November 13th, 2019 05:43 PM | |

Aakashd | Re: Tata Consultancy Services placement question papers pf previous years free downloAs you are looking for Tata Consultancy Services placement question papers, here we have given the complete details in attachment. Feel free to download. If you have any query, write in reply box. |

November 2nd, 2015 10:14 AM | |

Kesari | Re: Tata Consultancy Services placement question papers pf previous years free downloI have a sample paper of Tata Consultancy Services placement exam. So here I am providing you as you want. 1) Online Test 1. Verbal section: Antonyms and Synonyms 2. Verbal reasoning 3. Quantitative aptitude 4. Critical reasoning 5. Analytical Reasoning Interview 1) Technical + HR 2) MR (Managerial Round) Written Test: 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 - 2-3-5-7 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 = (n-1) ! = (5-1)! = 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 (100-40) % 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. |

November 2nd, 2015 10:13 AM | |

Unregistered | Re: Tata Consultancy Services placement question papers pf previous years free downloI want to appear in Tata Consultancy Services placement exam. so here I need a sample paper of this exam. So here can you provide me its pattern as well as paper? |

May 28th, 2014 10:13 AM | |

Sashwat |
I want to apply for the post in Tata Consultancy Services and for that I want to get the previous year placement question papers so can you provide me that? As you want to get the previous year placement question papers of Tata Consultancy Services so here is the information of the same for you: 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+(log0.318-log0.317) = 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 8-5hr 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 10---10/5=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=10-----eq4 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 300 300-210=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 3x-4y=1/4(3x) 12x-16y=3x 9x=16y y=9x/16 (3x-4(9x/16))/((4x-5(9x/16))) ans:12/19 11) If G(0) = -1 G(1)= 1 and G(N)=G(N-1) - G(N-2) 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 A-C, We will find 2's compliment of C and them add it with A, That will give us (A-C) 2's compliment of C=1's compliment of C+1 =11000101+1=11000110 A-C=11000101+11000110 =10001001 Now (A-C) U B is .OR. logic operation on (A-C) 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 two-dimensional 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=34.5-27 =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*53-x*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, 3000-1750 = 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 = 3000-50*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 (10-6) = 4 pens = (2-1.5)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 = 68-6 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) (1-1/6) (1-1/7).... (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 100-x=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->30-30 (2) 15-15 (3) 7-7 (4) 3-3 (5) 1-1 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 cann t 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 air-mail. He starts to go to airport on his motorbike. Plane comes early and the mail is sent by a horse-cart. 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. |

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