1.A and B undertake a piece of work for Rs. 250 . A alone can do that work in 5 days and B alone can do that work in 15 days . With the help of C, they finish the work in 3 days . If every one gets paid in proportion to work done by them, the amount C will get is
1) Rs.50 2) Rs.100 3) Rs.150 4) Rs.200
2.Dealer purchased an article for Rs. 900 and fixes the list price in such a way that he gains 20% after allowing 10% discount, then the list price is
1) Rs.1180 2) Rs.1080 3) Rs.1200 4) Rs.1100
3.Cost price of 28 articles is equal to Sale price of 21 articles, then percentage of profit is
1) 12% 2) 33⅓ % 3) 20% 4) 22%
4.If the speed of a boat in still water is 20 km/hr and the speed of the current is 5 km/hr, then the time taken by the boat to travel 100 km with the current is
1) 2 hrs 2) 3 hrs 3) 4 hrs 4) 7 hrs
5.If a sum of money doubles itself in 8 yrs, then the interest rate in percentage is ?
1) 8 ½% 2) 10% 3) 10 ½ % 4) 12 ½ %
6.If ( x - 5)2 + (y - 2)2 + (z - 9)2 = 0 , then value of (x + y - z) is
1) 16 2) -1 3) -2 4) 12
7.Points P , Q and R are on a circle such that ∠PQR = 40° and ∠QRP = 60° . Then the subtended angle by arc QR at the centre is
1) 80° 2) 120° 3) 140° 4) 160°
8.If α + β = 90° and α:β = 2:1, then the ratio of cosα to cosβ is
1) 1:√3 2) 1:3 3) 1:√2 4) 1:2
9.The least number to be added to 13851 to get a number which is divisible by 87 is
1) 18 2) 43 3) 54 4) 69
10.Two numbers are in the ratio 3:5. If 6 is added to both of them, the ratio becomes 2:3. The numbers are 1) 21 and 35 2) 30 and 50 3) 24 and 40 4) 18 and 30
11.Amir, Amit, Amrit and Amrut share some coins from an old collection. Amit takes half as many coins as Amir and Amrut take together. The ratio of number of coins Amrit takes to the remaining is 2:5. Amit gets two coins less than Amrut. Amrit gets twice as many as Amir. What is the total number of coins got by all 4 together?
(1) 21 (2) 42 (3) 28 (4) 43 (5) None of these
Solution:
Let the coins taken by Amir, Amit, Amrit and Amrut be denoted by a, b, c and d respectively. By using the given data, the following equations can be written:
i) b = (1/2) × (a + d) (1) ii) c = (2/5) × (a + b + d) (2) iii) b = (d - 2) (3)
iv) c = 2a (4) Using (1) and (3), we get,
2(d - 2) = a + d ⟹ d = a + 4 (5) Using (5) in (3), we get
b = a + 2 (6)
Using (4), (5) and (6) in (3), we get
c = (2/5) × (a + b + d) ⟹ 2a = (2/5) × (a + a + 2 + a + 4) Solving, we get a = 3, b = a + 2 = 5, c = 2a = 6, d = a + 4 = 7.
Thus, the total number of coins = a + b + c + d = 3 + 5 + 6 + 7 = 21. Hence, the correct answer is option 1.
12.If n is a positive integer greater than 4, then the highest number by which the expression: (n6 – n4)(2n2 + 3n + 2) (n + 1)2 is always divisible by is
a a b Wall (1) 96 (2) 24 (3) 144 (4) 36 (5) 48 Solution:
The terms of the given expression could be re-arranged to form: (n6-n4)(2n2+3n+2) (n+1)2
= n4 × (n2-1) × (n+1)(2n+1) × (n+1)2 = n(n+1)(n-1) × n(n+1)(2n+1) × n2(n+1)2
It is known that: (n-1).n.(n+1) is always divisible by 6
(Since 3 Consecutive integers will have a number divisible by 3 and since an even number would also be present, hence divisible by 2 as well).
Also n(n+1)(2n+1) is always divisible by 6
(Since, n(n+1)(2n+1) = n(n+1)(n-1) + n(n+1)(n+2), here both the terms are divisible by 6, hence their sum is also divisible by 6).
Also n2(n+1)2 is always divisible by 4 (Since, either n or (n+1)is even, and its square would be divisible by 4).
Hence the given expression is always divisible by: (n-1)n(n+1) × n(n+1)(2n+1)
= 6 × 6 × 4=144.
Hence, the correct answer is option 3.
13.In a three hour cooking session, chef Seethapathi continuously prepares one or the other of the two items, dosa and poori. The mean time taken to prepare a poori is twice the mean time taken to prepare a dosa. Also, seven times the total time spent on preparing pooris is equal to eight times the total time spent on preparing dosas. If the total number of items prepared is 33, how many pooris can be prepared in 72 minutes?
(1) 8 (2) 6 (3) 12 (4) 9 (5) None of these
Solution:
Let the number of dosas and pooris prepared be denoted by d and p respectively. Also, let the mean time taken to prepare adosa be denoted by t.
Then, the mean time taken to prepare a poori is given by 2t.
Total time = Total time spent on making dosas + Total time spent on making pooris = (t × d) + (2t × p) = t × (d + 2p) = 180 (1)
From given data, 7 × (2t × p) = 8 × (t × d) ⟹ p/d = 4/7 (2) If total number of items = 33, then (d + p) = 33 (3) Using (2) in (3), we get: d = 21, p = 12.
Using these values in (1), we get: t × (21 + 2 × 12) = 180 → t = 180/45 = 4 mins.
Thus, the number of pooris that can be prepared in 72 minutes = 72/ (2 × 4) = 9. Hence, the correct answer is option 4.
14.Ram has 100 ft of barbed wire to fence 3 sides of his rectangular garden, while its 4th side is completely bounded by a wall. For the largest possible area that can be generated, the perimeter of the rectangle is (in ft)
(1) 150 (2) 120 (3) 133 (4) 160 (5) 180
Solution:
Consider the following rectangle with sides ‘a’ and ‘b’. The length of the barbed wire
= 100 ft = 2a + b
Area of the rectangle = a × b.
For this value to be maximum, given the constraint condition, 2a = b. But it is known that: 2a + b = 100 = 2b.
Therefore, b = 50, a = 25.
Hence, the perimeter is = 2(a + b) = 2(25 + 50) = 150 ft. Hence, the correct answer is option 1.
15.The product of the first 4 terms of an Arithmetic Progression is 6160. Given that the common difference between the terms is equal to 3, find the 100th term.
Solution:
Let the terms of the A.P be: a, a + 3, a + 6, a + 9. (Since the common difference is 3)
It is given that the product of the first 4 terms is equal to 6160. → (a) × (a + 3) × (a + 6) × (a + 9) = 6160
→ (a2 + 9a) × (a2 +9a +18) = 6160 Let (a2 + 9a) = b,
→ b × (b + 18)=6160
Solving for the above equation, we get: b=70 → (a2 + 9a) = b = 70 → a(a+9) = 70
Solving for the above equation, we get: a = 5 Hence, the 100th term of this A.P = a + (100 - 1) × d = 5 + 99 × 3 = 302.
Alternately,
The product 6160 can be factorised as follows: 6160 = 2×2×2×2×5×7×11
= 2×2×2×5×7×2×11 = 8×5×14×11.
Thus, it can be observed that the numbers 5, 8, 11 and 14 are in Arithmetic Progression and their product is 6160. Here, a = 5 and d = 3
Hence, the 100th term of this A.P = a + (100 - 1) × d = 5 + 99 × 3 = 302.
Hence, the correct answer is option 5.
Directions for questions 16 and 17: Refer the following data to answer the questions given below.
Twelve chairs or seats (numbered 1 to 12 from left to right) are arranged in a row. If four men are to be seated such that 2 of them occupy seats 1 and 8, and no two men sit adjacent to each other.
16.What is the total number of ways in which the four men can be seated?
(1) 16 (2) 24 (3) 384 (4) 360 (5) None of these
Solution:
Given 12 chairs numbered from 1 to 12 in a row:
X ___ ___ ___ ___ ___ ___ X ___ ___ ___ ___ 1 2 3 4 5 6 7 8 9 10 11 12 The various ways in which the four men can be seated are:
(a) 1, 3, 8, _ : The 4th man can sit in one of the 5, 6, 10 , 11 or 12 numbered chair, hence 5 ways. (b) 1, 4, 8, _ : The 4th man can sit in one of the 6, 10 , 11 or 12 numbered chair, hence 4 ways. (c) 1, 5, 8, _ : The 4th man can sit in one of the 10 , 11 or 12 numbered chair, hence 3 ways. (d) 1, 6, 8, _ : The 4th man can sit in one of the 10 , 11 or 12 numbered chair, hence 3 ways. (e) 1, 8, 10, _ : The 4th man can sit in the 12 numbered chair, hence only 1 way.
Thus, there are 16 such combinations. In each of these combinations, the 4 men could be made to arrange themselves in 4! ways.
Hence, the required number of ways is given by: 16 × 4! = 16 × 24 = 384 ways.
Alternately: Since seats 1 and 8 are always occupied, there are 10 seats remaining, out of which no one can be seated in chairs 2, 7 and 9. This leaves 7 seats for the remaining 2 men. The number of ways they can be seated is given by 7C2 = 21 ways, out of which there are 5 cases (Seats 3, 4; 4, 5; 5, 6; 10, 11; 11, 12) in which the 2 men sit next to each other.
Thus there are 21 – 5 = 16 combinations in which the 4 men can be seated according to the given conditions. In each of these combinations, the 4 men could be made to arrange themselves in 4! ways.
Hence, the required number of ways is given by: 16 × 4! = 16 × 24 = 384 ways.
17.Now, 2 women are also to be seated (along with the 4 men) in such a way that there is a gap of atleast 2 seats between each of the 4 men, then what is the total number of ways in which these 6 people can be seated? Note that either of the 2 women can sit next to a man or the other woman.
(1) 2688 (2) 5376 (3) 2880 (4) 10752 (5) None of the above
Solution:
It is now given that there should be a gap of atleast 2 seats between each of the 4 men; it is also know that chairs 1 and 8 are already occupied by 2 of the 4 men. So, the other 2 men can now occupy only chairs 4, 5, 11 or 12. Also, it can be observed that only one of chairs 4 and 5, 11 and 12 can be occupied at a time. So the number of ways the 2 men can be seated is given by:
2C
1* 2C1 = 2×2 = 4 ways.
Now considering the 2 women, they can be seated in any of the remaining 8 chairs, so the number of ways is given by: 8C
2= 28 ways.
Hence, the total number of combinations in which the 4 men and the 2 women can be seated is given by: 4×28 = 112 combinations.
In each of these combinations, we can make the 4 men to arrange themselves in 4! ways and the 2 women to arrange themselves in 2! ways.
Hence, the total number of ways is given by: 112 × (4!) × (2!) = 112 × 24 × 2 = 5376 ways. Hence, the correct answer is option 2.
18.Given: ax + by = 320, where a, b, x, y are all positive integers greater than 1. What is the minimum value of the expression: (a + b + x + y)?
(1) 12 (2) 15 (3) 16 (4) 18 (5) 20
Solution:
Given: ax + by = 320 (1)
Since a, b, x, y >1; 320 should be split into a sum of squares, this can be done only as follows: 320 = 64+256= 82 + 162.
Also, 82 = 43 = 26 and 162 = 44 = 28.
Now since, both the terms can be written in 3 different ways, the equation (1) could have 3×3 = 9 different sets of values. Let s = a + x + b + y.
They are as listed below:
i) 320 = 82+162, ⇒ a=8, x=2, b=16, y=2, s=28. ii) 320 = 43+162, ⇒ a=4, x=3, b=16, y=2, s=25. iii) 320 = 26 +162 , ⇒ a=2, x=6, b=16, y=2, s=26. iv) 320 = 82+44, ⇒ a=8, x=2, b=4, y=4, s=18. v) 320 = 43+44, ⇒ a=4, x=3, b=4, y=4, s=15. vi) 320 = 26 +44, ⇒ a=2, x=6, b=4, y=4, s=18. vii) 320 = 82 +28, ⇒ a=8, x=2, b=2, y=8, s=20. viii) 320 = 43+28, ⇒ a=4, x=3, b=2, y=8, s=17. ix) 320 = 26 +28,, ⇒ a=2, x=6, b=2, y=8, s=16.
For set (v), s is found to be at a minimum value of 15. Hence, the correct answer is option 2.
19.All the capital letters of the English alphabet from A to Z followed by the lower case letters from a to z are placed one in each slot, in 52 slots around a circle in clockwise direction. All the slots are counted in the clockwise direction starting with the slot containing A and the letter in every fifth slot is removed. How many more slots will have letters left in them immediately after “m” is removed from its slot?
(1) 12 (2) 16 (3) 14 (4)13 (5) None of these
Solution:
There will totally be 52 letters out of which “m” is the 39th letter. If we count from A, in the first round the letters removed will be those positioned in the slots of
5, 10, 15, ….. 50 – Ten letters then, we have 3, 8, 13, ….. 48 – ten letters, then
1, 6, 11, … 51 – eleven letters, then we have
4, 9, 14, 19, 24, 29, 34, 39 – seven letters before 39.
Total number of letters struck off is 10 + 10 + 11 + 7 = 38 letters before 39. So, ‘m’ is the 39th letter to be removed.
Hence, 52 – 39 = 13 letters would be remaining. Hence, the correct answer is option 4.
20.What is the sum of the digits of a four-digit number whose leftmost digit is 3 less than rightmost digit, which is one more than the 3rd digit from the right, which is twice the 3rd digit from the left, which is not a prime number and is one more than 1/3 of 4th digit from the left?
(1) 24 (2) 18 (3) 25 (4) 30 (5) 27
Solution:
Let the four digit number be denoted by abcd. Following equations represent the data given:
a = d – 3 (1)
d = b + 1 (2) b = 2c (3) = (d/3) + 1 (4)
From (4), it is clear that d is a multiple of 3.
Then, d can take values of 3, 6 or 9; correspondingly using (2),
b should be either 2, 5 or 8, then correspondingly using (3),
c should be either 1, 2.5 or 4, since 2.5 is not an integer, possible value of c can be either 1 or 4.
Possible values
Set d b c a
1 3 2 1
2 6 5 Not valid
3 9 8 4 6
From above table, only 2 sets of values are possible for d and c; among them only the set d = 9, c = 4 satisfies (4). Thus, the digits are: a = d – 3 = 6, b = 2c = 8, c = 4, d = 9.
Hence, the sum of the digits are 6 + 8 + 4 + 9 = 27. Hence, the correct answer is option 5.
21.Select the related word/letters/numbers from the given alternatives: Line : Square :: Arc : ?
1) Ring 2) Sphere 3) Circle 4) Ball
22.Select the related word/letters/numbers from the given alternatives: FJSP : DLQR :: GMIL : ?
1) EOGN 2) JNIO 3) HOGN 4) IONG
23.Select the related word/letters/numbers from the given alternatives: 21 : 3 :: 574 : ?
1) 23 2) 82 3) 97 4) 113
24.For the following questions
Find the odd word/letters/number from the given alternatives.
1) Tabla 2) Tanpura 3) Sarod 4) Sitar
25.For the following questions
Find the odd word/letters/number from the given alternatives.
1) CBED 2) JILK 3) TSVU 4) VZXY
26.For the following questions
Find the odd word/letters/number from the given alternatives.
27.Arrange the following words as per order in the dictionary
1. Sinister 2. Sinuous 3. Sinhalese 4.Sinusitis 5. Sinecure
1) 5, 3, 2, 4, 1 2) 5, 2, 4, 3, 1 3) 5, 2, 1, 3, 4 4) 5, 3, 1, 2, 4
28.A series is given, with one term missing. Choose the correct alternative from the given ones that will complete the series. LMA, NOB, PQC, ? , STE
1) TUV 2) RSD 3) DOA 4) BRD
29.A series is given, with one term missing. Choose the correct alternative from the given ones that will complete the series. 41, 43, 47, 53, ?
1) 59 2) 63 3) 61 4) 65
30.Pointing towards a boy Meera said 'He is the son of the only son of my grandfather.' How is that boy related to Meera?
1) Cousin 2) Brother 3) Uncle 4) Brother-in-law
Directions for questions 31 to 35: Refer the following data to answer the questions given below.
The table below gives the details of the fuel efficiency experiment of a car from its metre reading 1440 km, to its metre reading 3805 km over a month.
Fuel efficiency of the car is defined as the number of kilometres that the car gives per litre of the fuel. Fuel was always filled up in the morning in the garage before the car was taken out.
Day Kilometre reading at the time filling of the fuel Fuel filled (in litres) Status of the tank after filling of the fuel in the tank
1st Jan 1440 0 Empty 2nd Jan 1440 5 - 4th Jan 1474 28 Full 7th Jan - 10 - 14th Jan 2139 25 Full 17th Jan 2689 25 Full 26th Jan 3114.5 22 Full 28th Jan 3724 3 - 31st Jan 3805 35 Full Detailed Solutions:
Solutions for questions 31 to 35: Day km reading at
filling of the fuel Fuel filled
Status of the tank
after fuel filled Fuel efficiency
1st 1440 0 Empty Based on 4th Jan Based on 14th Jan Based on 17th Jan Based on 26th Jan 2nd 1440 5 - 4th 1474 28 Full 7th - 10 - 14th 2139 25 Full 19 17th 2689 25 Full 20.25 22 26th 3114.5 22 Full 20 20.75 19.34 28th 3724 3 - - - - - 31st 3805 35 Full 19.425 19.6 18.6 18.17
31.Considering all the information available, what is the most accurate figure for the fuel efficiency of the car (in km/litre)? (1) 19.425 (2) 20.25 (3) 19.60 (4) 19.35 (5) 15.45
Solution:
Distance travelled = 3805 – 1474 = 2331 km
Petrol consumed = 10 + 25 + 25 + 22 + 3 + 35 = 120 litres
∴The most accurate fuel efficiency is =2331/120 = 19.425 km/litre Hence, the correct answer is option 1.
32.Considering all the information available, what is the best fuel efficiency of the car (in km/litre)?
(1) 20 (2) 20.75 (3) 20.25 (4) 22 (5) 19.6
Solution:
Referring to the above table,
On 14th Jan., the tank became full after filling in 25 litres and on 17th Jan., it was again full. Distance travelled = 2689 – 2139 = 550 km
Petrol consumed = 25 itres
∴Fuel efficiency = 550/25 = 22 km/litre Hence, the correct answer is option 4.
33.Considering all the information available, what is least fuel efficiency of the car (in km/litre) ? (1) 14.50 (2) 17.90 (3) 18.17 (4) 18.60 (5) 15.45
Solution:
Referring to the above table,
On 26th Jan., the tank became full after filling in 22 litres and on 31st Jan., it was again full. Distance travelled = 3805 – 3114.5 = 690.5 km
Petrol consumed = 3 + 35 = 38 itres
∴Fuel efficiency = 690.5/38 = 18.17 km/litre Hence, the correct answer is option 3.
34.What is the fuel efficiency given by the car between 4th Jan. and 28th Jan morning (in km/litre)? (1) 19.50 (2) 20.70 (3) 13.50 (4) 18.60 (5) Indeterminate Solution:
On 4th Jan., the tank was full. On 28th Jan., the status of the tank is not known.. Because the capacity of the tank is not known, the fuel efficiency cannot be found.
Hence, the correct answer is option 5.
35.What is the fuel efficiency given by the car between 14th Jan. and 31st Jan. morning (in km/litre)? (1) 22.21 (2) 19.60 (3) 20.70 (4) 18.60 (5) Indeterminate Solution:
Referring to the above table,
On 14th Jan., the tank became full after filling in 25 litres and on 31st Jan., it was again full. Distance travelled = 3805 – 2139 = 1666 km
Petrol consumed = 25 + 22 + 3 + 35 = 85 itres ∴Fuel efficiency = 666/85 = 19.6 km/litre Hence, the correct answer is option 2.
Directions for questions 36 to 40: Refer the following data to answer the questions given below.
A knockout tournament is divided into successive rounds; each player plays in at most one match per round. If a player gets a bye means that he is advancing to the next round in a tournament without playing an opponent. Each player plays matches till he loses a match and the first match he loses, he is out of the tournament. Each player who is not already out of the tournament plays one match in each round unless he gets a “bye” in the round. By winning his match in a particular round (or) by getting a “bye” he progress to the next round. There were 333 players in a knockout tournament all seeded – with first seed being the topmost and 333th seed being the lowest.
36.What is the maximum number of “byes” a player could have got in the tournament?
Solution: Roun d Teams playing Byes Teams surviving 1 333 1 166 + 1 2 167 1 83 + 1 3 84 0 42 4 42 0 21 5 21 1 10 + 1 6 11 1 5 + 1 7 6 0 3 8 3 1 1 + 1 9 2 0 1
A player can get maximum of 5 byes. Hence, the correct answer is option 5.
37.What is the least number of matches the winner of the tournament would have to play?
(1) 7 (2) 6 (3) 9 (4) 4 (5) 0
Solution:
Since there are 9 rounds and 5 byes, the minimum number of games that a winner would have to play is = 9 – 5 = 4 Hence, the correct answer is option 4.
38.If there were 3000 more players, then how many more rounds should have been played in the tournament?
(1) 3 (2) 4 (3) 11 (4) 5 (5) None of these
Solution:
Suppose there are N players. If N = 2k, then K rounds will be played, else if N ≠ 2k, then the power if 2 which will result in 2k > N will be taken, where K is the minimum possible number.
If N = 333, then 28 = 256 and 29 = 512. Now, we know that 28 < 333 < 29.
So, the number of rounds will be played is 9. If N = 3333, then 211 = 2048 and 212 = 4096. Now, we know that 211 < 3333 < 212.
So, the number of rounds will be played is 12. Hence, the correct answer is option 1.
39.Totally how many matches were played in the tournament?
(1) 333 (2) 332 (3) 331 (4) 334 (5) None of these
Solution:
If there are n players or teams in a knockout tournament, (n – 1) matches will be plated. Hence, the correct answer is option 2.
40.If you are allowed to organize the matches between players the way you desire and you can also determine the winner of each match subject to the following conditions:
A.Any player can beat any other player lower-seeded if they meet in a match
B.Any player can also beat any other player at most two seeds higher than him, if they meet in a match (i.e. seed No. 50 can beat seed No. 48 but not seed No. 47)
Who is the lowest seeded player who can win the tournament?
(1) 309 (2) 331 (3) 16 (4) 15 (5) 332
Solution:
We are interested in finding the lowest possible seed winner.
In the first round 3rd seed can beat 1st seed, 4th seed can beat 2nd seed, 5th seed can beat 3rd seed and so on. ∴ The top two seeds left after the rounds are as follows:
Round Top two Seeds left 1 3 and 4 2 5 and 6 3 7 and 8 4 9 and 10 5 11 and 12 6 13 and 14 7 14 and 15 8 15 and 16 9 16 (Final)
