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#2
 
 
Re: TCS Placement Papers with solutions
As you require the Latest Placement Papers of TCS Exam with solutions so here I am sharing the same with you 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 Solution: 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 Solution: 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 Solution: 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 Rest of the Questions are attached in below file which is free of cost
__________________ Answered By StudyChaCha Member 
#5
 
 
Re: TCS Placement papers with solutions
The placement paper of TCS with solutions is as follows: TCS Aptitudereasoning placement paper: 1) The water from one outlet, flowing at a constant rate, can fill the swimming pool in 9 hours. The water from second outlet, flowing at a constant rate can fill up the same pool in approximately in 5 hours. If both the outlets are used at the same time, approximately what is the number of hours required to fill the pool? ans: (1/A) + (1/B) = 1/T So when you solved, and got 14/45, that was equal to 1/T. To find T, you need to take the reciprocal; T = 45/14, (3.21) u 2) If 75 % of a class answered the first question on a certain test correctly, 55 percent answered the second question on the test correctly, and 20 percent answered neither of the questions correctly, what percentage answered both correctly? ans: n(a ^ b) =n(a)+n(b)n(a v b) = 0.75+0.550.8 = 0.5 (50%) 3) A student's average ( arithmetic mean) test score on 4 tests is 78. What must be the students score on a 5th test for the students average score on the 5th test to be 80? ans: 4 tests sum=78*4=312 5 tests sum=80*5=400 then 5th test score=400312=88 (88) 4) Rural households have more purchasing power than do urban households at the same income level, since some of the income urban and suburban households use for food and shelter can be used by the rural households for other needs. Which of the following inferences is best supported by the statement made above? (A) The average rural household includes more people than does the average urban or suburban household. (B) Rural households have lower food and housing costs than do either urban or suburban households. (C) Suburban households generally have more purchasing power than do either rural or urban households. (D) The median income of urban and suburban households is generally higher than that of rural households. (E) All three types of households spend more of their income on housing than on all other purchases combined. ans: B 5) Jose is a student of horticulture in the University of Hose. In a horticultural experiment in his final year, 200 seeds were planted in plot I and 300 were planted in plot II. If 57% of the seeds in plot I germinated and 42% of the seeds in plot II germinated, what percent of the total number of planted seeds germinated? ans: 200 * 57 % = 114 and 300* 42% = 126. Total = 240 out of 500. (48%). 6) A closed cylindrical tank contains 36 pie cubic feet of water and its filled to half its capacity. When the tank is placed upright on its circular base on level ground, the height of water in the tank is 4 feet. When the tank is placed on its side on level ground, what is the height, in feet, of the surface of the water above the ground? ans: What given Volume = 36 Height = 4ft Volume = Pi* r2 * h 36 = Pi*r2 * 4 r2 =9 r = 3 So the radius is 3 which means the diameter is 6. 7) The present ratio of students to teachers at a certain school is 30 to 1. If the student enrollment were to increase by 50 students and the number of teachers were to increase by 5, the ratio of the teachers would then be 25 to 1 What is the present number of teachers? ans: 30/1 = s/t Current student to teacher ratio s+50/t+5 = 25/1 Future student to teacher ratio solving the first equation first equation for s gives s = 30t. Substitute this value of s into the second equation, and solve for t. s+50/t+5 = 25/1 30t + 50 = 25t + 125 multiply both sides by t+5 5t = 75 simplify by subtraction t = 15 (15) 8) College T has 1000 students. Of the 200 students majoring in one or more of the sciences,130 are majoring in Chemistry and 150 are majoring in Biology. If at least 30 of the students are not majoring in either Chemistry or Biology, then the number of students majoring in both Chemistry and Biology could be any number from ans: total=200 neither>=30 chemistry=130 bio=130 maximum common=minimum of the two=130 if neither=30 then common=150+130+30200=110 range is 110130 9) Kelly and Chris are moving into a new city. Both of them love books and thus packed several boxes with books. If Chris packed 60% of the total number of boxes, what was the ratio of the number of boxes Kelly packed to the number of boxes Chris packed? ans: 10060=40 40:60=2:3 10) A drug that is highly effective in treating many types of infection can, at present, be obtained only from the bark of the ibora, a tree that is quite rare in the wild. It takes the bark of 5,000 trees to make one kilogram of the drug. It follows, therefore, that continued production of the drug must inevitably lead to the ibora's extinction. Which of the following, if true, most seriously weakens the argument above? ans: The ibora can be propagated from cuttings and grown under cultivation. 11) Machine A produces bolts at a uniform rate of 120 every 40 second, and Machine B produces bolts at a uniform rate of 100 every 20 seconds. If the two machines run simultaneously, how many seconds will it take for them to produce a total of 200 bolts? ans: Machine A produces 120/40 = 3 bolts in 1 seconds and machine B produces 100/20 = 5 bolts in one second. Hence, both of them will produce 8 bolts per second. Hence, they wil take 200/8 = 25 seconds to produce 200 bolts. (25) 12) Wood smoke contains dangerous toxins that cause changes in human cells. Because wood smoke presents such a high health risk, legislation is needed to regulate the use of openair fires and wood burning stoves ans: In valleys where wood is used as the primary heating fuel, the concentration of smoke results in poor air quality. 13) Analysing the good returns that Halocircle Insurance Pvt Ltd was giving, Ratika bought a 1year, Rs 10,000 certificate of deposit that paid interest at an annual rate of 8% compounded semiannually.What was the total amount of interest paid on this certificate at maturity? ans: @8%, for half year you get 400/ 10000x 8/100x1/2=400 for the next year there will be interest on your investment of 10000 and also on your accrue interest of Rs.400/ that means another 400/+ interest on 400/ @8% for half year.=400+ (400x8/100x1/2) 400+400+16=816 14) Juan is a gold medalist in athletics. In the month of May, if Juan takes 11 seconds to run y yards, how many seconds will it take him to run x yards at the same rate? ans: 11x/y 15) A certain company retirement plan has a rule of 70 provision that allows an employee to retire when the employee's age plus years of employment with the company total at least 70. In what year could a female employee hired in 1986 on her 32nd birthday first be eligible to retire under this provision? ans: The female employee must gain at least 70 points. Now she has 32 and every year gives her two more points: one for age and one for additional year of employment, so 32 + 2 * (number of years) = 70 (Number of years) = 19 So, 1986 + 19 = 2005 16) Homeowners aged 40 to 50 are more likely to purchase ice cream and are more likely to purchase it in larger amounts than are members of any other demographic group. The popular belief that teenagers eat more icecream than adults must, therefore, be false. The argument is flawed because the author _____ ans: fails to distinguish between purchasing and consuming 17) Andalusia has been promoting the importance of health maintenance. From January 1,1991 to January 1,1993, the number of people enrolled in health maintenance organizations increased by 15 percent. The enrollment on January 1,1993 was 45 million. How many million people(to the nearest million) was enrolled in health maintenance organizations on January 1,1991? ans: let there are 'x' population at jan1,1991 15% of x = 15/100*(x) (x)+[15/100*(x)]=45 /* as there is 15% increase from 1991 to 1993*/ x=39.13 18) What is the lowest possible integer that is divisible by each of the integers 1 through 7, inclusive? ans: L.C.M of(1,2,3,4,5,6,7)=2*2*3*5*7=420 19) If the area of a square region having sides of length 6 cms is equal to the area of a rectangular region having width 2.5 cms, then the length of the rectangle, in cms, is ans: Area of a Square s x s 6 x 6 = 36 Area of a Rectangle lw ?(2.5) =36 as it is equal to the area of a square 36/2.5 = 14.4 cm 20) A tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume. If 2500 gallons of water evaporate from the tank, the remaining solution will be approximately what percentage of sodium chloride? ans: Original =10,000 > sodium chloride = 5% ie 500 New = 10,0002,500 = 7500. New % = (500/7500) * 100 = 6.67% 21) After loading a dock, each worker on the night crew loaded 3/4 as many boxes as each worker on the day of the crew. If the night crew has 4/5 as many workers as the day crew, what fraction of all the boxes loaded by two crews did the day crew load? ans: Number of workers Day shift: 5 workers (this is an easy number to find 4/5 of) Night shift: 4 workers (4/5 of 5 = 4) Boxes loaded per worker Day shift: 4 boxes per worker Night shift: 3 boxes per worker (3/4 of 4 = 3) Total boxes loaded Day shift: 5 workers times 4 boxes per worker = 20 boxes Night shift: 4 workers times 3 boxes per worker = 12 boxes Combined total boxes for both shifts = 20 + 12 = 32 Of the 32 boxes, the day shift loaded 20 of them. 20/32 = 5/8 22) A bakery opened yesterday with its daily supply of 40 dozen rolls. Half of the rolls were sold by noon and 80 % of the remaining rolls were sold between noon and closing time. How many dozen rolls had not been sold when the bakery closed yesterday? ans: 4 rolls were not sold.............half of them were sold by noon.......so 20 rolls remain....and then 20% were not sold,, so 20% of the 20 rolls is 4 then it is the answer 23) If n=4p, where p is a prime number greater than 2, how many different positive even divisors does n have including n? ans: 4 24) A dealer originally bought 100 identical batteries at a total cost of q rupees. If each battery was sold at 50 percent above the original cost per battery, then, in terms of q, for how many rupees was each battery sold? ans: bought for q/100 dollars per battery each battery sold with profit 50% is q/100 + q/100*50/100 =q/100+q/200 =3q/200 25) The price of lunch for 15 people was 207 pounds, including a 15 percent gratuity of service. What was the average price per person, EXCLUDING the gratuity? ans: Let the net price excluding the gratuity of service = x pounds Then, total price including 15% gratuity of service = x + 0.15 x = 1.15 x pounds So, 1.15 x = 207 pounds ==> x = 207 / 1.15 = 180 pounds Net price of lunch for each person = 180 / 15 = 12 pounds 26) Of the following, which is the closest approximation of (50.2*0.49)/199.8 ? ans: For approximation (50.2*0.49)/199.8 can be taken as 50*0.5/200 = 25/200 = 1/8 = 0.125 27) How many prime numbers between 1 and 100 are factors of 7150? ans: 7,150 = 2 * 5 * 5 * 11 * 13 so there are 4 distinct prime numbers that are below 100 28) Guitar Strings often go dead  become less responsive and bright in tone  after a few weeks of intense use. A researcher whose son is a classical guitarist hypothesized that dirt and oil, rather than changes in the material properties of the string, were responsible. Which of the following investigations is most likely to yield significant information that would help evaluate the researcher's hypothesis? ans: Determining whether smearing various substances on new guitar strings causes them to go dead 29) Red blood cells in which the malarialfever parasite resides are eliminated from a person's body after 120 days. Because the parasite cannot travel to a new generation of red blood cells, any fever that develops in a person more than 120 days after that person has moved to a malaria free region is not due to malarial parasite. Which of the following, if true, most seriously weakens the conclusion above? ans: In some cases, the parasite that causes malarial fever travels to cells of the spleen, which are less frequently eliminated from a person’s body than are red blood cells. 30) Among a group of 2500 people, 35 percent invest in municipal bonds, 18 percent invest in oil stocks, and 7 percent invest in both municipal bonds and oil stocks. If 1 person is to be randomly selected from 2500 people, what is the probability that the person selected will be one who invests in municipal bonds but not in oil stocks ans: 7/25 31) When a polygraph test is judged inconclusive, this is no reflection on the examinee. Rather, such a judgment means that the test has failed to show whether the examinee was truthful or untruthful. Nevertheless, employers will sometime refuse to hire a job applicant because of an inconclusive polygraph test result. Which of the following conclusions can most properly be drawn from the information above? ans: An inconclusive polygraph test result is sometimes unfairly held against the examinee 32) Country Club has an indoor swimming club. Thirty percent of the members of a swim club have passed the lifesaving test. Among the members who have not passed the test, 12 have taken the preparatory course and 30 have not taken the course. How many members are there in the swim club? ans: Suppose there are x members, non pass members arre 7x/10 which are equal to 12 + 30 = 42 7x/10 = 42 x = 60 There are 60 members in that club 33) A necklace is made by stringing N individual beads together in the repeating pattern red bead, green bead, white bead, blue bead and yellow bead. If the necklace begins with a red bead and ends with a white bead, then N could be: ans: R G W B Y is the bead pattern and it repeats. Bead want to end with White. So, the 3rd, 8th, 13th, 18th... beads will be W. this can be expressed as 5n+3,where n is an integer.(counting no of white bead) Test each of the answer choices to determine which is multiple of 5 plus a value of 3.Of the options ,only 68=5(13)+3 can be written in the form 5n+3. So,the answer is 68. Posted by kamesh prajapati at 9:52 PM No comments: Email This BlogThis! Share to Twitter Share to Facebook 33 A dog taken four leaps for every five leaps of hare but three leaps of the dog is equal to four leaps of the hare. Compare speed? Let the distance covered by dog in 1 leap is x and hare covered 1 leap is y. then, 3x = 4y => x =(4/3) y => 4x =(16/3) y Then, The ratio of speeds of dog and hare = Ratio of distances covered by them in the same time = 4x : 5y = (16/3)y : 5y =(16/3) : 5 = 16:15 Ans...=16:15 34 There are two boxes,one containing 39 red balls & the other containing 26 green balls.you are allowed to move the balls b/w the boxes so that when you choose a box random & a ball at random from the chosen box,the probability of getting a red ball is maximized.this maximum probability is a)60 b)50 c)80 d)30 the probability of getting a red ball should be maximized so we will keep 1 red ball in the first box and transfer remaining 38 red balls to other box which has green. The reason for doing this only this combination will make the probability of getting red ball is maximum (i.e) 1. In the question its mentioned that first we have to choose a box at random. So 2 boxes are there, probability of selecting a bag is 1/2 Probability of getting red ball from first box is 1 Probability of getting green ball from second box is 38/64 So (1/2)*1 for first box and (1/2)*(38/64) for box 2 answer is (1/2)*1 + ((1/2)*(38/64) which is approximately 0.8 Next number in the given series 1, 7, 8, 49, 50, 56, 57, 343 344 because 1*7=7 7+18 7*7=49 49+1=50 8*7=56 56+1=57 49*7=343 343+1=344 ans= 344 q.1 in how many ways can 3 postcards can be posted in 5 postboxes? q.2.in how many ways can 5 postboxes hold 3 post cards? q.1 in how many ways can 3 postcards can be posted in 5 postboxes? first post card can be posted in any of the 5 post boxes =5c1 ways second post card can be posted in any of the 5 post boxes =5c1 ways third post card can be posted in any of the 5 post boxes =5c1 ways number of ways=5c1*5c1*5c1=125=5^3 q.2.in how many ways can 5 postboxes hold 3 post cards? first post box can hold 3 post cards in 3c1 ways second post box can hold 3 post cards in 3c1 ways third post box can hold 3 post cards in 3c1 ways fourth post box can hold 3 post cards in 3c1 ways fifth post box can hold 3 post cards in 3c1 ways so number of ways =3c1*3c1*3c1*3c1*3c1=3^5 35 A number has exactly 3 prime factors, 125 factors of this number are perfect squares and 27 factors of this number are perfect cubes. overall how many factors does the number have??? Answer for this question is 729 The explanation goes like this Let a,b,c be three prime no. since the no. of perfect square as a factor is 125 which can only be if every prime no. has power in the multiple of 2 including zeros(bcoz any no. to the power of zero is a perfect square) so power must be 0,2,4,6,8 thereby 5*5*5=125 Since no. of cube as a factor is 27 therefore power in the prime no. will be in the multiple of 3 so power are 0,3,6 hence the no. N=a^8*b^8*c^8 hence total no. of factor is (8+1)(8+1)(8+1)=729 there r three buckets..of 8,5 n 3 litres...out of which only 8 ltr bucket is fully filled...u hv to fill exact 44 ltr liquid in 8 and 5 litre bucket by using only these buckets in minimm num of steps...... 8ltr 5ltr 3ltr initially 8 0 0 step1 3 5 0 step2 3 2 3 step3 6 2 0 step4 6 0 2 step5 1 5 2 step 6 1 4 3 step 7 4 4 0 total 7 steps 36 The climb from foot to top of a hill 800 meters, Jack can climb at 16 meters per minute and rests for two minutes or 20meters per 2 minutes and rest for one minute. Paul can climb at 10 meters per one minute and rest for one minute or16 meters per minute and rest for 2 minutes. If take has to reach the top in exactly two hours. What is the maximum number of rests that he can take? a) 41 b) 42 c) 40 d) 43 total=800m 16m per min means for 1m=800/16 =50min to climb 10m per min means for 1m=800/10=80min to climb for paul .....the total time is 2hours==120min....so,time taken for rest by paul is 12080=40min length of minute hand is 5.4 cm, area covered by this in 10 min is ? a)50.97 b)57.23 c)55.45 d)59.14 t will be (3.14*5.4*5.4)*10/60 = 15.26 cm2. 37 Raj tossed 3 dices and there results are noted down then what is the probability that raj gets 10? a) 1/72 b) 1/9 c) 25/216 d)1/8 ossible event.......(1,3,6)(1,4,5)(1,5,4)(1,6,3)(2,2,6)(2, 3,5)(2,4,4)(2,5,3)(2,6,2)(3,1,6)(3,2,5)(3,3,4)(3,4 ,3)(3,5,2)(3,6,1)(4,1,5)(4,2,4)(4,3,3)(4,4,2)(4,5, 1)(5,1,4)(5,2,3)(5,3,2)(5,4,1)(6,1,3)(6,2,2)(6,3,1 ) total 27 possible event so probability = 27/216 i.e 1/8 answerd(1/8) Probability of a sum of 3: 1/216 = 0.5% Probability of a sum of 4: 3/216 = 1.4% Probability of a sum of 5: 6/216 = 2.8% Probability of a sum of 6: 10/216 = 4.6% Probability of a sum of 7: 15/216 = 7.0% Probability of a sum of 8: 21/216 = 9.7% Probability of a sum of 9: 25/216 = 11.6% Probability of a sum of 10: 27/216 = 12.5% Probability of a sum of 11: 27/216 = 12.5% Probability of a sum of 12: 25/216 = 11.6% Probability of a sum of 13: 21/216 = 9.7% Probability of a sum of 14: 15/216 = 7.0% Probability of a sum of 15: 10/216 = 4.6% Probability of a sum of 16: 6/216 = 2.8% Probability of a sum of 17: 3/216 = 1.4% Probability of a sum of 18: 1/216 = 0.5% 38 Apple costs L rupees per kilogram for first 30kgs and Q rupees per kilogram for each additional kilogram. If the price of 33 kilograms is 11.67and for 36kgs of Apples is 12.48 then the cost of first 10 kegs of Apples is a) 3.50 b) 10.53 c) 1.17 d)2.8 30L+3Q=11.67 30L+6Q=12.48  3Q=.81 Q= .27 from that L=0.362 cost of 10 kg apple is 10*.362=3.6 The letters in the word ABUSER are permuted in all possible ways and arranged in alphabetical order then find the word at position 49 in the permuted alphabetical order? a) ARBSEU b) ARBESU c) ARBSUE d) ARBEUS AB**** =4!= 24 ways AE****=4!=24 ways next word is 49th so AR**** in alphabetical order **** will be BESU ans is B 39 A is twice efficient than B. A and B can both work together to complete a work in 7 days. Then find in how many days A alone can complete the work? ans is 10.5 gn: A+B=(1/7) A=2B =>B=1/21 A=2/21 =>A take (21/2)=10.5 days to complete the work. In 8*8 chess board what is the total number of squares. 1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2=204.... formula=1^2+2^2+3^2+....+n^2 40 X,Y,W,Z ARE INTEGER THE EXPRESSION XYZ IS EVEN&YWZ ODD IF X IS EVEN THEN WHICH OF FOLLOWING IS TRUE(a)Y MUST BE ODD(b)YZ MUST BE ODD(c)W MUST BE ODD(d)Z MUST BE ODD Ans.(c)W must be ODD XYZ=even and X even no. hence (yz) must be even because (even even = even) and ywz=yzw=odd and (yz) is even so W must be odd because (even  odd=odd) hence option (c) is correct 41 Two cyclist begin training on oval racecourse at same timethe professional cyclist complete each lap in 4 sec noves take 6 mintue how many mintue after start will both cyclist pass at exactly same spot where they begin to cycle(a)10(b)8(C)14(D)12 ans is d. simply LCM of 4 and 6. which of following true:A occur only B or C occur B occur D&E occurf occur if Cnot occur g occur if A and F occur option A occur whenever F occur2)F never occur3) G not occur if D not occur4)none of these 3) g nt occure i d not occure a=>b or c b=>d & e f=>c not g =>a & f 1) a occure whenevr f : depen on c where as 'a' can occure if 'c' not ..so this is not true 2) f nevr occure => depnd on c only n c is indipendent ...if c not occure f will occure ....so this is also not true 3) g not occure if d not:g depnd on a & f ...if d not ocuure b will not occure (as b=>d&e) a will occure if c occure...if c occure f will not occure ...hence g will not occure...this is true House1 is older than H2, H3 is taller than H4,H4 is older than H1...(i)H1 is older than H3 (ii)H2 is taller than H4 (a)only 1 (b)both (c)neither (d)only2 c)neither, is the answer. 42 In how many ways can 20 identical pencils be distributed among three girls so that each gets at least 1 pencil? now A'+B'+C''=17, number of non negative integers of A',B',C' are (17+31)C(31) so total Combinations are 19c2=19*9 = 171, simple formula for this kind of problems is (n1)c(r1) so 19c2= 171 if p(x)=ax^4+bx^3+cx^2+dx+e has roots at x=1,2,3,4 & p(0)=48.what is p(5) if 1 is a root then x1 is a factor of p(x) similarly x2 is a factor,x3 ,x4 are factors but p(x) is 4th degree polynomial therefore it can be in the form p(x) = Q(x1)(x2)(x3)(x4) , where Q is a constant but given p(0) =48 therefore 24 Q =48 Q=2 p(x) =2(x1)(x2)(x3)(x4) p(5)= 2*4*3*2*1 = 48 1!+2!+3!...+50! divided by 5! remainder will be ? 1!+2!+3!...+50! divided by 5! remainder will be whatever is obtained by dividing 1!+2!+3!+4! with 5!. so remainder is obtained by dividing (1+2+6+24)= 33 with 5! ( 120). so remainder is 33. 43 If there are Six periods in each working day of a school, In how many ways can one arrange 5 subjects such that each subject is allowed at least one period? we have 5 sub and 6 periods so their arrangement is 6P5 and now we have 1 period which we can fill with any of the 5 subjects so 5C1 6P5*5C1=3600 44 An article manufactured by a company consists of two parts X and Y. In the process of manufacturing of part X, 9 out 100 parts many be defective. Similarly , 5 out of 100 are likely to be defective in the manufacturer of Y. Calculate the probability that the assembled product will not be defective? a)0.6485 b)0.6565 c)0.8645 d)none of these ans=c (0.8645) probablity of nondefective of x= 91/100=.91 probablity of non defective of y= 95/100=.95 so, probablity of nondefective product=.91*.95=0.8645 45 There are six multiple choice questions in the examination. How many sequences of anwers are possible, if the first two question have 3 choices each, the next two have 4 choice each and last two have 5 choices each? 3c1*3c1*4c1*4c1*5c1*5c1=3600 for any one of first three option of Q.1 we have three options in Q.2 4 option in Q.3.....this goes on... ind the area (in square units) of the triangle formed by 2x+3y=5, y=x and XAxis. the points of intersection for lines: 2x+3y=5 and x=y => 5x=5 => x=1 and y=1 =>(1,1) x=y and y=0(xaxis) => x=0 and y=0 =>(0,0) 2x+3y=5 and y=0(xaxis) => 2x=5 => x=5/2 and y=0 =>(5/2,0) area of triangle with vertices (x1,y1),(x2,y2),(x3,y3) is x1(y2y3)+x2(y3y1)+x3(y1y2)/2 =1(00)+0(01)+5/2(10)/2 =1+5/2/2 =7/4 sq units... 46 Arun was all bent on building a new house. He carefully got the blue print of his house designed buy his friend Ashwin, a civil engineer. He wanted to build a room of dimension 27 by 48 ft and lay tiles in this room. Each tile was of dimension 2 by 3 ft. How many tiles should Arun buy? 27*48/2*3=216 47 Roy is now 4 year older than Erik and half of that amount than Lewis.If in two years Roy will be twice as old as Erick,then in two year what would be Roy age multiplied by Lewis age?? ans is 48 firstly two equation from 1st sentence .. suppose roy=x, erik=y, lewis=z so x=y+4....(1) 2nd is x=z+2...(2) (Roy is now 4 year older than Erik and half of that amount than Lewis so 4/2=2) also after 2 years erik and roy relation wil b x+2=2(y+2).. (3) so put this into 1st equ. u will get y=2.. and then x=6,z=4(by substituting other) nw asking multiplication after two years of roy n lewis is (6+2)*(4+2)=48.. if there are maximum acute angles then how many convex hectagons are there there can be max 3 acute angles in a convex polygon which is a triangle. 48 The number of bacteria was growing in a city exponentially.at 4 pm yesterday , the number of bacteria was 400 and at 6 pm yesterday it was 3600.How many bacteria were there at 7pm yesterday? The number of bacteria was growing in a city exponentially.at 4 pm yesterday , the number of bacteria was 400 and at 6 pm yesterday it was 3600.How many bacteria were there at 7pm yesterday? .49 There are 12 children in a family .the youngest 1 is a boy.then find the probabality of finding boy in the family. a)4096 b)2048 c)2 d)2! Probability cannot be more than 1..it ranges from 0 to 1.. there can be two options..either a boy/girl in a family asked is to find a boy=1/2 given 12 children but it is already mentioned that the youngest is a boy so no need to count that.. hence,1/2*1/2*1/2......(11 times)=1/2048 option b..but the option is wrong..it should be 1/2048 50 A volume of 10936 l is in a conatiner of sphere . how many semispheres of volume 4 l each will be required to transfer all the water into small semispheres? v10936/4=2734.....is the ans a volume of A are having in a container of sphere. how many semi hemispheres of B volume each will be required to transfer all the A in to semi hemispheres? ans is A/B. 51 My name is PREET.But my son accidentally types the by interchanging a pair of letters in my name.What is the probability that despite this interchange, the name remains unchanged? a.5% b.20% c.25% d.12.5% 10% Selecting any two  5C2 = 10[ treating E AND E as distinct) Selecting EE = 1 1/10 * 100 = 10% TCS 2013 placement new pattern questions 02 1 Four friends namely Rahul, Ravi, Rajesh and Rohan contested for a dairy milk chocolate. To decide which friend will get the chocolate they decided to throw two dice. Every friend was asked to choose a number and if the sum of the numbers on two dice equals that number, the concerned person will get the chocolate. Rahul's choice was7, Ravi's choice was 9, Rajesh's choice was 10 and Rohan's choice was 11. Who has the maximum probability of winning the amount. a) Rahul b) Ravi c) Rajesh d) Rohan Rahul has chosen 7, similarly Ravi9,Rajesh10,Rohan11 (1,1)(1,2)(1,3)(1,4)(1,5)(1,6) Sum of the digits are= 2,3,4,5,6,7 (2,1)(2,2)(2,3)(2,4)(2,5)(2,6) Sum of the digits are= 3,4,5,6,7,8 (3,1)(3,2)(3,3)(3,4)(3,5)(3,6)Sum of the digits are= 4,5,6,7,8,9 (4,1)(4,2)(4,3)(4,4)(4,5)(4,6)Sum of the digits are= 5,6,7,8,9,10 (5,1)(5,2)(5,3)(5,4)(5,5)(5,6)Sum of the digits are= 6,7,8,9,10,11 (6,1)(6,2)(6,3)(6,4)(6,5)(6,6)Sum of the digits are= 7,8,9,10,11,12 Among these 7 comes 6 times, 9 comes 4times, 10 comes 3times,11 comes only 2 times. So the maximum probability goes to Rahul 2 Messrs. Siva Constructions, leading agents in Chennai prepared models of their lands in the shape of a rectangle and triangle. They made models having same area. The length and width of rectangle model are 24 inches and 8 inches respectively. The base of the triangle model is 16 inches. What is the altitude of triangle model from the base to the top? a) 24 inches b) 8 inches c) 20 inches d) 32 inches Area of a rectangle= Area of the Triangle So, l*b=1/2(b*h) 24*8=1/2(16*8) therefore 24inches will be the answer 3 There is a 12*6 grid..grid contains 1*1 squares..what is d total number of squares..(consider 1*1,2*2,3*3,4*4,5*5,6*6 size squares as well).Lady has 25 blues, 7 red,9 yellow gloves and hats.how many gloves of same colour. (max + min + 2) = 25 + 9 +2=36 1!+2!+.....=50!=? 3.10350532 Ã— 10^64 Anita make cube with dimension 5*5*5 using 1*1*1 cubes.find no. Of cubes to make it hollow of same shape. To make the 5*5*5 hollow, the inner cubic part must be removed. This would constitute one cube each from front,back,top,bottom,left and right ! Since the 5*5*5 is made up of 1*1*1 cubes, 3*3*3 cubes (i.e. 52*52*52) would have to be removed to make it hollow with the shape retained. 27 cubes would be removed, remaining 12527=98. 1) if 3y + x > 2 and x + 2y let 3y+x=2 y+2y+x=2 2y+x=2y(ans) 3y > 2 x 2y > 4/3  2x/3 x+2y > 4/3 + x/3 x + 2y > 1/3(4 + x) 4 Tim and Elan are 90 km from each other, they start to move each other simultaneously, Tim at speed 10 and Elan 5kmph, if every hour they double their speed what is the distance that Tim will pass until he meet Elan? Tim and Elan will cover the distance in ratio of 2: 1 in same time. so out of 90 kms, Tim will travel 60 kms and Elan will travel 30 mtrs. he length and breadth of a field is 300x400ft,if there are 3 ants on average per square inch of field,find the approximate number of ants in field as 300 ft=3600 inch and 400 ft=4800 inch.then area in sqr inch=17280000 1 sqr inch need 3 ant hence 17280000 sqr inch need 3*17280000 ants. .diagonal of a square is double the side of an equilateral triangle find the ratio of area of triangle to area of square let sides of equilateral triangle be 'a', therefore area= √3/4 * a^2 now, diagonal of square is 2a therefore sides of square will be √2*a and area of square = 2 a^2 hence, ratio = (√3/4* a^2)/(2* a^2) answer = √3/8 5 Given a collection of points P in the plane, a 1set is a point in P that can be separated from the rest by a line, .i.e the point lies on one side of the line while the others lie on the other side. The number of 1sets of P is denoted by n1(P). The minimum value of n1(P) over all configurations P of 5 points in the plane in general position(.i.e no three points in P lie on a line) is a)3 b)5 c) 2 d)1 The meaning of this question is little complicated. In easy language the question should be: 6 There are some points in the plane. We have to make a separation between these points by forming a line in such a way that one point could be separate from the others. No 3 points are in a line. Maximum no of methods for plotting the line and minimum no of methods for plotting the line will be (arrangement of points is your choice) Now Answer Is: For Maximize: The number of methods, Points should be drawn in the circumference of the CIRCLE So answer would be the number of total point in the plane. For minimize: The number of methods we should draw 3 points in a TRIANGLE and all other point inside this triangle in such a way that no 3 points could be in a line. If it will be allowed to draw the points in a line then minimum possibilities will be 2 only, because if we take all points in a line then you can separate only the corner points from the others. For example If we have 10 points then maximum possibilities = 10 and minimum = 3 similarly for 5 points Max = 5, Min = 3 for 19 max= 19 Min = 3 f(x)=f(f(x))=f(x^2)....how many such functions with degree >2 exist does not exist.... In a sequence of integers, A(n)=A(n1)A(n2),where A(n) is the nth term in the sequence, n is an integer and n>=3, A(1)=1,A(2)=1. Calculate S(1000), where S(1000) is the sum of first 1000 terms. check it now A(1)=1,A(2)=1. A(3)=A(2)A(1) acc to the given formula A(n)=A(n1)A(n2) A(3)=11=0. similarliy,A(4)=A(3)A(2)=01=1. A(5)=A(4)A(3)=10=1. A(6)=A(5)A(4)=1(1)=0. after 6th term seq is repeating like...1,1,0,1,1,0...after every 6 terms so the sum of these 6 terms are 1+1+0+(1)+(1)+0=0. now we have to find sum of 1000 terms... sum of first 600 terms is 0...(as consider 1,1,0,1,1,0 as one sequence,now consider 166 seq i.e 166*6=996 terms) so the sum of first 996 terms is 0...after that 4 terms come i.e 1,1,0,1 so the sum of last 4 terms = 1+1+0+(1)=1. 7 planet four firesides in 4dimensional space and thus the currency used by it's residentds are 3dimensional objects. the rupee notes are cubical in shape while their conis are spherical. however the coni minting machinery lays out some stipulations on the size of the coins. the diameter of the conins should be at least 64 mm and not exceed 512 mm. given a conin the diameter of the next larger coin is at least 50% greater . the diameter of the coin must allways be an integer you are asked to disign a set of coins of different diameters with these requirements and your goal is to desing as many coins as possible. how many coins can you desing 1st coin=64cm 2nd "=64+32(50% of 64)=96cm similarly 3rd=96+48=144 4th=144+72=216 5th=216+108=324 6th=324+112=436 7th not possible ans is 6 8 There are 3 boys A, B, C and 2 Girls D, E. E always sit right to A. Girls never sit in extreme positions and in the middle position. C always sits in the extreme positions. Who is sitting immediate right to E? ns is either B or C POSSIBLE WAYS TO SIT 1) B/C D A E C/B 2) A E B D C 9 What is reminder when 6^17+17^6 is devided by 7? (71)^17 + (7*2+3)^6 binomial theorem 1+3^6=728 728%7=0 (40*40* 4031*31*31)/(40*40+40*31+31*31)=?a simle calcutation (a^3b^3)/(a^2+ab+b^2) ab 4031=9 10 In base 7,a number is written only using the digits 0,1,....6.the number 135 in base 7 is 1x7^2+3x7+5=75 in base 10.what is the sum of the same of the base 7 numbers 1234 and 6543 in base 7? 11110 1234 6543  11110 which is 7^4+7^3+7^2+7= 2401+343 +49+7= 2800 11 There are 20 balls which are red,blue or green,If 7 balls are green and the sum of red balls and green balls is less than 13,atmost how many red balls are there 5, atmost there can be 5 red balls so that sum of green and red can be less than 13 (7+5=12) 3y+x>2 and x+2y1 y the ans is y>1. explanation: x+2y1=0 so x+2y=1 now 3y+x>2 given we can write in this way x+2y+y>2 then putting x+2y=1 we get 1+y>2 =>y>21=>y>1 12 Find the no of zeros in the product of 1^1*2^2*3^3*.....*49^49?? total zeros at the end of product 5^5 x 10^10 x 15^15 x 20^20 x 25^25 x 30^30 x 35^35 x 40^40 x 45^45 will be 5+10+15+20+50+30+35+40+45 = 250 zeros 13 n! has 13 zeros than wat is the higest and lowest value of n?? Here we have to consider only the 5,10,15,20.... these terms.. so form (5,10,15,20,30,35,40,45,55 ) we got 1 no of zero from each. Now from 25(5*5) and 50(5*5*2) we got 2 no of zeroes form each. so upto 55 factorial we get (9+4)= 13 zeroes. so 55! is the smallest value and 59! is the largest value. The sequence {A(n)} is defined by A(1)=2 and A(n+1)=A(n)+2n. What is the value of A(100). The sequence {A(n)} is defined by A(1)=2 and A(n+1)=A(n)+2n. The value of A(100)=A(99+1)=A(99)+2*99 =A(98)+2*98+2*99 =A(97)+2*97+2*98+2*99 A(100)=A(1)+2*1+2*2+2*3+......+2*98+2*99 = 2+(2*1+2*2+2*3+......+2*98+2*99) = 2+2(1+2+3+.....+98+99) = 2+2(sum of the first 99 natural numbers) = 2+2(99*100/2) [sum of the first n natural numbers={n*(n+1)/2}] = 2+9900 = 9902 14 A,B COMMON TO ALL THREE. THEN ATLEAST 1 ELEMNT COMMON TO 2 IE X1,X2,X2,X3,X3,X1. FINDMINIMUM ELEMENTS IN EACH GROUP. ANSWER IS 4. 15 Directions for Q. 1 to Q. 5: Refer the data: J, K, L, M and N collected stamps. They collected a total of 100 stamps. None of them collected less than 10. No two among them collected the same number. (i) 3 collected the same number as K and M together. (ii) L collected 3 more than the cube of an integer (iii) The no. collected by J was the square of an integer. (iv) Total no. collected by K was either the square or cube of an integer. 1. The no. collected by J was: (1) 27 (2) 49 (3) 36 (4) 64 2. The no. collected by K was: (1) 16 (2) 27 (3) 25 (4) 36 3. The difference of numbers collected by L & M was: (1) 3 (2) 2 (3) 5 (4) 9 in the question the number collected by J should be cube of the integer and l should collect 3 more than square of integer. J+L+N=K+M so the sum of J+L+N=K+M=50 l=3^2+3=12 so J=3^3=27 n=502712=11 now k is either square or cube so possibilty are 36 or 16.and m=14 or 34 so j=27 now value for k can come after solving question 3 ml=1412=2 or 3412=22 so there is 2 in option so value of m is 14 and k=36. 16 Bhanu spends 30% of his income on petrol on scooter. ? of the remaining on house rent and the balance on food. If he spends Rs.300 on petrol then what is the expenditure on house rent? let the income x,so in petrol he spends 3x/10 rs,remaining x3x/10=7x/10 on house rent and balance on.now 3x/10=300,so x=1000,on house rent he spends rs 700balane.. Let exp(m,n) = m to the power n. If exp(10, m) = n exp(2, 2) where to and n are integers then n = exp(m,n) = m to the power n= m^n exp(10, m) = n exp(2, 2) = n*2^2 = 4*n 10^m =4*n when m=2, n=25 so min integer values of m and n are 2 and 25 respectively. A Grocer bought 24 kg coffee beans at price X per kg. After a while one third of stock got spoiled so he sold the rest for $200 per kg and made a total profit of twice the cost. What must be the price of X? A. $33 1/3 B. 66 2/3 C.44 4/9 D.50 1/3 ns: C. 44 4/9 cost price= 24*X $, spoiled= 24*1/3=8 kg, now total amount= (248)=16 kg selling price= 16*200$ =3200$ profit= (320024*X) profit=2* cost , (320024*X)= 2*24*X, x=44 4/9 17 The age of two people is in the ratio 6:8. the sum of their ages is 77. after 2 years the ratio of their ages becomes 5:7. wat is their present age? The ratio of two people be 6x and 8x sum of their ages is 6x+8y=77 after 2 yrs,ratio is 5:7,so, (6x+2)/(8x+2)=5/7 on cross multiplying this, x=2 hence age of a person 6(2)=12 hence another person ages is 8(2)=16 answer is 12,16 MOTHER +DAUGHTER+INFANT AGE IS 74. MOTHER AGE IS 46 MORE THEN DAUGHTER AND INFANT. AND INFANT AGE IS 0.4 OF DAUGHTER. FIND DAUGHTERS AGE. M+D+I = 74 ..,.,.Equ..1 I=0.4D M=D+0.4D+46 M=1.4D+46 put in equ 1 1.4D+46+D+0.4D=74 as I=0.4D 2.8D=7446 D=28/2.8 D=10.,.,. , 16 Mr and mrs smith had invited 9 of their friend and their spouses for party at wiki beachresort.the stand for group photograph if mr smith never stand next to mrs smith then how many way group arrange in row(A)20!(B)19!+18!(C)18*19!(D)2*19! simplest... ans is 18*19! method ..: mr and mrs smith never b together so out of 20(with both them) 19 can b any place so total = 19! also 1 of them can stand with other so 18 places /.. so total is 18*19!.. 19!*18bcoz along with mr and mrs smith can be arranged in 19!*2 (2 bcoz of mr and mrs smith arrangement) all together can be arranged in 20! ways so no mr and mrs smith can be arranged in 20!19!*2 =19!*18 Is this solution Helpfull? Yes (8)  No (2) viresh (4 Day ago) 20!(19!*2) total possibility minus they will be together A father purchases dress for his three daughter. The dresses are of same color but of different size .the dress is kept in dark room .What is the probability that all the three will not choose their own dress... a dosesnot choose his dress 2/3 b dosesnot choose his dress 1/2 c dosesnot choose his dress 1 total 1/3 17 Leena cuts small cubes of 3 cubic cm each.She joined it to make a cuboid of length 10 cm,width 3cm, and depth 3 cm.How many more cubes does she need to make a perfect cube? a)910 b)250 c)750 d)650 ns is a)910. To make cube we need 10*10*10=1000. now we have only 10*3*3=90.so the more cube we need are 100090=910. 18 Q: A boy wants to make cuboids of dimension 5m,6m & 7m from small cubes of .03m^3.Later he realized that he can make some cuboids by making it hollow.Then it takes some cubes less.What is the number of cubes to be removed a)2000 b)5000 c)3000 d)7000 Volume of the cuboid= 5*6*7 No of cubes needed = (5*6*7)/0.03 = 7000 Volume of hollow cube= (52)(62)(72) No of cubes needed = (52)(62)(72)/0.03 = (3*4*5)/0.03= 2000 So, number of cubes to be removed = 7000  2000 = 5000 19 An empty tankbe filled with aninlet pipe 'A' in 42 minutes. after 12 minutes an outlet pipe 'B' is opened which can empty the tank in 30 minutes. After 6 minutes another inlet pipe 'C' opened into the same tank, which can fill the tank in 35 minutes and the tank is filled find the time taken to fill the tank? he milk and water in two vessels A and B are in the ratio 4 : 3 and 2: 3 respectively. In what ratio, the liquids in both the vessels be mixed to obtain a new mixture in vessel C containing half milk and half water? let c.p. of 1liter milk is 1RS. now milk in 1st mixer is 4/7 and in 2nd mixure is 2/5. milk in finle mixer is 1/2 now from mixure formula.. 4/7 2/5 1/2 1/10 1/14 so ratio is 14/10 =7/5 20 How many kgs. of wheat costing Rs. 8 per kg must be mixed with 86 kg of rice costing Rs. 6.40 per kg so that 20% gain may be obtained by Belling the mixture at Rs. 7.20 per kg ? Ex. 4. .How many kgs. of wheat costing Rs. 8 per kg must be mixed with 86 kg of rice costing Rs. 6.40 per kg so that 20% gain may be obtained by Belling the mixture at Rs. 7.20 per kg ? Sol. S.P. of 1 kg mixture = Rs. 7.20,Gain = 20%. ? C.P. of 1 kg mixture = Rs.[(100/120)*7.20]=Rs. 6. By the rule of alligation, we have: C_P. of 1 kg wheat of 1st kind C.P. of 1 kg wheat of 2nd kind (800p) . (540 p) Mean price ( 600 p) 60 200 Wheat of 1st kind: Wheat of 2nd kind = 60 : 200 = 3 : 10. Let x kg of wheat of 1st kind be mixed with 36 kg of wheat of 2nd kind. Then, 3 : 10 = x : 36 or lOx = 3 * 36 or x = 10.8 kg. 21 In what ratio must water be mixed with milk to gain 20 % by selling the mixture at cost price? let the sp op 1 l mix is =re 1 cp =(100/120)*1=5/6 cp of one let. water =0 so ratio=1(5/6)/(5/6)0=1/5 so required ratio =1:5 How much water must be added to 60 litres of milk at 1 Â½ litres for Rs. 2 So as to have a mixture worth Rs.10 2/3 a litre ? x=(sin nx/n)=? x=(sin nx/n) so, nx=sin nx sin is one to one function and there is only one case for φ=0 where sinφ=φ so nx=0 so x=0 (n!=0) ans will be 0
__________________ Answered By StudyChaCha Member 
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Re: Solved Placement Papers for TCS
Here are Tata Consultancy Services placement solved questions: 1) If log 0.317=0.3332 and log 0.318=0.3364 then find log 0.319 ? Sol) log 0.317=0.3332 and log 0.318=0.3364, then log 0.319=log0.318+(log0.318log0.317) = 0.3396 2) A box of 150 packets consists of 1kg packets and 2kg packets. Total weight of box is 264kg. How many 2kg packets are there ? Sol) x= 2 kg Packs y= 1 kg packs x + y = 150 .......... Eqn 1 2x + y = 264 .......... Eqn 2 Solve the Simultaneous equation; x = 114 so, y = 36 ANS : Number of 2 kg Packs = 114. 3) My flight takes of at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N70W. What is the local time when my plane landed? 6:00 am b) 6:40am c) 7:40 d) 7:00 e) 8:00 Sol) The destination place is 80 degree west to the starting place. Hence the time difference between these two places is 5 hour 20 min. (=24hr*80/360). When the flight landed, the time at the starting place is 12 noon (2 AM + 10 hours). Hence, the time at the destination place is 12 noon  5:20 hours = 6: 40 AM 4) A plane moves from 9°N40°E to 9°N40°W. If the plane starts at 10 am and takes 8 hours to reach the destination, find the local arrival time ? Sol) Since it is moving from east to west longitide we need to add both ie,40+40=80 multiply the ans by 4 =>80*4=320min convert this min to hours ie, 5hrs 33min It takes 8hrs totally . So 85hr 30 min=2hr 30min So the ans is 10am+2hr 30 min =>ans is 12:30 it will reach 5) The size of the bucket is N kb. The bucket fills at the rate of 0.1 kb per millisecond. A programmer sends a program to receiver. There it waits for 10 milliseconds. And response will be back to programmer in 20 milliseconds. How much time the program takes to get a response back to the programmer, after it is sent? Please tell me the answer with explanation. Very urgent. Sol) see it doesn't matter that wat the time is being taken to fill the bucket.after reaching program it waits there for 10ms and back to the programmer in 20 ms.then total time to get the response is 20ms +10 ms=30ms...it's so simple 6) A file is transferred from one location to another in 'buckets'. The size of the bucket is 10 kilobytes. Each bucket gets filled at the rate of 0.0001 kilobytes per millisecond. The transmission time from sender to receiver is 10 milliseconds per bucket. After the receipt of the bucket the receiver sends an acknowledgement that reaches sender in 100 milliseconds. Assuming no error during transmission, write a formula to calculate the time taken in seconds to successfully complete the transfer of a file of size N kilobytes. (n/1000)*(n/10)*10+(n/100)....as i hv calculated...~~!not 100% sure 7) A fisherman's day is rated as good if he catches 9 fishes, fair if 7 fishes and bad if 5 fishes. He catches 53 fishes in a week n had all good, fair n bad days in the week. So how many good, fair n bad days did the fisher man had in the week Ans:4 good, 1 fair n 2 bad days Sol) Go to river catch fish 4*9=36 7*1=7 2*5=10 36+7+10=53... take what is given 53 good days means  9 fishes so 53/9=4(remainder=17) if u assume 5 then there is no chance for bad days. fair days means  7 fishes so remaining 17  17/7=1(remainder=10) if u assume 2 then there is no chance for bad days. bad days means 5 fishes so remaining 1010/5=2days. Ans: 4 good, 1 fair, 2bad. ==== total 7 days. x+y+z=7 eq1 9*x+7*y+5*z=53 eq2 multiply eq 1 by 9, 9*x+9*y+9*z=35 eq3 from eq2 and eq3 2*y+4*z=10eq4 since all x,y and z are integer i sud put a integer value of y such that z sud be integer in eq 4 .....and ther will be two value y=1 or 3 then z = 2 or 1 from eq 4 for first y=1,z=2 then from eq1 x= 4 so 9*4+1*7+2*5=53.... satisfied now for second y=3 z=1 then from eq1 x=3 so 9*3+3*7+1*5=53 ......satisfied so finally there are two solution of this question (x,y,z)=(4,1,2) and (3,3,1)... 8) Y catches 5 times more fishes than X. If total number of fishes caught by X and Y is 42, then number of fishes caught by X? Sol) Let no. of fish x catches=p no. caught by y =r r=5p. r+p=42 then p=7,r=35 9) Three companies are working independently and receiving the savings 20%, 30%, 40%. If the companies work combinely, what will be their net savings? suppose total income is 100 so amount x is getting is 80 y is 70 z =60 total=210 but total money is 300 300210=90 so they are getting 90 rs less 90 is 30% of 300 so they r getting 30% discount 10) The ratio of incomes of C and D is 3:4.the ratio of their expenditures is 4:5. Find the ratio of their savings if the savings of C is one fourths of his income? Sol) incomes:3:4 expenditures:4:5 3x4y=1/4(3x) 12x16y=3x 9x=16y y=9x/16 (3x4(9x/16))/((4x5(9x/16))) ans:12/19 11) If G(0) = 1 G(1)= 1 and G(N)=G(N1)  G(N2) then what is the value of G(6)? ans: 1 bcoz g(2)=g(1)g(0)=1+1=2 g(3)=1 g(4)=1 g(5)=2 g(6)=1 12) If A can copy 50 pages in 10 hours and A and B together can copy 70 pages in 10 hours, how much time does B takes to copy 26 pages? Sol) A can copy 50 pages in 10 hrs. A can copy 5 pages in 1hr.(50/10) now A & B can copy 70 pages in 10hrs. thus, B can copy 90 pages in 10 hrs.[eqn. is (50+x)/2=70, where x> no. of pages B can copy in 10 hrs.] so, B can copy 9 pages in 1hr. therefore, to copy 26 pages B will need almost 3hrs. since in 3hrs B can copy 27 pages. 13) what's the answer for that : A, B and C are 8 bit no's. They are as follows: A > 1 1 0 0 0 1 0 1 B > 0 0 1 1 0 0 1 1 C > 0 0 1 1 1 0 1 0 (  =minus, u=union) Find ((A  C) u B) =? To find AC, We will find 2's compliment of C and them add it with A, That will give us (AC) 2's compliment of C=1's compliment of C+1 =11000101+1=11000110 AC=11000101+11000110 =10001001 Now (AC) U B is .OR. logic operation on (AC) and B 10001001 .OR . 00110011 The answer is = 10111011, Whose decimal equivalent is 187. 14) One circular array is given(means memory allocation tales place in circular fashion) diamension(9X7) and sarting add. is 3000, What is the address of (2,3)........ Sol) it's a 9x7 int array so it reqiure a 126 bytes for storing.b'ze integer value need 2 byes of memory allocation. and starting add is 3000 so starting add of 2x3 will be 3012. 15) In a twodimensional array, X (9, 7), with each element occupying 4 bytes of memory, with the address of the first element X (1, 1) is 3000, find the address of X (8, 5). Sol) initial x (1,1) = 3000 u hav to find from x(8,1)so u have x(1,1),x(1,2) ... x(7,7) = so u have totally 7 * 7 = 49 elementsu need to find for x(8,5) ? here we have 5 elements each element have 4 bytes : (49 + 5 1) * 4 = 212 ( 1 is to deduct the 1 element ) 3000 + 212 = 3212 16) Which of the following is power of 3 a) 2345 b) 9875 c) 6504 d) 9833 17) The size of a program is N. And the memory occupied by the program is given by M = square root of 100N. If the size of the program is increased by 1% then how much memory now occupied ? Sol) M=sqrt(100N) N is increased by 1% therefore new value of N=N + (N/100) =101N/100 M=sqrt(100 * (101N/100) ) Hence, we get M=sqrt(101 * N) 18) 1)SCOOTER  AUTOMOBILE A. PART OF 2.OXYGEN WATER  B. A Type of 3.SHOP STAFF FITTERS C. NOT A TYPE OF 4. BUG REPTILE D. A SUPERSET OF 1)B 2)A 3)D 4)C 19) A bus started from bustand at 8.00a m and after 30 min staying at destination, it returned back to the bustand. the destination is 27 miles from the bustand. the speed of the bus 50 percent fast speed. at what time it returns to the bustand this is the step by step solution: a bus cover 27 mile with 18 mph in =27/18= 1 hour 30 min. and it wait at stand =30 min. after this speed of return increase by 50% so 50%of 18 mph=9mph Total speed of returnig=18+9=27 Then in return it take 27/27=1 hour then total time in joureny=1+1:30+00:30 =3 hour so it will come at 8+3 hour=11 a.m. So Ans==11 a.m 20) In two dimensional array X(7,9) each element occupies 2 bytes of memory.If the address of first element X(1,1)is 1258 then what will be the address of the element X(5,8) ? Sol) Here, the address of first element x[1][1] is 1258 and also 2 byte of memory is given. now, we have to solve the address of element x[5][8], therefore, 1258+ 5*8*2 = 1258+80 = 1338 so the answer is 1338. 21) The temperature at Mumbai is given by the function: t2/6+4t+12 where t is the elapsed time since midnight. What is the percentage rise (or fall) in temperature between 5.00PM and 8.00PM? 22) Low temperature at the night in a city is 1/3 more than 1/2 high as higher temperature in a day. Sum of the low temperature and highest temp. is 100 degrees. Then what is the low temp? Sol) Let highest temp be x so low temp=1/3 of x of 1/2 of x plus x/2 i.e. x/6+x/2 total temp=x+x/6+x/2=100 therefore, x=60 Lowest temp is 40 23) In Madras, temperature at noon varies according to t^2/2 + 8t + 3, where t is elapsed time. Find how much temperature more or less in 4pm to 9pm. Ans. At 9pm 7.5 more Sol) In equestion first put t=9, we will get 34.5...........................(1) now put t=4, we will get 27..............................(2) so ans=34.527 =7.5 24) A person had to multiply two numbers. Instead of multiplying by 35, he multiplied by 53 and the product went up by 540. What was the raised product? a) 780 b) 1040 c) 1590 d) 1720 Sol) x*53x*35=540=> x=30 therefore, 53*30=1590 Ans 25) How many positive integer solutions does the equation 2x+3y = 100 have? a) 50 b) 33 c) 16 d) 35 Sol) There is a simple way to answer this kind of Q's given 2x+3y=100, take l.c.m of 'x' coeff and 'y' coeff i.e. l.c.m of 2,3 ==6then divide 100 with 6 , which turns out 16 hence answer is 16short cut formula constant / (l.cm of x coeff and y coeff) 26) The total expense of a boarding house are partly fixed and partly variable with the number of boarders. The charge is Rs.70 per head when there are 25 boarders and Rs.60 when there are 50 boarders. Find the charge per head when there are 100 boarders. a) 65 b) 55 c) 50 d) 45 Sol) Let a = fixed cost and k = variable cost and n = number of boarders total cost when 25 boarders c = 25*70 = 1750 i.e. 1750 = a + 25k total cost when 50 boarders c = 50*60 = 3000 i.e. 3000 = a + 50k solving above 2 eqns, 30001750 = 25k i.e. 1250 = 25k i.e. k = 50 therefore, substituting this value of k in either of above 2 eqns we get a = 500 (a = 300050*50 = 500 or a = 1750  25*50 = 500) so total cost when 100 boarders = c = a + 100k = 500 + 100*50 = 5500 so cost per head = 5500/100 = 55 27) Amal bought 5 pens, 7 pencils and 4 erasers. Rajan bought 6 pens, 8 erasers and 14 pencils for an amount which was half more than what Amal had paid. What % of the total amount paid by Amal was paid for pens? a) 37.5% b) 62.5% c) 50% d) None of these Sol) Let, 5 pens + 7 pencils + 4 erasers = x rupees so 10 pens + 14 pencils + 8 erasers = 2*x rupees also mentioned, 6 pens + 14 pencils + 8 erarsers = 1.5*x rupees so (106) = 4 pens = (21.5)x rupees so 4 pens = 0.5x rupees => 8 pens = x rupees so 5 pens = 5x/8 rupees = 5/8 of total (note x rupees is total amt paid byamal) i.e 5/8 = 500/8% = 62.5% is the answer 28) I lost Rs.68 in two races. My second race loss is Rs.6 more than the first race. My friend lost Rs.4 more than me in the second race. What is the amount lost by my friend in the second race? Sol) x + x+6 = rs 68 2x + 6 = 68 2x = 686 2x = 62 x=31 x is the amt lost in I race x+ 6 = 31+6=37 is lost in second race then my friend lost 37 + 4 = 41 Rs 29) Ten boxes are there. Each ball weighs 100 gms. One ball is weighing 90 gms. i) If there are 3 balls (n=3) in each box, how many times will it take to find 90 gms ball? ii) Same question with n=10 iii) Same question with n=9 to me the chances are when n=3 (i) nC1= 3C1 =3 for 10 boxes .. 10*3=30 (ii) 10C1=10 for 10 boxes ....10*10=100 (iii)9C1=9 for 10 boxes .....10*9=90 30) (11/6) (11/7).... (1 (1/ (n+4))) (1(1/ (n+5))) = ? leaving the first numerater and last denominater, all the numerater and denominater will cancelled out one another. Ans. 5/(n+5) 31) A face of the clock is divided into three parts. First part hours total is equal to the sum of the second and third part. What is the total of hours in the bigger part? Sol) the clock normally has 12 hr three parts x,y,z x+y+z=12 x=y+z 2x=12 x=6 so the largest part is 6 hrs 32) With 4/5 full tank vehicle travels 12 miles, with 1/3 full tank how much distance travels Sol) 4/5 full tank= 12 mile 1 full tank= 12/(4/5) 1/3 full tank= 12/(4/5)*(1/3)= 5 miles 33) wind blows 160 miles in 330min.for 80 miles how much time required Sol) 160 miles= 330 min 1 mile = 330/160 80 miles=(330*80)/160=165 min. 34) A person was fined for exceeding the speed limit by 10mph.another person was also fined for exceeding the same speed limit by twice the same if the second person was travelling at a speed of 35 mph. find the speed limit Sol) (x+10)=(x+35)/2 solving the eqn we get x=15 35) A sales person multiplied a number and get the answer is 3 instead of that number divided by 3. what is the answer he actually has to get. Sol) Assume 1 1* 3 = 3 1*1/3=1/3 so he has to got 1/3 this is the exact answer 36) A person who decided to go weekend trip should not exceed 8 hours driving in a day average speed of forward journey is 40 mph due to traffic in Sundays the return journey average speed is 30 mph. How far he can select a picnic spot. 37) Low temperature at the night in a city is 1/3 more than 1/2 hinge as higher temperature in a day. Sum of the low temp and high temp is 100 c. then what is the low temp. ans is 40 c. Sol) let x be the highest temp. then, x+x/2+x/6=100. therefore, x=60 which is the highest temp and 100x=40 which is the lowest temp. 38) car is filled with four and half gallons of oil for full round trip. Fuel is taken 1/4 gallons more in going than coming. What is the fuel consumed in coming up. Sol) let feul consumed in coming up is x. thus equation is: x+1.25x=4.5ans:2gallons 39) A work is done by the people in 24 min. One of them can do this work alone in 40 min. How much time required to do the same work for the second person Sol) Two people work together in 24 mins. So, their one day work is (1/A)+(1+B)=(1/24) One man can complete the work in 40mins one man's one day work (1/B)= (1/40) Now, (1/A)=(1/24)(1/40) (1/A)=(1/60) So, A can complete the work in 60 mins. 40) In a company 30% are supervisors and 40% employees are male if 60% of supervisors are male. What is the probability? That a randomly chosen employee is a male or female? Sol) 40% employees are male if 60% of supervisors are male so for 100% is 26.4%so the probability is 0.264 41) In 80 coins one coin is counterfeit what is minimum number of weighing to find out counterfeit coin Sol) the minimum number of wieghtings needed is just 5.as shown below (1) 80>3030 (2) 1515 (3) 77 (4) 33 (5) 11 42) 2 oranges, 3 bananas and 4 apples cost Rs.15. 3 oranges, 2 bananas, and 1 apple costs Rs 10. What is the cost of 3 oranges, 3 bananas and 3 apples? 2x+3y+4z=15 3x+2y+z=10 adding 5x+5y+5z=25 x+y+z=5 that is for 1 orange, 1 bannana and 1 apple requires 5Rs. so for 3 orange, 3 bannana and 3 apple requires 15Rs. i.e. 3x+3y+3z=15 43) In 8*8 chess board what is the total number of squares refers Sol) odele discovered that there are 204 squares on the board We found that you would add the different squares  1 + 4 + 9 + 16+ 25 + 36 + 49 + 64. Also in 3*3 tic tac toe board what is the total no of squares Ans 14 ie 9+4(bigger ones)+1 (biggest one) If you ger 100*100 board just use the formula the formula for the sum of the first n perfect squares is n x (n + 1) x (2n + 1) ______________________ 6 if in this formula if you put n=8 you get your answer 204 44) One fast typist type some matter in 2hr and another slow typist type the same matter in 3hr. If both do combinely in how much time they will finish. Sol) Faster one can do 1/2 of work in one hourslower one can do 1/3 of work in one hourboth they do (1/2+1/3=5/6) th work in one hour.so work will b finished in 6/5=1.2 hour i e 1 hour 12 min. 45) If Rs20/ is available to pay for typing a research report & typist A produces 42 pages and typist B produces 28 pages. How much should typist A receive? Here is the answer Find of 42 % of 20 rs with respect to 70 (i.e 28 + 42) ==> (42 * 20 )/70 ==> 12 Rs 46) An officer kept files on his table at various times in the order 1,2,3,4,5,6. Typist can take file from top whenever she has time and type it.What order she cannt type.? 47) In some game 139 members have participated every time one fellow will get bye what is the number of matches to choose the champion to be held? the answer is 138 matches Sol) since one player gets a bye in each round,he will reach the finals of the tournament without playing a match. therefore 137 matches should be played to detemine the second finalist from the remaining 138 players(excluding the 1st player) therefore to determine the winner 138 matches shd be played. 48) One rectangular plate with length 8inches, breadth 11 inches and 2 inches thickness is there. What is the length of the circular rod with diameter 8 inches and equal to volume of rectangular plate? Sol) Vol. of rect. plate= 8*11*2=176 area of rod=(22/7)*(8/2)*(8/2)=(352/7) vol. of rod=area*length=vol. of plate so length of rod= vol of plate/area=176/(352/7)=3.5 49) One tank will fill in 6 minutes at the rate of 3cu ft /min, length of tank is 4 ft and the width is 1/2 of length, what is the depth of the tank? 3 ft 7.5 inches 50) A man has to get airmail. He starts to go to airport on his motorbike. Plane comes early and the mail is sent by a horsecart. The man meets the cart in the middle after half an hour. He takes the mail and returns back, by doing so, he saves twenty minutes. How early did the plane arrive? ans:10min:::assume he started at 1:00,so at 1:30 he met cart. He returned home at 2:00.so it took him 1 hour for the total jorney.by doing this he saved 20 min.so the actual time if the plane is not late is 1 hour and 20 min.so the actual time of plane is at 1:40.The cart travelled a time of 10 min before it met him.so the plane is 10 min early. 51) Ram singh goes to his office in the city every day from his suburban house. His driver Mangaram drops him at the railway station in the morning and picks him up in the evening. Every evening Ram singh reaches the station at 5 o'clock. Mangaram also reaches at the same time. One day Ram singh started early from his office and came to the station at 4 o'clock. Not wanting to wait for the car he starts walking home. Mangaram starts at normal time, picks him up on the way and takes him back house, half an hour early. How much time did Ram singh walked? 52) 2 trees are there. One grows at 3/5 of the other. In 4 years total growth of the trees is 8 ft. what growth will smaller tree have in 2 years. Sol) THE BIG TREE GROWS 8FT IN 4 YEARS=>THE BIG TREE GROWS 4FT IN 2 YEARS.WHEN WE DIVIDE 4FT/5=.8*3=>2.4 ans: 1.5 mt 4 (x+(3/5)x)=88x/5=2x=5/4 after 2 years x=(3/5)*(5/4)*2 =1.5 53) There is a six digit code. Its first two digits, multiplied by 3 gives all ones. And the next two digits multiplied by 6 give all twos. Remaining two digits multiplied by 9 gives all threes. Then what is the code? sol) Assume the digit xx xx xx (six digits) First Two digit xx * 3=111 xx=111/3=37 ( first two digits of 1 is not divisible by 3 so we can use 111) Second Two digit xx*6=222 xx=222/6=37 ( first two digits of 2 is not divisible by 6 so we can use 222) Thrid Two digit xx*9=333 xx=333/9=37 ( first two digits of 3 is not divisible by 9 so we can use 333) 54) There are 4 balls and 4 boxes of colours yellow, pink, red and green. Red ball is in a box whose colour is same as that of the ball in a yellow box. Red box has green ball. In which box you find the yellow ball? ans is green... Sol) Yellow box can have either of pink/yellow balls. if we put a yellow ball in "yellow" box then it wud imply that "yellow" is also the colour of the box which has the red ball(becoz acordin 2 d question,d box of the red ball n the ball in the yellow box have same colour) thus this possibility is ruled out... therefore the ball in yellow box must be pink,hence the colour of box containin red ball is also pink.... =>the box colour left out is "green",,,which is alloted to the only box left,,,the one which has yellow ball.. 55) A bag contains 20 yellow balls, 10 green balls, 5 white balls, 8 black balls, and 1 red ball. How many minimum balls one should pick out so that to make sure the he gets at least 2 balls of same color. Ans:he should pick 6 ball totally. Sol) Suppose he picks 5 balls of all different colours then when he picks up the sixth one, it must match any on of the previously drawn ball colour. thus he must pick 6 balls 56) What is the number of zeros at the end of the product of the numbers from 1 to 100 Sol) For every 5 in unit palce one zero is added so between 1 to 100 there are 10 nos like 5,15,25,..,95 which has 5 in unit place. Similarly for every no divisible by 10 one zero is added in the answer so between 1 to 100 11 zeros are added for 25,50,75 3 extra zeros are added so total no of zeros are 10+11+3=24 57) 10 Digit number has its first digit equals to the numbers of 1's, second digit equals to the numbers of 2's, 3rd digit equals to the numbers of 3's .4th equals number of 4's..till 9th digit equals to the numbers of 9's and 10th digit equals to the number of 0's. what is the number?.(6marks) ans:2100010006 2shows that two 1's in the ans 1shows that one 2 in ans 0shows no 3 in the ans 0shows no 4 in the ans 0shows no 5 in the ans 1shows one 6 in the ans 0shows no 7 in the ans 0shows no 8 in the ans 0shows no 9 in the ans 6shows six 0's in the ans 58) There are two numbers in the ratio 8:9. if the smaller of the two numbers is increased by 12 and the larger number is reduced by 19 thee the ratio of the two numbers is 5:9. Find the larger number? sol) 8x:9x initialy 8x+ 12 : 9x  19 = 5x:9x 8x+12 = 5x > x = 4 9x = 36 not sure about the answer .. 59) There are three different boxes A, B and C. Difference between weights of A and B is 3 kgs. And between B and C is 5 kgs. Then what is the maximum sum of the differences of all possible combinations when two boxes are taken each time AB = 3 Bc = 5 ac = 8 so sum of diff = 8+3+5 = 16 kgs 60) A and B are shooters and having their exam. A and B fall short of 10 and 2 shots respectively to the qualifying mark. If each of them fired atleast one shot and even by adding their total score together, they fall short of the qualifying mark, what is the qualifying mark? ans is 11 coz each had atleast 1 shot done so 10 + 1 = 11 n 9 + 2 = 11 so d ans is 11 61) A, B, C, and D tells the following times by looking at their watches. A tells it is 3 to 12. B tells it is 3 past 12. C tells it is 12:2. D tells it is half a dozen too soon to 12. No two watches show the same time. The difference between the watches is 2,3,4,5 respectively. Whose watch shows maximum time? sol) A shows 11:57, B shows 12:03, C shows 12:02, and D shows 11:06 therefore, max time is for B 62) Falling height is proportional to square of the time. One object falls 64cm in 2sec than in 6sec from how much height the object will fall. Sol) The falling height is proportional to the squere of the time. Now, the falling height is 64cm at 2sec so, the proportional constant is=64/(2*2)=16; so, at 6sec the object fall maximum (16*6*6)cm=576cm; Now, the object may be situated at any where. if it is>576 only that time the object falling 576cm within 6sec .Otherwise if it is situated<576 then it fall only that height at 6sec. 63) Gavaskar average in first 50 innings was 50. After the 51st innings his average was 51 how many runs he made in the 51st innings Ans) first 50 ings. run= 50*50=2500 51st ings. avg 51. so total run =51*51=2601. so run scored in that ings=26012500=101 runs. 64) Anand finishes a work in 7 days, Bittu finishes the same job in 8 days and Chandu in 6 days. They take turns to finish the work. Anand on the first day, Bittu on the second and Chandu on the third day and then Anand again and so on. On which day will the work get over? a) 3rd b) 6th c) 9th d) 7th Ans is d) 7th day Sol) In d 1st day Anand does 1/7th of total work similarly, Bithu does 1/8th work in d 2nd day hence at d end of 3 days, work done = 1/7+1/8+1/6=73/168 remaining work = (16873)/168 = 95/168 again after 6 days of work, remaining work is = (9573)/168 = 22/168 and hence Anand completes the work on 7th day.(hope u understood.) 65) A man, a women and a child can do a piece of work in 6 days,man can do it in 14 days, women can do it 16 days, and in how many days child can do the same work? The child does it in 24 days 66) A: 1 1 0 1 1 0 1 1 B: 0 1 1 1 1 0 1 0 C: 0 1 1 0 1 1 0 1 Find ( (AB) u C )==? Hint : 109 AB is {A}  {A n B} A: 1 1 0 1 1 0 1 1 B: 0 1 1 1 1 0 1 0 by binary sub. ab = 01100001 (10=1, 11=0,00=0, n for the 1st 3 digits 110011=011) now (ab)uc= 01100001 or 01101101 gives 1101101... convert to decimal equals 109 1) In a college ,1/5 th of the girls and 1/8 th of the boys took part in a social camp.What of the total number of students in the college took part in the camp? a)2/11 b)2/10 c)2/13 d)3/13 Answer is 2/13 2) A zookeeper counted the heads of the animals in a zoo and found it to be 80. When he counted the legs of the animals he found it to be 260. If the zoo had either pigeons or horses, how many horses were there in the zoo? a)36 b)65 c)45 d)50 Answer is 50 3) A three digit number consists of 9,5 and one more number . When these digits are reversed and then subtracted from the original number the answer yielded will be consisting of the same digits arranged yet in a different order. What is the other digit? a)7 b)4 c)3 d)None of these Answer is 4 4) A man sells an article at a profit of 25%. If he had bought it at 20 % less and sold it for Rs.10.50 less, he would have gained 30%. Find the cost price of the article? a)Rs.50 b)Rs.49.50 c)Rs.50.50 d)Rs.51 Answer is Rs.50.
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Re: TCS Solved Placement papers with
Tata Consultancy Services (TCS) placement papers pattern help the fresher as well as experienced candidates who are waiting to get place in TCS. Sample Questions: 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) If log 0.317=0.3332 and log 0.318=0.3364 then find log 0.319 ? Sol) log 0.317=0.3332 and log 0.318=0.3364, then log 0.319=log0.318+(log0.318log0.317) = 0.3396 2) A box of 150 packets consists of 1kg packets and 2kg packets. Total weight of box is 264kg. How many 2kg packets are there ? Sol) x= 2 kg Packs y= 1 kg packs x + y = 150 .......... Eqn 1 2x + y = 264 .......... Eqn 2 Solve the Simultaneous equation; x = 114 so, y = 36 ANS : Number of 2 kg Packs = 114. 3) My flight takes of at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N70W. What is the local time when my plane landed? 6:00 am b) 6:40am c) 7:40 d) 7:00 e) 8:00 Sol) The destination place is 80 degree west to the starting place. Hence the time difference between these two places is 5 hour 20 min. (=24hr*80/360). When the flight landed, the time at the starting place is 12 noon (2 AM + 10 hours). Hence, the time at the destination place is 12 noon  5:20 hours = 6: 40 AM 4) A plane moves from 9°N40°E to 9°N40°W. If the plane starts at 10 am and takes 8 hours to reach the destination, find the local arrival time ? Sol) Since it is moving from east to west longitide we need to add both ie,40+40=80 multiply the ans by 4 =>80*4=320min convert this min to hours ie, 5hrs 33min It takes 8hrs totally . So 85hr 30 min=2hr 30min So the ans is 10am+2hr 30 min =>ans is 12:30 it will reach 5) The size of the bucket is N kb. The bucket fills at the rate of 0.1 kb per millisecond. A programmer sends a program to receiver. There it waits for 10 milliseconds. And response will be back to programmer in 20 milliseconds. How much time the program takes to get a response back to the programmer, after it is sent? Please tell me the answer with explanation. Very urgent. Sol) see it doesn't matter that wat the time is being taken to fill the bucket.after reaching program it waits there for 10ms and back to the programmer in 20 ms.then total time to get the response is 20ms +10 ms=30ms...it's so simple.... 6) A file is transferred from one location to another in 'buckets'. The size of the bucket is 10 kilobytes. Each bucket gets filled at the rate of 0.0001 kilobytes per millisecond. The transmission time from sender to receiver is 10 milliseconds per bucket. After the receipt of the bucket the receiver sends an acknowledgement that reaches sender in 100 milliseconds. Assuming no error during transmission, write a formula to calculate the time taken in seconds to successfully complete the transfer of a file of size N kilobytes. (n/1000)*(n/10)*10+(n/100)....as i hv calculated...~~!not 100% sure 7) A fisherman's day is rated as good if he catches 9 fishes, fair if 7 fishes and bad if 5 fishes. He catches 53 fishes in a week n had all good, fair n bad days in the week. So how many good, fair n bad days did the fisher man had in the week Ans:4 good, 1 fair n 2 bad days Sol) Go to river catch fish 4*9=36 7*1=7 2*5=10 36+7+10=53... take what is given 53 good days means  9 fishes so 53/9=4(remainder=17) if u assume 5 then there is no chance for bad days. fair days means  7 fishes so remaining 17  17/7=1(remainder=10) if u assume 2 then there is no chance for bad days. bad days means 5 fishes so remaining 1010/5=2days. Ans: 4 good, 1 fair, 2bad. ==== total 7 days. x+y+z=7 eq1 9*x+7*y+5*z=53 eq2 multiply eq 1 by 9, 9*x+9*y+9*z=35 eq3 from eq2 and eq3 2*y+4*z=10eq4 since all x,y and z are integer i sud put a integer value of y such that z sud be integer in eq 4 .....and ther will be two value y=1 or 3 then z = 2 or 1 from eq 4 for first y=1,z=2 then from eq1 x= 4 so 9*4+1*7+2*5=53.... satisfied now for second y=3 z=1 then from eq1 x=3 so 9*3+3*7+1*5=53 ......satisfied so finally there are two solution of this question (x,y,z)=(4,1,2) and (3,3,1)... 8) Y catches 5 times more fishes than X. If total number of fishes caught by X and Y is 42, then number of fishes caught by X? Sol) Let no. of fish x catches=p no. caught by y =r r=5p. r+p=42 then p=7,r=35 9) Three companies are working independently and receiving the savings 20%, 30%, 40%. If the companies work combinely, what will be their net savings? suppose total income is 100 so amount x is getting is 80 y is 70 z =60 total=210 but total money is 300 300210=90 so they are getting 90 rs less 90 is 30% of 300 so they r getting 30% discount 10) The ratio of incomes of C and D is 3:4.the ratio of their expenditures is 4:5. Find the ratio of their savings if the savings of C is one fourths of his income? Sol) incomes:3:4 expenditures:4:5 3x4y=1/4(3x) 12x16y=3x 9x=16y y=9x/16 (3x4(9x/16))/((4x5(9x/16))) ans:12/19 11) If G(0) = 1 G(1)= 1 and G(N)=G(N1)  G(N2) then what is the value of G(6)? ans: 1 bcoz g(2)=g(1)g(0)=1+1=2 g(3)=1 g(4)=1 g(5)=2 g(6)=1 12) If A can copy 50 pages in 10 hours and A and B together can copy 70 pages in 10 hours, how much time does B takes to copy 26 pages? Sol) A can copy 50 pages in 10 hrs. A can copy 5 pages in 1hr.(50/10) now A & B can copy 70 pages in 10hrs. thus, B can copy 90 pages in 10 hrs.[eqn. is (50+x)/2=70, where x> no. of pages B can copy in 10 hrs.] so, B can copy 9 pages in 1hr. therefore, to copy 26 pages B will need almost 3hrs. since in 3hrs B can copy 27 pages. 13) what's the answer for that : A, B and C are 8 bit no's. They are as follows: A > 1 1 0 0 0 1 0 1 B > 0 0 1 1 0 0 1 1 C > 0 0 1 1 1 0 1 0 (  =minus, u=union) Find ((A  C) u B) =? To find AC, We will find 2's compliment of C and them add it with A, That will give us (AC) 2's compliment of C=1's compliment of C+1 =11000101+1=11000110 AC=11000101+11000110 =10001001 Now (AC) U B is .OR. logic operation on (AC) and B 10001001 .OR . 00110011 The answer is = 10111011, Whose decimal equivalent is 187. 14) One circular array is given(means memory allocation tales place in circular fashion) diamension(9X7) and sarting add. is 3000, What is the address of (2,3)........ Sol) it's a 9x7 int array so it reqiure a 126 bytes for storing.b'ze integer value need 2 byes of memory allocation. and starting add is 3000 so starting add of 2x3 will be 3012. 15) In a twodimensional array, X (9, 7), with each element occupying 4 bytes of memory, with the address of the first element X (1, 1) is 3000, find the address of X (8, 5). Sol) initial x (1,1) = 3000 u hav to find from x(8,1)so u have x(1,1),x(1,2) ... x(7,7) = so u have totally 7 * 7 = 49 elementsu need to find for x(8,5) ? here we have 5 elements each element have 4 bytes : (49 + 5 1) * 4 = 212 ( 1 is to deduct the 1 element ) 3000 + 212 = 3212 16) Which of the following is power of 3 a) 2345 b) 9875 c) 6504 d) 9833 17) The size of a program is N. And the memory occupied by the program is given by M = square root of 100N. If the size of the program is increased by 1% then how much memory now occupied ? Sol) M=sqrt(100N) N is increased by 1% therefore new value of N=N + (N/100) =101N/100 M=sqrt(100 * (101N/100) ) Hence, we get M=sqrt(101 * N) 18) 1)SCOOTER  AUTOMOBILE A. PART OF 2.OXYGEN WATER  B. A Type of 3.SHOP STAFF FITTERS C. NOT A TYPE OF 4. BUG REPTILE D. A SUPERSET OF 1)B 2)A 3)D 4)C 19) A bus started from bustand at 8.00a m and after 30 min staying at destination, it returned back to the bustand. the destination is 27 miles from the bustand. the speed of the bus 50 percent fast speed. at what time it returns to the bustand this is the step by step solution: a bus cover 27 mile with 18 mph in =27/18= 1 hour 30 min. and it wait at stand =30 min. after this speed of return increase by 50% so 50%of 18 mph=9mph Total speed of returnig=18+9=27 Then in return it take 27/27=1 hour then total time in joureny=1+1:30+00:30 =3 hour so it will come at 8+3 hour=11 a.m. So Ans==11 a.m 20) In two dimensional array X(7,9) each element occupies 2 bytes of memory.If the address of first element X(1,1)is 1258 then what will be the address of the element X(5,8) ? Sol) Here, the address of first element x[1][1] is 1258 and also 2 byte of memory is given. now, we have to solve the address of element x[5][8], therefore, 1258+ 5*8*2 = 1258+80 = 1338 so the answer is 1338. 21) The temperature at Mumbai is given by the function: t2/6+4t+12 where t is the elapsed time since midnight. What is the percentage rise (or fall) in temperature between 5.00PM and 8.00PM? 22) Low temperature at the night in a city is 1/3 more than 1/2 high as higher temperature in a day. Sum of the low temperature and highest temp. is 100 degrees. Then what is the low temp? Sol) Let highest temp be x so low temp=1/3 of x of 1/2 of x plus x/2 i.e. x/6+x/2 total temp=x+x/6+x/2=100 therefore, x=60 Lowest temp is 40 23) In Madras, temperature at noon varies according to t^2/2 + 8t + 3, where t is elapsed time. Find how much temperature more or less in 4pm to 9pm. Ans. At 9pm 7.5 more Sol) In equestion first put t=9, we will get 34.5...........................(1) now put t=4, we will get 27..............................(2) so ans=34.527 =7.5 24) A person had to multiply two numbers. Instead of multiplying by 35, he multiplied by 53 and the product went up by 540. What was the raised product? a) 780 b) 1040 c) 1590 d) 1720 Sol) x*53x*35=540=> x=30 therefore, 53*30=1590 Ans 25) How many positive integer solutions does the equation 2x+3y = 100 have? a) 50 b) 33 c) 16 d) 35 Sol) There is a simple way to answer this kind of Q's given 2x+3y=100, take l.c.m of 'x' coeff and 'y' coeff i.e. l.c.m of 2,3 ==6then divide 100 with 6 , which turns out 16 hence answer is 16short cut formula constant / (l.cm of x coeff and y coeff) 26) The total expense of a boarding house are partly fixed and partly variable with the number of boarders. The charge is Rs.70 per head when there are 25 boarders and Rs.60 when there are 50 boarders. Find the charge per head when there are 100 boarders. a) 65 b) 55 c) 50 d) 45 Sol) Let a = fixed cost and k = variable cost and n = number of boarders total cost when 25 boarders c = 25*70 = 1750 i.e. 1750 = a + 25k total cost when 50 boarders c = 50*60 = 3000 i.e. 3000 = a + 50k solving above 2 eqns, 30001750 = 25k i.e. 1250 = 25k i.e. k = 50 therefore, substituting this value of k in either of above 2 eqns we get a = 500 (a = 300050*50 = 500 or a = 1750  25*50 = 500) so total cost when 100 boarders = c = a + 100k = 500 + 100*50 = 5500 so cost per head = 5500/100 = 55 27) Amal bought 5 pens, 7 pencils and 4 erasers. Rajan bought 6 pens, 8 erasers and 14 pencils for an amount which was half more than what Amal had paid. What % of the total amount paid by Amal was paid for pens? a) 37.5% b) 62.5% c) 50% d) None of these Sol) Let, 5 pens + 7 pencils + 4 erasers = x rupees so 10 pens + 14 pencils + 8 erasers = 2*x rupees also mentioned, 6 pens + 14 pencils + 8 erarsers = 1.5*x rupees so (106) = 4 pens = (21.5)x rupees so 4 pens = 0.5x rupees => 8 pens = x rupees so 5 pens = 5x/8 rupees = 5/8 of total (note x rupees is total amt paid byamal) i.e 5/8 = 500/8% = 62.5% is the answer 28) I lost Rs.68 in two races. My second race loss is Rs.6 more than the first race. My friend lost Rs.4 more than me in the second race. What is the amount lost by my friend in the second race? Sol) x + x+6 = rs 68 2x + 6 = 68 2x = 686 2x = 62 x=31 x is the amt lost in I race x+ 6 = 31+6=37 is lost in second race then my friend lost 37 + 4 = 41 Rs 29) Ten boxes are there. Each ball weighs 100 gms. One ball is weighing 90 gms. i) If there are 3 balls (n=3) in each box, how many times will it take to find 90 gms ball? ii) Same question with n=10 iii) Same question with n=9 to me the chances are when n=3 (i) nC1= 3C1 =3 for 10 boxes .. 10*3=30 (ii) 10C1=10 for 10 boxes ....10*10=100 (iii)9C1=9 for 10 boxes .....10*9=90 30) (11/6) (11/7).... (1 (1/ (n+4))) (1(1/ (n+5))) = ? leaving the first numerater and last denominater, all the numerater and denominater will cancelled out one another. Ans. 5/(n+5) 31) A face of the clock is divided into three parts. First part hours total is equal to the sum of the second and third part. What is the total of hours in the bigger part? Sol) the clock normally has 12 hr three parts x,y,z x+y+z=12 x=y+z 2x=12 x=6 so the largest part is 6 hrs 32) With 4/5 full tank vehicle travels 12 miles, with 1/3 full tank how much distance travels Sol) 4/5 full tank= 12 mile 1 full tank= 12/(4/5) 1/3 full tank= 12/(4/5)*(1/3)= 5 miles 33) wind blows 160 miles in 330min.for 80 miles how much time required Sol) 160 miles= 330 min 1 mile = 330/160 80 miles=(330*80)/160=165 min. 34) A person was fined for exceeding the speed limit by 10mph.another person was also fined for exceeding the same speed limit by twice the same if the second person was travelling at a speed of 35 mph. find the speed limit Sol) (x+10)=(x+35)/2 solving the eqn we get x=15 35) A sales person multiplied a number and get the answer is 3 instead of that number divided by 3. what is the answer he actually has to get. Sol) Assume 1 1* 3 = 3 1*1/3=1/3 so he has to got 1/3 this is the exact answer 36) A person who decided to go weekend trip should not exceed 8 hours driving in a day average speed of forward journey is 40 mph due to traffic in Sundays the return journey average speed is 30 mph. How far he can select a picnic spot. 37) Low temperature at the night in a city is 1/3 more than 1/2 hinge as higher temperature in a day. Sum of the low temp and high temp is 100 c. then what is the low temp. ans is 40 c. Sol) let x be the highest temp. then, x+x/2+x/6=100. therefore, x=60 which is the highest temp and 100x=40 which is the lowest temp. 38) car is filled with four and half gallons of oil for full round trip. Fuel is taken 1/4 gallons more in going than coming. What is the fuel consumed in coming up. Sol) let feul consumed in coming up is x. thus equation is: x+1.25x=4.5ans:2gallons 39) A work is done by the people in 24 min. One of them can do this work alone in 40 min. How much time required to do the same work for the second person Sol) Two people work together in 24 mins. So, their one day work is (1/A)+(1+B)=(1/24) One man can complete the work in 40mins one man's one day work (1/B)= (1/40) Now, (1/A)=(1/24)(1/40) (1/A)=(1/60) So, A can complete the work in 60 mins. 40) In a company 30% are supervisors and 40% employees are male if 60% of supervisors are male. What is the probability? That a randomly chosen employee is a male or female? Sol) 40% employees are male if 60% of supervisors are male so for 100% is 26.4%so the probability is 0.264 41) In 80 coins one coin is counterfeit what is minimum number of weighing to find out counterfeit coin Sol) the minimum number of wieghtings needed is just 5.as shown below (1) 80>3030 (2) 1515 (3) 77 (4) 33 (5) 11 42) 2 oranges, 3 bananas and 4 apples cost Rs.15. 3 oranges, 2 bananas, and 1 apple costs Rs 10. What is the cost of 3 oranges, 3 bananas and 3 apples? 2x+3y+4z=15 3x+2y+z=10 adding 5x+5y+5z=25 x+y+z=5 that is for 1 orange, 1 bannana and 1 apple requires 5Rs. so for 3 orange, 3 bannana and 3 apple requires 15Rs. i.e. 3x+3y+3z=15 43) In 8*8 chess board what is the total number of squares refers Sol) odele discovered that there are 204 squares on the board We found that you would add the different squares  1 + 4 + 9 + 16+ 25 + 36 + 49 + 64. Also in 3*3 tic tac toe board what is the total no of squares Ans 14 ie 9+4(bigger ones)+1 (biggest one) If you ger 100*100 board just use the formula the formula for the sum of the first n perfect squares is n x (n + 1) x (2n + 1) ______________________ 6 if in this formula if you put n=8 you get your answer 204 44) One fast typist type some matter in 2hr and another slow typist type the same matter in 3hr. If both do combinely in how much time they will finish. Sol) Faster one can do 1/2 of work in one hourslower one can do 1/3 of work in one hourboth they do (1/2+1/3=5/6) th work in one hour.so work will b finished in 6/5=1.2 hour i e 1 hour 12 min. 45) If Rs20/ is available to pay for typing a research report & typist A produces 42 pages and typist B produces 28 pages. How much should typist A receive? Here is the answer Find of 42 % of 20 rs with respect to 70 (i.e 28 + 42) ==> (42 * 20 )/70 ==> 12 Rs 46) An officer kept files on his table at various times in the order 1,2,3,4,5,6. Typist can take file from top whenever she has time and type it.What order she cannt type.? 47) In some game 139 members have participated every time one fellow will get bye what is the number of matches to choose the champion to be held? the answer is 138 matches Sol) since one player gets a bye in each round,he will reach the finals of the tournament without playing a match. therefore 137 matches should be played to detemine the second finalist from the remaining 138 players(excluding the 1st player) therefore to determine the winner 138 matches shd be played. 48) One rectangular plate with length 8inches, breadth 11 inches and 2 inches thickness is there. What is the length of the circular rod with diameter 8 inches and equal to volume of rectangular plate? Sol) Vol. of rect. plate= 8*11*2=176 area of rod=(22/7)*(8/2)*(8/2)=(352/7) vol. of rod=area*length=vol. of plate so length of rod= vol of plate/area=176/(352/7)=3.5 49) One tank will fill in 6 minutes at the rate of 3cu ft /min, length of tank is 4 ft and the width is 1/2 of length, what is the depth of the tank? 3 ft 7.5 inches 50) A man has to get airmail. He starts to go to airport on his motorbike. Plane comes early and the mail is sent by a horsecart. The man meets the cart in the middle after half an hour. He takes the mail and returns back, by doing so, he saves twenty minutes. How early did the plane arrive? ans:10min:::assume he started at 1:00,so at 1:30 he met cart. He returned home at 2:00.so it took him 1 hour for the total jorney.by doing this he saved 20 min.so the actual time if the plane is not late is 1 hour and 20 min.so the actual time of plane is at 1:40.The cart travelled a time of 10 min before it met him.so the plane is 10 min early. 51) Ram singh goes to his office in the city every day from his suburban house. His driver Mangaram drops him at the railway station in the morning and picks him up in the evening. Every evening Ram singh reaches the station at 5 o'clock. Mangaram also reaches at the same time. One day Ram singh started early from his office and came to the station at 4 o'clock. Not wanting to wait for the car he starts walking home. Mangaram starts at normal time, picks him up on the way and takes him back house, half an hour early. How much time did Ram singh walked? 52) 2 trees are there. One grows at 3/5 of the other. In 4 years total growth of the trees is 8 ft. what growth will smaller tree have in 2 years. Sol) THE BIG TREE GROWS 8FT IN 4 YEARS=>THE BIG TREE GROWS 4FT IN 2 YEARS.WHEN WE DIVIDE 4FT/5=.8*3=>2.4 ans: 1.5 mt 4 (x+(3/5)x)=88x/5=2x=5/4 after 2 years x=(3/5)*(5/4)*2 =1.5 53) There is a six digit code. Its first two digits, multiplied by 3 gives all ones. And the next two digits multiplied by 6 give all twos. Remaining two digits multiplied by 9 gives all threes. Then what is the code? sol) Assume the digit xx xx xx (six digits) First Two digit xx * 3=111 xx=111/3=37 ( first two digits of 1 is not divisible by 3 so we can use 111) Second Two digit xx*6=222 xx=222/6=37 ( first two digits of 2 is not divisible by 6 so we can use 222) Thrid Two digit xx*9=333 xx=333/9=37 ( first two digits of 3 is not divisible by 9 so we can use 333) 54) There are 4 balls and 4 boxes of colours yellow, pink, red and green. Red ball is in a box whose colour is same as that of the ball in a yellow box. Red box has green ball. In which box you find the yellow ball? ans is green... Sol) Yellow box can have either of pink/yellow balls. if we put a yellow ball in "yellow" box then it wud imply that "yellow" is also the colour of the box which has the red ball(becoz acordin 2 d question,d box of the red ball n the ball in the yellow box have same colour) thus this possibility is ruled out... therefore the ball in yellow box must be pink,hence the colour of box containin red ball is also pink.... =>the box colour left out is "green",,,which is alloted to the only box left,,,the one which has yellow ball.. 55) A bag contains 20 yellow balls, 10 green balls, 5 white balls, 8 black balls, and 1 red ball. How many minimum balls one should pick out so that to make sure the he gets at least 2 balls of same color. Ans:he should pick 6 ball totally. Sol) Suppose he picks 5 balls of all different colours then when he picks up the sixth one, it must match any on of the previously drawn ball colour. thus he must pick 6 balls 56) What is the number of zeros at the end of the product of the numbers from 1 to 100 Sol) For every 5 in unit palce one zero is added so between 1 to 100 there are 10 nos like 5,15,25,..,95 which has 5 in unit place. Similarly for every no divisible by 10 one zero is added in the answer so between 1 to 100 11 zeros are added for 25,50,75 3 extra zeros are added so total no of zeros are 10+11+3=24 57) 10 Digit number has its first digit equals to the numbers of 1's, second digit equals to the numbers of 2's, 3rd digit equals to the numbers of 3's .4th equals number of 4's..till 9th digit equals to the numbers of 9's and 10th digit equals to the number of 0's. what is the number?.(6marks) ans:2100010006 2shows that two 1's in the ans 1shows that one 2 in ans 0shows no 3 in the ans 0shows no 4 in the ans 0shows no 5 in the ans 1shows one 6 in the ans 0shows no 7 in the ans 0shows no 8 in the ans 0shows no 9 in the ans 6shows six 0's in the ans 58) There are two numbers in the ratio 8:9. if the smaller of the two numbers is increased by 12 and the larger number is reduced by 19 thee the ratio of the two numbers is 5:9. Find the larger number? sol) 8x:9x initialy 8x+ 12 : 9x  19 = 5x:9x 8x+12 = 5x > x = 4 9x = 36 not sure about the answer .. 59) There are three different boxes A, B and C. Difference between weights of A and B is 3 kgs. And between B and C is 5 kgs. Then what is the maximum sum of the differences of all possible combinations when two boxes are taken each time AB = 3 Bc = 5 ac = 8 so sum of diff = 8+3+5 = 16 kgs 60) A and B are shooters and having their exam. A and B fall short of 10 and 2 shots respectively to the qualifying mark. If each of them fired atleast one shot and even by adding their total score together, they fall short of the qualifying mark, what is the qualifying mark? ans is 11 coz each had atleast 1 shot done so 10 + 1 = 11 n 9 + 2 = 11 so d ans is 11 61) A, B, C, and D tells the following times by looking at their watches. A tells it is 3 to 12. B tells it is 3 past 12. C tells it is 12:2. D tells it is half a dozen too soon to 12. No two watches show the same time. The difference between the watches is 2,3,4,5 respectively. Whose watch shows maximum time? sol) A shows 11:57, B shows 12:03, C shows 12:02, and D shows 11:06 therefore, max time is for B 62) Falling height is proportional to square of the time. One object falls 64cm in 2sec than in 6sec from how much height the object will fall. Sol) The falling height is proportional to the squere of the time. Now, the falling height is 64cm at 2sec so, the proportional constant is=64/(2*2)=16; so, at 6sec the object fall maximum (16*6*6)cm=576cm; Now, the object may be situated at any where. if it is>576 only that time the object falling 576cm within 6sec .Otherwise if it is situated<576 then it fall only that height at 6sec. 63) Gavaskar average in first 50 innings was 50. After the 51st innings his average was 51 how many runs he made in the 51st innings Ans) first 50 ings. run= 50*50=2500 51st ings. avg 51. so total run =51*51=2601. so run scored in that ings=26012500=101 runs. 64) Anand finishes a work in 7 days, Bittu finishes the same job in 8 days and Chandu in 6 days. They take turns to finish the work. Anand on the first day, Bittu on the second and Chandu on the third day and then Anand again and so on. On which day will the work get over? a) 3rd b) 6th c) 9th d) 7th Ans is d) 7th day Sol) In d 1st day Anand does 1/7th of total work similarly, Bithu does 1/8th work in d 2nd day hence at d end of 3 days, work done = 1/7+1/8+1/6=73/168 remaining work = (16873)/168 = 95/168 again after 6 days of work, remaining work is = (9573)/168 = 22/168 and hence Anand completes the work on 7th day.(hope u understood.) 65) A man, a women and a child can do a piece of work in 6 days,man can do it in 14 days, women can do it 16 days, and in how many days child can do the same work? The child does it in 24 days 66) A: 1 1 0 1 1 0 1 1 B: 0 1 1 1 1 0 1 0 C: 0 1 1 0 1 1 0 1 Find ( (AB) u C )==? Hint : 109 AB is {A}  {A n B} A: 1 1 0 1 1 0 1 1 B: 0 1 1 1 1 0 1 0 by binary sub. ab = 01100001 (10=1, 11=0,00=0, n for the 1st 3 digits 110011=011) now (ab)uc= 01100001 or 01101101 gives 1101101... convert to decimal equals 109
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