-4 Divide 12 in such a way that its product is 2.

Divide 12in 3:4

Divide 20 into two parts such that the product of the cube of one and the square of the other shall be maximum. a)10,10 b)16,4 c)8,12 d)12,8

Divide 124 into four parts in such a way that they are in AP and the product of the first and the 4 th part is 128 less than the product of the 2nd and the 3 rd parts.

Divide 12x^3 by 4x .

Divide 63 in three parts in such a way that they form a GP and the product of the first two parts is equal to (3)/(4) times third part.

Divide 34 in to two parts in such a way that ((4)/(7))^(t)h of one part is equal to ((2)/(5))th of the other

Divide 34 in to two parts in such a way that (4/7)^(th) of one part is equal to (2/5)^(th) of the other

4. (a) Divide 20 into 4 parts which are in A.P. and such that the product of the first and fourth to the product of the second and third are in the ratio 2:3.

Let 5 be a set of first ten natural numbers. What is the possible number of pairs (a, b) where a,b in S and a ne b such that the product ab ( gt 12 ) leaves remainder 4 when divided by 12 ?

Divide : a^(2) + 7a + 12 by a + 4