Free GATE CE mock test — 65 questions, GATE marking 🇮🇳

8 original practice questions written by Myndaq in the GATE CE style. They are not taken from this mock paper or from official past papers. Try each one, then open the answer.

1. Question 1General AptitudeOne correct answer

   A crew of 6 workers completes 40% of a road-marking job in 8 days. Assuming every worker works at the same constant rate, how many ADDITIONAL workers must join the crew so that the remaining work is finished in exactly 6 more days?

   1. A. 4
   2. B. 6
   3. C. 8
   4. D. 12
   Show answer and explanation

   Answer: B

   The crew puts in 6 × 8 = 48 worker-days for 40% of the job, so 1 worker-day does 40/48 = 5/6 % of the job.

   The remaining 60% needs 60 ÷ (5/6) = 72 worker-days. To finish in 6 days the crew must have 72 ÷ 6 = 12 workers, so 12 - 6 = 6 more workers must join.

   12 is the total crew size, not the number of additional workers. 8 would be the total crew if the remaining work were wrongly taken as 40%, and 4 extra workers (10 in all) would take 7.2 days.

2. Question 2Engineering MathematicsType a numeric answer

   Using the trapezoidal rule with a step size h = 0.5, the approximate value of the integral of x³ from x = 0 to x = 2 is ______.

   Show answer and explanation

   Answer: 4.25

   The nodes are x = 0, 0.5, 1, 1.5, 2 with f(x) = x³ values 0, 0.125, 1, 3.375 and 8.

   Trapezoidal rule: I ≈ (h/2)[f0 + 2(f1 + f2 + f3) + f4] = 0.25 × [0 + 2(0.125 + 1 + 3.375) + 8] = 0.25 × (9 + 8) = 0.25 × 17 = 4.25.

   The exact value is 2⁴/4 = 4. The trapezoidal rule overestimates here because x³ is convex on [0, 2]. Simpson's 1/3 rule would give exactly 4, since it integrates cubics without error, but the question asks for the trapezoidal value.

3. Question 3Structural EngineeringType a numeric answer

   A simply supported beam of span 6 m has a rectangular cross-section 300 mm wide and 600 mm deep. It carries a uniformly distributed load of 20 kN/m over the whole span (self-weight included) and a concentrated load of 30 kN at midspan. The maximum bending stress in the beam, in N/mm², is ______.

   Show answer and explanation

   Answer: 7.5

   Both loads produce their largest moment at midspan, so the moments add there.

   M = wL²/8 + PL/4 = 20 × 6²/8 + 30 × 6/4 = 90 + 45 = 135 kN·m = 135 × 10^6 N·mm.

   Section modulus Z = bd²/6 = 300 × 600²/6 = 1.8 × 10^7 mm³.

   Maximum bending stress = M/Z = 135 × 10^6 / (1.8 × 10^7) = 7.5 N/mm². Using wL²/8 alone (5.0 N/mm²) or PL/8 for the point load (6.25 N/mm²) are the usual slips.

4. Question 4Geotechnical EngineeringOne correct answer

   A fully saturated clay sample has a water content of 30% and a specific gravity of solids of 2.65. Taking the unit weight of water as 9.81 kN/m³, the saturated unit weight of the clay is closest to

   1. A. 9.02 kN/m³
   2. B. 14.48 kN/m³
   3. C. 18.83 kN/m³
   4. D. 24.35 kN/m³
   Show answer and explanation

   Answer: C

   For a saturated soil Se = wG with S = 1, so e = 0.30 × 2.65 = 0.795.

   γsat = (G + e)γw / (1 + e) = (2.65 + 0.795) × 9.81 / 1.795 = 3.445 × 9.81 / 1.795 ≈ 18.83 kN/m³.

   14.48 kN/m³ is the dry unit weight Gγw/(1 + e), and 9.02 kN/m³ is the submerged unit weight γsat - γw. 24.35 kN/m³ comes from wrongly taking e = w/G = 0.113.

5. Question 5Water Resources EngineeringType a numeric answer

   A rectangular concrete channel 4 m wide carries uniform flow at a depth of 1.5 m. The bed slope is 0.0009 and Manning's roughness coefficient is 0.015. The discharge in the channel, in m³/s (rounded to two decimal places), is ______.

   Show answer and explanation

   Answer: 10.78 to 10.88

   Flow area A = 4 × 1.5 = 6 m². Wetted perimeter P = 4 + 2 × 1.5 = 7 m. Hydraulic radius R = A/P = 6/7 = 0.857 m.

   Manning's equation: V = (1/n) R^(2/3) S^(1/2) = (1/0.015) × 0.857^(2/3) × √0.0009 = 66.67 × 0.9023 × 0.03 = 1.805 m/s.

   Q = A × V = 6 × 1.805 ≈ 10.83 m³/s.

   A common error is to treat the channel as wide and use R = depth = 1.5 m, which overestimates the discharge (about 15.7 m³/s) because the side walls add a lot of wetted perimeter in a 4 m channel.

6. Question 6Environmental EngineeringOne or more correct answers

   Which of the following statements about water and wastewater treatment are CORRECT?

   1. A. The 5-day BOD of a wastewater sample at 20 °C is always greater than its ultimate carbonaceous BOD.
   2. B. In breakpoint chlorination, a free chlorine residual builds up only after the dose passes the breakpoint, by which point most of the chloramines have been oxidised.
   3. C. In the activated sludge process, a longer mean cell residence time (sludge age) generally lowers the observed biomass yield, because more cell mass is lost to endogenous decay.
   4. D. Adding alum as a coagulant raises the alkalinity and pH of the water being treated.
   Show answer and explanation

   Answer: B, C

   B is correct: below the breakpoint the added chlorine reacts with ammonia to form chloramines and then oxidises them, and only after the breakpoint does extra chlorine remain as free chlorine.

   C is correct: the observed yield is Yobs = Y/(1 + kd θc), so a larger θc (sludge age) gives a smaller Yobs.

   A is wrong: BOD5 = BODu(1 - e^(-kt)) is only a fraction of the ultimate BOD, so it is always smaller. D is wrong: alum hydrolysis releases acidity, which consumes alkalinity and lowers the pH; lime or soda ash is often added to make up for this.

7. Question 7Transportation EngineeringOne or more correct answers

   Which of the following statements are CORRECT?

   1. A. Stopping sight distance on a level road is the sum of the lag (perception-reaction) distance and the braking distance.
   2. B. For a given design speed and superelevation, assuming a higher coefficient of lateral friction increases the minimum radius of a horizontal curve.
   3. C. A rigid pavement spreads wheel loads over a wide area mainly through the flexural (slab) action of the concrete, while a flexible pavement spreads them through grain-to-grain contact in its layers.
   4. D. For a traffic stream, the time mean speed is always greater than or equal to the space mean speed.
   Show answer and explanation

   Answer: A, C, D

   A is correct: SSD = vt + v²/(2gf) on level ground, the first term being the distance covered during the reaction time and the second the braking distance.

   B is wrong: the minimum radius is R = V²/[127(e + f)], so a higher f gives a smaller, not larger, radius.

   C is correct: this is the basic difference in load transfer between rigid and flexible pavements.

   D is correct: time mean speed exceeds space mean speed by roughly the speed variance divided by the space mean speed, so the two are equal only when every vehicle travels at the same speed.

8. Question 8Geomatics EngineeringOne correct answer

   A levelling run starts on a benchmark of reduced level 100.000 m. With the first instrument setup, the backsight on the benchmark is 1.425 m and the foresight on a change point is 0.865 m. The instrument is then moved; the backsight on the change point is 2.140 m and the foresight on point P is 1.975 m. The reduced level of P is

   1. A. 99.275 m
   2. B. 100.725 m
   3. C. 101.590 m
   4. D. 102.700 m
   Show answer and explanation

   Answer: B

   Height of instrument at setup 1 = 100.000 + 1.425 = 101.425 m, so the change point is at 101.425 - 0.865 = 100.560 m.

   Height of instrument at setup 2 = 100.560 + 2.140 = 102.700 m, so RL of P = 102.700 - 1.975 = 100.725 m.

   Check: ΣBS - ΣFS = (1.425 + 2.140) - (0.865 + 1.975) = 0.725 m, and 100.000 + 0.725 = 100.725 m.

   102.700 m is the second height of instrument, 101.590 m ignores the foresight on the change point, and 99.275 m applies the rise as a fall.