Question

Jamila sold a table and a chair for Rs.1050, there by making a profit of 10% on table and 25% on the chair. If she had taken a profit of 25% on the table and 10% on chair she would have got Rs.1065. Find the cost price of each.

Answer 1

Given :

The selling price of 1 table and 1 chair = Rs 1050

The profit % on table = 10%

The profit % on chair = 25%

Again

The selling price of 1 table and 1 chair = Rs 1065

The profit % on table = 25%

The profit % on chair = 10%

To Find :

The cost price of 1 chair and 1 table

Solution :

Let The cost price of 1 chair = Rs c

     The cost price of 1 table = Rs t

∵ profit % = $\dfrac{s.p}{c.p}$ - 1

From first statement

Let s.p of 1 table = Rs x

So, s.p of 1 chair = Rs (1050 - x )

For Table

   10 %  = $\dfrac{x}{t}$ - 1

Or,  $\dfrac{10}{100}$ = $\dfrac{x}{t}$ - 1

or,  $\dfrac{x}{t}$ = 1 + $\dfrac{10}{100}$

Or,  $\dfrac{x}{t}$ = 1.1

i.e    x = 1.1 t                         .........1

Again

For chair

   25 %  = $\dfrac{1050-x}{c}$ - 1

Or,  $\dfrac{25}{100}$ = $\dfrac{1050-x}{c}$ - 1

or,  $\dfrac{1050-x}{c}$  = 1 + $\dfrac{25}{100}$

Or,  $\dfrac{1050-x}{c}$  = 1.25

i.e    x = 1050 - 1.25 c                         .........2

From eq 1 and 2

1050 - 1.25 c = 1.1 t

or,  1.1 t + 1.25 c = 1050                    ...........3

Similarly

From second statement

Let s.p of 1 table = Rs y

So, s.p of 1 chair = Rs (1065 - y )

For Table

   25 %  = $\dfrac{y}{t}$ - 1

Or,  $\dfrac{25}{100}$ = $\dfrac{y}{t}$ - 1

or,  $\dfrac{y}{t}$ = 1 + $\dfrac{25}{100}$

Or,  $\dfrac{y}{t}$ = 1.25

i.e   y = 1.25 t                         .........4

Again

For chair

   10 %  = $\dfrac{1065-y}{c}$ - 1

Or,  $\dfrac{10}{100}$ = $\dfrac{1065-y}{c}$ - 1

or,  $\dfrac{1065-y}{c}$  = 1 + $\dfrac{10}{100}$

Or,  $\dfrac{1065-y}{c}$  = 1.1

i.e    y = 1065 - 1.1 c                         .........5

From eq 1 and 2

1065 - 1.1 c = 1.25 t

or,  1.25 t + 1.1 c = 1065                   ...........6

Now,

Solving eq 3 and eq 6

1.1 × ( 1.25 t + 1.1 c ) - 1.25 × (1.1 t + 1.25 c)= 1.1 × 1065 - 1.25 × 1050

Or,  0.88 t + 1.21 c - 0.88 t - 1.5625 c = 1171.5 - 1312.5

Or,  - 0.3525 c = - 141

i.e     c = $\dfrac{141}{0.3525}$

∴ c.p of chair = c = Rs 400

And, put the value of c into eq 6

i.e 1.25 t + 1.1 × Rs 400 = 1065

Or,  1.25 t = Rs 1065 - 440

Or, 1.25 t = Rs 625

i.e    t = $\dfrac{625}{1.25}$

∴ c.p of table = t = Rs 500

Hence, The cost price of 1 table is Rs 500

And The cost price of 1 chair is Rs 400  . Answer

Answer 2

Answer:

e

Step-by-step explanation:

e