The numerical ability and logical reasoning questions in CSAT, Paper II for the year 2017 were a good mix of easy and tough questions. There were certain questions which required candidate to know complex mathematical concepts while at the same time there were certain questions which were pretty easy
We have tried to use simple numerical methods and logic to solve most questions except in cases where complex mathematics was the only way out but luckily such questions were very few in number. The numerical ability and logical reasoning question have followed the same trend over the years with good number of easy and medium questions interspersed with tough nuts
Number system, Time-Speed-Distance, Work and Average have been recurring themes over the years and it shall be beneficial for the candidates to solve the questions from related chapters of Class IX and Class X. It shall also come handy to solve all the previous year CSAT Paper II questions as sometimes the concept remains the same, only the presentation changes in questions being asked
So here are the solutions for the numerical ability and logical reasoning questions for the year 2017:
- Certain 3-digit numbers following characteristics:
- All the three digits are different.
- The number is divisible by 7.
- The number on reversing the digits is also divisible by 7.
How many such 3-digit numbers are there?
(a) 2
(b) 4
(c) 6
(d) 8
Solution:
Let the number be 100x + 10y + z, where x, y and z are single digit numbers and x > 0.
Now the reverse number shall be 100z + 10 y + x. Now it is given that both of them are divisible by 7 and hence their difference shall also be divisible by 7.
Thus (100x + 10y + z) – (100z + 10 y + x) = 7m, where m is an integer.
This implies that 99(x – z) = 7m.
Now since 99 is not divisible by 7 therefore x-z should be divisible by 7. Again since x and z are single digit numbers their difference which is divisible by 7 would actually be 7 only, because for no value of x or z shall x-z be equal to 14 or 21 or 28 etc.
Thus we have x – z = 7 and the values that x and z can take are (9,2) and (8, 1)
Thus the numbers are of the form 1_8 and 2_9.
Now in 1_8, number at unit’s place is 8 so we should get a carry-over of 2 from the division of number formed by the digits at hundred’s and ten’s place to make the number become completely divisible by 7 (we want to make it 28 which is divisible by 7). Thus the ten’s digit should be 6 and number shall be 168. The reverse number 861 is also divisible by 7. Thus two of our required numbers are 168 and 861.
Now in 2_9, number at unit’s place is 9 so we should get a carry-over of 4 from the division of number formed by the digits at hundred’s and ten’s place to make the number become completely divisible by 7 (we want to make it 49 which is divisible by 7). Thus the ten’s digit should be 5 and number shall be 259. The reverse number 952 is also divisible by 7. Thus other two of our numbers are 259 an 952.
Thus there are four numbers satisfying all three conditions and hence the answer is
(b) 4
- Examine the following statements:
- All colours are pleasant.
- Some colours are pleasant.
- No colour is pleasant.
- Some colours are not pleasant.
Given that statement 4 is true, what can be definitely concluded?
(a) 1 and 2 are true.
(b) 3 is true.
(c) 2 is false.
(d) 1 is false.
Solution:
It is given that statement 4 is true i.e. some colours are not pleasant. Now if some colour are pleasant then its corollary is that some of them are pleasant.
Now let us evaluate rest of the three statements:
- All colours are pleasant….This is in clear contravention to statement 4 so Statement 1 is false.
- Some colours are pleasant….Yes this is true as we found out in the corollary of statement 4.
- No colour is pleasant…..No this is false as we found out in the corollary of statement 4.
Thus we know that statement 1 is false, 2 is true and 3 is false.
And hence answer choice is (d).
(d) 1 is false.
- How many numbers arc there between 99 and 1000 such that the digit 8 occupies the units place?
(a) 64
(b) 80
(c) 90
(d) 104
Solution:
We have to find out number between 99 and 1000 i.e. all three digit numbers such that unit’s digit is 8. Nothing has been said of digits at hundred’s or ten’s place.
Now number of digits that can occupy hundred’s place = 9 (1,2,3,4,5,6,7,8,9) [0 not included as it will make the number a two digit number]
Number of digits that can occupy ten’s place = 10 (0,1,2,3,4,5,6,7,8,9) [0 included as well]
Now number of digits that can occupy one’s place = 1 (only 8 can occupy this position as stated in the question)
Therefore total numbers between 99 and 1000 where unit’s place is occupied by 8 = 9 × 10 ×1 = 90.
Thus the answer option is
(c) 90
- If for a sample data
Mean < Median < Mode
then the distribution is
(a) symmetric
(b) skewed to the right
(c) neither symmetric nor skewed
(d) skewed to the left
Solution:
A direct question from basic statistics, For a distribution skewed to the left Mean < Median < Mode. Thus the answer is:
(d) skewed to the left
- The age of Mr. X last year was the square of a number and it would be the cube of a number next year. What is the least number of years he must wait for his age to become the cube of a number again?
(a) 42
(b) 38
(c) 25
(d) 16
Solution:
Last year the age was square of a number therefore it may have been 1, 4, 9, 16, 25, 36, 49 etc.
Now let us check in which of the above cases the age next year i.e. after 2 years shall be a cube. After 2 years the ages shall be 3, 6, 11, 18, 27, 38, 51 etc. In only one case, that is 64, the age shall be cube of a number. Thus we get the combination of ages we are looking for. Last year age was 25 (a square), this year age is 26 and next year age shall be 27 (a cube).
Now his age will be cube of number again when he is 64. So he must wait for 64 – 26 = 38 years. Thus the answer is option (b).
(b) 38
- P works thrice as fast as Q, whereas P and Q together can work four times as fast as R. If P, Q and R together work on a job, in what ratio should they share the earnings?
(a) 3 : 1 : 1
(b) 3 : 2 : 4
(c) 4 : 3 : 4
(d) 3 : 1 : 4
Solution:
They will share earnings in the ratio of the amount of work they do. Now P and Q together can work four times as fast as R which means in the time taken by R to finish one unit of work, P and Q together shall finish 4 units of work.
Let us assume there are a total of 15 units of work to be finished. When the work is finished, P and Q together would have finished 12 units (4/5th of 15 units) and R would have finished 3 units. Out of these 12 units, since P is 3 times more efficient, P would have finished 9 units (3/4th of 12 units) of work. Therefore the units of work done by P, Q and R is respectively 9 : 3 : 3. Therefore their earnings shall be shared in the same ratio of 9 : 3 : 3 i.e. 3 : 1 : 1. Therefore, answer shall be option (a):
(a) 3 : 1 : 1
[Note: If A can work x times faster than B, then A’s share of work shall be x/(x+1)th part of the total work]
- Consider the following relationships among members of a family of six persons A, B, C, D, E and F:
- The number of males equals that of females.
- A and E are sons of F.
- D is the mother of two, one boy and one girl.
- B is the son of A.
- There is only one married couple in the family at present.
Which one of the following inferences can be drawn from the above?
(a) A, B and C are all females.
(b) A is the husband of D.
(c) E and F are children of D.
(d) D is the daughter of F.
Solution:
Number of males = 3 and Number of females = 3.
A and E are males/sons of F (who can be male/female)
B is son of A. This means A, B and E are the males in the family and C, D and F are females.
F (female)
A (son/male) E (son/male)
B (son/male)
Now it is given that there is only one married couple and D is mother of 2 (a boy and a girl). D is married to either A or E or B. D cannot be married to B because then with 2 children (of which one is son) number of males shall become 4 and it shall violate given condition. D cannot be married to E because then again number of males in family would increase to 4 and that shall violate given condition. That means D is married to A and has B as son, C as daughter and A as husband. So out of the options given, only option (b) can be inferred. Thus the answer is
(b) A is the husband of D.
- A bag contains 20 balls. 8 balls are green, 7 are white and 5 are red. What is the minimum number of balls that must be picked up from the bag blindfolded (without replacing any of it) to be assured of picking at least one ball of each colour?
(a) 17
(b) 16
(c) 13
(d) 11
Solution:
We have to find out minimum number of balls that should be drawn to be sure to get at least one ball of each of the three colours. Now let us consider the worst case scenario:
If we start drawing balls and all the balls turn out to be green, then all balls turn out to be white. So we have drawn 8 + 7 = 15 balls by now. Now if we draw one more ball it will definitely be red and we would have at least one ball of each colour with us. So total balls drawn = 15 + 1 = 16. Thus the answer is option (b).
(b) 16
[Note : It is possible that first ball we draw is green, second we draw is white and third we draw is red i.e. we have three draws and three balls, one each of different colour, has been picked up as required. But in this case it is luck that it happened in three picks. But the question is that we have to be sure that we get at least one ball of each colour out and for that we have to consider the worst case scenario as explained above.]
- If 2 boys and 2 girls are to be arranged in a row so that the girls are not next to each other, how many possible arrangements are there?
(a) 3
(b) 6
(c) 12
(d) 24
Solution:
Instead of using permutation and combination let us try to count the arrangements as the number of boys and girls is less.
Let the seats be numbered 1, 2, 3 and 4 and boys and girls be represented as B1, B2, G1 and G2.
First of all let us fix the seat of G1 at seat no 1 and count the arrangements we need:
1 | 2 | 3 | 4 |
G1 | B1 | G2 | B2 |
G1 | B1 | B2 | G2 |
G1 | B2 | G2 | B1 |
G1 | B2 | B1 | G2 |
So total arrangement = 4. Similarly let us place G1 at seat 2, 3 and 4 in turn and count the arrangements:
1 | 2 | 3 | 4 |
B1 | G1 | B2 | G2 |
B2 | G1 | B1 | G2 |
Number of arrangement = 2
1 | 2 | 3 | 4 |
G2 | B1 | G1 | B2 |
G2 | B2 | G1 | B1 |
Number of arrangement = 2
1 | 2 | 3 | 4 |
B1 | G2 | B2 | G1 |
G2 | B1 | B2 | G1 |
B2 | G2 | B1 | G1 |
G2 | B2 | B1 | G1 |
Number of arrangement = 4
Thus the total number of arrangements = 4 + 2 + 2 + 4 = 12
Hence the answer is option (c):
(c) 12
- The outer surface of a 4 cm x 4 cm x 4 cm cube is painted completely in red. It is sliced parallel to the faces to yield sixty four 1 cm x 1 cm x 1 cm small cubes. How many small cubes do not have painted faces?
(a) 8
(b) 16
(c) 24
(d) 36
Solution:
The question involves a bit of imagery on the part of the candidates and a careful counting. If the top and bottom levels of smaller cubes (all of which shall have at least one painted face) are removed, we shall be left with 32 cubes. Out of these 32 cubes, which form the remaining two levels, the cubes on the outer periphery shall have one painted face. At each level there shall be 12 such cubes that will have one face painted. Thus removing these 12 cubes from each of the remaining levels we are left with 32 – 12 – 12 = 8 cubes who would not have any painted face. Therefore the answer is option (a):
(a) 8
- Consider the following:
A, B, C, D, E, F, G and H are standing in a row facing North.
B is not neighbour of G.
F is to the immediate right of G and neighbour of E.
G is not at the extreme end.
A is sixth to the left of E.
H is sixth to the right of C.
Which one of the following is correct in respect of the above?
(a) C is to the immediate left of A.
(b) D is immediate neighbour of B and F.
(c) G is to the immediate right of D.
(d) A and E are at the extreme ends.
Solution:
Let us assume we are facing them and then imagine the arrangement:
A is sixth to the left of E, which means there are 5 others between A and E.
A | E |
Or
A | E |
H is sixth to the right of C, which means there are 5 others between H and C.
C | H |
Or
C | H |
If we combine above arrangements, we have following possibilities:
C | A | H | E |
Or
A | C | E | H |
Since F is neighbor of E, we can have only of the above arrangements as the possibility and which is as under:
A | C | F | E | H |
Putting G to the immediate right of F we get:
A | C | G | F | E | H |
B is not neighbour of G, which means final position is as under:
A | C | B | D | G | F | E | H |
Now let us check the options:
(a) C is to the immediate left of A…..ANSWER
(b) D is immediate neighbour of B and F…..NO
(c) G is to the immediate right of D……NO
(d) A and E are at the extreme ends…..NO
- In a certain code, ‘256’ means ‘red colour chalk’, ‘589’ means ‘green colour flower’ and ‘254’ means ‘white colour chalk’. The digit in the code that indicates `white’ is
(a) 2
(b) 4
(c) 5
(d) 8
Solution:
Let us solve this question using elimination. 256 means red colour chalk (thus white is none of 2 or 5 or 6). 589 means ‘green colour flower’ (thus white is none of 5, 8 or 9). Thus white is none of 2, 5, 6, 8 or 9 and hence we get only one option (b) which can represent white. Thus answer is option (b).
(b) 4
- The average rainfall in a city for the first four days was recorded to be 0.40 inch. The rainfall on the last two days was in the ratio of 4 : 3. The average of six days was 0.50 inch. What was the rainfall on the fifth day?
(a) 0.60 inch
(b) 0.70 inch
(c) 0.80 inch
(d) 0.90 inch
Solution:
The average rainfall in first four days = 0.40 inch.
Thus total rainfall in first four days = 4 × 0.40 = 1.6 inch……(1)
The average rainfall six days = 0.50 inch.
Thus total rainfall in six days = 6 × 0.50 = 3.0 inch……………(2)
Subtracting (1) from (2) we get total rainfall in fifth and sixth day = 3.0 – 1.6 = 1.4 inch….(3)
Now the rainfall on fifth and sixth day was in the ratio 4 : 3. Let us say rainfall on fifth day = 4x inch and that on sixth day = 3x inch. Total rainfall = 4x + 3x = 7x = 1.4 inch [as found out in (3)].
Hence x = 0.2 inch and rainfall on fifth day = 4x = 4 × 0.2 = 0.8 inch. Thus answer is option (c).
(c) 0.80 inch
Directions for the following 3 (three) items : consider the given information and answer the three items that follow.
A, B, C, D, E, F and G are Lecturers from different cities—Hyderabad, Delhi, Shillong, Kanpur, Chennai, Mumbai and Srinagar (not necessarily in the same order) who participated in a conference. Each one of them is specialized in a different subject, viz., Economics, Commerce, History, Sociology, Geography, Mathematics and Statistics (not necessarily in the same order). Further
- Lecturer from Kanpur is specialized in Geography
- Lecturer D is from Shillong
- Lecturer C from Delhi is specialized in Sociology
- Lecturer B is specialized in neither History nor Mathematics
- Lecturer A who is specialized in Economics does not belong to Hyderabad
- Lecturer F who is specialized in Commerce belongs to Srinagar
- Lecturer G who is specialized in Statistics belongs to Chennai
Let us solve the case and fill up what is clearly given in the table first:
Lecturer City Specialization
Kanpur Geography
D Shillong
C Delhi Sociology
F Srinagar Commerce
G Chennai Statistics
Now we have made us of all statements in above table except the following two:
- Lecturer B is specialized in neither History nor Mathematics
- Lecturer A who is specialized in Economics does not belong to Hyderabad
Now if we combine our table and above two statements we can see that B does not specialize in History and Mathematics (as is given) or Sociology, Commerce or Statistics (they are specializations of C, F and G) or Economics (A specializes in it). Thus only specialization left for B is Geography. Our table now looks as under:
Lecturer City Specialization
B Kanpur Geography
D Shillong
C Delhi Sociology
F Srinagar Commerce
G Chennai Statistics
Now A specializes in Economics but does not belong to Hyderabad (given) or Kanpur, Shillong, Delhi, Srinagar, Chennai (they are cities of B, D, C, F and G). Thus A must belong to Mumbai. So now our table is as under:
Lecturer City Specialization
B Kanpur Geography
D Shillong
C Delhi Sociology
F Srinagar Commerce
G Chennai Statistics
A Mumbai Economics
Now only city of Hyderabad is left out and thus E must belong to it. Thus our table now looks like this:
Lecturer City Specialization
B Kanpur Geography
D Shillong
C Delhi Sociology
F Srinagar Commerce
G Chennai Statistics
A Mumbai Economics
E Hyderabad
Now we can answer the questions:
- Who is specialized in Geography?
(a) B
(b) D
(c) E
(d) Cannot be determined as data are inadequate
Solution:
(a) B
- To which city does the Lecturer specialized in Economics belong?
(a) Hyderabad
(b) Mumbai
(c) Neither Hyderabad nor Mumbai
(d) Cannot be determined as data are inadequate
Solution:
(b) Mumbai
- Who of the following belongs to Hyderabad?
(a) B
(b) E
(c) Neither B nor E
(d) Cannot be determined as data are inadequate
Solution:
(b) E
- In a school, there are five teachers A, B, C, D and E. A and B teach Hindi and English. C and B teach English and Geography. D and A teach Mathematics and Hindi. E and B teach History and French. Who teaches maximum numb, of subjects?
(a) A
(b) B
(c) D
(d) E
Solution:
A – Hindi, English
B – Hindi, English, Geography, History, French
C – English, Geography
D – Mathematics, Hindi
E – History, French
Thus the answer is option (b):
(b) B
- A 2-digit number is reversed. The larger of the two numbers is divided by the smaller one. What is the largest possible remainder?
(a) 9
(b) 27
(c) 36
(d) 45
Solution:
Since we want maximum remainder, we shall try to find a 2 digit number which is as high as possible and try to divide it by its reverse and check the remainder:
Let us begin with 99. The reverse shall be 99 as well. When 99 is divided by 99, the remainder is 0.
Take 98. Reverse is 89 and remainder shall be 9.
Take 97. Reverse is 79 and remainder shall be 18.
Take 96. Reverse is 69 and remainder shall be 27.
Take 95. Reverse is 59 and remainder shall be 36.
Take 94. Reverse is 49. When 94 is divided by 49, the remainder is 45, which is the largest value given in the options and hence our answer. Thus the answer is
(d) 45
- The monthly incomes of X and Y are in the ratio of 4 : 3 and their monthly expenses are in the ratio of 3: 2. However, each saves Rs. 6,000 per month. What is their total monthly income?
(a) Rs. 28,000
(b) Rs. 42,000
(c) Rs. 56,000
(d) Rs. 84,000
Solution:
Let their incomes be 4x and 3x respectively and their savings be 3y and 2y respectively.
Thus savings of X and Y are respectively 4x – 3y and 3x – 2y.
Since both have same savings of Rs. 6000 per month that means : 4x – 3y = 3x – 2y, which means x = y.
Now saving of X = 4x-3y = 4x – 3x = x = Rs. 6000 (given).
Thus their total monthly income = 4x + 3x = 7 = Rs 42,000. Thus answer is option (b):
(b) Rs. 42,000
- Two walls and a ceiling of a room meet at right angles at a point P. A fly is in the air 1 m from one wall, 8 m from the other wall and 9 m from the point P. How many meters is the fly from the ceiling?
(a) 4
(b) 6
(c) 12
(d) 15
Solution:
If the candidates can again use a bit of imagination they will realize that question is basically about finding the height of a cuboid whose length, breadth and the length of the diagonal joining opposite corners across the cuboid is given. The fly and point P are at the opposite corners across the cuboid in this case. We know that if a, b and c are length, breadth and height of a cuboid then the length of the diagonal, d, joining the opposite corners across the cuboid is equal to a²+b²+c² .
In this case, a = 8 m, b = 1 m, d = 9 m. Putting these values in above formula we get, c = 4 m. Thus the fly is 4 meters away from the ceiling. Therefore the answer is:
(a) 4
Directions for the following 3 (three) items : Consider the given information and answer the three items that follow.
Eight railway stations A, B, C, D, E, F, G and H are connected either by two-way passages or one-way passages. One-way passages are from C to A, E to G, B to F, D to H, G to C, E to C and H to G. Two-way passages are between A and E, G and B, F and D, and E and D.
Let us show the ways through arrows and draw the figure:
C A
E G
B F
D H
- While travelling from C to H, which one of the following stations must be passed through?
(a) G
(b) E
(c) B
(d) F
Solution:
Now while travelling from C to H, we can have the following paths:
D H
C A E
G B F D H
We can see that of the given options E must be passed through while travelling from C to H and therefore answer is option (b):
(b) E
- In how many different ways can a train navel from F to A without passing through any station more than once?
(a) 1
(b) 2
(c) 3
(d) 4
Solution: G C A
C A
E A
F D
H G C A
Thus there are four distinct paths to reach C from F without passing through any station twice and therefore the answer is option (d):
(d) 4
- If the route between G and C is closed, which one of the following stations need not be passed through while travelling from H to C?
(a) E
(b) D
(c) A
(d) B
Solution:
If the route between G and C is closed, there shall be only one route to travel from H to C as follows:
H G B F D E C
Thus E, D and B must be passed through to reach C from H. Therefore answer is option (c):
(c) A
- There are certain 2-digit numbers. The difference between the number and the one obtained on reversing it is always 27. How many such maximum 2-digit numbers are there?
(a) 3
(b) 4
(c) 5
(d) None of the above
Solution:
Let the number be 10x + y. Thus the number obtained by reversing its digits shall be 10y + x.
Now as per the given condition, (10x + y) – (10y + x) = 27
Thus 9(x-y) = 27 and x – y = 3
Now (x, y) can take following values: (9,6), (8, 5), (7, 4), (6, 3) (5, 2) and (4, 1). Let us validate our answer:
96 – 69 = 27
85 – 58 = 27
74 – 47 = 27
63 – 36 = 27
52 – 25 = 27
41 – 14 = 27
Thus there are 6 such 2 digit numbers and hence the answer is
(d) None of the above
- What is the total number of digits printed, if a book containing 150 pages is to be numbered from 1 to 150?
(a) 262
(b) 342
(c) 360
(d) 450
Solution:
From 1 to 150, there are
9 single digit numbers (1 to 9) + 90 two digit numbers (10 to 99) + 51 three digit numbers (100 to 150)
Thus total digits to be printed = (9 × 1) + (90 × 2) + (51 × 3) = 342. Therefore the answer is option (b):
(b) 342
- Suppose the average weight of 9 persons is 50 kg. The average weight of the first 5 persons is 45 kg, whereas the average weight of the last 5 persons is 55 kg. Then the weight of the 5th person will be
(a) 45 kg
(b) 47.5 kg
(c) 50 kg
(d) 52.5 Kg
Solution:
Average weight of first 5 persons = 45 kg
Therefore, total weight of first 5 persons = 5 × 45 kg = 225 kg……(1)
Average weight of last 5 persons = 55 kg
Therefore, total weight of last 5 persons = 5 × 55 kg = 275 kg…….(2)
If we add (1) and (2) we will get the total weight of 9 persons + weight of 5th person (because weight of 5th person will be present in the total weight of first 5 persons as well as in the total weight of last five persons).
This means (1) + (2) = Total weight of 9 person + weight of 5th person….(3)
Total weight of 9 persons = 9 × 50kg = 450 kg….(4)
Putting values from (1), (2) and (4) in equation (3) we get:
Weight of 5th person = (1) + (2) – (4) = 225 + 275 – 450 = 50 kg. Therefore the answer is option (c):
(c) 50 kg
- In a group of six women, there are four tennis players, four postgraduates in Sociology, one postgraduate in Commerce and three bank employees. Vimala and Kamla are the bank employees while Amala and Komala are unemployed. Komala and Nirmala are among the tennis players. Amala, Kamla, Komala and Nirmala are postgraduates in Sociology of whom two are bank employees. If Shyamala is a postgraduate in Commerce, who among the following is both a tennis player and a bank employee?
(a) Amala
(b) Komala
(c) Nirmala
(d) Shyamala
Solution:
Let us put the given information with each of the women:
Vimala : Bank employee
Kamla : Bank employee, Sociology
Amala : Unemployed, Sociology
Komala : Unemployed, Tennis, Sociology
Nirmala: Tennis, Sociology
Shyamala :Commerce
It is given that Amala, Kamla, Komala and Nirmala are postgraduates in Sociology of whom two are bank employees and since it is also given that Amala and Komala are unemployed, therefore Nirmala and Kamla should be Bank employees (something which is already given in the case of Kamla).
Thus we have Nirmala who plays tennis, is post-graduate in Sociology and is also a bank employee. Thus the answer is option (c):
(c) Nirmala
- P = (40% of A) + (65% of B) and Q = (50% of A) + (50% of B), where A is greater than B. In this context, which of the following statements is correct?
(a) P is greater than Q.
(b) Q is greater than P.
(c) P is equal to Q.
(d) None of the above can be concluded with certainty.
Solution:
Here nothing has been said about the relative magnitudes of A and B or whether they are both positive, both negative or one positive and one negative. Therefore, let us assume some value for A and B and see whether we can make any progress:
Let A = 100 and B = 50. In this case P shall come out to be 72.5 and Q shall be 75. That is P shall be smaller to Q.
Now Let A = 100 and B = 99. In this case P shall be 104.35 and Q shall become 99.5. That is P shall be greater to Q.
Thus with different values of A and B, we get different results about magnitudes of P and Q. Thus the answer shall be
(d) None of the above can be concluded with certainty.
- A watch loses 2 minutes in every 24 while another watch gains 2 minutes, in 24 hours. At a particular instant, the two watches showed an identical time. Which of the following statements is correct if 24-hour clock is followed
(a) The two watches show the identical time again on completion of 30 days.
(b) The two watches show the identical time again on completion of 90 days.
(c) The two watches show the identical time again on completion of 120 days.
(d) None of the above statements correct.
Solution:
First clock loses 2 minutes per 24 hours i.e. it loses 1/30 hours per 24 hours. Thus for t hours it loses t/(30×24) = t/720 hours.
The second one gains 2 minutes in 24 hours. AAs above, in t hours it gains t/720 hours.
Now suppose they showed same time at 12 midnight, then after t hours their respective times shall be t + t/720 and t – t/720. Now since it is 24 hour clock, they will show same time again when difference between their times becomes 24.
That is [t + t/720] – [t – t/720] = 24
Thus t = 360 × 24 hours
The number of days after which they will show same time, therefore, shall be (360 × 24) / 24 days = 360 days.
Thus the answer shall be
(d) None of the above statements correct.
- In a city, 12% of households earn less than Rs. 30,000 per year, 6% households earn more than Rs. 2,00,000 per year, 22% households earn more than Rs. 1,00,000 per year and 990 households earn between Rs. 30,000 and Rs. 1,00,000 per year. How many households earn between Rs. 1,00,000 and Rs. 2,00,000 per year?
(a) 250
(b) 240
(c) 230
(d) 225
Solution:
It is given that 6% households earn more than Rs. 2,00,000/- and 22% earn more than Rs. 1,00,000/-. These statements means that this 22% also includes the 6% that earn more than Rs. 2,00,000/-. Therefore, those who earn more than Rs. 1,00,00/- but less than Rs. 2,00,000/- is 22 – 6= 16%
< Rs, 30, 000 | 12% |
Rs. 30, 000 – Rs. 1, 00, 000 | 990 |
Rs. 1,00,000 – Rs. 2,00,000 | 16% |
> Rs. 2, 00, 000 | 6% |
Now in percentage terms, number of households earning between Rs. 30, 000 – Rs. 1, 00, 000 will be 100 – 12 – 16 – 6 = 66% (which is equal to 990, as given). Now households earning between Rs. 1,00,000 – Rs. 2,00,000 is 16% i.e. (990/66) × 16 = 240
Therefore the answer is
(b) 240
31.** A clock strikes once at 1 o’clock, twice at 2 o’clock and thrice at 3 o’clock, and so on. If it takes 12 seconds to strike at 5 o’clock, what is the time taken by it to strike at 10 o’clock?
(a) 20 seconds
(b) 24 seconds
(c) 28 seconds
(d) 30 seconds
Solution:
The clock will strike 5 times at 5 o’ clock. These five strikes are interspersed with four intervals as under:
Strike – Strike – Strike – Strike – Strike
It is these intervals that take time. When clock strikes 5 times, there are 4 intervals. Total time taken by them is 12 seconds. Thus each interval is of duration 3 seconds.
When the clock will strike 10 times at 10 o’ clock there shall be 9 such intervals and thus time taken shall be 9 ×3 = 27 seconds.
Since this is the ideal approach for the question but the answer 27 seconds is not given therefore this question may be dropped by UPSC from evaluation.
Alternatively, if they want to keep this question (which is unlikely) they would take the answer as option (b) 24 seconds because time taken for 5 strikes = 12 seconds and thus time taken for 10 strikes shall be double of it i.e. 12 × 2 = 24 seconds.
(b) 24 seconds
- Consider the given statement and the two conclusions that follow:
Statement:
Morning walk is good for health.
Conclusions:
- All healthy people go for morning walk.
- Morning walk is essential for maintaining good health.
What is/are the valid conclusion/ conclusions?
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
Solution:
Let us evaluate the conclusions:
- All healthy people go for morning walk………Morning walk is good for health but that does not mean ALL healthy people go for walk. Some might go for other activities. And some may not indulge in any activity and still stay fit and healthy. Thus this conclusion is not valid.
- Morning walk is essential for maintaining good health….’Essential’ means without morning walk one cannot be healthy but we just proved above that one can be healthy without morning walk too. Thus this conclusion is also not valid.
Thus answer is option (d):
(d) Neither 1 nor 2
- There are thirteen 2-digit consecutive odd numbers. If 39 is the mean of the first five such numbers, then what is the mean of all the thirteen numbers?
(a) 47
(b) 49
(c) 51
(d) 45
Solution:
Let us assume that the first five required odd number are x – 4, x – 2, x, x + 2, x + 4.
Now their mean will be [(x – 4) + (x – 2) + x + (x + 2) + (x + 4)]/5 = 39, as given.
Thus x = 39 and first five numbers become 35, 37, 39, 41, 43.
Thus the thirteen numbers shall be 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59.
Their mean shall be (35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 + 51 + 53 + 55 + 57 + 59)/13 = 47.
Therefore the answer is
(a) 47
Alternatively,
For an odd number of consecutive numbers the mean is the central number. Here mean of first five number is 39 which means that the third number is 39. For 13 consecutive numbers, mean shall be 7th number. Now third number is 39, fourth shall be 41, fifth shall be 43, sixth shall be 45 and seventh shall be 47.
- Six boys A, B, C, D, E and F play a game of cards. Each has a pack of 10 cards. F borrows 2 cards from A and gives away 5 to C who in turn gives 3 to B while B gives 6 to D who passes 1 to E. Then the number of cards possessed by D and E is equal to the number of cards possessed by
(a) A, B and C
(b) B, C and F
(c) A, B and F
(d) A, C and F
Solution:
Here just remember every give and take shall affect two members: One who gains and one who loses. Now let us proceed with each of the with 10 cards at the start and then adjusting the transactions:
A : 10 – 2 = 8
B : 10 + 3 – 6 = 7
C : 10 + 5 – 3 = 12
D : 10 + 6 – 1 = 15
E : 10 + 1 = 11
F : 10 + 2 – 5 = 7
Now cards possessed by D and E = 15 + 11 = 26.
Now lets us evaluate the options:
(a) A, B and C = 8 + 7 + 12 = 27
(b) B, C and F = 7 + 12 + 7 = 26
(c) A, B and F = 8 + 7 + 7 = 22
(d) A, C and F = 8 + 12 + 7 = 27
Thus answer is option (b):
(b) B, C and F
- There is a milk sample with 50% water in it. If 1/3rd of this milk is added to equal amount of pure milk, then water in the new mixture will fall down to
(a) 25%
(b) 30%
(c) 35%
(d) 40%
Solution:
Since we have to take 1/3 of the solution therefore let us assume that the volume of solution was 90 L (i.e. something which we can easily divide in 3 parts).
Now one-third of this solution is taken. That means 30 L is taken. This 30 L will have 15 L milk and 15 L water (because water and milk are 50% and 50%).
Now this solution of 30 L (15 L milk + 15 L water) is added to equal amount of pure milk. Thus 30 L of pure milk is added. Thus finally we have in the new mixture: 45 L milk + 15 L water. Thus water is 15/60 parts = ¼ parts which is 25%.
Thus the answer is option (a):
(a) 25%
- There are 4 horizontal and 4 vertical lines, parallel and equidistant to one another on a board. What is the maximum number of rectangles and squares that can be formed?
(a) 16
(b) 24
(c) 36
(d) 42
Solutions:
When a group of vertical and horizontal lines, which also happen to be parallel equidistant, intersect they form quadrilaterals, each of which is either a rectangle or a square.
Now to make a quadrilateral we should choose 2 vertical lines and 2 horizontal lines. From 4 vertical lines 2 lines can be chosen in C(4,2) ways = 4! / (2! × 2!) = 6 ways.
Similarly out of 4 horizontal lines, 2 lines can be chosen in C(4,2) ways = 4! / (2! × 2!) = 6 ways.
Thus total number of quadrilateral (which would invariably be either square or rectangle as required by the question) shall be 6 × 6 = 36. Thus the answer is option (c):
(d) 42
- A freight train left Delhi for Mumbai at an average speed of 40 km/hr. Two hours later, an express train left Delhi for Mumbai, following the freight train on a parallel track at an average speed of 60 km/hr. How far from Delhi would the express train meet the freight train?
(a) 480 km
(b) 260 km
(c) 240 km
(d) 120 km
Solution:
Two hours later the freight train from Delhi to Mumbai must have covered a distance of 40 × 2 = 80 km. Now in this question, we know that the express train is running faster than the freight train so it will eventually catch up with the freight train. If we assume that after 2 hours the freight train has stopped then basically we have to find out in how much time the express train (whose relative speed will be 60-40 = 20 km/hr) cover the distance of 80 km that separates the two trains. Now with 20 km/hr the express train would cover 80 km in 4 hours. Thus in four hours the distance covered by the express train shall be 4 × 60 = 240 km. Hence the answer is option (c).
[This is similar to the policeman and thief question asked in one of the previous years.]
(c) 240 km
Alternatively,
When express train starts, the freight train has already covered 80 km. After one more hour, the trains would have covered 60 km and 120 km respectively. After yet another hour: 120 km and 160 km. After yet another hour: 180 km and 200 km. And finally after one more hour they would have covered distances of 240 km and 240 km. Thus after covering 240 km from Delhi both train shall be at the same position. This method can be used if the above concept explained earlier is not clear and the figures of distance and speed are pretty simple as in this question.
- In a test, Randhir obtained more marks than the total marks obtained by Kunal and Debu. The total marks obtained by Kunal and Shankar are more than those of Randhir. Sonal obtained more marks than Shankar. Neha obtained more marks than Randhir. Who amongst them obtained highest marks?
(a) Randhir
(b) Neha
(c) Sonal
(d) Data are inadequate
Solution:
Let us list the contenders first:
Randhir, Kunal, Debu ,Shankar, Sonal and Neha
Now Randhir > Kunal + Debu this means Randhir has got more marks than Kunal and Debu. Thus Kunal and Debu are out of race.
Sonal > Shankar and Neha > Randhir. This means Shankar and Randhir are out of race. And we are left with Sonal and Neha. Now there is no condition that can help us compare the marks of Sonal and Neha. Therefore, the answer is
(d) Data are inadequate
Directions for the following 2 (two) items: – Consider the given information and answer the two items that follow.
No supporters of ‘party X’, who knew Z and supported his campaign strategy, agreed for the alliance with ‘party Y’; but some of them had friends in ‘party Y’.
- With reference to the above information, which one among the following statements must be true?
(a) Some supporters of ‘party Y’ did not agree for the alliance with the `party X’.
(b) There is at least one supporter of `party Y’ who knew some supporters of ‘party X’ as a friend.
(c) No supporters of ‘party X’ supported Z’s campaign strategy.
(d) No supporters of ‘party X’ knew Z.
Solution:
Let us evaluate the options:
(a) Some supporters of ‘party Y’ did not agree for the alliance with the `party X’….No mention of alliance between X and Y…OUT
(b) There is at least one supporter of `party Y’ who knew some supporters of ‘party X’ as a friend…Statement has some supporters of party X has friends in party Y which means that there were some (at least one) supporter of party Y who knew some supporter (at least one) of party X as friend…ANSWER
(c) No supporters of ‘party X’ supported Z’s campaign strategy….The statement is that no supporters of party X who knew Z and supported his campaign agreed for alliance with Party Y, which means that there were some supporters of party X who did support Z and his campaign strategy…OUT
(d) No supporters of ‘party X’ knew Z….The statement begin with those supporters of party X who knew Z…OUT
- With reference to the above information, consider the following statement
- Some supporters of ‘party X’ knew Z
- Some supporters of ‘party X’, who opposed Z’s campaign strategy, knew Z.
- No supporters of ‘party X’ supported Z’s campaign strategy.
Which of the statements given above, is/are not correct?
(a) 1 only
(b) 2 and 3 only
(c) 3 only
(d) 1, 2 and 3
Solution:
Let us evaluate the options:
- Some supporters of ‘party X’ knew Z…… This statement is correct.
- Some supporters of ‘party X’, who opposed Z’s campaign strategy, knew Z…Statement is silent about those supporters of party X who knew Z but still would have supported his campaign…So this statement is not correct.
- No supporters of ‘party X’ supported Z’s campaign strategy….Statement says there were some supporters of party X who knew Z and supported his campaign…. So this statement is not correct.
Thus statement 2 and 3 are not correct and answer is option (b):
(b) 2 and 3 only
- If second and fourth Saturdays and all the Sundays are taken as only holidays for an office, what would be the minimum number of possible working days of any month of any year?
(a) 23
(b) 22
(c) 21
(d) 20
Solution:
Since we have to find out minimum number of working days therefore let us chose a month with minimum number of days. That would be February of 28 days. These 28 days shall have 4 days of each of the 7 days of the week i.e. 4 Mondays, 4 Tuesdays and so on. Now out of these 28 days 4 Sundays shall be holiday and 2 Saturdays (second and fourth) shall be holiday. So number of working days = 28 – 4 – 2 = 22. Therefore the answer is:
(b) 22
- If there is a policy that 1/3rd of population of a community has migrated every year from one place, to some other place, what is the leftover population of that community after the sixth year, if there is no further growth in the population during this period?
(a) 16/243rd part of the population
(b) 32/243rd part of the population
(c) 32/729th part of the population
(d) 64/729th part of the population
Solution:
1/3rd population is migrating every year therefore the population will continue to become 2/3rd of the previous year.
So after one year the population will be 2/3 × original population.
After 2 years it will be 2/3 × 2/3 × original population and so on.
So after six years remaining population shall be 2/3 × 2/3 ×2/3 × 2/3 ×2/3 × 2/3 × original population i.e. 64/729th part of population. Thus answer is option (d):
(d) 64/729th part of the population
- Four tests—Physics, Chemistry, Mathematics and Biology are to be conducted on four consecutive days, not necessarily in the same order. The Physics test is held before the test which is conducted after Biology. Chemistry is conducted exactly after two tests are held. Which is the last test held?
(a) Physics
(b) Biology
(c) Mathematics
(d) Chemistry
Solution:
Chemistry is third to be held as is given. Now Biology cannot be fourth because passage says there is a test that is to be held after Biology. Now Biology cannot be first as well since Physics is to be held before Biology. Thus Biology is second. Since Physics is before Biology thus Physics is first and the remaining one i.e. Mathematics is last. And hence the answer is:
(c) Mathematics
Alternatively,
Since Physics is to be held before the test (say T) which is conducted after Biology that means we have this order Physics, Biology, T (we are not considering the final order yet). This means Physics and Biology are not the last tests to be held and thus option (a) and (b) are not the answer. Chemistry is held exactly after two tests are held i.e. Chemistry is the third test to be held and thus even Chemistry is not the last test. And hence the answer is Mathematics.
- The sum of income of A and B is more than that of C and D taken together. The sum of income of A and C is the same as that of B and D taken together. Moreover, A earns half as much as the sum of the income of B and D. Whose income is the highest?
(a) A
(b) B
(c) C
(d) D
Solution:
Let A, B, C and D represent their respective incomes.
A + B > C + D….(1)
A + C = B + D
A = (B+D)/2 and thus C = (B+D)/2 which implies that A = C….(2)
Putting A = C in (1) we get, A + B > A + D which means B > D….(3)
Now adding C + D to both sides of (1), we get A + B + C + D > 2C + 2D
Or (A+C) + (B+D) > 2C + 2D
Or 2B + 2D > 2C + 2D (because A + C = B + D)
Or B > C…. (4)
From (2), (3) and (4) we get B > C = A and B > D, thus B is the greatest.
Therefore answer is
(b) B
- Consider the following:
Statement:
Good voice is a natural gift but one has to keep practising to improve and excel well in the field of music.
Conclusions:
- Natural gifts need nurturing and care.
- Even though one’s voice is not good, one can keep practising.
Which one of the following is correct, in respect of the above statement and conclusions?
(a) Only conclusion I follows from the statement.
(b) Only conclusion II follows from the statement.
(c) Either conclusion I or conclusion II follows from the statement.
(d) Neither conclusion I nor conclusion II follows from the statement.
Solution:
The Statement basically means that good voice is natural gift but to excel in field of music one still needs to practice in order to improve and excel. That means natural gift alone is not sufficient, it needs nurturing and care FOR music. But the statement nowhere proffers that every natural gift needs nurturing and caring as the statement is specific to the field of music. Thus conclusion I does not necessarily follows from the statement. Conclusion II is not connected to the statement as the statement is about natural gifts and not about those who do not have good voice and thus does not even mention what people with voice which is not good should or should not do. Consequently conclusion II does not follow from the statement. Therefore the answer is option (d):
(d) Neither conclusion I nor conclusion II follows from the statement.
- There are three pillars X, Y and Z of different heights. Three spiders A, B and C start to climb on these pillars simultaneously. In one chance, A climbs on X by 6 cm but slips down 1 cm. B climbs on Y by 7 cm but slips down 3 cm. C climbs on Z by 6.5 cm but slips down 2 cm. If each of them requires 40 chances to reach the top of the pillars, what is the height of the shortest pillar?
(a) 161 cm
(b) 163 cm
(c) 182 cm
(d) 210 cm
Solution:
After one chance:
actual height covered by X = 6 – 1 = 5 cm,
actual height covered by Y = 7 – 3 = 4 cm,
actual height covered by Z = 6.5 – 2 = 4.5 cm.
Now in 40th attempt they will reach the top means till 39th chance they will climb up and slip but in 40th chance they will only climb and reach the top (if they slip they will not reach the top).
Thus after 39th chance:
total height covered by X = 39 × 5 = 195 cm
total height covered by Y = 39 × 4 = 156 cm
total height covered by Z = 39 × 4.5 = 175.5 cm
and after 40th climb:
total height covered by X = 195 + 6 = 201 cm
total height covered by Y = 156 + 7 = 163 cm
total height covered by Z = 175.5 + 6.5 = 182 cm
Thus shortest pillar is 163 cm in height and the answer is option (b):
(b) 163 cm
- “Rights are certain advantageous conditions of social well being indispensable to the true development of the citizen.”
In the light of this statement, which one of the following is the correct understanding of rights?
(a) Rights aim at individual good only.
(b) Rights aim at social good only.
(c) Rights aim at both individual and social good.
(d) Rights aim at individual good devoid of social well-being.
Solution:
Statements can be understood as meaning that the Rights are conditions of social well being which is indispensable for individual development as well thus Rights target both social and individual well being and thus the answer is option (c):
(c) Rights aim at both individual and social good.
- 15 students failed in a class of 52. After removing the names of failed students, a merit order list has been prepared in which the position of Ramesh is 22nd from the top. What is his position from the bottom?
(a) 18th
(b) 17th
(c) 16th
(d) 15th
Answer. C
Solution:
15 students failed out of 52. After removing the students who failed we are left with 52 – 15 = 37 students. In the list of 37 students Ramesh is 22 from the top that means there are 37 – 22 = 15 students below it and thus from the bottom his position is 16th. Thus the answer is option (c):
(c) 16th
- Consider the following:
A+ B means A is the son of B.
A – B means A is the wife of B.
What does the expression P + R – Q mean?
(a) Q is the son of P.
(b) Q is the wife of P.
(c) Q is the father of P.
(d) None of the above
Solution:
P + R = P is son of R which means P is male and R is male or female.
So P + R – Q = R is wife of Q (If we take R is male then she cannot be in the role of wife thus we can say that R is female).
Thus we get that R (wife) and Q (husband) have a son P.
Thus answer is option (c):
(c) Q is the father of P.
- Gopal bought a cell phone and sold it to Ram at 10% profit. Then Ram wanted to sell it back to Gopal at 10% loss. What will be Gopal’s position if he agreed?
(a) Neither loss nor gain
(b) Loss 1%
(c) Gain 1%
(d) Gain 0.5%
Solution:
Suppose Gopal got the cell phone for Rs. 100/-. He sold it to Ram at 10% profit i.e. he sold it to Ram for Rs. 110/-. Now Ram wants to sell it back to Gopal at 10% loss i.e. only for 90% of the amount for which is got it. Thus he will sell it to Gopal for Rs. 0.9 × 110 = Rs. 99/-
Thus Gopal makes a profit of Rs. 1 over an initial investment of Rs. 100/- Thus the answer is
(c) Gain 1%
* * * * * * *