Hi
The solution for 1st problem:

Number of stamps with Kay = K
Number of stamps with Alberto = A

Asked: K - A = ??

Given:
K:A = 5:3 i.e., k/A = 5/3 ==> 3K = 5A ==> 3K - 5A = 0

After Kay gives 10 stamps to Alberto => Kaye has (K - 10) stamps and Alberto has (A + 10) stamps

Therefore, (k - 10)/(A + 10) = 7/5 ==> 5(k - 10) = 7(A + 10) ==> 5K - 50 = 7A + 70 ==> 5K - 7A = 120

So now we have 2 equations and 2 variable

3K - 5A = 0
5K - 7A = 120

Solving these we get : K = 150 & A = 90

Therefore, (K - 10) - (A + 10) = 140 - 100 = 40
Answer: As a result Kaye has 40 more stamps than Alberto.

I am not very sure about the 2nd problem, but i think the answer is none.

I hope this helps.

K/A=5/3--A

K-10/A+10=7/5 --B

5K-50=7A+70

15K-150=21A+210

From equation A,

3K=5A -->15K=25A

-->25A-150=21A+210-->A=90 &K=150

After Kay gave alberto 10 stamps,A=100,K=140

-->Kay had 40 more stamps than alberto

The number of stamps that kay and Alberto had were in the ratio 5:3. After Kaye gave 10 stamps to Alberto, the ratio is 7:5. As a result, kaye had how many more stamps than Alberto
let the total number of stamp be X
so kay had 5x stamps
alberto had 3x stamp
after kay gave 10 stamp to alberto
now kay has 5x-10 stamp
alberto has 3x+10 stamp
so we know 5x-10/3x+10=7/5
solving we get x as 30
case 1)kaye had how many more stamps than Alberto before the transaction of stamps took place
i.e kay had 5*30=150
albert had 3*30=90
kay had 60 stamps more
case 2)kaye had how many more stamps than Alberto after the stamp transaction took place
kaye had (5x-10) - (3x+10)=40 stamps more

problem 2)
450y=n^3. i.e
5*5*3*3*2*y= n^3 (now since y and n are positive integer )
to make L.H.S a cube y will have to take value of 5*3*2*2
so now we have
if y is 5*3*2*2 then n will be 30
now taking this value of Y and N
we can say
(i) y/ 3*2^2*5
will only be integer
remaining two option will give fraction value

AKS
"ideation without execution is delusion"