The cost of 100 fruits is rupees 500. Cost of 1 watermelon ,1 guava,1 banana respectively rupees 50,10 and 1 find the no.of fruit of each type

Answer 1

# of watermelons ---- x
# of guavas -------- y
# of bananas ------ 100-x-y

50x + 10y + 1(100-x-y) = 500
50x + 10y + 100 - x- y = 500
49x + 9y = 400
remember , x and y must be whole numbers
if y = 0 , x = 400/49 , closest integer is 8
if x = 0, y = 400/9 , closest integer is 44

y = (400 - 49x)/9

trying x = 0,1,2,...,8,9,10
the first solution I get is when x = 10
then y = -10 , which makes no sense

your question appears to have no integer solution.
Check my algebra, I sometimes make errors at my age.

Answer 2

#number of watermelons - (x)
#number of guavas - (y)
#number of bananas - (z)(100-x-y)
x+y+z=100
50x+10y+(100-x-y)=500
49x+9y=400
9y=400-49x
y=(400-49)/9
x=1
y=39
z=60

EZ :)