Question

I want to buy 100 things for 100 rupees , I want to buy plate,glass and spoon.1 plate for 5 rupees,1 glass for 3 rupees and 40 spoon for 1 rupee. So how many things and how many rupees in my hand

Answer 1

Given :

Rate of plate= 1 for 5 rupees.

Rate of glass = 1 for 3 rupees.

Rate of spoon = 40 for 1 rupee.

Total in hand money = 100 rupees.

To find :

Number of each item bought and money spent on it.

Solution :

Rate of 40 spoons = Rs. 1

Rate of 1 plate = Rs. 5

Rate of 1 glass = Rs. 3

So, Rate of 1 spoon :

$= \frac{1}{40} = 0.025 \: rupees$

Now let,

Number of plates = a

Number of glasses = b

Number of spoons = c

As it is given that,

Total number of items = 100

So,

$\bf \: a + b + c = 100$

(equation 1)

As it is also given that,

Total rupees to spend = Rs. 100

So, by given amount of each item above, we can say :

$\bf5a + 3b + \frac{c}{40} = 100$

(equation 2)

We got 2 equations now, but there are three unknowns to find.So,

We have to take every possible value of at least one unknown.

We know that spoons can be eaither 0, 40 or 80 because in 1 rupee 40 spoons can be baught (no less)

and at most 80 (not 120 and more) only because total 100 items to bought.

So, c = 0, 40 or 80.

Now it is given that each item is necessary so

$\bf \: c \neq \: 0$

Putting c = 40 in equation 1 and 2.

then,

$\bf \: a + b + 40 = 100 \\ \bf a + b = 60$

and

$\bf \: 5a + 3b + (\frac{40}{40} ) = 100 \\ \bf \: 5a + 3b = 99$

Now there are two equations and two unknowns.

Solving these equations, we get :

$\bf \: a = - \frac{2}{81}$

which is not allowed because quantities can not be negative and must be integral value.

So,

$\bf \: c \neq40$

Putting c = 80 in equation 1 and 2,

then,

$\bf \: a + b + 80 = 100 \\ \bf a + b = 20$

and

$\bf \: 5a + 3b + (\frac{80}{40} ) = 100 \\ \bf \: 5a + 3b = 98$

Now there are two equations and two unknowns.

Solving these equations, we get :

$\bf a = 19 \\ \bf b = 1 \\ \bf \: c = 80$

So,

Number of plates = 19

Number of glasses = 1

Number of spoons = 80