UGC/CSIR Net level Maths Solution Set 3

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 third set of answer and structured solution of 15 Net level questions follow.

This is a set of 15 questions for practicing UGC/CSIR Net exam: ANS Set 3
Answer all 15 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. Rainfall is measured in mm. One day there was a rainfall of 4mm at Kolkata. To collect the rainwater a hemispherical glass bowl of radius 1cm was kept in an open area. How much percent of the bowl was filled up by rainwater?

  1. 40
  2. 50
  3. 60
  4. 75

Solution: 4mm rainfall means if the rainfall is collected in a flat bottom container with straight rising sides, the depth of water collected will be 4mm. In other words, rainfall per unit horizontal area is of 4mm column height.

In our case the collecting surface area is $A={\pi}10^2=100{\pi}$ sq mm. The volume of water will be, $V_{1}=A\times{4mm}=400{\pi}$ cu mm. The volume of the hemipherical bowl is $V_{2}=\frac{2}{3}{\pi}10^3$ cu mm. So, $V_{1} : V_{2} = \frac{400{\pi}\times{3}}{2\times{\pi}\times{1000}}=3 : 5=60{\%}$.

Answer: Option c: $60{\%}$.

Key concept used: Rainfall concept -- visualizing of the rainfall being collected in a cylindrical vessel of radius of mouth 1cm -- ratio of the volume of cylindrical column of water 4mm deep to the volume of the hemisphere.

Q2. Two friends decided to start a race on a bright sunny day. Friend A with usual speed of 8m/sec ran faster than his friend B, whose usual speed was 6m/sec and so it was agreed that the race would be run with a handicap of 40m to A. How much distance would B cover before A overtook him?

  1. 120m
  2. 160m
  3. 80m
  4. 100m

Solution: When the speedier man starts the race 40m behind the slower man, he will catch up when he covers the intial gap running at a relative speed of 2m/sec which he will do in 20sec. In this time B will cover a distance of 120m.

Answer: Option a: 120m. 

Key concept used: Concept of race -- make the slower man standstill and correspondingly reduce the speed of speedier man to their relative speed -- calculate how much time he takes to cover the initial gap at this relative speed. This is the time taken to catch up. In reality, running at their original speeds, the slower man will cover 120m distance in this 20secs time.

This is the easiest way to deal with races. Races may be between various objects moving at different speeds. It may even be between the hour hand and the minute hand of a clock.

Q3. One day Minti started down in a lift from the 10th floor of a building at the same time as her friend Pinki started up from the ground floor in another lift. They had decided to meet in between rather than wait longer. If Minti's lift covered 3 floors by the time Pinki's covered 2 floors and if they knew it, on which floor will they meet wasting minimum time?

  1. 3
  2. 4
  3. 5
  4. 2

Solution: The two lifts will cover a total of 10 floors at a relative speed of 5 floors per unit time embodied in the ratio. It would take two unit time to cover 10 floors and meet. In this time Pinki's lift will be on 4th floor and Minti's lift will come down 6 floors to meet her.

Answer: Option b: 4th floor.

Key concept used: Relative speed of two moving objects approaching each other is the summation of their individual speeds -- At this speed they will cover the distance between them when they meet -- From ratios of two quantities, their sum can obtained as the sum of the two ratio numbers in terms of ratio unit, actual quantities are not required to be known.

Q4. At the end of a year in a business, profit decreased by 15% with cost price remaining the same. If the previous year profit was 20% what is the percent decrease in sale price this year?

  1. 12.5%
  2. 15%
  3. 10%
  4. 20%

Solution: Profit is always expressed as a percent of cost price or investment. Assuming the cost price as 100 with last year profit as 20%, last year sale price was 120. Profit decreasing by 15%, this year sale price decreased to 105 a decrease of 15 from 120, which is a fall of one eighth or 12.5%.

Answer: Option a: 12.5%

Key concept used: Clear idea of profit expressed always as a percent of cost price or investment -- but when decrease of sale price was wanted it was on the basis of the original sale price as the basis, not the cost price obviously.

In percentage sums, three things are important. The base of percentage expressed is to be carefully identified and secondly, the percentage is always equivalent to a number uaually a decimal being divided by 100. For example, 20% profit means, this profit is on the base of cost or investment price, that is, 20% of the cost price or one fifth of the cost price.

Thirdly, In absence of actual figures of the quantities, the base quantity of the first percentage expressed can always be taken as 100. All other values will be in terms of this base value.

Q5. $\left(\frac{1}{5}\right)^{3x} = 0.008.$ The value of $(0.25)^x$ is,

  1. 22.5
  2. 0.25
  3. 0.0625
  4. 0.5

Solution: $\left(\frac{1}{5}\right)^{3x} = 0.008 = \frac{8}{1000} = \frac{2^3}{10^3}=\left(\frac{2}{10}\right)^3=\left(\frac{1}{5}\right)^3$.

So $x=1$, and $(0.25)^x=0.25$.

Answer: Option b: 0.25.

Key concept used: Concept of indices -- RHS analysis to make base $\frac{1}{5}$ equal on both sides so that we get a value of $x$.

Use of base equalization technique in most of the indices sums solve the problem.

Q6. Five boys are sitting in a row. Pintu is to the left of Khokon. Kalua is to the right of Apu. Sunny is in the middle of Kalua and Pintu. Who is to the extreme right of the row?

  1. Khokon
  2. Apu
  3. Pintu
  4. Kalua

Solution: Examining the last statement we find it carries maximum information by forming a trio of Kalua -- Sunny -- Pintu where Kalua may be on the right or left Sunny. In first statement Pintu appears to the left of Khokon. In this case, Khokon should be on the right of the trio, that is on the right of Pintu.

As Kalua is on the right of Apu from the second statement, Apu has to be on the left of the trio. Thus, Khokon is left to be the rightmost. Notice, Kalua may be on the right or left of the trio, but both will satisfy the conditions.

The third statement forms the trio -- the first statement puts Khokon on the right of the trio and the second statement puts Apu on the left of the trio.

Alternatively, from first two statements, Khokon and Kalua are the initial candidates to be the rightmost. As Khokon has to be on the right of Pintu from the first statement, the whole of the trio including Kalua has to be on the left of Khokon from the third statement. So Khokon is the rightmost. The relative position of Khokon and Pintu is transferred to the relative position of Khokon and the trio and hence to Khokon and Kalua. This is a better reasoning as it focuses on the end requirement all along.

This approach is another example of applying End State Analysis Approach.

Answer: Option a: Khokon.

Key concept used: Reasoning with respect to relative position in a row -- end requirement targetted analysis.

Q7. A can do a work in 3 days while B can do the same work in in 4 days. If they work together for a total wage of Rs. 2800 how much does A get?

  1. Rs. 1200
  2. Rs. 1900
  3. Rs. 1300
  4. Rs. 1600

Solution: In work wage problem wage earned is in proportion of amount of work done in a day. As A does work faster than B he should get more wage. In 1 day A does $\frac{1}{3}rd$ of the total work and B does $\frac{1}{4}th$ of the total work. Their daily wage will be in proportion of their amount of work in a day, that is, in ratio $\frac{1}{3} : \frac{1}{4} = 4 : 3$, it will be the inverse of number of days in which they do the total work. So total wage unit is 7 which is Rs. 2800. Thus 1 wage unit is Rs. 400 and A will get 4 units that is Rs. 1600.

Answer: Option d: Rs. 1600.

Key concept used: Wage work concept: the more one works the more wage he will get -- wage of a person is fixed in rupees per day -- wage will be in direct proportion of work amount they do per day or in inverse proportion of number of days they take to do same amount of work.

Q8. Remainder of $7^{57}$ divided by 10 is,

  1. 7
  2. 9
  3. 3
  4. 1

Solution: Remainder after division by 10 is the unit's digit. Powers of 7 follow a 4 digit cycle of 7, 9, 3, 1. As 57 modulo 4 is 1 (remainder after dividing by 4), the answer is 7.

Answer: Option a: 7.

Key concept used: Place value system -- remainder concept -- unit's digit cycle of powers.

Q9. In $\triangle{PQR}$, S and T are two points on the sides PR and PQ respectively. $\angle{PQR}=\angle{PST}$. If PT = 5 cm, PS = 3 cm and TQ= 3 cm, the length of line SR is,


  1. $\frac{31}{3}$ cm
  2. 5 Cm
  3. 3 cm
  4. $\frac{41}{3}$ cm

Solution: In triangles PQR and PST, the three angles are same. $\angle{PQR}=\angle{PST}$, $\angle{QPR}=\angle{TPS}$, and so the third pair of angles $\angle{PRQ}=\angle{PTS}$. Thus these two are similar triangles. Thus ratios of corresponding pair of sides are same which leads to $\frac{PR}{PT}=\frac{PQ}{PS}$, or, $PR=\frac{8}{3}\times{5}=\frac{40}{3}$.

So, $SR = PR - PS = \frac{31}{3}$.

Answer: Option a: $\frac{31}{3}$.

Key concept used: Two triangles are similar if three pairs of angles are equal - similar triangle property : ratio of all three pairs of corresponding sides of the two triangles are equal -- rule of identifying corresponding sides in two similar triangles ABC and PQR, where angles A, B and C are equal to angles P, Q and R respectively : take the side opposite to angle A in triangle ABC as numerator of first ratio and the side opposite to its equal angle P in triangle PQR as denominator. This forms the first ratio. Similarly the second and third ratios are formed. All these three ratios will be equal to each other.

Q10. Growth of a type of fungus was monitored at regular intervals of time and is shown in the graph below. Around which time is the rate of growth zero?

Fungus mass

  1. Near day 1
  2. On day 3
  3. Between days 3 and 5
  4. Between days 5 and 7

Solution: The fungus mass is plotted on y-axis and the slope of the curve at any point gives its rate of growth. The slope at any point is the angle the tangent at the point makes with the x-axis. The more vertical the tangent is more is the rate of growth. Between days 3 and 5 the slope is zero and so growth also is zero.

Answer: Option c: Between days 3 and 5.

Key concept used: Growth is change per unit time and thus in a growth curve rate of growth is given by the slope of the tangent at the point under consideration -- in any inverted bell shaped curve growth is zero at its maximum value, that is, at the top point where the tangent of horizontal.

Q11. There are seven sisters in a house in a village where there is no electricity or any gadget. The activities of the seven sisters in one day are,

  • Sister-1: Reading Novel
  • Sister-2: Cooking
  • Sister-3: Playing Chess
  • Sister-4: Playing Sudoku
  • Sister-5: Washing clothes
  • Sister-6: Gardening

What is Sister-7 doing?

  1. Sleeping
  2. Taking bath
  3. Out marketing
  4. Playing chess

Solution: She could have been doing umpteen number of things, but as there is no gadget or electricity and as chess usually can't be played alone, she should be playing chess with her other sister.

Answer: Option d: Playing chess.

Key concept used: Analyzing given facts for any clue rather than leaving the problem altogether as possibilities are uncertain.

Q12. How many squares are there in the following figure?


  1. 13
  2. 10
  3. 4
  4. 5

Solution: The four smaller squares and the larger square, in total 5. As the other sides are not completed, the squares are not formed.

Answer: Option c: 5.

Key concept used: Observation -- square detection.

Q13. In a bicycle wheel if there were 10 more spokes, the angle between them would have reduced by six degrees. Can you find out the number of spokes in the wheel?

  1. 10
  2. 20
  3. 30
  4. 40

Solution: If $n$ be the number of spokes now, then $\frac{360}{n+10}=\frac{360}{n} - 6=\frac{360 - 6n}{n}$,

Or, $360n = 360n + 3600 - 6n^2 -60n$, Or, $6n^2 + 60n = 3600$, or, $n(n+10)=600.$

n = 20 satisfies this condition.

Answer: Option b: 20.

Key concept used: Formation of algebraic relationship -- simplification -- trial and error.

Alternatively, $n(n+10)=600$, or, $n^2 +10n - 600=0$, or, $(n + 30)(n - 20) = 0$.

As $n$ can't be negative -30, it must be 20.

Q14. How many numbers are there from 20 to 40 that have no factor among the numbers from 2 to 10?

  1. 3
  2. 4
  3. 5
  4. 6

Solution: It is basically equivalent to the question, "how many prime numbers are there between 20 to 40?." the number is 4: 23, 29, 31 and 37.

Answer: Option b: 4.

Key concept used: Factor -- prime number - divisibility.

Q15. If $2^{4x} + 2^{5x} = 1280$, then x is,

  1. 4
  2. 5
  3. 3
  4. 2

Solution: There are two ways to solve the problem. You can test the value of expression with choice values of x. Actually x will be 2 and if you have a good estimate of powers of 2, you may get this answer quickly.

But the more elegant way is to simplify the expression. $2^{4x} + 2^{5x} = 2^{4x}(1 + 2^{x})=1280=2^{8}\times{5}$. Thus, both of the equalities $2^{4x}=2^{8}$ and $(1 + 2^{x})=5$ must be true, both of which give x=2.

Answer: Option d: 2.

Key concept used: Algebraic simplification -- factorization and factor equality.