John buys 5 shirts and 3 trousers from a garment store at ₹9000. In the following week, when the store offers a discount of 10% on the same brand of the shirt and a discount of 25% on the same brand of the trouser, Jacob buys 5 shirts and 2 trousers for ₹6300. Find the price at which John buys a trouser.
Answers
★ Given:-
- Cost of 5 shirts and 3 trousers = ₹9000.
- Discount % available on shirts = 10%
- Discount % available on trousers = 25%
- Cost of 5 shirts and 2 trousers after discount = ₹6300.
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★ To Find:-
- Price at which John buys a trouser.
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★ Solution:-
☞ Let the cost of a shirt be x
Hence, Cost of 5 shirts paid by John = 5x
☞ Let the cost of a trouser be y
Hence, Cost of 3 trousers paid by John = 3y
☞ So,According to given conditions;
Cost of 5 shirts and 3 trousers paid by John = ₹9000
➟ Cost of 5 shirt + Cost of 3 trouser= ₹9000
➟ 5x + 3y = ₹9000 ------ eq. 1
☞ Cost of a shirt after discount
= Price - Discount %
= x - 10% of x
= ₹9x / 10
☞ Cost of 5 shirts paid by Jacob
= 5 × (9x/10)
= (5 × 9x) ÷ 10
= 45x ÷ 10
= 9x / 2
☞ Cost of a trouser after discount
= Price - Discount%
= y - 25% of y
= ₹3y / 4
☞ Cost of 2 trousers paid by Jacob
= 2 × (3y/4)
= (2×3y) ÷ 4
= 6y ÷ 4
= ₹3y / 2
☞ So,According to given conditions;
Cost of 5 shirts and 3 trousers paid by Jacob = ₹6300
➟ Cost of 5 shirt + Cost of 2 trouser = ₹6300
➟ (9x/2) + (3y/2) = ₹6300
(Multiplying 2 on both sides)
➟ 2 × ( ( 9x + 3y ) / 2 ) = 2 × ₹6300
➟ 9x + 3y = ₹12600 --------- eq.2
☞ Subtracting eq.1 from eq.2
9x + 3y - (5x + 3y) = ₹12600 - ₹9000
➟ 9x + 3y - 5x - 3y = ₹3600
➟ 9x - 5x + 3y - 3y = ₹3600
➟ 4x = ₹3600
➟ x = ₹3600 ÷ 4
∴ x = Cost of shirt = ₹900
☞ Putting the value of x in eq. 1
5x + 3y = ₹9000
➟ 5 × ₹900 + 3y = ₹9000
➟ ₹4500 + 3y = ₹9000
➟ 3y = ₹9000 - ₹4500
➟ y = ₹4500 ÷ 3
➟ y = ₹1500
∴ y = Cost of trouser = ₹1500
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★ Answer:-
Cost of a trouser paid by John is ₹1500.
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Answer:
The above answer is the bast answer with explanation