30x + x^2
12x = 24
X = 2

Let us prove our answer bu putting 2 in place of x
(5 + 2) : (11 + 2) :: (19 + 2) : (37 + 2)
7 : 13 :: 21 : 39
7*39 = 13*21
273 = 273
Thus the required number is 2.

Solution:

Taking the LCM of 8 and 5 i.e. 40,
We make the ratio of B in both A:B and B:C as 40
Now the new ratio will be 24:40 (A:B) and 40:25 (B:C)
Therefore A:B:C = 24:40:25
and 35*(40/89) = 15.73

Solution:

Let us assume that the purchase price is Rs. 100, then the selling price at the loss of 12 1/2 % will be Rs. 87.5.

Thus the ratio of purchasing price and selling price when the loss is 12 1/2 %
= 100/87.5 = 8/7 or 8:7.

Solution:

7x + 8x + 6x = 4200
x = 200
Now,
Pulkit = 7x = 7*200 = Rs. 1400
Rajesh = 8x = 8*200 = Rs. 1600
Tanmay = 6x = 6*200 = Rs. 1200

If we add Rs. 200 to each, the new amount will be
Pulkit = 1400 + 200 = 1600
Rajesh = 1600 + 200 = 1800
Tanmay = 1200 + 200 = 1400

Thus the new ratio will be 1600:1800:1400
Or 8:9:7

Solution:

Let the four nos. in A.P. are (a-d), (a), (a + d) and (a + 2d).
Now as per the given conditions
4a + 2d = 28
2a+d=14

Also a*(a + 2d) / ((a + d)*(a - d)) = 8/15
15a^2 + 30ad = 8a^2 - 8d^2
7a^2 + 30ad + 8d^2 = 0

Or (a + 4d) (7a + 2d) = 0
a + 4d = 0 or 7a + 2d = 0
Solving a + 4d = 0 and 2a + d = 0 we obtain a = 8 and d = (-2)

Thus the numbers are
4, 6, 8 and 10.

Solution:

2:3

Explanation:
100-60 = 40
Ration => 40:60 = 2:3

Solution:

3X/200

Explanation:
The price at which each micro SD card was bought = X/100
Now each card was sold with 50% profit
Thus, cost of each battery when sold = X/100 + X/100*50/100
= X/100 + X/200
= 3X/200