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Civil Services Paper II - (200 marks) CSAT
• Comprehension
• Interpersonal skills including communication skills;
• Logical reasoning and analytical ability
• Decision-making and problem-solving
• 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)
• English Language Comprehension skills (Class X level).
Note  : The questions will be of multiple choice, objective type.

We are solving previous years' question papers of CSAT (Quantitative Mathematics only) in this section. Please see the way the questions have been solved and explained.

Shun the fear of Mathematics and Logic.

Every student has a hidden knack for mathematics. This section is an attempt of regenerating confidence in you to solve Mathematics so as to lead you to the path of success by extracting and unleashing the unbridled potential from within him of mastering Mathematics.

Average rainfall in a city for the first 4 days was recorded to be 0.4 inch. The rainfall on the last two days was in the ratio 4:3. The average of 6 days was 0.5 inch. What was the rainfall on fifth day? (A) 0.6 inch (B) 0.7 inch (C) 0.8 inch (D) 0.9 inch. CSAT 2017

Sol: Obviously the total rainfall for the first 4 days was 4 x 0.4 = 1.6 inch. Let the rainfall on the last two days be 4P and 3P.

Then total rainfall for 6 days = 1.6+4P+3P = 1.6+7P, which is given to be was 6x0.5 = 3 inch.

Thus, 7P = 3 - 1.6 = 1.4, i.e. P = 0.2. Hence the rainfall on fifth day = 4P = 4 x 0.2 = 0.8 inch.

The monthly incomes of X and Y are in the ratio 4:3, and their monthly expenses are in the ratio of 3:2. However, each saves Rupees 6000 per month. What is their total monthly income? (A) Rs. 28000 (B) Rs. 42000 (C) Rs. 56000 (D) Rs. 84000. CSAT 2017

Sol: Let the monthly income of X and Y be 4P and 3P; and their expenditures 3Q and 2Q.

Given, 4P-3Q = 6000 and 3P-2Q = 6000.

Since we need to work out 7P, we shall multiply first equation by 2 and second equation by (-3),

we get 8P-6P = 12000 and -9P+6P = -18000. We add these two equations, and get:

(-P) = -6000, i.e., P = 6000 and 7P = Rs. 42000

A SHORT-CUT

Since 4P-3Q = 6000 and 3P-2Q = 6000 (given) , we have 4P-3Q = 3P-2Q (= 6000)

This gives, P=Q. This gives, 4P-3P = 6000 or P = 6000 or 7P = 42000.

There are certain two 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. CSAT 2017

Sol: Obviously the numbers are 10P+Q and 10Q+P where P and Q are whole numbers from 0 to 9. Given, (10P+Q) - (10Q+P) = 27 ==> 9P-9Q = 27 ==> 9(P-Q) = 27 ==> P-Q = 3

Hence P can have values from 3 to 9 (If P = 0, 1 or 2; we cannot have some Q whose difference with P is 3). Thus P can have 7 values (3, 4, 5, 6, 7, 8, 9). Hence (D) is the answer.

How many numbers are there between 99 and 1000 such that digit 8 occupies the unit place? (A) 64 (B) 80 (C) 90 (D) 104. CSAT 2017

Sol: The unit place is fixed for 8. The tens place can be filled up in 10 ways (0 to 9) and since hundreds place cannot occupy 0 (you cannot have a three digit number like 098), hundreds place can be filled up in 9 ways (1 to 9). Thus we shall have 10x9 = 90 numbers. Hence (C) is the answer.

The age of Mr X last year was the square of a number and 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. CSAT 2017

Sol: This is rather a very simple but tricky question.

Let the present age of X is P. Thus (P-1) must be a square and (P+1) must be a cube. Or there should be difference of 2 between a cube number and a square number. Since (P+1) is greater than (P-1), we shall take into consideration ‘cubes’ first.

Cube of 2 is 8 but 8-2 = 6 is not a square.

Cube of 3 is 27 and 27-2 = 25 is a square (of 5).

Thus present age of X is 26 years.

The next cube number is 64 (of 4). X will have to wait for 38 years as 64-26 = 38. Option (B) is the answer.

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 CSAT 2017

Sol: Let P do 3A units of work in one day, then, as given, Q will do A units of work per day. Both will do 4A units of work per day. As given, R will do A (1/4th of 4A) units of work per day. Thus units of work done by P, Q and R are 3, 1, 1 units per day respectively which is the ratio of earnings amongst them. Options (A) is the answer.

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 color? (A) 17 (B) 16 (C) 13 (D) 11. CSAT 2017

Sol: This question is not from Probability, rather is a logical problem. Understand it like this. Suppose you pick up 3 balls, may be that you are picking up all green or all white or all red balls. Similarly, suppose you pick up 10 balls, may be that you are picking up all green and two white or no red balls; or you are picking up 5 green and three white or two red balls. Thus there may be several permutations and combinations upto you pick up 15 balls (because you may be picking up 8 greens and 7 white balls). Hence, as given, of assuring of picking up at least one ball of each color, you must pick up 16 balls at the least (8 greens at most + 7 white at most + 1 red at least). And remember, here may also be various permutations and combinations (like 5 green + 6 white + 4 red; and many more). Option (B) is the answer.

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%. CSAT 2017

Sol: An easy question. Supposing the CP of phone Rs. 100, the SP for Gopal is Rs. 110. It is sold again at 10% loss, i.e. for 110 – 11 = 99 Rs. Thus there is a loss of Re. 1. Since this loss is at Rs. 100, there is 1% loss. Option (B) is the answer.

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. CSAT 2017

Sol: Let us represent the nine persons by 1, 2, 3, 4, 5, 6, 7, 8, 9

Given weight of all nine persons = 9 x 50 = 450

Given, weight of first five persons, i.e. 1+2+3+4+5 = 5x45 = 225

Given, weight of last five persons, i.e. 5+6+7+8+9 = 5x5 = 275

Adding these two: weight of (1+2+3+4+5) + (5+6+7+8+9) = 225 + 275 = 500

This is: weight of (1+2+3+4+5+6+7+8+9) + 5 =  500

Hence weight of 5 = 500 - weight of (1+2+3+4+5+6+7+8+9) = 500 – 450 = 50 Kg. Option (C) is the answer.

The average monthly income of a person in a certain family of 5 is Rs. 10,000. What will be the average monthly income of a person in the same family if the income of one person increased by Rs. 1,20,000 per year? (a) Rs. 12,000 (b) Rs. 16,000 (c) Rs. 20,000 (d) Rs. 34,000. CSAT 2016

Sol: Obviously, the total income of 5 members = 5x10000 = 50000

After adding additional income of Rs. 120000 per year, i.e. Rs. 10000, it becomes, Rs. 60000

Average income of one member = 60000/5 = 12000. Option (d) is the answer.

A cylindrical overhead tank of radius 2 m and height 7 m is to be filled from an underground tank of size 5.5 m × 4 m × 6 m. How much portion of the underground tank is still filled with water after filling the overhead tank completely? (a) 1/3 (b) 1/2 (c) 1/4 (d) 1/6. CSAT 2016

Sol: The volume V of overhead tank = pi x r x r x h, where r is the radius of tank and h is the height.

The volume W of underground tank = l x b x d, where l, b, d are the three dimensions of the tank.

V = pi x 2 x 2 x 7 = 88 (as pi = 22/7) and W = 5.5 x 4 x 6 = 132 (unit of volume is cubic meter)

The filled portion of underground tank = 132 - 88 = 44, which is 44/132 = 1/3. Option (a) is the answer.

A person allows a 10% discount for cash payment from the marked price of a toy and still he makes a 10% gain. What is the cost price of the toy which is marked Rs. 770? (a) Rs. 610 (b) Rs. 620 (c) Rs. 630 (d) Rs. 640. CSAT 2016

Sol: Obviously, as given, SP = 90% of MP and SP = 110% of CP

Thus 110% of CP = 90% of MP ==> CP = (90 x MP)/110 = (9 x MP)/11 = (9x770)/11 = 9x70 = 630 Rs. Option (c) is the answer.

A piece of tin is in the form of a rectangle having length 12 cm and width 8 cm. This is used to construct a closed cube. The side of the cube is: (a) 2 cm (b) 3 cm (c) 4 cm (d) 6 cm. CSAT 2016

Sol: The area of rectangular sheet = 12x8 = 96 sq. cm. Since this sheet is used to construct a closed cube, this area shall be the total surface area of this closed cube.

Surface area of closed cube = 6x4a = 24a, where 'a' is the side of the cube.

24a = 96, a = 4. Option (c) is the answer.

Two numbers X and Y are respectively 20% and 28% less than a third number Z. By what percentage is the number? (a) 12 (b) 10 (c) 9 (d) 8. CSAT 2016

Sol: Let Z = 100P. Then X = 80P and Y = 72P.

Thus Y is 8P less than X. The required percentage = {8Px100/80P}% = 10%. Option (b) is the answer.

Ram and Shyam work on a job together for four days and complete 60% of it. Ram takes leave then and Shyam works for eight more days to complete the job. How long would Ram take to complete the entire job alone? (a) 6 days (b) 8 days (c) 10 days (d) 11 days. CSAT 2016

Sol: Given that Shyam does 40% of the work, i.e. 2/5 of the work in 8 days. Thus Shyam alone would complete the work in (5/2)x8 = 20 days. Thus Shyam does 1/20 part of the work in one day. In 4 days he does 4/20 or 1/5 part of the work.

Given that both Ram and Shyam do 60%, i.e. 3/5 work out of which 1/5 is done by Shyam.

Thus in 4 days, Ram does 2/5 work. Thus Ram alone complete the work in 4x(5/2) = 10 days.

AB is a vertical trunk of a huge tree with A being the point where the base of the trunk touches the ground. Due to a cyclone, the trunk has been broken at C which is at a height of 12 meters, broken part is partially attached to the vertical portion of the trunk at C. If the end of the broken part B touches the ground at D which is at a distance of 5 meters from A, then the original height of the trunk is: (a) 20m (b) 25m (c) 30m (d) 35m. CSAT 2016

Sol:

If R and S are different integers both divisible by 5, then which of the following is not necessarily true? (a) R - S is divisible by 5 (b) R + S is divisible by 10 (c) R × S is divisible by 25 (d) R2 + S2 is divisible by 5. CSAT 2016

Sol: Given R and S are multiples of 5, let R = 5X and S = 5Y. Then R-S = 5(X-Y) which is divisible by 5, hence (a) is true. R+S = 5(X+Y) which may not be divisible by 5, hence (b) is false. RxS = 25XY which is divisible by 25, hence (c) is true. R square + S square = 25(X square + Y square) is divisible by 5 hence divisible by 5, hence (d) is true. Option (b) is the answer.

How many numbers are there between 100 and 300 which either begin with or end with 2? (a) 110 (b) 111 (c) 112 (d) None of the above. CSAT 2016

Sol: How many numbers are there between 100 and 300 which either begin with or end with 2? (a) 110 (b) 111 (c) 112 (d) None of the above. Two options are possible (x represents the digit):

1. 1x2. The number of such numbers shall be 10 (from 101 to 199, i.e. 102, 112, 122, ..., 192).

2. 2xx. The number of such numbers shall be 100 (all numbers from 200 to 299).

Thus total number of required ways = 110. Option (a) is the answer.

W can do 25% of a work-in 30 days, X can do 1/4 of the work in 10 days, Y can do 40% of the work in 40 days and Z can do 1/3 of the work in 13 days. Who will complete the work first? (a) W (b) X (c) Y (d) Z. CSAT 2016

Sol:

There is an order of 19000 quantity of a particular product from a customer. The firm produces 1000 quantity of that product per day out of which 5% are unfit for sale. In how many days will the order be completed? (a) 18 (b) 19 (c) 20 (d) 22. CSAT 2016

Sol: As given, 95 articles are produced every day. 19000 articles will be produced in 19000/95 = 20 days. Option (c) is the answer.

A class starts at 11:00 am and lasts till 2:27 pm. Four periods of equal duration are held during this interval. After every period, a rest of 5 minutes is given to the students. The exact duration of each period is: (a) 48 minutes (b) 50 minutes (c) 51 minutes (d) 53 minutes. CSAT 2016

Sol: The total duration is 3 hours and 27 minutes, means, 207 minutes. There shall be 3 intervals (between first and second class; between second and third class; between third and fourth class) of 5 minutes each totaling to 15 minutes. Total duration of class periods = 207-15 = 192 minutes. Duration of one class = 192/4 = 48 minutes. Option (a) is the answer.

30g of sugar was mixed in 180 ml water in a vessel A, 40 g of sugar Was mixed in 280 ml of water in vessel B and 20 g of sugar was mixed in 100 ml of water in vessel C. The solution in vessel B is: (a) sweeter than that in C (b) sweeter than that in A (c) as sweet as that in C (d) less sweet than that in C. CSAT 2016

Sol: Content of sugar in A = 30g/180 ml = 1/6 g/ml
Content of sugar in B = 40g/280 ml = 1/7 g/ml
Content of sugar in C = 20g/100 ml = 1/5 g/ml

Since 5 < 6 < 7, we have, (1/5) > (1/6) > (1/7) or (1/7) < (1/6) < (1/5)

This means: Content of sugar in B < Content of sugar in A < Content of sugar in C
Hence sugar content in vessel C is highest, followed by vessels A and then B. Option (d) is the answer.

30g of sugar was mixed in 180 ml water in a vessel A, 40 g of sugar Was mixed in 280 ml of water in vessel B and 20 g of sugar was mixed in 100 ml of water in vessel C. The solution in vessel B is: (a) sweeter than that in C (b) sweeter than that in A (c) as sweet as that in C (d) less sweet than that in C. CSAT 2016

Sol: Let total number of students in the class be P. Since every student contributes Rs. P, total contribution is P^2. Given, P^2 + 2 = 443. P^2 = 441. P = 21. Option (b) is the answer.

Each of A., B, C and D has Rs 100. A pays Rs 20 to B, who pays Rs 10 to C, who gets Rs 30 from D. In this context, which one of the following statements is not correct? (a) C is the richest (b) D is the poorest. (c) C has more than what A and D have together. (d) B is richer than D Answer. CSAT 2015. CSAT 2015

Sol:

We did it by creating a matrix. This method does away with any ambiguity in framing solution. From the given options, it is clear that option (c) is the answer.

In a town, 45% population read magazine A, 55% read magazine B, 40% read magazine C, 30% read magazines A and B, 15% read magazines B and C, 25% read magazines A and C; and 10% read all the three magazines. What percentage does not read any magazine? (a) 10% (b) 15% (c) 20% (d) 25%. CSAT 2015

Sol: It is given that n(A) = 45%; n(B) = 55%; n(C) = 40%

Further given that 30% read magazines A and B, hence n(A ∩ B) = 30%. Similarly, n(B ∩ C) = 15%, n(A ∩ C) = 25% and n(A ∩ B ∩ C) = 10%

We know that:

n(A U B U C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C)

Hence, n(A U B U C) = 45 + 55 + 40 - 30 - 25 - 15 + 10 = 80.

80% population read any magazine. Hence, 20% population does not read any magazine. The option (d) is the answer.

Shahid and Rohit start from the same point in opposite directions. After each 1 km, Shahid always turns left and Rohit always turns right. Which of the following statements is correct? (a) After both have travelled 2 km, the distance between them is 4 km. (b) They meet after each has travelled 3km. (c) They meet for the first time after each has travelled 4 km. (d)They go on without ever meeting again. CSAT 2015

Sol:

Sol:  Both start from O. And goes like this:

OA -> AB -> BM; and OC->CD->DM. They meet at M. Obviously both travel three km each. Option B is the answer.

In a 500 meter race, B starts 45 meter ahead of A, but A wins the race while B is still 35 meter behind. What is the ratio of the speeds of A to B assuming that both start at the same time? (a) 25: 21 (b) 25: 20 (c) 5:3 (d) 5:7. CSAT 2015

Sol: Obviously in the same time, A covers 500 meter, and B covers (500-45-35) meter, i.e. 420 meters. Speed = distance/time.

Since time is same for both A and B, we have:

Speed of A : Speed of B = Distance covered by A : Distance covered by B

== 500:420 = 25:21. Option (a) is the answer.

Two equal glasses of same type are respectively 1/3 and 1/4 full of milk. They are then filled up with water and the contents are mixed in a pot. What is the ratio of milk and water in the pot? (a) 7 : 17 (b) 1 : 3 (c) 9 : 21 (d) 11 : 23. CSAT 2015

Sol: If we have 120 ml in each of the glass, as per the parts filled up by milk, we have following matrix:

If quantity in both glasses is filled in one glass, the ratio of milk and water = 70:170 = 7:17. Option (a) is the answer.

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