Top 100 questions on QA & DI
In absence of any prescribed syllabus aspirants get intrigued on exams and it is pertinent to note that QA & DI are the two main sections which overlap with all exams and since these two sections are most difficult hence, our experts have prepared important question bank with solutions for you who aspire to crack forthcoming CAT/XAT/IIFT/NMAT/CMAT/SNAP and MAT exams
SECTION – I : Quantitative Aptitude
1.  Of five letters mailed today, each have a 2/5 probability of arriving in three days or sooner. What is the probability that exactly two of the five letters will arrive three days or sooner?  


2.  Given: Min(a, b) = Least of a, b Max(a, b) = Larger of a, b G(a, b) = GCD of a, b L(a, b) = LCM of a, b P(a, b) = Product of a, b Then the value of P(7, Max(3, Min(L(3, 5), P(17, 1)))) is 


Stay informed, Stay ahead and Stay inspired with MBA Rendezvous
For any query you may also mail us at [email protected]
1. Quantitative Aptitude

SECTION – II : Data Interpretation
DIRECTIONS FOR QUESTIONS 53 to 57 :
Refer to the data below and answer the questions that follow. 

53.  What was the percentage increase in imports between 199596 and 19992000?  


54.  What was the approximate percentage change in trade gap between 199899 and 19992000?  


55.  What was the approximate percentage increase in trade gap between 199697 and 199798?  


56.  Which of the following is true?  


57.  If oil imports constituted 20% of the total imports in 199798, then what percent of the trade gap was due to oil (assuming that no oil is exported)?  


DIRECTIONS FOR QUESTIONS 58 to 62 :


RAILWAY TIMETABLE : NEW DELHI – BHUBANESWAR RAJDHANI EXPRESS




58.  The longest run for the train between the two successive halts is  


59.  The average speed that the train maintained between two successive stations was the highest between  


60.  The average speed that the train maintained between New Delhi and Bhubaneswar was nearly equal to  


61.  If we consider a journey that begins in New Delhi and ends in Bhubaneswar, the train has the longest halt at  


62.  The train begins its return journey from Bhubaneswar to New Delhi Seventeen hours after it has arrived at Bhubaneswar. If the train left New Delhi on Tuesday on what day will it have returned to New Delhi? (Assume that on the return journey that train maintains the same average speed as on the onward journey).  


DIRECTIONS FOR QUESTIONS 63 to 64 : 

Answer the questions on the basis of the information given below. Two stock traders – Hasan and Sajid – trade in the shares of Pineapple Group only. Each of them follows a different strategy for selling and buying the stocks. Sajid sells an equal number of shares at 9 a.m. and 10 a.m. and buys them back in equal numbers at 1 p.m. and 2 p.m. Hasan sells some shares at 11 a.m. and the rest at 12 noon such that the amount he gets on the two occasions is the same. He buys back the whole lot at 3 p.m. The chart given below displays the performance of Pineapple Group stock on two particular days – Day 1 and Day 2. The profit/loss made by a trader on a particular day is the difference between the amount that he obtains by selling the shares and the amount that he spends in buying the shares. “Margin” for a day is the profit/loss expressed as a percentage of the total amount obtained by selling the shares on that day. 

63. 
If Sajid and Hasan sold an equal number of shares on Day 1, then which of the following statements would definitely be true? I. Hasan made more profit than Sajid on Day 1. 



64.  What was the ratio of Hasan’s Margin to Sajid’s Margin on Day 2?  


DIRECTIONS FOR QUESTIONS 65 to 67 : 

Answer the questions on the basis of the information given below. The graph given below shows the statistics of five companies – Perfitti VM, UB Group, ITC, Unilever and Wipro – in India. Each point on the graph indicates the Total Revenue generated by a different department of one of these companies and the Average Revenue generated per Unit Energy Consumed by that particular department. 

65.  If the departments represented in the graph are the only departments in the five companies where energy is consumed, then for which company is the Energy Consumption the highest?  


66.  How many of the represented departments across the five companies have consumed less than 100Whrs of energy?  


67.  Which of the five companies has the highest number of departments that generate more than Rs. 600 crores as the Total Revenue and consume less than 200Whrs of energy?  


DIRECTIONS FOR QUESTIONS 68 to 69 : 

Answer the questions on the basis of the information given below. The graph given below shows the total Exports and Imports of a country for four different years Please note: 

68.  Which of the following experiences the highest increase?  


69.  Which of the following experiences the lowest percentage change?  


DIRECTIONS FOR QUESTIONS 70 to 71 : 

Answer the questions on the basis of the information given below. The performance appraisal of the employees of Hondai Motors Pvt Ltd. was done three times in the year 2010. The first appraisal was done in January, the second in July and the third in November. Only the employees who were appraised in January were eligible for appraisal in July and only those who were appraised in July were eligible for appraisal in November. The table given below shows the number of employees appraised in each of the three appraisal months in 2010 for different performance areasâ€‹


70.  How many employees of Hondai Motors Pvt Ltd. were appraised on exactly one performance area in 2010?  


71.  Among the employees of Hondai Motors Pvt Ltd. who were appraised in 2010, how many were not appraised on Individual Performance?  


DIRECTIONS FOR QUESTIONS 72 to 74 : 

Answer the following questions on the basis of information given below: The graph given below shows the statistics of 12 Cricket players. Each point on the graph indicates the average score per match of a player and the number of matches played by that player. Each of the players plays for one of the four teams  Team 1, Team 2, Team 3 and Team 4.


72.  If only the runs scored by the above mentioned 12 players are considered then which team has got the maximum aggregate score?  


73.  How many players are there who have played more than 200 matches and have scored less than 9,000 runs?  


74.  What is the overall average score of those players of Team 4, whose averages are better than the average of that player who has played the second highest number of matches for Team 3?  


DIRECTIONS FOR QUESTIONS 75 to 77 : 

Answer the following questions on the basis of information given below: The following table shows the marks obtained by four students in three Mock CAT tests. The four students Abhishek, Akshay, Amitabh and Aamir are disguised as A, B, C and D in no particular order.
Additional information:


75.  How many different total scores are possible for Akshay?  


76.  What is the name of the person disguised as B?  


77.  How many different values are possible for ‘a’?  


DIRECTIONS FOR QUESTIONS 78 to 80 : 

Answer the following questions on the basis of information given below: Five actresses Careena, Catrina, Carishma, Coena and Cashmira are disguised as U, V, X, Y and Z, not necessarily in this order. Director Bhakti Kapoor interviewed the actresses in the order Z, Y, X, U and V for his film “Bhoot Aunty”. Additional Information:


78.  Who is disguised as Y?  


79.  What is the fee (in Rs.) demanded by Coena?  


80.  Who was interviewed last ?  


DIRECTIONS FOR QUESTIONS 81 to 83 : 

Answer the questions on the basis of the information given below. In a country called Khelabad, each sportsman plays either Cricket or Hockey, but not both. Each of them plays for one of the two teams – Under19 or Above19. All the sportsmen of Khelabad who had passed in a Physical Fitness Examination conducted by the Sports Ministry recently are certified as “fit” and the rest are called “unfit”. It is also known that: 

81.  What is the total number of sportsmen in Khelabad?  


82.  What is the ratio of the number of unfit sportsmen who play Hockey for Above19 team to the number of unfit sportsmen who play Hockey for Under19 team?  


83.  What is the total number of fit sportsmen who play for Above19 team?  


DIRECTIONS FOR QUESTIONS 84 to 86 : 

Answer the questions on the basis of the information given below. The table given below shows the marks scored by six students of a School in Physics, Chemistry, Mathematics and English in an exam. Each subject is assigned a Credit as mentioned along with the subject name in bracket. E.g. the Credit assigned to Physics is 2.5.â€‹
The Grade assigned to a student against the marks scored by him/her in a subject is calculated as per the table given below.
The points received by a student against the grade assigned to him/her in a subject are calculated as per the table given below.
The CGPA (Cumulative Grade Point Average) of a student is calculated using the formula given below.
(‘Y’ represents one of the four subjects.)â€‹ 

84.  Who among the six students got the highest CGPA?  


85.  Which two students got the same CGPA?  


86.  What was the CGPA of Fardeen?  


DIRECTIONS FOR QUESTIONS 87 to 89 : 

Answer the questions on the basis of the information given below. The breakup of the financial budget of a country called Chaupatland for FY 201011 is represented by the pie charts given below. Pie chart A represents Budgeted Revenue and pie chart B represents Budgeted Expenditure.
Budgeted Revenue was less than Budgeted Expenditure by Rs. 625 thousand crores. Interest Payment exceeded NonPlanned Capital Account by 12% of Budgeted Expenditure. NonTax Revenue was Rs. 285 thousand crores less than Tax Revenue.â€‹ 

87.  The difference between Budgeted Expenditure and Budgeted Revenue was what percentage of Budgeted Revenue?  


88.  By how much did the sum of Planned Revenue Account and Planned Capital Account exceed NonTax Revenue?  


89.  In FY 201112, if Budgeted Revenue is increased by 52% over that of the previous year, then what will be the new percentage share of Tax Revenue in Budgeted Revenue?  


DIRECTIONS FOR QUESTIONS 91 to 92 : 



90.  In Census data based estimates (014 Yrs.), which year saw the maximum percentage change in Full Time Child Workers as compared to the previous Census?  


91.  In Nonworkers & Nonstudents (514 Yrs.) category, the two estimates are based on Census and MHRD & NSS respectively. For which year is the absolute difference between the two estimates the highest?  


DIRECTIONS FOR QUESTIONS 92 to 94 : 

Answer the questions on the basis of the information given below. The table given below shows the marks obtained by 20 students of a school in their annual exam. The Performance Score of a student is calculated as: Performance Score = Academic Score + 3 × Extra Curricular Score
The school calculates ‘Happy Growth Index’ for every student to determine the relative happiness of its students. Students are classified as either High or Low on Happy Growth Index based on the criteria shown in the table given below.


92.  How many students are Low on Happy Growth Index at Primary Stage of Education?  


93.  What is the ratio of the number of male students at Senior Secondary Stage who are High on Happy Growth Index to the number of female students at Primary Stage who are High on Happy Growth Index?  


94.  How many female students whose Academic Score is less than 50 have the same Happy Growth Index as that of Renuka?  


DIRECTIONS FOR QUESTIONS 95 to 97 : 

Answer the questions on the basis of the information given below. The political parties mentioned in the graph given below participated in a constituency election. The graph shows the number of votes received by the candidates of six major parties and other smaller parties in Round 1 of the election. The election is held as follows: In Round 1, if a candidate gets more than 50% of the total number of votes cast in that round, (s)he is declared the winner of the election, else, top 5 candidates based on the number of votes move to Round 2. In Round 2, if a candidate gets more than 50% of the total number of votes cast in that round, (s)he is declared the winner of the election, else, top 3 candidates based on the number of votes move to Round 3. In Round 3, the candidate who gets the maximum number of votes in that round is declared the winner of the election. Assume that there are no ties in deciding top 5, top 3 (if Round 2 happens) candidates or the winner (if Round 3 happens) in Round 3.


95.  Among the five candidates who moved to Round 2, the candidate with the maximum number of votes in Round 1 got what percentage of the total number of votes in that round?  


96.  If the votes received by the candidates of Samajik Party and Hindu Muslim Party in Round 1 are not taken into account, then by what percentage points does the share of “Others” increase?  


97.  If 1 Million votes are cast in Round 2 and 3 Million votes are cast in Round 3, then what is the minimum total number of votes needed by a candidate to be declared the winner of the election?  


DIRECTIONS FOR QUESTIONS 98 to 100 : 

Answer the questions on the basis of the information given below.


98.  What is the ratio of men’s contribution from Group D to women’s contribution from Group C to the town’s income?  


99.  In 2011 the number of women in Group C increases by 15%. By what percentage will the total number of women in the town increase?  


100.  If the average contribution of men to the town’s income is Rs 35,000, then what is the average contribution of women to the town’s income?  


Hints & Solutions for Top 40 Questions on QA & DI
SECTION – I : Quantitative Aptitude
1.  Number of ways in which exactly 2 of 5 letters will arrive in three days or sooner = 5! ÷ [2! X (5  2)!] = 10. Now, the probability of a letter failing to arrive in 3 days or sooner = 1 – 2/5 = 3/5. The probability of the first 2 letters arriving in 3 days or sooner and the other 3 letters arriving later than that = (2/5)2 x (3/5)3 = 108/3125. So, probability of exactly two letters arriving in three days or sooner = (108/3125) x 10 = 216/625. 
2.  P(17, 1) = 17, L(3,5) = 15, Min(15, 17) = 15, Max(3, 15) = 15, So, P(7, 15) = 105 
3.  Let E = ab + bc + ca= ab + c(a + b), Taking a, b Ëƒ 0 and c = a very large negative value, E = A very large negative vale. Therefore as c decreases E decreases. Thus E tends to negative infinity. 
4.  Total distance travelled = 20 + 2[20(1/2) + 20(1/2)2 + ……. Up to infinity] = 20 + 2 x 1/2 x 20 x [1÷ (1 – 1/2)] = 60m 
5.  12N = 22 X 31 X 111 X 131 X 171 = 2a X 3b X 11c X 13d X 17e Therefore, 0 ≤ a ≤ 2, 0 ≤ b ≤ 1, 0 ≤ c ≤ 1, 0 ≤ d ≤ 1, 0 ≤ e ≤ 1 So, total number of factors of 12N = 3 X 2 X 2 X 2 X 2 = 48, which includes 1 and 12N Thus the number of factors of 12N which does not include 1 and 12n = 48 – 2 = 46. 
6.  Let one side of the triangle is ‘a’ then the other sides are (5 – a) and 3. s = Semi perimeter = 8/2 = 4. Therefore, Area of Triangle, A = [ s(s – a)(s –b)(s – c)]1/2 ⇒ A = [4(4 – 3)(4 – a)(4 – 5 + a )]1/2 = 4(4 – a)(a – 1) ⇒ A2 =  4a2 + 20a – 16 =  4( a2  5a + 4) =  4[ (a – 5/2)2 – 9/4] So, for maximum area(A), a = 5/2. Thus the other two sides are 5/2 and 5/2. 
7.  Since the ratio of their circumference is directly proportional to their diameter, therefore ratio of their diameter = 6 : 5. Hence ratio of their volume = (6/5)3 = 216/125 = 1.73 S0, if it were priced according to its size or weight, bigger one should cost 173% as much as the smaller one. ⇒ Bigger one is 73% more than the smaller one. Now as it is 50% costlier, it is more profitable to buy the bigger melon. 
8.  Let the first pipe alone can fill the tank in x hours Then the second pipe can fill the tank in (x5) hours and the third pipe can fill the tank in (x5)4 = (x9) hours Therefore, part filled by first pipe and second pipe together in 1 hr = part filled by third pipe in 1 hr ⇒ 1/ x + 1/ (x – 5) = 1/(x – 9) Solving we get, x = 15 or 3. But x cannot be 3 otherwise (x – 9) becomes negative. So, x = 15 
9.  Total number of ways of advertising = C(4, 2) X C(3, 1) X C(2, 1) = 36 
10.  Here we can have four cases: (i) a is even and b is even (ii) a is even and b is odd (iii) a is odd and b is even (iv) a is odd and b is odd Out of the four cases, in case (ii) and (iii), the sum will be odd. Therefore, the required probability = 2/4 = 1/2 
11.  Let the numbers in ascending orders be x, y and z. ⇒ x + y + z = 59/24 and z/x = 7/6 and y = 7/6 – 1/3 = 5/6 Since z/x = 7/6, so, x = 6z/7, ⇒ 6z/7 + 5/6 + z = 59/24, ⇒ z = 7/8, x = 3/4, y = 5/6 
12.  log10 5 + log10 (5x + 1) = log10 (x + 5) + 1 ⇒ log10 5 + log10 (5x + 1) = log10 (x + 5) + log10 10 ⇒ log10 [5(5x+1)] = log10 [10(x+5)] ⇒ 5(5x+1) = 10(x+5), => 5x+1 = 2(x+5) ⇒ 5x + 1 = 2x + 10, => 3x = 9, => x = 3 
13.  Draw an obtuse angled triangle ABC. Draw CE perpendicular to extended AB. In âˆ† CEA, tan A = CE/EA = CE/2EB = 1/2 tan CBE = 1/2 tan (π – B) =  1/2 tan B ⇒ 2 tan A + tan B = 0 
14.  Let the work done by A, B and C in 1 day be x, y and z respectively. ⇒ x + y = 1/30, ïƒÅ¾ y + z = 1/24, ïƒÅ¾ z + x = 1/20 On adding we get, 2(x + y + z) = 1/30 + 1/24 + 1/20 ⇒ x + y + z = 1/16 ⇒ In 1 day work done by A, B and C = 1/16 So, in 10 day work done by A, B and C = 5/8 So, remaining work = 1  5/8 = 3/8 Now, work done by A in 1 day = 1/16 – 1/24 = 1/48 So, the number of days required by A complete the remaining work = 3/8 ÷ 1/48 = 18 
15.  f(g(h(x))) = 1/x2 – 3 …………….. (1) f(h(g(x))) = 1/x2 – 3 …………….. (2) h(f(g(x))) = 1/(x2 – 3) ……………(3) So, (a) and (c) are not true. Now, suppose f(g(h(x))) = h(f(g(x))), then x2 = 0.3 or 3.6 (approximately). The graphs of (1) and (3) intersect at x = 0.5 and x = 1.9. Hence, the answer is (d) 
16.  l.c.m.( 4, 8, 10, 12, 15, 20) = 120 Hence all the six bells will ring together in every 120 seconds or 2 minutes. So, the number of times they will ring together in 60 minutes = 60/2 + 1 = 31. 
17.  The first term is 21 and the third term is 23. So, the 40th term is 240. Now, Let x = 240, => log x = 40(log2), => log x = 40 x 0.3(Approx)= 12.0 (Approx) So, x = AL(12.0..), So, 240 has 13 digits. 
18.  For n = odd, (xn + 1) is always divisible by (x+1). ⇒ (6767 + 1) is divisible by (67 + 1) ⇒ (6767 + 1) is divisible by 68 ⇒ (6767 + 1) ÷ 68 gives a remainder of 0 ⇒ [(6767 + 1) + 66] ÷ 68 gives a remainder of 66 ⇒ (6767 + 67) ÷ 68 gives a remainder of 66. 
19.  Manu is 26th from the left and Hari is 10 places to the left of Manu. So, Hari is 16th from left. Now, there are 3 boys between Hari and Ram. Thus the position of Ram is either 12th or 20th from the left. Thus the exact position of Ram cannot be ascertained. Hence to find the position of Shyam in the row, given data is insufficient. 
20.  Let the quantity of pure copper be x kg. By given condition, the quantity of copper = 4/5 x 10 + 1/4 x 16 + x ⇒ 3/5(10 + 16 + x) = 8 + 4 + x ⇒ x = 9 Thus, the weight of the new alloy = 10 + 16 + 9 = 35 kg. 
21. 
Let the speed of A, B and C be 5v, v and v respectively. Let the circumference of the track be 12vt. Instance when A meets: B – 3t, 6t, 9t, 12t, 15t, 18t .... C – 2t, 4t, 6t, 8t, 10t, 12t .... In every 6t units of time A meets B twice while A meets C thrice. So in 36t A would give: B – 2 × 6 = 12 cards C – 3 × 6 = 18 cards. (total 30) Now, at 38t A gives C a card (his 31st) at 39t A gives B a card (his 32nd) and at 40t A gives C a card (his last) So the required difference = (18 + 2) – (12 + 1) = 7 
22. 
S = 1/100 + 1/101 + 1/102 .... 1/998 + 1/999 + 1/1000 , then
In the first 450 terms, the sum of the two numbers in the denominator is constant. Thus, their product will be maximum when they are equal

23. 
Let the coordinates of A and B be (0,0) and (a,0) respectively. Let the coordinates of point C be (h, k). Given: 2AC = BC
The equation written above is that of a circle with center (say O) at (a/3 – 0) and radius 2a/3, The center O (a/3 – 0) divides AB externally such that 4AO = BO 
24.  We need to check for a base where 677 must have 5 digits if converted.We also know that in Base 3, the decimal value of a 5digit number must lie between 80 and 243 (as ‘81 signifies 10000’ and ‘242 signifies 22222’ – the minimum and maximum possible 5digit number in base 3).So, the base cannot be 3.Next we check for Base 4. The decimal value of a 5digit number must lie between 255 and 1024. As 256 < 677 ≤ 1023, this must be the minimum possible value of n. 
25. 
a_{1} = 1 a_{2} = 2 a_{3} = 1 (1) = 1 a_{4} = 2 (0) = 0 a_{5} = 1 (– 1) = – 1 a_{6} = 0 (– 2) = 0 a_{7} = – 1 (– 1) = 1 a_{8} = 0 (0) = 0 a_{9} = 1 (– 1) = – 1 After a_{1} and a_{2}, there is a cyclicity of ‘4’ in the remaining terms a_{1000} = a_{2 + 4 × 249 + 2 }= 0.â€‹ 
26. 
The minutehand of a normal clock covers 360/ 60 =6° per minute. The hourhand of a normal clock covers 30/60 =1°/ 2 per minute. So once they are together, in every minute the minute hand gains 6 − 1/2 =11°/2 over the hour hand. So, time between two meetings = 360/11/2 = 720/11 minutes. So, in any clock the hourhand and the minutehand meet after every 720/11 minutes. If 60 minutes have passed in a normal clock then time passed in the faulty clock is 70 minutes. If 24 hrs (or 24 × 60 minutes) have passed in a normal clock then time passed in the faulty clock must be 24 × 70 = 1680 minutes. Number of times the hands meet –

27. 
If Brazil never gets a lead over India, the first goal of the match should be scored by India. At this stage India leads by 1 – 0, and the only way in which Brazil can take a lead is by scoring the next two goals i.e. a scoring pattern like IBBIII. In all other cases Brazil would never be able to lead India. Total cases where India scores the first goal 5!/3!2! =10 Different scoring patterns possible = 10 – 1 = 9 
28. 
Now, three cases arise: Case–1 P(x) = 1 and Q(x) may be anything. ∴ (x−7) = 1 or x = 8 But, for x = 8, Q(x) is not defined. Case–2 P(x) = – 1 and Q(x) is an even exponent. (x – 7) = – 1 or x = 6 For x = 6, Q(x) = – 4, which satisfies the given equation. Case–3 Q(x) = 0 and P(x) 0
⇒ x = 7 or x = 22 But, for x = 7, P(x) = 0, for which the given equation is not defined. 
29. 
When N = 1 M = 1 × 6 N = 2 M = 6 × 11 N = 200 M = 996 × 1001 
30.  
31. 
Construction: Extend BC to cut the circle at E and join AE.
âˆ†AEB is right angled at E. (Angle in a semicircle is a right angle). Hence, both AE and OD are perpendicular to EB. By AA property: âˆ† AEB ~ âˆ†ODB Hence, OB/AB = OD/AE or OD/AE = 1/2 ⇒ AE = 2OD = 3cm. In right âˆ†AEC : ∠ACE = 180^{o }−∠ACB = 180^{o} − 120^{o} = 60â€‹^{o} 
32.  All the flowers can be given to one person and then the fruits distributed among all others in such a way that all of them get at least one fruit. 
33. 
As per the information given in the question, we can conclude that
⇒ 0.05c + 0.15d = 0.05b + 0.15a ⇒ c + 3d= b + 3a The ratio 5 : 2 : 7 : 3 does not satisfy the given relation. 
34. 
3y^{2 }= x^{2} – 1376 As we can see L.H.S. is definitely a multiple of 3 and in R.H.S. 1376 leaves a remainder of 2 when divided by 3. There are three possibilities for x in R.H.S: (i) If x is multiple of 3, so is x^{2}, and R.H.S. will leave a remainder of 1 when divided by 3. (ii) If x is of the form 3m + 1, x^{2} will be of the form 3n + 1 and R.H.S will leave a remainder of 2. m, n ∈ N (iii) If x is of the form 3m + 2, x^{2} will be of the form 3n + 1 and R.H.S. will leave a remainder of 2. m, n ∈ N So R.H.S. can never be a multiple of 3, while L.H.S. is always a multiple of 3. Hence no real solution exists. 
35. 
f(0) > 0 and f(1) < 0 implies that one root for f(x) = 0 lies between x = 0 and x = 1. f(6) + f(8) = 0 implies that f(6) and f(8) are of opposite sign but same absolute value. Hence another root for f(x) = 0 must lie between x = 6 and x = 8. As f(1) < 0, f(6) must also be less than zero, otherwise we’ll have more than 2 roots for f(x) = 0. Hence f(8) > 0 and f(6) < 0. Further f(7).f(9) > 0 implies that both f(7) and f(9) are greater than zero. So the second root for f(x) = 0 must lie between x = 6 and x = 7. So f(x) would look like :â€‹ As f(1), f(2) and f(3) are less than zero, f(1).f(2).f(3) < 0 is true. As f(3), f(5) < 0 and f(7), f(9) > 0, f(3).f(5).f(7).f(9) > 0 is true. As f(7), f(8) > 0, f(7).f(8) < 0 is false. f(0), f(9), f(10) > 0 and f(1) < 0, but since we don’t know the magnitude of any of these four we cannot judge if f(0) + f(1) + f(9) + f(10) is greater than zero or not.â€‹ 
36. 
The maximum total work done for three consecutive days will be when A, B and C work in distinct pairs (for example ‘A and B’, ‘B and C’ and ‘C and A’). Hence on these three days fraction of total work done will be The remaining 4/120 fraction of work can be done in least amount of time if A and B work together on the 7^{th} day. A and B together can do 1/8 +1/12 = 5/24 + = fraction of work in a day. So, they will complete the 4/120 fraction of work in 4/120 ÷ 5/24 = 4/25 day 
37. 
By Pythagoras theorem: Since diagonals of a rectangle are equal and bisect each other: AG = DG = AC/2 = a Hence, âˆ†DGA is equilateral and since AF is perpendicular to the base DG we can say that AF must divide âˆ†DGA into two equal halves. Area of equilateral âˆ†DGA = √3/4 a^{2} square units 
38. 
From Statement A: This statement is clearly insufficient. The answer can given only when exact values of a, b and K are known. For example let (a, b) = (4, 2) or a/b = 2 Case 1: K = 0 (a + K)/(b + K) = a/b Case 2: K = 2 (a + K)/(b + K) = 6/4 = 3/2 > a/b Case 3: K = –1 (a + K)/(b + K) = 3/1 = 3 > a/b From Statement B: As explained above for Statement A, this statement is also insufficient to answer without knowing the exact values of a, b and K. Combining Statement A and Statement B: Even the two statments combined cannot give a unique answer. 
39. 
If f(0) is nonnegative, roots of f(x) = 0 cannot be of opposite signs. So the possibilities for the two roots are (0, 3), (0, 2), (0, 1), (1, 2), (1, 3), (2, 3), (0, –3), (0, –2), (0, –1), (–1, –2), (–1, –3) and (–2, –3). Hence the required answer is 12. 
40. 
Let the quantity of total solution be 800 units. So, quantity of milk is 500 units and water is 300 units. Now x percent of 800 = 8x units mixture is removed. The quantity (in units) of milk and water removed is 5x and 3x respectively. Final concentration of milk (in percentage) is: (500 − 5x)/800 × 100 So, 30 < (500 − 5x)/8 < 50 ⇒ 240 < 500 − 5x < 400 ⇒ − 260 < −5x < −100 ⇒ 20 < x < 52. 
41.  Required number of ways = ^{4}C_{2} X ^{4}C_{3} = 6 X 4 = 24 
42.  Days of the year on which Kamla fasted: 1, 3, 6, 10, 15, 21, 28…. Days of the year on which Bimla fasted: 1, 4, 8, 13, 19, 26, 34…. It can be analysed that if Kamla fasted on the Kth day, Bimla must have fasted on the (K – 2)th day. Hence for both of them to fast on the same day, Kamla must fast on some (K – 2)th day and Kth day as well. This is impossible as in the first series no two consecutive terms will have a difference of 2 after the first two days. 
43. 
Distance of origin (0, 0) from the line 3y – 4x – 15 = 0: Let the new lines drawn parallel to 3y – 4x – 15 = 0 
44. 
We can see that the difference between the divisor and the respective remainder is the same in each division i.e. 2 – 1 = 4 – 3 = 6 – 5 = 8 – 7 = 1 Hence the general form of such numbers will be LCM(2, 4, 6 and 8).K – 1 = 24K – 1, where ‘K’ is any natural number. Hence the numbers are 23, 23 + 24, 23 + 2 × 24, ......, 23 + 40 × 24 A total of 41 such numbers are there between 0 and 1000. 
45. 
a^{15} + a^{16} + a^{17} + ....... +a^{50} Sum = a^{15} {1 + a + a2 +  + a^{35}} Since a is the root of equation x^{5} – 1 = 0,â€‹ a^{5} – 1 = 0 ⇒ a^{5} = 1 
46. 
Let’s assume that ‘I’ denotes the integral part and ‘F’ denotes the fraction part of (5 + √19)^{n}. ⇒ I + F = 5^{n} = ^{n}C_{1}.5^{n1}.√19 +^{n}C_{2}.5^{n2}.19 + ^{n}C_{3}.5^{n3}.19√19 + ........â€‹ Nowâ€‹ ( 5 − √19 )^{n }is a proper fraction as ( 5 − √19 ) < 1 Let’s assume that F′ = ( 5 − √19 )^{n} ⇒ F′ = 5n − ^{n}C_{1}.5^{n1}.√19 + ^{n}C_{2}.5^{n2}.19 + ^{n}C_{3}.5^{n3}.19√19 + ........â€‹ Hence,â€‹ I + F + F′ = 2 [ 5^{n} + ^{n}C_{2}.5^{n2}.19 + ^{n}C_{4}.5^{n4}.19^{2}+ ........â€‹ ] Evidently, F + F = 1 ′ as the sum of two proper fractions is always less than 2 and greater than 0. Also the right hand side is an integer and hence F + F′ should also Therefore I + 1 is an even number and I is an odd number. Alternative method: Putting n = 1, we get 5+ √19 whose integral part is 9. 
47. 
Let A involves 30 units, B involves 15 units and C involves 20 units of work. Two welders would take 5 days to complete A. They finish A at the end of day 5. 
48. 
∠BAD = 90° Extend BA and CD to meet at E. âˆ†EBC becomes an equilateral triangle. So BE = 4 cm In âˆ†ADE, ∠ADE = 30° tan30° = AE/2 = 1/√3 ⇒ AE = 2/√3 cm ∴â€‹ AB = BE − AE = 4 − 2/√3 = (4√3 − 2)/√3 cm 
49. 
Discount to Saral 20 + 25 –(20 × 25)/100 = 40% Discount to Himanshu = 40 + 40 – (40 × 40)/100 = 64% Let a, b, c be the number of units sold at 20%, 40% and 64% respectively ⇒ (a × 20)/100 + (b × 40)/100 + (c × 64)/100 = {(a+b+c)50}/100 ⇒ 7c/50 – b/10 – 3a/10 = 0 ⇒ 7c – 5b – 15a = 0 So ‘c’ has to be a multiple of 5. Values of a, b, c satisfying the equation are (2, 1, 5); (1, 4, 5); (4, 2, 10); (3, 5, 10); (2, 8, 10). Hence the possible values for x are 8, 10, 16, 18 and 20.â€‹

50. 
Area of âˆ†DEC = Area of âˆ†DBC ……..( âˆ†'s between same parallel line and same base) = 1/2 × Area of parallelogram ABCDâ€‹ Area of âˆ†EFG = 1/2 × Area of parallelogram EBCFâ€‹ = 1/2 × 3/5 × Area of parallelogram ABCDâ€‹ 
51. 
Case 1: Case 2: ^{4}C_{2}/2! = 3 ways It is divided by 2! because the boxes are identical. Case 3: ^{4}C_{2} = 6 ways Note: We need not select a box for the remaining two Case 4: ^{4}C_{3} = 4 ways 
52.  The average of the three numbers will be 15. Let the numbers be 15 – d, 15 and 15 + d. ⇒ (15 – d)^{2 } + 15^{2} + (15 + d)^{2} = 683 ⇒ d = 2 So the numbers are 13, 15 and 17. 
SECTION – II
DIRECTIONS FOR QUESTIONS 53 to 57 : 

53.  Imports in 199596 = 7000 crore Imports in 19992000 = 14000 crore Percentage increase in imports = (14000 – 7000)/7000x100 = 100 % 

54.  Trade gap in 199899 = 13500  7000 = 6500 Trade gap in 19992000 = 14000 – 7800 = 6200 Percentage decrease in trade gap = (6500 – 6200)/6500x100 = 4.6 % 

55.  Trade gap in 199697 = 9000  6600 = 2400 Trade gap in 199798 = 13000 – 6400 = 6600 Percentage increase in trade gap = (6600 – 2400)/2400x100 = 175 % 

56.  The trade gap first increases and then exhibits a decreasing trend.  
57.  Oil Imports in 199798 = 20/100 x 13000 = Rs. 2600 crore Trade gap in 199798 = Rs. 6600 crore Hence, percentage trade gap due to oil = 2600/6600 X 100 = 39.39 % ~ 40 % 

DIRECTIONS FOR QUESTIONS 58 to 62 :




58.  With the run is of 440 km, the longest run is between New Delhi  Kanpur Central.  
59.  The average speed of 93.22 kmph is the highest between New Delhi  Kanpur Central.  
60.  The average speed that the train maintained between New Delhi and Bhubaneswar = 1800 km/25 hrs and 25 min = 70.82 kmph.  
61.  The train has the longest halt of 15 minutes at Kharagpur JN.  
62.  Friday  
DIRECTIONS FOR QUESTIONS 63 to 64 : 

63. 
Let the number of shares sold by Sajid and Hasan on Day 1 be 36x each. Investment made by Sajid Investment made by Hasan 

64. 
For Hasan: Margin (loss) = (6750 − 4000)/4000 = 68.75% For Sajid: Margin (loss) = (1125 − 500)500 = 125% Required ratio = 68.75 : 125 = 11 : 20 

DIRECTIONS FOR QUESTIONS 65 to 67 : 

65.  The Energy Consumption of a department can be obtained by dividing the Total Revenue of that department by the Average Revenue per Unit Energy Consumed by that department. Among the five companies, the Energy Consumption is the highest for Perfitti VM at approximately 1900Whrs.  
66.  There are six departments in all whose Energy Consumption is less than 100Whrs. They include two departments of UB Group, one of Unilever and three of Wipro.  
67.  Unilever has two departments whose Total Revenue is more than Rs. 600 crores and Energy Consumption is less than 200Whrs. The only other department that satisfies the given criteria is of ITC Ltd.  
DIRECTIONS FOR QUESTIONS 68 to 69 : 

68.  Both Oil Trade Balance and Total Trade Balance decreased from 199091 to 200001. The increase in Oil Imports from 199091 to 200001 was more as compared to the increase in Oil Exports from 200001 to 201011.  
69.  NonOil Exports increased by approximately 273% from 200001 to 201011. Total Exports increased by approximately 377% from 199091 to 200001. Oil Imports increased by 175% from 198081 to 199091. Total Imports increased by approximately 55% from 200001 to 201011. 

DIRECTIONS FOR QUESTIONS 70 to 71 : 

70.  The total number of employees who were appraised in January was 71 + 67 + 97 i.e. 235. These were the employees who were appraised on at least one performance area. The total number of employees who were appraised in July was 30 + 22 + 29 i.e. 81. These were the employees who were appraised on at least two performance areas. The number of employees who were appraised on exactly one performance area is 235 – 81 i.e. 154 

71.  The number of employees who were not appraised on Individual Performance in January was 67 + 97 i.e. 164. The employees who were appraised on Individual Performance in July and November were among these 164 employees only. So the number of employees who were not appraised on Individual Performance in 2010 was 164 – (30 + 9) = 125.  
DIRECTIONS FOR QUESTIONS 72 to 74 : 

72.  Team 4 has scored the maximum runs through the mentioned players.â€‹  
73.  Only 3 players have scored less than 9,000 runs even after playing more than 200 matches  
74. 
Player 1 and Player 2 of Team 4 have better averages than Player 2 of Team 3. Runs scored by Player 1 and Player 2 of Team 4 = 13600 + 10800 = 24400 Matches played by them = 340 + 240 = 580 Average score = 24400/580 = 42.07 

FOR DIRECTIONS FOR QUESTIONS 75 to 77 : Aamir is either A or B. Accordingly, Abhishek would be either B or A. Therefore, Amitabh and Akshay are disguised as C and D. As the maximum possible marks in each Mock Cat were 75, the total marks obtained by C are definitely more than the total marks obtained by D. Therefore, C is Amitabh and D is Akshay. Let us assume that Abhishek is disguised as B and the marks scored by Abhishek in Mock 1 are x. ∴ x + 87 > 138 or x > 51 which would mean that Abhishek could not score the lowest in Mock 1. Hence, the assumption is incorrect and so Abhishek must be disguised as A and Akshay as Bâ€‹ 

75. 
Let’s assume that Akshay, who is disguised as D, score p marks in Mock 3. So his total score d = 63 + p As we know that d < c 63 + p < 138 or p < 75 All the scores from –25 to +75 are possible in any of the tests, except 70, 73 and 74. Therefore, different scores possible for Akshay = 101 – 3 – 1 = 97 

76.  a  
77. 
Abhishek is disguised as A. Let’s assume that Abhishek scores q marks in Mock 2. As Aamir scored the highest in Mock 2, q < 62 and 47 + 72 + q > 51 + 59 + 28. ∴ 19 < q < 62 Abhishek’s score is a multiple of 6. ∴ 119 + q needs to be a multiple of 6. ∴ q = 6K + 1, where K is a whole number. 19 < 6K + 1 < 62 ∴ 4 ≤ K ≤ 10 Therefore, 7 different values are possible for a. 

FOR DIRECTIONS FOR QUESTIONS 78 to 80 : Let the fee (in Rs. lakhs) demanded by Careena, Carishma, Cashmira and Catrina be a, b, c and d (not necessarily in that order). Let e = 1123. The following cases are possible:â€‹
Note: 

78.  câ€‹  
79.  dâ€‹  
80.  c  
FOR DIRECTIONS FOR QUESTIONS 81 to 83 : Let the number of fit sportsmen who play Hockey be x. From statement (ii), the number of unfit sportsmen who play Hockey is also x. From statements (v) and (vii), the number of fit sportsmen who play Cricket is 1250 and the number of unfit sportsmen who play Cricket is 0. From statement (i), we can deduce that x is equal to 2500. From statement (iv), the number of sportsmen who play Hockey for Under19 team is 4000 and the number of sportsmen who play Hockey for Above19 team is 1000. Further analysis leads to the tables given below.â€‹


81.  The total number of sportsmen in Khelabad is 6250.â€‹  
82.  The required ratio is 1 : 4.â€‹  
83. 
The total number of fit sportsmen who play for Above19 team is 1000.â€‹ 

DIRECTIONS FOR QUESTIONS 84 to 86 : The table given below shows the grades received by the students in different subjects.â€‹


84.  Esha got the highest CGPA.  
85.  Bipin and David got the same CGPA  
86.  The CGPA of Fardeen was 6.96  
DIRECTIONS FOR QUESTIONS 87 to 89 : Let Interest Payment be x% and NonPlanned Capital Account be y% of Budgeted Expenditure. From pie chart B, x + y = 22% and x – y = 12%. Thus, x = 17 and y = 5. 

87. 
The difference between Budgeted Expenditure and Budgeted Revenue when expressed as a percentage of Budgeted Revenue = 625/500 × 100 = 125% 

88.  The sum of Planned Revenue Account and Planned Capital Account (Rs. 281.25 thousand crores) exceeded the NonTax Revenue (Rs. 95 thousand crores) by Rs. 186.25 thousand crores.  
89.  Since nothing is mentioned about the breakup of Budgeted Revenue in FY 201112, the percentage share of Tax Revenue cannot be determined.  
DIRECTIONS FOR QUESTIONS 90 to 91 : 

90. 
Percentage change for the year 1971 is the highest and the value is


91.  The absolute difference for the year 1981 is clearly much more than that for the other four years and the value is 12188903.  
DIRECTIONS FOR QUESTIONS 92 to 94 : 

92.  Among males, Performance Scores of Kaushik and Maneet don’t fall in the interval [70, 100] and so they are Low on Happy Growth Index. Among females, Performance Scores of Asmita and Amit don’t fall in the interval [60, 90] and so they are Low on Happy Growth Index. So 4 students in all are Low on Happy Growth Index at Primary Stage of Education.  
93.  Among males at Senior Secondary Stage, Saurabh and Mudit are High on Happy Growth Index. Among females at Primary Stage, Meenal, Manjari and Renuka are High on Happy Growth Index. So the required ratio is 2 : 3.  
94.  Renuka is High on Happy Growth Index. Meenal and Manjari are the only female students whose Academic Score is less than 50 and who are High on Happy Growth Index.  
DIRECTIONS FOR QUESTIONS 95 to 97 : 

95.  In Round 1, the total number of votes cast is 1,00,00,000. The candidate with the maximum number of votes gets 16,52,754 votes or 16.5% of the total number of votes.  
96.  The total number of votes becomes 67,23,323. Then “Others” will have 1,51,325/67,23,323 = 2.25% share. This means an increase of 0.74  
97.  Among the candidates who enter Round 2, the Kongress candidat 