For the second question is the unit digit 1?
gautamgomzi SaysFor the second question is the unit digit 1?
no..its 3..
Q. The power of 5 that will exactly divide 123! is ?
a. 28 b. 30 c. 31 d.59 e. 29
Q. Find the smallest natural no. n such that n! is divisible by 990.
a. 3 b. 5 c. 11 d. 12 e. none of these
Q. A hundred and twenty digit no. is formed by writing the first x natural no. infront of each other as 12345678910111213. Find the remainder when this no. is divided by 8.
a. 6 b. 7 c. 2 d. 0
Q. What is the greatest no. of four digits that when divided by any of the numbers 6, 9, 12, 17 leaves a remainder of 1 ?
a.9997 b. 9793 c. 9895 d. 9487 e. 9897
well this one can b solved through options but cn anyone xplain me the process involved?
Regards
Gautam :grab:
a. 28 b. 30 c. 31 d.59 e. 29
Q. Find the smallest natural no. n such that n! is divisible by 990.
a. 3 b. 5 c. 11 d. 12 e. none of these
Q. A hundred and twenty digit no. is formed by writing the first x natural no. infront of each other as 12345678910111213. Find the remainder when this no. is divided by 8.
a. 6 b. 7 c. 2 d. 0
Q. What is the greatest no. of four digits that when divided by any of the numbers 6, 9, 12, 17 leaves a remainder of 1 ?
a.9997 b. 9793 c. 9895 d. 9487 e. 9897
well this one can b solved through options but cn anyone xplain me the process involved?
Regards
Gautam :grab:
great!!! i was finding it very difficult to solve this book and finally i hv a place where i can clarify my doubts
man u deserve appreciation.........
man u deserve appreciation.........
guys i am havin a few problems solving the chapter of time speed and distance..plz help me with it..
hemant and ajay start a two-length swimming race at the same moment but from opposite ends of the pool.they swim in lane and at uniform speeds,but hemant is faster than ajay..they first pass at apoint 18.5m from the deep end and having completed one length,each one is allowed to rest on the edge for exactly 45 seconds..after setting off on the return length,the swimmers pass for the second time just 10.5m from the shallow end..how long is the pool????
a)55.5m b)45m c)66m d)49m
plz post the explanation as well
Q. The power of 5 that will exactly divide 123! is ?
a. 28 b. 30 c. 31 d.59 e. 29
Q. Find the smallest natural no. n such that n! is divisible by 990.
a. 3 b. 5 c. 11 d. 12 e. none of these
Q. A hundred and twenty digit no. is formed by writing the first x natural no. infront of each other as 12345678910111213. Find the remainder when this no. is divided by 8.
a. 6 b. 7 c. 2 d. 0
Q. What is the greatest no. of four digits that when divided by any of the numbers 6, 9, 12, 17 leaves a remainder of 1 ?
a.9997 b. 9793 c. 9895 d. 9487 e. 9897
well this one can b solved through options but cn anyone xplain me the process involved?
Regards
Gautam :grab:
Q.1.
see, there are 24 multiples of 5 (since 5x24=120)..but amongst those 24, there are 4 multiples of 25 also...so the number 5 occurs twice in those 4 numbers...hence, total no. of 5 in 123! = 24+4 = 28
answer = A.
Q.2
990 = 3x3x11x2x5....so the smallest factorial divisible by 990 must contain two 3's, one 11, 2, 5..the only number is 11!
answer is C.
Q.3.
there are 9 one digit nos.
then, we require 55 two digit numbers to form 110 digit number
appending this to the 9 one digit numbers, we get a 119 digit number
so the 120th digit would be the first digit of the 56th two digit number...the first two digit number is 10, so the 56th number would be 65
so the last three digits of the 120 digit number would be 646, which, if divided by 8, would leave a remainder of 6
answer is A.
Q.4.
i dont know the process that u r talkin about, but solving by evaluating the options seems to be the fastest way in such problems even if u know the method...so i feel u shud use the fastest method..answer is B.
thnx bro.
OK for the first ques even I was gettin 28 but the answer was 31 which I suppose is misprinted.
For the second ques. how did u come to the conclusion as 11? Do we take the largest no.??(amongst all d factors)
GAUTAM........
hey guys can nebody help me with the followin ques...
1) From 4 gentleman n 4 ladies a committee of 5 is to be formed. i.e 1 president , 1 vice president and 3 secretaries.
What will be the number of ways of selecting a committee with atleast 3 women such that atleast 1 women holds the post of either a president or a vice president....?
Please explain the approach if possible...
guys i am havin a few problems solving the chapter of time speed and distance..plz help me with it..
hemant and ajay start a two-length swimming race at the same moment but from opposite ends of the pool.they swim in lane and at uniform speeds,but hemant is faster than ajay..they first pass at apoint 18.5m from the deep end and having completed one length,each one is allowed to rest on the edge for exactly 45 seconds..after setting off on the return length,the swimmers pass for the second time just 10.5m from the shallow end..how long is the pool????
a)55.5m b)45m c)66m d)49m
plz post the explanation as well
answer is 45.
let the speed is a & b (b>a)
total distance =x
two equation will form
18.5/a=(x-18.5)/b (Deep end equation)
x=18.5(a+b)/a..................1
also,
x/b+45+10.5/b=x/a+45+(x-10.5)/a (swallow end equation)
x=10.5(a+b)/(2a-b).........2
equating 1 & 2
we get a/b=37/53
so, x=18.5*90/37=45
so, option 2
i think there must be a short way to solve it also.........
wrong post !!!
hey guys can nebody help me with the followin ques...
1) From 4 gentleman n 4 ladies a committee of 5 is to be formed. i.e 1 president , 1 vice president and 3 secretaries.
What will be the number of ways of selecting a committee with atleast 3 women such that atleast 1 women holds the post of either a president or a vice president....?
Please explain the approach if possible...
Will post in after an hr....plz don't mind..goin for work...i have solved that Q. trying to post a form of solution that can elucidate clearly...but if u trying to do then i can give u few lines of solution
Hint
P(1), V.P(2),S(3),S(4),S(5)
Case I
Lady at [ position 1 ]and [ any 2 or any 3 positions at the positions 3,4,5]
Case II
Lady at [ position 1 & 2]and [ any 1 or any 2 positions at the positions 3,4,5]
P(1), V.P(2),S(3),S(4),S(5)
Case I
Lady at [ position 1 ]and [ any 2 or any 3 positions at the positions 3,4,5]
Case II
Lady at [ position 1 & 2]and [ any 1 or any 2 positions at the positions 3,4,5]
Case I
a)for any 2--- 2!*[4C1*4C1]*[3C2*3C1]=288
b)for any 3--- 2!*[4C1*4C1]*3C3=32
Case II
c) for any 1----4P2*2C1*4C2=144
d) for any 2----4P2*2C2*4C1=48
#Note: bold is for lady and unbold is for gentlemen .
Adding all 512 is ans
Plz ask if still have any doubt !!
Case I
Lady at [ position 1 ]and [ any 2 or any 3 positions at the positions 3,4,5]
Case II
Lady at [ position 1 & 2]and [ any 1 or any 2 positions at the positions 3,4,5]
Case I
a)for any 2--- 2!*[4C1*4C1]*[3C2*3C1]=288
b)for any 3--- 2!*[4C1*4C1]*3C3=32
Case II
c) for any 1----4P2*2C1*4C2=144
d) for any 2----4P2*2C2*4C1=48
#Note: bold is for lady and unbold is for gentlemen .
Adding all 512 is ans
Plz ask if still have any doubt !!
1. find the number of zeros in the prdt (LOD 2 number theory)
1^1*2^2*3^3*.....*100^100
Q. Define a no. K such that it is the squares of the sum of the first M natural numbers.(i.e. K = 1^2 + 2^2+.M^2) where M
a. 10 b. 11 c. 12 d. none of these
Q. M is a two digit no. which has the property that: The product of the factorials of its digits> sum of the factorials of its digits.
How many values of M exist?
a. 56 b. 64 c. 63 d. none of these
Q. Find the 28383rd term of the series 123456789101112..
a. 3 b. 4 c. 9 d. 7
Q. If you form a subset of integers chosen between 1 to 3000, such that no two integers add up to a multiple of 9, what can be the maximum number of elements in the subset.
a. 1668 b. 1332 c. 1333 d. 1334
Q. The series of numbers (1, 1/2, 1/3, 1/4..1/1972) is taken. Now two numbers are taken from this series (the first two) say x,y. Then the operation x+y + x.y is performed to get a consolidated number. The process is repeated. What will be the value of the set after all the numbers are consolidated into one number.
a. 1970 b. 1971 c. 1972 d. None of these
Plz xplain the process.
regards
Gautam.................................:robot:
a. 10 b. 11 c. 12 d. none of these
Q. M is a two digit no. which has the property that: The product of the factorials of its digits> sum of the factorials of its digits.
How many values of M exist?
a. 56 b. 64 c. 63 d. none of these
Q. Find the 28383rd term of the series 123456789101112..
a. 3 b. 4 c. 9 d. 7
Q. If you form a subset of integers chosen between 1 to 3000, such that no two integers add up to a multiple of 9, what can be the maximum number of elements in the subset.
a. 1668 b. 1332 c. 1333 d. 1334
Q. The series of numbers (1, 1/2, 1/3, 1/4..1/1972) is taken. Now two numbers are taken from this series (the first two) say x,y. Then the operation x+y + x.y is performed to get a consolidated number. The process is repeated. What will be the value of the set after all the numbers are consolidated into one number.
a. 1970 b. 1971 c. 1972 d. None of these
Plz xplain the process.
regards
Gautam.................................:robot:
1. find the number of zeros in the prdt (LOD 2 number theory)
1^1*2^2*3^3*.....*100^100
1300 zeroes........
only multiple of 5's will contribute to the no. of zeroes.
Now,
The no. of 5's can b counted as -> from 5 to 95 (common diff = 10)
here no. of terms is 10. Hence the sum will b 10*100/2 = 500
The 10's can b counted as -> from 10 to 90 (common diff = 10)
here the no of terms is 9. Hence the sum will b 9*100/2 =450
100 will contribute 200 zeroes.
but in the above calulation 25,50,75 each wud have contributed 2 zeroes and v have counted only one till now..........therefore 25+50+75 zeroes more.
total adds-up to = 500+450+200+150 = 1300 zeroes.
i was in a fix...as to where to start in this forum...i have also just started the book by arun sharma...will definitely look fwd for assistance in this thread..:satisfie:
Hi Guys,
Please give me solution of the following questions...
Chapter 18 -> Quadratic Equations -> LOD II
Q 7... If both the roots of the quadratic equation ax^2 + bx + c=0 lie in the interval (0,3), then a lies in
a. (1,3)
b. (-1,-3)
c. (-sr. root(121)/91, - sq root( 8 )
d. None of these.
Q 23. if a^2 b^2 + c^2 = 1 then which of the following can not be the value of (ab + bc + ca)?
a. 0
b. 1/2
c. -1/4
d. -1
Thanks in advance for the help :satisfie:
answer is 45.
let the speed is a & b (b>a)
total distance =x
two equation will form
18.5/a=(x-18.5)/b (Deep end equation)
x=18.5(a+b)/a..................1
also,
x/b+45+10.5/b=x/a+45+(x-10.5)/a (swallow end equation)
x=10.5(a+b)/(2a-b).........2
equating 1 & 2
we get a/b=37/53
so, x=18.5*90/37=45
so, option 2
i think there must be a short way to solve it also.........
thanks man for solving it...i hv got d idea now..
guys there r two more...a bus left point X for point Y..two hours later a car leftpoint X for point Y and arrived at Y at the same time as the bus..if the car and the bus left simultaneously from the opposite ends X and Y towards each other,they would meet 1.33 hours after the start..how much time did it take the bus to travel from X to Y..a)2h b)4h c)6h d)8hamitabh covered a distance of 96km two hours faster than he had planned to..this he achieved by travelling 1km more every hour than he intended to cover every 1 hour 15 minutes..what was the speed at which amitabh travelled during the journey???a)16km/h b)26km/h c)36km/h d)none of these
sorry there seems to be some problem..its not taking spaces