UGC/CSIR Net level Maths Solution set 2

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 second 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 2
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. The angles of a right-angled shaped garden are in arithmetic progression and the smallest side is 10.00 m. The total length of the fencing of the garden in m is

  1. 60.00

  2. 47.32

  3. 12.68

  4. 22.68

Solution: Let the three angles in a right angle triangle shaped garden be $A$, $B$ and $C$. If three angles in a right triangle are in arithmetic progression, the middle valued angle must be less than $90^0$ (three times the middle valued angle = $180^0$).

So the angles are, 900 + (900 - x) + (900 - 2x) = 1800,

or, 3x = 900,

or, x=300.

The angles are then, $30^0, 60^0, 90^0$.


The smallest 10m side is opposite to the smallest angle of $30^0$. So, longest side hypotenuse $AC = \displaystyle\frac{AB}{\sin30^0}=20m$ and the base,

$BC = AC\cos30^0$

or, $BC=20\times{\frac{\sqrt{3}}{2}}\cong20\times{.85}\cong17m$.

The perimeter is,

$P=AB + CA + BC$

Or, $P\cong10m + 20m +17\cong47m$

Answer: The value near 47m, that is Choice b: 47.32.

Key concepts used: Average, Triangle angle property, Basics of trigonometry, measurement of perimeter, approximation technique.

Q2. 25% of 25% of a quantity is x% of a quantity where x is

  1. 6.25%

  2. 12.5%

  3. 25%

  4. 50%

Solution: By the statement of the problem assuming the quantity to be $y$, we have,

25% of 25% of y = x% of y,

or, $0.25\times{0.25}y$ = x% of y,

or, 0.0625y = x% of y,

or, 6.25% of y = x% of y.

So, x = 6.25%.

Answer: Choice a: 6.25%.

Key concept used: Percentage, arithmetic simplification.

Q3. Cucumber contains 99% water. Ramesh buys 100kgs of cucumber. After 30 days of storing, the cucumbers lose some water. They now contain 98% water. What is the total weight of the cucumber now?

  1. 99kg

  2. 50kg

  3. 75kg

  4. 2kg

Solution: Let the cucumber without water be $x$ kg. This does not reduce, only the water reduces. First case,

99 kg of water + $x$ kg = 100 kg

or, $x=1$ kg

In second case let $y$ kg be the water left,

Total wt  = $y+1$,

or, $0.98(y + 1) = y$

or, $y = \frac{0.98}{.02} = 49$ kg

Answer: Option b: 50 kg (total weight=49+1)

Key concept used: Percentage, arithmetic simplification, and clarity of representation.

In the second case, though the statement seems to be simple, the equation formulation should be done carefully, as there is some degree of difficulty embedded therein.

Q4. What is the minimum number of days between one Friday the 13th and the next Friday the 13th? (Assume that the year is a leap year).

  1. 58

  2. 56

  3. 91

  4. 84

Solution: Between 13th to next 13th the minimum number of days is 28 and being multiple of 7, that would also be a Friday if the difference between the two dates were 28. This would have been the case only if the starting month were February and the year not a leap year. But the year being a leap year, from February 13th to next month 13th difference is 29 days, one day extra and so February starting point does not give us any advantage towards the minimum number of days. We need to adopt a more general approach.

We know each 30 day month gives 2 days extra after dividing by 7 and 31 day month gives 3 days extra while the 29 day February gives 1 day extra. These extra number of days are called odd days and is the most important concept in calendar sums as this is directly related to weekdays. For example, if odd days between two dates is 3 days and the starting day was a Sunday, the ending date would be Sunday + 3 days, that is, a Wednesday. Thus we need to collect the odd days of each month and go on dividing the sum of odd days by 7 again to always know the resultant number of odd days.

In this problem as February failed as the starting month, in a general approach, we need to collect number of days of each intervening month from the starting month to detect when the sum of odd days becomes 7 with minimum number of intervening months. In other words, the problem is transformed to breaking up 7 into a sum of minimum number of integers that are the odd days of the intervening months.

With known values of odd days of months, this can only be, 7 = 2 + 3 + 2, that is, starting month a 30 day month and next two months a 31 day month and a 30 day month. Thus to have the earliest next 13th the Friday from the starting at 13th the Friday, the number of intervening days would be,

30 + 31 + 30 = 91.

Answer: Choice c: 91 days.

Key concept used: Calendar concepts specifically number of days between dates on the basis of odd days as remainder after dividing a number of days by 7. If number of odd days is 0 or a multiple of 7, after the intervening number of days, we get the same day (Monday or Tuesday etc).

Q5. A 16.2m long wooden log has a uniform diameter of 2m. To what length the log should be cut to obtain a piece of 22m$^3$ volume?

  1. 3.5m

  2. 7.0m

  3. 14.0m

  4. 22.0m

Solution: Area of cross section = $\pi r^2=22/7$ sqm. If target volume is 22m$^3$, the length required would be,

$L = \frac{\text{Volume}}{\text{Area of cross section}}$

or, $L = \frac{22}{\displaystyle\frac{22}{7}}=7m$

Answer: Choice b: 7m.

Key concept used: Relation between area of cross section and volume of a cylinder.

Q6. A bee leaves its hive in the morning and after flying for 30 minutes due south reaches a garden and spends 5 minutes collecting honey. Then it flies for 40 minutes due west and collects honey in another garden for 10 minutes. Then it returns to the hive taking the shortest route. How long was the bee away from its hive? (Assume that the bee flies at constant speed).

  1. 85 minutes

  2. 155 minutes

  3. 135 minutes

  4. less than an hour

Solution: After moving south when the bee flies west and stops at a point, it has traversed two perpendicular sides of a right triangle. To return to the starting point along the shortest path then it would lastly traverse the hypotenuse of the right triangle.

This part of the analysis is the mapping the problem to a geometric problem involving a right triangle.

In this triangle, the bee has traversed the vertical side in 30 mins and the base side in 40 mins both at constant speed. As time is proportional to distance with speed remaining constant (Distance=Speed$\times{\text{Time}}$), the hypotenuse length would be $L = \sqrt{30^2 + 40^2} = 50$ units that would be covered by the bee in 50 mins (because of proportionality, we need not bother to find speed; the distance unit would be distance covered in 1 minute).

So total time = 30 mins + 5 mins + 40 mins + 10 mins + 50 mins = 135 mins.

Answer: Choice c: 135 mins.

Key concept used: Mapping the problem to a geometric problem, -- clear application of proportionality principle, -- Pythagoras theorem.

Q7. What is the last digit of 7$^{73}$?

  1. 7

  2. 9

  3. 3

  4. 1

Solution: You can't calculate the value. So let's try out the unit's digits for values of various powers of 7 that we can work out easily and see the result.

$7^1=7$, $7^2=49$ - unit's digit 9; $7^3=343$ - unit's digit 3; $7^4$ gives unit's digit $7\times{3}$, that is, 1; $7^5$ gives unit's digit $7\times{1}$, that is 7 again. After this if we continue with powers of 6, 7 and 8, the unit's digit will follow the same sequence of 9, 3 and 1.

It means the unit's digit for powers of 7 follows a cyclic pattern sequence of 7, 9, 3, 1 and repeats every 4 values of powers.

Remainder of 73 divided by 4 being 1, the required unit's digit would be 1st number in the sequence 7, 9, 3, 1, that is, 7.

Answer: Choice a: 7.

Key concept used: Analysis of what can be done -- enumeration -- pattern identification -- using the pattern to get the solution.

Q8. A bird perched at the top of a 12m high tree sees a centipede moving towards the base of the tree from a distance equal to twice the height of the tree. The bird flies along a straight line to catch the centipede. If both the bird and the centipede move at the same speed, at what distance from the base of the tree will the centipede be picked up by the bird?

  1. 16m

  2. 9m

  3. 12m

  4. 14m

Solution: The 12m high tree is the vertical side of a right triangle with its base 24m. As the bird and the centipede moves at the same speed and in straight lines, they meet at a point on the base after traveling equal distances. The following triangle depicts the scenario.


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

$AB^2 + BD^2$ = $AD^2$

or, $12^2 + (24 - x)^2$ = $x^2$

or, $12^2 + 24^2 - 48x + x^2$ = $x^2$

or, cancelling 12 we get, $12 + 2\times{24} - 4x$ = 0

or, $4x = 60$

or, $x = 15$

So, $BD = 9m$

Answer: So answer is Option b: 9m.

Key concept used: The most basic concept of Pythagoras theorem.

Wherever possible stick to Pythagoras theorem.

Q9. What is the angle between the minute and hour hands of a clock at 7.35?

  1. $0^0$

  2. $17.5^0$

  3. $19.5^0$

  4. $20^0$

Solution: In a clock the minute hand moves full circle, that is 360$^0$, in 1 hour and the hour hand moves more slowly and moves 360$^0$ in 12 hours. In 1 hour or 60 minutes the hour hand moves $360^0\div{12}=30^0$. This movement is from one hour mark to the next hour mark.


At 7' O clock, the hour hand stood at 7 mark and the minute hand stood at 12 mark. As time progressed, in 35 minutes the minute hand stood at 7 mark (each mark represents 5 minutes) and the hour hand moved forward by $35\times{30}\div{60}=17.5^0$ from the 7 mark.

Answer: Option b : $17.5^0$.

Key concept used: Clock hand movement with relation to speed -- relative movement of the hands -- angular movement -- unitary method.

Q10. A large tank filled with water is to be emptied by removing half of the water present in it every day. After how many days will there be closest to 10% water left in the tank?

  1. 1

  2. 2

  3. 3

  4. 4

Solution: In general: there are two types of pipes in a water tank, one filling and the other emptying the tank. Rates of filling or emptying are specified in terms of "fills in F hours (or empties in E hours)"; questions are asked in terms of in how many hours the two pipes running together will fill up or empty the tank.

Key technique to answer this question is first to get the per hour rates as $\displaystyle\frac{1}{F}$ or $\displaystyle\frac{1}{E}$ and then get the effective per hour rate of filling (or emptying) the tank as $\left(\displaystyle\frac{1}{F} - \displaystyle\frac{1}{E}\right)$. This will give the number of hours to fill the tank as inverse of this rate.

Our problem: Water emptying rate per day is half of the water present previous day end, that is, 0.5 or $\displaystyle\frac{1}{2}$ of the water present. In two days water present would be $\displaystyle\frac{1}{2^2} = \frac{1}{4}$. After 3 days it would be $\displaystyle\frac{1}{2^3} = \frac{1}{8} = 12.5$% of water present and after 4 days it would be $\displaystyle\frac{1}{2^4} = \frac{1}{16} = 6.25$% of water present.

Answer: Option c : 3.

Key concept used: Understanding the problem clearly -- water emptying in a tank - percentage.

Q11. The distance between two oil rigs is 6 km. What will be the distance between these two rigs in maps of 1 : 50000 and 1 : 5000 scales respectively?

  1. 12 cm and 1.2 cm

  2. 2 cm and 12 cm

  3. 120 cm and 12 cm

  4. 12 cm and 120 cm

Solution: In first map, 50000 cm in real world = 1 cm in map, or, 0.5 km in real world = 1 cm in map, or, 6 km in real world = 12 cm in map.

In second map reduction rate is 10 times less, so any distance in real world would be represented in the map by 10 times more separation. For second map then separation is 120 cm.

Answer: Option d : 12 cm and 120 cm.

Key concept used: Map reading concept -- unitary method.

Q12. The capacity of the conical vessel shown below is V. It is filled with water up to half its height. The volume of water in the vessel is,


  1. $\displaystyle\frac{V}{2}$

  2. $\displaystyle\frac{V}{4}$

  3. $\displaystyle\frac{V}{8}$

  4. $\displaystyle\frac{V}{16}$

Solution: Volume of a cone is $\displaystyle\frac{1}{3}\pi r^2h$ where $r$ is the radius of circular base and $h$ the height. Here, water height is half of $h$ and so the radius is also half of $r$.

So, $\frac{V_{water}}{V}=\frac{\displaystyle\frac{r^2}{4}\times{\frac{h}{2}}}{r^2h} = \frac{1}{8}$

or, $V_{water} = \frac{V}{8}$

Answer: Option c: $\displaystyle\frac{V}{8}$.

Key concept used: Volume of cone -- proportionality applied to basic parameters height and radius -- ratio.

Q13. A stream of ants goes from point A to point B and return to A along the same path. All the ants move at a constant speed and from any given point 2 ants pass per second one way. It takes 1 minute for an ant to go from A to B. How many returning ants will an ant meet in its journey from A to B?

  1. 120

  2. 60

  3. 240

  4. 180

Solution: Rate of ant movement moving forward or returning is 2 ants per second. In 1 minute there are 60 seconds. So in forward journey an ant will meet $60\times{2}=120$ ants.

Answer: Option a : 120.

Key concepts used: Flow of objects in an uniform rate -- basic reasoning.

Q14. A king ordered that a golden crown be made for him from 8 kg of gold and 2kg of silver. The goldsmith took away some amount of gold and replaced it with an equal amount of silver and the crown when made weighed 10kgs. Archimedes knew that under water gold lost 1/20th of its weight while silver lost 1/10th. When the crown was weighed under water it was 9.25 kgs. How much gold was stolen by the goldsmith?

  1. 0.5 kg

  2. 1 kg

  3. 2 kg

  4. 3 kg

Solution: Under water gold weighs 95% or 0.95 of its original weight and silver weighs 90% or 0.9 of its original weight. Assuming that the crownmaker replace $x$ kgs of gold with silver, gold weight in the crown underwater is, $0.95(8 - x) = 7.6 - 0.95x$ and silver weight in the crown underwater is, $0.9(2 + x)= 1.8 + 0.9x$. Adding the two we get, $9.4 - 0.05x = 9.25$. This gives us, $0.05x = 0.15$, or $x = 3$ kgs.

Answer: Choice d : 3 kgs.

Key concepts used: Formulation of the problem in its most basic form -- percentage -- algebra.

Q15. A point was chosen at random from a circular disc shown below. What is the probability that the point lies in the sector OAB where Angle AOB = x radians?


  1. $\displaystyle\frac{x}{\pi}$

  2. $\displaystyle\frac{2x}{\pi}$

  3. $\displaystyle\frac{x}{2\pi}$

  4. $\displaystyle\frac{x}{4\pi}$

Solution: The required probability,

$ P = \frac{\text{No. of points in area of x radians at centre}}{\text{No. of points in the disc}}$

Or, $P =\frac{\text{Area of x radians held at the centre}}{\text{Area of disc}}$

Or, $P =\frac{\text{Area of x radians}}{\text{Area of 2$\pi$ radians}} =\frac{x}{2\pi} $

Answer: Choice c: $\displaystyle\frac{x}{2\pi}$.

Key concepts used: Basic concept of probability -- points make up an area -- proportionality of areas and angles in radians in a circular disc.

Q16. A ray of light after getting reflected twice from a hemispherical mirror of radius R emerges parallel to the incident ray. The separation between the original incident ray and final reflected ray is,


  1. $R$

  2. $R\sqrt{2}$

  3. $2R$

  4. $R\sqrt{3}$

Solution: The law of reflection states the the angle of incidence is equal to angle of reflection. Let the two points of incidence be $A$ and $B$.


Angle of incidence at $A$, $\angle SAO$ will be equal to angle of reflection $\angle OAB$. Same at point $B$. The surface of reflection will be the planes passing through the tangents to the points of reflection so that the perpendiculars AO and BO will meet at the centre (perpendicular to a tangent passes through the centre).

Perpendicular OP bisects the chord AB (chord property). If length of AB is L,

$\frac{AP}{AO}=\cos {45^0}=\frac{1}{\sqrt{2}}$

Or, $\frac{\displaystyle\frac{L}{2}}{R}=\frac{1}{\sqrt{2}}$

Or, $L = R\sqrt{2}$

Answer: Choice b: $R\sqrt{2}$.

Key principle used: Laws of reflection -- tangent property in a circle -- chord property -- Trigonometry basics.

Q17. An ant goes from A to C in the figure crawling only on the lines and taking the least length of path. The number of ways it can do so is,


  1. 2

  2. 4

  3. 5

  4. 6

Solution: In the first move, if the ant moves right by one section, there will be three possibilities of path length additional 3 sections. Same happens if it moves down by one section.

Answer: Choice d : 6.

Key concepts used: Identifying similar length units of movement -- taking the first option of movement from A and enumerating all possible minimum length paths -- doing same for the second option. This is enumerative method. Problem being small this would work.

Q18. In the figure below number of circles in the blank rows must be,


  1. 12 and 20

  2. 13 and 20

  3. 13 and 21

  4. 10 and 11

Solution: The number of circles bottom upwards forms a Fibonacci sequence where nth term = sum of previous two terms. It is a famous series that starts with 1, 1 as the first two terms. The terms here are, 1, 1, 2, 3, 5, 8, 13 and 21

Answer: Choice c: 13 and 21.

Key concepts used: Identification of pattern in a series by observing relationship between two consecutive terms.

Q19. In the figure $\angle ABC = \pi/2$ and $AD = DE = EB$. What is the ratio of the area of the $\triangle ADC$ to that of $\triangle CDB$?


  1. 1 : 1

  2. 1 : 2

  3. 1 : 3

  4. 1 : 4

Solution: Area of the triangles having same height are proportional to their base lengths (area of a triangle = half of base multiplied by height). $\triangle ADC$ has half the base length of $\triangle CDB$. So area ratio is 1 : 2.

Answer: Choice b: 1 : 2.

Q20. The area of the shaded region in cm$^2$ is


  1. $(\pi - \sqrt{2})$

  2. $(\pi - 2)$

  3. $\displaystyle\left(\frac{\pi}{4} - \frac{\sqrt{2}}{2}\right)$

  4. $(\pi + 2)$

Solution: Let $r$ be the radius of the circle and $L$ be the side length of the square. Area of the circle is $\pi r^2=4\pi$. Area of square $A = L^2 =(2^2 + 2^2)=8$. Area of shaded region,

$A_{shaded} = \frac{1}{4}\left(\text{Circle - Square}\right)$

Or, $A_{shaded} = \frac{1}{4}\left(4\pi - 8\right) = (\pi - 2)$

Answer: Choice b: $(\pi - 2)$

Key concepts used: Transforming target area to a simple relation between Circle area and the Square area -- deriving square area by using Pythagoras theorem.