Math problem !!!!!!!!!!!!???????

Question

My question is , I went to the market with 100 dollars . It was not a dollar note . It was coins , They were 50 cents , 5 dollars and 10 dollars . I bought many things . When I reached home all my 100 dollars were spent and 100 things were also there . Now tell me how much things I bought from 50 cents , 5 dollars and 10 dollars ????

Tell me the correct answer .

Hint :
100 cents = 1 dollar .

The first correct answer will recieve 10 points . Try .

Answer 1

I'm assuming you want 100 items costing either 50cents $5 or $10 and coming to a total of $100

Each item must cost an average of $1

So for each $10, you must have 18 items costing $0.50 to maintain the average i.e. 19 items coming to a total of $19
For each $5, you must have 8 items cosing $0.50 to maintain the average i.e. 9 items coming to a total of $9

Now you need numbers that add up to 100, using only 19 and 9

9*9 + 19 = 81 + 19 = 100
So you have 1(10 + (18*0.5)) + 9(5 + (8*0.5))
1(10) + 18(0.5) + 9(5) + 72(0.5)

1 item for $10
9 items for $5
90 items for 50 cents

Answer 2

Well , This question is just assuming question .
First lets take 50 cents . (Let us take 50 cents as 50 c)

If we take 50 things in 50c then it's 25$ , but we need a maximum of 35+ . If we assume .

Now let's take , 90 things then it is 45$ .

Let's take 5$
What we need is 10 things and 55$ . The rule is it has to be 5$ and 10$ . So , we can't take 10 things and finish because the dollars add up to 75$ only . Then we take the next number 9 .

9*5=45$

90+9=99

45$+45$=90$ .

Now let's take 10$ ,

We need 1 thing and 10$ . So , we finish the problem by adding one thing to 99 , which costs 10$ .

we add:

99+1=100

and

90$+10$=100$ .

Thus , the one and only solution is 50 cents = 90 things , 5 dollars = 9 things and 10 dollars = 1 thing .

Thank you for giving an opportunity to prove my skills in maths and to use my brain for some time .

Thank you once again .

-----------------------------------*---------------------------*-------------------

Answer 3

These are called Diophantine equations in the first degree( solvable only in integers). Since u bought 100 things and assuming of course that u are a sweet girl who never cheats, we can assume that the total number of coins u had was 100 and their monetary value was hundred too. Let us say that the number of 50 cent coins u had was l, 5 dollars m and 10 dollars n.
then we get the equations,
0.5l+5m+10n=100
and
l+m+n=100
l=100-n-m.
Go back and substitute n in the first equation and after simplification,
You get
9m+19n=100
m=9 and n=1 is one of the solutions.
By a popular result in Diophantine analysis, the other solutions are given by
m=9+19t and n=1-9t
where t is an integer. It is easy to see that for any value of t other than 0, either m or n will be negative.
So,
m=9, n=1,l=100-9-1=90.
so u initially had 90 50 cent cions, 9 5dollar coins and 1 10dollar coin. By the logic i had used before, u must have also purchased 90 items from 50cent coins, 9 items from 5dollar coins and 1 from 10$ coin.

Answer 4

Let us denote the number of 50¢ things you bought as a, the number of $5 things as b, and the number of $10 things you bought as c. Then:

a∈N, b∈N, c∈N
a+b+c=100
a/2 + 5b + 10c = 100

Solving the system of equations:

a+b+c -2(a/2 + 5b + 10c) = 100 - 200
-9b - 19c = -100
9b + 19c = 100
9b = 100-19c
b=(100-19c)/9

Now b∈N, so 100-19c must be an exact multiple of 9. Thus:

100-19c ≡ 0 mod 9
100≡19c mod 9
c≡ 1 mod 9

Since clearly c=10 would not work, we have c=1. Thus:

b=(100-19)/9 = 81/9=9

And finally:
a=100-9-1=90

Thus a=90, b=9, c=1

Answer 5

90 items for 50 cents,9 items for 5$ and 1 item for 10$

Answer 6

Things purchase 1 for 10 Doller, things purchase 9 for 5 Dollers each and 90 things purchase of 50 cents each

Answer 7

Things rate($) $
1 X 10 = 10
9 X 5 = 45
90 X 0.5 = 45
---------------------------------------------
100 things for 100 dollar
=========================

i think you got correct solution.

Answer 8

That's right.. no single answer..

x - number of things you bought for 10c
y - for 5$
z - for 10$..

x+y+z= 100 things
x*0.1 + y*5 + z*10 = 100$

.. no single answer... you need to predefine one of the variables... the it will be solvable..

Answer 9

I dont understand!!

Answer 10

no single solusion(?)