This Day in History

Quote of the day

Monday, April 14, 2008

Satyam placement papers

Recent Aptitude Satyam Papers

31. What was the day of the week on 12th January, 1979?
Ans: Friday
Sol: Number of odd days in (1600 + 300) years = (0 + 1) = 1 odd day.
78 years = (19 leap years + 59 ordinary years) = (38 + 59) odd days = 6 odd days
12 days of January have 5 odd days.
Therefore, total number of odd days= (1 + 6 + 5) = 5 odd days.
Therefore, the desired day was Friday.
32. Find the day of the week on 16th july, 1776.
Ans: Tuesday
Sol: 16th july, 1776 means = 1775 years + period from 1st january to 16th july
Now, 1600 years have 0 odd days.
100 years have 5 odd days.
75 years = 18 leap years + 57 ordinary years
= (36 + 57) odd days = 93 odd days
= 13 weeks + 2 odd days = 2 odd days
Therefore, 1775 years have (0 + 5 + 2) odd days = 0 odd days.
Now, days from 1st Jan to 16th july; 1776
Jan Feb March April May June July
31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days
= (28 weeks + 2 days) odd days
Therefore, total number of odd days = 2
Therefore, the day of the week was Tuesday
33 .Find the angle between the minute hand and hour hand of a click when the time is
7.20?
Ans: 100deg
@=11/2 M-30H
Sol: Angle traced by the hour hand in 12 hours = 360 degrees.
Angle traced by it in 7 hrs 20 min i.e. 22/3 hrs = [(360/12) * (22/3)] = 220 deg.
Angle traced by minute hand in 60 min = 360 deg.
Angle traced by it in 20 min = [(360/20) * 60] = 120 deg.
Therefore, required angle = (220 - 120) = 100deg.
34.The minute hand of a clock overtakes the hours hand at intervals of 65 min of the correct time. How much of the day does the clock gain or lose?
Ans: the clock gains 10 10/43 minutes
Sol: In a correct clock, the minute hand gains 55 min. spaces over the hour hand in 60
minutes.
To be together again, the minute hand must gain 60 minutes over the hour hand.
55 minutes are gained in 60 min.
60 min. are gained in [(60/55) * 60] min == 65 5/11 min.
But they are together after 65 min.
Therefore, gain in 65 minutes = (65 5/11 - 65) = 5/11 min.
Gain in 24 hours = [(5/11) * (60*24)/65] = 10 10/43 min.
Therefore, the clock gains 10 10/43 minutes in 24 hours.


35.A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours. What will be the true time when the clock indicates 1 p.m. on the following day?


Ans. 48 min. past 12.


Sol: Time from 8 a.m. on a day to 1 p.m. on the following day = 29 hours.
24 hours 10 min. of this clock = 24 hours of the correct clock.
145/6 hrs of this clock = 24 hours of the correct clock.
29 hours of this clock = [24 * (6/145) * 29] hrs of the correct clock
= 28 hrs 48 min of the correct clock.
Therefore, the correct time is 28 hrs 48 min. after 8 a.m.
This is 48 min. past 12.


36. At what time between 2 and 3 o' clock will the hands 0a a clock together?
Ans: 10 10/11 min. past 2.
Sol: At 2 o' clock, the hour hand is at 2 and the minute hand is at 12, i.e. they are 10
min space apart.
To be together, the minute hand must gain 10 minutes over the other hand.
Now, 55 minutes are gained by it in 60 min.
Therefore, 10 min will be gained in [(60/55) * 10] min = 10 10/11 min.
Therefore, the hands will coincide at 10 10/11 min. past 2.

37. A sum of money amounts to Rs.6690 after 3 years and to Rs.10035 after 6 years on compound interest. Find the sum.
Ans: Rs. 4460


Sol: Let the Sum be Rs. P. Then
P [1 + (R/100)]^3 = 6690………..(i)
P [1 + (R/100)]^6 = 10035………..(ii)
On dividing, we get [1 + (R/100)]^3 = 10035/6690 = 3/2.
P * (3/2) = 6690 or P = 4460.
Hence, the sum is Rs. 4460.


38. Simple interest on a certain sum is 16/25 of the sum. Find the rate percent and time, if both are numerically equal.
Ans: Rate = 8% and Time = 8 years
Sol: Let sum = X. Then S.I. = 16x/25
Let rate = R% and Time = R years.
Therefore, x * R * R/100 = 16x/25 or R^2 = 1600/25, R = 40/5 = 8
Therefore, Rate = 8% and Time = 8 years.
39. Find
1. i. S.I. on RS 68000 at 16 2/3% per annum for 9 months.
2. ii. S.I. on RS 6250 at 14% per annum for 146 days.
3. iii. S.I. on RS 3000 at 18% per annum for the period from 4th Feb 1995 to 18th April 1995.
Ans: i. RS 8500.
ii. RS 350.
iii. RS 108.


Sol:
1. i. P = 68000, R = 50/3% p.a. and T = 9/12 year = ¾ years
Therefore, S.I. = (P * Q * R/100)
= RS (68000 * 50/3 * ¾ * 1/100) = RS 8500.
2. ii. P = RS 6265, R = 14% p.a. and T = (146/365) year = 2/5 years.
Therefore, S.I. = RS (6265 * 14 * 2/5 *1/100) = RS 350.
3. iii. Time = (24 + 31 + 18) days = 73 days = 1/5 year
P = RS 3000 and R = 18% p.a.
Therefore, S.I. = RS (3000 * 18 * 1/5 * 1/100) = RS 108
40. A sum at simple interest at 13 ½% per annum amounts to RS 2502.50 after 4 years. Find the sum.
Ans: sum = RS 1625
Sol: Let sum be x. Then,
S.I. = (x * 27/2 * 4 * 1/100) = 27x/50
Therefore, amount = (x + 27x/50) = 77x/50
Therefore, 77x/50 = 2502.50 or x = 2502.50 * 50 / 77 = 1625
Hence, sum = RS 1625

41. A sum of money doubles itself at C.I. in 15 years. In how many years will it become eight times?
Ans.45 years.
Sol: P [1 + (R/100)]^15 = 2P  [1 + (R/100)]^15 = 2……….(i)
Let P [1 + (R/100)]^n = 8P  P [1 + (R/100)]^n = 8 = 2^3
= [{1 + (R/100)}^15]^3.
 [1 + (R/100)]^n = [1 + (R/100)]^45.
 n = 45.
Thus, the required time = 45 years.
42. A certain sum amounts to Rs. 7350 in 2 years and to Rs. 8575 in 3 years. Find the sum and rate percent.
Ans: Sum = Rs. 5400,Rate=16 2/3 %.
Sol: S.I. on Rs. 7350 for 1 year = Rs. (8575-7350) = Rs. 1225.
Therefore, Rate = (100*1225 / 7350*1) % = 16 2/3 %.
Let the sum be Rs. X. then, x[1 + (50/3*100)]^2 = 7350.
 x * 7/6 * 7/6 = 7350.
 x = [7350 * 36/49] = 5400.
Therefore, Sum = Rs. 5400.
43. A, B and C start a business each investing Rs. 20000. After 5 months A withdrew Rs. 5000, B withdrew Rs. 4000 and C invests Rs. 6000 more. At the end of the year, a total profit of Rs. 69,900 was recorded. Find the share of each.
Ans. A's share = Rs. 20,500
B's share = Rs. 21200
C's share = Rs. 28200
Sol: Ratio of the capitals of A, B and C
= (20000*5+ 15000*7) : (20000*5+16000*7): (20000*5+26000*7)
=205000: 212000 : 282000 = 205:212:282
Therefore, A's share = Rs. ( 69900*205/699) = Rs. 20,500
B's share = Rs. (69900*212/699) = Rs. 21200,
C's share = Rs. (69900*282/699) = Rs. 28200.


44. Sanjiv started a business by investing Rs. 36000. After 3 months Rajiv joined him by investing Rs. 36000. Out an annual profit of Rs. 37100, find the share of each?


Sol: Ratio of their capitals= 36000*12:36000*9 = 4:3
Sanjiv's share= Rs. ( 37100*4/7) = Rs. 21200.
Rajiv's share = Rs. ( 37100*3/7) = Rs.15900.


45. If 20 men can build a wall 56m long in 6 days, what length of a similar wall can be built by 35 men in 3 days?
Ans. Length=49m.
Sol: Since the length is to be found out, we compare each item with the length as shown below.
More men, more length built (Direct).
Less days, less length built (Direct).
Men 20:35 :: 56: x
Similarly, days 6:3 :: 56: x.
Therefore, 20*6*x= 35*3*56 or x= 49.
Hence, the required length= 49m.


46.If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours a day; how much coal will be required for 8 engines, each running 13 hours a day, it being given that 3 engines of the former type consume as much as 4 engines of latter type.


Ans:26metric tonnes.


Sol: We shall compare each item with the quantity of coal.
Less engines, less coal consumed (direct)
More working hours, more coal consumed (direct)
If 3 engines of former type consume 1 unit, then 1 engine will consume 1/3 unit.
If 4 engines of latter type consume 1 unit, then 1 engine will consume 1/4 unit.
Less rate of consumption, less coal consumed (direct).
Therefore, number of engines 9:8 :: 24:x
Working hours 8:13 :: 24:x
Rate of consumption 1/3:1/4 :: 24:x.
9*8*1/3*x= 8*13*1/4*24 or x= 26.
Therefore, required consumption of coal 26 metric tonnes.


47. A contract is to be completed in 46 days and 117 men were set to work, each working 8 hours a day. After 33 days 4/7 of the work is completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day?


Ans.81


Sol: Remaining work = 1-4/7 =3/7.
Remaining period = (46-33) days =13 days.
Less work, less men (direct)
Less days, more men (indirect).
More hours per day, less men (indirect)
Therefore, work 4/7:3/7 ::117/x
Days 13:33 :: 117/x
Hrs/day 9:8:: 117/x
Therefore, 4/7*13*9*x= 3/7*33*8*117 or x= 198.
Therefore, additional men to be employed =(198-117) =81.


48. A garrison of 3300 men had provisions for 32 days, when given at the rate of 850gms per head. At the end of 7 days, reinforcement arrives and it was found that the provisions will last 17 days more, when given at the rate of 825gms per head. What is the strength of the reinforcement?


Ans: 1700


Sol: The problem becomes:
3300 men taking 850gms per head have provisions for (32-7) or 25 days. How many
men taking 825gms each have provisions for 17 days?
Less ration per head, more men (indirect).
Less days, more men (indirect)
Ration 825:850::3300:x
Days 17:25::3300:x
Therefore, 825*17*x= 850*25*3300 or x= 5000.
Therefore, strength of reinforcement = 5000-3300 = 1700.


49. Find the slant height, volume, curved surface area and the whole surface area of a cone of radius 21 cm and height 28 cm.


Sol: Slant Height, l = √(r^2 + h^2) =√(21^2 + 28^2) = √1225 = 35 cm
Volume = 1/3пr^2h = (1/3 * 22/7 * 21 * 21 * 28) cm^3 = 12936 cm^3
Curved surface area = пrl = 22/7 * 21 *35 cm^3 = 2310 cm^2
Total Surface Area = (пrl + пr^2) = (2310 + 22/7 * 21 * 21) cm^2 = 3696 cm^2


50. If the radius of the sphere is increased by 50%, find the increase percent in volume and the increase percent in the surface area.


Sol: Let the original radius = R. Then, new radius = 150/100 R = 3R/2
Original Volume = 4/3пR^3, New volume = 4/3п(3 R/2)^3 = 9пR^3/2
Original surface area = 4пR^2 , New surface area = 4п(3R/2)^2 = 9пR^2
Increase % in surface area = (5пR^2/4пR^2 * 100)% = 125%


Satyam REASONING Papers Latest
1. 1. If in a certain language MYSTIFY is coded as NZTUJGZ, how is NEMISES coded in that code?
(a) MDLHRDR (b) OFNJTFT (c) ODNHTDR (d) PGOKUGU
Sol. Clearly, in the code, each letter is the alphabet next to the corresponding letter in
the word


M Y S T I F Y




N Z T U J G Z
So, for NEMISES, N will be coded as O,E as F, M as N and so on.
Answer is (b).




1. 2. If TAP is coded as SZO, then how is FREEZE coded?
(a) ESDFYF (b) GQFDYF (c) EQDFYG (d) EQDDYD
Sol. Each letter in the code is the alphabet before the corresponding letter in the word.


S Z O




T A P




3. If CROWN is coded as BSNXM, how is BOARD coded?
(a) ANZQC (b) APZSC © CPBSE (d) CNBQE
Sol. Each letter is alternately one before and one ahead than the corresponding letter in the word.
So, B is coded as A, O as P, A as Z, R as S, D as C. hence, the answer is (b).
Thus, in FREEZE , F is coded as E, R as Q, E as D and Z as Y.




4. If TAIL is coded as VCKN, how is PEACE coded?
Sol.In the code each letter is two ahead than the corresponding letter in the word.
Thus, P is coded as R, E as G, A as C and C as E. so the answer is (a)




5. If LIGHT is coded as LJGIT , how is FLAMES coded?
(a) GLBNET (b) FKALER (c) FMANET (d) GLBMFS
Sol In the code, each letter is alternatively the same and one ahead than the
Corresponding letter in the word. So, in FLAMES, F is coded as F, L as M, A as
A, M as N, E as E and S as T.
The answer is (C)








1. 6. In a certain code, SIKKIM is written as THLJJL how is TRAINING written in that code?
Sol: Clearly, the letters in the word SIKKIM are moved alternatively one step
forward and one step backward to obtain the letters of the code.
UQBHOHOF




1. 7. In a certain code, FORGE is written as FPTJI how is CULPRIT written in that
code?
Sol:Clearly, the first letter in the word FORGE remains as it is and the second, third,
fourth and fifth letters are respectively moved one, two, three and four steps
forward to obtain the corresponding letters of the code.
CVNSVNZ




============================================================




In each questions below are given two statements followed by two
conclusions numbered I and II. You have to take the two given statements to be
true even if they seem to be at variance from the commonly known facts and
then decide which of the given conclusion logically follows from the two given
statements, disregarding commonly known facts.


Give answer (a) if only conclusion I follows; (b) if only conclusion II follows; © if either I or II follows; (d) if neither I nor II follows; and (e) if both I and II follows.






8. Statements : All cakes are ice-creams.
All ice-creams are toffees.
Conclusions: I . All cakes are toffees.
II. All toffees are ice-creams.
Sol (a) Since both statements are affirmative, the conclusions must be
affirmative. Conclusion II cannot follow as it contains the middle term. So only
conclusion I follows




9. Statements : All tigers are ships
some ships are cupboards.
Conclusions: I some tigers are cupboards.
II some cupboards are tigers.
Sol. (d) the first premise is A type and distributes the subject. So, the middle
term 'ships', which forms its predicate, is not distributed.
The second premise is I type and does not distribute either subject or predicate.
So, middle term 'Ships' forming its subject is not distributed. Since the middle
term is not distributed at least once in the premises, no conclusion follows.




10. Statements : Some pearls are gems.
Some gems are ornaments.
Conclusions: I. Some gems are pearls.
II Some ornaments are gems.
Sol. (d) Since both the premises are particular, no conclusion follows.

No comments:

some useful links