Question
Vikalp, Nilay and Nilabh and visited a shop to buy LED TV and each one of them bought at least one unit of the TV. The shopkeeper gave a discount of 20% to Vikalp. Nilay, a good bargainer got two successive discounts of 20% and 60% . Nilabh, the hardest bargainer of all got three successive discounts of 20% , 40% and 50% respectively. After selling total of 'x' units to three of them, the shopkeeper calculated that he had given an overall discount of 56% . If
10≤x≤20 , then how many values of 'x' are possible? (Assume every one bought at least one unit).
10≤x≤20 , then how many values of 'x' are possible? (Assume every one bought at least one unit).
Answer: Option B
:
B
Discount toNilay =1−0.8×0.4=68%
Discount to Nilabh =1−0.8×0.6×0.5=76%
Let a,b,c be the number of units sold to Vikalp, Nilay, and Nilabh respectively.
So, 0.2a+0.68b+0.76c=0.56(a+b+c)
So, 5c=3.(3a-b),
So,c is multiple of 3.
If,c=3 then (a,b)= (2,1),(3,4), (4,7)
If, c=6 then (a,b)= (4,2), (5,5), (6,8)
Hence 6 solutions.
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:
B
Discount toNilay =1−0.8×0.4=68%
Discount to Nilabh =1−0.8×0.6×0.5=76%
Let a,b,c be the number of units sold to Vikalp, Nilay, and Nilabh respectively.
So, 0.2a+0.68b+0.76c=0.56(a+b+c)
So, 5c=3.(3a-b),
So,c is multiple of 3.
If,c=3 then (a,b)= (2,1),(3,4), (4,7)
If, c=6 then (a,b)= (4,2), (5,5), (6,8)
Hence 6 solutions.
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