An alternative keyboard in Windows....

An alternative keyboard in Windows
In case your keyboard or some keys stop working
Microsoft provides you with an alternative way to type in using the mouse
To work this tool go to
Start menu
Then
Select Run
Type in OSK
Press OK

A keyboard will appear that you can use both in Arabic and English

Prepared by,

Raj Shah,

Founder,

Techyuth.

Mobile- 09825867994.

Quantitative preparation for GRE/CAT/GMAT

1) col a:(2^8)*(15^5)+(2^8)*(15^5)
col b:(5^10)*(8^2)

Answer B

2) one given dat x*=(x-1)/x condition is x<0 x="x*..." 99 =" 3^99" 33 =" 9^33" cube="X" cube="Xroot2" surface="Xroot2*X=" 2="main"> 0 & X>Y

Col A Col B
XZ. YZ.

Sol: Case I: X=3, Y=2, Z=1 then col A is Greater.
Case II: X=5, Y= -2, Z= -3 then Col B is Greater

Therefore the answer is D

3) cola:0.0001/0.9 + 2.456/1.294
colb:5.563/2.78 + 13.569/5.265
Answer B

4) (x+2)^(x+2)<-1

col a x col b -1

5). If a man works for an entire year, he earns an amount...But if he works for 6 months and completes his work, what was his ratio of efforts that he put in the 6 months to his expected effort?

Ans. 1/2

6) COL A: 33^99 = 3^99 * 11^99 COL B: 99^33 = 9^33 * 11^33

ANS:A

7) in 1979 the average home heating cost in central region was same as avg for entire country, in 1980 central avg cost incrsd by $260 per home, an incrs of 25% over 1979. if in 1080 avg cost of of entire cntry incrsd by 22%, wat was avg increase in cost per home for entire country?

57.2 65 228.8 317 325

8) there was a cube figure which was cut thru the centre diagonal .. had to find the T.S.A(TOTAL SURFACE AREA) of the either of d figure (which is the same)

Sol) Let side of cube=X Diagonal of side of a cube=Xroot2

Area of the diagonal surface=Xroot2*X=X^2root2

T.S.A of the cube(which is cut diagonally) =x^2+x^2+x^2/2+x^2/2+ X^2root2

9) in rectangular cubic the face diagonal was given along with the main diagonal and also the length ¡K. Had to find the volume ¡K..

ans: just apply phythagoras and also that l^2+b^2+k^2=main diognal^2.

10). XYZ > 0 & X>Y

Col A Col B
XZ. YZ.

Sol: Case I: X=3, Y=2, Z=1 then col A is Greater.
Case II: X=5, Y= -2, Z= -3 then Col B is Greater
Therefore the answer is D

11). The mean of the three numbers 2X, 7X, X^2 ( X square) is 12 what is the range of the given data ( X<0). 3 =" 12" 2="36" 36="0" x=" -12" range =" Highest" commodity =" 144-" 3y="4" 4 =" 1" 3 =" 1" circle =" Pi" circle =" Pi" probability ="number" chance =" (X^2)" television =" 84." pager =" 84-X" pager =" 72-" x =" X-" 2y="5">

Sol:
90 rupees interest„³For 600 rupees
X„³For 100 rupees

X= 9000/600 = 15 rupees for 100 , which means 15%

12)2X+3Y=4 what is the X- intercept of the line

Sol: 2X/4 + 3Y/4 = 1
X/2 + Y/4/3 = 1
Therefore the X- intercept is 2

13)The Three vertices of a parallelogram is given as (0,0) (1,2) (3,3) and asked to find out the fourth vertices.

Sol: The answer is (2,1) , Figure is given

14) A circle is drawn inside an other circle, the center of both circles are the same, the radius of the first circle is X and the radius of the second circle is X+Y , we are asked to find out the probability that if a point is taken from the circle the point should lie in the smaller circle of radius X.

Sol: Area of smaller circle = Pi ( X^2)
Area of larger circle = Pi ( X+Y)^2
The probability =number of favorable chance / Total number of chance
= (X^2) / (X+Y) ^ 2

15) There are 84 people who watch television and 72 members use pagers and X members watch only Television .
Col A Col B
Number of people use only pagers X ¡V 12

Sol: Draw a Venn diagram
Total number of ppl who watch television = 84.
No of ppl who watch TV and use pager = 84-X
Number of ppl who use only pager = 72- ( 84 ¡V X )
= 72- 84 + X
= X- 12
Therefore C is the answer.

16). Find the Y ¡V intercept of the equation 3X-2Y=5

Sol: X/ (5/3) + Y/(-5/2) = 1, Therefore the Y intercept is : -5/2

17). The amount of Rs.600/- is given at a rate of X% per annum , If he receives additional Rs.90/- after an year then what is the rate of interest ( i.e- X )

Sol:
90 rupees interest„³For 600 rupees
X„³For 100 rupees

X= 9000/600 = 15 rupees for 100 , which means 15%

18). 90Col A Col B

The perimeter of the Triangle 24

Sol: If X = 90 then triangle is a RAT (Right angle triangle ), which satisfies Pythagorean theorem from which third side is 10 , then perimeter of triangle is 24 ( 6 + 8 + 10 ), but it is clearly given that 90

19) A parking garage rents parking spaces for $10 per week or $30 per month. How much does a person save in a year by renting by the month rather than by the week?
(A) $140
(B) $160
(C) $220
(D) $240
(E) $260

20) trains 100 m and 150m in length long are going in same direction on parallel tracks. faster train speed is 60 km ph and slower train speed 42 km/hour. at what time faster train over take slower train.

ans: 250/5 = 50secs

21)1979 approximately of the 37.3 million airline passengers traveling to or from the United States used Kennedy Airport. If the number of such passengers that used Miami Airport was the number that used Kennedy Airport and 4 times the number that used Logan Airport, approximately how many millions of these passengers used Logan Airport that year?

(A) 18.6
(B) 9.3
(C) 6.2
(D) 3.1
(E) 1.6

22)certain basketball team that has played of its games has a record of 17 wins and 3 losses. What is the greatest number of the remaining games that the team can lose and still win at least of all of its games?
(A) 7
(B) 6
(C) 5
(D) 4
(E) 3


answer: 4 ( 2/3 is 20 so total they will play 30 games. 3/4 should be winning ratio so they can max loose 7 games. 3 they have already lost so now they can max loose 4 games...

23)If n is an integer and then which of the following could be the value of k?
(A) 22
(B) 26
(C) 35
(D) 54
(E) 60

answer: 26


24) There were 36.000 hardback copies of a certain novel sold before the paperback version was issued. From the time the first paperback copy was sold until the last copy of the novel was sold. 9 times as many paperback copies as hardback copies were sold. If a total of 441.000 copies of the novel were sold in all, how many paperback copies were sold?
(A) 45.000
(B) 360.000
(C) 364.500
(D) 392.000
(E) 396.900

answer: 364,500

25) symbolic representation problem very time consumed one given dat x*=(x-1)/x condition is x<0>

a)2x*=2(x*)
b)-x*=x*
c)(x-1)*/x=x*...
d)(x*)*=x

26) cola:0.0001/0.9 + 2.456/1.294
col b:5.563/2.78 + 13.569/5.265

Answer B

27) THE DECORATING COMMITTEE FOR A DANCE PLANS TO FIRNGE THE 3 INCH WIDE END OF A STREAMER BY MAKIN SMALL CUTS EVERY 1/16TH INCH. HOW MANY CUTS MUST BE MADE TO FRINGE THE END?

A)45 B)46 C )47 D) 48 E) 49


consider e.g. :- u have 4 inch fringe, and want to cut in each 1 inch. so we will get not 4/1, but 4/1-1, coz the last part will be detached automatically. so the answer wil not b 4, but 3. so n-1...

so the actual answer is (16*3)-1 = 48-1 = 47

28). x+5 when divided by 3 leaves a remainder 2..
col A: remainder whn x is divided by 2
col B: 1

Ans. D

29) a car goes around a circle of radius of 14km in the first round it travels with a speed of 11km/hr in sud sequent rounds its speed increases by 11km/hr in each round how much time does the car take to cover a dist of 440km

a)14*8/5 b)15*14/18 c)18*4/15 d)4*15/18 e)7*4/15

30)THE DECORATING COMMITTEE FOR A DANCE PLANS TO FIRNGE THE 3 INCH WIDE END OF A STREAMER BY MAKIN SMALL CUTS EVERY 1/16TH INCH. HOW MANY CUTS MUST BE MADE TO FRINGE THE END?

A)45 B)46 C )47 D) 48 E) 49


consider e.g. :- u have 4 inch fringe, and want to cut in each 1 inch. so we will get not 4/1, but 4/1-1, coz the last part will be detached automatically. so the answer will not be 4, but 3. so n-1...

so the actual answer is (16*3)-1 = 48-1 = 47

31) Roger sold a watch at a profit of 10%. If he had bought it at 10% less and sold it for $13 less he would have made a profit of 15%. What is the cost price of the watch?

$220
$300
$400
$480
$540

Ans. let cost price b x

so profit 15% that means selling price is 1.1x
Now purchasing at 10% less that means 0.9x

sold for 13$ less i.e. (1.1x-13) & got profit of 15% means (15/100*90/100*x)=0.135x

selling price-purchase=profit

so (1.1x-13)-0.9x=0.135x
0.2x-13=0.135x
0.065x=13
x=200 $

so by increasing it 10% means $20 gives $200+$20 = $220......

32). the number of 2 digit even numbers that can be formed from the digits 1,2,3,4 and 5 if repetitions are not allowed is :
A) 8!
B) 5!
C) 2^6
D) 16
E) 16!


33). the number of ways in which 8 different flowers can be seated to form a garland so that 4 particular flowers are never separated is :
A) 4!4!
B) 288
C) 8!/4!
D) 5!4!
E) 8!4!

ans d)

34)an examination paper contains 8 questions of which 4 have 3 possible answers,3 have 2 possible answers each and the remaining one question has 5 possible answers. The total number of possible answers to all the questions is :
A) 3240
B) 94
C) 78
D) 2880
E) 36


35). the number of 2 digit even numbers formed from the digits 1,2,3,4,5 and 6 if repetitions of digits is not allowed
A) 6
B) 9
C) 5
D) 15
E) 17

ANS -15)

36). a box contains 2 white balls,3 black balls and 4 red balls. The number of ways in which 3 balls can be drawn from the box so that atleast one of the balls is black is
A) 74
B) 64
C) 94
D) 20
E) 24

ANS ¡V 64)

37). the number of 5 digit telephone numbers having atleast one of their digits repeated is
A) 90000
B) 100000
C) 30240
D) 69760
E) 69000

ANS C)

38). the number of 4 digit even numbers that can be formed using 0,1,2,3,4,5,6 without repetition is
A) 120
B) 300
C) 420
D) 20
E) 42

ANS-B)

39) if xyz=0, x>y than
col a: xz
col b: yz

ans.: d

40)comparison of SD of following

seq1 : 50,50,50,50,50
seq2 : 5,10,0,5,15

Solution:
no. elements in both series is same, i.e. 5

Range of first= max - min value = 0
Range of second = max-min = 10

41) there is a class of 90. avg of people who passed is 84. avg of people who failed is 60. Total class avg is 80. how many people passed ?
ans: 75

Prepared by,

Riddhi Shah,

U. V. Patel College of Engineering.

Mock lunar lander contest ends with a bang...!!!

Armadillo Aerospace's Module 1 rocket came close to winning the $350,000 level 1 prize in the Lunar Lander Challenge (Image: David Shiga)

The engine of an experimental Moon lander exploded on the launch pad on Sunday, dashing Armadillo Aerospace's hopes of winning up to $1.35 million in NASA prize money.

The 2007 Northrop Grumman Lunar Lander Challenge was the main event at the X Prize Cup rocket expo this past weekend near Alamogordo, New Mexico, US. The competition is intended to promote innovative ideas for a new generation of vehicles that could land humans on the Moon.

Nine teams had registered to compete, but only Armadillo Aerospace, based in Mesquite, Texas, US, and headed by Doom video game creator John Carmack, was ready in time for the event.

The competition is divided into two levels. In order to be eligible for the lesser $350,000 level 1 prize, a rocket must rise 50 metres from its launch pad, move 100 metres horizontally and land on a concrete pad, staying in the air for a total of at least 90 seconds. It must then repeat the feat, returning to its starting point within 2.5 hours.

To qualify for the $1 million level 2 prize, a rocket must make a similar flight, but this time it must stay aloft for 180 seconds and land on a bumpy surface that simulates that of the Moon.

Armadillo had planned to compete in both levels, but in the end attempted only the easier level 1 flights. It used a rocket called Module 1, whose two spherical fuel tanks were stacked vertically (see image at right).

Cracked engine

On Saturday, Module 1's first attempted flight failed when a blockage in a fuel line prevented the vehicle's rocket engine from starting.

The following flight was successful, meeting all the requirements for the prize and putting Armadillo in place to win the level 1 prize money, if it could make a successful return flight.

The return flight came very close to winning the prize for Armadillo, with Module 1 rising 50 metres above the ground and making it across the 100 metres to hover over the landing pad.
But the vehicle ran out of fuel, which had leaked out of a crack that had formed in its engine during ignition. Module 1 then plummeted to the ground and tipped over on its side 83 seconds into the flight – just 7 seconds short of the required 90-second minimum.
Loud bang

After replacing the cracked engine overnight, Armadillo made a good first flight early on Sunday. But at the start of the second flight, the engine cracked and sprang leaks again, forcing Armadillo to land the vehicle back on its launch pad after it had barely risen from the ground.

In a last-ditch attempt to win the prize, Armadillo took the engine out of another, more powerful rocket called Pixel – which it had planned to fly in the more difficult level 2 challenge – and used it as a replacement for Module 1's cracked engine.

But the engine exploded in a fireball at launch, producing a bang that was audible from more than a kilometre away. Armadillo declared an emergency and fire trucks rushed to the scene, but the fire quickly burned itself out without any intervention or injuries.

Stand down

The engine itself was destroyed in the explosion. "There were pieces on the pad," said Ken Davidian, who heads NASA's Centennial Challenges programme, which is providing the prize money for the contest.

"It was unrecoverable."

It is not clear how much damage was done to the rest of the vehicle, but it was not totally destroyed, as it was still standing on the launch pad for some time following the fire.

This ended Armadillo's hopes of winning the level 1 challenge, since the company had used up all four of its allowed attempts. There was still the possibility that Armadillo could try to win the more difficult level 2 challenge, by flying Pixel with a spare engine, but company founder John Carmack decided not to make the attempt.

Excess fuel

"This weekend, we've had more problems than we've had in the last six months. We know what went wrong, but not why," said Armadillo vice president Neil Milburn. "We know we'll be back again, and we'll nail it next time."

One reason the engines may have cracked is that fuel used for the "outbound" leg of the competition flights did not burn up completely before the return trip, said Armadillo team member Phil Eaton. Excess fuel in an engine can explode during ignition, cracking the engine.
That particular problem did not come up in previous tests because Armadillo usually let more time pass between flights, speculates Eaton. That lag allowed excess fuel to boil out of the vehicle, producing an audible gurgling sound.

Some of the other lunar lander teams at the event expressed sympathy for Armadillo. "It's painful – even for us competitors," said David Masten of Masten Space Systems, based in Mojave, California, US.

'Gut-wrenching'

"It's just got to be gut-wrenching," added Paul T Breed of Unreasonable Rocket, based in Solana Beach, California, US. "I know how hard they worked."

The $2 million pledged by NASA for the first and second prizes in the level 1 and level 2 challenges will now be up for grabs in the 2008 challenge.

"We've got all these other teams," said Becky Ramsey of the X Prize Foundation, which organises the X Prize Cup. "Some of them just need a little more time and I would expect next year we'll have a real race on our hands."

Armadillo was also the only competitor in the first year of the challenge in 2006, but its hopes for the prize ended with a crash landing that year.

NASA has expressed interest in potentially using Armadillo vehicles on Earth to test an automatic hazard avoidance system for its future lunar lander, which would use sensors and software to detect and avoid threats like craters and boulders.

Courtesy: http://technology.newscientist.com/home.ns

Prepared by,

Riddhi shah,

U. V. Patel college of Engineering