thanks

could u tell abt the eulers theorem

euler's theorem
example:
51^8/7
now 51 and 7 are co-primes to each other.
and 7 = 7 *1
therefore, 7 * ( 1- 1/7) = 6
according to euler's theorem 51^6/7 will give remainder as 1.
therefore, (51^6 * 50^2)/7
= 51^2/7 as 51^6 gives 1 as remainder.
= 2*2/7
= 4 is remainder.
this is euler's theorem.
hope u understand.

mbafrmfms Says

Find the maximum value of the function 1/(x^2-3x+2). Ans - N.O.T

1/(x-3x+2)= 1/(x-2)(x-1)
Now at x=2 and x=1 the value reaches infinity hence a maximum value cannot be found

The domain of definition of y=sqrt{log(5x-x^2/4} Ans:

For the function log x, x>0
Hence for this case (5x-x)/4>0
Also for the square root function to be defined the value present inside the root should be positive.
hence log{(5x-x)/4}>0
=> (5x-x)/4>1 ; because log function is always positive when it is taken >1 (You can see the log graph for this)
Hence x-5x+4<0
=>Value ranges between 1 and 4 (both included)

f(x).f(1/x)=f(x)+f(1/x) and f(4)=65 ........find f(6) Ans:217

Substituting x=4 we get the value of f(1/4)=65/64
Also for x=1, f(1)=0 or 2
Now check the pattern
f(4)= 4^3+1
f(1/4)=(1/4)^3+1
f(1)= 1^3+1 (The other value is not considered as it does not follow the same pattern)
Hence,
f(x)= x^3+1
So f(6)= 216+1=217

At what integral value of x will the function (((x^2+3x+1))/((x^2-3x+1))) Ans:-3

I think the question is incomplete

50^51^52 / 11
by euler's theorem,
50^10 when divided by 11 will give a remainder of 1.
therefore, 50^10k shod give 1 as remainder.
so, we need to get the expression in the form of 50^(10k+r).
thats why we divide 51^52 by 10 and find remainder to be 1.
therefore, it is of the form 50^(10k+1)/11
= (50^10k * 50^1) / 11
= remainder(6) as 50/11 gives 6 as remainder and 50^10k gives 1 as remainder.
so, the answer is 6.

thanks

could u tell abt the eulers theorem

50^51^52 / 11
by euler's theorem,
50^10 when divided by 11 will give a remainder of 1.
therefore, 50^10k shod give 1 as remainder.
so, we need to get the expression in the form of 50^(10k+r).
thats why we divide 51^52 by 10 and find remainder to be 1.
therefore, it is of the form 50^(10k+1)/11
= (50^10k * 50^1) / 11
= remainder(6) as 50/11 gives 6 as remainder and 50^10k gives 1 as remainder.
so, the answer is 6.

couldnt get kindly elaborate

whats the remainder of
50^51^52 when divided by 11

p.s.-plz elaborate the approach

50^51^52 / 11
by euler's theorem,
50^10 when divided by 11 will give a remainder of 1.
therefore, 50^10k shod give 1 as remainder.
so, we need to get the expression in the form of 50^(10k+r).
thats why we divide 51^52 by 10 and find remainder to be 1.
therefore, it is of the form 50^(10k+1)/11
= (50^10k * 50^1) / 11
= remainder(6) as 50/11 gives 6 as remainder and 50^10k gives 1 as remainder.
so, the answer is 6.

guys according to arun sharma's book... what is the reminder of 18/4??

is it 2 or 1??

it is 2 and not 1; go by the basic remainder theorem, rather divide the conventional way you will get the answer

:thumbsup:

guys according to arun sharma's book... what is the reminder of 18/4??

is it 2 or 1??

its 2
when we divide the num and denm by 2 thn the remainder is multiplied by 2(1*2)

whats the remainder of
50^51^52 when divided by 11

p.s.-plz elaborate the approach

hi everybody

i m little confused in arun sharmas quant question frm number system page - 20 ,ques-15

the greatest num which will divide 4003, 4126, 4249 leaving the same remainder.
ans given in book is 41

isnt it shud be the hcf of abs val of ( 4003-4126) (4126 - 4249 ) (4003 -4249 )....
which is 123.
plz help

Given ans is ryt...req no is Hcf of diff b/w ny two nos and not 3 as u did...:-)
No...it

guys according to arun sharma's book... what is the reminder of 18/4??

is it 2 or 1??

hi everybody

i m little confused in arun sharmas quant question frm number system page - 20 ,ques-15

the greatest num which will divide 4003, 4126, 4249 leaving the same remainder.
ans given in book is 41

isnt it shud be the hcf of abs val of ( 4003-4126) (4126 - 4249 ) (4003 -4249 )....
which is 123.
plz help

you are perfectly correct, it should be 123:cheerio:

hi everybody

i m little confused in arun sharmas quant question frm number system page - 20 ,ques-15

the greatest num which will divide 4003, 4126, 4249 leaving the same remainder.
ans given in book is 41

isnt it shud be the hcf of abs val of ( 4003-4126) (4126 - 4249 ) (4003 -4249 )....
which is 123.
plz help

I agree with everything you have done, except the part where you say that the qsns appear in three papers. The question reads ' .. question appears in some other paper' and not papers. Therefore, it can't be three papers.Please do correct me if i am wrong .

:)

Thanks!

My take on this one:

Answer: 1062

Explanation:
Number of unique questions = 35*15. This is very easy to understand.

Now, 15 questions remain for each set.
To get maximum number of questions, each repeated question should be shared by only 2 papers.
However, we have 15*25 questions. So, we have off number of questions. So, one question needs to be shared by 3 papers.

So, number of remaining questions = (25*15 - 1) / 2 = 187

Possible arrangement:
Say, first 24 papers form groups of two.
5 groups i.e. 10 papers share all 15 questions.
Remaining 7 groups i.e 14 papers share 14 questions. And each paper shares remaining question with 25th paper, ging 25th paper 14 questions. 15th question of 15th paper will be the only question which is repeated for 3rd time and can be taken form any repeated question.

my answer is also 1062 but if look at the option given in the book 1055 is max questions possible.

LOD2 TIME,SPEED AND DISTANCE QUESTION

33)
ravi,who lives in the countryside,caught a train for home earlier than usual yesterday.his wife normally drive to the station to meet him.but yesterday he set out from the station to meet his wife on the way.he reached home 12 minutes earlier than he would have done had he waited at the station for his wife.the car travels at the uniform speed,which is five times the ravis speed on the foot.ravi reached home at exactly 6"0 clock.at what time would he have reached home if his wife ,forewarned of his plan,had met him at the station?

a)5:48 b)5:24 c)5:00 d)5:36

Answer: (d) 5:36

Explanation:
Car travels 5 times faster than Ravi.
Say, Ravi started walking and met car at point A.
If Ravi has stayed at station, he had taken 12 more mins to reach home. This extra time would have been for car coming to station from point and back to point A again. So, from point to station takes 12/2 = 6 mins for car.
For Ravi it will take 6*5 = 30 mins.
It means Ravi walked for 30 mins.

Now, if wife had already been present at station, Ravi would not have walked for 30 mins. That time they would have covered by car in 6 mins. Thus, they would have saved 24 mins and reached at 5:36.

Now someone solve my doubt

Q no. 42 pg 50

Suppose the product of n consecutive integers is x.

(x+1)(x+2)(x+3)(x+n-1)=1000

Then which of the following canNOT be true abt n

1.no of terms can be 16
2.no of terms can be 5
3.no of terms can be 25
4.no of terms can be 20


Yaar in this sum tell me how can the product of n consecutive numbers equal 1000?
1000 has only 2 factors 2 and 5 ..now n consecutive numbers will have other factors as well. Eg 3,4,5,6,7 has to have 3 as a factor..

So thoda gyan on this one please.

Are u sure this is the question ?
(x+1),(x+2) & (x+3) are three consecutive integers and we cannot have three consecutive factors of 1000 . At best we can have 2*4*5*25. But certainly that does not help much !

One more...and this time I am sure Arun Sharma is gonna be on the backfoot :p


Q45 Lod 3 pg 51

Twenty five sets of problems for 25 papers. one set has 50 q. out of 50, 35 are unique and 15 are repeated in some other paper.What is the max possibl no of qs

My answer-1055
35*25 + 24*15


any one for this one?


Regards
Meet_the_man

My take on this one:

Answer: 1062

Explanation:
Number of unique questions = 35*15. This is very easy to understand.

Now, 15 questions remain for each set.
To get maximum number of questions, each repeated question should be shared by only 2 papers.
However, we have 15*25 questions. So, we have off number of questions. So, one question needs to be shared by 3 papers.

So, number of remaining questions = (25*15 - 1) / 2 = 187

Possible arrangement:
Say, first 24 papers form groups of two.
5 groups i.e. 10 papers share all 15 questions.
Remaining 7 groups i.e 14 papers share 14 questions. And each paper shares remaining question with 25th paper, ging 25th paper 14 questions. 15th question of 15th paper will be the only question which is repeated for 3rd time and can be taken form any repeated question.

LOD2 TIME,SPEED AND DISTANCE QUESTION

33)
ravi,who lives in the countryside,caught a train for home earlier than usual yesterday.his wife normally drive to the station to meet him.but yesterday he set out from the station to meet his wife on the way.he reached home 12 minutes earlier than he would have done had he waited at the station for his wife.the car travels at the uniform speed,which is five times the ravis speed on the foot.ravi reached home at exactly 6"0 clock.at what time would he have reached home if his wife ,forewarned of his plan,had met him at the station?

a)5:48 b)5:24 c)5:00 d)5:36

Kudos!!! to u my frend for bringing up this thread ........

now itll b really helpful if u can solve one of doubts ....:)
im having some probs in solving the inequalities chap of the book
can u give a detailed solution for Q.13 in tat chap for LOD1.

cheers:

--------------
Get busy living or get busy dying.

HI guys...
Its a good idea to start such a thread bcoz dis book has very good questions but very poor explanations...I am in Block1 wherein i find certain questions wrng....anyone to help me on dis????please

One more...and this time I am sure Arun Sharma is gonna be on the backfoot :p


Q45 Lod 3 pg 51

Twenty five sets of problems for 25 papers. one set has 50 q. out of 50, 35 are unique and 15 are repeated in some other paper.What is the max possibl no of qs

My answer-1055
35*25 + 24*15


any one for this one?


Regards
Meet_the_man

Now someone solve my doubt

Q no. 42 pg 50

Suppose the product of n consecutive integers is x.

(x+1)(x+2)(x+3)(x+n-1)=1000

Then which of the following canNOT be true abt n

1.no of terms can be 16
2.no of terms can be 5
3.no of terms can be 25
4.no of terms can be 20


Yaar in this sum tell me how can the product of n consecutive numbers equal 1000?
1000 has only 2 factors 2 and 5 ..now n consecutive numbers will have other factors as well. Eg 3,4,5,6,7 has to have 3 as a factor..

So thoda gyan on this one please.

N = 32*101*10001*100000001*...*100000..1 (31 zeors)
= 32(10^2 + 1)(10^4 + 1)(10^8 + 1)(10^16 + 1)(10^32 + 1)
= 32(10^62 + 10^60 + .... + 10^2 + 1)
= 3232323232....32 (32 times 32)

=> Sum of digits will be (3 + 2)*32 = 160

One more way of solving this could be

N= 32*101*10001*100000001*...*100000..1 (31 zeors)

32*(101010101010......1-----63 times)

(we have 63/2 +1 =32 ones in there)

=323232323232.......(each one is replaced by 32)

sum contributed by of one pair of 32= 3+2 =5

In total we have 32 pairs

therefore-32*5 =160:cheerio:

Regards
meet_the_man

Hi friends!!!!!!!!

This question is from number system......

LOD 3

Q) N = 202 X 20002 X 200000002 x 20000000000000002 x 2000000000000........2( 31 zeros) the sum of digits in this multiplication will be:

a) 112 b) 160 c)144 d) cannot be determined


need help...........

N = 32*101*10001*100000001*...*100000..1 (31 zeors)
= 32(10^2 + 1)(10^4 + 1)(10^8 + 1)(10^16 + 1)(10^32 + 1)
= 32(10^62 + 10^60 + .... + 10^2 + 1)
= 3232323232....32 (32 times 32)

=> Sum of digits will be (3 + 2)*32 = 160

done.. please let me know if there is any other efficient approach

Accroding to my calculations answer is B - 160

Details

202 * 20002 = 4040404
4040404 * 200000002 = 808080808080808

with above two calculations we can lead to two observations

1. The answer of multiplication is in multiple of 2
2. The non zero part in answer appears one time more than number of zeros in the 20002 format numbers.

so, with this observations we can conclude that for this problem multiplication result would be

32 times 32 as there are 31 zeros

now solution

(3+2)*32 = 160

trying to solve in ...

Hi friends!!!!!!!!

This question is from number system......

LOD 3

Q) N = 202 X 20002 X 200000002 x 20000000000000002 x 2000000000000........2( 31 zeros) the sum of digits in this multiplication will be:

a) 112 b) 160 c)144 d) cannot be determined


need help...........

Guys Incase Any Of You Are Going To Buy New Book For Quants By Arun Sharma, Wait For A While Because A Fourth Edition Release Is Round THe COrner It Should Be Availaible Any Time Soon.....
- - McGraw-Hill Education

Tc....
CHeerz Good Luck...

Hi Puys !!
Dis is a question from TSD chapter..
LOD 2

goes lik dis.."The JK train travels from delhi to srinagar was delayed by snowfall for 16 minutes and made up for the delay on a section of 80 km travelling with a speed 10 km per hour higher than its normal speed.Find the original speed of the train ?"
a)60 b)66.66 c)50 d)40 e)none

How to go abt it ?

SIMPLE BRO!!!!
original speed is given as 50
Take----80/50=1.6 hrs comes upto 1hr 36 mints
speed increases by 10km
here total time he takes= 16+ 80/60= 16 MIN+ 1hr 20 min= 1 hr 36 mins

Therefore 50(answer)

A sum of money placed at compound interest doubles itself in 3 yrs.In how many yrs will it amount to 8 times itself?


Is there any shortcut method for these types of questions which does not involve finding out the rate of interest.

2^n=8
n=3
ans=3*3=9 years
formula a principal amounts to x times in t years c.i. in how many years will it becomes y times.
years=t*n
where x^n=y:thumbsup:

please post the answer of this ques with sol. as i m finding answer wrong in book
Q In 4 throws with a pair of dices , what is probab. of throwing a double twice?
thanx in advance

A sum of money placed at compound interest doubles itself in 3 yrs.In how many yrs will it amount to 8 times itself?


Is there any shortcut method for these types of questions which does not involve finding out the rate of interest.

Here are some of the Qs from our dear arun sharma's book which baffle me either because of wrong answers or my lallugiri!

1.It takes the pendulum of a clock 7 seconds to strike 4 o'clock.How much time will it take to strike 11 o' clock(Q18 pg 84)
----My ans is 19.25 but the book says it's 23.33.

At 4 o'clock, the 1st strike is when the second hand strikes 12, the next 3 strikes take 7 seconds. So each strike takes 7/3 = 2.33 seconds

So at 11 o'clock, the 1st strike would be when the second strikes 12, the next 10 strikes would take 10*2.33 = 23.33 seconds

4.an intelligence agency decides on code of 2 digits.It is put on a a slip which can be read upside down as well. How many code are there such that there is ambiguity( eg 91 can be read as 16 but 45 can not be read as 54) Q8 pg 83 My ans:=75

According to me, the following digits can be read upside down:
0
1
2 (can be read as 5)
5
6
8
9

So total 7 digits can be read upside down and considering the first digit of the code can also be 0, the total 'ambiguous' combinations can be 7*6 = 42, i.e. each of the 7 digits can be paired by the other 6 digits so as to create an ambiguity.

5. How many three digit numbers have the property that the digits taken from left to right form an Arthmetic or a GP.(Q41 Pg73)
My Ans:56 The book says 18

In the case of AP, numbers can be:
123, 234..... 789 = 7
135, 246, 579 = 5
147, 258, 369 = 3
159 = 1

In case of GP, there can be the following numbers:
124, 248 = 2
139 = 1

Total numbers = 19, the answer in the book has to be wrong, until I have missed a very obvious step

6. If a man saves 1000 each year and invests at the end of the year at 5% compound interest ,how much will the amount be at the end of 15 years?
I think it should be 1000*(1+1.05+1.05^2 + 1.05^3..+1.05^14)
But not sure about this one.

Your answer is correct

Here are some of the Qs from our dear arun sharma's book which baffle me either because of wrong answers or my lallugiri!

1.It takes the pendulum of a clock 7 seconds to strike 4 o'clock.How much time will it take to strike 11 o' clock(Q18 pg 84)
----My ans is 19.25 but the book says it's 23.33.

2.A young girl counted in the following way on the fingers of her left hand....she ended on her.(Q10 pg 83)
---I think it is b> index finger- just want to confirm my answer..sorry for being lazy here but I was just wanted to confirm if my ans is right or wrong.

3.Is it possible to find distinct Positive intergers x,y such that (x+y)/xy is a positive integer?(Q9 pg 83)
- My ans:No
Explanation:
x+y>=xy only when (x or y is a 1--> then xy and x+y differ by 1 making them co-prime, hence indivisible) or (x=2 and y=2--> not acceptable as we need distinct integers)

4.an intelligence agency decides on code of 2 digits.It is put on a a slip which can be read upside down as well. How many code are there such that there is ambiguity( eg 91 can be read as 16 but 45 can not be read as 54) Q8 pg 83 My ans:=75


5. How many three digit numbers have the property that the digits taken from left to right form an Arthmetic or a GP.(Q41 Pg73)
My Ans:56 The book says 18

6. If a man saves 1000 each year and invests at the end of the year at 5% compound interest ,how much will the amount be at the end of 15 years?
I think it should be 1000*(1+1.05+1.05^2 + 1.05^3..+1.05^14)
But not sure about this one.


A BIG thank you for helping!

-meet_the_man

Find the maximum value of the function 1/(x^2-3x+2). Ans - N.O.T

The domain of definition of y=sqrt{log(5x-x^2/4} Ans:

f(x).f(1/x)=f(x)+f(1/x) and f(4)=65 ........find f(6) Ans:217

At what integral value of x will the function (((x^2+3x+1))/((x^2-3x+1))) Ans:-3

What would be the sum of the series 1/1, 1/3, 1/6, 1/10, 1/15,...

I understand that the difference between the denominators of consecutive terms is in A.P. But I dont know how to go about solving such series. Could anyone explain?

the general term of the series is 2/[n(n+1)] , which can be written as :
2[1/n - 1/(n+1)]

now to get the sum, we can put successive values of n ( i.e., n = 1, 2, 3, ...)
so the series will be like :
2[ (1-1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + ... ]
removing all the brackets, we get the answer as 2 * 1 = 2

What would be the sum of the series 1/1, 1/3, 1/6, 1/10, 1/15,...

I understand that the difference between the denominators of consecutive terms is in A.P. But I dont know how to go about solving such series. Could anyone explain?

The best way is to go by options..

if his original speed is 50, he will take..80/50=1.6 hours, i.e 1hr 36 minutes..

now his speed increases by 10km..here total time he takes= 16min(due to delay) + 80/60hr= 16 min + 1hr 20 min= 1 hr 36 mins

hence 50 km is the answer

Should we not form a quadratic equation and then check for the options?

Hi Puys !!
Dis is a question from TSD chapter..
LOD 2

goes lik dis.."The JK train travels from delhi to srinagar was delayed by snowfall for 16 minutes and made up for the delay on a section of 80 km travelling with a speed 10 km per hour higher than its normal speed.Find the original speed of the train ?"
a)60 b)66.66 c)50 d)40 e)none

How to go abt it ?

The best way is to go by options..

if his original speed is 50, he will take..80/50=1.6 hours, i.e 1hr 36 minutes..

now his speed increases by 10km..here total time he takes= 16min(due to delay) + 80/60hr= 16 min + 1hr 20 min= 1 hr 36 mins

hence 50 km is the answer

at burger king- a famous fast food center on main street in Pune,burgers r made only on an automatic burger making machine.d machine conntinuously makes different sorts of burger by adding different sorts of filling. d machine makes burger at d rate of 1 burgers per half minute.the various fillings are added in d following manner. the 1st,5th,9th.... burgers r filled with chicken patty; d 2nd,9th,16th.... burgers with vegetable patty ; d 1st,5th,9th.... burgers with mashroom patty and the rest with plain cheese and tomato fillings.
d machine makes exactly 660 burgers per day

1) how many burgers] per day are made with cheee n tomato fillings
a) 424
b) 236
c) 237
d) none of these[/QUOTE



..srry..being first time here..just now posted a blank reply
..nyways..answer goes this way..first ap(chicken) has cd as 4, it is the same as the third ap(mushroom)..2nd ap n first has common terms(the problem basically involves number theory tactics as well), we first need to take care of the common terms.. for first ap 1+164*4=657 i.e; this has 165 terms..it is 1,5,9,13,17,21,25,29,33,37....u see 9 and 37 along with other terms which are equivalent to some multiple of 7*4+9(even 9 can be written as 0*7*4+9) are common in this series with 2nd series..so while calculating the number of terms in 2nd ap, we will subtract this number of common terms from the final sum..i have 9+4*7*23 i.e; 24 such terms common in 1st n 2nd aps..and 2nd ap has 2+7*94 i.e; 95 terms..so in total i have 165+95-24 terms i.e; burgers without cheese n tomato..now u do the rest..

mbafrmfms Says

Sleeping beauties w8ing for ur reply

Please post the questions. I dont carry the book to my office..!!

Hi Mr/Ms,

Kindly check ur language before posting or dont post when ur :drinking:.
Puys on this thread are not being paid to post solutions to ur questions.

You should be thankful and hence respectful to the people helping you in this journey, and this shall be evident in your posts.

Else Mods will better take ur care

Reply for the query ............oly I dint mean that jus was restless.........No personal comments plzz u shld refrain from ur do's which u r doin and plzz don waste my tyme if u have answers plz pour dem else kindly stay away

My Problems My way:

Chapter:Progressions

LOD1: Q23,Q29

LOD2: Q16,Q20,Q44

LOD3: Q1,Q7,Q15,Q18,Q19


N many more to come :P

mbafrmfms Says

Sleeping beauties w8ing for ur reply

Hi Mr/Ms,

Kindly check ur language before posting or dont post when ur :drinking:.
Puys on this thread are not being paid to post solutions to ur questions.

You should be thankful and hence respectful to the people helping you in this journey, and this shall be evident in your posts.

Else Mods will better take ur care

My Problems My way:

Chapter:Progressions

LOD1: Q23,Q29

LOD2: Q16,Q20,Q44

LOD3: Q1,Q7,Q15,Q18,Q19


N many more to come :P

Sleeping beauties w8ing for ur reply

My Problems My way:

Chapter:Progressions

LOD1: Q23,Q29

LOD2: Q16,Q20,Q44

LOD3: Q1,Q7,Q15,Q18,Q19


N many more to come

soln 33
We need to find the general term of the series.

On analysis of the series u will find, it can be written as
2/(n(n+1))
= 2

so the series can be rewritten as
2

so finally the sum will be
2 ----------------------------------------

here n=infinity so n+1 also = infinity

and 1/infinity ->0

so the answer will be 2


soln 35
the series can be rewritten as
1/2
which is same as the above series
so the general term will be
1/n*(n+1)

the last term of the series is 1/182
so
1/n(n+1)=1/182

on solving n=13

from
the sum of the series would be
1/2 * 2
=1-1/(n+1)

put n=13

=13/14

Thx for d solution mate I was desperately w8ing for the solution jus one more doubt which I feel for me is that how to approach to general term like there are not many questions to approach to general term in Arun sharma so I take lot of time approaching general term by hit n trial..

Two requests:
1.How to approach to general term after seeing the series
2.Source for some more questions like dat?

Regs

hiiiiiiiiii
i hve 1 prob. also......
if cosz=2 thn wth the value of z??
solve it ........

bhai cos ka graph bana le, 2 kahan se le aooge cos ke liye,

Chapter 2 :Progressions

LOD-2

Q.33
Q.35

soln 33
We need to find the general term of the series.

On analysis of the series u will find, it can be written as
2/(n(n+1))
= 2

so the series can be rewritten as
2

so finally the sum will be
2 ----------------------------------------

here n=infinity so n+1 also = infinity

and 1/infinity ->0

so the answer will be 2


soln 35
the series can be rewritten as
1/2
which is same as the above series
so the general term will be
1/n*(n+1)

the last term of the series is 1/182
so
1/n(n+1)=1/182

on solving n=13

from
the sum of the series would be
1/2 * 2
=1-1/(n+1)

put n=13

=13/14

Chapter 2 :Progressions

LOD-2

Q.33
Q.35

hiiiiiiiiii
i hve 1 prob. also......
if cosz=2 thn wth the value of z??
solve it ........

Cosine of something will always be between -1 and 1. CosZ=2 is not possible.



PS:- Please don't spam!

hiiiiiiiiii
i hve 1 prob. also......
if cosz=2 thn wth the value of z??
solve it ........

i think Arun Sharma is best for DI as well

hi can u suggest me a good book for preparing DI. am very weak in that... i wanna learn from basics...

Hi karthick

consider the terms in g.p to be a/r,a,ar

so the sum a/r + a + ar=14
=> r+1/r = 14/a - 1 ------------------

now sum of the squares
(a/r)^2 + a^2 + (ar)^2= 84
=>r^2+(1/r)^2= 84/a^2 - 1----------------------

now (a+b)^2= a^2 + b^2 +2ab
so a^2+b^2 = (a+b)^2 - 2ab

applying this formula to we get
(r+1/r)^2 -2=84/a^2 -1

from
(14/a -1)^2 - 2 =84/a^2 -1

now expand and solve the equation
u will get a=4
and r=2 or 1/2

a/r=2 or 8
a=4
ar=8 or 2

so the answer is 8

btw
My name is vivek sinha
and i am a software engineer in cognizant kolkata.

wow.. thats really cool.. was dying how to solve after getting he eqns..
btw am karthik working in LNT ryt now staying in vizag.. nice to meet you.. thank you so much for ur answers..

karthick vs Says

sum of three terms of a GP is 14 and the sum of their squares is 84. wat is the biggest term of the series??

Hi karthick

consider the terms in g.p to be a/r,a,ar

so the sum a/r + a + ar=14
=> r+1/r = 14/a - 1 ------------------

now sum of the squares
(a/r)^2 + a^2 + (ar)^2= 84
=>r^2+(1/r)^2= 84/a^2 - 1----------------------

now (a+b)^2= a^2 + b^2 +2ab
so a^2+b^2 = (a+b)^2 - 2ab

applying this formula to we get
(r+1/r)^2 -2=84/a^2 -1

from
(14/a -1)^2 - 2 =84/a^2 -1

now expand and solve the equation
u will get a=4
and r=2 or 1/2

a/r=2 or 8
a=4
ar=8 or 2

so the answer is 8

btw
My name is vivek sinha
and i am a software engineer in cognizant kolkata.

Ok so you can find the area of each rectangular plank, hts are 6,12,18..24 terms, but since area =l*width and the width is a constant it's easier to find the summation of the lengths and then multiply with the width
summation(6,12,18..24 terms) is na + /2
that gives you 75 * 24inches or 150 feet
area will be 150 * (9/12)feet or 112.5 sq feet

hey thanks.. ve got one more doubt.. solve for me if possible..
qn goes like this.. its all from arun sharma...

sum of three terms of a GP is 14 and the sum of their squares is 84. wat is the biggest term of the series??

sum of three terms of a GP is 14 and the sum of their squares is 84. wat is the biggest term of the series??

hey yeah.. u wer exactly ryt.. but shudn multiply by 81/144.. shud multiply by 9/144....!!!! thanks a lot... i ve one more qn.. will post now.. wru by the way and wat u doing..

thanks vivek...
lemme tell u wat i tried actually.. if possible expalin wats wrong in that..
i used the same AP formula.. but did the entire pbm without converting to feet. after finding the area in inches i converted to feet but the ans was not matching.. anything wrong in doing so?

:

I think, for converting, rather than multiplying the answer with 1/144 u were multiplying it with 1/12 ( since the answer is in square inches (1/12)^2 should be multiplied). Even i did the same for the first time

Let me know if u were doing this.