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Pair of Linear Equations in Two Variables | Pair of Linear Equations in Two Variables - Exercise 3.4

Question 10

A shopkeeper sells a saree at 8% profit and a sweater at 10% discount; thereby getting a sum of ₹1008. If she had sold the saree at 10% profit and the sweater at 8% discount. She would have got ₹1028, then find the cost price of the saree and the list price (price before discount) of the sweater.

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Let the cost price of a saree = ₹x

and the list price of sweater = ₹y

Case I: (S. P. of saree at 8% profit) + (S.P. of a

sweater at 10% discount) = ₹1008

\frac{(100+8)}{100} x+\frac{(100-10)}{100} y=1008 \\ 108 \mathrm{x}+90 \mathrm{y}=100800 \\ 6 \mathrm{x}+5 \mathrm{y}=5600 \ldots(\mathrm{i})

Case II: (S.P. of saree at 10% profit) + (S.P. of a

sweater at 8% discount) = ₹1028

\frac{(100+10)}{100} x+\frac{(100-8)}{100} y=1028 \\ \text { 110x + 92y = } 102800 \text { ...(ii) }

⇒ Dividing

(ii) by 2, we have 55x + 46y = 51400 …(iii)

Again, multiplying (iii) by

5, we get 275x + 230y = 257000 …(iv) Multiplying (i) by 46, we get 276x + 230y

= 257600 …(v) Subtracting (v) from (iv), we get

⇒ –x = –600

⇒ x = 600

Now, 6x + 5y = 5600 (From (i))

⇒ 6 × 600 + 5y =

5600 ( x = 600)

⇒ 5y = 5600 –

3600

y= 2000/5

⇒ y = 400

Hence,

the cost price of a saree and list price of sweater are 600, 400 respectively.

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Our top 5% students will be awarded a special scholarship to Lido.

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Question 10

A shopkeeper sells a saree at 8% profit and a sweater at 10% discount; thereby getting a sum of ₹1008. If she had sold the saree at 10% profit and the sweater at 8% discount. She would have got ₹1028, then find the cost price of the saree and the list price (price before discount) of the sweater.

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

Let the cost price of a saree = ₹x

and the list price of sweater = ₹y

Case I: (S. P. of saree at 8% profit) + (S.P. of a

sweater at 10% discount) = ₹1008

\frac{(100+8)}{100} x+\frac{(100-10)}{100} y=1008 \\ 108 \mathrm{x}+90 \mathrm{y}=100800 \\ 6 \mathrm{x}+5 \mathrm{y}=5600 \ldots(\mathrm{i})

Case II: (S.P. of saree at 10% profit) + (S.P. of a

sweater at 8% discount) = ₹1028

\frac{(100+10)}{100} x+\frac{(100-8)}{100} y=1028 \\ \text { 110x + 92y = } 102800 \text { ...(ii) }

⇒ Dividing

(ii) by 2, we have 55x + 46y = 51400 …(iii)

Again, multiplying (iii) by

5, we get 275x + 230y = 257000 …(iv) Multiplying (i) by 46, we get 276x + 230y

= 257600 …(v) Subtracting (v) from (iv), we get

⇒ –x = –600

⇒ x = 600

Now, 6x + 5y = 5600 (From (i))

⇒ 6 × 600 + 5y =

5600 ( x = 600)

⇒ 5y = 5600 –

3600

y= 2000/5

⇒ y = 400

Hence,

the cost price of a saree and list price of sweater are 600, 400 respectively.

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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