Mathematical Reasoning and Aptitude Questions and Answers

Q: In a certain cricket tournament 45 matches were played. Each team played once against each of the other teams. The number of teams participated in the tournament is

The number of teams was 10.If each of n teams plays every other team once, the number of matches is the number of ways of choosing 2 teams from n.\[ \frac{n(n-1)}{2} = 45 \]\[ n(n-1) = 90 = 10 \times 9 \]So n = 10. Check with the others: 8 teams give 28 matches, 12 teams give 66 and 14 teams give 91.Another way to see it: every team plays 9 others, 10 teams times 9 matches is 90 team-appearances, and each match involves two teams, so 45 matches in all.

Q: The missing number in the series 40, 120, 60, 180, 90, ?, 135 is

The missing number is 270.The series alternates between multiplying by 3 and dividing by 2: 40 times 3 gives 120, then 120 divided by 2 gives 60, 60 times 3 gives 180, and 180 divided by 2 gives 90.\[ 90 \times 3 = 270 \]\[ 270 \div 2 = 135 \]The last term, 135, confirms that the missing term is 270. The values 110, 105 and 210 break the pattern.Any alternate-operation series is solved by writing the ratios between successive terms, and here they are 3, one half, 3 and one half in turn.

Q: The odd numbers from 1 to 45 which are exactly divisible by 3 are arranged in an ascending order. The number at 6th position is

The number at the 6th position is 33.The odd numbers divisible by 3 between 1 and 45 are the odd multiples of 3, which go up in steps of 6.\[ 3,\ 9,\ 15,\ 21,\ 27,\ 33,\ 39,\ 45 \]The sixth number in this ascending list is 33. The numbers 18, 24 and 36 are even, so they are not in the list.A quick check confirms the position: the first odd multiple of 3 is 3 and each next one is 6 more, so the sixth is 3 plus 5 times 6, which is 33.

Q: The mean of four numbers a, b, c, d is 100. If c = 70, then the mean of the remaining numbers is

The mean of the remaining three numbers is 110.The four numbers have a mean of 100, so their sum is 400. Taking out c, which is 70, leaves the sum of the other three.\[ a + b + d = 400 - 70 = 330 \]\[ \text{mean} = \frac{330}{3} = 110 \]The values 30, 85/2 and 170/3 come from dividing by the wrong number of items or subtracting wrongly.A faster check is to note that the three remaining numbers must make up 330 out of the total of 400, so their average is a little above the overall mean of 100 and clearly cannot be 30.

Q: If the radius of a circle is increased by 50%, the perimeter of the circle will increase by

The perimeter will increase by 50%.The perimeter of a circle is 2 pi r, which is directly proportional to the radius. If r grows by 50%, the new radius is 1.5 r.\[ 2\pi (1.5r) = 1.5 \times (2\pi r) \]The perimeter becomes 1.5 times the old perimeter, an increase of 50%. The area, by contrast, would grow by 125%, which is why a different option is not chosen.This result holds for any circle, whatever the starting radius, because the perimeter depends on the radius to the first power only.

Q: Choose the proper alternative given in the codes to replace the question mark. Bee - Honey, Cow - Milk, Teacher - ?

The key completes the analogy with wisdom.A bee produces honey and a cow gives milk, so each item on the left is paired with the valuable product it yields. By the same relation, a teacher yields wisdom, the lasting result of teaching.Intelligence is an inborn capacity a teacher does not produce. Marks are given by an examiner or teacher as an assessment, and lessons are what is delivered, not the product, so these do not fit the relation as well.

Q: P is the father of R and S is the son of Q and T is the brother of P. If R is the sister of S, how is Q related to T ?

Q is the sister-in-law of T.R and S are sister and brother. P is the father of R, so P is also the father of S. S is the son of Q, so Q is S's mother and therefore P's wife. T is the brother of P, so Q is the wife of T's brother, which makes her T's sister-in-law.She is not T's wife, since she is married to P. She is not a brother-in-law, because Q is a woman, nor a daughter-in-law, since she is not T's son's wife.

Q: The world population growth rate at a certain reference year was 3.5%. Assuming exponential growth of population, after how many years, the population of the world would have increased by a factor 16 ?

The population grows by a factor of 16 in about 80 years.With exponential growth \(P = P_{0}e^{rt}\) and a factor of 16, we need \(e^{rt} = 16\):\[ t = \frac{\ln 16}{r} = \frac{2.77}{0.035} \]\[ t \approx 79 \text{ years} \approx 80 \text{ years} \]Equivalently, 16 is four doublings, and the doubling time at 3.5 per cent is about 20 years (70 divided by 3.5), which gives 80 years. The values 40, 160 and 320 years correspond to two, eight and sixteen doublings and do not give a factor of 16.

Q: A person writes all the numbers from 0 to 99. The number of times digit 3 will be written is

The digit 3 is written 20 times from 0 to 99.Count the units place and the tens place separately.\[ \text{Units digit 3: } 3, 13, \ldots, 93 = 10 \]\[ \text{Tens digit 3: } 30 \text{ to } 39 = 10 \]\[ 10 + 10 = 20 \]The number 33 is counted in both lists because it carries two 3s. The values 18, 19 and 21 come from leaving out or double counting a number.

Q: Starting from point A, Ajit walks 14 metres towards west, he then turns to his right and walks 14 metres and then turns to his left and walks 10 metres. He again turns to his left and walks 14 metres and reaches to the point E. The shortest distance between A and E is

The shortest distance between A and E is 24 metres.Place A at the origin and take east as positive x and north as positive y. Facing west, a right turn points north and a left turn from north points west, and a left turn from west points south.\[ A(0,0) \to B(-14,0) \to C(-14,14) \to D(-24,14) \to E(-24,0) \]E is directly below D at the same y-value as A, so the distance is the horizontal gap.\[ AE = 24 \text{ m} \]

71.

In a certain cricket tournament 45 matches were played. Each team played once against each of the other teams. The number of teams participated in the tournament is

A.8
B.10
C.12
D.14

Exam Year(s)

UGC NET - 2013,

Answer: B

The number of teams was 10.

If each of n teams plays every other team once, the number of matches is the number of ways of choosing 2 teams from n.

\[ \frac{n(n-1)}{2} = 45 \]

\[ n(n-1) = 90 = 10 \times 9 \]

So n = 10. Check with the others: 8 teams give 28 matches, 12 teams give 66 and 14 teams give 91.

Another way to see it: every team plays 9 others, 10 teams times 9 matches is 90 team-appearances, and each match involves two teams, so 45 matches in all.

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72.

The missing number in the series 40, 120, 60, 180, 90, ?, 135 is

A.110
B.270
C.105
D.210

Exam Year(s)

UGC NET - 2013,

Answer: B

The missing number is 270.

The series alternates between multiplying by 3 and dividing by 2: 40 times 3 gives 120, then 120 divided by 2 gives 60, 60 times 3 gives 180, and 180 divided by 2 gives 90.

\[ 90 \times 3 = 270 \]

\[ 270 \div 2 = 135 \]

The last term, 135, confirms that the missing term is 270. The values 110, 105 and 210 break the pattern.

Any alternate-operation series is solved by writing the ratios between successive terms, and here they are 3, one half, 3 and one half in turn.

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73.

The odd numbers from 1 to 45 which are exactly divisible by 3 are arranged in an ascending order. The number at 6th position is

A.18
B.24
C.33
D.36

Exam Year(s)

UGC NET - 2013,

Answer: C

The number at the 6th position is 33.

The odd numbers divisible by 3 between 1 and 45 are the odd multiples of 3, which go up in steps of 6.

\[ 3,\ 9,\ 15,\ 21,\ 27,\ 33,\ 39,\ 45 \]

The sixth number in this ascending list is 33. The numbers 18, 24 and 36 are even, so they are not in the list.

A quick check confirms the position: the first odd multiple of 3 is 3 and each next one is 6 more, so the sixth is 3 plus 5 times 6, which is 33.

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74.

The mean of four numbers a, b, c, d is 100. If c = 70, then the mean of the remaining numbers is

A.30
B.\(\frac{85}{2}\)
C.\(\frac{170}{3}\)
D.110

Exam Year(s)

UGC NET - 2013,

Answer: D

The mean of the remaining three numbers is 110.

The four numbers have a mean of 100, so their sum is 400. Taking out c, which is 70, leaves the sum of the other three.

\[ a + b + d = 400 - 70 = 330 \]

\[ \text{mean} = \frac{330}{3} = 110 \]

The values 30, 85/2 and 170/3 come from dividing by the wrong number of items or subtracting wrongly.

A faster check is to note that the three remaining numbers must make up 330 out of the total of 400, so their average is a little above the overall mean of 100 and clearly cannot be 30.

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75.

If the radius of a circle is increased by 50%, the perimeter of the circle will increase by

A.20%
B.30%
C.40%
D.50%

Exam Year(s)

UGC NET - 2013,

Answer: D

The perimeter will increase by 50%.

The perimeter of a circle is 2 pi r, which is directly proportional to the radius. If r grows by 50%, the new radius is 1.5 r.

\[ 2\pi (1.5r) = 1.5 \times (2\pi r) \]

The perimeter becomes 1.5 times the old perimeter, an increase of 50%. The area, by contrast, would grow by 125%, which is why a different option is not chosen.

This result holds for any circle, whatever the starting radius, because the perimeter depends on the radius to the first power only.

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76.

Choose the proper alternative given in the codes to replace the question mark. Bee - Honey, Cow - Milk, Teacher - ?

A.Intelligence
B.Marks
C.Lessons
D.Wisdom

Exam Year(s)

UGC NET - 2013,

Answer: D

The key completes the analogy with wisdom.

A bee produces honey and a cow gives milk, so each item on the left is paired with the valuable product it yields. By the same relation, a teacher yields wisdom, the lasting result of teaching.

Intelligence is an inborn capacity a teacher does not produce. Marks are given by an examiner or teacher as an assessment, and lessons are what is delivered, not the product, so these do not fit the relation as well.

Discuss in Forum

77.

P is the father of R and S is the son of Q and T is the brother of P. If R is the sister of S, how is Q related to T ?

A.Wife
B.Sister-in-law
C.Brother-in-law
D.Daughter-in-law

Exam Year(s)

UGC NET - 2013,

Answer: B

Q is the sister-in-law of T.

R and S are sister and brother. P is the father of R, so P is also the father of S. S is the son of Q, so Q is S's mother and therefore P's wife. T is the brother of P, so Q is the wife of T's brother, which makes her T's sister-in-law.

She is not T's wife, since she is married to P. She is not a brother-in-law, because Q is a woman, nor a daughter-in-law, since she is not T's son's wife.

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78.

The world population growth rate at a certain reference year was 3.5%. Assuming exponential growth of population, after how many years, the population of the world would have increased by a factor 16 ?

A.~ 80 years
B.~ 40 years
C.~ 160 years
D.~ 320 years

Exam Year(s)

UGC NET - 2013,

Answer: A

The population grows by a factor of 16 in about 80 years.

With exponential growth \(P = P_{0}e^{rt}\) and a factor of 16, we need \(e^{rt} = 16\):

\[ t = \frac{\ln 16}{r} = \frac{2.77}{0.035} \]

\[ t \approx 79 \text{ years} \approx 80 \text{ years} \]

Equivalently, 16 is four doublings, and the doubling time at 3.5 per cent is about 20 years (70 divided by 3.5), which gives 80 years. The values 40, 160 and 320 years correspond to two, eight and sixteen doublings and do not give a factor of 16.

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79.

A person writes all the numbers from 0 to 99. The number of times digit 3 will be written is

A.18
B.19
C.20
D.21

Exam Year(s)

UGC NET - 2013,

Answer: C

The digit 3 is written 20 times from 0 to 99.

Count the units place and the tens place separately.

\[ \text{Units digit 3: } 3, 13, \ldots, 93 = 10 \]

\[ \text{Tens digit 3: } 30 \text{ to } 39 = 10 \]

\[ 10 + 10 = 20 \]

The number 33 is counted in both lists because it carries two 3s. The values 18, 19 and 21 come from leaving out or double counting a number.

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80.

Starting from point A, Ajit walks 14 metres towards west, he then turns to his right and walks 14 metres and then turns to his left and walks 10 metres. He again turns to his left and walks 14 metres and reaches to the point E. The shortest distance between A and E is

A.38
B.42
C.52
D.24

Exam Year(s)

UGC NET - 2013,

Answer: D

The shortest distance between A and E is 24 metres.

Place A at the origin and take east as positive x and north as positive y. Facing west, a right turn points north and a left turn from north points west, and a left turn from west points south.

\[ A(0,0) \to B(-14,0) \to C(-14,14) \to D(-24,14) \to E(-24,0) \]

E is directly below D at the same y-value as A, so the distance is the horizontal gap.

\[ AE = 24 \text{ m} \]

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