## Re: [Ascent CAT] Doubt in Ratio & Proportio

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• x,y,z are in continued proportion implies........ x/y=y/z *hence*, xz=y^2 (cross multiplication) xz/x^2=y^2/x^2 (dividing both sides of equation
Message 1 of 21 , Feb 19, 2007
x,y,z are in continued proportion implies........
x/y=y/z

hence, xz=y^2        (cross multiplication)

xz/x^2=y^2/x^2        (dividing both sides of equation by x^2)

z/x=y^2/x^2

thatz it,
z:x=y^2:x^2

THANX..

On 2/13/07, Ganesan MahendraBabu <babumathi246@...> wrote:

Hi all,

I have a doubt in the follow question. can anyone solve this equation and tell me the logic

1. There are 3 unequal quantities x,y&z in continued proportion. Which of the following equals   z:x?

ans a)y^2:x^2  b)(z^2-y^2)/(y^2-x^2)  c) z^2:y^2  d) all of these

Regards
Ganesan m

--
RRUULLII
• Can anyone please solve the question given below? A certain city with a population of 132,000 is to be divided into 11 voting districts, and no district is to
Message 2 of 21 , Mar 1, 2007
Can anyone please solve the question given below?

A certain city with a population of 132,000 is to be
divided into 11 voting districts, and no district is
to have a population that is more than 10% greater
than the population of any other district. what is the
minimum possible population that the least populated
district could have?

10,700
10,800
10,900
11,000
11,100

Thanks,
Rajib

--- raju dictator <rajudictator@...> wrote:

> have a doubt in the follow question. can anyone
> solve this equation and
> tell me the logic
>
> 1. There are 3 unequal quantities x,y&z in continued
> proportion. Which of
> the following equals z:x?
> ans a)y^2:x^2 b)(z^2-y^2)/(y^2-x^2) c) z^2:y^2 d)
> all of these
>
>
> well the ans to this question is quite easy
> now,we know that x,y and z are i cont.prop
> so,x/y=y/z
> now,z:x=z/x=(z/y)\(x/y)=(y/x)\(x/y)=y^2/x^2=z^2/y^2
> also as y^2/x^2=z^2/y^2=z^2-y^2)/(y^2-x^2) (refer
> any standard book on
> ratios for the proof of this prop.)
> so,ans is (D)
>

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• let x = minimum possible population other districts max possible population = 1.1x so 10 other ditricts pop = 10*1.1x = 11x total population = x+11x = 132000
Message 3 of 21 , Mar 5, 2007
let x = minimum possible population

other districts max possible population  = 1.1x
so 10 other ditricts pop = 10*1.1x = 11x
total population = x+11x = 132000
x= 11000

rajib samanta <rajibsam@...> wrote:
Can anyone please solve the question given below?

A certain city with a population of 132,000 is to be
divided into 11 voting districts, and no district is
to have a population that is more than 10% greater
than the population of any other district. what is the
minimum possible population that the least populated
district could have?

10,700
10,800
10,900
11,000
11,100

Thanks,
Rajib

--- raju dictator wrote:

> have a doubt in the follow question. can anyone
> solve this equation and
> tell me the logic
>
> 1. There are 3 unequal quantities x,y&z in continued
> proportion. Which of
> the following equals z:x?
> ans a)y^2:x^2 b)(z^2-y^2)/(y^2-x^2) c) z^2:y^2 d)
> all of these
>
>
> well the ans to this question is quite easy
> now,we know that x,y and z are i cont.prop
> so,x/y=y/z
> now,z:x=z/x=(z/y)\(x/y)=(y/x)\(x/y)=y^2/x^2=z^2/y^2
> also as y^2/x^2=z^2/y^2=z^2-y^2)/(y^2-x^2) (refer
> any standard book on
> ratios for the proof of this prop.)
> so,ans is (D)
>

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• ... Solution: answer is 11,000 say least possible population is x, which is possible when all other districts hav max. possible population, that is, just 10%
Message 4 of 21 , Mar 5, 2007
--- In ascent4cat@yahoogroups.com, rajib samanta <rajibsam@...> wrote:
>
> Can anyone please solve the question given below?
>
> A certain city with a population of 132,000 is to be
> divided into 11 voting districts, and no district is
> to have a population that is more than 10% greater
> than the population of any other district. what is the
> minimum possible population that the least populated
> district could have?
>
>
> 10,700
> 10,800
> 10,900
> 11,000
> 11,100
>
> Thanks,
> Rajib
>

Solution:

say least possible population is x,
which is possible when all other districts hav max. possible
population, that is, just 10% more thn x, i.e 1.1x
so, total population= 1.1x*10+x = 132,000
=>12*x = 132,000;
=>x = 11,000
• ... so
Message 5 of 21 , Mar 5, 2007
On 3/2/07, rajib samanta <rajibsam@...> wrote:

Can anyone please solve the question given below?

A certain city with a population of 132,000 is to be
divided into 11 voting districts, and no district is
to have a population that is more than 10% greater
than the population of any other district. what is the
minimum possible population that the least populated
district could have?

10,700
10,800
10,900
11,000
11,100

the answer is 11000 because when we take equal populations for all districts the population comes out to be 12000.so,now if you consider the option given only 11000 leaves us with a population of 121000 which has to be divided into 10 districts.so,if we take equal distribution the population of remaining district comes out to be 12100 which is exactly 10% greater than trhe min and hence just acceptable.any lower option wont satisfy this criteria

so

• a1, a2, a3, ...., a15 In the sequence shown, a[n] = a[n-1] + k, where 2
Message 6 of 21 , Mar 14, 2007
a1, a2, a3, ...., a15

In the sequence shown, a[n] = a[n-1] + k, where
2<=n<=15 and k is a non-zero constant. How many of the
term in the sequence are greater than 10?

Given: a[8] = 10

Thanks,
Rajib

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• On Tue, 06 Mar 2007 07:49:55 +0530, raju dictator
Message 7 of 21 , Mar 22, 2007
On Tue, 06 Mar 2007 07:49:55 +0530, raju dictator <rajudictator@...>
wrote:

> On 3/2/07, rajib samanta <rajibsam@...> wrote:
>>
>> Can anyone please solve the question given below?
>>
>> A certain city with a population of 132,000 is to be
>> divided into 11 voting districts, and no district is
>> to have a population that is more than 10% greater
>> than the population of any other district. what is the
>> minimum possible population that the least populated
>> district could have?
>>
>>
>> 10,700
>> 10,800
>> 10,900
>> 11,000
>> 11,100
>>
>> the answer is 11000 because when we take equal populations for all
>> districts the population comes out to be 12000.so,now if you consider
>> the
>> option given only 11000 leaves us with a population of 121000 which has
>> to
>> be divided into 10 districts.so,if we take equal distribution the
>> population of remaining district comes out to be 12100 which is exactly
>> 10%
>> greater than trhe min and hence just acceptable.any lower option wont
>> satisfy this
>> criteria
>>
>>
>
>
>
>
> so
>
>
>
>
>
>>
• Since k can be either +ve or -ve we can say that 1.if k 0 ,then a9,a10,.....a15 10 2.if k 10 Either case,7 numbers are greater than
Message 8 of 21 , Apr 9, 2007
Since k can be either +ve or -ve
we can say that
1.if k > 0 ,then a9,a10,.....a15>10
2.if k < 0 ,then a7,a6,a5....a1>10
Either case,7 numbers are greater than 10

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• Solution: Given series is: a1,a2,a3,...........a15 and, a[n] = a[n-1] + k which clearly indicates that the given series is in A.P. Also given that, a8=10 so,
Message 9 of 21 , Apr 10, 2007
Solution:

Given series is: a1,a2,a3,...........a15
and, a[n] = a[n-1] + k
which clearly indicates that the given series is in A.P.
Also given that, a8=10
so, a8=a7+ k
i.e  10 = a7+k ......(1)

also, a9=a8+k
i.e.  a9=10 +k........(2)

Solving these two equations we get, a7 - a9 = 20
which means the geiven series is of decreasing A.P. type

From above we can figure out that vaule of k would be= -10
and a9 = 0 , a7= 20

a6=30, a5=40, a4=50, a3=60, a2=70, a1=80

Hence, 7 values in the given series will be greater than 10.

rajib samanta <rajibsam@...> wrote:
a1, a2, a3, ...., a15

In the sequence shown, a[n] = a[n-1] + k, where
2<=n<=15 and k is a non-zero constant. How many of the
term in the sequence are greater than 10?

Given: a[8] = 10

Thanks,
Rajib

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• Dear Rajib, Ans - 7 Explaination - a(8) = a(7) + k or, 10 = a(7) +K 10-k = a(7) Like wise, following are value for a1 to a7 a1 = 10-7k a2 = 10- 6k....
Message 10 of 21 , Apr 10, 2007
Dear Rajib,
Ans - 7
Explaination -

a(8) = a(7) + k
or, 10 = a(7) +K
10-k = a(7)

Like wise, following are value for a1 to a7
a1 = 10-7k
a2 = 10- 6k....                       Total 7 elements a(1) to a(7)
......
a(7) = 10-k

Values for a(9) to a(15) are followings -
a(9) = 10+k
a(10) = 10+2k...
.....                                        Total 7 elements a(9) to a(15)
a(15) = 10 +7k

CASE 1 -Now if k is positive a(9) to a(15) will be more than 10.
CASE 2 - If K is -ve a(1) to a(7) will be more than 10

In both cases it is 7 elements.

So for sure there are 7 elements in series more than 10.
rajib samanta <rajibsam@...> wrote:
a1, a2, a3, ...., a15

In the sequence shown, a[n] = a[n-1] + k, where
2<=n<=15 and k is a non-zero constant. How many of the
term in the sequence are greater than 10?

Given: a[8] = 10

Thanks,
Rajib

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• Hi All I have got one question in prob solving which am facing difficulty in getting it solved. Plz help me. Problem Statement: If xy = 1 then what is the
Message 11 of 21 , Apr 29, 2007
Hi All

I have got one question in prob solving which am facing difficulty in getting it solved. Plz help me.

Problem Statement:
If xy = 1 then what is the value of

{2* (x+y)^2} / {2* (x-y)^2}

The answer options are:  a) 2, b) 4, c) 8, d) 16, e) 32

Thanks,
Sathish

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• Hi All This is yet another question which also bothers me on solving it, For every positive even integer n, the function h(n) is defined to be the product of
Message 12 of 21 , Apr 29, 2007
Hi All

This is yet another question which also bothers me on solving it,

For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is

a) between 2 and 10
b) between 10 and 20
c) between 20 and 30
d) between 30 and 40
e) greater than 40

Thanks,
Sathish

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• The Function h(n) is given by h(n)+1=2^(n/2) x (n/2)!+1 So by substituting various even integer n, we can get consecutive Values .So by checking for prime
Message 13 of 21 , May 6, 2007
The Function h(n) is given by
h(n)+1=2^(n/2) x (n/2)!+1
So by substituting various even integer n, we can get consecutive Values .So by checking for prime numbers. I got the answer to be
Ans: Greater than 40
For the term h(100)+1 = (2^50 + 50! )+1
Iam not sure with the answer

Rajkumar

sathish vijayaraghavan <sathish_vijayaraghavan@...> wrote:
Hi All

This is yet another question which also bothers me on solving it,

For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is

a) between 2 and 10
b) between 10 and 20
c) between 20 and 30
d) between 30 and 40
e) greater than 40

Thanks,
Sathish

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• Hi All, The number in context here is 2^50*50!+1 let 2^50*50! be y so we want the divisibility of y+1 now we can clearly see that y is divisible be all the
Message 14 of 21 , May 9, 2007
Hi All,
The number in context here is
2^50*50!+1
let 2^50*50! be y
so we want the divisibility of y+1 now we can clearly see that y is divisible be all the numbers till 50 including the prime numbers so obviously y+1 will not be divisibile by those numbers. Hence any number less than 50 will not divide y+1 so the smallest prime number has to be greater than 50, thus the option greater than 40 (e) is the answer.
Regards
uday

Funda: y and y+1 are coprimes, 100 n 101

sathish vijayaraghavan <sathish_vijayaraghavan@...> wrote:
Hi All

This is yet another question which also bothers me on solving it,

For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is

a) between 2 and 10
b) between 10 and 20
c) between 20 and 30
d) between 30 and 40
e) greater than 40

Thanks,
Sathish

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• h(100)+1 = 2*4*6*8*...*100 + 1 = 2 * (1*2*3*...*50) + 1 Look at this number carefully. The quantity inside the brackets is divisible by each of the numbers
Message 15 of 21 , May 14, 2007
h(100)+1 = 2*4*6*8*...*100 + 1 = 2 * (1*2*3*...*50) + 1

Look at this number carefully. The quantity inside the brackets is divisible by each of the numbers from 2 to 50. So when you multiply it by 2, the quantity obtained [2 * (1*2*3*...*50)] will also be ivisible by each of the numbers from 2 to 50. So, the next integer, i.e., 2 * (1*2*3*...*50) + 1, cannot be divisible by any of the numbers from 2 to 50.

Therefore, the option e) must be the correct one.

Regards,
Kaushik

On 4/30/07, sathish vijayaraghavan <sathish_vijayaraghavan@...> wrote:

Hi All

This is yet another question which also bothers me on solving it,

For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is

a) between 2 and 10
b) between 10 and 20
c) between 20 and 30
d) between 30 and 40
e) greater than 40

Thanks,
Sathish

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• Hi, h(n) = 2.4.6.....100 + 1 = 2(50!)+1 I think the answer should be (e). Expl : any no can not divide the number obtained by adding 1 to the multiple of that
Message 16 of 21 , May 27, 2007
Hi,

h(n) = 2.4.6.....100 + 1
= 2(50!)+1

I think the answer should be (e).

Expl : any no can not divide the number obtained by adding 1 to the multiple of that number. e.g. since 25 is divisible by 5 so 25+1 is not divisible by 5.

since 2(50!) is divisible by all the no from 2 to 50 including prime no. so 2(50!)+1 can not be divisible by any prime no between 2 to 50. so the first prime no which can divide h(100) should be more than 50.

raj kumar <dv_rajkumar@...> wrote:
The Function h(n) is given by
h(n)+1=2^(n/ 2) x (n/2)!+1
So by substituting various even integer n, we can get consecutive Values .So by checking for prime numbers. I got the answer to be
Ans: Greater than 40
For the term h(100)+1 = (2^50 + 50! )+1
Iam not sure with the answer

Rajkumar

sathish vijayaraghavan <sathish_vijayaragha van@yahoo. com> wrote:
Hi All

This is yet another question which also bothers me on solving it,

For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is

a) between 2 and 10
b) between 10 and 20
c) between 20 and 30
d) between 30 and 40
e) greater than 40

Thanks,
Sathish

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• Dear All, Sheetal here .I want suggestion as to wat to do regarding verbal cutoff it worries me like a night mare. All possible suggestion and advices are
Message 17 of 21 , Jun 4, 2007
Dear All,

Sheetal here .I want suggestion as to wat to do regarding verbal
cutoff it worries me like a night mare.
All possible suggestion and advices are welcome.
Completely lost .No confidence at all.

Regards
Sheetal
• This is not addition that u will be taking common out here. On taking 2 common from each term it becomes 2^50. then u can follow rest of ur steps. Kaushik Bar
Message 18 of 21 , Jun 7, 2007
This is not addition that u will be taking common out here.
On taking 2 common from each term it becomes 2^50.
then u can follow rest of ur steps.

Kaushik Bar <kaushikbar@...> wrote:
h(100)+1 = 2*4*6*8*...* 100 + 1 = 2 * (1*2*3*...*50) + 1

Look at this number carefully. The quantity inside the brackets is divisible by each of the numbers from 2 to 50. So when you multiply it by 2, the quantity obtained [2 * (1*2*3*...*50) ] will also be ivisible by each of the numbers from 2 to 50. So, the next integer, i.e., 2 * (1*2*3*...*50) + 1, cannot be divisible by any of the numbers from 2 to 50.

Therefore, the option e) must be the correct one.

Regards,
Kaushik

On 4/30/07, sathish vijayaraghavan <sathish_vijayaragha van@yahoo. com> wrote:
Hi All

This is yet another question which also bothers me on solving it,

For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is

a) between 2 and 10
b) between 10 and 20
c) between 20 and 30
d) between 30 and 40
e) greater than 40

Thanks,
Sathish

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• Hey Sheetal 3 things can help you... 1. Start reading A LOT. It sounds cliched... and maybe even a little too late. But it should help. Make sure that
Message 19 of 21 , Jul 17, 2007
Hey Sheetal

1. Start reading A LOT. It sounds cliched... and maybe even a little too late. But it should help. Make sure that everything you read, you do give it a deep introspection. Like what one line could summarize the whole idea, whats the tone of the author etc. You might want to have someone to verify all that you make out of the paragraph, but even if you don't you'll be just fine.
2. Practice. Take loads of "reputed" (to avoid wrong answers) material.... and keep practicing. You'll get better with time. Many a times the problem with the verbal section is that you never read the stuff properly owing to the "time running away" factor. Take your time. Read properly. Solve. And then check your answers. Dont expect this to give you a leap in your marks. You might fair just as bad. But then when you'll analyze you would understand better. And fair better in the following tests. Do not forget to learn all the new words you encounter. Surprisingly, contrary to what many of us believe, they ARE very common words.
3. Improve your reading speed. Its a byproduct of the first point so you dont really have to make a special effort for this. Just make sure theres a small sense of urgency while reading. It will help in subduing the negative effect of the "time running away" factor.
Just try. And be patient. Above all, be confident. Your mindset should be - If I am not getting to solve this question, probably there's no one else either.

All the best.
Tarun

On 6/4/07, sheetal_feb <sheetal_feb@...> wrote:

Dear All,

Sheetal here .I want suggestion as to wat to do regarding verbal
cutoff it worries me like a night mare.
All possible suggestion and advices are welcome.
Completely lost .No confidence at all.

Regards
Sheetal

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