aggregate is 60% so,total of 150 marks.
now,the marks are,
a+b+c+d=150.
let,
p=50-a
q=50-b
r=50-c
s=100-d
p+q+r+s=100
103c3=176851.
none of these.
is that correct??
can we apply this theorem
(150+4-1)C(4-1) ?
can we apply this theorem
(150+4-1)C(4-1) ?
can we apply this theorem
(150+4-1)C(4-1) ?
nope,we cannot apply as there might be some values,where,a,b,c>50 or d>100.so,we have to modify it slightly.
does any body have a question bank in pdf format?
sagaciouslass Saysdoes any body have a question bank in pdf format?
hi,this is not the proper place to ask for material.this thread is for queries related to Arun Sharma's Quant book.
for CAT and related material,please try here:
http://www.pagalguy.com/forum/prep-resources/11984-materials-related-cat-xat-other-33.html#post1475188
http://www.pagalguy.com/discussions/cat-preperation-download-materials-25034373
hope that helps.
Hi Guys I did not understand the question fully, below is the question
Arun, Bikas and Chetakar have a total of 80 coins among them. Arun triples the number of coins with the others by giving them some coins from his own collection. Next Bikas repeats the same process. After this Bikas now has 20 coins. Find the number of coins he had at the begining.
Can any one please explain the sentence in bold?
The sole purpose of thisThread is to bring together the CAT aspirants who are solving the Quant book by Arun Sharma. This will be a common place to discuss any doubts while solving the questions. As there are many questions in the book where either the Hint/Solution is missing or the answer itself is wrong. It will be very nice of you all to help people clear their doubts and also they could learn various methods of problem solving.
:grab:
Thread is to bring together the CAT aspirants who are solving the Quant book by Arun Sharma. This will be a common place to discuss any doubts while solving the questions. As there are many questions in the book where either the Hint/Solution is missing or the answer itself is wrong. It will be very nice of you all to help people clear their doubts and also they could learn various methods of problem solving.HEY I AM REFFERING TO R.S AGRAWAL,,,IS IT WORTH PREPARING???
OR SHOULD I GO FOR ARUN SHARMA???WELL I FOUND THE BOOK QUITE NICE...GOOD STRATEGIES ARE GIVEN THERE...
BY THE WAY WHAT ELSE EXCEPT ARUN SHARMA IN QUANTS...
JIMMY
Hi Guys I did not understand the question fully, below is the question
Can any one please explain the sentence in bold?
Arun triples the number of coins with the others by giving them some coins from his own collection.
it means
when arun distribute some portion of money to others , others coins became threetimes instead of singletimes
means if arun,bikas,chetakr have A,B,C coins respectively
if x coin arun give from his money
then arun have A-x
bikas have 3B
chetakar have 3C
Hey, Another doubt in question 89. just read the following
Question:
The super computer at Ram Mohan Roy Seminary takes an input of a number N and a X where X is a factor of the number N. In a particular case N is equal to 83p796161q and X is equal to 11where 0
Choices : a) 6 b) 3 c) 2 d) 9
My doubt is if N = 83p796161q then
8 + p + 9 + 1 + 1 = 3 + 7 + 6 + 6 + q (as X = 11)
if this is true then 0
Arun triples the number of coins with the others by giving them some coins from his own collection.
it means
when arun distribute some portion of money to others , others coins became threetimes instead of singletimes
means if arun,bikas,chetakr have A,B,C coins respectively
if x coin arun give from his money
then arun have A-x
bikas have 3B
chetakar have 3C
Forward from QN 88 NS(number systems)
If that is correct then how the answer choices are
a) 11 b) 10 c) 9 d) 12 all are less than 20 for B, if B gives A & C and becomes 20, then B must have more than 20 coins.
Forward from QN 88 NS(number systems)
If that is correct then how the answer choices are
a) 11 b) 10 c) 9 d) 12 all are less than 20 for B, if B gives A & C and becomes 20, then B must have more than 20 coins.
if u search in previous post then its already solved and ans is 20
a , b and c be the inititals of the guys above
Now initial case : a+b+c = 80
After a gives some x coins the status is
(a-x) + 3b + 3c = 80
And After b does the same thing , viz : gives back y coins
3(a-x) + (3b -y) + 9c = 80
3(a-x) + 20 + 9c = 80 (Given)
3(a-x) + 9c = 60 => (a-x) + 3c = 20
put this value in
(a-x) + 3b + 3c = 80 => 3b = 60 => b= 20
Arun triples the number of coins with the others by giving them some coins from his own collection.
it means
when arun distribute some portion of money to others , others coins became threetimes instead of singletimes
means if arun,bikas,chetakr have A,B,C coins respectively
if x coin arun give from his money
then arun have A-x
bikas have 3B
chetakar have 3C
if u search in previous post then its already solved and ans is 20
a , b and c be the inititals of the guys above
Now initial case : a+b+c = 80
After a gives some x coins the status is
(a-x) + 3b + 3c = 80
And After b does the same thing , viz : gives back y coins
3(a-x) + (3b -y) + 9c = 80
3(a-x) + 20 + 9c = 80 (Given)
3(a-x) + 9c = 60 => (a-x) + 3c = 20
put this value in
(a-x) + 3b + 3c = 80 => 3b = 60 => b= 20
Thanks buddy, most of the answers in Arun Sharma are either wrong or incorrect.
Hey, Another doubt in question 89. just read the following
Question:
The super computer at Ram Mohan Roy Seminary takes an input of a number N and a X where X is a factor of the number N. In a particular case N is equal to 83p796161q and X is equal to 11where 0
Choices : a) 6 b) 3 c) 2 d) 9
My doubt is if N = 83p796161q then
8 + p + 9 + 1 + 1 = 3 + 7 + 6 + 6 + q (as X = 11)
if this is true then 0
if we see u r explanation then its absolutely right
0
Hey, Another doubt in question 89. just read the following
Question:
The super computer at Ram Mohan Roy Seminary takes an input of a number N and a X where X is a factor of the number N. In a particular case N is equal to 83p796161q and X is equal to 11where 0
Choices : a) 6 b) 3 c) 2 d) 9
My doubt is if N = 83p796161q then
8 + p + 9 + 1 + 1 = 3 + 7 + 6 + 6 + q (as X = 11)
if this is true then 0
the divisibility by 11 suggests that,the resultant from subtracting the terms should be divisible by 11.
so,
there are 2 terms:One is 19+p and other is q+22
as,q>p,
q+22>p+19
so,
q>p-3
so,q-p+3=11k
q-p=11k-3
so,k cannot be 0.it has to be 1.
q-p=8
q=9,p=1=>number is 8317961619
r1=9,r2=0=>r1+r2=9
option (D)
hope that helps.
Here is another question:
What is the remainder when 2(8!) - 21(6!) divides 14(7!) + 14(13!) ?
Answers : a) 1 b) 7! c) 8! d)9! Given ans = (b)
But my answer is coming as 1 i.e. (a). Here is how
2(8!) - 21(6!) = 7! * 13
Hence, [14(7!) + 14(13!)] / [7! * 13]
= [14* (7!)] / [(7!)*13] + 14(13!) /7!*13 [ the bold part is divisible]
Hence remainder is 1.
Am I Correct?
Here is another question:
What is the remainder when 2(8!) - 21(6!) divides 14(7!) + 14(13!) ?
Answers : a) 1 b) 7! c) 8! d)9! Given ans = (b)
But my answer is coming as 1 i.e. (a). Here is how
2(8!) - 21(6!) = 7! * 13
Hence, [14(7!) + 14(13!)] / [7! * 13]
= [14* (7!)] / [(7!)*13] + 14(13!) /7!*13 [ the bold part is divisible]
Hence remainder is 1.
Am I Correct?
14(7!)[1+13*12*11*10*9*8] mod 13(7!)
canceling 7! off
14(1+13k)mod 13
14 mod 13
remainder is 1.
as we had canceled 7! earlier,we have to multiple the resultant by the same.
so,remainder is 7!
hope that helps.
I think u made a mistake shashank,
as we had canceled 7! earlier,we have to multiple the resultant by the same.
does not come into picture as we have already cancelled out 7!(by division).
Secondly, I have calculated manually this to confirm my answer here.
My answer is correct the remainder is 1.
14(1+13k)mod 13itself shows that later part is divisible and no remainder from that side. so only 14 remains to be divided by 13 which has remainder 1.
Hello again the question is
How many integer valus of x and y are there such that 4x + 7y = 3, while x Answers a) 144 b) 141 c) 143 d) 142
I have calculated this way,
I found out that 4's multiplication in the series of 5,2,5,2... with the 7's multiplication of 2,2,2,2... satisfy the result like
4*(-1) + 7*(+1) = 3
4*(6) + 7*(-3) = 3
4* + 7*(5) = 3
4*(13) + 7*(-7) = 3 ...etc
More details:
1. Diff between 4's multiplicant 1 & 6 is 5 where 7's multiplicant 1 and 3 is 2
2. This combination goes with alternate - and + way as you can see from the example above.
Hence by this way I found 35 ( x , y) numbers which satisfy the equation.
How can the answer be like that range? I don't understand.
Is it talking about number of x's + number of y's ?
I think u made a mistake shashank,
does not come into picture as we have already cancelled out 7!(by division).
Secondly, I have calculated manually this to confirm my answer here.
My answer is correct the remainder is 1.
itself shows that later part is divisible and no remainder from that side. so only 14 remains to be divided by 13 which has remainder 1.
1 cannot be the remainder here,as the terms end in 0 each
on manual checking,
(70560+87178361760)=87178361760=>numerator
(80640-15120)=65520=>denominator
on subtracting 7! i.e. 5040,we get, 87178356720 which is divisible by 65520.
please check again.tell me if i am missing something.
and,if we cancel out a term,we have to multiply the same after division.
Hello again the question is
How many integer valus of x and y are there such that 4x + 7y = 3, while x Answers a) 144 b) 141 c) 143 d) 142
I have calculated this way,
I found out that 4's multiplication in the series of 5,2,5,2... with the 7's multiplication of 2,2,2,2... satisfy the result like
4*(-1) + 7*(+1) = 3
4*(6) + 7*(-3) = 3
4* + 7*(5) = 3
4*(13) + 7*(-7) = 3 ...etc
More details:
1. Diff between 4's multiplicant 1 & 6 is 5 where 7's multiplicant 1 and 3 is 2
2. This combination goes with alternate - and + way as you can see from the example above.
Hence by this way I found 35 ( x , y) numbers which satisfy the equation.
How can the answer be like that range? I don't understand.
Is it talking about number of x's + number of y's ?
7y+4x=3
y=1=>x=-1
y=5=>x=-8
y=9=>x=-15
.
.
.
x=-498=>y=285
total of 72 terms.
also,
x=6=>y=-3
x=13=>y=-7
.
.
.
.
x=496
71 terms.
total of 143 terms.
option (C)
1 cannot be the remainder here,as the terms end in 0 each
on manual checking,
(70560+87178361760)=87178361760=>numerator
(80640-15120)=65520=>denominator
on subtracting 7! i.e. 5040,we get, 87178356720 which is divisible by 65520.
please check again.tell me if i am missing something.
and,if we cancel out a term,we have to multiply the same after division.
yes I understand now, that's right. Thanks for digging out my mistake.