A trader sells 10 litres of a mixture of paints A and B, where the amount of B in the mixture does not exceed that of A. The cost of paint A per litre is Rs. 8 more than that of paint B. If the trader sells the entire mixture for Rs. 264 and makes a profit of 10%, then the highest possible cost of paint B, in Rs. per litre, is
26
16
20
22
Answers
Answer:
The correct option is C.
Explanation:
Let the quantities of the paints A and B in the mixture sold be a litres and b litres respectively.
Value at which the entire mixture is sold = Rs. 264
Profit percent made = 10%
Value at which the entire mixture is bought = 264 × 100/110 = Rs. 240
Price at which the entire mixture is bought = Rs. 24 per litre
Let the cost of B be Rs. x per litre.
Cost of A = Rs. (x + 8) per litre
[(x + 8)a + xb]/10 = 24
Maximum cost of B will occur when a is minimum.
Since, b <= a. So, minimum value of a is 5.
Accordingly b = 5.
Then,
(x + 8)(5) + x(5) = 240
10x + 40 = 240
x = 20
Answer:
The correct option is A.
Explanation:
Lakshmi's brother and her daughter are in opposite teams and directly across the net.
Lakshmi's son and worst player's sibling are in opposite teams. Since, they are diagonally across, either Lakshmi's brother or her daughter is worst player's sibling. So, worst player can be either Lakshmi or her son. But, son is diagonally across the worst player's sibling. Thus, Lakshmi is the worst player.
This means that Lakshmi's son and daughter are on the same side. Lakshmi and her brother are on the same side. As the best and the worst player are on the same side, her brother is the best player.
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Explanation: