antodaya Says

If S1 = {1, 2, 3, 4, ... , 23} and S2 = {207, 208, 209, 210, 211, ... , 691}, how many elements of the set S2 are divisible by at least four distinct prime numbers that are elements of the set S1?

is it 8 bhai



??

mani0303 Says

A hexagon with sides of length 2, 7, 2, 11, 7, 11 is inscribed in a circle. Find the radius of the circle.

I guess this has been done before. Was a typical one and answer was coming out to be R = 7 cm.

antodaya Says

If S1 = {1, 2, 3, 4, ... , 23} and S2 = {207, 208, 209, 210, 211, ... , 691}, how many elements of the set S2 are divisible by at least four distinct prime numbers that are elements of the set S1?

Only 8?

Numbers - 210,420,630,330,660,390,462,546?

antodaya Says

If S1 = {1, 2, 3, 4, ... , 23} and S2 = {207, 208, 209, 210, 211, ... , 691}, how many elements of the set S2 are divisible by at least four distinct prime numbers that are elements of the set S1?

i think 8...may be more...
(2,3,5,7)
(2,3,5,11)
(2,3,5,13)
(2,3,5,17)
(2,3,5,17)..
(2,3,5,,19)
(2,3,5,23)
(2,3,7,11)..
(2,3,7,13)...

A hexagon with sides of length 2, 7, 2, 11, 7, 11 is inscribed in a circle. Find the radius of the circle.

If S1 = {1, 2, 3, 4, ... , 23} and S2 = {207, 208, 209, 210, 211, ... , 691}, how many elements of the set S2 are divisible by at least four distinct prime numbers that are elements of the set S1?

your explanation was a bouncer bhai ;)


thoda detail mein explain kar do plz

check mudwel post below my post .. I think that should help ..

basically initially we get 580/29 .. though it simplifies to 20 , initially we dont get an integer

Bro ,

Since the given number will not be in its simplest form, we can say that GCD of 29 and (n + 5) is not 1. Since 29 is prime, n + 5 should be a multiple of 29 .

so we first 69 multiples as answer .. does it solve your query ?

your explanation was a bouncer bhai ;)


thoda detail mein explain kar do plz

Approach:

3^1024 - 1 = (3^512+1)(3^512 - 1) = (3^512 + 1)(3^256 +1)(3^256 - 1) and so on expand karne pa --

=> (3^512 + 1)(3^256 + 1)(3^128 + 1)(3^64 + 1).....(3+1)(3-1)

Here, all factors are even (Total 11 factors) and also, 2nd last term is 4. So, one more factor of 2. Thus, n should be 12.

3^1024 - 1=(2+1)^1024 - 1

expand using binomial and you will get 11 as the answer.
How come you are getting 12 as the answer??
I just glance sometimes over this thread and came across your solution.
Correct me if i am wrong.
Just count your factors again;)

Mojo Sir, I have mentioned it there that all factors are even so we get 11 powers of 2. BUT, second last term is 3+1 = 4 and here, we have TWO powers of 2 in it. => Total 12 powers of 2 in the expression.

In an isosceles triangle ABC AB = AC = 7 cm . AD produced on BC such that D is closer to B , AD= 5 cm . The sum of lengths of BD and CD , if BD and CD (in cms) are distinct integers could be

A) 11 cm B) 28 cm C) 14 cm D) 9 cm

Regards,

rafaelnadal Says

A train after travelling 50 km from A meets with an accident and proceeds at 4/5 th of it's former speed and reaches D 45 mins late.had the accident occurred 20 kms further on,it would have arrived 12 mins sooner.Find the original speed and distance.

50/x + (d-50)/(4x/5) = d/x + 45/60

70/x + (d-70)/(4x/5) = d/x + 12/60

solving dis,, we get speed=x=100/19

but i guess,,m wrong

P.S.>> Mani anna,,just have a look at dis,,wat r u getting??

This might help:)


D - 50 = (4/5)*S * (5/4)*t1

1/4*t1 = 45 mins which he has taken 180 mins for D - 50 km - >1

D - 70 = (4/5)*S * (5/4)*t2

1/4*t2 = 33 mins(not 12 mins)

t2 = 132 min,so he has taken 132 mins for D -70 km->2

From 1 and 2,we can say 20 units of distance would be traveled in 48 mins

=>25 km in 1 hr

Therefore the distance is 125km as he is traveling for 5 hrs from condition 1

@Naga,
that was the calculation mistake
but my point is that for first value it does simplify and for other values 29*2 onwards it does not

but we are considering first value too.why?

for n=24 n^2+4/n+5 = 580/29 which is definitely not in the simplest form..
we cancel 29 out and then we get 20...

hence first value 24 is included in the answer...

__//\\__ naga & mudwel bhai

when n=24

(n-5) + 29/(n+5) =20 which is in the simplest form so, should we include this

similarly, when n=2,3,4,etc the eqtn is again not in the simplest form so why arn't we including that :splat::splat:

Bro ,

Since the given number will not be in its simplest form, we can say that GCD of 29 and (n + 5) is not 1. Since 29 is prime, n + 5 should be a multiple of 29 .

so we first 69 multiples as answer .. does it solve your query ?

Triangle ABC is right angled at A . DEFG is a square of side 4 cm such that DE lies on the hypotenuse of this triangle . Find the sum of BD and EC if it is known that their lengths in cm are distinct integers greater than 1 .

A)10 B) 8 C) 6 D) 12

Regards,

Guys anybody for this one ?

if n + 5 = 29 , then n = 24

subs we get 19 + 29/29 = 20

like wise for 53,82 , etc

@Naga,
that was the calculation mistake
but my point is that for first value it does simplify and for other values 29*2 onwards it does not

but we are considering first value too.why?

(n^2 + 4)/(n+5) = n -5 + 29/(n+5)....
so 29/(n+5) shoud be in its simplest form..
we have to find integers for which n+5=29k...
that is n= 24,53....
so 24+(k-1)29 <=2009....k<=69.xx so k=69...
hence 69 terms r there...

(n^2+4)/(n+5)=(n-5) + 29/(n+5)

The fraction won't be in the simplest form when n+5 is a multiple of 29

29*1,29*2.............29*69 (2001)

Corresponding n=24,53.................1996

So 69 such values.

__//\\__ naga & mudwel bhai

when n=24

(n-5) + 29/(n+5) =20 which is in the simplest form so, should we include this

similarly, when n=2,3,4,etc the eqtn is again not in the simplest form so why arn't we including that :splat::splat:

but when n+5 =29*1=29

the fraction is in simple form

(n-5)+(29/29)=(23+1)=24

am i making any silly mistake here?

if n + 5 = 29 , then n = 24

subs we get 19 + 29/29 = 20

like wise for 53,82 , etc

sparklingaubade Says

please explain what this ques is?

The question asks values of n for which the given fraction does not simplify and remains in P/q form means fractional form

Q.For how many integers between 1 and 2009 is the improper fraction (n^2+4)/(n+5) not in simplest form?

I am getting answer as 68, but answer given is 69..Please share the approach

please explain what this ques is?

(n^2+4)/(n+5)=(n-5) + 29/(n+5)

The fraction won't be in the simplest form when n+5 is a multiple of 29

29*1,29*2.............29*69 (2001)

Corresponding n=24,53.................1996

So 69 such values.

but when n+5 =29*1=29

the fraction is in simple form

(n-5)+(29/29)=(23+1)=24

am i making any silly mistake here?

Approach:

3^1024 - 1 = (3^512+1)(3^512 - 1) = (3^512 + 1)(3^256 +1)(3^256 - 1) and so on expand karne pa --

=> (3^512 + 1)(3^256 + 1)(3^128 + 1)(3^64 + 1).....(3+1)(3-1)

Here, all factors are even (Total 11 factors) and also, 2nd last term is 4. So, one more factor of 2. Thus, n should be 12.

3^1024 - 1=(2+1)^1024 - 1

expand using binomial and you will get 11 as the answer.
How come you are getting 12 as the answer??
I just glance sometimes over this thread and came across your solution.
Correct me if i am wrong.
Just count your factors again;)

Q.For how many integers between 1 and 2009 is the improper fraction (n^2+4)/(n+5) not in simplest form?

I am getting answer as 68, but answer given is 69..Please share the approach

(n^2+4)/(n+5)=(n-5) + 29/(n+5)

The fraction won't be in the simplest form when n+5 is a multiple of 29

29*1,29*2.............29*69 (2001)

Corresponding n=24,53.................1996

So 69 such values.

Q.For how many integers between 1 and 2009 is the improper fraction (n^2+4)/(n+5) not in simplest form?

I am getting answer as 68, but answer given is 69..Please share the approach

(n^2 + 4)/(n+5) = n -5 + 29/(n+5)....
so 29/(n+5) shoud be in its simplest form..
we have to find integers for which n+5=29k...
that is n= 24,53....
so 24+(k-1)29 <=2009....k<=69.xx so k=69...
hence 69 terms r there...

invincible2910
if 15x67y is divisible by 56, find x+y
options : 8,6,4,2

Ans: 6;
For divisibility of 56, check for 8 and 7 divisibility
for divisblty of 8, y =2
and for 7 among the options only 4 fits for x; so ans is 6

Q.For how many integers between 1 and 2009 is the improper fraction (n^2+4)/(n+5) not in simplest form?

I am getting answer as 68, but answer given is 69..Please share the approach

rkshtsurana Says

chevy bhai, is it not possible that if one of terms is divisble by 4?? then it ll exceed 12..or if we take euler of 2^11 then it is 2^10..and 3 and 2^11 are coprime so 3^(2^10) = 1 mod 2^11..then it must be 11 na..

Obviously, 3^(2^10) = 1 mod 2^11
which implies that --> 3^(2^10) -1 = 0 mod 2^11.

But then this doesn't mean that 2^11 is the highest possible power available in it. If the highest possible power is 2^12, then also the number will give ZERO as remainder with 2^11 as 2^12 is divisible by 2^11.

So, here you cannot just try out Euler. If you expand it, you will get one "4" in it which will contribute one more power of 2 to the result.

Estallar12 ,upar meri approach me kya flaw he??

What is the highest possible value of n for which (3^1024) 1 is divisible by 2^n?
(a) 13 (b) 10 (c) 11 (d) 12

Approach:

3^1024 - 1 = (3^512+1)(3^512 - 1) = (3^512 + 1)(3^256 +1)(3^256 - 1) and so on expand karne pa --

=> (3^512 + 1)(3^256 + 1)(3^128 + 1)(3^64 + 1).....(3+1)(3-1)

Here, all factors are even (Total 11 factors) and also, 2nd last term is 4. So, one more factor of 2. Thus, n should be 12.

chevy bhai, is it not possible that if one of terms is divisble by 4?? then it ll exceed 12..or if we take euler of 2^11 then it is 2^10..and 3 and 2^11 are coprime so 3^(2^10) = 1 mod 2^11..then it must be 11 na..

What is the highest possible value of n for which (3^1024) 1 is divisible by 2^n?
(a) 13 (b) 10 (c) 11 (d) 12

12??

3^1024-1^1024= (3^512+1^512) ( 3^512-)1^512)
= (3^512+1^512)(3^256+1^256) (3^256-1^256)
nd keep on expanding

u'll get 12 terms all divisible by
so highest possible value of n=12

What is the highest possible value of 'n' for which (3^1024) - 1 is divisible by 2^n?
(a) 13 (b) 10 (c) 11 (d) 12

Answer is 25km/hr and 125km
i don't know how to approach.:(

(20/(4x/5))-(20/x )= 1/5
solve to get x=25km/hr
also
(d-50)/20 - (d-50)/ 25 = 3/4
solve to get d=125 km

I'm getting Distance=290 kms
Speed =80km/hr

original speed is 25km/hr
total dis. is 125km

I'm getting Distance=290 kms
Speed =80km/hr

Answer is 25km/hr and 125km
i don't know how to approach.:(

rafaelnadal Says

A train after travelling 50 km from A meets with an accident and proceeds at 4/5 th of it's former speed and reaches D 45 mins late.had the accident occurred 20 kms further on,it would have arrived 12 mins sooner.Find the original speed and distance.

I'm getting Distance=290 kms
Speed =80km/hr

q) number of zeroes at end of product of all prime nos between 1 and 1111?

A) 222 b) 21 c) cant be determined d) none of these

None f dese,, dere will be only one 0,,

as dere is only one prime number dat is multiple of 2,, i.e. 2 itself
and only one prime number dat is multiple of 5 i.e. 5 itself

2*5=10

one zero
my take: none f dese

q) number of zeroes at end of product of all prime nos between 1 and 1111?

A) 222 b) 21 c) cant be determined d) none of these

no prime no. ends in 0.
also 2 and 5 are prime nos and their product is 10.
so solution will be 1

There are 8436 balls each wid radius of 1 cm stacked in a pile with 1 ball on top,3 balls in 2nd layer,6 in 3rd layer,10 in 4th and so on,the no. of horizontal layers in the pile is
a)30
b)42
c)36
d)45

Tn= n(n+1)/2...

now sum(tn) = n(n+1)(n+2)/6 = 8436...
hence n = 36.....

rafaelnadal Says

A train after travelling 50 km from A meets with an accident and proceeds at 4/5 th of it's former speed and reaches D 45 mins late.had the accident occurred 20 kms further on,it would have arrived 12 mins sooner.Find the original speed and distance.

50/x + (d-50)/(4x/5) = d/x + 45/60

70/x + (d-70)/(4x/5) = d/x + 12/60

solving dis,, we get speed=x=100/19

but i guess,,m wrong

P.S.>> Mani anna,,just have a look at dis,,wat r u getting??

There are 8436 balls each wid radius of 1 cm stacked in a pile with 1 ball on top,3 balls in 2nd layer,6 in 3rd layer,10 in 4th and so on,the no. of horizontal layers in the pile is
a)30
b)42
c)36
d)45

if n is a natural number such that (10^12) < n < (10^13),and sum of the digits of n is 2,then number of values n can take
a) 13
b)12
c)11
d)10

I kno the ans and solution but a litlle confused
the no can be of this type also na 100000000091 = 11=1+1=2
this no also satisfy the condition
but they r not including it
help puys

this should be 12...
n can be 1 (12 digits in which there is single 1 and rest r zeoes)...
so we can have 1 in any of the 12 places...

q) number of zeroes at end of product of all prime nos between 1 and 1111?

A) 222 b) 21 c) cant be determined d) none of these

there can be only one zero...as there is only one 2 and one 5 in the product of primes..

ao none of these...

Is that 6?

Approach:

BD + DE + EC = largest value of the one of Pythagoras triplets

BD + ED = largest value of the one of Pythagoras triplets - 4

largest value of the one of Pythagoras triplets can be 10 from the options

so,BD + ED = 6

Thanks for solving it but the answer given is 10 . Even i dont know the way to solve this .

q) number of zeroes at end of product of all prime nos between 1 and 1111?

A) 222 b) 21 c) cant be determined d) none of these

Triangle ABC is right angled at A . DEFG is a square of side 4 cm such that DE lies on the hypotenuse of this triangle . Find the sum of BD and EC if it is known that their lengths in cm are distinct integers greater than 1 .

A)10 B) 8 C) 6 D) 12

Regards,

Is that 6?

Approach:

BD + DE + EC = largest value of the one of Pythagoras triplets

BD + ED = largest value of the one of Pythagoras triplets - 4

largest value of the one of Pythagoras triplets can be 10 from the options

so,BD + ED = 6

if n is a natural number such that (10^12) < n < (10^13),and sum of the digits of n is 2,then number of values n can take
a) 13
b)12
c)11
d)10

I kno the ans and solution but a litlle confused
the no can be of this type also na 100000000091 = 11=1+1=2
this no also satisfy the condition
but they r not including it
help puys

A train after travelling 50 km from A meets with an accident and proceeds at 4/5 th of it's former speed and reaches D 45 mins late.had the accident occurred 20 kms further on,it would have arrived 12 mins sooner.Find the original speed and distance.

Triangle ABC is right angled at A . DEFG is a square of side 4 cm such that DE lies on the hypotenuse of this triangle . Find the sum of BD and EC if it is known that their lengths in cm are distinct integers greater than 1 .

A)10 B) 8 C) 6 D) 12

Regards,

If a, b, c are positive real numbers such that a^2 + b^2 + c^2 = 48, then the maximum value of a+b+c is
(1) 9 (2) 16 (3) 12 (4) 24

Maximum occurs when a = b = c as the expression is symmetric

so, a + b + c = 12

gajab solution.
but is there any way u can solve it using alligation??

Here you go,


100 --------------------------------- 200

------------------ 110 --------------


90 ----------------------------------- 10

Ratio of matches he played before and after an average score of 100 is 9:1

he played 5 matches after,the number of matches is 50

Homer1 Says

but how can u decide that the c.d for the AP IS 1??

Because if you increased the difference,the number of matches played would be decreased..so we wouldn't have got the maximum;)" />

sachin had an average score of 100 in some matches.In the next 5 matches his runs follow an AP and max he scores is 202 runs.If his score in no 2 matches is same and the average at the end of those 5 matches is 110,what is the max number of matches that he has played.
a)25
b)55
c)50

let no. of matches be x...
so his total is 100x...
now in bext 5 matches he makes a-2d,a-d,a,a+d,a+2d...
so total in x+5 matches is 100x + 5a...
(100x+5a)/(x+5)=110...
100x + 5a = 110x + 550...
a=2x + 110....
x=(a-110)/2....now for max value of x we have to find max a...
a can be max 200 becoz his runs can be 198,199,200,201,202....
so x = 200-110/2 = 45...
so no. of matches sachin played = 45 + 5=50...

yeah made a mistake , have edited my post

thanks to culdip and mani303

If a, b, c are positive real numbers such that a^2 + b^2 + c^2 = 48, then the maximum value of a+b+c is
(1) 9 (2) 16 (3) 12 (4) 24

Is it 12???

gajab solution.
but is there any way u can solve it using alligation??

but how can u decide that the c.d for the AP IS 1??" />

50?

Let the number of matches played be m

=>110*(m+5) - 100m = 202 + 201 + 200 + 199 + 198 = 1000

=>10m + 550 = 1000

=> m = 45

Total matches played,m+5 = 50




Typo,bro;)

BTW,Thanks for pointing out:)

gajab solution.
but is there any way u can solve it using alligation??" />

yeah made a mistake , have edited my post

thanks to culdip and mani303

If a, b, c are positive real numbers such that a^2 + b^2 + c^2 = 48, then the maximum value of a+b+c is
(1) 9 (2) 16 (3) 12 (4) 24

12 hai kya?

maximum value of a+b+c will be when all three are equal
a=b=c
so 3a^2 = 48
a^2=16
a=b=c=4

4+4+4 = 12???

yeah made a mistake , have edited my post

thanks to culdip and mani303

If a, b, c are positive real numbers such that a^2 + b^2 + c^2 = 48, then the maximum value of a+b+c is
(1) 9 (2) 16 (3) 12 (4) 24

using cauchy schwartz...

(a+b+c)^2 <= (a^2+b^2+c^2)(1^2 + 1^2 + 1^2)...
(a+b+c)^2 <= 48 x 3 <=144...
a+b+c <= 12......hence option 3...

50?

Let the number of matches played be m

=>110*(m+5) - 100m = 202 + 201 + 200 + 199 + 198 = 1000

=>10m + 550 = 1000

=> m = 45

Total matches played=45+5=50
gajab solution.:wow:
but is there any way u can solve it using alligation??

A 50-digit number containing only 7's when divided by 74, the remainder is...??

74=2*37
77777...50 times/2 = 2a+1

nd we know any number repeated 6 times is completely divisible by 37

777..(48 times)is compeletely divisible by 7,,
we r left wid 77,,
77/37 = 37b+3
2a+1=37b+3

remainder will be 3

Homer1 Says

answer is 50

yeah made a mistake , have edited my post

thanks to culdip and mani303

If a, b, c are positive real numbers such that a^2 + b^2 + c^2 = 48, then the maximum value of a+b+c is
(1) 9 (2) 16 (3) 12 (4) 24