- Created by @vishal.das

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a train and car start at same time ...car at rear end reaches front end and come back to rear end . in the mean time train travels 1 km and speed of train is 1 km /hr ...wat is distance covered by car

:banghead:

:banghead:

all the quant enthusiasts

please continue here.

http://www.pagalguy.com/forum/quantitative-questions-and-answers/78107-official-quant-thread-cat-2012-a.html#post3235962mods plz close this thread

pls everyone, continue on the thread jain bhai opened (link -above). A thread needs to be closed when it has completed around 1000 pages.

Thanx.

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Learn MoreGetting 15000.

18k*10/12 = 15k.

=====================Q) A credit card number has 6 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?

digits 1 2 3 4 5 & 6

numbers 1 2 ( 7,8,9) (3,6,9) (31,62,93)

Ans would be 3*3*3= 27.

Correct me if i am wrong

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ans= 27

3rd digit could be 7,8 and 9

4th could be only 3,6,9

and 5 could be only 3, 6 and 9

and 6th digit be fixed according to 5th digit

so 3x3x3= 27

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Getting 15000.

18k*10/12 = 15k.

=====================Q) A credit card number has 6 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?

my take

third digit can take 3 values

4th digit can take 3,6,9 = 3 values

5th digit = 3 times 6th. 6th can take only 1,2,3 and corresponding three values for 5th digit

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sorry for repost but i think its worthy of a shot

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please continue here.

http://www.pagalguy.com/forum/quantitative-questions-and-answers/78107-official-quant-thread-cat-2012-a.html#post3235962

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A truck having ten tyres travels from P to Q and then back to P. If there are 2 additional spare tyres and the distance between P and Q is 9000 kilometres (km), then the maximum distance travelled (in km) by any tyre is at least?

(1) 12000 (2) 18000 (3) 9000 (4) 16000 (5) 15000

sol:

so clearly 1st tyre will travel 9000km and so by the 2nd tyre so total distance

travelled by individual tyre

=9000*10(forward)+9000*10(backward)

=18000*10

=180000km

now when there are twelve tyres=(10*18000)/12=

catch here is this that adding two tyre will not increase the truck length

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A truck having ten tyres travels from P to Q and then back to P. If there are 2 additional spare tyres and the distance between P and Q is 9000 kilometres (km), then the maximum distance travelled (in km) by any tyre is at least?

(1) 12000 (2) 18000 (3) 9000 (4) 16000 (5) 15000

Getting 15000.

18k*10/12 = 15k.

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

- 1 Like

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A 250 m long train is approaching an unmanned railway crossing. The train is

running at a uniform speed of 75 km/h and is 3 km away from the crossing. At the same instant, a scooter is also approaching the crossing and is 1300 m away from it. The scooter is moving at a uniform speed of A. In which of the following ranges, A exists such that the scooter would pass the crossing by just escaping the collision?

(1) 24km/h to 28 km/h

(2) 29km/h to 33km/h

(3) 35km/h to 40km/h

(4) 41km/h to 45km/h

(5) 45 km/h to 50km/h

(3/75)=(1.3/A)

=>A=32.5~33

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anujgupta21 Sayskindly solve this question...GUY!!!

bhai d ans is x-8

(kx-6)(x^2+2x-

=x^3-ax^2-bx+48

comparing a=4

putting a in d exp.

x^2-4x-32.

div by x-8

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A truck having ten tyres travels from P to Q and then back to P. If there are 2 additional spare tyres and the distance between P and Q is 9000 kilometres (km), then the maximum distance travelled (in km) by any tyre is at least?

(1) 12000 (2) 18000 (3) 9000 (4) 16000 (5) 15000

my take

total distance travelled ( sum of distance travelled by individual tyres) = 18000*10

we have used 12 tyres for this so per tyre it come sot 18000*10/12 = 15000

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why did you take 25 instead of 50 on the LHS and RHS mei why did you multiply by 2/5f(1000) = 999

f(1000)*f(f(1000))= 1

or, 999*f(999) = 1

or, f(999) = 1/999

as f(x) is continuous, there must be a real value 'c' in the range (999,1000)

for which,

f(c) = 891

now,

f(c) * f(f(c)) = 1

or, 891 * f(891) = 1

or, f(891) = 1/891

Please revert back, in case you have doubt/queries with the solution

I am also getting the same answer with the same method .I think ans would be 1/891

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A 250 m long train is approaching an unmanned railway crossing. The train is

running at a uniform speed of 75 km/h and is 3 km away from the crossing. At the same instant, a scooter is also approaching the crossing and is 1300 m away from it. The scooter is moving at a uniform speed of A. In which of the following ranges, A exists such that the scooter would pass the crossing by just escaping the collision?

(1) 24km/h to 28 km/h

(2) 29km/h to 33km/h

(3) 35km/h to 40km/h

(4) 41km/h to 45km/h

(5) 45 km/h to 50km/h

Any speed> 32.5k/hr so ans is option 2

running at a uniform speed of 75 km/h and is 3 km away from the crossing. At the same instant, a scooter is also approaching the crossing and is 1300 m away from it. The scooter is moving at a uniform speed of A. In which of the following ranges, A exists such that the scooter would pass the crossing by just escaping the collision?

(1) 24km/h to 28 km/h

(2) 29km/h to 33km/h

(3) 35km/h to 40km/h

(4) 41km/h to 45km/h

(5) 45 km/h to 50km/h

Any speed> 32.5k/hr so ans is option 2

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why did you take 25 instead of 50 on the LHS and RHS mei why did you multiply by 2/5

Because Work is inversly proportional to the Men and days.

I have taken 25 because additional men are not required till the 25th day and after that they require another men ( assume x) so total ( 105 + x) men.

But they have to finish the work in 50 days but only 25 days are left ( to complete 3/5th work ) so both side 25 days

I hope you are clear now .

Correct me if i am wrong

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