tomanerjin
contestada

boxes A and B contain some counters
box A contains more counters than box B
box A: 4k counters
Box B: k+9 counters
what is the smallest possible value of K

Respuesta :

Mibo

Step-by-step explanation:

If we take 4 as k then

4k = 4(4) = 16 boxes

k + 9 = 4 + 9 = 13 boxes

ujalakhan18

Answer:

4

Step-by-step explanation:

Box A = 4k

Box B = k+9

Box A > Box B

=> 4k > k+9

Subtracting k to both sides

=> 4k-k > 9

=> 3k > 9

Dividing both sides by 3

=>  k > 3

So, The smallest possible value of k is 4

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