#2
 
 
Re: L&T Aptitude Questions
As you require the Latest L&T aptitude questions so here I am sharing the same with you Q1. Cost of an item is x. It's value increases by p% and decreases by p% Now the new value is 1 rupee, what is the actual value ? Q2. A right circular cylinder and a cone are there. Base radius of cone is equal to radius of cylinder. What is the ratio of height to slant side if their volume are the same? Q3. Distance between two poles is 50 meters. A train goes by 48 at a speed of kmph. In one minute how many poles will be crossed by the train ? Q4. A pole seen from a certain distance at an angle of 15 degrees and 100 meters ahead by 30 degrees. What is the height of pole ? Q5. For 15 peopleeach has to pay Rs.20.For 20 peopleeach has to pay Rs.18. For 40 peoplehow much has each to pay ? Q6. If p=2q then q=r*r , if podd then q is even, whether r is even or odd ? a) first condition is sufficient b) second condition is sufficient c) both are sufficient d) both are not sufficient Rest of the Questions are attached in below file which is free of cost for you 
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#3
 
 
Re: L&T Aptitude Questions
As you want to get L&T Aptitude Question paper so here I am giving you some questions of that paper: 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)
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