UGC/CSIR Net level Maths Solution set 1


UGC conducts Jointly with CSIR, the Net exam twice a year for PhD and Lecturership aspirants. For science subjects such as Life Science, Chemical Science etc, the question paper is divided into three parts - Part A, B and C. Part A is on Maths related topics and contains 20 questions any 15 of which have to be answered. The other two parts are on the specific subject chosen by the student.

The questions in Maths are tuned towards judging the problem solving capability of the student using the basic knowledge in maths and not the procedural competence in maths.

The first full set of answer and structured solution of 20 Net level questions follow.


This is a set of 20 questions for practicing UGC/CSIR Net exam: ANS Set 1
Answer any 15 out of 20 questions. Each correct answer will add 2 marks to your score and each wrong answer will deduct 0.5 mark from your score. Total maximum score 30 marks. Time: Proportionate (35 mins).

Q1. In solving a quadratic equation of the form $x^2 + ax + b=0$, one student took the wrong value of $a$ and got the roots as 6 and 2, while another student took the wrong value of $b$ and got the roots as 6 and 1. What are the correct values of $a$ and $b$ respectively?

  1. 7 and 12

  2. 3 and 4

  3. -7 and 12

  4. 8 and 12

Solution: If the two roots of a quadratic equation are $r_{1}$ and $r_{2}$, we have, \[ x^2 + ax + b = (x - r_{1})(x - r{2})\] Thus always we get, $b=r_{1}\times{r_{2}}$. In the first case of the question, the student took wrong value of $a$ and the roots are 6 and 2. As $b$ is independent of other parameters in a quadratic equation, we have correct value of  $b=6\times{2}=12$.

In the second case, the student took wrong value of $b$ and got the roots 6 and 1. The correct value value of $a$ does not depend on $b$ and is, $a = -(r_{1} + r_{2} ) = -(6+1) = -7$.

Answer: Option c: -7 and 12.

Key concepts used: Think from the most basic level. We always have, $a = -(r_{1} + r_{2} )$ and $b=r_{1}\times{r_{2}}$, in a quadratic equation with roots $r_{1}$ and $r_{2}$, \[ x^2 + ax + b = (x - r_{1})(x - r{2})\].

Q2. Suppose you expand the product $(x_{1} + y_{1})(x_{2} + y_{2})....(x_{20} + y_{20})$. How many terms will have only one $x$ and rest $y$’s?

  1. 1

  2. 5

  3. 10

  4. 20

Solution: A polynomial in $x$ and $y$ of degree 2 is, \[a_{0}x^2 + a_{1}xy + a_{2}y^2\]

A polynomial of degree 3 is \[a_{0}x^3 + a_{1}x^2y + a_{2}xy^2 + a_{3}y^3\]

You would notice that, in any of the polynomials above, there is always one term each which is purely in powers of $x$ or $y$. Similarly, there also is one term each where the powers of $x$ (or $y$) is 1 and the rest $y$'s (or, $x$'s). By inductive reasoning this would be true for any polynomial of degree $n$.

Thus in our problem also, the answer is 1.

Answer: option 1: 1

Key concept used: The nature of the powers of $x$ and $y$ in a polynomial in $x$ and $y$.

You have to ignore here the subscripts of $x$ and $y$ for clarity of analysis. 

Q3. $n$ is a natural number. If $n^5$ is odd, which of the following is true?

A. n is odd

B. $n^3$ is odd

C. $n^4$ is even 

  1. A only

  2. B only

  3. C only

  4. A and B only

Solution: Truth: For any odd natural number $n$, any of its powers would also result in an odd number and vice versa.

Answer: A and B both are true. So answer is d.

Q4. In a museum there were old coins with their respective years engraved on them, as follows,

P. 1837 AD

Q. 1907 AD

R. 1947 AD

S. 200 BC

Identify the fake coins, 

  1. Coin P

  2. Coin S

  3. Coins P and Q

  4. Coin R

Solution: From only the year engraved how can the genuineness be ascertained? The only possible way is to examine the year engraved itself. When you examine the years you may be surprised by the engraved year 200 BC. Then you would realize the basic contradiction:

To engrave the year the engraver had to engrave according the calendar available to him or her. The calendar we use stopped backwards at 0 AD. None in the before Christ era was aware that Christ would be born at 0 AD.

Answer: Option $b$ : Coin S : 200 BC.

Key concept used: Context and main action (engraving and calendar use) analysis and identifying contradiction. 

Q5. A vertical pole of length ‘a’ stands at the centre of a horizontal regular hexagonal ground of side ’a’. A rope that is fixed taut in between a vertex on the ground and the tip of the pole has a length,

  1. $a^2$

  2. $\sqrt{2}a$

  3. $\sqrt{3}a$

  4. $\sqrt{6}a$

Solution: The pole height is $a$ and the base is a regular hexagon of side also of length $a$. A regular hexagon has the total of six internal angles = $(6 - 2)\pi=4\times{180^0}=720^0$. So each internal angle of the regular hexagon is $120^0$ and it forms at its centre, six equilateral triangles each of side $a$ (internal angle of a triangle$=\frac{1}{2}\times{120^0}=60^0$). The rope length is the slanted side made by the top of the pole with one of the vertices. This is the hypotenuse of a vertical right triangle with height and base each of length $a$. Thus the rope length is$=\sqrt{a^2 + a^2}=\sqrt{2}a$.

Answer: Option b: $\sqrt{2}a$.

Key concept used: Properties of angles of a regular polygon in 2D - creation of regular shapes at its centre - creation of a vertical right triangle with the pole as the vertical side -- hypotenuse gives the solution: deductive linked reasoning based on geometric concepts.

You need to visualize the relations between the shapes in your mind quickly.

             poleatcentre

Q6. The rabbit population in community A increases at 25% per year while that in B increases at 50% per year. If the present populations of A and B are equal, the ratio of number of rabbits in B to that in A after 2 years will be

  1. 1.44

  2. 1.72

  3. 1.90

  4. 1.25

Solution: After 1 year the population in A and B would be 1.25 and 1.5 of starting population P. At the end of second year, population in A would be, (1.25 + .25 of 1.25) = 1.25$^2$ of P. Similarly, population at B would be 1.5$^2$ of P. Ratio of B to A, \[ \frac{1.5^2}{1.25^2}=\frac{150^2}{125^2}=\frac{6^2}{5^2}=1.44 \]

Answer: Option a: 1.44.

Key concept used: Percentage increase -- simplification -- ratio basics.

Q7. A peacock perched on the top of a 12m high tree spots a snake moving towards its hole at the base of the tree from a distance equal to thrice the height of the tree. The peacock flies towards the snake in a straight line and they both move at the same speed. At what distance from the base of the tree will the peacock catch the snake? 

  1. 16m

  2. 18m

  3. 14m

  4. 12m

Solution: The 12m high tree is the vertical side of a right triangle with its base 36m. As the peacock and the snake moves at the same speed and in straight lines, they meet at a point on the base after travelling equal distances. The following triangle depicts the scenario.

triangleq7

Let the distance travelled be $x$ metre. So $AD=x$ and in right triangle $\triangle ABD$,

             $AB^2 + BD^2$ = $AD^2$

        or, $12^2 + (36 - x)^2$ = $x^2$

        or, $12^2 + 36^2 - 72x + x^2$ = $x^2$

        or, cancelling 12 we get, $12 + 3\times{36} - 6x = 0$

        or, $6x = 120$

        or, $x = 20$

So, $BD = 16m$

Answer: So answer is Option a: 16m.

Key concept used: The most basic concept of Pythagoras theorem. Wherever possible stick to Pythagoras theorem.

Q8. An overweight person runs 4 km everyday as an exercise. After losing 20% of his body weight, if he has to run the same distance in the same time, the energy expenditure would be,

  1. 20% more

  2. The same as earlier

  3. 20% less

Solution: Energy expenditure, \[ E=\frac{1}{2}mv^2 \] Where $m$ is the mass and $v$ is the velocity or speed. Here the speed remains same in two cases. So, energy expenditure $E$ is proportional to mass or weight of the body. Thus, $E_{1}:E_{2}=W_{1}:W_{2}=1:0.8$. So, $E_{2}$ energy expenditure in the second case would also be 20% less.

Answer: Option c: 20% less.

Key concept used: Proportionality in energy formula followed by concept of percentage decrease and ratio handling.

Note: Distance run 4km is superfluous data.

Q9. A solid cube of side L floats on water with 20% of its volume under water. Cubes identical to it are placed one by one on it. Assume that the cubes do not topple or slip and contact between their surfaces is perfect. How many cubes are required to submerge one cube completely?

  1. 4

  2. 5

  3. 3

  4. 6

Solution: 20% or one fifth of the volume of the cube is submerged. We know by Archimedes' Principle, loss of weight is equal to the weight of the fluid displaced by the cube. In other words, one cube displaces one fifth of one cube volume of water. To submerge one cube completely, one complete cube volume of water is to be displaced. This is possible only if five cubes are placed one top of another.

Again there will still be four cubes above the water. That is, there still would be 80% of five cubes above water. To submerge these 5 cubes completely, again 20 more cubes are to be placed on top of each other.

Answer: Option b: 5.

Key concept used: Proportionality from Archimedes' Principle and Unitary method that works only in case of proportional quantities.

Q10. In triangle ABC angle A is larger than angle C, and smaller than angle B by the same amount. If angle B is $67^0$, angle C is

  1. $67^0$

  2. $53^0$

  3. $60^0$

  4. $57^0$

Solution: By the statement of the problem, $B > A > C$ and the difference between them is same. So $A$ is the average of $B$ and $C$ and so, total of the three angles is

$A + B + C = 180^0 =3A$.

Or, $A = 60^0$. Thus,

\begin{align} B + C & = 180^0 - 60^0 = 120^0 \\ & = 67^0 + C . \\ \end{align}

So, $C = 60^0 - 7^0 = 53^0$.

Answer: Option b: $53^0$.

Key concept used: The concept of averages, arithmetic progression and angle property of triangles.

Q11. A physiological disorder X always leads to the disorder Y. However disorder Y may occur by itself. A population shows 4% incidence of disorder Y. Which of the following inference is valid?

  1. 4% of the population suffers from both X and Y.

  2. Less than 4% of the population suffers from X.

  3. At least 4% of the population suffers from X.

  4. At most 4% of the population suffers from X.

Solution: X will always lead to Y amount of second disorder, but the reverse is not a certainty as Y may occur by itself. If we observe an amount A of disorder Y, there may be some amount of Y by itself in it; it may not be wholly by X. By observing an amount A of disorder Y then we can state: at most A portion of population suffers from X.

By this reasoning it can be concluded that option d of the choices given is correct.

Answer. Option d: At most 4% of the population suffers from X.

Key concept used: Logical reasoning, implications and set operation.

Q12. See the following mathematical manipulations,

P. let $X = 5$

Q. let $X^2 – 25 = X – 5$

R. $(X – 5)(X + 5) = X – 5$

S. $X + 5 = 1$ [cancelling X – 5 from both sides]

T. $10 = 1$ [putting X = 5]

Which of the above is the wrong step?

  1. P to Q

  2. Q to R

  3. R to S

  4. S to T

Solution: Upto equation R there is no problem. But from R you can't arrive at equation S because to do that you have to divide both sides by X - 5 which is 0. Division by 0 is prohibited. S would be $0=0$.

Answer: Option c.

Key concept used: Awareness of algebraic manipulation and division by 0 principle.

Q13. What is the angle $\theta$ in the quadrant of a circle shown below?

circleq13

  1. $135^0$

  2. $90^0$

  3. $120^0$

  4. May have any value between $90^0$ and $120^0$

Solution: We place $\theta$ at the mid-point $P$ of the arc $AB$ and draw a third radius $OP$ that bisects both $\angle \theta$ and $\angle APB$.

We do this because of the principle that a chord $AB$ will always subtend the same angle $\theta$ at the periphery (On the opposite portion of the periphery the angle will be different but value constant for all vertex points). In two isosceles triangles, $\triangle OPA$ and $\triangle OPB$ all the base four angles are equal.

Thus,

\begin{align} \angle OAP & =\angle OPA = \angle OPB \\ & = \angle OBP=x \end{align}

This is true because two triangles are isosceles and the radius OP bisects $\angle \theta$.

So in quadrilateral $OAPB$, $4x = 360^0 - 90^0=270^0$. Thus $\angle \theta=2x=135^0$.

circleq132

Answer: Option a: $135^0$.

Key concepts used: Chord angle subtending property -- angle bisection principle (if you bisect an arc, its angle at the centre also gets bisected) -- triangle angle property (total $180^0$) -- quadrilateral angle property (total angle $360^0$).

Note: There is a quicker and shorter solution. Try for it.

Q14. AB is the diameter of semicircle as shown in the diagram. If AQ = 2AP, then which of the following is correct?

circleq14

  1. $\angle APB = \displaystyle\frac{1}{2}\angle AQB$

  2. $\angle APB = 2\angle AQB$

  3. $\angle APB = \angle AQB$

  4. $\angle APB = \displaystyle\frac{1}{4}\angle AQB$

Solution: Diameter is a chord and by chord angle principle diameter AB subtends the same angle of $90^0$ at every point of the semicircle.

The angle subtended at the centre by the diameter is $180^0$ (invisible) which is double the angle subtended at the periphery.

Answer: Option c: $\angle APB = \angle AQB$.

Key concepts used: Arc of a chord that is a diameter is a semicircle -- chord angle subtending property.

Note: AQ = 2AP is superfluous information.

Q15. Which of the following is indicated by the accompanying diagram?

squares15

  1. $a + ab + ab^2 + ... = a / (1 – b)$, for $|b| < 1$

  2. $a > b$ implies $a^3 > b^3$

  3. $(a+ b)^2 = a^2 + 2ab + b^2$

  4. $a > b$ implies $–a < -b$

Solution: There are three squares. The outer square area = (a + b)$^2$. Inner larger square area = a$^2$ and the inner smaller square area = b$^2$. Area of the other two rectangles are same and total = 2ab.

Thus the figure embodies geometrically the algebraic relation, $ (a + b)^2 = a^2 + 2ab + b^2$.

Answer: Option c: $(a+ b)^2 = a^2 + 2ab + b^2$.

Key concepts used: Area of 2D shapes -- algebraic formulae -- domain mapping.

Q16. A string of diameter 1mm is kept on the table in the shape of a close flat spiral, i. e., a spiral with no gap between the turns. The area of the table occupied by the spiral is $1m^2$. Then the length of the string is

  1. $10m$

  2. $10^2m$

  3. $10^3m$

  4. $10^6m$

Solution: At first glance it seems pretty difficult to relate the surface area of a circle and the length of a coil or closed flat spiral shaped rope. But the the spiral can be straightened to a straight line form and its total horizontal area coverage area would be = length $\times$ diameter = L mm $\times{1mm}$ = L sqmm. This equals 1 m$^2$. So, L $\times{10^{-6}} sqm = 1 sqm$, So, $L=10^6 mm=10^3m$.

Answer: Option c: $10^3m$.

Key principle used: Transformation of a spiral into a straight rope -- area covered side by side by the rope across itself is its diameter multiplied by its length. 

Q17. A granite block $2m\times{5m}\times{3m}$ is cut into 5cm thick slabs of $2m\times{5m}$ size. These slabs are laid over a 2m wide pavement. What is the length of the pavement that can be covered with these slabs?

  1. 100m

  2. 200m

  3. 300m

  4. 500m

Solution: The block is a cuboid of base area $2m\times{5m}$ and of height $3m$. The height of $3m$ is sliced into slabs of $5cm$ thickness. Total number of slabs obtained is thus = $\displaystyle\frac{3m}{5cm}=\displaystyle\frac{300}{5}=60$.

These 60 slabs are laid 5m as length and 2m as width. Total length covered would thus be, $L=60\times{5m}=300m$

Answer: Option c: 300m.

Key concepts used: Manipulation of length, breadth and width of solid bodies - 3D manipulation.

Q18. The cities of a country are connected by intercity roads. If a city is directly connected to odd number of other cities, it is called an odd city. If a city is directly connected to an even number of other cities, it is called an even city. Then which of the following is impossible?

  1. There are an even number of odd cities

  2. There are an odd number of odd cities

  3. There are an even number of even cities

  4. There are an odd number of even cities

Solution: As each road connects two cities, connectedness value (number of cities it is connected to) of each of the two cities increase by one. In other words, each road or each edge in a connected graph adds to the total value of connectedness by 2. If there are $n$ edges, total connectedness value would be $2n$. 

If in a network of nodes, the number of nodes with odd number of incident edges is odd, the total of connectedness values of those nodes would be odd which is impossible.

Answer: Option b: There are an odd number of odd cities.

Key concepts used: Concepts of a graph comprising of nodes connected by edges. This is a general and powerful set of concepts and can be used in varieties of situations.

Q19. What is the next number in the “see and tell” sequence?

1, 11, 21, 1211, 111221, ----

  1. 312211

  2. 1112221

  3. 1112222

Solution: In "see and tell" mode, a 11 will result in seeing two 1s and thus would result in: 21. Similarly, a 121 would result in, one 1, one 2 and one 1, resulting in: 111211. So to "see and tell" 111221 would result in, 312211.

Answer: Option a: 312211.

Q20. In the figure below $\angle ABC = \pi/2$. I, II and III are areas of semicircles on the sides opposite to angles $\angle B$, $\angle A$ and $\angle C$ respectively. Which of the following is true always?

halfcircles20

  1. II$^2 +$ III$^2$ = I$^2$

  2. II + III = I

  3. II$^2$ + III$^2 >$ I$^2$

  4. II + III $<$ I

Solution: Let $r_{1}, r_{2}, r_{3}$ be the radii of the semicircles of areas I, II and III. By Pythagoras theorem, $r_{1}^2 = r_{2}^2 + r_{3}^2$ (diameters are proportional to corresponding radii). Multiplying both sides by $\frac{1}{2}\pi$, we get, $\frac{1}{2}\pi r_{1}^2 = \frac{1}{2}\pi r_{2}^2 + \frac{1}{2}\pi r_{3}^2$, or, $I^2 = II^2 + III^2$.

Answer: Option a: II$^2 +$ III$^2$ = I$^2$.

Key concepts used: Relating sides of a right triangle to areas of circles using Pythagoras theorem.