A businessman bought some items for `3000. He kept 10 items for
himself and sold the remaining at a profit of `25 per item. From the
amount he received in this deal, he could buy 15 more items. Find the
original price of each item
Answers
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Given :- A businessman bought some items for 3000. He kept 10 items for himself and sold the remaining at a profit of 25 per item. From the amount he received in this deal, he could buy 15 more items. Find the original price of each item. ?
Solution :-
Let us assume that, the businessman buy total x items .
now,
→ Total CP of x items = Rs. 3000
then,
→ CP of each item = Rs.(3000/x)
Now, he kept 10 items for himself sold the remaining at a profit of 25 per item .
So,
→ items left = (x - 10)
→ Profit on each item = Rs.25
then,
→ SP = CP + Profit
→ Total SP = (x - 10) * [(3000/x) + 25)] = Rs. (x-10)[(3000 + 25x)/x]
Now, given that, he could buy 15 more items from this profit.
then we can conclude that, he can buy total (x + 15) items from given SP .
therefore,
→ (x - 10)[(3000 + 25x)/x] = (3000/x) * (x + 15)
→ (x - 10)(3000 + 25x) = (3000) * (x + 15)
→ 3000x + 25x² - 30000 - 250x = 3000x + 45000
→ 25x² - 250x - 30000 - 45000 = 0
→ 25x² - 250x - 75000 = 0
→ x² - 10x - 3000 = 0
→ x² - 60x + 50x - 3000 = 0
→ x(x - 60) + 50(x - 60) = 0
→ (x - 60)(x + 50) = 0
→ x = 60 or (-50) .
since items cant be in negative..
hence,
→ Total items he bought = 60 .
∴
→ The original price of each item = 3000/60 = Rs.50 (Ans.)
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Answer:
75 is the price of each article
Step-by-step explanation:
let the no of items be x so
for x items he gave 3000
on each=3000/x
he kept 10
so x-10 items
one item would be of 3000/x-10
profit of 25 per item so
(3000/x-10)-(3000/x)=25
3000[(1/x-10)-1/x]=25
3000[x-x+10/x²-10x]=25
3000[10/x²-10x]=25
cut 3000 by 25 so
120×10=x²-10x
x²-10x=1200
x²-10x-1200=0
x²-40x+30x-1200=0
x(x-40)+30(x-40)=0
(x+30)(x-40)=0
x can't be -30 coz items cannot be in negative no
so x=40
items were 40
one item price =3000/40
=75
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