GATE 2007 Civil Engineering (CE) Question Papers with Answer Key PDFs

GATE 2007 Civil Engineering (CE) Question Paper with Answer key PDFs are available for download. GATE 2007 CE Paper was conducted on February 11, 2007 by the Indian Institute of Technology (IIT) Kanpur. The questions of GATE Civil Engineering (CE) 2007 were from the General Aptitude and core Civil (CE) Engineering section. The total questions in the GATE 2007 CE Paper were 85 out of which the General Aptitude sections comprise 20 questions and the Civil (CE) Engineering section consisted of 65 questions and the total mark of GATE CE is 150. The overall difficulty of the GATE 2007 CE paper was rated as difficult.

GATE 2007 Civil Engineering (CE) Question Paper with Answer Key PDFs

GATE 2007 Civil Engineering (CE) Question Paper	GATE 2007 Civil Engineering (CE) Answer Key
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GATE 2007 Civil Engineering (CE) Detailed Paper Analysis

The detailed paper analysis of GATE Civil Engineering (CE) is mentioned below.

There were questions from two sections which are General Aptitude and Civil Engineering in the GATE 2007 CE paper.
In the GATE 2007 CE paper, there is a negative marking of 0.25 for 1 mark questions and 0.50 for two marks questions.
The total section and weightage of marks in the GATE 2007 CE paper are given below.

Types of Question	General Aptitude		Civil Engineering Core Discipline
Types of Question	Number of Question	Marks	Number of Question	Marks
Number of 1 mark MCQs	10	10	10	10
Number of 2 mark MCQs	10	20	55	110

GATE Previous Year Question Paper

The candidates who are appearing for the GATE examination must solve the previous year’s CE question paper to analyze their overall performance and boost confidence. Moreover, if the candidates solve the question paper it will help them to develop accuracy and speed which will help in the time management of the examination. GATE previous year's question paper link is mentioned below.

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1.
Write a balanced equation for the reaction of molecular nitrogen (N₂) and oxygen (O₂) to form dinitrogen pentoxide. $(f(x)=x^2)$

N₂ + O₂ → N₂O₅ (unbalanced equation)
2N₂ + 5O₂ → 2N₂O₅
option 3

View Solution

2.
In p-type semiconductor density of mobile holes exceeds that of conduction electrons. Hence, minority carriers in p -type semiconductor are conduction (free) electrons.

View Solution

3.
sample text 2

dsfdg
fdgd

View Solution

4.
M. Hamel is introduced as a ruler-wielding teacher. This demonstrates that:

he is adamant.
he is a hard taskmaster.

View Solution

5.
For the reaction, $H_{2} + I_{2} {\rightleftharpoons} 2HI, K= 47.6.$ If the initial number of moles of each reactant and product is 1 mole then at equilibrium

$\left[I_{2}\right]=\left[H_{2}\right], \left[I_{2}\right] > \left[HI\right]$
$({\frac{x^3}{9}})$
$\left[I_{2}\right]>\left[H_{2}\right], \left[I_{2}\right] = \left[HI\right]$
$\omega\propto\,n^{\frac{1}{3}}$

View Solution

6.
$(1+i)^{2}\div i(2i-1)$

View Solution

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Women	900
sc	900
pwd	900

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GATE 2007 Civil Engineering (CE) Question Papers with Answer Key PDFs

GATE 2007 Civil Engineering (CE) Question Paper with Answer Key PDFs

GATE 2007 Civil Engineering (CE) Detailed Paper Analysis

Other PG Exam Question Papers:

GATE Questions

1.
Write a balanced equation for the reaction of molecular nitrogen (N₂) and oxygen (O₂) to form dinitrogen pentoxide. \((f(x)=x^2)\)

2.
In p-type semiconductor density of mobile holes exceeds that of conduction electrons. Hence, minority carriers in p -type semiconductor are conduction (free) electrons.

3.
sample text 2

4.
M. Hamel is introduced as a ruler-wielding teacher. This demonstrates that:

5.
For the reaction, $H_{2} + I_{2} {\rightleftharpoons} 2HI, K= 47.6.$ If the initial number of moles of each reactant and product is 1 mole then at equilibrium

6.
\((1+i)^{2}\div i(2i-1)\)

Fees Structure

Structure based on different categories

Comments

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GATE 2007 Civil Engineering (CE) Question Paper with Answer Key PDFs

GATE 2007 Civil Engineering (CE) Detailed Paper Analysis

Other PG Exam Question Papers:

GATE Questions

1.Write a balanced equation for the reaction of molecular nitrogen (N2) and oxygen (O2) to form dinitrogen pentoxide. \((f(x)=x^2)\)

2.In p-type semiconductor density of mobile holes exceeds that of conduction electrons. Hence, minority carriers in p -type semiconductor are conduction (free) electrons.

3.sample text 2

4.M. Hamel is introduced as a ruler-wielding teacher. This demonstrates that:

5.For the reaction, $H_{2} + I_{2} {\rightleftharpoons} 2HI, K= 47.6.$ If the initial number of moles of each reactant and product is 1 mole then at equilibrium

6.\((1+i)^{2}\div i(2i-1)\)

Fees Structure

Structure based on different categories

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Write a balanced equation for the reaction of molecular nitrogen (N₂) and oxygen (O₂) to form dinitrogen pentoxide. \((f(x)=x^2)\)

2.
In p-type semiconductor density of mobile holes exceeds that of conduction electrons. Hence, minority carriers in p -type semiconductor are conduction (free) electrons.

3.
sample text 2

4.
M. Hamel is introduced as a ruler-wielding teacher. This demonstrates that:

5.
For the reaction, $H_{2} + I_{2} {\rightleftharpoons} 2HI, K= 47.6.$ If the initial number of moles of each reactant and product is 1 mole then at equilibrium

6.
\((1+i)^{2}\div i(2i-1)\)