MP PSC Civil Services Paper II Prelims  (CSAT)
• Comprehension
• Interpersonal skills including communication skills;
• Logical reasoning and analytical ability
• Decisionmaking and problemsolving
• General mental ability
• Basic numeracy (numbers and their relations, orders of magnitude, etc.) (Class X level), Data interpretation (charts, graphs, tables, data sufficiency etc.  Class X level)
• Hindi Language Comprehension skills (Class X level).
Note : The questions will be of multiple choice, objective type.
Sample questions with explanations
Qu. 01. If the number 7538m59n is divisible by 24, find the maximum value of m + n. A. 2 B. 5 C. 8 D. 11
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Sol: The factors of 24 are 8 and 3. First we shall check of 8 because it will give us the value of n. Obviously, for m=8, we have n = 2 because then 592 shall be divisible by 8, and hence the given number shall be divisible by 8.
Since the given number 7538m592 should be divisible by 3,the sum of the digits of given number should be divisible by 3, means 7+5+3+8+x+5+9+2 should be divisible by 3, hence 39+m should be divisible by 3, hence m = 0.
Hence, m + n = 0 + 2 = 2. Option A is the answer.
Qu. 02. If the HCF of 65 and 117 is expressible in the form 65m117, the value of m is: A. 4 B. 2 C. 1 D. 3
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Sol: The HCF of 65 and 113 is 13.
Hence, as given, 65m – 117 = 13, 65m = 130, m= 2. Option B is the answer.
Qu. 03. The largest number which divides 70 and 115 and leaves remainder 5 and 8 respectively is: A. 13
B. 35 C. 65 D. 55
Sol: Obviously the numbers whose HCF is to be obtained are: 705 = 65, and 1158 = 117. The HCF of 65 and 117 is 13. Option A is the answer.
Qu. 04. The difference between a twodigit number and the number obtained by interchanging the position of the digit is 45. What is the difference between the digits of the number? Options are: (A) 4 B) 5 (C) 6 (D) 7 (Reserve Bank of India Grade B Exam 2015)
Sol: Let the number be 10x+y. Its reverse number shall be: 10y+x. Their difference is: 10x+y 10yx = 9x9y = 45 è xy = 5. Option B is the answer.
Qu. 05: Given 7^113, what is the digit at unit place?
Sol: We get unit place equal to 1 when the index or power of 7 is 4 or multiple of 4.
Unit place of 7^113= 7^(28x4) + 1 = 1 x 7 = 7.
Use of Modular Arithmetic {Chapter Modular Arithmetic}:
7 == 7 mod 10, hence 7^2 = 49 == 49 mod 10, this is:
7^2 = 49 == (1) mod 10. Thus, 7^112 = (7^2)^56 = 1 mod 10. Thus 7^113 = 1x7 mod 10 = 7 mod 10.
Hence 7 is the required unit digit.
Questions from previous papers with explanations
4. A man sitting in an express train which is travelling at 60 kmph observes that a goods train, travelling in opposite direction, takes 9 seconds to pass him. If the goods train is 250 meters long, the speed of goods train in kmph is: (A) 20 (B) 30 (C) 40 (D) 50. MPPSC CSAT 2015.
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Sol: Let the required speed of goods train be P kmph.
Relative speed of express train = 60+P {trains are running in opposite directions}
Now; Distance = 250 meter = ¼ km, and Time elapsed in crossing = 9 sec = 9/3600 = 1/400 hrs.
Since, speed = distance/time, we have:
60+P = (1/4) / (1/400). This gives: 60 + P = 400/4. This is: 60 + P = 100
P = 40 kmph
{Please note that there is no relevance of length of express train, as the question does not ask as to the speed of goods train in being crossed by the express train. The question is with reference to time taken by the man (as if the man is running) in crossing the goods train}
5. In how many ways, a committee of 5 members comprising 3 men and 2 women can be selected from 6 men and 5 women? (A) 150 (B) 200 (C) 900 (D) 1800. MPPSC CSAT 2015
Sol: The selection of 3 men and 2 women out of 6 men and 5 women comprises of two independent events, i.e. selection of 3 men out of 6 men; and selection of 2 women out of 5 women.
Hence, the required answer is: C(6, 3) x C(5,2)
6. A square and a triangle have equal areas. If the ratio of side of square to height of triangle is 2/3, the ratio of height to base of triangle is: (A) 2/3 (B) 2/5 (C) 9/8 (D) 9/11
MPPSC CSAT 2015.
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Sol: Let the side of square be L and height of triangle be H. Given, L/H = 2/3. Hence, L = 2H/3.
Hence if base of triangle is B, then as given that square and a triangle have equal areas, we have:
(1/2) x B x H = L x L => (1/2) x B x H = (2H/3) x (2H/3) ==> B = (8/9) H
Then, H/B = 9/8. Option (C) is the answer.
7. A number consists of two digits “PQ”, whose sum is equal to 11. A new number “QP” can be generated by interchanging the digits of the original number. On subtracting “”QP” from “PQ”, we get 9. The original number “PQ” is: (A) 83 (B) 47 (C) 74 (D) 65. MPPSC CSAT 2015
Sol: Though the question can be solved by calculating: (10A+B) – (10B+A) = 9, i.e. AB = 1. Option (D) is the answer, as 65 = 1, I shall suggest you to solve by observing the options. Option (A) is not the answer as the difference between 83 and 38 is not 9. Similarly, options (B) and (C) can be ruled out. Option (D) is the correct answer as 6556 = 9.
â€‹â€‹Qu. Directions: Consider the following data on the number of Languages known by people in a sample size of 150. Based on this data, answer the following questions:
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8. The number of persons who know at least languages: (A) 50 (B) 55 (C) 80 (D) 120
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9. The number of persons who know at most 4 languages: (A) 15 (B) 30 (C) 105 (D) 135
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10. The fraction of the number of persons knowing at least 2 languages to the persons knowing at most 2 languages: (A) 27/14 (B) 11/30 (C) 7/8 (D) 3/10. MPPSC CSAT 2015
Solutions:
8. The number of persons who know at least 3 languages = 50+15+10+5 = 80. Option (B) is the answer.
9. The number persons who know at most 4 languages = 15+55+50+15 = 135. Option (D) is the answer.
10. The fraction of the number of persons knowing at least 2 languages to the persons knowing at most 2 languages = (55+50+15+10+5)/(15+55) = 135/70 = 27/14. Option (A) is the answer.
11. A father is thrice as old as his son today. Ten years later, father will be twice as old as his son. What will be the age of the son when his age will be 3/4th of his father’s age? (A) 30 (B) 45 (C) 60 (D) 75. MPPSC CSAT 2015
Sol: Let the present age of son be P years today, then the age of father is 3P.
Ten years later, it is given that (3P+10) = 2(P+10). This gives, P=10. Thus the present ages of father and son are 30 and 10 years respectively.
Let it be after M years when son’s age will be 3/4th of his father’s age. Then; 10+M = (3/4)(30+M)
This is: 40+4M = 90+3M. This gives: M = 50. Son’s age shall be 10+50 = 60 years when his father shall be of 30+M = 80 years of age (Obviously, 60 is 3/4th of 80). Option (C) is the answer.
12. Rohan spends 25% of his monthly scholarship on books. Out of the balance amount, he spends 75% on the hostel and college fees. If he is left with Rs. 120 at the end of a month, what is Rohan’s monthly scholarship? (A) 640 (B) 850 (C) 1000 (D) 1260. MPPSC CSAT 2015
Sol: Let the scholarship be Rs. 1200P.
As given, he spends 25% of it on books, he has a balance amount of 75%, i.e., Rs. 900P, and he spends 75% of it on hostel etc., i.e. he is left with 25% of Rs. 900P, i.e. Rs. Rs. 225P.
But given that 225P = 120. Thus 1200P = 1200 x (120/225) = (48x120)/9 = 16 x 40 = 640. Option (A) is the answer. {I presumed the amount of Rs. 1200P, as it was less cumbersome to work out its 75%; and 25% of balance amount}
13. Kumar spends 30% of his monthly income on food, 20% on electricity bill and 70% of the remaining income on house rent. After all these expenses, he has Rs. 300 left with him. Kumar’s monthly income in Rs. is: (A) 2000 (B) 3000 (C) 2500 (D) 3500. MPPSC CSAT 2015
Sol: Let the income be Rs. 1200P.
As given, he spends 30% and 20% on food and electricity respectively, thus spends 50%. He has a balance amount of 50%, i.e., Rs. 600P, and its 70% is spent on house rent and he is left with 30% of Rs. 600P, i.e. Rs. 180P.
But given that 180P = 300. Thus 60P = 100. Thus 1200P = 2000. Option (A) is the answer.
14. Vandana has less than Rs. 200 and decides to give half of the amount to her friend. She continues to give half of the money she has in her hand at any point of time to her other friends during this process till she has a balance in her hand, which is equal to the first two digit prime number. The amount given by Vandana to her first friend divided by the amount given to her last friend is equal to: (A) 2 (B) 8 (C) 4 (D) 16. MPPSC CSAT 2015
Sol: The first two digit prime number is 11.
We have, as given in the question, this number sequence: 22, 44, 88, 176
Thus Vandana has Rs. 176 with her (less than Rs. 200) and her first friend whom she gives half of 176 gets Rs. 88. Obviously, as given, her last friend gets Rs. 22.
The answer is 88/22 = 4. Option (C) is the answer.
15. In a basketball match, all ten players shake hands with each other once after the match. How many handshakes will be there? (A) 20 (B) 45 (C) 55 (D) 90. MPPSC CSAT 2016
Sol: We shall do it with two methods:

Of course, all 10 players will shake hands with remaining 9 players hence there shall be 90 handshakes, but no one will shake hands with oneself, hence there are a total of 45 handshakes.

We are selecting 2 players out of 10 who will shake hands with each other. Hence answer is: C(10,2) = 10!/(8! X 2!) = 5 x 9 = 45
16. Shyam buys pineapples and coconuts. A pineapple costs Rs. 70 and a coconut costs Rs. 50. If Shyam spends a total amount of Rs. 380, how many pineapples does he purchase? (A) 2 (B) 3 (C) 4 (D) Data is insufficient. MPPSC CSAT 2016
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Sol: If a pineapple and a coconut costs Rs. P and C respectively, then, as given, 70P + 50C = 380. Then 7P + 5C = 38
Now putting C = 1, 2, 3, .. we have, P = 33/7, 4, 23/7, .
Obviously for C=2, we get an integral value of P = 4. Thus option (C) is the answer.
Qu. 18. If A+B = C+D and A+D > B+C, then which one of the following is definitely wrong? (A) A>B (B) A>C (C) C>D (D) B>D. MPPSC CSAT 2016
Sol: Given, A+B = C+D, or A – D = C – B …. (1)
Given, A+D > B+C ……(2)
Adding (1) and (2), 2A > 2C, A >C. But this option is already given, hence discarded as we have to work out the impossible situation.
Thus, subtracting (1) from (2), we have, 2D > 2B, or D>B. It is in contradiction with given option (D), hence option (D) is the answer.
17. Rohit scored 30% marks and failed by 15 marks. Mohan 40% marks and obtained 35 marks more than those required to pass. The pass percentage is: (A) 28% (B) 33% (C) 35% (D) 42%. MPPSC CSAT 2016
Sol: Let the maximum marks be 100P. Then as given, 30P + 15 = 40P – 35
10P = 50 or P = 5. Thus maximum marks are 500, and pass marks are 30P + 15 = 165.
Thus pass percentage is: 165x100/500 = 33%. Thus option (B) is the answer.
Qu. 19. In a local starting from station A, the number of female passengers is half the number of male passengers. At the next station, 10 male passengers leave the train and 5 female passengers board the train, which makes the number of male and female passengers equal. How many passengers entered the local train at station A? (A) 15 (B) 30 (C) 36 (D) 45. MPPSC CSAT 2016
Sol: Let there be 100P male passengers and 50P female passengers initially (as given), thus there were 150P passengers at station A.
Given, 100P – 10 = 50P + 5; or, 50P = 15. Hence, 150P = 45. Option (D) is the answer.
Qu. 20. A group of friends organized a party and planned to spend Rs. 960 for total expenses. Four of them could not make it. Hence, the remaining people had to pay Rs. 40 per head to cover the expenses. The number of people who attended the party was: A) 8 (B) 12 (C) 16 (D) 24. MPPSC CSAT 2016
Since () value is not feasible, P = 8. Option (A) is the answer.
Note: I shall advise not to calculate
that much. You must put the
options in equation (1), and
find out the correct answer.
Since () value is not feasible, P = 8. Option (A) is the answer.
Note: I shall advise not to calculate
that much. You must put the
options in equation (1), and
find out the correct answer.
Qu. 21.
Let the age of Rahul 8 years ago be 4P years.
Then, as given, age of Rahul’s son 8 years ago = P
Their present ages: Rahul = 4P+8; Rahul’s son = P+8
Their ages after 8 years:
Rahul = 4P+8+8; Rahul’s son = P+8+8 …. (1)
Their ages after 8 years: Rahul = 4P+16; Rahul’s son = P+16
Given, 4P+16 = 2(P+16), or, 2P = 16, P = 8
Present ages of Rahul and his son {from (1)}= 40, 16
Option (B) is the answer.
Qu. 22. The following operations are carried out sequentially on a number to yield the same starting number: First divide be two, then take square root and finally take the cube. What is this number? A) 64 (B) 27 (C) 8 (D) 216. MPPSC CSAT 2016
Sol: There is no need of any calculation. We take the options one by one. First, 64 when divided by 2 and getting 32 does not yield a square root, hence ruled out. Second, 27 is not halved. Third, 8 when halved is 4 whose square root is 2 and whose cube is 8, again the same number. Hence, option (C) is the answer.
Fourth, 216 when halved yields 108 which is not a perfect square.
Otherwise, doing calculation, if the number is 2P, we have: P^(3/2) = P. This is: P^3 = P^2 or (P^2)(P – 1) = 0. Since P 0, P = 1. Hence, 2P = 2.
The required number is 2.
Qu. 23: The average weight of three persons is equal to 140 kg. The ratio of these three persons’ weights is 1:2:4. What is the weight (in Kg) of the heaviest person? A) 140 (B) 120 (C) 60 (D) 240. MPPSC CSAT 2016
Sol: Obviously the sum of weights of three persons is equal to 140x3 = 420 kg. Given the ratio of weights 1:2:4, the weight of heaviest person = 420 x (4/7) = 240 kg. Option (D) is the answer.
Qu. 24: The square root of a two digit number gives a prime number. The sum of the two digits of such a number is equal to: (a) 8 (b) 13 (c) 10 (d) 15. MPPSC CSAT 2017
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Sol: This is a misleading question, as square root of 25 (which is a two digit number) is 5 which is a prime number and 2+5=7. Again, square root of 49 (which is a two digit number) is 7 which is again a prime number and 4+9=13. But since there is no option for 7, the correct answer is 13, hence option (b) is the choice.