Important word problems on Linear Equations in two variables 6
Question for practice
The following questions are more helpful for class tenth students for all boards.
*Questions 6. A man purchases 5 pen drives and 10 memory cards for Rs. 10,000. He sells the
pen drives at 15% profit and the
memory cards at 10% loss. Thus, he gets Rs. 375 as profit. Find the cost of 1
pen drive and 1
memory card separately.
Submit your answer here in the comments or wait for answer
*HOTS
LET THE COST OF 1 PEN DRIVE BE = x
ReplyDeleteLET THE COST OF 1 MEMORY CARD BE = Y
5x+10y=10,000...(1)
AFTER A PROFIT OF 15% ON PEN DRIVES AND A LOSS OF 10% ON MEMORY CARD
5(100x+15x/100)+10(100y-10y/100)=10375...(2)
SOLVING (1)AND(2) BY ELIMINATION WE GET
x=1100
y=450
THEREFORE THE COST OF 1 PEN DRIVE IS RS 1100 AND THAT OF 1 MEMORY CARD IS RS 450
let the cost of 1 pen drive be x and the cost of 1 memory card be y.
ReplyDelete5x+10y=10000(1)
he got a loss of 10% on memory cards and a profit of 15% on pen drives.
so,
5(100x+15x/100)+10(100y-10y/100)=10375(2)
by elimination we get
x=1100
y=450
we get the cost of 1 pen drive as Rs.1100 and cost of 1 memory card as Rs.450
Ans. Let cost of one pen drive = Rs. x
ReplyDeleteLet cost of one memory card = Rs. y
5x +10y = 10000 .....(1)
profit of 15% on pen drives & loss of 10% on memory cards gives profit of Rs.375
therefore, selling price,
5(100x+15x)/100 + 10(100y-10y)/100 = 10000+375
or 5(115x)/100 + 10(90y)/100 = 10375
or 575x + 900y = 1037500 .....(2)
By elimination from (1) and (2),
we get x = 1100
y = 450
therefore, Cost of one pen drive = Rs.1100
Cost of one memory card = Rs.450
let cost of 1 pen drive = Rs. x
ReplyDeletelet cost of 1 memory card = Rs. y
so, 5x + 10y = 10,000
=> x + 2y = 2,000
=> 23x + 46y = 46,000........(1)
also, 5[(100+15)/100]x + 10 [(100-10)/100]y = 10,000 + 375
=> (23/4)x + 9y = 10,375
=> 23x + 36y = 41,500 ........(2)
eliminating (1) & (2) we get
y = 450 and x = 1,100
THANKS!!!