First find out how many 2`s and how many 3`s are available in 200! 200 when successively divided by 2 given 100, 50, 25, 12, 6, 3, 1, as the quotients and the sum of all these given the number of 2’s which is 197. 200 when successively divided by 3 gives 66 22,7,2as the quotion and the sum of all these gives the number of 3`s which is 97. For 12100 ., we have (2 2 x 3 )100 i.e. 2200 x 3100 we need 200 two`s and 100three `s where as we have 197 two`s and 97 3`s so, we need another three 2`s I.e. we need to multiply 200! by 23 x 3 3 = 8 x 27 = 216.

From 1 to 100, we have number like 3, 13, 23, 33 etc. which are 10 in number. Then number 30 to 39 have one 3 extra each giving 10 threes. So, upto hundred, we have 10 + 10 = 20 threes. Since we have 700 pages we have number of 3’s as 7x20 = 140 Then all the 100 numbers from 300 to 399 have one 3 extra each giving a hundred 3’s So totally, 240 threes.

The new product must be a multiple of 53. Only one choice fulfills this requirement.

Solution:

S.P of 1 litre of mixture = Rs.10.80 and Gain 20%.

Therefore, C.P of 1 litre of mixture = Rs.100/120 x 10.80 = Rs.9 ...(1)

C.P of 1 litre milk of 1st kind = d = Rs.12

C.p of 1 litre milk of 2nd kind = c = Rs.8

Mean Price = m = Rs.9 (from equation 1)

Then d - m = 3 , m - c = 1

Applying above values to our allegations formula (refer introduction before question)

Ratio of quantities of 1st and 2nd kind = 1:3 .

Let X litres of milk of 1st be mixed with 60litres of 2nd kind.

Then, 1 : 3 = X : 60

1/3 = X/60

X = 60/3 = 20.