A motorboat can go 10 miles downstream on a river in 20 min. It takes 30 min for this boat to go back at the same 10 miles. Find the speed of the current ?

Part filled in 5 min. = 5 x (1/12 + 1/15) = 5 x 9/60 = 3/4

Part emptied in 1 min. (when all the pipes are opened) = 1/6 - (1/12 + 1/15) = (1/6 - 3/20) = 1/60

Now, 1/60 part is emptied in 1 min. 3/4 part will be emptied in (60 x 3/4) = 45 min.

First of all we write the fraction of milk present in three mixtures. In A : 7 /12
In B : 17 /24
In combination of A and B : 5 /8
We now apply allegation rule on these fractions from figure.
So, Ratio of A: B = 2 : 1


2nd Method
Let us assume P mixture taken from first vessel and Q mixture taken from second vessel to form a new mixture.
Part of Milk in P mixture from first vessel = 7P/12
Part of Milk in Q mixture from Second vessel = 17Q/24
Part of Water in P mixture from first vessel = 5P/12
Part of Milk in Q mixture from Second vessel = 7Q/24
According to question,
After mixing the P and Q, we will get mixture.
Milk in New Mixture / Water in New Mixture = 5/3
{(7P/12) + (17Q/24 ) } /{ (5P/12) + (7Q/24) } = 5/3
{(14P + 17Q)/24 } /{ (10P + (7Q)/24 } = 5/3
{(14P + 17Q) } /{ (10P + (7Q)} = 5/3
(14P + 17Q) x 3 = 5 x (10P + (7Q)
42P + 51Q = 50P + 35Q
51Q - 35Q = 50P - 42P
8P = 16Q
P = 2Q
P/Q = 2
P : Q = 2 : 1

Quantity of milk @ Rs.10 per liter / Quantity of milk @ Rs. 16 per liter = 1 / 2
So, quantity of milk @ Rs. 10 per liter = 26 / 2 = 13 liter.

2nd Method
Let us assume shopkeeper buy P liter milk of price @ Rs. 10 per liter.
Buy price of 26 liter of milk @ Rs. 16 per liter = 26 x 16
Buy price of P liter of milk @ Rs. 10 per liter = P x 10
Sell price of total milk ( P + 26 ) @ Rs. 14 per liter = 14 x ( P + 26 )
According to question there is no loss or no profit.
Then Buy Price = Sell Price
26 x 16 + P x 10 = 14 x ( P + 26 )
? 26 x 16 + 10P = 14P + 14 x 26
? 26 x 16 - 26 x 14 = 14P - 10P
? 2 x 26 = 4P
? 4P = 2 x 26
? P = 2 x 26 / 4 = 13

So, quantity of milk @ Rs. 10 per liter = 13 liter.

Let the number be 27a and 27b
Then, 27a + 27b = 216
or a + b = 216/27 = 8
? Values of co-primes with sum 8 are (1, 7) and (3, 5)
So, the number are (27 x 1 , 27 x 7) i.e. (27, 189)

Required length = H.C.F of 700, 385, 1295 cm
= 35 cm.

Let us assume the number of boys = B and number of girls = G.
According to question,
B + G = 30
Lets us assume total weight of boys = W1 and total weight of girls = W2
average weight of boys = total weight of boys/number of boys
total weight of boys/number of boys = 20
W1/B = 20
W1 = 20B

average weight of girls = total weight of girls/number of girls
25 = W2/G
W2 = 25G

Data is not sufficient to solve the equation.
since we do not know either the average weight of the whole class or the ratio of no. of boys to girls.

Given, a = 12, b =16 , c =18, k = 5
According to the formula,
Required number =(LCM of a, b and c) + k
= (LCM Of 12, 16, 18) + 5

LCM of 12, 16, 18 is
? LCM = 2 x 2 x 3 x 4 x 3 = 144

? Required number =144 + 5 = 149

Let R's capital = 1
Then, Q's capital = 4
2 (P's capital) = 3 (Q's capital) = 3 x 4 = 12
? P's capital = 12/2 = 6
? P's share : Q's share : R's share = 6 : 4 : 1
Thus, Q's share profit = { 4/(6 + 4 + 1)} x 148500
= (4/11) x 148500
= 4 x 13500
= ? 54000

Let numbers are 2N and 3N.
According to the question, 6N = 48
? N = 8 ( ? LCM = 6N )
? Required sum = ( 2N + 3N ) = 5N
= 5 x 8 = 40

Given Ratio = 7/2 : 4/3 : 6/5 = 105 : 40 : 36
Let them initially invest Rs. 105, Rs. 40 and Rs. 36 respectively.
Ratio investment = [105 x 4 + (150% of 105) x 8] : (40 x 12) : (36 x 12)
=1680 : 480 : 432 = 35 : 10 : 9
? B's share = Rs. 21600 x (10/54) = Rs. 4000