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X, Y and Z had taken a dinner together. The cost of the meal of Z was 20% more than that of Y and the cost of the meal of X was \( \Large \frac{5}{6} \) as much as the cost of the meal of Z. If Y paid Rs.100, then what was the total amount that all the three of them had paid?

Correct Answer:

A) Rs.285 |
B) Rs.300 |

C) Rs.355 |
D) 320 |

Correct Answer:

D) 320 |

Description for Correct answer:

Given that,

The cost of meal of y = Rs.100

Now, according to the question,

The cost of the meal of z = 20% more than that of y

\( \Large \left(100 + \frac{20}{100} \times 100\right)= \left(100 + 20\right) = Rs.120 \)

and the cost of the meal of x = \( \Large \frac{5}{6} \) as much

as the cost of the meal of z

\( \Large \frac{5}{6} \times 120 \) = Rs.100

Therefore, Total amount that all the three of them has paid

= 100 + 120 + 100 = Rs.320

Given that,

The cost of meal of y = Rs.100

Now, according to the question,

The cost of the meal of z = 20% more than that of y

\( \Large \left(100 + \frac{20}{100} \times 100\right)= \left(100 + 20\right) = Rs.120 \)

and the cost of the meal of x = \( \Large \frac{5}{6} \) as much

as the cost of the meal of z

\( \Large \frac{5}{6} \times 120 \) = Rs.100

Therefore, Total amount that all the three of them has paid

= 100 + 120 + 100 = Rs.320

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