UPSC Exam  >  UPSC Tests  >  UPSC Previous Year Question Papers and Video Analysis  >  UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - UPSC MCQ

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - UPSC MCQ


Test Description

30 Questions MCQ Test UPSC Previous Year Question Papers and Video Analysis - UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT)

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) for UPSC 2024 is part of UPSC Previous Year Question Papers and Video Analysis preparation. The UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) questions and answers have been prepared according to the UPSC exam syllabus.The UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) MCQs are made for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) below.
Solutions of UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) questions in English are available as part of our UPSC Previous Year Question Papers and Video Analysis for UPSC & UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) solutions in Hindi for UPSC Previous Year Question Papers and Video Analysis course. Download more important topics, notes, lectures and mock test series for UPSC Exam by signing up for free. Attempt UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) | 80 questions in 120 minutes | Mock test for UPSC preparation | Free important questions MCQ to study UPSC Previous Year Question Papers and Video Analysis for UPSC Exam | Download free PDF with solutions
UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 1

The number of times the digit 5 will appear while writing the integers from 1 to 1000 is

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 1

First, you need to know how many 5's are there from 1 to 100.

5, 15, 25, ... , 95 = ten 5s at the unit's place.

50, 51, ..., 59 = ten 5s at the ten's place.

So, occurrence of 5 from 1 to 100 = 20 times.

So, from 1 to 1000, 5's in tens and units place would be 20 x 10 = 200 times

Occurrence of 5 at hundredth place from 500 to 600 = 100 times

Therefore, total number of times 5 digit would appear = 200 + 100 = 300.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 2

A solid cube is painted yellow, blue and black such that opposite faces are of same colour. The cube is then cut into 36 cubes of two different sizes such that 32 cubes are small and the other four cubes are Big. None of the faces of the bigger cubes is painted blue. How many cubes have only one face painted?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 2

Size of bigger cubes is twice the size of smaller cubes. Also, as none of the faces of the bigger cubes is painted blue, so each one of the bigger cube has one face painted yellow, one is black and all other faces are unpainted.

Number of cubes with one face painted = 8

Number of cubes with two faces painted = 8 smaller cubes on each of blue painted surface + 4 bigger cubes

= (8 x 2) + 4 = 20

The correct option is C.

1 Crore+ students have signed up on EduRev. Have you? Download the App
UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 3

A and B are two heavy steel blocks. If B is placed on the top of A, the weight increases by 60%. How much weight will reduce with respect to the total weight of A and B, if B is removed from the top of A?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 3

Step-by-step explanation:

Step 1- assume the weight of A=100%

Step 2-if you place B on A 60% increases

i.e, 100%+60%= 160%

Step 3- If you remove B with respect to total weight of A and B

i.e, total [A+B]-B= 160% - 60%

Reduction in weight with respect to total weight of A and B

(A/A+B) × [A+B]-B = 100/160 × 60 %

=37.5% is the answer

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 4

Mr. 'X' has three children. The birthday of the first child falls on the 5th Monday of April that of the second one falls on the 5th Thursday of November. On which day is the birthday of his third child, which falls on 20th December?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 4

5th Monday of April can be

either 29th April or 30th April

Case 1 : 29th April is 5th Monday

then 1 + 31 + 30 + 31 + 31 + 30 + 31 + 1 = 186 Days = 26*7 + 4

=> so 1 November will be Friday

=>  which means 29th , 30th Nov are  Friday & Saturday

so there's no 5th Thursday in November when 5th Monday of April is on 29th .

Case 2 :  30th April Monday is 5th Monday

then 31 + 30 + 31 + 31 + 30 + 31 + 1 = 185 Days = 26*7 + 3

=> so 1 November will be Thursday

=> which means 29th Nov is 5th Thursday, which is our case

=> so 6th Dec is also Thursday

6 + (2 * 7) = 20

=> So, 20th Dec is also Thursday (3rd Thursday of December)

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 5

Consider the following Statements and Conclusions:
Statements:
1. Some rats are cats.
2. Some cats are dogs.
3. No dog is a cow.
Conclusions:
I. No cow is a cat.
II. No dog is a rat.
III. Some cats are rats.
Q. Which of the above conclusions is/are drawn from the statements?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 5

The conclusion of 'Some rats are cats' is 'Some cats are rats'. So conclusion III is valid. No conclusion can be drawn in terms of rats and dogs as statement I and II both are starting with 'some'. So, conclusion II is not valid. That eliminates all options except option (c).

The only conclusion in terms of cat and cow will be 'Some cats are not cows'. So conclusion I is not valid.

 

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 6

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of four parallel lines, is

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 6

In the diagram, let’s count the parallelograms one by one.

Case I - Parallelograms of 1 × 1 (ABFE type) - ABFE, BCGF, CDHG, EFJI, FGKJ, GHLK, IJNM, JKON, KLPO – total 9.

Case II - Parallelograms of 1 × 2 (ACGE type) - ACGE, BDFH, EGKI, FHLJ, IKOM, JLPN – total 6.

Case III - Parallelogram of 2 × 1 (ABJI type) - ABJI, EFNM, BCKJ, FGON, CDLK, GHPO – total 6.

Case IV - Parallelograms of 1 × 3 (ADHE type) - ADHE, EHLI, ILPM – total 3.

Case V - Parallelograms of 3 × 1 (ABNM type) - ABNM, BCON, CDPO – total 3.

Case VI - Parallelograms of 2 × 2 (ACKI type) - ACKI, BDLJ, EGOM, FHPN – total 4.

Case VII - Parallelograms of 3 × 2 (ADLI type) - ADLI, EHPM – total 2.

Case VIII - Parallelograms of 2 × 3 (ACOM type) - ACOM, BDPN – total 2.

Case IX -Parallelograms of 3 × 3 (ADPM type) – ADPM – total 1.

Total 36.

A much shorter method is by using permutations and combinations.

Select any two of the first set of 4 lines. That can be done in 4C2 ways.

Now select any two of the second set of 4 lines. That can also be done in 4C2 ways.

So the total number of ways of doing it = 4C2  x 4C2 = 6 x 6 = 36 ways.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 7

In a school every student is assigned a unique identification number. A student is a football player if and only if the identification number is divisible by 4, whereas a student is a cricketer if and only if the identification number is divisible by 6. If every number from 1 to 100 is assigned to a student, then how many of them play cricket as well as football?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 7

The required number should be completely divisible by both 4 and 6. That means it should be divisible by LCM of 4 and 6, which is 12. Such 8 numbers are possible which are completely divisible by 12. They are 12, 24, 36, 48, 60, 72, 84 and 96.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 8

When a runner was crossing the 12 km mark, she was informed that she had completed only 80% of the race. How many kilometres was the runner supposed to run in this event?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 8

This can be solved mentally. 12 km is 80% of whole race

i.e  8/10 = 4/5 of the whole race. So whole race must be 15 km.

In this question 12 km is 80% of the total race.

=> 12 km = 0.8 R => R = 12 / 0.8 = 15.

So total race will be of 15 km. 

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 9

Raju has Rs. 9000 with him and he wants to buy a mobile handset; but he finds that he has only 75% of the amount required to buy the handset. Therefore, he borrows 2000 from a friend. Then

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 9

In this question Rs. 9000 is 75% of the cost of mobile phone. So total cost of mobile phone is Rs. 12000 ( = 9000 / 0.75). Now, as he borrows Rs. 2000, he will be still short of Rs.1000 to buy the phone.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 10

In 2002, Meenu's age was one-third of the age of Meera, whereas in 2010, Meenu's age was half the age of Meera. What is Meenu's year of birth?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 10

Answer:

Meenu's year of birth - 1994

Step-by-step explanation:

in 2002 Meenu's age was one third of the age of Meera where as in 2010,Meenu's age was half of the age of Meera.

Let say Meenu's age in 2002 = M Years

Meenu's age was one third of the age of Meera

so, Meera's age in 2002 = 3M

in 2010 i.e 8 years after 2002

Meenu's age was half of the age of Meera

Meenu age = M + 8

Meera age = 3M + 8

2(M + 8) = 3M + 8

=> 2M + 16 = 3M + 8

=> M = 8

Meenu was 8 years old in 2002

so Meenu was born in 2002-8  = 1994

Meenu's Year of Birth = 1994

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 11

Rakesh and Rajesh together bought 10 balls and 10 rackets. Rakesh spent 1300 and Rajesh spent 1500. If each racket costs three times a ball does, then what is the price of a racket?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 11

Let the cost of each ball is Rs. X. Then cost of each racket will be 3X.

Cost of 10 balls = 10X, and cost of 10 rackets = 30X.

So total cost = 10X + 30X = 40X.

By the condition given in question, we have

40X = 1300 + 1500 or 40X = 2800 or X = 70. Price of each racket = Rs. 210.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 12

In a conference, out of a total 100 participants, 70 are Indians. If 60 of the total participants are vegetarian, then which of the following statements is/are correct?
1. At least 30 Indian participants are vegetarian.
2. At least 10 Indian participants are non-vegetarian.
Q. Select the correct answer using the codes given below:

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 12

Let’s try to maximise the number of Indian-Vegetarians. Out of 70 Indians, all vegetarians (i.e, 60) can be Indians. So, at least 10 Indians will be there who will be non-vegetarians. This number can increase depending on the number of vegetarian-Indians.

Let’s try to minimise the number of Indian-Vegetarians. For that we have maximise the number of non-Indian-Vegetarians. Out of 30 Non-Indians, at max all can be vegetarian. Still 30 vegetarians remain which will fall under Indian category. So, at least 30 Indians will be there who will be vegetarians. Hence both statements are correct.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 13

Passage — 1
What stands in the way of the widespread and careful adoption of 'Genetic Modification (GM)' technology is an `Intellectual Property Rights' regime that seeks to create private monopolies for such technologies. If GM technology is largely corporate driven, it seeks to maximize profits and that too in the short run. That is why corporations make major investments for herbicide-tolerant and pest-resistant crops. Such properties have only a short window, as soon enough, pests and weeds will evolve to overcome such resistance. This suits the corporations. The National Farmers Commission pointed out that priority must be given in genetic modification to the incorporation of genes that can help impart resistance to drought, salinity and other stresses.
Q. Which one of the following is the most logical, rational and crucial message conveyed by the above passage?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 13

Option (b) is not relevant to this passage. Option (c) is an extreme, and is not stated thus, in the passage. Option (d) is wrong, as the passage indicates otherwise (last part). Best answer is option (a), as the passage criticizes the approach of private corporations, and indicates what India really needs (which private firms won’t do, and hence only public enterprises can or have to)

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 14

Passage — 1
What stands in the way of the widespread and careful adoption of 'Genetic Modification (GM)' technology is an `Intellectual Property Rights' regime that seeks to create private monopolies for such technologies. If GM technology is largely corporate driven, it seeks to maximize profits and that too in the short run. That is why corporations make major investments for herbicide-tolerant and pest-resistant crops. Such properties have only a short window, as soon enough, pests and weeds will evolve to overcome such resistance. This suits the corporations. The National Farmers Commission pointed out that priority must be given in genetic modification to the incorporation of genes that can help impart resistance to drought, salinity and other stresses.
Q. On the basis of the above passage, the following assumptions have been made:
​1. The issue of effects of natural calamities on agriculture is not given due consideration by GM technology companies.
2. In the long run, GM technology will not be able to solve agricultural problems arising due to global warming. Which of the above assumptions is/are valid?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 14

Assumption 1 is correct, as that is clearly mentioned in the passage. Assumption 2 is too broad, as it assumes that GM technology will “never” be able to do it, which may be wrong. Hence, option (a) is best.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 15

Passage - 2
Most invasive species are neither terribly successful nor very harmful. Britain's invasive plants are not widespread, not spreading especially quickly, and often less of a nuisance than vigorous natives such as bracken. The arrival of new species almost always increases biological diversity in a region; in many cases, a flood of newcomers drives no native species to extinction. One reason is that invaders tend to colonise disturbed habitats like polluted lakes and post-industrial wasteland, where little else lives. They are nature's opportunists.
Q. Which one of the following is the most logical and rational inference that can be made from the above passage?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 15

Option (d) uses the term “foreign plants” which makes in unsuitable. Same holds for options (b) and (c) too.

Option (d) uses the term “foreign plants” which makes in unsuitable. Same holds for options (b) and (c) too. Hence, option (a) is best, which is correct also as per the passage. Even if we do not use this logic (trick), option (b) is too extreme as in some cases we may need such laws. Option (c) is not the most logical inference (due to “sometimes”). Option (d) seems a strange way to go about it (we should rather protect our own indigenous biodiversity than destroy it). Hence (a) is best.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 16

Passage - 3
Diarrhoeal deaths among Indian children are mostly due to food and water contamination. Use of contaminated groundwater and unsafe chemicals in agriculture, poor hygiene in storage and handling of food items to food cooked and distributed in unhygienic surroundings; there are myriad factors that need regulation and monitoring. People need to have awareness of adulteration and ways of complaining to the relevant authorities. Surveillance of food-borne diseases involves a number of government agencies and entails good training of inspection staff. Considering the proportion of the urban population that depends on street food for its daily meals, investing in training and education of street vendors is of great significance.

On the basis of the above passage, the following assumptions have been made:
1. Food safety is a complex issue that calls for a multi-pronged solution.
2. Great investments need to be made in developing the manpower for surveillance and training.
3. India needs to make sufficient legislation for governing food processing industry.

Q. Which of the above assumptions is/are valid?

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 17

Passage - 4
The interests of working and poor people have historically been neglected in the planning of our cities. Our cities are increasingly intolerant,', unsafe and unlivable places for large numbers of citizens and yet we continue to plan via the old ways — the static Development Plan — that draws exclusively from technical expertise, distanced from people's live experiences and needs, and actively excluding large number of people, places, activities and practices that are an integral part of the city.
Q. The passage seems to argue

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 17

Options (a) and (b) are clearly wrong. Option (c) is not wrong, as that is one logic given right at the start. But option (d) is best as it covers option (c) as well. If we start involving peoples’ groups in city planning, then automatically we will have the interests of working class and poor people taken care of. In fact, much more than that is possible then.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 18

Passage - 5
A vast majority of Indians are poor, with barely 10 percent employed in the organised sector. We are being convinced that vigorous economic growth is generating substantial employment. But this is not so. When our economy was growing at 3 percent per year, employment in the organised sector was growing at 2 percent per year. As the economy began to grow at 7 - 8 percent per year, the rate of growth of employment in the organised sector actually declined to 1 percent per year.

The above passage seems to imply that
1. Most of modern economic growth is based on technological progress.
2. Much of modern Indian economy does not nurture sufficient symbiotic relationship with labour-intensive, natural resource-based livelihoods.
3. Service sector in India is not very labour intensive.
4. Literate rural population is not willing to enter organised sector.
Q. Which of the statements given above are correct?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 18

Statement 4 is not mentioned anywhere. So, options (b) and (d) are ruled out. Now check statement 3. It is a big assumption we need to make to equate service sector with organized sector keeping in mind the growth of economy coming largely from services. That is nowhere mentioned in the passage. Hence, 3 is not correct. Hence, option (c) is ruled out. Hence, (a) is best.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 19

Passage — 6
India has banking correspondents, who help bring people in the hinterland into the banking fold. For them to succeed, banks cannot crimp on costs. They also cannot afford to ignore investing in financial education and literacy. Banking correspondents are way too small to be viewed as a systemic risk. Yet India's banking regulator has restricted them to serving only one bank, perhaps to prevent arbitrage.'Efforts at banking outreach may succeed only if there are better incentives at work for such last-mile workers and also those providers who ensure not just basic bank accounts but also products such as accident and life insurance and micro pension schemes.
Q. Which one of the following is the most logical, rational and crucial inference that can be derived from the above passage?

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 20

A five-storeyed building with floors from I to V is painted using four different colours and only one colour is used to paint a floor. Consider the following statements:
1. The middle three floors are painted in different colours.
2. The second (II) and the fourth (IV) floors are painted in different colours.
3. The first (I) and the fifth (V) floors are painted red.
​To ensure that any two consecutive floors have different colours

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 20

Precondition - five floors are painted with 4 different colours.

Statement 1 – The middle three floors are painted in different colours does not guarantee any two consecutive floors of different colours as the colour of “I and II” or “IV and V” can be same.

Statement 2 – Second (II) and the fourth (IV) floors are painted in different colours does not guarantee any two consecutive floors of different colours as the colour of I and II or II and III, III and IV or IV and V can be same.

Statement 3 will ensure that any two consecutive floors have different colours as there are only four colours to be used. The remaining three floors will have different colours

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 21

P, Q and R are three towns. The distance between P and Q is 60 km, whereas the distance between P and R is 80 km. Q is in the West of P and R is in the South of P. What is the distance between Q and R?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 21

A direct application of Pythagoras theorem.

Refer to the diagram given :

Using Pythagoras theorem –

Distance between Q and R
= √( (80)2+ (60)2 )=√(6400+3600)=√10000=100 km

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 22

All members of a club went to Mumbai and stayed in a hotel. On the first day, 80% went for shopping and 50% went for sightseeing, whereas 10% took rest in the hotel. Which of the following conclusion(s) can be drawn from the above data?
1. 40% members went for shopping as well as sightseeing.
​2. 20% members went for only shopping.

Q. Select the correct answer using the code given below:

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 22

Percentage of members went for shopping  = 80%

percentage of members went for sightseeing = 50%

Percentage of members took rest in the hotel = 10%

Members who either went for shopping or sightseeing or both

= 100% - 10% = 90%

Members who went for both shopping and sightseeing

= 80% + 50% - 90% = 40%

Members who went only for shopping = 80% - 40% = 40%

40% members went for shopping as well as sightseeing.

and 40% members went for only shopping.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 23

In a school, 60% students play cricket. A student who does not play cricket, plays football. Every football player has got a twowheeler. Which of the following conclusions cannot be drawn from the above data?
1. 60% of the students do not have two wheelers.
2. No cricketer has a two-wheeler.
3. Cricket players do not play football.

​Q. Select the correct answer using the code given below:

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 23

60% students play cricket. So, 40% students play football. All football players have two-wheelers. It doesn't conclude that 60% students do not have two-wheelers. Statement 1 is false.
It is not mentioned whether cricketer has two-wheelers or not. So, statement 2 is false.
A student who does not play cricket, plays football. Nothing is mentioned about students who play cricket. Statement 3 is also false.
The correct option is A.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 24

The ratio of a two-digit natural number to a number formed by reversing its digits is 4: 7. The sum of number of such pairs is

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 24

Let the two digit number be 10a + b and the number formed by reversing  its digits be 10b + a.

70a + 7b = 40b + 4a
66a = 33b
Therefore, 
So, let us list down all possible values for a and b.

Hence, the sum of all the numbers would be, 
12 + 21 + 24 + 42 + 36 + 63 + 48 + 84 = 330.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 25

In an examination, A has scored 20 marks more than B. If B has scored 5% less marks than A, how much has B scored?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 25

Check options directly.

Start with (a). If B is 360, A will be 380. Now, 5% of 380 = 19. So B will become 380 – 19 = 361. Hence this option is wrong (B is 360, not 361).

If B is 380, A will be 400. Now, 5% of 400 = 20. So B will be 400 – 20 = 380. Hence (b) is correct. You do not need to check options (c) and (d) at all.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 26

Seeta and Geeta go for a swim after a gap of every 2 days and every 3 days respectively. If on 1st January both of them went for a swim together, when will they go together next?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 26

Seeta next - 4th | 7th | 10th | 13th
Geeta next - 5th | 9th | 13th
Seeta went on 1st so 2nd 3rd off & next class on 4th.
Similarly
Geeta went on 1st so 2nd 3rd 4th off & next class on 5th and so on.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 27

X, Y and Z are three contestants in a race of 1000 m. Assume that all run with different uniform speeds. X gives Y a start of 40 m and X gives Z a start of 64 m. If Y and Z were to compete in a race of 1000 m, how many metres start will Y give to Z?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 27

In the 1000 m race, X gives Y 40 m start. That means Y starts race 40 m ahead of X.

In the 1000 m race, X gives Z 64 m start. That means Z starts race 64 m ahead of X

So, this means Y gives a lead of 24 m in a 1000 – 40 = 960 m race.

So, by unitary method, lead given by Y in 1000 m = 24/960×1000 = 25 m.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 28

​If x is greater than or equal to 25 and y is less than or equal to 40, then which one of the following is always correct?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 28

Given that x is greater than or equal to 25. Also, y is less than or equal to 40.

Let’s try to find the various values of y – x.

If y = 40, x can take various values like 25, 26, 27, ....... (not necessarily integers – not given in question) In that case y – x will take values like 15, 14, 13, 12,......0, - 1 etc.

If y = 39, x can take various values 25, 26, 27.... (not necessarily integers) In that case y – x will take values like 14, 13, 12, 11 ..... 0, - 1 etc.

Similarly, if y = 38, values of y – x will be 13, 12, 11, ..... 0, -1 etc. and so on.

So, we can say that value of y – x is less than or equal to 15 in all cases.

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 29

Ena was born 4 years after her parents' marriage. Her mother is three years younger than her father and 24 years older than Ena, who is 13 years old. At what age did Ena's father get married?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 29

Ena's present age = 13

Ena's mother's present age = 13 + 24 = 37.

Ena's father's present age = 37 + 3 = 40.

Now at present Ena is 13 years old and her parents got married 4 years before she was born.

So, Ena's parents got married 13 + 4 = 17 years earlier.

So at the time of marriage her father would be of 40 – 17 = 23 years old

UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 30

Rakesh had money to buy 8 mobile handsets of a specific company. But the retailer offered very good discount on that particular handset. Rakesh could buy 10 mobile handsets with the amount he had. What was the discount the retailer offered?

Detailed Solution for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) - Question 30

Discount = Marked Price – Selling Price.
Let the Marked Price of 10 articles = Rs. 100.

So, Marked Price of 8 articles = Rs. 80. (so each was marked at Rs.10)

So, as per the question, Rakesh purchased 10 mobile phones for Rs. 80

So, the selling price of 10 mobile phones = Rs. 80 (so each bought at Rs.8)

The discount is 2 rupees on 10 which is 20%.

[Also, discount % = Discount × 100/ MP = 20 × 100/100 = 20%]

View more questions
183 videos|402 docs|21 tests
Information about UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) Page
In this test you can find the Exam questions for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT) solved & explained in the simplest way possible. Besides giving Questions and answers for UPSC Prelims Past Year Paper 2019: Paper 2 (CSAT), EduRev gives you an ample number of Online tests for practice

Top Courses for UPSC

Download as PDF

Top Courses for UPSC