A) Rs. 285
B) Rs. 300
C) Rs. 355
D) None of these
Correct Answer: D
Solution :
Given that, The cost of meal of y = Rs. 100 Now, by condition The cost of the meal of z = 20% more than that of \[y=(100+\frac{20}{100}\times 100)=(100+20)\] = Rs. 120 and the cost of the meal of x \[=\frac{5}{6}\]as much as the cost of the meal of \[z=\frac{5}{6}\times 120\] = Rs. 100 \[\therefore \]Total amount that all the three of them has pan: \[=100+120+100=\]Rs. 320You need to login to perform this action.
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