The hall had 160 seats. After the reconstruction every row had one seat added to them, and the number of rows doubled. How many seats were before and after the reconstruction, if the number of seats increased by 38?

2 Answers
May 27, 2017

Answer:

Solution 2 of 2
Assumption the declared count of seats is the initial count before the change.

#color(red)("Does not work")#

Explanation:

The beginning of the question clearly states a count of seats as being 160. What is not clear is if this is the initial count or the count after the change. If you assume it is the initial count the numbers go wrong.

Let initial count of seats per row be #S_r#
Let the initial count of rows be #R_0#
Assumed that the initial total count of seats be 160

#color(blue)("Initial condition: ")#

#S_rR_0=160" ".................Equation(1)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Change in condition")#

Every row has 1 seat added#" "->(S_r+1)#
The count of rows is doubled#" "->2R_o#
The count of seats increases by 38 #" "->160+38= 198#

Giving:
#(S_r+1)2R_0 =198#

Divide both sides by 2

#(S_r+1)R_0 =99#

#S_rR_0+R_0=99" ".....................Equation(2)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Using #Equation(1)# substitute for #S_rR_0# in #Equation(2)#

#160+R_0=99#

Subtract 160 from both sides

#R_0=-61 larr color(red)(" Negative answer is not logical")#

May 27, 2017

Answer:

Solution 1 of 2
#color(red)("Something wrong with the question")#

Explanation:

Let initial count of seats per row be #S_r#
Let the initial count of rows be #R_0#
Let the initial total count of seats be #T_0#

Assumed that the final total count of seats is 160

#color(blue)("Initial condition: ")#

#S_rR_0=T_0" ".................Equation(1)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Change in condition")#

Every row has 1 seat added#" "->(S_r+1)#
The count of rows is doubled#" "->2R_o#
The count of seats increases by 38 #" "->T_0+38= 160#
Thus #color(brown)(T_0=160-38=122)#

#(S_r+1)2R_0=160#

Divide both sides by 2

#(S_r+1)R_0=80#

#S_rR_0+R_0=80" ".............Equation(2)#
#S_rR_0" "=color(brown)(122)" "....................Equation(1_a)#

#Equation(1_a)-Equation(2)#

#R_0=42#

Substitute for #R_0# in #Equation(1_a)#

#S_r42=122#

#S_r=122/42=2.90.. color(red)(larr" This is not logical")#

You will not have parts of a seat