Q) A boy took a 7 digit no. ending in 9 and raised it to a even power greater than 2000. He then took number 17 and raised it to a power which leaves the remainder 1 when divided by 4. If he now multiplies both the no.s , what will be the unit's digit of the no. he so obtains? a) 7 b)9 c)3 d) cannot be determined
now my doubt is if a no. ending with 9 is raised to even power then the units digit will always be 1 and also 17 raised to any power(whether even or odd) when divided by 4 leaves remainder 1 .Therefore if those 2 digits are multiplied then the 7 digit no. ending with 1 will get multiplied to 7 or 9 or 3 or 1 Therefore the unit's digit could be 7 or 9 or 3 or 1. At the back no explanation is given and the answer is (a)
sometimes arun sharma's book brings huge disappointment , its a pity that i share the same surname :-(
9^even = unit digit will be always 1. (9^2 =81, 9^4 =6561)
Given, Which leaves the remainder 1 when divided by => can be 1,5...
Ok guys, I have been struggling with this "permutation & combination" problem for quite a while. "The crew of an 8 member rowing team is to be chosen from 12 men, of which 3 can row on one side only & 2 can row on the other side only. Find the number of ways of arranging the crew with 4 on each side" The answer provided is 60480, which I guess is wrong because , the answer assumes those 3 & 2 guys who can row on either side are always chosen, but is nowhere explictly mentioned in the problem I guess this question needs to be tackled in whole different view. please provide your opinions, whether I am treading the wrong path?
3 and 2 people are fixed here.
Now, 7C1*6C2 for other 3 men needed. Arrangements = 4!*4!
Bro, I know how to work out the solution. It is based on assumption that those 3 & 2 people are to selected always, which is nowhere explicitly mentioned. Please tell me which part of the question actually allows us to draw that assumption.
The problem can be worked out without this assumption, though it would be a long drawn process. I wasted a lot of time in arriving at correct solution, to be aghast to find it has solved in a completely brainless manner.
The way I solved the problem was: Scenario 1: One side out of the boat- all the 3 specialist rovers are chosen & 1 guy is chosen from rest (12-3-2) of 7 guys Other side of the boat - 2 rovers are chosen from rest 8 guys
Scenario 2: One side of the boat- only 2 specialist rover is chosen is chosen (out of possible 3) & 2 guys are chosen from rest 7 guys Other side- 2 rovers are chosen from 7 guys(8 guys are left, however 1 cannot row on this side of the boat)
So on & so forth, which would present 4 possible scenarios, which seems to be more apt solution.I guess I'm over thinking a lot (not a bad think, haha)
The book is excellent, no doubt, but the framing of questions leaves a lot to be desired.
9^even = unit digit will be always 1. (9^2 =81, 9^4 =6561)
Given, Which leaves the remainder 1 when divided by => can be 1,5...
7^1 = unit digit =>7
so 7*1 = 7
But still 17 raised to even or odd power and then divided by 4 will always give remainder as 1 ; i.e R(17^1/4) = R(17^2/4)=R(17^3/4)=R(17^4/4)= 1 , so the 1 in the unit's place of the 9 digit no. can get multiplied by 17^1 or 17^2 or 17^3 or 17^4 resulting the final answer of unit's place as 7 or 9 or3 or 1 (as you these are the unit's digits for various powers of 17)
Could you please help me to know the procedure solving the below questions
Topic:Allegations and Mixtures
Question 1: How many Kilos of sugar worth rs 3.6 per kg should be mixed with 8 kg of sugar worth rs 4.2 per kg.such that by selling the mixture at rs 4.4 per kg,there may be a gain of 10%......Answer is 4 kg
Question 2: A Dishonest Milkman purchased milk at 10 per litre and mixed 5 litres of water in it.by selling the mixture at the rate of rs 10 per litre he earns a profit of 25%.the quantity of the amount of the mixture that he had was?..............ans is 25 litres
Question 3: In what ratio should water be mixed with soda costing rs 12 per litre so as to make a profit of 25% by selling the diluted liquid at rs 13.75 per litre?.......ans is 1:11
question 4:
A grocer professes to sell pure butter at cost price,but he mixes it with adulterated fat and thereby gains 25%.find the % of adulterated fat in the mixture assuming that adulterated fast is freely available........ans is 20%
If this question comes in any exam (not CAT as it is not of that level) then what to do .This question is from one of our "FAVOURITE" AUTHOR Mr. Sharma
Q) 16^5 + 2^15 is divisible by a) 31 b) 13 c) 27 d)33 e) 12
i know we can just write (16^5 + 2^15) as ; a^n +b^n is divisible a+b when n= odd , therefore 16+8=24 ; 16^5+2^15 is divisible by 24 and amongst the options 24 is divisible by only 12 so the answer should be option (e) But another approach of sharma ji says 16^5+2^15= 2^15 x (2^5+1) =2^15x33 hence it is divisible by 33 . so now which method to follow to get a single answer ; i think in exams of quality (other than SNAP) it will be mentioned to choose largest no. or smallest no.
Question 1: How many Kilos of sugar worth rs 3.6 per kg should be mixed with 8 kg of sugar worth rs 4.2 per kg.such that by selling the mixture at rs 4.4 per kg,there may be a gain of 10%......Answer is 4 kg
how to solve such question :
as the value of Sugar is 4.4 with 10% gain hence Cost price = 4.4/110%=4 rs/kg
moreover
we have to get 4 as resultant with mixture of two different quantities
one causes an increase of .2*8=1.6 ruppes
so to compensate the other should be mixed as 1.6/.4=4 kg
so by mixing 4 kg of 3.6rs/kg sugar with 8 kg of 4.2rs/kg u will get 12kg of 4rs/kg sugar mixture
Question 2: A Dishonest Milkman purchased milk at 10 per litre and mixed 5 litres of water in it.by selling the mixture at the rate of rs 10 per litre he earns a profit of 25%.the quantity of the amount of the mixture that he had was?..............ans is 25 litres
to find amount of mixture :
for 25 % profit
as cost of water is zero hence 25% of milk should be equal to quantity of water =5litre
hence milk = 20 litres
and mixture =20+5=25
Question 3: In what ratio should water be mixed with soda costing rs 12 per litre so as to make a profit of 25% by selling the diluted liquid at rs 13.75 per litre?.......ans is 1:11
sp =13.75
cost to the seller will be =13.75/1.25 = as profit is 25%
hence cost =11 rs
so to find the ratio of mixing again cost of water is 0
11-0=11 and 12-11=1
so ratio will be 1:11
question 4:
A grocer professes to sell pure butter at cost price,but he mixes it with adulterated fat and thereby gains 25%.find the % of adulterated fat in the mixture assuming that adulterated fast is freely available........ans is 20%
take butter as 80rs per kg
to get 25% profit that means selling will be at 100 rs
now actually cost =80 becoz adulterated fat is free
, so the 1 in the unit's place of the 9 digit no. can get multiplied by 17^1 or 17^2 or 17^3 or 17^4 resulting the final answer of unit's place as 7 or 9 or3 or 1 (as you these are the unit's digits for various powers of 17)
Buddy kindly read the question carefully... it says 17 has been raised to such power that will leave R=1 when divided by 4
i know we can just write (16^5 + 2^15) as ; a^n +b^n is divisible a+b when n= odd , therefore 16+8=24 ; 16^5+2^15 is divisible by 24 and amongst the options 24 is divisible by only 12 so the answer should be option (e) But another approach of sharma ji says 16^5+2^15= 2^15 x (2^5+1) =2^15x33 hence it is divisible by 33 . so now which method to follow to get a single answer ; i think in exams of quality (other than SNAP) it will be mentioned to choose largest no. or smallest no.
here i am not able to understand what you are trying to say... a number which is divisible by 12 is not necessarily is divisible by 24..
Me and my roomie have been struggling with this progressions question for quiet some time now... Rohit drew a rectangular grid of 529 cells,arranged in 23 rows and 23 columns, and filled each cell with a no. The numbers with which he filled each cell were such that the numbers of each row taken from left to right form an arithmetic series and the numbers of each column taken from top to bottom also form a arithmetic series.The 7th and 17th no. of the 5th row were 47 and 63 resp., while the 7th and 17th no. of 15th row were 53 and 77.What is the sum of all the numbers in the grid? a)32798 b)65596 c)52900 d)none of these
@gudda 1122 here i am not able to understand what you are trying to say... a number which is divisible by 12 is not necessarily is divisible by 24..
But a no. which is divisible by 24 is always divisible by 12 , u can check it for any multiple . Also in the question after solving we get that (16^5 + 2^15) is divisible by 24 and by above mentioned theory in "bold" if any no. is divisible by 24 then it has to be divisible by 12.
Buddy kindly read the question carefully... it says 17 has been raised to such power that will leave R=1 when divided by 4
But dude 17 raised to any power when divided by 4 will always always leave remainder =1 so the unit's digit inthe answer could be 7 or 9 or 3 or 1 ....... u know all the recurring unit's digit of 17
But dude 17 raised to any power when divided by 4 will always always leave remainder =1 so the unit's digit inthe answer could be 7 or 9 or 3 or 1 ....... u know all the recurring unit's digit of 17
mate... atleast have a other look at the question
17 and raised it to a power which leaves the remainder 1 when divided by 4
IT means it is raised to such power which leave reminder 1..when divided by 4....
101 is prime so E(101) = 100 and using Fermant's Law we have the remainder of M^(N-1) when divided by N, when N being prime is 1... so 2^100/101 = 1
For question number 1... 12341234.. 400digits each digit will come for 100 times.. (400/4) so all we have to do is 100(1+2+3+4) = 100*10 = 1000/909 = 91 pls confirm
what is this ???? ..... what logic are u using ????? Referring to the question (Block 1 : Review Test 5), that => Find the remainder when (123412341234 ..... upto 400 times) is divided by 909 ? Options >> (a) 623 (b) 650 (c) 685 (d) 675 (e) none of these We have known Arun Sharma for providing wrong answers in his answer-keys, well for this question he has left the answer-key blank and given no answer. ... !! Please any1 ... some help here !!!!
In a certain examination paper there are n questions. for j=1,2...n , there are 2^(n-1) students who answered j or more questions wrongly. If total no of wrong answers is 4095.then the value of n is : a.12 b.11 c.10 d.9
In a certain examination paper there are n questions. for j=1,2...n , there are 2^(n-1) students who answered j or more questions wrongly. If total no of wrong answers is 4095.then the value of n is : a.12 b.11 c.10 d.9
Ans= a
Help !!!
here it is : when j=1
then no of students who answered 1 question wrong is =1 = 2^n-1
when j=2
then no of students who answered 2 question wrong is =2 = 2^n-1
when j=3
then no of students who answered 1 question wrong is =4 = 2^n-1
similarly when j=12 then n=12
then no of students who answered 12 question wrong is =2048 = 2^n-1
so total no of wrong quesions =J(1)+J(2)+J(3)+J(4)+..........................j(12)
In a certain examination paper there are n questions. for j=1,2...n , there are 2^(n-1) students who answered j or more questions wrongly. If total no of wrong answers is 4095.then the value of n is : a.12 b.11 c.10 d.9
Ans= a
Help !!!
Number of students who answered one or more ques wrongly = 2^n-1. Number of students who answered two or more ques wrongly = 2^n-2. Number of students who answered exactly one ques wrongly = 2^n-1 - 2^n-2. Likewise, for exactly 2 = 2^n-2 - 2^n-3 and so on.. => (2^(n-1) - 2^(n-2)) + 2*(2^(n-2) - 2^(n-3)) + .... + (n-1)*(2^(n-(n-1)) - 2^(n-n)) + n*2^(n-n) = 4095. => 2^(n-1) + 2^(n-2) + .... + 2^1 + 1 = 4095. => 2^n - 1 = 4095. => 2^n = 4096. Hence, n = 12.
the sum of the squares of the fifth and the eleventh terrm of an AP is 3 and the product of the second and fourteenth term is equal to P. Find the product of the first and fifteenth term of the AP.
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AUTHOR Mr. Sharma
