mklchandra, brother, thanks a lot for the try, but as per the source, the answer is 5/6 😞 😞
Find the max. Value of n such that
570*60*30*90*100*500*700*343*720*81
is perfectly divisible by 30^n.
A. 12
b. 11
c. 10
d. 13
The ans is (1) 12 , right ?
30=5*2*3 so in order to find the highest power of 30 that divides 570*60*30*90*100*500*700*343*720*81 , we simply need to find highest power of 5 as powers of 2 or 3 will be greater than that of 5 here.
570*60*30*90*100*500*700*343*720*81
=5*..*5..*..*5*..5*..*5^2..*..*5^3*..5^2*..*5^1..*..*5^0*..*5^0*
=5*12.
hence 30^12 will divide the given expression completely.
off topic , but i have 2 Edition of this book , and current edition is 4th i guess, is there any need to buy new edition ?
dishwalla Saysoff topic , but i have 2 Edition of this book , and current edition is 4th i guess, is there any need to buy new edition ?
No.
:biggrin:
Please help me... got stuck..
Q. A shrewd milkman mixes water and milk in the ratio of 2:3. What part of this mixture should be removed and replaced with water so that the solution contains water and milk in the ratio 1:1?
1)1/6
2) 1/4
3) 1/3
4) 5/6
Yaar layman's logic dekh lo, kuchh galat hai toh batao. Arun Sharma's answers at times are incorrect (personal exp. :)) -
Let Water = 200, Milk = 300 - Total mixture = 500 ml.
For 1:1, both should be 250.
So, we need to remove 50 ml of milk and pour 50 ml of water. Now, in order to take out 50 ml of milk we need to remove 250/3 ml of the mixture .
Now, well have to add ml of water to it, 50 + 100/3 = 250/3 .
Therefore, fraction of mixture to be taken out will be = 1/6.
The ans is (1) 12 , right ?
30=5*2*3 so in order to find the highest power of 30 that divides 570*60*30*90*100*500*700*343*720*81 , we simply need to find highest power of 5 as powers of 2 or 3 will be greater than that of 5 here.
570*60*30*90*100*500*700*343*720*81
=5*..*5..*..*5*..5*..*5^2..*..*5^3*..5^2*..*5^1..*..*5^0*..*5^0*
=5*12.
hence 30^12 will divide the given expression completely.
yup...it is correct...!!
Btw...thnk u very much....
karan5 Saysanswer for question 2 is 2...will xplain if i am rite....
Hey Karan ur ans is correct.
Dude pls explain me the method.
N if u can get the ans for other Questions then pls let me know.
Here's a question from Number System(Pg-46)-
What is the highest power of 3 available in the expression 58!-38!?
a) 17 b) 18 c) 19 d) None of these.
Help!
is it 17???
The ans is (1) 12 , right ?
30=5*2*3 so in order to find the highest power of 30 that divides 570*60*30*90*100*500*700*343*720*81 , we simply need to find highest power of 5 as powers of 2 or 3 will be greater than that of 5 here.
570*60*30*90*100*500*700*343*720*81
=5*..*5..*..*5*..5*..*5^2..*..*5^3*..5^2*..*5^1..*..*5^0*..*5^0*
=5*12.
hence 30^12 will divide the given expression completely.
Buddy i think it will be 11. The highest power of 3 is 11
The ans is (1) 12 , right ?
30=5*2*3 so in order to find the highest power of 30 that divides 570*60*30*90*100*500*700*343*720*81 , we simply need to find highest power of 5 as powers of 2 or 3 will be greater than that of 5 here.
570*60*30*90*100*500*700*343*720*81
=5*..*5..*..*5*..5*..*5^2..*..*5^3*..5^2*..*5^1..*..*5^0*..*5^0*
=5*12.
hence 30^12 will divide the given expression completely.
yup...it is correct...!!
Btw...thnk u very much....
But the highest power of 3 is 11 then why 12??????
ksgr32 Saysmklchandra, brother, thanks a lot for the try, but as per the source, the answer is 5/6 😞 :(
Bro i have cross-verified .The ans must be 1/6.
btw is this a qn from arun sharma ?
Some Questions puys.
Q2. Find Infinite sum of the series 1/1+1/3+1/6+1/10+1/15.............
Ans = 2, 2.25, 3, 4
N = 1/1 + 1/3 + 1/6 ...
= Summation of 1/n(n+1)/2 for n = 1 to infinity
= Summation of 2/n(n+1)
= Summation of 2(1/n - 1/n+1)
= 2{(1/1 - 1/2) + (1/2 - 1/3) + .... + (1/infinity - 1/infinity+1)}
= 2 (1 - 1/infinity+1)
= 2 (1 - 0)
= 2
Hope this helps
For the question of highest power of 3 dividing 58! - 38!
ans is 17
right?
Some Questions puys.
Q2. Find Infinite sum of the series 1/1+1/3+1/6+1/10+1/15.............
Ans = 2, 2.25, 3, 4
N = 1/1 + 1/3 + 1/6 ...
= Summation of 1/n(n+1)/2 for n = 1 to infinity
= Summation of 2/n(n+1)
= Summation of 2(1/n - 1/n+1)
= 2{(1/1 - 1/2) + (1/2 - 1/3) + .... + (1/infinity - 1/infinity+1)}
= 2 (1 - 1/infinity+1)
= 2 (1 - 0)
= 2
Hope this helps
Thanks for the solution
Some More Questions puys.
Q1. The sum of series
1/(sqrt2+sqrt1) + 1/(sqrt2+qrt3) +.......................+ 1/(sqrt120+sqrt121)
Ans = 10, 11 , 12, None of these
Q2. How many 3 digit nos have the property that their digits taken from left to right form an AP or a GP?
Ans = 15, 36, 20, 42
Q3. An AP P consists of n erms.
From the progressions three different progressions P1, P2, & P3 re created such that P1 is obtained by the 1st,4th,7th....terms of P, P2 has the 2nd,5th,8th .......terms of P and P3 hs the 3rd,6t,9th....terms of P. It is found that of P1,P2 and P3 two progression have the propertythat their average is ielfa term of the original Progression P. Find possible values of n.
Ans = 20, 26, 36, both 1 & 2
Waiting for the Explanation puys.
The Questions is of Progression from Arun Sharma in LOD2 & LOD 3
i am in graduation, 3rd year and want to start preparation for CAT from now onwards. I would like to know where should i start from.
Need to form a committee of 5 from 4 men and 4 women. Committee consists of a president, a vice president and 3 secretaries.
How many ways can the committee be selected with a maximum of 2 women and with a maximum of 1 woman holding one of the two posts on the committee?
a) 16 b)512 c)608 d)256 e)324
(Arun sharma PnC lod2 q9)
For the question of highest power of 3 dividing 58! - 38!
ans is 17
right?
IMO 17
Regards,
Never Back Down
Thanks for the solution
Some More Questions puys.
Q1. The sum of series
1/(sqrt2+sqrt1) + 1/(sqrt2+qrt3) +.......................+ 1/(sqrt120+sqrt121)
Ans = 10, 11 , 12, None of these
Q2. How many 3 digit nos have the property that their digits taken from left to right form an AP or a GP?
Ans = 15, 36, 20, 42
1: multiply (a-b) to the numerator and denominator of each fraction, for eg. multiply sqrt2-sqrt1 to the num and den of the first term. You will get (sqrt2-sqrt1)+(sqrt3-sqrt2)+....+(sqrt121-sqrt120).
This will reduce to sqrt121-sqrt1, i.e, 11-1=10
2: first, ap terms:
CD of 1: 123,234,345,....,789
CD of 2: 135,246,357,468,579
CD of 3: 147,258,369
CD of 4: 159
Total terms = 7+5+3+1=16; then 16X2 ( for reverse order numbers like 321, 642 etc)= 32
4 more numbers ending in 0= 210,420,630,840
Net total of ap terms=36
Now GP terms:
CD=2: 124,248; CD=3:139
3 termsX2 (for reverse)=6
Thus GP terms=6
Total numbers (ap/gp)= 36+6=42
Need to form a committee of 5 from 4 men and 4 women. Committee consists of a president, a vice president and 3 secretaries.
How many ways can the committee be selected with a maximum of 2 women and with a maximum of 1 woman holding one of the two posts on the committee?
a) 16 b)512 c)608 d)256 e)324
(Arun sharma PnC lod2 q9)
Is the answer 512.
If yes then will explain it, so let me know as i dont have the book.
mba4mheart SaysBut the highest power of 3 is 11 then why 12??????
sorry....
Checked the answr frm book...it is 11