by selling a tea set for Rs4000 A shopkeeper made a profit of 25% how much did he pay for the tea set
Answers
Step-by-step explanation:
Let the cost price of the tea-set and the lemon-set be Rs x and Rs y respectively.
CaseI: When tea set is sold at 5% loss and lemon-set at 15% gain.
Loss in tea-set = Rs.
100
5x
= Rs.
20
x
Gain on lemon-set = Rs.
100
15y
= Rs.
20
3y
∴ Net gain = Rs.
20
3y
−
20
x
⇒
20
3y
−
20
x
=7
⇒3y−x=140
⇒x−3y+140=0 (i)
CaseII: When tea-set is sold at 5% gain and the lemon-set at 10% gain.
Gain on tea-set = Rs.
100
5x
= Rs.
20
x
Gain on lemon-set = Rs.
100
10y
= Rs.
100
y
∴ Total gain = Rs.
20
x
+
10
y
⇒
20
x
+
10
y
=13
⇒x+2y=260
⇒x+2y−260=0 .(ii)
Subtracting equation (ii) from equation (i), we get
−5y+400=0⇒y=80
Putting y=80 in equation (i), we get
x−240+140=0⇒x=100
Hence, the cost prices of tea-set and lemon-set are Rs. 100 and Rs. 80 respectively.
Answer:
Let the cost price of the tea-set and the lemon-set be Rs x and Rs y respectively.
CaseI: When tea set is sold at 5% loss and lemon-set at 15% gain.
Loss in tea-set = Rs.
100
5x
= Rs.
20
x
Gain on lemon-set = Rs.
100
15y
= Rs.
20
3y
∴ Net gain = Rs.
20
3y
−
20
x
⇒
20
3y
−
20
x
=7
⇒3y−x=140
⇒x−3y+140=0 (i)
CaseII: When tea-set is sold at 5% gain and the lemon-set at 10% gain.
Gain on tea-set = Rs.
100
5x
= Rs.
20
x
Gain on lemon-set = Rs.
100
10y
= Rs.
100
y
∴ Total gain = Rs.
20
x
+
10
y
⇒
20
x
+
10
y
=13
⇒x+2y=260
⇒x+2y−260=0 .(ii)
Subtracting equation (ii) from equation (i), we get
−5y+400=0⇒y=80
Putting y=80 in equation (i), we get
x−240+140=0⇒x=100
Hence, the cost prices of tea-set and lemon-set are Rs. 100 and Rs. 80 respectively.
Step-by-step explanation:
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