CSAT (Civil Services Aptitude Test) | UPSC | CIVIL SERVICES | Practice Question with Solution | Previous Year question | 2020 | part 1

CSAT Practice Question

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Question no. 1

A frog tries to come out of a dried well 4.5 m deep with slippery walls. Every time frog jumps 30 cm , slides down 15cm , What is the number of jumps he required to come out of the well ? [CSAT 2020]

looks_one 28

looks_two 29

looks_3 30

looks_4 31

Option looks_two 29

Solution :
Every jump cover distance = 30 cm .
Slip distance = 15 cm .
he covers 15 cm after each slip.
at 28th jump , he covers = 28 * 15 = 420 cm .
At 29th jump , he will be at the top of well ( no slipping)
So answer will be 29.

Question no. 2

A vessel full of water weighs 40 kg. If it is one third filled , it weight becomes 20kg. What is the weight of empty vessel ? [CSAT 2020]

looks_one 10 kg

looks_two 15kg

looks_3 20 kg

looks_4 25kg

Option looks_one 10 kg

Solution :
weight of vessel = v kg
weight of full water = w kg
According to question
v + w = 40 kg ..........[1]
v + w/3 = 20 kg ...........[2]
solve the equation [1] and [2]
v = 10 kg.

Question no. 3

If 1 litre of water weighs 1 kg , then how many cubic millimeters of water will weigh 0.1 gm ? [CSAT 2020]

looks_one 1

looks_two 10

looks_3 100

looks_4 1000

Option looks_3 100

Solution :
Weight of 1 litre = 1 kg
1 litre = 1000 cm³
1 kg = 1000 cm³
1000 gm = 1000 cm³
1 gm = 1 cm³
0.1 gm = 0.1 cm³
0.1 gm = 0.1 *(1000)mm³ = 100 mm³

Question no. 4

A digit n > 3 is divisible by 3 but not divisible by 6 , Which one of the following is divisible by 4 ? [CSAT 2020]

looks_one 2n

looks_two 3n

looks_3 2n + 4

looks_4 3n + 1

Option looks_4 3n + 1

Solution :
In such question , assume n some value according to question
n > 3 , which is divisible by 3 but not divisible by 6 . assume n = 9
according to option
1). 2n = 18
3n = 27
2n + 4 = 22
3n + 1 = 28
option 4 is satisfying above condition.

Question no. 5

Which one of the following will have minimum change in its value if 5 is added to both numerator and the denominator of the fractions 2/3 , 3/4 , 4/5 and 5/6 ? [CSAT 2020]

looks_one 2/3

looks_two 3/4

looks_3 4/5

looks_4 5/6

Option looks_4 5/6

Solution :
larger the digit in numerator and denominator , lesser will be the difference and it will move towards one.

Question no. 6

How many pairs of natural number are there such that the difference of whose squares is 63 ? [CSAT 2020]

looks_one3

looks_two 5

looks_3 2

looks_4 4

Option looks_one3

Solution :
let the numbers are a and b
a² - b² = 63
(a-b)(a+b)= 7*9 ......[case 1]
(a-b)(a+b)= 21*3 .......[ case 2]
(a-b)(a+b)= 63 * 1 .......[ case 3]
Case 1 : a-b = 7
a+b = 9
a = 8 , b = 1
Case 2 : a-b = 3
a+b = 21
a = 12 , b = 9
Case 3 : a-b = 1
a+b = 63
a = 32 , b = 31

Question no. 7

A man takes half time in rowing a certain distance downstream than upstream. What is ratio of the speed in still water to the speed of current ? [CSAT 2020]

looks_one 1:2

looks_two 2 : 1

looks_3 1:3

looks_4 3:1

Option looks_3 1:3

Solution :
let the speed of current = u
speed of man in still water = v
during downstream :
D (distance) = (u+v) * t₁
during upstream :
D = (v-u) * t₂
According to Question :
t₁ = t₂/2
D/(u+v) = D/2(v-u)
2v - 2u = u+v
v = 3u

Question no. 8

Consider the following statements :
1. The minimum number of points of intersection of a square and a circle is 2 .
2. The maximum number of points of intersection of a square and a circle is 8 .
Which one of the following is correct? [CSAT 2020]

looks_one Only 1

looks_two Only 2

looks_3 both 1 and 2

looks_4 Neither 1 and 2

Option looks_two Only 2

Solution :
Minimum point is 1

Maximum Point is 8.

Question no. 9

A car travels from a place X to place Y at an average speed of v km/hr , from Y to X at an average speed of 2v km/hr , again from X to Y at an average speed of 3v km/hr and again from Y to X at an average speed of 4v km/hr . then the average speed of the car for the entire journey [CSAT 2020]

looks_one is less than v km/hr

looks_two lies betwwen v and 2v km/hr

looks_3 lies between 2v and 3v km/hr

looks_4 lies between 3v and 4v km/hr

Option looks_two lies betwwen v and 2v km/hr

Solution :
let D is the distance
Average Velocity = Total Distance / Total time

Question no. 10

A person X can complete 20% of work in 8 days and another person Y can complete 25% of the same work in 6 days. If they work together , in how many days will 40 % of the work be completed ? [CSAT 2020]

looks_one 6

looks_two 8

looks_3 10

looks_4 12

Option looks_one 6

Solution :
20 % of work = 8 days
complete work by X = 8 * 5 = 40 days
X do work in 1 day = 1/40 of work.
25% of work By Y = 6 days
complete work by Y = 6 * 4 = 24 days
Y do work in 1 day = 1/24 of work.
if both work together work in 1 day = 1/24 + 1/40 = 1/15
15 days require to complete = 100% work
for 40% work = .4 * 15 = 6 days

Question no. 11

As a result of 25% hike in the price of rice per Kg , a person is able to purchase 6 Kg less rice for Rs 1200. What was the original price of rice per kg ? [CSAT 2020]

looks_one Rs 30

looks_two Rs 40

looks_3 Rs 50

looks_4 Rs 60

Option looks_two Rs 40

Solution :
Original price = x
Number of kg of rice = 1200/x
Now price hike = 1.25 x
According to Question

Question no. 12

The average score of a batsman after his 50th innings was 46.4. after 60th innings , his average score increases by 2.6. What was his average score in the last ten innings ? [CSAT 2020]

looks_one 122

looks_two 91

looks_3 62

looks_4 49

Option looks_3 62

Solution :
(x₁ + x₂ +.....x₅₀)/50 = 46.4
x₁ + x₂ +.....x₅₀ = 46.4 * 50
(x₁ + x₂ +.....x₆₀)/60 = 49
x₁ + x₂ +.....x₅₀ + x₅₁ + .... x₆₀ = 60 * 49
x₅₁ + .... x₆₀ = 49*60 - 46.4 * 50 = 620
average score in the last ten innings = 620 / 10 = 62

Question no. 13

A bottle contains 20 litres of liquid A. 4 litres of liquid A is taken out of it and replaced by same quantity of liquid B. again 4 litres of the mixture is taken out and raplaced by same quantity of liquid B. what is the ratio of quantity of liquid A to that of liquid B in the final mixture ? [CSAT 2020]

looks_one 4:1

looks_two 5:1

looks_3 16:9

looks_4 17:8

Option looks_3 16:9

Solution :
A = 20 litre
After first operation :
A = 16 litre
B = 4 litre
After second operation :
4 litre of mixture is taken out , So 1/5 of 4 litre Of B = 0.8 litre and 1/5 of 16 litre of A = 3.2 litre
A = 16-3.2 = 12.8 litre
B = 4 - 0.8 = 3.2 Litre
After third operation :
A = 12.8
b = 7.2
A/B = 12.8/7.2 = 16/9

Question no. 14

How many different 5 letter words can be constructed using all the letters of the word 'DELHI' so that each word has to start with D and end with I ? [CSAT 2020]

looks_one24

looks_two 18

looks_3 12

looks_4 6

Option looks_4 6

Solution :
letters require = D 2 3 4 I
number of letter that can be used at position 2 = 3
number of letter that can be used at position 3 = 2
number of letter that can be used at position 4 = 1
number of ways that can be formed = 3 * 2 *1 = 6

Question no. 15

How many different sums can be formed with the denominations Rs 50 , Rs 100 , Rs 200 , Rs 500 and Rs 2000 taking at least three denomination at a time ? [CSAT 2020]

looks_one 16

looks_two 15

looks_3 14

looks_4 10

Option looks_one 16

Solution :
No. of different sum = C(5,3) + C(5,4) + C(5,5) = 10 + 5 + 1 = 16

Question no. 16

In a class , there are three groups A ,B and C . If one studentfrom group A and two students from group B are shifted to group C , then what happens to the average weight of the students of the class ? [CSAT 2020]

looks_one it increases

looks_two it decreases

looks_3 it remains the same

looks_4 No conclusion

Option looks_3 it remains the same

Solution :
It remains the same beacuse only students shifted only in group no to other class , So average of class will reamin same.

Question no. 17

For What value of n , the sum of digits in the number (10ⁿ + 1 ) is 2 ? [CSAT 2020]

looks_one For n = 0 only

looks_two For any whole number n

looks_3 for any positive integer n only

looks_4 for any real number n

Option looks_two For any whole number n

Solution :
For any whole number n

Question no. 18

A sum of Rs 2500 is distributed among X, Y and Z in the ratio 1/2 : 3/4 : 5/6 . What is the difference between the maximum share and the minimum share ? [CSAT 2020]

looks_one 300

looks_two 350

looks_3 400

looks_4 450

Option looks_3 400

Solution :
X : Y : Z = 1/2 : 3/4 : 5/6
simplify the ratio by multiply the lcm of denominator
X : Y : Z = 6 : 9 : 10
Share of X = (6 * 2500)/25 = 600
Share of Y = (9 * 2500)/25 = 900
Share of Z = (10 * 2500)/25 = 1000
difference between the maximum share and the minimum share = 1000 - 600 = 400

Question no. 19

What is the remainder when 51✕27✕35✕62✕75 is divided by 100 ? [CSAT 2020]

looks_one 50

looks_two 25

looks_3 5

looks_4 1

Option

Solution :
51✕27✕35✕62✕75
because of 62 ✕ 75 , last digit will be 0
51✕27✕35✕ will make ten's place digit '5'
So last two digit will be 50.
So , remainder will be 50.

Question no. 20

In adult population of a city , 40% men and 30 % women are married. What is the percentage of married adult population if no man marries more than one woman and no woman marries more than one man and there are no widows and widowers ? [CSAT 2020]

looks_one (33)1/7 %

looks_two 34%

looks_3 (34)2/7%

looks_4 35%

Option looks_3 (34)2/7%

Solution :
let population of men = x
let polulation of women = y
according to question
percentage of married male = 40% of x = 0.4 x
percentage of married female = 30% of y = 0.3 y
Number of married males = number of married females , i.e 0.4 x = 0.3 y
percentage of married adult population = Number of married / total population of adult
percentage of married adult population = (0.3y + 0.4x)/(x+y)
= (0.4x + 0.4x )/(x +0.4x/0.3 )
= 0.24/0.7 = (34)2/7%