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Re: TCS Interview Questions MBA HR
According to sources these are the some basis interview question for TCS MBA Hr: 1) Market yourself 2) Why TCS? 3) Will you switch over to any other company after joining TCS? If No then why? 4) Are you ready to shift to any other TCS office worldwide? 5) Are you ready to go to places with extreme temperatures? What do u think about it? 6) What are the requirements for being a good leadership? 7) Why you switch over to software from your own back ground? What are the essential qualities of a software engineer and a Team leader? 9) What are your good qualities? 10) Do u wish to pursue higher studies? 11) Give an instance of a situation which u think u could have handled much better? 12) Do you have any queries? 
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Re: TCS Interview Questions MBA HR
Here I am providing some question papers of Aptitude Test of TCS Placement: TCS Placement Aptitude Test Paper 1 TCS Placement Aptitude Test Paper 2 1. Ray writes a two digit number. He sees that the number exceeds 4 times the sum of its digits by 3. If the number is increased by 18, the result is the same as the number formed by reversing the digits. Find the number. a) 35 b) 42 c) 49 d) 57 Solution: Let the two digit number be xy. 4(x + y) +3 = 10x + y .......(1) 10x + y + 18 = 10 y + x ....(2) Solving 1st equation we get 2x  y = 1 .....(3) Solving 2nd equation we get y  x = 2 .....(4) Solving 3 and 4, we get x = 3 and y = 5 2. a, b, c are non negitive integers such that 28a+30b+31c = 365. a + b + c = ? a) Greater than 14 b) less than or equal to 11 c) 13 d) 12 In a calender, Number of months having 28 days = 1 Number of months having 30 days = 4 Number of months having 31 days = 7 28 x 1 + 30 x 4 + 31 x 7 = 365 Here, a = 1, b = 4, c = 7. a+b+c = 12 3. George can do a piece of work in 8 hours. Paul can do the same work in 10 hours, Hari can do the same work in 12 hours. George, paul and hari start the same work at 9 am, while george stops at 11 am, the remaining two complete the work. What time will the work complete? a) 11.30 am b) 12 noon c) 12.30 pm d) 1 pm Let the total work = 120 units. As George completes this entire work in 8 hours, his capacity is 15 units /hour Similarly, the capacity of paul is 12 units / hour the capacity of Hari is 10 units / hour All 3 started at 9 am and worked upto 11 am. So total work done upto 11 am = 2 x (15 + 12 + 10) = 74 Remaining work = 120  74 = 46 Now this work is to be done by paul and hari. 46 / (12 + 10) = 2 hours (approx) So work gets completed at 1 pm 4. If x^y denotes x raised to the power y, Find last two digits of (1141^3843) + (1961^4181) a) 02 b) 82 c) 42 d) 22 Remember 1 raised to any power will give 1 as unit digit. To find the digit in the 10th place, we have to multiply, 10th digit in the base x unit digit in the power. So the Last two digits of the given expression = 21 + 61 = 82 5. J can dig a well in 16 days. P can dig a well in 24 days. J, P, H dig in 8 days. H alone can dig the well in How many days? a) 32 b) 48 c) 96 d) 24 Assume the total work = 48 units. Capacity fo J = 48 / 16 = 3 units / day Capacity of P = 48 / 24 = 2 units / day Capacity of J, P, H = 48 / 8 = 6 units / day From the above capacity of H = 6  2  3 = 1 So H takes 48 / 1 days = 48 days to dig the well 6. If a lemon and apple together costs Rs.12, tomato and a lemon cost Rs.4 and an apple costs Rs.8 more than a lemon. What is the cost of lemon? L + A = 12 ...(1) T + L = 4 .....(2) L + 8 = A Taking 1 and 3, we get A = 10 and L = 2 7. 3 mangoes and 4 apples costs Rs.85. 5 apples and 6 peaches costs 122. 6 mangoes and 2 peaches costs Rs.144. What is the combined price of 1 apple, 1 peach, and 1 mango. a) 37 b) 39 c) 35 d) 36 3m + 4a = 85 ..(1) 5a + 6p = 122 ..(2) 6m + 2p = 144 ..(3) (1) x 2 => 6m + 8a = 170 4 x (3) => 6m + 2p = 144 Solving we get 8a  2p = 56 ..(4) (2) => 5a + 6p = 122 3 x (4) = 24a  6p = 168 Solving we get a = 10, p = 12, m = 15 So a + p + m = 37 8. An organisation has 3 committees, only 2 persons are members of all 3 committee but every pair of committee has 3 members in common. what is the least possible number of members on any one committee? a) 4 b) 5 c) 6 d) 1 Total 4 members minimum required to serve only on one committee. 9. There are 5 sweets  Jammun, kaju, Peda, Ladu, Jilebi which can be consumed in 5 consecutive days. Monday to Friday. A person eats one sweet a day, based on the following constraints. (i) Ladu not eaten on monday (ii) If Jamun is eaten on Monday, Ladu should be eaten on friday. (iii) Peda is eaten the day following the day of eating Jilebi (iv) If Ladu eaten on tuesday, kaju should be eaten on monday based on above, peda can be eaten on any day except a) tuesday b) monday c) wednesday d) friday From the (iii) clue, peda must be eaten after jilebi. so Peda should not be eaten on monday. 10. If YWVSQ is 25  23  21  19  17, Then MKIGF a) 13  11  8  7  6 b) 1  2357 c) 9  8  7  6  5 d) 7  8  5  3 MKIGF = 13  11  9  7  6 Note: this is a dummy question. Dont answer these questions 11. Addition of 641 + 852 + 973 = 2456 is incorrect. What is the largest digit that can be changed to make the addition correct? a) 5 b) 6 c) 4 d) 7 641 852 963  2466 largest among tens place is 7, so 7 should be replaced by 6 to get 2456 12. Value of a scooter depriciates in such a way that its value at the end of each year is 3/4th of its value at the beginning of the same year. If the initial value of scooter is 40,000, what is the value of the scooter at the end of 3 years. a) 23125 b) 19000 c) 13435 d) 16875 value of the scooter at the end of the year = 40000×(34)3 = 16875 13. At the end of 1994, R was half as old as his grandmother. The sum of the years in which they were born is 3844. How old R was at the end of 1999 a) 48 b) 55 c) 49 d) 53 In 1994, Assume the ages of GM and R = 2k, k then their birth years are 1994  2k, 1994  k. But given that sum of these years is 3844. So 1994  2k + 1994  k = 3844 K = 48 In 1999, the age of R is 48 + 5 = 53 14. When numbers are written in base b, we have 12 x 25 = 333, the value of b is? a) 8 b) 6 c) None d) 7 Let the base = b So, (b+2)(2b+5) = (b+2)(2b+5)=3b2+3b+3 2b2+9b+10=3b2+3b+3 b2−6b−7=0 Solving we get b = 7 or 1 So b = 7 15. How many polynomials of degree >=1 satisfy f(x2)=[f(x)]2=f(f(x) a) more than 2 b) 2 c) 0 d) 1 Let f(x) = x2 f(x2)=[x2]2=x4 (f(x))2=[x2]2=x4 f(f(x))=f(x2)=[x2]2=x4 Only 1 16. Figure shows an equilateral triangle of side of length 5 which is divided into several unit triangles. A valid path is a path from the triangle in the top row to the middle triangle in the bottom row such that the adjacent triangles in our path share a common edge and the path never travels up (from a lower row to a higher row) or revisits a triangle. An example is given below. How many such valid paths are there? a) 120 b) 16 c) 23 d) 24 Sol: Number of valid paths = (n1) ! = (51)! = 24 17. In the question, A^B means, A raised to power B. If x*y^2*z < 0, then which one of the following statements must be true? (i) xz < 0 (ii) z < 0 (iii) xyz < 0 a) (i) and (iii) b) (iii) only c) None d) (i) only As y^2 is always positive, x*y^2*z < 0 is possible only when xz < 0. Option d is correct. 18. The marked price of a coat was 40% less than the suggested retail price. Eesha purchased the coat for half the marked price at the fiftieth anniversary sale. What percentage less than the suggested retail price did Eesha pay? a) 60 b) 20 c) 70 d) 30 Let the retail price is Rs.100. then market price is (10040) % of 100 = 60. Eesha purchased the coat for half of this price. ie., 30 only. which is 70 less than the retail price. So Option C is correct. 1. The wages of 24 men and 16 women amounts to Rs.11600 per day. Half the number of men and 37 women earn the same amount per day. What is the daily wage of a man? Let the wage of a man is m and woman be w. 24m+16w=11600 12m+37w = 11600 Solving we get m = 350 2. The sum of three digits a number is 17. The sum of square of the digits is 109. If we substract 495 from the number, the number is reversed. Find the number. Let the number be abc. Then a + b + c= 17 .....(1) a2+b2+c2=109 .....(2) 100a+10b+c 495 = 100c+10b+a ......(3) From 3, we get a  c = 5 So the possibilities for (a, c, b) are (6,1,10), (7,2,8), (8,3,6), (9,4,4) From the above, (8,3,6) satisfies the condition. 3. A calculator has a key for squaring and another key for inverting. So if x is the displayed number, then pressing the square key will replace x by x^2 and pressing the invert key will replace x by 1/x. If initially the number displayed is 6 and one alternatively presses the invert and square key 16 times each, then the final number displayed (assuming no roundoff or overflow errors) will be Evern number of inverse key has no effect on the number. By pressing the square key, the value got increased like 2, 4, 8, .... Which are in the format of 2n. So after the 16 pressings the power becomes 216 So the final number will be 6216=665536 4. How many two digit numbers are there which when substracted from the number formed by reversing it's digits as well as when added to the number formed by reversing its digits, result in a perfect square. Let the number xy = 10x + y Given that, 10x+y  (10y  x) = 9(xy) is a perfect square So xy can be 1, 4, 9.  (1) So given that 10x+y +(10y +x) = 11(x+y) is a perfect square. So x+y be 11. Possible options are (9,2), (8,3),(7,4),(6,5) (2) From the above two conditions only (6,5) satisfies the condition Only 1 number 56 satisfies. 5. Find the 55th word of SHUVANK in dictionary Sol: Arranging the letters in alphabetical order we get : A H K N S U V Now Total words start with A are 6! Total words start with AH are 5! = 120 Now Total words start with AHK are 4! = 24 Total words start with AHN are 4! = 24 Total words start with AHSK are 3! = 6 Now AHSNKUV will be the last word required. 6. Car A leaves city C at 5pm and is driven at a speed of 40kmph. 2 hours later another car B leaves city C and is driven in the same direction as car A. In how much time will car B be 9 kms ahead of car A if the speed of car is 60kmph Relative speed = 60  40 = 20 kmph Initial gap as car B leaves after 2 hours = 40 x 2 = 80 kms Car B should be 9 km ahead of the A at a required time so it must be 89 km away Time = 89 / 20 = 4.45 hrs or 267 mins 7. Find the average of the terms in the series 12+34+5....+199200 Sol12) +(34) + (56) +........(199200) = 100 Average = 100 / 200 = 0.5 8. n is a natural number and n^3 has 16 factors. Then how many factors can n^4 have? Total factors of a number N=ap.bq.cr... is (p+1)(q+1)(r+1)... As n3 has 16 factors n3 can be one of the two formats given below n3 =a15 n3 = a3.b3 If n3 =a15 then n = a5 and number of factors of n4 = 21 n3 = a3.b3 then n = ab and number of factors n4 = 25 9. Two cars start from the same point at the same time towards the same destination which is 420 km away. The first and second car travel at respective speeds of 60 kmph and 90 kmph. After travelling for sometime the speeds of the two cars get interchanged. Finally the second car reaches the destination one hour earlier than the first. Find the time after which the speeds get interchanged? Let the total time taken by the cars be a and b Let the time after which the speed is interchanged be t For car A, 60t+90(at) = 420, 90a  30t = 420 .......(1) For car B, 90t + 60(bt) = 420, 60b + 30t = 420 ....(2) Using both (1) and (2), we get 90a + 60b = 840 But as a  b =1, 90a + 60(a1) = 840. Solving a = 6. Substituting in equation 1, we get t = 4 TCS Placement Aptitude Test Paper 3 1. The value of diamond varies directly as the square of its weight. If a diamond falls and breaks into two pieces with weights in the ratio 2:3. what is the loss percentage in the value? Sol: Let weight be “x” the cost of diamond in the original state is proportional to x2 when it is fallen it breaks into two pieces 2y and the 3y x = 5y Original value of diamond = (5y)2 = 25y2 Value of diamond after breakage = (2y)2 +(3y)2 =13y2 so the percentage loss will be = 25y2 −13y225y2×100=48% 2. Five college students met at a party and exchanged gossips. Uma said, “Only one of us is lying”. David said, “Exactly two of us are lying”. Thara said, “Exactly 3 of us are lying”. Querishi said, “Exactly 4 of us are lying”. Chitra said “All of us are lying”. Which one was telling the truth? a)David b)Querishi c)Chitra d)Thara Sol: As all are contradictory statements, it is clear that ONLY one of them is telling the truth. So remaining 4 of them are lying. Querishi mentioned that exactly 4 are lying. So, he is telling the truth. Explanation: Let us 1st assume that Uma is telling the truth. Then according to her only one is lying. But if only one is lying then all the others’ statements are contradicting the possibility. In the same way all the other statements should be checked. If we assume the Querishi is telling the truth, according to him exactly 4 members are lying. So all the others are telling lies and he is the one who is telling the truth. This case fits perfectly. 3. Cara, a blue whale participated in a weight loss program at the biggest office. At the end of every month, the decrease in weight from original weight was measured and noted as 1, 2, 6, 21, 86, 445, 2676. While Cara made a steadfast effort, the weighing machine showed an erroneous weight once. What was that. a) 2676 b) 2 c) 445 d) 86 SOL: This is a number series problem nothing to do with the data given. 1x 1+1=2 2 x 2+2=6 6 x 3+3=21 21 x 4+4=88 and not 86 88 x 5+5 = 445 445*6+6 = 2676 4. The letters in the word ADOPTS are permuted in all possible ways and arranged in alphabetical order then find the word at position 42 in the permuted alphabetical order? a) AOTDSP b) AOTPDS c) AOTDPS d) AOSTPD SOL: In alphabetical order : A D O P S T A _ _ _ _ _ : the places filled in 5! ways = 120, But we need a rank less than 120. So the word starts with A. A D _ _ _ _ : empty places can be filled in 4!=24 A O _ _ _ _ : the places filled with 4! ways = 24. If we add 24 + 24 this total crosses 42. So We should not consider all the words starting with AO. A O D _ _ _ : 3!= 6 A O P _ _ _ : 3!=6 Till this 36 words are obtained, we need the 42nd word. AOS _ _ _ : 3!= 6 Exactly we are getting the sum 42. So last 3 letters in the descending order are TPD. So given word is AOSTPD 4. A man who goes to work long before sunrise every morning gets dressed in the dark. In his sock drawer he has 6 black and 8 blue socks. What is the probability that his first pick was a black sock, but his second pick was a blue sock? SOL: This is a case of without replacement. We have to multiply two probabilities. 1. Probability of picking up a black sock, and probability of picking a blue sock, given that first sock is black. 6C114C1×8C113C1=2491 5. There are 6 red balls,8 blue balls and 7 green balls in a bag. If 5 are drawn with replacement, what is the probability at least three are red? Sol: At least 3 reds means we get either : 3 red or 4 red or 5 red. And this is a case of replacement. case 1 : 3 red balls : 6/21 x 6/21 x 6/21 x 15/21 x 15/21 case 2 : 4 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 15/21 case 3 : 5 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 6/21 Total probability = = (6/21 x 6/21 x 6/21 x 15/21 x 15/21)+(6/21 x 6/21 x 6/21 x 6/21 x (15 )/21)+ (6/21 x 6/21 x 6/21 x 6/21 x 6/21) = 312/16807 6. Total number of 4 digit number do not having the digit 3 or 6. Sol: consider 4 digits _ _ _ _ 1st blank can be filled in 7C1 ways (0,3,6 are neglected as the first digit should not be 0) 2st blank can be filled in 8C1 ways (0 considered along with 1,2,4,5,7,8,9) 3st blank can be filled in 8C1 ways 4st blank can be filled in 8C1 ways Therefore total 4 digit number without 3 and 6 is 7 x 8 x 8 x 8=3584 7. Find the missing in the series: 70, 54, 45, 41,____. Sol: 40 7054 = 16 = 42 5445 = 9 = 32 4541 = 4 = 22 4140 = 1 = 12 8. A school has 120, 192 and 144 students enrolled for its science, arts and commerce courses. All students have to be seated in rooms for an exam such that each room has students of only the same course and also all rooms have equal number of students. What is the least number of rooms needed? Sol: We have to find the maximum number which divides all the given numbers so that number of roots get minimized. HCF of 120,192 & 144 is 24. Each room have 24 students of the same course. Then rooms needed 12024+19224+14424 = 5 +8 + 6 = 19 9. A farmer has a rose garden. Every day he picks either 7,6,24 or 23 roses. When he plucks these number of flowers the next day 37,36,9 or 18 new flowers bloom. On Monday he counts 189 roses. If he continues on his plan each day, after some days what can be the number of roses left behind? (Hint : Consider number of roses remaining every day) a)7 b)4 c)30 d)37 SOL: let us consider the case of 23. when he picks up 23 roses the next day there will be 18 new, so in this case., 5 flowers will be less every day. So when he counts 189, the next day 184, 179,174,169,................ finally the no. of roses left behind will be 4. 10. What is the 32nd word of "WAITING" in a dictionary? Sol: Arranging the words of waiting in Alphabetical Order : A,G,I,I,N,T,W Start with A_ _ _ _ _ _ This can be arranged in 6!/2! ways=720/2=360 ways so can't be arranged starting with A alone as it is asking for 32nd word so it is out of range AG_ _ _ _ _then the remaining letters can be arranged in 5!/2! ways so,120/2=60 ways. Out of range as it has to be within 32 words. AGI_ _ _ _ Now the remaining letters can be arranged in 4! ways =24 AGN _ _ _ _ can be arranged in 4!/2! ways or 12 ways so,24+12 =36th word so out of range. So we should not consider all the words start with AGN now AGNI_ _ _can be arranged in 3! ways =6 ways so 24+6=30 within range Now only two word left so, arrange in alphabetical order. AGNTIIW  31st word AGNTIWI  32nd word 1. A manufacturer of chocolates makes 6 different flavors of chocolates. The chocolates are sold in boxes of 10. How many “different” boxes of chocolates can be made? Sol: If n similar articles are to be distributed to r persons, x1+x2+x3......xr=n each person is eligible to take any number of articles then the total ways are n+r−1Cr−1 In this case x1+x2+x3......x6=10 in such a case the formula for non negative integral solutions is n+r−1Cr−1 Here n =6 and r=10. So total ways are 10+6−1C6−1 = 3003 2. In a single throw with two dice, find the probability that their sum is a multiple either of 3 or 4. a. 1/3 b. 1/2 c. 5/9 d. 17/36 Sol: Their sum can be 3,4,6,8,9,12 For two dice, any number from 2 to 7 can be get in (n1) ways and any number from 8 to 12 can be get in (13  n) ways. Then possible ways are 2 + 3 + 5 + 5 + 4 + 1 = 20 possible cases. So probability is (20/36)=(5/9) 3. B alone can do piece of work in 10 days. A alone can do it in 15 days. If the total wages for the work is Rs 5000, how much should B be paid if they work together for the entire duration of the work? a. 2000 b. 4000 c. 5000 d. 3000 Sol: Time taken by A and B is in the ratio of = 3:2 Ratio of the Work = 2 : 3 (since, time and work are inversely proportional) Total money is divided in the ratio of 2 : 3 and B gets Rs.3000 4. On a 26 question test, 5 points were deducted for each wrong answer and 8 points were added for right answers. If all the questions were answered how many were correct if the score was zero. a. 10 b. 11 c. 13 d. 12 Sol: Let x ques were correct. Therefore, (26 x) were wrong 8x−5(26−x)=0 Solving we get x=10 5. Arun makes a popular brand of ice cream in a rectangular shaped bar 6cm long, 5cm wide and 2cm thick. To cut costs, the company had decided to reduce the volume of the bar by 19%. The thickness will remain same, but the length and width will be decreased by some percentage. The new width will be, a. 5.5 b. 4.5 c. 7.5 d. 6.5 Sol: Volume =l×b×h = 6×5×2 = 60 cm3 Now volume is reduced by 19%. Therefore, new volume = (100−19)100×60=48.6 Now, thickness remains same and let length and breadth be reduced to x% so, new volume: (x100×6)(x100×5)2=48.6 Solving we get x =90 thus length and width is reduced by 10% New width = 5(10% of 5)=4.5 6. If all the numbers between 11 and 100 are written on a piece of paper. How many times will the number 4 be used? Sol: We have to consider the number of 4's in two digit numbers. _ _ If we fix 4 in the 10th place, unit place be filled with 10 ways. If we fix 4 in units place, 10th place be filled with 9 ways (0 is not allowed) So total 19 ways. Alternatively: There are total 9 4's in 14, 24, 34...,94 & total 10 4's in 40,41,42....49 thus, 9+10=19. 7. If twenty four men and sixteen women work on a day, the total wages to be paid is 11,600. If twelve men and thirty seven women work on a day, the total wages to be paid remains the same. What is the wages paid to a man for a day’s work? Sol: Let man daily wages and woman daily wages be M and W respectively 24M+16W=11600 12M+37W=11600 solving the above equations gives M=350 and W=200 8. The cost price of a cow and a horse is Rs 3 lakhs. The cow is sold at 20% profit and the horse is sold at 10% loss. Overall gain is Rs 4200. What is the cost price of the cow? Sol: Profit = 4200 Profit =SP  CP 4200=SP  300000 therefore SP=304200 x+y = 300000 1.2x + 0.9y = 304200 Solving for x = 114000 = CP of cow. 9. 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4...... In the above sequence what is the number of the position 2888 of the sequence. a) 1 b) 4 c) 3 d) 2 Sol: First if we count 1223334444. they are 10 In the next term they are 20 Next they are 30 and so on So Using n(n+1)2×10≤2888 For n = 23 we get LHS as 2760. Remaining terms 128. Now in the 24th term, we have 24 1's, and next 48 terms are 2's. So next 72 terms are 3's. The 2888 term will be “3”. 10. How many 4digit numbers contain no.2? Sol: Total number of four digit numbers =9000 (i.e 1000 to 9999 ) We try to find the number of numbers not having digit 2 in them. Now consider the units place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9) Tens place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9) Hundreds place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9) Thousands place can be selected in 8 ways (i.e 1,3,4,5,6,7,8,9) here '0' cannot be taken Total number of numbers not having digit 2 in it =9 x 9 x 9 x 8 =5832 Total number of numbers having digit 2 in it = 90005832 =3168 1. 2ab5 is a four digit number divisible by 25. If a number formed from the two digits ab is a multiple of 13, then ab is a. 52 b. 45 c.10 d.25 Sol: For a number to be divisible by 25, last two digits of that number should be divisible by 25. So b must be either 2 or 7 it is given that ab must be divisible by 13 and in the options only 52 is divisible by 13. 2. The average temperature of Tuesday Wednesday and Thursday was 37 C. The average temperature of Wednesday and Thursday and Friday was 38 C. if the temperature on Friday was 39 C. Find the temperature on Tuesday. a. 37.33 b. 38.33 c. 36 d. None of the above Sol: (tues + wed + thurs)/3=37 tues + wed + thurs=111...(1) (wed + thurs + fri)/3=38 (wed + thurs + fri) =114...(2) Given friday is 39. then, (2)  (1) Fri  Tues = 3 So 39  Tues = 3 Tuesday =36 3. There are 5 boxes in a cargo. The weight of the 1st box is 200 KG, the weight of the 2nd box is 20% higher than the third box, whose weight is 25% higher than the 1st box weight. The 4th box which weighs 350 KG is 30% lighter than the 5th box. Find the difference in average weight of the 4 heaviest boxes and the four lightest boxes. Sol: weight of 1st box=200 weight of 3rd box=(125/100)*200=250 weight of 2nd box=(120/100)*250=300 weight of 4th box =350 weight of 5th box=(10/7)*350=500 average of 4 highest weighted boxes=(500+350+300+250)/4=350 average of 4 lightest boxes=(350+300+250+200)/4=275 therefore difference=350275=75 4. The length, breadth and height of a room are in the ratio 3:2:1. If the breadth and height are halved, while the length is doubled. Then the total area of the 4 walls of the room will be decreased by a. 30% b. 18.75% c. 15% d. 13.6% Sol: Given l:b:h=3:2:1 let h=10, b = 20, and l = 30 area = 2(l+b)h area= 2*(3x+2x)*x = 2(30+20)10=1000 Now after those adjustments in the measurements, l=60, b=10, h=5 area= 2(l+b)h = 2(60+10)5=700 Percentage decrease= 1000−7001000×1000=30% 5. A circle circumscribes three unit circles that touch each other. What is the area of the larger circle? Note that p is the ratio of the circumference to the diameter of a circle ( 3.14159265). Sol: By joining centres of 3 unit circles we will get an equilateral triangle of length 2 unit. We have to find the length of the orange line. And center of the equilateral triangle will be the center of the big circle. So radius of the big circle will be = (1 + Circum radius of the equilateral triagle) Circum radius of equilateral triangle = 23×3√2×2=23√ Area of big circle will be = πr2=3.14×(1+23√) 6. Rajesh calculated his average over the last 24 tests and found it to be 76. He finds out that the marks for three tests have been inverted by mistake. The correct marks for these tests are 87, 79 and 98. What is the approximate percentage difference between his actual average and his incorrect average? Sol: No Change Incorrect value is: 78, 97, 89 correct values are: 87, 79, 98 difference between correct and incorrect value is= 9 + 9 18=0 7. Joke is faster than Paul, Joke and Paul each walk 24 KM. The sum of their speed is 7 Km per hour. And the sum of times taken by them is 14 hours. Then, Joke speed is a. 3 KM/Hr b. 4 KM/Hr c. 5 KM/Hr d.7 KM/Hr Sol: Speed=Timedistance let the speed of joke x then speed of paul will be 7x 24x+247−x=14 Try to plugin the values from the options. If Joke speed is 4 the paul is 3. 8. The crew of a rowing team of 8 members is to be chosen from 12 men (M1, M2, …., M12) and 8 women (W1, W2,…., W8), such that there are two rows, each row occupying one the two sides of the boat and that each side must have 4 members including at least one women. Further it is also known W1 and M7 must be selected for one of its sides while M2, M3 and M10 must be selected for other side. What is the number of ways in which rowing team can be arranged. SoL: We need two person for one side and 1 women for the another side. We select that women in 7 ways. Now that second side people can sit in 7x4! ways. Now for the first side we need two people from the remaining 14. So this can be done in 14C2 ways and this side people can sit in 4C2×4! ways. Again the first group may take any of the two sides. So total ways are 2×7×4!×14C2×4! 9. In a certain city, 60% of the registered voters are congress supporters and the rest are BJP supporters. In an assembly election, if 75% of the registered congress supporters and 20% of the registered BJP supporters are expected to vote for candidate A, what percent of the registered voters are expected to vote for candidate A? Sol: let the people in the city be 100 Congress supporters = 60% of 100 = 60 40% are BJP=40% of 100 = 40 out of 60,75% voted for congress=75%(60)=45 out of 40%,20% voted for congress=20%(40)=8 Total=45 + 8 = 53 Total percent= 53% 10. Anusha, Banu and Esha run a running race of 100 meters. Anusha is the fastest followed by Banu and then Esha. Anusha, Banu and Esha maintain constant speeds during the entire race. When Anusha reached the goal post, Banu was 10m behind. When Banu reached the goal post Esha was 10m behind. How far was behind Anusha when the latter reached the goal post. option a) 70 b) 81 c) 90 d) 80 Sol: By that time Anusha covered 100m, Bhanu covered 90m. So ratio of their speeds = 10 : 9 By that time Bhanu reached 100m, Esha covered 90m. So ratio of their speeds = 10 : 9 Ratio of the speed of all the three = 100 : 90 : 81 By that time Anusha covered 100m, Esha Covers only 81. 11. Seven different objects must be divided among three persons. In how many ways this can be done if at least one of them gets exactly one object. Sol: Division of m+n+p objects into three groups is given by (m+n+p)!m!×n!×p! But 7 = 1 + 3 + 3 or 1 + 2 + 4 or 1 + 1 + 5 So The number of ways are (7)!1!×3!×3!×12!+(7)!1!×2!×4!+(7)!1!×1!×5!×12! = 70 + 105 + 21 = 196 12. George while driving along the highway saw road markers which are at equal distances from each other. He crosses the markers every 20 seconds. If he increases his speed by x meters per second, he crosses the markers at every 15 seconds. But if he increases his speed by y meters per second, he crosses the marker at every 10th second. If yx = 40 meters per second, then what is the distance between two markers. Sol: Let speed be =z m/s then Distance= 20z m (z+x)15=20z; (z+y)10=20z Also given that y  x = 40 solving we get 20z=1200 13. How many different 9 digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even position? Sol: Odd places are 4 and these are occupied by 3355. So this can be done in 4!/ (2! 2!) = 6 There are 5 even numbers which have to be placed at 5 odd places. So 5!/(2!3!) = 10 ways so total number of ways of arranging all these numbers are 10 * 6 = 60 ways 14. In a vessel, there are 10 litres of alcohol. An operation is defined as taking out five litres of what is present in the vessel and adding 10 litres of pure water to it. What is the ratio of alcohol to water after two operations? a) 1 : 5 b) 2 : 3 c) 1 : 6 d) 3 : 2 Sol: Final concentration = Initial concentration(1−replacement quantityFinal volume) Final concentration = 1×(1−1015)=13 Final concentration = 13×(1−1020)=16 So ratio of alcohol : water = 1 : 5
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