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palagg
Joined: 22 Feb 2006
Posts: 12
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 Posted: Fri Feb 24, 2006 8:29 pm Post subject: number system doubts.. |
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hii plz help me solvin these ques..
1. x is a number between 1 and 12000, such dat it is neither divisible by 3 nor by 5?
a) 5600 b) 6400 c) 5800 d) 6200
2. there are 2 lights, one red and other green. the red flashes 3 times every minute and the green one flashes 5 times every 2 minutes. if the lights start flashing together, then the total number of times both will hav flashed togethe in an hour is
a) 30 b) 24 c) 48 d) none
3. let "a" denote the set of positive integers, each of which when divided by 2,3,4,5,6 leaves a remainder of 1,2,3,4,5 respectively.. how many numbers in "a" are between 0 and 100?
a) 0 b) 1 c) 2 d)none
4. manisha has to multiply 2 numbers together, but she uses 53 instead of 35 in this product, and finds that the product has increased by 540. wat is this increased product?
a) 1050 b) 540 c) 1440 d) 1590
5. how many 2 digit numbers are there such that the sum of the digits constituting the number is not less than 7; the sum of the squares of the digits is not greater than 30; the number consisting of the same digits written in the reverse order is not larger than half the given number.
a) 1 b) 2 c) 3 d)4
waiting for replies...................... |
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CZ@@@H@@@R
Serious about CAT
Joined: 14 Jan 2006
Posts: 64
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| Posted: Sat Feb 25, 2006 1:06 pm Post subject: Re: number system doubts.. |
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| palagg wrote: |
1. x is a number between 1 and 12000, such dat it is neither divisible by 3 nor by 5?
a) 5600 b) 6400 c) 5800 d) 6200
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(Ans) b) 6400
NM(3 or 5) = NM(3) + NM(5) - NM(5 and 6) = 4000 + 2400 - 800 = 5600, Hence answer 6400. NM is number of multiples |
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CZ@@@H@@@R
Serious about CAT
Joined: 14 Jan 2006
Posts: 64
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| Posted: Sat Feb 25, 2006 1:08 pm Post subject: Re: number system doubts.. |
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| palagg wrote: |
2. there are 2 lights, one red and other green. the red flashes 3 times every minute and the green one flashes 5 times every 2 minutes. if the lights start flashing together, then the total number of times both will hav flashed togethe in an hour is
a) 30 b) 24 c) 48 d) none
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(Ans) (a) 30 times
Both flash together after every 120 seconds i.e 2 minutes. Hence 30 times. |
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CZ@@@H@@@R
Serious about CAT
Joined: 14 Jan 2006
Posts: 64
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| Posted: Sat Feb 25, 2006 1:13 pm Post subject: Re: number system doubts.. |
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| palagg wrote: |
3. let "a" denote the set of positive integers, each of which when divided by 2,3,4,5,6 leaves a remainder of 1,2,3,4,5 respectively.. how many numbers in "a" are between 0 and 100?
a) 0 b) 1 c) 2 d)none
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(Ans) 1 hence (b)
a = {(60 N - 1)}, where N is an integer. 59 is the first number. |
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CZ@@@H@@@R
Serious about CAT
Joined: 14 Jan 2006
Posts: 64
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| Posted: Sat Feb 25, 2006 1:15 pm Post subject: Re: number system doubts.. |
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| palagg wrote: | 4. manisha has to multiply 2 numbers together, but she uses 53 instead of 35 in this product, and finds that the product has increased by 540. wat is this increased product?
a) 1050 b) 540 c) 1440 d) 1590
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(Ans) Increased product = 53 x 30 = 1590, Hence (d) |
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CZ@@@H@@@R
Serious about CAT
Joined: 14 Jan 2006
Posts: 64
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| Posted: Sat Feb 25, 2006 1:17 pm Post subject: Re: number system doubts.. |
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| palagg wrote: |
5. how many 2 digit numbers are there such that the sum of the digits constituting the number is not less than 7; the sum of the squares of the digits is not greater than 30; the number consisting of the same digits written in the reverse order is not larger than half the given number.
a) 1 b) 2 c) 3 d)4
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(Ans) a) 1
Number be 10x + y
(Maximum value of x or y is 5) & (x >= 2.375 y) & (x+y >= 7)
52 is only possible, Hence a) 1 |
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palagg
Joined: 22 Feb 2006
Posts: 12
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| Posted: Sat Feb 25, 2006 4:28 pm Post subject: Re: number system doubts.. |
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hiiiiii
thanx a lot for ur reply... i got all the answers... i saw dat u gave reply to sum1 earlier laso.. how come u no all the problems.. i mean which books u hav practised.....
biiiiiii |
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CZ@@@H@@@R
Serious about CAT
Joined: 14 Jan 2006
Posts: 64
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| Posted: Sun Feb 26, 2006 9:02 am Post subject: Re: number system doubts.. |
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| palagg wrote: | hiiiiii
thanx a lot for ur reply... i got all the answers... i saw dat u gave reply to sum1 earlier laso.. how come u no all the problems.. i mean which books u hav practised.....
biiiiiii |
I prefer the books I am convenient with. |
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shaji
Serious about CAT
Joined: 22 May 2005
Posts: 71
Location: UK
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| Posted: Sun Feb 26, 2006 2:09 pm Post subject: Re: number system doubts.. |
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| CZ@@@H@@@R wrote: | | palagg wrote: |
1. x is a number between 1 and 12000, such dat it is neither divisible by 3 nor by 5?
a) 5600 b) 6400 c) 5800 d) 6200
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(Ans) b) 6400
NM(3 or 5) = NM(3) + NM(5) - NM(5 and 6) = 4000 + 2400 - 800 = 5600, Hence answer 6400. NM is number of multiples |
Hi;
6400 is the # of numbers satisfying the criteria but not the number itself as the question poses.
Regards;
Shaji |
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CZ@@@H@@@R
Serious about CAT
Joined: 14 Jan 2006
Posts: 64
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| Posted: Sun Feb 26, 2006 2:30 pm Post subject: Re: number system doubts.. |
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| palagg wrote: | hii plz help me solvin these ques..
1. x is a number between 1 and 12000, such dat it is neither divisible by 3 nor by 5?
a) 5600 b) 6400 c) 5800 d) 6200
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I agree, None of the option is correct in true sense. However, I feel the question would be to find the number of Numbers satisfying the criteria. |
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banter
Joined: 19 Mar 2006
Posts: 18
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| Posted: Sun Mar 19, 2006 2:09 am Post subject: Re: number system doubts.. |
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[quote="CZ@@@H@@@R"][quote="palagg"]
1. x is a number between 1 and 12000, such dat it is neither divisible by 3 nor by 5?
a) 5600 b) 6400 c) 5800 d) 6200
[/quote]
[b][color=blue](Ans) b) 6400 [/color][/b]
NM(3 or 5) = NM(3) + NM(5) - NM(5 and 6) = 4000 + 2400 - 800 = 5600, Hence answer 6400. NM is number of multiples[/quote]
NM(5 and 6) is 400 and not 800 |
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CZ@@@H@@@R
Serious about CAT
Joined: 14 Jan 2006
Posts: 64
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| Posted: Sun Mar 19, 2006 10:24 am Post subject: Re: number system doubts.. |
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| banter wrote: | | CZ@@@H@@@R wrote: | | palagg wrote: |
1. x is a number between 1 and 12000, such dat it is neither divisible by 3 nor by 5?
a) 5600 b) 6400 c) 5800 d) 6200
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(Ans) b) 6400
NM(3 or 5) = NM(3) + NM(5) - NM(5 and 6) = 4000 + 2400 - 800 = 5600, Hence answer 6400. NM is number of multiples |
NM(5 and 6) is 400 and not 800 |
6 should read 3 |
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