Question

the price of variety of commodity is rupees 5 per kg and that of another is rupees 8 per kg in that ratio this true varieties been mixed so that the price is 7 per kg​

Answer 1

Answer:

Step-by-step explanation:

Let the given polynomial be p(x) = x3 + ax2 + bx + 6 Given p(x) is divisible by (x – 2), hence p(2) = 0 Put x = 2 in p(x) p(2) = 23 + a(2)2 + b(2) + 6 ⇒ 0 = 8 + 4a + 2b + 6 ⇒ 4a + 2b = – 14 ⇒ 2a + b = – 7 → (1) It is also given that when p(x) is divided by (x – 3) the remainder is 3 That p(3) = 3 Put x = 3 in p(x) p(3) = 33 + a(3)2 + b(3) + 6 ⇒ 3 = 27 + 9a + 3b + 6 ⇒ 9a + 3b = – 30 ⇒ 3a + b = – 10 → (2) Subtract (2) from (1) 2a + b = – 7 3a + b = – 10 ------------------- – a = 3 ∴ a = – 3 Substitute a = – 3 in (1), we get 2(– 3) + b = –7 ∴ b = –1

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Answer 2

Answer:

Step-by-step explanation:

Let the amount taken from variety A be "x" and that from variety B be "y".

The first condition of the question is for weight of the new mixture.It clearly says that x and y should be mixed in such a way that it's weight is 1kg.

Therefore x+y=1.(eq1)

Now the second condition of the question says that amount of x and y should be mixed in such a way that its price is rs.7. So, if the price of 1kg of 1st variety is rs.5, then the price of amount extracted (x) from this will be x×5=5x.

Similarly, price of amount extracted from 2nd variety will be y×8=8y.

Also these two should be combined in such a way that its price is rs.7.

So, 5x+8y=7.(eq2)

Now we have 2 equations.

x+y=1

5x+8y=7

on solving we get x=1/3 and y=2/3

So, x:y=1/3:2/3

=1:2.

so, the two varieties should be mixed in the ratio of 1:2.