GPSC (Gujarat) - MCQ Practice Questions

Volume of box = 5 × 4 × 3 = 60 cubic cm. Volume of each 1 cm cube = 1 cubic cm. Number of cubes = 60.

x : y = 3 : 4 and y : z = 8 : 9. To make y the same in both, multiply first by 2: x : y = 6 : 8. So x : y : z = 6 : 8 : 9.

Work completed per day = $\frac{1}{12}$. Work completed in 3 days = 3 × ($\frac{1}{12}$) = $\frac{3}{12}$ = $\frac{1}{4}$.

Total balls = 10 + 15 = 25. Probability of red ball = $\frac{10}{25}$ = $\frac{2}{5}$.

At 3:15, minute hand is at 3 (90°). Hour hand moves 30° per hour + 0.5° per minute = 90° + 7.5° = 97.5°. Angle = 97.5° - 90° = 7.5°

All others are perfect squares of consecutive integers (11², 12², 13², 14², 15²). 256 = 16², breaking the consecutive pattern.

B's father = A's grandfather. Therefore A is B's grandson.

Pattern: 1×2, 2×3, 3×4, 4×5, 5×6, 6×7, 7×8 = 56

Check constraints: P<Q ✓, S<R ✓, R<T ✓. Only option B satisfies all conditions.

Adding all three: 2(A+B+C) = 50, so A+B+C = 25

Each number is multiplied by 3: 8×3=24, 24×3=72, 72×3=216, 216×3=648, 648×3=1944

Work remains constant: 5 × 12 = X × 3. Therefore X = $\frac{60}{3}$ = 20 men

Downstream speed = 15 km/h, Upstream speed = 10 km/h. Current speed = (15-10)/2 = 2.5 km/h

Favorable outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6. Total outcomes = 36. Probability = $\frac{6}{36}$ = $\frac{1}{6}$

Each letter is shifted forward by 1 position in the alphabet. S→T, Q→R, U→V, A→B, R→S, E→F.

Time1 = $\frac{240}{60}$ = 4 hours. Time2 = $\frac{180}{90}$ = 2 hours. Total distance = 420 km, Total time = 6 hours. Average speed = $\frac{420}{6}$ = 70 km/h.

From given conditions: A > B, D > C, B > C. Combining: A > B > C and D > C. So C is definitely the shortest.

From x + 2y = 8, x = 8 - 2y. Substitute in 3x + 5y = 21: 3(8-2y) + 5y = 21. 24 - 6y + 5y = 21. y = 3, x = 2.

Pattern: 1×2, 2×3, 3×4, 4×5, 5×6, 6×7. So the next term is 6×7 = 42.

P(both red) = ($\frac{4}{10}$) × ($\frac{3}{9}$) = $\frac{12}{90}$ = $\frac{2}{15}$.