!!!!!!!!!!!!!!!!!!!!!!!!!!
7^2008...divide 2008 by 4 and chk if it is divisible by 4...if it it divible take 7^4 and find it's last digit otherwise take it's reaminder...like for 7^2007..take it as 7^3..and find it's last digit..:)
Its 2......
1000 has three 0's......so the product has to have three multiples of 5.....
Now product of n consecutive integers is x=n!*k......x will have a zero at the end for all n greater than or equal to 5......but we need 0 or 5 at the end of at least three terms among (x+1),(x+2),....(x+(n-1))......the first three terms will be x+5, x+10 and x+15.......so n-1 has to be at least 15.....or n has to be at least 16......
just a small query...
in the series "(x+1)(x+2)(x+3).....(x+(n-1))=1000" as we are taking consecutive integers, we are bound to have multiples of 3...but 1000 is not divisible by 3...how is it possible? what is the thing I am missing?
just a small query...
in the series "(x+1)(x+2)(x+3).....(x+(n-1))=1000" as we are taking consecutive integers, we are bound to have multiples of 3...but 1000 is not divisible by 3...how is it possible? what is the thing I am missing? :lookround:
Oops...my mistake....misread the question.....thought 1000 is a factor.....
Q) Find all two digit numbers such that the sum of the digits constituting the number is not less than 7; the sum of the square of the digits is not greater than 30; the number consisting of the same digits written in the reverse order is not larger than half of the given number?
ans a) 52 b)51 c)49 d) 53 e)50
ans a) 52
Q) Find all two digit numbers such that the sum of the digits constituting the number is not less than 7; the sum of the square of the digits is not greater than 30; the number consisting of the same digits written in the reverse order is not larger than half of the given number?
ans a) 52 b)51 c)49 d) 53 e)50
I am not able to buy this question
geetikagreat Saysans a) 52
hey can u please explain how u arrived at it
hellorajesh Sayshey can u please explain how u arrived at it
its given that the sum of 2 digit no. is not less than 7 .. eliminating from options
52 , 49 , 53 fulfill dis condition.. second condition is sum of square of digits is not more than 30.. accord to this eliminating from above 3 nos. only 52 fits accord to the 2nd condition
finally if 52 is reversed it is 25 which is less than half of 52 i.e 26.
katariags SaysI am not able to buy this question
Is the answer B....not sure..ill explain it if it's B
I am new to CAT preparation
I have a query regarding quant by arun sharma
Are the options and the solutions given in the book correct?
I cant find many of the solutions at the back specially of difficlut questions..
Plz help
Thanks
Alok Biyani, Kolkata
I am new to CAT preparation
I have a query regarding quant by arun sharma
Are the options and the solutions given in the book correct?
I cant find many of the solutions at the back specially of difficlut questions..
Plz help
Thanks
Alok Biyani, Kolkata
Hiie.There are solutions and hints to selective problems only,not all.Also,there are errors in the answer keys of Arun Sharma.You can discuss the controversial ones in this thread.:thumbsup:
Can someone help me out with this question?
Probability - LOD 2
In four throws with a pair of dices,what is the chance of throwing a double twice?
a)11/216
b)25/216
c)35/126
d)45/216
e)41/216
Can someone help me out with this question?
Probability - LOD 2
In four throws with a pair of dices,what is the chance of throwing a double twice?
a)11/216
b)25/216
c)35/126
d)45/216
e)41/216
I think it should be B
25/216
For one throw possible outcomes= 36
Favourable = 6
So P(double)= 1/6
P(other)= 5/6
In four throws, 2 doubles P= (1/6)^2*(5/6)^2
Also these four outcomes can be arranged in c(4,2) ways=6
So answer is= 25/216
yes i too agree its B
LOD1 q43 pg457 In how many ways 4 socks
can be selected out of 5
blue socks, 4 red socks and
3 green socks. Options
245, 120, 495, 60
OA 495
my approach-
A selected sock can be
blue red or green. So 1st
sock can be selected in 3
ways. Similarly, second
third and fourth can be
selected in 3 ways each.
But selection of 4 green
socks is not possible. So
no. of ways=3^4-1=80
Puys, please tell where i'm
wrong..
LOD1 q43 pg457 In how many ways 4 socks
can be selected out of 5
blue socks, 4 red socks and
3 green socks. Options
245, 120, 495, 60
OA 495
my approach-
A selected sock can be
blue red or green. So 1st
sock can be selected in 3
ways. Similarly, second
third and fourth can be
selected in 3 ways each.
But selection of 4 green
socks is not possible. So
no. of ways=3^4-1=80
Puys, please tell where i'm
wrong..
I don't think u need to consider the cases separately...
there r 12 socks(5+4+3) and u need to select 4 in tat.. so 12C4 = 495
A simple one... But getting wrong answer!!!!!!!!!
if f(x)=f(x-2)-f(x-1)
and f(1)=0 and f(2)=1
1)what is the value of f(
a)0 b)-5 c)13 d)-9 e)-8
2) f(7)+f(4)
please explain ur approach..
i calculated f(3) f(4) and so on till f(..
A simple one... But getting wrong answer!!!!!!!!!
if f(x)=f(x-2)-f(x-1)
and f(1)=0 and f(2)=1
1)what is the value of f(
a)0 b)-5 c)13 d)-9 e)-8
2) f(7)+f(4)
please explain ur approach..
i calculated f(3) f(4) and so on till f(..
just going by ur approach as a straight forward method, we can calculate the corresponding f(x) values
where we are getting f( 8 ) = 13
f(7)=-8 & f(4)=2
so f(7) + f(4) = -6+2 = -4
just going by ur approach as a straight forward method, we can calculate the corresponding f(x) values
where we are getting f( 8 ) = 13
f(7)=-8 & f(4)=2
so f(7) + f(4) = -6+2 = -4
yeah thanku 😃 some silly mistakes which i always do ... btw is tat the normal approach??? or any other approach??
chandrakant.k Saysyeah thanku 😃 some silly mistakes which i always do ... btw is tat the normal approach??? or any other approach??
thr may b some other approaches as well but this is too easy a method to miss... 😉