UGC/CSIR Net level Maths Solution Set 4


Twice a year UGC Jointly with CSIR conducts the Net exam 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 ability of the student using basic concepts in maths rather than procedural competence in maths.

The fourth set of answer and structured solution of 15 Net level questions follow. To gain maximum benefits from this resource, the student must first answer the corresponding 15 question set and then only refer to this solution set.


This is a set of 15 questions for practicing UGC/CSIR Net exam: ANS Set 4
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: (25 mins).

Q1. The unit's digit of the product $(693\times{694}\times{695}\times{698})$ is,

  1. 2
  2. 8
  3. 0
  4. 4

Solution: In a product of numbers, the unit's digit is always determined by the unit's digit of the product of unit's digits of the numbers.

To further simplify, if we find a 5 and a multiple of 2 in the unit's digits of the numbers, the final unit's digit will always be 0, as once a 5 is multiplied with a power of 2, it results in unit's digit 0, and any further multiplication stays at 0 value of the unit's digit.

Answer: Option c: 0.

Key concepts used: Unit's digit in product of numbers concept -- use of 0 multiplication concept to cut short solution time.

Q2. In a city average age of men and women are 72.4 years and 67.4 years respectively, while the average of the total number of citizens is 69.4 years. The percentage of men in the city is,

  1. 40
  2. 50
  3. 60
  4. 66.7

Solution: If $m$ and $n$ are number of men and women, $m\times{72.4} + n\times{67.4} = (m + n)\times{69.4}$

Or, $3m = 2n$,

Or, $\displaystyle\frac{m}{n} = \frac{2}{3} = \frac{2x}{3x}$, where $m = 2x$,  $n=3x$ and $x$ is the HCF of $m$ and $n$.

Thus, total number of citizens = $5x$ and the percentage of men is = $\displaystyle\frac{2x}{5x}\times{100}=40$.

Alternatively, $\displaystyle\frac{m}{n} = \frac{2}{3}$

Or, $\displaystyle\frac{n}{m} = \frac{3}{2}$. Adding 1 to both sides,

$\displaystyle\frac{m + n}{m} = \frac{5}{2}$

Or, $\displaystyle\frac{m}{m + n} = \frac{2}{5}$, that is, the percentage of men is 40.

Answer: Option a: 40.

Key concepts used: Average concept -- concept that in a ratio, original quantities can be obtained by multiplying each of the ratio elements by the HCF of the original quantities and this enables to get the total.

Conceptual solution: The positive contribution of men to the combined average age is 3 years and the negative contribution of women is 2 years. It follows, contribution to the total combined age by the men is, $3\times{Men}$ and same by women, $2\times{Women}$ which must nullify each other (as the average is between two values) and so are equal. This results in a ratio of Men : Women as 2 : 3.

With any such ratios, experience guides us to the final result, Men are 40%, Women 60%, ratio 2 : 3 and total 100%. This solution is conceptual, intuitive and hence can be reached quickly.

Q3. Two angles of a triangle are 40% and 60% of the largest angle. The largest angle is,

  1. $120^0$
  2. $90^0$
  3. $80^0$
  4. $100^0$

Solution: Ratios of the two angles to the largest angle in a triangle are 2 : 5 and 3 : 5 giving a ratio of 2 : 5 : 3 between the three angles. If ratio of a : c is p : q and b : c is r : q, we have to equalize the middle value of the ratios to combine three ratios. Inverting second ratio, c : b is q : r, we get, a : c : b = p : q : r.

With this proportion 2 : 5 : 3 between the three angles let's assume the angles as 2x, 5x and 3x where x is the HCF between the values of the three angles. As total angle measure of a triangle is $180^0$, we have, $2x + 5x + 3x = 10x = 180^0$.

Or, $x = 18^0$ giving the largest angle as $5x = 5\times{18^0} = 90^0$.

Answer: Option b: $90^0$.

Key concepts used: Conversion of percentage to ratios -- Combining two ratios -- total value of three angles of a triangle $180^0$ -- each quantity in a ratio can be multiplied by their HCF to get the individual quantities and their total.

This solution is the conventional mathematical solution. Let's see how we can proceed conceptually.

Conceptual solution:

One angle is 40% of the largest and the second 60% of the largest. Together these two angles are equal to the largest (100%), forming a second equal part of the total angle $180^0$. Largest angle then is, half of $180^0$ = $90^0$.

Q4. A bird sitting at the top of a pole spots a centipede 36m away from the base of the pole. If the bird catches the centipede at 16m from the base of the pole both moving at the same speed, the height of the pole is,

  1. 18m
  2. 15m
  3. 12m
  4. $12\sqrt{2}$m

Solution: AB is the vertical side of the right angled triangle ABC with BC as 36m. At point D 16m away from the base, the bird catches the centipede. As the bird and the centipede moves at the same speed for the same time when they meet, AD = DC = 20m.

So, $AB^2 = AD^2 - BD^2 = 20^2 - 16^2 = 144$ and so, $AB = 12m$.

UGC/CSIR Net level Maths Solution set 4 triangle Q4

Answer: Option c: 12m.

Key concepts used: Right angle triangle geometry -- time and speed concept -- Pythagoras theorem.

Q5. In the circle with centre at $O$, $AB$ is a chord. $\angle{ACB} + \angle{OAB}$ =

UGC/CSIR Net level Maths Solution Set 4

  1. $120^0$
  2. $60^0$
  3. $90^0$
  4. $180^0$

Solution: In $\triangle{AOB}$,

$\angle{OAB} = 180^0 - (\angle{AOB} + \angle{ABO})$

Or, $ \angle{OAB} = 180^0 - (2\times{\angle{ACB}} + \angle{ABO})$

Or, $2\angle{ACB} + \angle{ABO} + \angle{OAB} = 180^0$

Or, $2(\angle{ACB} + \angle{OAB}) = 180^0$, as $\angle{ABO} = \angle{OAB}$ in isosceles $\triangle{AOB}$

Or, $\angle{ACB} + \angle{OAB} = 90^0$

Answer: Option c: $90^0$.

Key concepts used: Total of angles in a triangle $180^0$ -- recognizing that two same length radii creates an isosceles triangle -- angle held by a chord at centre is double the angle held by the chord at any point on the periphery on the same side of the chord.

Q6. A car travels uphill to a point at a speed of 20km/hr and returns back to its original position at a speed of 60km/hr. The average speed in which the car covered the whole distance is,

  1. 30km/hr
  2. 40km/hr
  3. 45km/hr
  4. 35km/hr

Solution: Assuming one way distance as d, time taken downhill is, $T_{1} = \displaystyle\frac{d}{60}$ and time taken uphill $T_{2} = \displaystyle\frac{d}{20} = 3T_{1}$

Total time taken $T = 4T_{1} = \displaystyle\frac{d}{60} + \displaystyle\frac{d}{20} = \displaystyle\frac{4d}{60} = \displaystyle\frac{d}{15}$,

Or, $d=15\times{T}$.

Total distance covered being $2d$, the average speed is,

$\displaystyle\frac{2d}{T} = 30km/hr$.

Alternate method: without going into formulas, let's assume distance as the LCM of speeds, that is 60km. So 120km is covered in 3 + 1 = 4 hours at an average speed of 30km/hr.

Note that, in this approach, with distance as any multiple of the LCM 60km, the same result will hold.

Answer: Option a: 30 km/hr.

Key concepts used: Speed time concept -- averaging concept.

Q7. If $5^{\sqrt{x}} + 12^{\sqrt{x}} - 13^{\sqrt{x}} = 0$, then x is,

  1. $\displaystyle\frac{25}{4}$
  2. 4
  3. 9
  4. 6

Solution: Initial examination of the equation reveals the third term higher than the first two and being in negative shifted to the right side of equation thus simplifying the equation giving,

$5^{\sqrt{x}} + 12^{\sqrt{x}} = 13^{\sqrt{x}}$

As there is no apparent relationship between the three terms, trial and error approach should be the best one.

The important point to make here is that the trial values should be for the power, that is $\sqrt{x}$ and not just $x$. Immediately at trial value 2 we reach the answer. Thus, $\sqrt{x} = 2$.

So at $x = 4$, the equation is satisfied.

Answer: Option b: 4.

Key concepts used: Analysis of expression and simplification, wherever possible use the right side of equation to simplify an equation -- trial and error approach -- idea of squares of integers.

Q8. The angle between the hour and minute hands of a clock at 10 past 10 is,

  1. $120^0$
  2. $115^0$
  3. $65^0$
  4. $55^0$

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 10' O clock, the hour hand stood at 10 hour mark and the minute hand stood at 12 hour mark. As time progressed, in 10 minutes the minute hand stood at 2 hour mark (each hour mark represents 5 minutes and 30 degrees angular separation) and the hour hand moved forward by $10\times{30}\div{60}=5^0$ from the 10 hour mark.

So the angle between the two hands at 10 past 10 is, $120^0 - 5^0 = 115^0$.

Answer: Option b : $115^0$.

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

UGC/CSIR Net level Maths Solution Set 4 clock face Q8

Q9. In a river, a boat traveling at a still water speed of 6m/sec overtook a second boat traveling in the same direction and of same length at a still water speed of 4m/sec in 10secs. The length of each boat is,

  1. 100m
  2. 5m
  3. 10m
  4. Cannot be determined

Solution: At first glance it may seem that the silence about the river water flow speed should make the problem unsolvable. But, as the boats are traveling in the same direction, relative speed will be the difference in actual speeds.

Actual speed of each boat will be its still water speed $\pm$ river flow speed. When subtracting the actual speed of the slower boat from that of the faster boat, the river flow speed will cancel out.

Thus the relative speed will turn out to be $6m/sec - 4m/sec = 2m/sec$.

To overtake the slower boat, the faster boat has to cover double its length (both lengths being same). At relative speed 2m/sec in 10secs the faster boat covers 20m. So length is half of 20m, that is, 10m.

Answer: Option c: 10m.

Key concepts used: Boat in river concept -- overtaking concept -- in overtaking, be it train or boat or car, faster moving object has to cover the sum of the two lengths.

For a man or pole, length is taken as zero.

In same direction overtaking or passing, relative speed is the difference in speeds whereas in opposite direction overtaking, speeds are added up to form the relative speed.

Q10. 16 men can do a piece of work in 20 days. How many extra percentage of men will have to be brought in to complete the work in 40% of the original time?

  1. 50%
  2. 250%
  3. 80%
  4. 150%

Solution: Total work amount is 320 man-days. This work is to be finished $\frac{2}{5}$th of 20 days, that is, in 8 days. To do it a total of $\frac{320}{8} = 40$ men will be required. This is an extra 24 men, that is, extra 150% men.

Answer: Option c: d : 150%.

Key concepts used: Man-days concept to measure amount of work in work-day problems -- careful estimation of extra percentage.

Q11. A cube of ice of volume $100cm^3$ floats in water with $\frac{9}{10}$ths of its volme under water. If its portion above water melts at a rate of $10{\%}$ per minute, what will be its approximate volume above water after 3 minutes?

  1. $7cm^3$
  2. $7.29cm^3$
  3. $9.7cm^3$
  4. $7.3cm^3$

Solution: By physical laws of buoyancy in water, whatever be the volume of an ice cube, $10{\%}$ of its volume will float above water. In this case, loss due to melting is $10{\%}$ of $10{\%}$, that is, $1{\%}$ of its volume every minute.

This is a case of compound loss (instead of growth) and the formula applicable is simply,

End volume = $100(1- .01)^3$. As the rate of loss is very small, compounding it will not increase the loss significantly compared to linear loss of $1{\%}$ per minute.

Thus in 3 minutes loss will about $3cm^3$, the total volume will be $97cm^3$ and volume above water will be $9.7cm^3$.

Answer: Option c: $9.7cm^3$.

Key concepts used: Buoyancy -- linear and compound growth/loss -- approximation.

Q12. The minimum value of $2p^2 + 3q^2$, where $p^2 + q^2=1$ and $-1 {\leq} p,q {\leq} 1$ is,

  1. 0
  2. 2
  3. 1
  4. 2.5

Solution: As $p^2 + q^2=1$, $2p^2 + 3q^2=1$ will be transformed to, $2(p^2 + q^2) + q^2 = 2 + q^2.$

As q is in square, its negative values also will increase the value of the expression. So the minimum value of the expression will arise only when $q^2$ = 0.

Answer: Option b: 2.

Key concepts used: Expression simplification -- relation of value of expression with the values of its variables.

Q13. From a bag containing 1 green, 4 blue and 5 red balls, if a ball is picked up blindfolded, what is the probability of picking up a blue ball?

  1. $\frac{1}{4}$
  2. $\frac{2}{5}$
  3. $\frac{1}{10}$
  4. $\frac{2}{3}$

Solution: Probability of an event happening,

$P = \displaystyle\frac{\text{Number of favorable outcomes of the event}}{\text{Number of total possible outcomes of the event}}$.

For example in a coin toss total possible outcomes is 2 and so, probability of getting Head in a coin toss is $\frac{1}{2}$.

Here total possible outcomes is the total number of balls and total number of favorable outcomes is number of blue balls which is 4. Thus the desired probability is, $\frac{4}{10}=\frac{2}{5}$.

Answer: Option b: $\frac{2}{5}$.

Key concepts used: Total and favorable outcomes -- randomness because of blindfolding -- probability concept.

Q14. A rubber mat of thickness 1cm is rolled tightly with no gaps between layers into a solid cylindrical shape that stood on ground with a base area of $1m^2$ and height 1m. The total surface area of the sheet is,

  1. $2.02m^2$
  2. $202.02m^2$
  3. $101.01m^2$
  4. $200.02m^2$

Solution: As the base surface area itself represents the lengthwise thin-edge surface area of the sheet and there are two such thin edges, lengthwise thin-edge areas together is, $2m^2$. Rolling out the sheet, the thin edge surface area is its length multiplied by thickness. So its length = $\frac{1}{.01}m = 100m$.

The height or breadth wise thin-edge surface area = $2\times{1\times{.01}}=.02m^2$ and its broad surface area is two times length by breadth or, $200m^2$. This gives a total surface area as $202.02m^2$.

Answer: Option b: $202.02m^2$.

Key concepts used: Concept of rolling or spiraling and its effect on areas of shapes -- areas of rectangular shapes.

Q15. In a queue of girls Lakshmi stood at the 5th position from the front and 15th position from the end. In another queue girls Veeny stood at the 18th position from the front and 7th from the end. The total number of girls in two queues together is,

  1. 20
  2. 38
  3. 40
  4. 43

Solution: If front and back position counts of a person in row or queue added up, the person is counted twice in the sum. Total number of persons in the queue then = Sum of positions from back and front - 1.

So total number of girls in two queues is, (5 + 15 - 1) + (18 + 7 -1) = 43.

Answer: Option d: 43.

Key concepts used: Ranking with respect to a row or queue analysis.