hii....i did the chapter Averages from Arun Sharma.....i got an accuracy of 80% in LOD 1, 75% in LOD 2 and 60% in LOD 3........
wat do u think of the above..i m preparing for CAT 2011...
thx in advance......!!!
hii....i did the chapter Averages from Arun Sharma.....i got an accuracy of 80% in LOD 1, 75% in LOD 2 and 60% in LOD 3........
wat do u think of the above..i m preparing for CAT 2011...
thx in advance......!!!
hello...ur score is fine as far as CAT'11 is concerned..u can easily work on ur scores since u hav an ample of time.
chapter1
lod2
30)find d no. of zeros in d product 1^1 * 2^2 * 3^3 -- 100^100???
a 1200
b 1300
c1050
d 1225
e none of his
chapter1
lod2
30)find d no. of zeros in d product 1^1 * 2^2 * 3^3 -- 100^100???
a 1200
b 1300
c1050
d 1225
e none of his
is the answer 1300
find the powers of 5
5+10+15+ .....+ 100 + (25+50+75+100)
is the answer 1300
find the powers of 5
5+10+15+ .....+ 100 + (25+50+75+100)
Mentally do this:
5^5 - 5 fives
10^10 - 10 fives
15^ 15 - 15 fives
20^20 - 20 fives
25^25=(5*5)^25 - 50 fives
30^30 - 30 fives
35^35 - 35 fives
40^40 - 40 fives
45^45 - 45 fives
50^50=(5*5*2)^50 - 100 fives
55^55 - 55 fives
60^60 - 60 fives
65^65 - 65 fives
70^70 - 70 fives
75^75=(5*5*3)^75 - 150 fives
80^80 - 80 fives
85^85 - 85 fives
90^90 - 90 fives
95^95 - 95 fives
100^100=(5*5*4)^100 - 200 fives
On paper it would be:
So, using AP,
a=5
d=5
l=100
n=20
S=n*(a+l)/2=20*(105)/2=1050
To this will be added 25 (from 25^25), 50 (from 50^50), 75 (from 75^75), 100 (from 100^100) additional fives. So,
S=1050+250=1300
1^1*2^2*3^3...................100^100
what v do is dat v take all those numbers wid 5 as the prime factor(since 5 contributes to the number of zeroes)
->5^5*10^10*15^15---------100^100
->5^5*(2*5)^10*(3*5)^15-------(5*20)^100
->5^5*5^10*5^15-----------5^100
->5^(5+10+15------------100)
no of terms=20
5^(20/2(5+100)) {since ths is in AP}
5^1050
so we do hav 1050 zeroes..bt in case of 25,50,75 and 100,we do hav two 5's,we hav included one 5 already,for the second 5
->5^25*5^50*5^75*5^100
solvng ths in the similar manner as above,we get 250 zeroes..
so in total we hav 1050+250 ie.1300 zeroes
correct me if I am wrong. I think the answer should be 13.
summation of n^2=n(n+1)(2n+1)/6. For this to be divisible by 4 ;
n(n+1)(2n+1) should have 2^3 and 3. Since only one the 3 factors can be even always, n should take values which are multiples of 7 and multiples of 8 only.
multiples of 8 multiples of 7 I feel it should be 13..
m9 is also cumin 13..Wots da rite ans?
the correct answer is 12..
k=7,8 (M=1-10)
k=15,16(M=11-20)
k=23,24,31,32,39,40,47,48(M=20-50)
in total values =13(sinceM
u need not calculate the whole,need to simply identify the pattern..i hope its clear..
the correct answer is 12..k=7,8 (M=1-10)
k=15,16(M=11-20)
k=23,24,31,32,39,40,47,48(M=20-50)
in total values =13(sinceM
u need not calculate the whole,need to simply identify the pattern..i hope its clear..
hii....i did the chapter averages from arun sharma.....i got an accuracy of 80% in lod 1, 75% in lod 2 and 60% in lod 3........
Wat do u think of the above..i m preparing for cat 2011...
thx in advance......!!!
its learning from these questions that you should concentrate on.
the correct answer is 12..
k=7,8 (M=1-10)
k=15,16(M=11-20)
k=23,24,31,32,39,40,47,48(M=20-50)
in total values =13(sinceM
u need not calculate the whole,need to simply identify the pattern..i hope its clear..
how did u arrive at the pattern??? and how come n(n+1)(2n+1) should have 2^3 and 3??? dint understand the concept
2 goats are tethered to diagonally opposite vertices of a field formed by joining mid -points of adjacent sides of another square field of side 20(2)^0.5. what is total gazing area of 2 goats?
a.100(3.14) b.50((2)^0.5 -1)(3.14) c. 100*3.14*(3-2*2^0.5) d.200*3.14*(2-2^0.5)e. none of these
ans=a
pls help me solve this!
Anindo0788 Sayshow did u arrive at the pattern??? and how come n(n+1)(2n+1) should have 2^3 and 3??? dint understand the concept
summation of n^2= n(n+1)(2n+1)/6 . For this to be divisible by 4 =>n(n+1)(2N+1)/6/4 which means, n(n+1)(2n+1) should be divisible by 24 which can be split as 2^3 and 3.
Now out of the 3 factors n, n+1 and 2n+1 only one can be even for any value of n which makes it impossible to have 8 in the numerator for any value of n other than the multiples of 8.
So when n=multiples of 8, the term n(n+1)(2n+1) will have 8 as well as 3 in it.
there are 6 multiples of 8 within 55.
Similarly when n=(multiples of
-1 => n+1 would be a multiple of 8 and then it would also have a 3 in it.so we have to consider all the multiples of 8 and their predecessors which total out to 12.
Hope this gives more clarity
Anindo0788 Sayshow did u arrive at the pattern??? and how come n(n+1)(2n+1) should have 2^3 and 3??? dint understand the concept
its simple for 1-10,i calculated and then the next 2 numbers were consecutive as the first 2 with a difference of 8 from the first number.the pattern was thus figured out.
LOD II
There are 42 students in a hostel. 13 new people joined in. As such, the mess expenses increased by Rs 31 per day while the average per head expenses decreased by Rs 3. Find the original mess expenses.
a.633.23
b.583.3
c.623.3
d.632
e.none
LOD II
There are 42 students in a hostel. 13 new people joined in. As such, the mess expenses increased by Rs 31 per day while the average per head expenses decreased by Rs 3. Find the original mess expenses.
a.633.23
b.583.3
c.623.3
d.632
e.none
Let the original average expenditure be Rs. x.
55(x-3)-42x=31
55x-165-42x=31
13x=196
x=196/13
expenditure=42x196/13=633.23
Rahim sets out to cross a forest .On the first day, he completes 1/10 th of the journey .On the second day ,he covers 2/3rd of the distance travelled the first day .He continues in this manner,alternating the days in which he travels 1/10th of the distance still to be covered , with days on which he travels 2/3 of the total distance already covered .At the end of seventh day,he finds that 22 1/2 km more will see the end of his journey.How wide is the forest????/
a 66 2/3
b 100
c 120
d 150
first day - 1/10th=>1/10
second day-2/30=>1/10x2/3=>1/15
total distance(td)=1/6
third day=>(1-1/6)x1/10=>1/12-------td=>1/4
fourth day=>tdx2/3=>1/6-------------td=>5/12
fifth day=>(1-5/12)x1/10=>7/120-----td=>57/120
sixthday=>tdx2/3=>38/120-----------td=>95/120
seventh day=>(1-td)x1/10=>25/1200--td=975/1200
its given that 1-975/1200=>225/1200=>22 1/2
So total distance = 120KM
The metro services has a train going from Mumbai to Pune and Pune to Mumbai every hour ,the first one at 6 am .The trip from one city to other takes 4 1/2 hours and all trains travel at the same speed .How many trains will you pass while going from Mumbai to Pune if you start at 12 noon???
a 8
b 10
c 9
d 13
plz post the approach
The metro services has a train going from Mumbai to Pune and Pune to Mumbai every hour ,the first one at 6 am .The trip from one city to other takes 4 1/2 hours and all trains travel at the same speed .How many trains will you pass while going from Mumbai to Pune if you start at 12 noon???
a 8
b 10
c 9
d 13
plz post the approach
9 is the answer
Assume that trian is leaving for pune from mumbai... Now its given that a train leaves every hour..
So you shd consider trains that u would encounter from pune to mumbai..(opposite direction)
when u leave mumbai at 12 noon u would reach pune at 4.30 pm so u would encounter trains from pune which leave at 8am,9am,10am,11am,12am,1am,2am,3am,4am..
you dont consider train at 6,7 becoz it would have reached mumbai before 12 noon