UGC/CSIR Net level Maths Solution set 5


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 solution and explanation to the fifth set of 18 Net level Math questions follows.


This is a set of 18 questions for practicing for UGC/CSIR Net exam: ANS Set 5
Answer any 15 out of 18 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: 30 mins.

Q1. The units digit of $2^{79} + 7^{35}$ is,

  1. 1
  2. 3
  3. 5
  4. 7

Solution:

Units digit of powers of 2 follow a 4 step cycle of 2, 4, 8 and 6, whereas for 7 it is again a four step cycle of 7, 9, 3 and 1. Power 79 of 2 produces a remainder of 3 when divided by 4 - it is actually the third step in the 20th cycle. Units digit of $2^{79}$ is then the third value of the cycle which is 8.

For power 35 of 7, it is also the third step of a cycle and so units digit of $7^{35}$ is the third value of the 4 step cycle which is 3.

Units digit of a sum of two numbers is the units digit of the sum of units digits of the numbers being summed. Here the sum is 8 + 3 = 11. Thus final units digit is 1.

Answer: Option a: 1.

Key concepts used: Concept of repeating cycle of values of units digits for various powers of a number - concept of units digit value of a sum of two numbers.

Q2. $2^{4x}.4^{10x}=8^{16}$. Find $x$.

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

Solution:

$2^{4x}.4^{10x}=2^{4x}.2^{20x}=2^{24x}=2^{48}$. So $x=2$

Answer: Option b: 2.

Key concepts used: Product of numbers with same base and different powers is base to the power of sum of powers -- Two bases have to be brought to the same value to equate the powers. Same way base 8 on the right is converted to base 2 so that the powers on both sides of the equation can be equated.

Q3. A tank is losing water 50% per hour. The percentage of water left after 2 hour will be,

  1. 50%
  2. 0%
  3. 25%
  4. 75%

Solution: It is a loss in compound rate. Loss per hour is 50% or 0.5 of volume at the start of hour. First hour loss is 0.5 of originial volume and the second hour loss is 0.5 of 0.5 left which is 0.25 of original volume. Total loss is $0.5 + 0.25=0.75$ of original volume. Left will be 0.25 or 25% of original volume.

Alternatively, volume suffering a compound loss at the rate of 50% per hour will be after 2 hours, $(1 - 0.5)^2=0.25$ or 25% of original volume.

Answer: Option c: 25%.

Key concepts used: Compound loss -- conversion of percentage to a decimal.

Q4. Sumi is 4th from the beginning of a row of girls and 20th from the end of a row while Ruma is 10th from the end of the same row and 14th from the beginning of the row. How many girls are between them?

  1. 6
  2. 7
  3. 9
  4. 10

Solution: At first glance we identify the problem as a problem of finding rank or position in a row. The usual concept of finding the length of the row when positions from end and beginning are given is to sum up the positions and subtract 1 (as the person is counted twice in summing up the two positions). By applying this concept, we find that the length of the row is 23. The second set of positions must also produce the same result.

If we go through the problem carefully now, we find that row length was not asked for. Instead separation between two members in the row was wanted. To find it we have to identify their position values from the same point - beginning or end. In this case positions from beginning are 4th and 14th. Subtracting one from the other and reducing by 1 you will get the number of row members between them as 9. (if two positions are 1st and 11th, in a total of 11 persons they are 2, so there are 9 persons between them. Be careful of subtracting 1 after subtracting the lower position from the higher one).

Answer: Option c: 9.

Key concepts used: Rank, position and separation concepts in a row.

Q5. The angle between the hour and the minute hands at 30 minutes past 5 is,

  1. $15^0$
  2. $20^0$
  3. $25^0$
  4. $45^0$

Solution: At 5 the hour hand was at 5 and minute hand at 12. It takes an hour for the hour hand to move from 5 to 6, while in 30 minutes it will move halfway. But in these 30 minutes the minute hand moved to 6 and is ahead by half the angle between two marks, that is, half of $30^0$, that is $15^0$.

Answer: Option a: $15^0$.

Key concepts used: Hour hand moves one hour mark to the next, that is, $30^0$ in an hour while the minute hand moves 12 times more, that is 12 hour marks, that is, $360^0$ in an hour. This difference in speeds, these actual speeds and the concept that at each hour, the hour hand is at the corresponding hour mark and the minute hand is at the zero or 12 hour mark form the basis of all clock sums. The crossing of hands can be dealt with by race concept of using the difference in speeds while holding the slower hand still.

Q6. To cover a distance upstream a boat takes 7 hours while to cover half the distance downstream it takes 2 hours. The ratio of still water boat speed and stream speed is,

  1. 5 : 3
  2. 11 : 3
  3. 11 : 7
  4. 7 : 3

Solution:

$Time = \displaystyle\frac{\text{Distance}}{\text{Speed}}$.

UpstreamĀ  travel time in hours,

$7 = \displaystyle\frac{d}{V_{1} - V_{2}}$.

Downstream, to cover the distance $d$ the boat will take 4 hours. So,

$4=\displaystyle\frac{d}{V_{1} + V_{2}}$, where $V_{1},$ and $V_{2}$ are the still water boat speed and stream speed respectively and $d$ is the distance covered.

Dividing one expression by the other,

$\displaystyle\frac{7}{4} = \displaystyle\frac{V_{1} + V_{2}}{V_{1} - V_{2}}$,

Adding 1 to both sides of the equation we have,

$\displaystyle\frac{11}{4} = \displaystyle\frac{2V_{1}}{V_{1} - V_{2}}$,

And subtracting 1 from both sides of the equation we have,

$\displaystyle\frac{3}{4} = \displaystyle\frac{2V_{2}}{V_{1} - V_{2}}$.

Taking ratio again we finally get, $\displaystyle\frac{V_{1}}{V_{2}} = \displaystyle\frac{11}{3}$.

Answer: Option b : 11 : 3.

Key concepts used: In a river,

Upstream speed = (Still water boat speed - stream speed), and

Downstream speed = (still water boat speed + stream speed.)

Upstream, the boat has to go against the stream and so its effective speed reduces by the stream speed. Conversely, downstream, the stream aids the boat in moving and the speeds are added.

Algebraically, if $\displaystyle\frac{x}{y} = \displaystyle\frac{a + b}{a - b}$, we can always get $\displaystyle\frac{a}{b} = \displaystyle\frac{x + y}{x - y}$, by adding 1, subtracting 1 and taking the ratio. This is rationalization of fractions.

Q7. The radius of a solid cylinder is 8 cm and height 20 cm while the diameter of a second cylinder is 16 cm while the height 5 cm. What is the ratio of volumes of the second to the first cylinder?

  1. 1 : 2
  2. 7 : 22
  3. 1 : 4
  4. 1 : 8

Solution: Diameters of both cylinders are same. As height of the second is one fourth of that of the first, volume ratio is 1 : 4.

Answer: Option c: 1 : 4.

Key concepts used: Careful observation to note diameter is twice the radius -- Volume of a cylinder = ${\pi}r^2h$ where $r$ is the base radius and $h$ the height -- as base radii are same, volume ratio will be in ratio of their heights.

Q8. 4 women can do a work in 8 days while 8 men do the same work in 2 days. In how many days 2 men and 4 women working together will finish the work?

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

Solution: Work amount = 32 womandays = 16 mandays. So 1 manday = 2 womanday. Thus 2 men and 4 women working together is equivalent to 4 + 4 = 8 women working together. It will take then 4 days to finish the work.

Answer: Option a: 4.

Key concepts used: To sum up working together of two or more types agents of different work rates, one needs first to find the per day (or per unit time) amount of work for each agent, sum these and get the amount of work done by all working together in a day.

By unitary method, we would then get the number of days for the agents to finish the work.

This is the way pipe and cistern problems are also handled. This is the standard conventional method.

Mandays concept encapsulates total work amount and also the work rate of a man in a single figure. A work of 16 mandays means 1 man will take 16 days or 4 men will take 4 days or 8 men will take 2 days to finish the work. It is a more powerful concept.

Q9. A cycle 2 m long running at 40m per sec overtakes another cycle of same length running at half the speed of the first. How long would it take to overtake?

  1. 10 secs
  2. 5 secs
  3. 4 secs
  4. 0.2 secs

Solution: Relative speed 20 m per sec. To cover 4 m. Time then will be 0.2 sec.

Answer: Option d: 0.2 sec.

Key concepts used: Overtaking case will have relative speed as difference in speeds - to overtake length to be covered is the total length of the two objects -- time is distance divided by speed.

Q10. Probability of a even digit coming up in a six faced die in a single throw is,

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

Solution: Probability is number of favourable outcomes divided by the total possible outcomes. Here favourable outcomes are, 2, 4 and 6, that is, 3 in number while total outcomes is 6. So probability is $\displaystyle\frac{1}{2}$.

Answer: Option a: $\displaystyle\frac{1}{2}$ .

Key concepts used: Probability concept -- proper counting of number of favourable outcomes.

Q11. 5 men do a work in 30 days. How many more men are required to finish the job in 25 days?

  1. 11
  2. 6
  3. 2
  4. 1

Solution: Work amount = 150 mandays. To finish the job in 25 days 6 men, that is, 1 more man will be required.

Answer: Option d: 1.

Key concepts used: Mandays give the amount of work in terms of work capability of a man -- work amount if divided by the days required you will get the men required directly. Total work 150 mandays, so to finish the work in 10 days 15 men will be required. To finish the work in 15 days 10 men will be required whereas 1 man finishes the work in 150 days.

Q12. A car covers certain distance in 2 hours and 30 minutes. But on a certain day, halfway to the distance it developed a fault and covered the rest of the distance at half its original speed. How much total time it took?

  1. 3 hours 30 minutes
  2. 3 hours 45 minutes
  3. 3 hours 15 minutes
  4. 4 hours

Solution: Careful examination of the problem reveals that in the second case, the car covers half the distance at its previous speed and so it takes half the time, that is, 1 hour 15 minutes. Rest of the distance is half the original distance, but its speed also has reduced to half and so it will take the full time of 2 hours 30 minutes to cover this half distance. Total time is then, 3 hours 45 minutes.

Answer: Option b : 3 hours 45 minutes.

Key concept used: Distance travelled = Time of travel$\times{}$Speed.

If speed is constant, time of travel is directly proportional to distance travelled and vice versa. Again, if time of travel is kept constant, distance travelled is directly proportional to the speed of travel. If you halve the speed the distance you would travel at the same time will be half the previous distance.

Q13. Manish got 65% marks after announcement of score in 3 subjects. When marks of the fourth subject was announced he found his average to increase to 67%. All subjects having same full marks, how much was his score in the fourth subject?

  1. 70%
  2. 72%
  3. 73%
  4. 69%

Solution: To increase the percentage by 2% for four subjects from earlier average of 65% increase over 65% should be $4\times{2}=8{\%}$. He got 73% in the fourth subject. As full marks in all subjects is same, we may well deal with individual subject scores in percentages.

Answer: Option c: 73%.

Key concepts used: Average concept: total of $n$ values divided by $n$. Alternatively, total of $n$ values = $n\times{average}$ -- increment analysis produces quicker results.

Q14. 20 men doing a work can finish a job in 30 days. If after a few days one fourth of the men leave and still they could finish the job in 35 days, for how many days the reduced number of men had to work?

  1. 20 days
  2. 18 days
  3. 15 days
  4. 25 days

Solution: Work is = 600 mandays. After $x$ days 5men left. So work completed in $x$ days was $20x$ mandays and work left = $600 - 20x$ mandays. 15 men did this work in $35-x$ days. So, $600 - 20x = 15(35 - x)$, or, $5x = 75$, or, $x=15$. So for 20 days the reduced number of men had to work together.

Answer: Option a: 20 days.

Key concepts used: Mandays = work amount, initial or left. Assumption of variable as the number of days the 20 men worked together. It could have been number of days reduced number men worked together.

Q15. Chord BD subtends an $\angle{BAD}=70^0$ on the circumference above the chord. Then $\angle{BCD}$ subtended by the chord on the lower circumference below the chord is,

ugc net 5 triangle circle q15

  1. $140^0$
  2. $120^0$
  3. $110^0$
  4. $130^0$

Solution: As $\angle{BAD}$ will be same wherever on the upper circumference A lies, and as $\angle{BCD}$ is same wherever on the lower circumference C lies, let us position A and C at two ends of the diameter AC.

Furthermore, as diameter AC subtends an angle of $90^0$ on the circumference, both the angles CBA and CDA are right angles that totals $180^0$. So, in the quadrilateral ABCD, sum of opposite angles held by a chord BD is always $180^0$. Thus, angle BCD is =$180^0 - 70^0=110^0$.

Answer: Option c: $110^0$ .

Key concepts used: A chord holds the same angle at any point on the upper circumference and also on the lower circumference -- the sum of these two angles is always $180^0$.

Q16. A chord AB of length $3\sqrt{2}$ cm subtends an angle $90^0$ at the centre O of a circle. Area of the sector AOBC is,

ugc net 5 triangle circle q16

  1. $\displaystyle\frac{9\pi}{4} cm^2$
  2. $10 cm^2$
  3. $5\sqrt{2}cm^2$
  4. $3\sqrt{2}cm^2$

Solution: An angle of $2\pi$ at the centre corresponds to the whole area of the circle, that is,

$A={\pi}r^2$.

So an angle $90^0=\displaystyle\frac{\pi}{2}$ at the centre corresponds to a sector area of $\displaystyle\frac{A}{4}$.

Now in right angle triangle AOB, AO=OB=radius r and $\angle{AOB}=90^0$, while hypotenuse AB=$3\sqrt{2}$cm.

So $2r^2=18$,

Or $r=3cm$ and area of the circle $A={\pi}r^2=9{\pi}sqcm$ and

area of the sector is

$\displaystyle\frac{9\pi}{4} cm^2$.

Answer: Option a : $\displaystyle\frac{9\pi}{4} cm^2$.

Key concepts used: Pythagoras theorem to get the radius of the circle and the area -- ratio of area of a sector or arc length of a circle to the whole area or circumference is proportional to the ratio of the angle it holds to $2{\pi}$.

For example if the sector angle is $60^0$, it is one sixth of $2{\pi}$ and so it will have a sector area one sixth of the total area and an arc length also one sixth of total circumference. -- use of unitary method.

Q17. ABCD is a quadrilateral inscribed in a circle with center at O.

ugc net 5 circle q17

If $\angle{COD}=120^0$ and $\angle{BAC}=30^0$, then $\angle{BCD}$ is,

  1. $75^0$
  2. $120^0$
  3. $90^0$
  4. $60^0$

Solution: Chord CD subtends an angle at center $\angle{COD}=120^0$ that is twice the angle $\angle{CAD}$ the chord subtends at the circumference. So $\angle{CAD}=60^0$. As $\angle{BAC}=30^0$, $\angle{BAD}=30^0 + 60^0=90^0$. Thus BOD must be a diameter (as OD is a radius, and only a diameter can subtend an angle of $90^0$ at the circumference). Then the other angle the diameter subtends at the circumference, $\angle{BCD}$ is also $90^0$.

Answer: Option c : $90^0$.

Key concepts used: A chord subtends an angle at the centre twice the angle it subtends at the circumference -- a diameter only can subtend an angle of $90^0$ at the circumference -- a diameter subtends two such right angles on two sides of it.

Q18. A die is rolled. If the outcome is an odd number, the probability that it is a prime is.

  1. $\displaystyle\frac{2}{3}$
  2. $\displaystyle\frac{1}{3}$
  3. $\displaystyle\frac{1}{2}$
  4. $\displaystyle\frac{1}{4}$

Solution: Out of three possible outcomes of 1, 3 and 5, only two are favorable outcomes. So required probability is $\displaystyle\frac{2}{3}$.

Note two points:

First: probability of a particular value coming up out of 6 values is not anymore important here as already event as an odd number has happened. We are to find the probability that the number is a prime out of these 3 possible odd numbers.

The second important point to note that 1 is not considered a prime number by definition.

Answer: Option a : $\displaystyle\frac{2}{3}$.

Key concepts used: Adhering strictly to the probability concept of: probability is number of favorable outcomes of an event divided by the total possible outcomes of the event. As we know that an odd number has come up, event starts from this point on -- definition of a prime number.