Hey is this the Sharma tread for 2010??
stigmata_007 SaysHey is this the Sharma tread for 2010??
It's the thread to discuss questions from Arun Sharma, not necessarily for this year only. You can post whatever queries you have here.
stigmata_007 SaysHey is this the Sharma tread for 2010??
this thread has been going on since 07-06-2007 and is still going strong !!!
Hi,
Please help solve the following:
1.What is the remainder when 128^1000 is divided by 153?
(a) 103 (b) 145 (c)118 (d)52
2.What is the remainder when 50^51^52 is divided by 11?
(a) 6 (b) 4 (c)7 (d)3
3.A young girl counted in the follwoing way on the finger of her left hand.She started calling the thumb 1,the index finger 2,middle finger 3,ring finger 4,little finger 5,then reversed direction,calling the ring finger 6,middle finger 7,index finger 8,thumb 9,then back to the index finger for 10,middle finger for 11 and so on.She counted upto 1994.She ended on her
(a) thumb (b) index finger (c) middle finger (d)ring finger
4.A five digit number is formed using digits 1,3,5,7 and 9 without repeating any one of them.What is the sum of all such possible numbers?
(a) 6666600 (b) 6666660 (c) 6666666 (d)None
5.It takes the pendulum of a clock 7 seconds to strike 4 0'clock.How much time will it take to strike 11 o'clock?
(a) 18 seconds (b) 20 sec (c) 19.25 sec (d)23.33 sec
6.P and Q are two integers such that 1:1 PQ=8.Which of the following cannot be the value of P+Q?
(a) 20 (b) 65 (c) 16 (d)35
7.A,B ,C are defined as follows:
A=(2.000004) / [(2.000004)^2 + (4.000008
B=(3.000003) / [(3.000003)^2 + (9.000009)
C=(4.000002) / [(4.000002)^2 + (8.000004)
(a)All of them lie between .18 and .20
(b)A is twice C
(c)C is the smallest
(d)B is the smallest
8.The infinite sum 1+4/7 + 9/7^2+16/7^3+25/7^4+...........
(a) 27/14 (b)29/13 (c) 49/27 (d)256/147
9.In acertain examinations paper,there are n questions.For j=1,2,3.............n there are 2^(n-1) students who answered j or more questions wrongly.If the total number of wrong answers is 4095,then the value of n is:
(a) 12 (b)11 (c) 10 (d)9
Hi,
Please help solve the following:
3.A young girl counted in the follwoing way on the finger of her left hand.She started calling the thumb 1,the index finger 2,middle finger 3,ring finger 4,little finger 5,then reversed direction,calling the ring finger 6,middle finger 7,index finger 8,thumb 9,then back to the index finger for 10,middle finger for 11 and so on.She counted upto 1994.She ended on her
(a) thumb (b) index finger (c) middle finger (d)ring finger
Now the thumb is 1 and then again it's 9, then it's 17 and so on. So it's an AP with a = 1 and d = 8. So after 1, we still need to count 1993 more to get to 1994. Since 1992 is divisible by 8, we would reach the count of 1993 (1 + 1992) at the thumb. Thus, 1994 would end up on her index finger.
5.It takes the pendulum of a clock 7 seconds to strike 4 0'clock.How much time will it take to strike 11 o'clock?
(a) 18 seconds (b) 20 sec (c) 19.25 sec (d)23.33 sec
The pendulum strikes first at the 0th second and the next three strikes take 7 seconds. Thus, time taken for each strike, 7/3 sec.
Now at 11 o'clock, the first strike will be at the 0th second. Next 10 strikes would take 7/3*10 seconds = 23.33 sec
i cant understand in your answer wher 15% passed in phy & 5% in chemy. will u explain me?
yathish Saysi cant understand in your answer wher 15% passed in phy & 5% in chemy. will u explain me?
ehen we get 65% as passed both in p & c applying n(AUB)=n(A)+n(B)-n(A()B)
so in physics the pass% is 80 so in only physics the pass% will be=80-65=15
so also in chem =70-65=5
Hi,
Please help solve the following:
2.What is the remainder when 50^51^52 is divided by 11?
(a) 6 (b) 4 (c)7 (d)3
4.A five digit number is formed using digits 1,3,5,7 and 9 without repeating any one of them.What is the sum of all such possible numbers?
(a) 6666600 (b) 6666660 (c) 6666666 (d)None
6.P and Q are two integers such that 1:1 PQ=8.Which of the following cannot be the value of P+Q?
(a) 20 (b) 65 (c) 16 (d)35
7.A,B ,C are defined as follows:
A=(2.000004) / [(2.000004)^2 + (4.000008
B=(3.000003) / [(3.000003)^2 + (9.000009)
C=(4.000002) / [(4.000002)^2 + (8.000004)
(a)All of them lie between .18 and .20
(b)A is twice C
(c)C is the smallest
(d)B is the smallest
8.The infinite sum 1+4/7 + 9/7^2+16/7^3+25/7^4+...........
(a) 27/14 (b)29/13 (c) 49/27 (d)256/147
9.In acertain examinations paper,there are n questions.For j=1,2,3.............n there are 2^(n-1) students who answered j or more questions wrongly.If the total number of wrong answers is 4095,then the value of n is:
(a) 12 (b)11 (c) 10 (d)9
Please help me with these!!
What is the method for the following:
Find the 28383rd term of the series: 123456789101112........?
What is the method for the following:
Find the 28383rd term of the series: 123456789101112........?
This question has been asked a number of times in this very thread. Just search using the "Search this thread" tool and you will find the solution. Anyway, just to save you some time, here's the solution that I posted earlier:
http://www.pagalguy.com/forum/quantitative-questions-and-answers/23813-quant-by-arun-sharma-167.html#post1636356
1)Define a number K such that it is d sum of the squares of the first M natural numbers(i.e K=1^2+2^2+3^2+.........+M^2)where Ma)10 b)11 c)12 d)none....
plz answer dis and also tell me d thought process
2)M is a 2-digit number which has the property dat:
the product of factorials of it's digits > sum of factorials of it's digits
how many values of M exist:
a)56 b)64 c)63 d)none
plz also tell me d thought process
3)find the 28383rd term of the series 12345678910111213...
a)3 b)4 c)9 d)7
plz also tell me d thought process
3)find the 28383rd term of the series 12345678910111213...
a)3 b)4 c)9 d)7
You don't even care to read the last post before posting your own set of queries.... You surely can't expect the same question to be answered over and over again..... hope this isn't the only question left in QA nowadays :-P
If you've still not understood which question I'm referring to, just scroll above and read the post prior to your own.
1)Define a number K such that it is d sum of the squares of the first M natural numbers(i.e K=1^2+2^2+3^2+.........+M^2)where Ma)10 b)11 c)12 d)none....
plz answer dis and also tell me d thought process
2)M is a 2-digit number which has the property dat:
the product of factorials of it's digits > sum of factorials of it's digits
how many values of M exist:
a)56 b)64 c)63 d)none
plz also tell me d thought process
3)find the 28383rd term of the series 12345678910111213...
a)3 b)4 c)9 d)7
for
Q1. 1^2+2^2+3^2+.........+M^2=M(M+1)(2M+1)/6
now we now that 2M+1 cannot be even for any natural no.
so M(M+1)/2 must be divisible by 4 or M(M+1) must be divisible by 8
either M should be divisible by 8 or M+1 must be divisible by 8 bcoz cannot simultanously.
so we have solution for M divisible by 8: 8,16,24,32,40,48
and M+1: 7,15,23,31,39,47
there r total 12 solutions
so option (c) is ryt.....
anyone plz solve the prob.
find the remainder when (10^3+9^3)^752 is divided by 12^3??:splat::splat:
anyone plz solve the prob.
find the remainder when (10^3+9^3)^752 is divided by 12^3??:splat::splat:
The Remainder is 1
expand the numerator..it is equal to 1729
1729^x \1728 = 1^x = 1
guys pls gv a solution
it"s from the lod2
28383rd term of the series 1234567891011....
guys its prlm no 65
not two integers in between 1 to 3000 which add up to a multiple of 9 then the no elements in the subset possible
guys pls gv a solution
it"s from the lod2
28383rd term of the series 1234567891011....
1* 9 =9(single digit nos)
2*90=180(2 digit)
3*900=2700(3 digit)
28383-2889
= 25494(4 digit nos)
6372 forms 25489-25492
6373
so 25494 th no s '3'
Hi All,
I am new to this thread .
Here i will like two have my two question that i am no able to solve:
1)Find the remainder when 73*75*78*57*197 is divided by 34 .
The option's are:
a)32 b)30 c)15 d)28
Thr problem is the method that i am using method of arun sharma itself. so i am getting the answer as 22
. So it will be better if anyone can provide the sol related to this method only
2)Find the remainder when 43^197 us divided by 7
The option is a)2 b)4 c)6 d)1
The answer that i am getting is d) but it is not correct.
Can someone please help me with this:banghead:
Hi All,
I am new to this thread .
Here i will like two have my two question that i am no able to solve:
1)Find the remainder when 73*75*78*57*197 is divided by 34 .
The option's are:
a)32 b)30 c)15 d)28
Thr problem is the method that i am using method of arun sharma itself. so i am getting the answer as 22. So it will be better if anyone can provide the sol related to this method only
2)Find the remainder when 43^197 us divided by 7
The option is a)2 b)4 c)6 d)1
The answer that i am getting is d) but it is not correct.
Can someone please help me with this:banghead:
You've got both the answers right. The first equation indeed gives a remainder of 22. Arun Sharma has a lot of incorrect answers, so better is to check it using a calculator.
Similarly, there's no way the 2nd question can have any other remainder apart from 1. So rest assured on that. If your methodology is right, your answers would be right too. Don't trust Arun Sharma's answers too much.