Test Index

MAT Exam 2025 Paper

Question : 87 of 150

Marks: +1, -0

80 \mathrm{~cm} \times 45 \mathrm{~cm} \times 40 \mathrm{~cm}

35 \%

Solution:

Concept:

The oil removed from the tanker equals the rise in oil volume in the storage tank.

Explanation:

Total height of storage tank

= 40\ \mathrm{cm}

35\%

of height

= 0.35 \times 40 = 14\ \mathrm{cm}

Volume of oil in storage tank at this height:

V = 80 \times 45 \times 14 = 50400\ \mathrm{cm}^3 = 50.4\ \mathrm{L}

Existing oil in tank

= 15.4\ \mathrm{L}

Oil transferred from tanker

= 50.4 - 15.4 = 35\ \mathrm{L}

After transferring,

\frac{8}{15}

of tanker is still oil.

So, transferred oil

= \left(1 - \frac{8}{15}\right) = \frac{7}{15}

of the initial oil.

Let initial oil in tanker be

x\ \mathrm{L}

\frac{7}{15}x = 35

x = 35 \times \frac{15}{7} = 75\ \mathrm{L}

Answer:

A. 75 L

Go to Question:

Prev Question

Next Question