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Monday, 31 August 2015

Founders of OS and technology

Important List of Founders / Inventors of Famous Programming Languages, Operating Systems and other technology related software:--

C ----- Dennis Ritchie

C++ ----- Bjarne Stroustrup

JAVA ------ James Gosling

JavaScript -----Brendan Eich

COBOL ---- Grace Murray Hopper

BASIC ---- John G. Kemeny and Thomas E. Kurtz

FORTRAN ---- John Backus

Algorithm --- alKhwarizmi

DOS ----- Tim Paterson

Linux OS ---- Linus Torvalds

Unix ----- Dennis Ritchie

Windows ---- Bill Gates

Oracle --- Ed Oates, Larry Ellison, Bob Miner

Microsoft ---- Paul Allen and Bill Gates

IBM ------ Thomas J. Watson

Yahoo ---- Jerry Yang , David Filo

Google --- Larry Page, Sergey Brin

Gmail ----- Paul Buchheit

WWW --- Tim Berners-Lee

RDBMS ---- Edgar F. Codd

Mouse --- Douglas Engelbart

First general-purpose computer, the IBM 701. ------ Cuthbert Hurd

ENIAC, UNIVAC ----- John Mauchly

father of Computer Science and Artificial Intelligence. ----- Alan Turing

Facebook ------- Mark Elliot Zuckerberg

Twitter --------- Jack Dorsey, Evan Williams, Biz Stone

Whasapp ---- Jan Koum and Brian Acton

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DAILY MATHS QUIZ.

KNOWLEDGE PHILIC PRESENTS......

DAILY QUIZ.





1. A labour was employed for 40 days. He gets Rs. 10 for every working days and Rs. 2 was deducted from his wages on every leave. At the end he got total Rs. 220. Find the number of working days ?
(a)    16 days
(b)   25 days
(c)    36 days
(d)   26 days
(e)   None of these
SOLUTION. (b) 25 days
If he works for 40 days , he will get = 400
But , he got Rs 220
400 – 220 = 180.
Now, on every leave he losses Rs 10+ Rs 2 = Rs 12
So, Leave days = 180/12 = 15 days.
Working Days = 40- 15 = 25 days.

2. Rajesh was asked to calculate average of ten, two digit number. By mistakes he replaced digits of a number therefore the obtained average will decrease by 3.6. Find difference of digits of this number ?
(a)    8
(b)   3
(c)    10
(d)   4
(e)   None of these
SOLUTION. (d) 4
[(10x+y) – (10y+x)] / 10 = 3.6
9x-9y = 36
(x-y) = 4

3. Bowling rate a bowler was 15.4 per wicket. In the next match he took 5 wickets by giving 31 runs, therefore his bowling rate improved by 0.4 runs. Find number of wickets taken by him after this match ?
(a)    110 wicket
(b)   115 wickets
(c)    105 wickets
(d)   120 wickets
(e)   None of these
SOLUTION. (b) 115 wickets.
Let no. of wickets taken after this match = X
Now, acc to question,
15.4(X-5) +31 = 15X
0.4X-77+31 = 0
0.4X = 46
X=115

Also,
If we let the number of wickets before this match be X
Then, acc to question
15.4X +31 = (X+5)15
0.4X = 75-31= 44
X= 110
Hence, after this match wickets = 110+5 = 115.



4. A solution of salt and water contains 15% salt by weight. Of it 30 kg. water evaporates and the solutions now contains 20% of salt. Find the original quantity of solution ?
(a)    72 kg
(b)   80 kg
(c)    90 kg
(d)   120 kg
(e)   None of these.
SOLUTION. (d) 120 kg
Salt:Water = 15:85 = 3:17, total = 20 units
Now, after evaporating 30 kg, Ratio = 1:4
Now, since water evaporates, salt remains same
Therefore, 1:4 = 3:12
Water  evaporated = 17-12 = 5 units.
But, given water evaporated = 30 kg
Therefore, 1 unit = 6 kg
Hence, total solution = 6x(3+17) = 7x20 = 120kg.

5. 25 men can take 30 days to make 200 toys by working 6 hours/day. Find another group of 20 men can take how many days to make 400 toys by working 5 hours/day but efficiency of the first group is 3/2 times of the 2nd group ?
(a)    60 days
(b)   45 days
(c)    135 days
(d)   40 days
(e)   None of these
SOLUTION. 135 days
M1D1H1E1W2 = M2D2H2E2W1
25x30x6x1.5 x400 = 20xD2x5x1x200
D2 = 135 days.

6. A certain amount becomes triple in 5 years. This amount becomes 27 times in how many years if the interest is same ?


(a)    15 years
(b)   32.5 years
(c)    65 years
(d)   45 years
(e)   60 years
SOLUTION. (c) 65 years
Interest in 5 years = 2P
Required Interest = 26P.
2P/26P = 5/X
X= 65 years.

7. A group of 4 boys and 2 men can finish a work in 6 days, while another group of 5 boys and 6 men can finish the work in 4 days. In how many days 2 boys and 4 men will finish the work ?
(a)    5 days
(b)   12 days
(c)    6 days
(d)   2 days
(e)   None of these
SOLUTION. (e) None of these 8 (2/5) days
(4B+2M)6 = (5B+6M)4
4B = 12M
1B= 3M.
Now,
(4B+2M)6 = (2B+4M)D
4B+2M = 12M+2M = 14M and 2B+4M = 10M
14Mx6 = 10MxD
D= 84/10 = 8(2/5)days.

8. In a printing shop, 25 men can print a book of counting 200 pages in 30 days while working 6 hours per day, this book contains, 10 lines per page and each line contains 20 words. Find how many days another group of 30 men will take to print a book of 400 pages while working 5 hours per day. While this book contains 20 lines per page and each line contains 10 words. The efficiency of first group is 50% more than the second group ?
(a)    45 days
(b)   90 days
(c)    60 days
(d)   40 days
(e)   None of these
SOLUTION. (b) 90 days
25x30x6x1.5/ (200x10x20) = 30x5xDx1 / (400x20x10)
D= 5x6x2x1.5 = 90 days.

9. Tap A can empty a tank in 12 hours, while tap B can fill 5 liters water/minute in tank. If both taps open together the tank emptied in 15 hours. Find the capacity of the tank ?
(a)    20,000 litres
(b)   18,000 litres
(c)    19,000 litres
(d)   17,500 litres
(e)   None of these
SOLUTION. (b) 18,000 litres
Let the capacity be Y litres.
And Efficiency of A be X/hr
Therefore, Efficiency of B = 5/min = 300/hr
Efficiency of A+(-B) = X-300/hr.
Now, Capacity when A works alone = 12hrs* X/hr = 12X = Y
Capacity When Aand B worked together = 15hrs*(X-300)/hr = 15(X-300)= Y
But, Capacity is same. Therefore,
12X= 15(X-300) = Y
3X= 4500
X= 1500
Y = 12X= 12*1500 = 18,000 litres.

10. When price of milk increased by 25% by what percentage a household should reduce consumption of the milk, so that total expenditure increased by only 10% ?
(a)    15%
(b)   16%
(c)    18%
(d)   12%
(e)   None of these
SOLUTION (d) Expenditure = Price*Consumption
New price = 1.25 of Original Price
New Expenditure = 1.10 of Original Expenditure.
New Consumption = ?
1.10Exp = 1.25Price*Consumption
110/125= 0.88 = New Consumption
So reduction = 1-0.88 = .12= 12%

11. A person sells his table at profit of 25% and chair at a loss of 20%. But, on the whole he gains Rs. 18. On the other hand, if he sells the table at 20% loss and chair at 25% gain, he neither gains nor losses. Find the Cost Price of Table and Chair.
(a)    200,160
(b)   160,200
(c)    250,180
(d)   210,170
(e)   None of these
SOLUTION(a) 200,160
Let CP of Table and Chair be X and Y respectively.
1.25X + 0.80Y = X+Y+18
25X-20Y = 1800.… (i)
Also,
0.80X+1.25Y = X+Y
25Y = 20X
Y = 4X/5…(ii)
Putting (ii) in (i), we get
25X-20x4X/5= 1800
9X = 1800
X= 200
Y= 160




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Time And Work

knowledge philic presents...

Time And Work

1. A can do a work in 50 days and B in 40 days . They work together for 10 days. and then A leaves B to finish the work alone. How long will B take to finish it??
(a)    11 days
(b)   18 days
(c)    22 days
(d)   26 days
(e)   None of these

2. 30 men, working 4 hrs a day can do a piece of work in 10 days. Find the number of days in which 45 men working 8 hrs a day can do twice the work. Assume that 2 men of the first group do as much wrk in 2 hrs as 4 men of the second group do in 1 hr.
(a)    6(1/3) days
(b)   6(2/3) days
(c)    5(3/6) days
(d)   3(1/6) days
(e)   None of these

3. A alone would take 27 hrs more to complete the job than if both A and B would together. If B worked alone, he took 3 hrs more to complete the job than A and B worked together. What time, would they take if both A and B worked together?
(a)    8 hours
(b)   10 hours
(c)    9 hours
(d)   6 hours
(e)   None of these

4. A and B together can do a piece of work in 12 days which B and C together will do in 16 days. After A has been working on it for 5 days, and B for 7 days, C finishes it in 13 days. In how many days A,B and C alone will do the work ?
(a)    16, 48 and 26 days respectively
(b)   16, 48 and 24 days respectively
(c)    26, 48 and 24 days respectively
(d)   16, 46 and 24 days respectively
(e)   None of these



5. Two women Ganga and Jamuna, working separately can mow a field in 8 and 12 hours respectively. If they work for an hour alternately, Ganga beginning at 9 am, when will the mowing be finished?
(a)    9:30 PM
(b)   8:30 PM
(c)    6:00 AM
(d)   7:00 PM
(e)   None of these

6. A, B and C together can do a work in 12 days. A alone can do the work in 36 days and B alone can do the same work in 54 days. Find in what time C alone can do that work?
(a)    9 days
(b)   18 days
(c)    24 days
(d)   27 days
(e)   None of these

7. A, B and C together can do a work in 4 days. A alone can do the work in 12 days B alone can do the same work in 18 days. Find in what time C alone can do the same work alone?
(a)    9 days
(b)   18 days
(c)    27 days
(d)   8 days
(e)   None of these

8. A can complete a work in 35 days and B can do the same work in 28 days. If A after doing 10 days, leaves the work , find in how many days B will do the remaining work?
(a)    15 days
(b)   10 days
(c)    27 days
(d)   24 days
(e)   None of these

9. A can complete a work in 24 days and B can complete the same work in 18 days. If A after doing 4 days leaves the work find in how many days B will complete the remaining work?
(a)    11 days
(b)   15 days
(c)    12 days
(d)   10 days
(e)   None of these


10. A and B together can do a piece of work in 6 days, B alone could do it in 8 days. Supposing B works at it for 5 days, in how many days A alone could finish the remaining work?
(a)    9 days
(b)   8 days
(c)    24 days
(d)   12 days
(e)   None of these

11. A and B can do a piece of work in 20 days and 30 days. both starts the work together for some time,but B leaves the job 5 days before the work is completed. Find the time in which work is completed.
(a)    7 days
(b)   12 days
(c)    14 days
(d)   16 days
(e)   None of these

Answers :-
1. (c) 22 days
Let the total work be 200 work
Efficiency of
A = 200/50 = 4 work/day
B = 200/40 = 5 work/day
A+B’s efficiency = 9/day
A+B’s 10 days work = 9*10 = 90
Remaining work = 200-90 = 110
Time taken by B alone to finish the remaining wrk = 110/5 = 22days

2. 6(2/3) days
<i> Solution:- M1D1H1E1W2 = M2D2H2E2W1 (From MDH Rule)
Efficinecy of first grp : 2nd grp = 2*2 :4*1 = 1:1
Now, D2 = M1D1H1E1W2 / M2H2E2W1
D2 = 30*4*10*1*2 / 45*8*1*1
D2 = 20/3 = 6(2/3) days

3. (c) 9 Hours
Let A+B together takes X hours
A will take X+27 hrs
B will take X+3 hrs
Let the total work be (X+27)(X+3)
Efficiency of A= X+3
B = X+27
Total efficiency = 2X+30
Time working together = (X+27)(X+3) / 2X+30 = X
==> X^2 +30X + 81 = 2X^2 + 30X
or, X^2 = 81 or X= 9 hrs (neglecting –Ve time )

4.  (b) 16, 48 and 24 days respectively
Let the total work be 48
Efficiency of
A+B = 4/day…… (i)
B+C = 3/day……..(ii)
Now, A works for 5 days, B works for 7 days and C works for 13 days and completes the total work of 48.
This can be rewritten as
A+B for 5 days + B+C for 2 days + C for 11 days completes the total work of 48
Now, A+B’s 5 days work = 20
B+C’s 2 days work = 6
Therefore, 20+6+ C’s 11 days work = 48
C’s  11 days work = 48-26 = 22
C’s efficiency = 2/day.. (iii)
From (i),(ii),(iii)
C’s efficiency = 2
B’s Efficiency = 1
A’s efficiency = 3
Time taken by
A= 16 days, B= 48 days, and C= 24 days

5. (e) None of these (6:30PM)
Let the total work be 24
Efficiency of Ganga = 24/8 = 3/hr
Efficiency of Jamuna= 24/12 = 2/hr
They work alternately starting from Ganga
First 2 hrs work = 3+2 = 5
First 8 hrs work = 20
Remaining = 24-20 = 4
9th hr work to be done by Ganga = 3
Remaining work = 4-3 = 1 to be done by Jamuna in 1/2 hr.
Total time = 8+1+(1/2) hrs = 9.5 hrs or 9 Hr 30 minutes
So work will be completed by 9AM + 9 hrs 30 minutes = 18 hrs 30 minutes or 6:30 PM

 6. (d) 27 Days
Let the total work be 108 (Common Multiple of 12,36 and 54)
Efficiency of A+B+C =108/12=  9,
of A alone = 108/36 = 3 and
of B alone = 108/54 = 2
Therefore of C alone = 9-(3+2) = 4
Time taken by C = 108/4 = 27 days

7. (a) 9 Days
Let the total work be 36 ( Can take any value Preferably Common Multiple )
Efficiency of
A+B+C = 36/4 = 9
A alone= 36/12 = 3
B alone = 36/18 = 2
C alone = A+B+C- (A+B) = 9-(3+2) = 4
Time taken by C alone = 36/4 = 9 days

8. (e) None of these (20 days)
Let the total work be 140
Efficiency of A = 4
Efficiency of B = 5
A works for 10 days = 4*10 = 40
Remaining work = 140-40 = 100 to be done by B
B will do it in 100/5 = 20 days

9. (b) 15 days
Let the total work be 72
Efficiency of A = 3 and Of B = 4
A’s 4 days work = 3*4 = 12 remaining work = 72 -12 = 60
Work completed by B in 60/4 = 15 days

10. (a) 9 days
Let the total work be 24
Efficiency of A+B = 4
Efficiency of B = 3
Efficiency of A = 1 as A+B = 4 and B= 3
Work done by B in 5 day = 3*5 = 15
Remaining work = 24-15 = 9
Remaining work to be done by A in 9/1 = 9 days

11. (c) 14 days
Let the total work be 60
Efficiency of A = 3 and of B = 2
Efficiency of A+B = 3+2 = 5
Suppose B never left the work then if the time taken remains same then work done by B in those 5 days will be added to original work.
Therefore, Now, works become = 60 + B’s 5 days work = 60+10 = 70
Time taken = 70/5 = 14 days

Other way :- B leaves the work 5 days before means A did work alone for that 5 days
Work Done by A in that 5 day = 5/20 = 1/4
Remaining work = 3/4
To complete the work together A+B would have taken  1/ (1/20+ 1/30)  = 600/50 = 12 days
3/4th of the work together will be completed in 12*3/4 = 9 days
Total time = 5+9 = 14 days




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