Important word problems on Linear Equations in two variables 4
Question for practice
The following questions are more helpful for class tenth students for all boards.
*Questions 4. I buy two tables for Rs. 1350. I sell one so as to lose 6% and the other so as to gain (7.5)%. On the whole I neither lose nor gain. What did each table cost?
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*HOTS
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ReplyDeletelet the CP of 1st table be x and of second be y.
Deleteso, x+y = 1350 rs. (1)
x-6x/100 + y+7.5y/100 = 1350
=> 47x/50 + 43y/100 = 1350
=> 188x + 215y = 270000 (2)
from (1) and (2) , by elimination
x = 750 rs.
therefore, y = 600rs
HENCE, the cost of two tables was rs.750 and rs.600 resp
correct ...
ReplyDelete:)
let the cp of 1 table be x and the cost of second table be y.
ReplyDeleteso,x+y=1350(1)
thererfore,
x-6x/100+y+7.5y/100=1350
47x/50+43y/100=1350
188x+215y=270000(2)
from (1) and (2) by elimination we get
x=750
y=600
therefore,the cost of one table is Rs.750 and the cost of other table is Rs.600
Ans. let cost price of one table = Rs.x
ReplyDeletelet cost price of other table = Rs.y
x+y = 1350 .....(1)
x-6/100 x + y+7.5/100 y =1350
or 100x-6x/100 + 100y+7.5y/100 =1350
or 94x-107.5y = 135000 .....(2)
by elimination from (1) and (2),
y = 600
x = 750
therefore, cost of one table is rs.750
cost of other table is rs.600
let the Cost Prize of 1st table be x and of second table be y.
ReplyDelete(1).... x+y = 1350 rs.
x-6x/100 + y+7.5y/100 = 1350
=> 47x/50 + 43y/100 = 1350
(2)...... 188x + 215y = 270000
by elimination from (1) and (2) ,
x = 750 rs.
therefore y = 600rs
HENCE, the cost of two tables was rs.750 and rs.600
Cost of 1st table = x
ReplyDeleteCost of 2nd table = y
x+y = Rs. 1350 ...i
x-6x/100 + y+75y/1000 = 1350
=> 47x/50 + 43y/100 = 1350
=> 188x + 215y = 270000 ...ii
from i and ii
x = Rs 750
=>y = Rs 600