Quantitative Aptitude Test - 1

Mean \(=\) Sum of data \(\div\) Cumulative Frequency 

Sum \(=1 \times 5+2 \times 4+3 \times 3+4 \times 2+5 \times 1=35\)

Cumulative Frequency \(=5+4+3+2+1\)

\(\therefore\) Mean \(=\frac{35}{15}=2 \frac{1}{3}\)

(1!)² = 1 (1! = 1)
(2!)² = 2² = 4 (2! = 1.2 = 2)
(3!)² = 6² = 36 (3! = 1.2.3 = 6)
(4!)² = 24² = 576 (4! = 1.2.3.4 = 24)
Last digit = Last digit of 1 + 4 + 6 + 6 = 7.

Part of the tank filled by pipe \(A\) in 1 minute \(=\frac{1}{30}\) 

Part of the tank filled by pipe \(\mathrm{B}\) in 1 minute \(=\frac{1}{20}\) 

Part of the tank filled by pipe \(\mathrm{C}\) in 1 minute \(=\frac{1}{10}\)

Here we have to find the proportion of the solution \(\mathrm{R}\).

Pipe C discharges chemical solution \(\mathrm{R}\).

Part of the tank filled by pipe \(\mathrm{C}\) in 3 minutes \(=3 \times \frac{1}{10}=\frac{3}{10}\)

Part of the tank filled by pipe \(\mathrm{A}, \mathrm{B}, C\) together in 1 minute \(=\frac{1}{30}+\frac{1}{20}+\frac{1}{10}=\frac{11}{60}\)

Part of the tank filled by pipe \(A, B, C\) together in 3 minute \(=3 \times \frac{11}{60}=\frac{11}{20}\)

Required proportion \(=\frac{\left(\frac{3}{10}\right)}{\left(\frac{11}{20}\right)}=\frac{3 \times 20}{10 \times 11}=\frac{6}{11}\)

Let the radius of S₄ be r cm.
Volume of S₁ = 4/3 π(3³)
Volume of S₂ = 4/3 π(4³)
Volume of S₃ = 4/3 π(5³)
Volume of S₄ = 4/3 πr³ = 4/3(3³) + 4/3 π(4³) + 4/3 π(5³).
4/3π(3³ + 4³ + 5³) = 4/5πr³
r³ = 216 => r = 6
Total surface areas of S₄ = 4πr² = 4π(6²) = 144πsq.cm

Number of cubes formed