GATE 2021 Civil Engineering | Question Paper with Solution

APSEd brings you solutions of GATE 2021 Civil Engineering for both forenoon and afternoon session conducted on 6th Feb. This year GATE was conducted by IIT Bombay

In this article, I will share some memory based questions and their solution hints with you. The key highlights and subject wise weightage of paper 1 and paper 2 can be found here.

Special Announcement!

Before you move on to see the solution, we have something for you.

  • Top 5 students of APSEd, which include all our subscribers will be rewarded hugely on the basis of GATE 2021 results.

  • Other than this, all the GATE top 1000 rankers are eligible to get the APSEd interview course for FREE.

Note: A special surprise reward is waiting for the students who get a top 100 AIR.

For any sort of queries, just hit us an email to support@apsed.in

GATE Civil 2021 Forenoon Session

As the questions are memory-based, for most of the questions, it is not sure to say whether the question is MCQ or NAT. But, the question is correct to the best of our knowledge.

We are keeping on updating its article based on memory-based questions. Keep coming back for more. The official key will be out soon on the GATE official website.

Question 1

Two matrices Q and R are given. Find the matrix that represents Transpose(Q) x Transpose(R).

Matrix R = (1, 2, 3, 4) Matrix Q = (0, 1, 1 , 0)


This is a very simple question. There is no trick and formula involved.

  • You know how to find the transpose of a matrix?

  • You know how to multiply two matrices?

Yeah, that is all you need to know for this question.

Question 2

What is the value of the given limit?


  • Step 1: Now the limit is in infinity - infinity form, which is undefined form. We will transfer it to zero by zero form. The 0/0 form can be brought just by subtracting the two bracketed terms simply.

  • Step 2: Put L'Hospital rule till the time it remains in 0/0 form.

The complete solution can be found here-

Question 3

In a confined aquifer of width 30 m, the radius of influence is 245 m, diameter of well is 20 cm, discharge through it is 40 l/sec, the drawdown at the well is 4 m. The hydraulic conductivity of the well will be ___ m/day.


This is just using the formula and getting the answer.

The hydraulic conductivity, k = [Q ln(R/r)] / [2πB(H2 - H1)]

Putting, Q = 40 x 10^-3 m³/sec

R = 245 m, r = 0.20/2 = 0.10 m

B = 30 m, H2 - H1 = 4 m

So, the final answer would be, k = 35.884 m/day

Question 4

For the given aquifer, find the number of days the flow will take to reach downstream (Take, k = 25 m/day and porosity of soil = 0.3)


The head loss, ΔH = 20 m

Seepage velocity, Vs = ki/n = (25 x (20/2000)) / 0.3 = 0.833 m/day

Time = distance/velocity = 2000/0.833 = 2400 days

Question 5

Which of the following options is/are correct? (MSQ)

(A) The back bearing is calculated by fore bearing ±180 degree

(B) The boundary of the calm water pond represents a contour line

(C) The stadia interval increases as we move closer towards the staff

(D) For a WCB of 270 degrees, the reduced bearing is given as 90 NW


Option A: This is correct

Option B: It is correct, as all the points on its surface have the same reduced levels

Option C: The stadia interval decreases as we move closer towards the staff

Option D: This is also correct

Question 6

The shape of the most commonly designed vertical curve is

(A) Spiral

(B) Elliptical

(C) Parabolic

(D) Circular


The parabolic curve is the most commonly designed vertical curve. In the case of valley curves, we design cubic parabolic and in the case of summit curves, we design square parabolic.

Question 7

PQRS is a closed traverse. The internal angles P, Q, R and S are 92, 68, 123 and 77 degrees respectively. The fore bearing of the line PQ is given as 27 degrees. Find the fore bearing of the line RS.


The question is very simple. The concept is here.

FB of PQ = 27, BB of PQ = 180 + 27 = 207

FB of QR = BB of PQ - 68 = 207 - 68 = 139

BB of QR = 139 + 180 = 319

Now, FB of RS = BB of QR - 123 = 319 - 123 = 196 degrees

Question 8

Which of the following statements is incorrect?

(A) The first reading from the station is foresight

(B) Planimeter is used to measure the area

(C) Contour lines can intersect in case of an overhanging cliff

(D) The basic principle of surveying is to work from whole to parts


Option A is incorrect as the first reading is always a backsight(BS).

Option B is correct. A planimeter is used to measure the area of an irregular body.

Option C: Contour lines can intersect in case of an overhanging cliff. This is a property of contour line.

Option D: The basic principle of surveying is to work from part to whole, not whole to part.

Question 9

The triple integration of (8xyz dx dy dz) within (2, 3) x (1, 2) x (0, 1).


The question was from volumetric integration. Just take the limit for x from 2 to 3. Limit of y is from 1 to 2 and the limit of z is from 0 to 1.

Question 10


This is a repeat here. It has been asked many times in GATE.

It is a homogeneous differential equation. If you know how to find the IF, the problem got way for you.

Question 11

The linear traffic model is given by an equation, V = 70 -0.7k. Determine the time headway at maximum traffic volume.


The speed density relation is,

v = Vsf(1-k/kj)

Vsf: free flow speed, k: density, kj: jam density

So, the given relationship can be written as,

V = 70(1-k/(70/0.7))

So, Vsf = 70 and kj = 100

Q = Vsf x Kj = 4 = 1750

Time headway = 3600/Q = 3600/1750 = 2.1 s

Question 12

A signalised intersection operates in two phases. Loss time = 3 seconds per phase. The maximum ratio of approach flow to saturation flow for the two phases are 0.37 and 0.40 respectively.


Q/S are given. We are asked to find the optimum cycle length using the Webster method.

C0 = (1.5L+5)/(1-Y)

C0 = ((1.5 x 6) + 5) / (1-0.37-0.40) = 60.869

Question 13

The spot speeds of a vehicle observed on a point on a highway are 40, 55, 60, 65 and 80 kmph. Space mean speed in kmph is __.


5/V = 1/40 + 1/55 + 1/60 + 1/65 + 1/80

So, V = SMS = 56.999 kmph

Question 14

If water is flowing at the same depth is the most hydraulically efficient triangular and rectangular triangle section. The ratio of the hydraulic radius of the triangular section to that of the rectangular channel section is ___.


We will find the ratio of Area/perimeter of triangle section to that of Area/perimeter of rectangular section.

The required ratio = (Y/2√ 2) / (Y/2) = 1/√ 2

Question 15

For a soil

C = 15 kPa