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Analysis of data: - Urea: cow manure = 2 : 5

Therefore proportional part of urea in the mixed fertilizer = 2 / 7

And that of cow manure in the mixed fertilizer = 5 / 7

In mathematical language the problem is

i) 28 x (2 / 7)

ii) 28 x (5 / 7)

i) 28 x (2 / 7)= 4 x 2 = 8

ii) 28 x (5 / 7)= 4 x 5 = 20

1

2

i) H.C.F. of the coefficients is 2

ii) The highest common powers of the variables are 2 for p, 1 for q and 3 for r.

Therefore, the required H.C.F. = 2 p^2 q r^3

i) The L.C.M. of the co-efficient = 2 x 2 x 3 = 12

ii) The highest powers of the variables are 3 for p, 2 for q and 4 for r.

Therefore, the required L.C.M. = 12 p^3 q^2 r^4