Question

The cost of 2 tables and 3 chairs is 340/- but a table cost 20 more than a chair. Find the cost of each.

with clearly and explain​

Answer 1

Answer:

Also, given that the cost of two tables and three chairs = 340. Also, it is given that the cost of a table is Rs 20 more than the cost of a chair. Hence, we have x = y+20 (ii). Hence, we have y = 60.

Answer 2

$\underline{ \underline{ \Large \pmb{\sf { {Given:}} }} }$

• The cost of 2 tables and 3 chairs is 340.

• Cost of a table = 20 more than the cost of a chair.

$\underline{ \underline{ \Large \pmb{\sf { {To \; calculate:}} }} }$

• The cost of each table and chair.

$\underline{ \underline{ \Large \pmb{\sf { {Calculation:}} }} }$

✰ Here, we are provided that the cost of 2 tables and 3 chairs is Rs. 340. And the condition given here is that the cost of a table is 20 more than the cost of a chair. We have to find the cost of each table and chair. In order to find the cost of each table and chair, we have to form an algebraic equation and by solving that equation we'll find the cost of each table.

⠀⠀⠀⠀⠀_____________

Let,

• Cost of 2 tables = p

• Cost of 3 chairs = q

According to the question,

• The cost of 2 tables and 3 chairs is 340.

$\longrightarrow \sf { {Cost}_{(2 \: tables)} + Cost_{(3 \: chairs)} = 340}$

$\longrightarrow \sf { p + q = 340}$

Let it be the equation (i).

Also, according to the question the cost of a table is 20 more than the cost of a chair.

• If the cost of 2 tables is p. Then,

$\longrightarrow \sf { Cost_{(1 \: table)} = \dfrac{p}{2} }$

• And, if the cost 3 chairs is q. Then,

$\longrightarrow \sf { Cost_{(1 \: chair)} = \dfrac{q}{3} }$

As the cost of a table is 20 more than the cost of a chair, so linear equation formed is :

$\longrightarrow \sf{Cost_{(1 \: table)} = 20 + Cost_{(1 \: chair)} }$

$\longrightarrow \sf{ \dfrac{p}{2} = 20 + \dfrac{q}{3} }$

Let it be equation (ii)

Now, we'll write the value p in the form of q to find the value of q and to solve the equation easily.

Coming to the equation (i) :

$\longrightarrow \bf { p + q = 340}$

$\longrightarrow \bf { p = 340 - q}$

Now, substitute the value of p in the equation (ii).

$\longrightarrow \sf{ \dfrac{p}{2} = 20 + \dfrac{q}{3} }$

$\longrightarrow \sf{ \dfrac{(340-q) }{2} = 20 + \dfrac{q}{3} }$

$\longrightarrow \sf{\dfrac{(340-q) }{2} =\dfrac{60 +q}{3} }$

$\longrightarrow \sf{ 3(340-q) = 2(60 +q) }$

By using distributive property we get,

$\longrightarrow \sf{ 3(340) + 3( -q) = 2(60 ) +2(q) }$

$\longrightarrow \sf{1020 - 3q= 120 + 2q }$

$\longrightarrow \sf{1020 - 120= 2q + 3q}$

$\longrightarrow \sf{900= 5q}$

$\longrightarrow \sf{\dfrac{900}{5}= q}$

$\longrightarrow \boxed{\sf{ 180= q} }$ $\green {\bigstar }$

Is this our answer? No!! It is cost of 3 chairs. We have to find the cost of each chair. That means we have to divide the cost of 3 chairs by 3 to find the cost of each chair. So,

$\longrightarrow \sf{ Cost_{(Each \: chair)}=\dfrac{180}{3}}$

$\longrightarrow \boxed{ \pmb{\rm \red{ Cost_{(Each \: chair)}= Rs. \: 60 }}}$

Also,

From the equation (i) , we have

$\longrightarrow \bf { Cost_{(2 \: tables)} = 340 - q}$

$\longrightarrow \sf { p = 340 - q}$

$\longrightarrow \sf { p = 340 - 180}$

$\longrightarrow \sf { p = 160}$

p is the cost of 2 tables. In order to find the cost of each table, we have to divide the cost of 2 tables by 2.

So,

$\longrightarrow \sf{ Cost_{(Each \: table)}=\dfrac{160}{2}}$

$\longrightarrow \boxed{ \pmb{\rm \red{ Cost_{(Each \: table)}= Rs. \: 80 }} }$

Therefore, cost of each table is Rs. 80 and cost of each chair is Rs. 60.