Question

37 pens and 53 pencils cost rs.320, while 53 pens and 37 pencils cost rs.400. Find the cost of a pen and that of pencil

Answer 1

Let the cost of pen=x=37x
Let the cost of pencil=y=53y
ATP---- 37x+53y=320------------(1)
ATP---- 53x+37y=400-----------(2)

Adding (1) (2)
90x++90y=720
90(x+y)=720
x+y=720/90
x+y=8---------(3)
Subtract (1) (2)
16x-16y=80
16(x-y)=80
x-y=80/16
x-y=5-----------(4)
Adding (3) (4)
x+y=8
x-y=5
----------
2x=13
x=13/2=Rs6.5
Pen =Rs6.5
purting value of x in (3)
x+y=8
6.5+y=8
y=8-6.5
y=2.5
Pen=Rs2.5

Answer 2

The cost of the pen is Rs.6.5

The cost of the pencil is Rs.1.5

Explanation:

Given:

1. 37 pens and 53 pencils cost rs.320

2. 53 pens and 37 pencils cost rs.400.

To find:

The cost of a pen and pencil

Solution:

==> Let pen = x

==> Pencil =y

==> 37x and 53y cost Rs.320

==> 37x+53y = Rs.320

==> 37x +53y = 320 ==>1

==> 53x and 37y cost rs.400

==> 53x+37y = Rs.400

==> 53x+37y=400 ==> 2

==> Using Substitution Method,

==> Rewrite the equation 2

==> 53x = 400-37y

==> $x = \frac{400-37y}{53}$ ==>3

==> Substitute equation 3 in 1

==> 37x+53y = 320

==> 37( $\frac{400-37y}{53}$ )+53y = 320

==> $\frac{37(400-37y)}{53} +53y = 320$

==> $\frac{37(400-37y)}{53} +53y = 320$

==> $\frac{37(400-37y)+}{53} +\frac{53\times53y}{53} = 320$

==> $\frac{37(400-37y)+}{53} +\frac{2809y}{53} = 320$

==> 37(400-37y)+2809y = 320×53

==> 14800-1369y +2809y =16960

==> 14800+1440y = 16960

==> 1440y = 16960-14800

==> 1440y = 2160

==> y = 2160÷1440

==> y =1.5

==> Substitute y in equation 3

==> $x = \frac{400-37y}{53}$

==> $x = \frac{400-37(1.5)}{53}$

==> $x = \frac{400-55.5}{53}$

==> $x = \frac{344.5}{53}$

==> x =6.5

The cost of the pen is Rs.6.5

The cost of the pencil is Rs.1.5

==> Substitute the x and y value in either equation 1 or 2

==> To check whether the answer is correct or not

==> 37x+53y =320

==> 37(6.5)+53(1.5)=320

==> 240.5 + 79.5 =320

==> 320 =320

==> Hence Proved