Puys help....
The remainder when 2^2 + 22^2 + 222^2 +.......+(222...49 times)^2 is divided by 9...
The options are : 2,5,6,7...
ans is 7 coz any squre no wen divded by 9 leaves a remainder 1,4,7 or 0 so 7 fits the ans..
ans is 7 coz any squre no wen divded by 9 leaves a remainder 1,4,7 or 0 so 7 fits the ans..
Puys help....
The remainder when 2^2 + 22^2 + 222^2 +.......+(222...49 times)^2 is divided by 9...
The options are : 2,5,6,7...
So,I tried it again and got it partially correct...Help me plz...
Here is what i did:
2^2 % 9 = 2^2
22^2 % 9 = 4^2
222^2 % 9 = 6^2
2222^2 %9 = 8^2
then,22222^2 % 9= 1^2
222222^2 % 9 = 3^2....
so this will give way to : (1^2 + 2^2 + 3^2 +...+ 8^2) 5times % 9
so 8*9*17/6 = 12*17= 204 (Sum of Squares formula)
Now 204 when divided by 9 gives a remainder 6..which is the answer...
But what is not that 204 * 5 % 9 shud have given the answer ? i mean after all the above series is being repeated 5 times so shud we not take it into consideration?
hi guys plz help me out in some of the problems of P&C; which is always my weak area.....
Q> the number of natural no.s' of 2 or more than 2 digits in which digits from left to right are in increasing order is?
(a) 127
(b) 128
(c) 502
(d) 501
(e) 512
Lets try to stop the motor of the World
Q> Three variants of CAT paper are to be given to 12 students(so that each variant is used for 4 students).In how many ways can students be placed in two rows of six each so that their should be no identical variants side by side and the students sitting one behind the other.Find no of ways this can be done??
(a) (6!)^2
(b) 6*6!*6!
(c) (6!)^3
(d) 30*12C6*6!*6!
Lets try to stop the motor of the World
I had posted this query few days back but no one replied for it.....
Seven different objects must be divided among three people. In how many ways can this be done if one or two of them can get no objects?
(a) 15120
(b) 2187
(c) 3003
(d) 792
here ans is 2187 (3^7) but i think in this ans there are cases in which all people will get at least one.....
if I am wrong plz help me out.....
Seven different objects must be divided among three people. In how many ways can this be done if one or two of them can get no objects?
(a) 15120
(b) 2187
(c) 3003
(d) 792
here ans is 2187 (3^7) but i think in this ans there are cases in which all people will get at least one.....
if I am wrong plz help me out.....
Q> how many 6 digits no.s' contains exactly 4 different digits?
(a) 4536
(b) 294840
(c) 191520
(d) N.O.D
Lets try to stop the motor of the World
So,I tried it again and got it partially correct...Help me plz...
Here is what i did:
2^2 % 9 = 2^2
22^2 % 9 = 4^2
222^2 % 9 = 6^2
2222^2 %9 = 8^2
then,22222^2 % 9= 1^2
222222^2 % 9 = 3^2....
so this will give way to : (1^2 + 2^2 + 3^2 +...+ 8^2) 5times % 9
so 8*9*17/6 = 12*17= 204 (Sum of Squares formula)
Now 204 when divided by 9 gives a remainder 6..which is the answer...
But what is not that 204 * 5 % 9 shud have given the answer ? i mean after all the above series is being repeated 5 times so shud we not take it into consideration?
dude I feel u r missing something....
the series (1^2 + 2^2 + 3^2 +...+ 8^2) is repeated 5 times but there are also 4 remaining terms as these above form only till (222.... 45 times) .
so now to consider the other 4 terms we get the first four terms of the series
i.e, (2^2 + 4^2 + 6^2 + 8^2) is the additional remainders to be considered.
so it implies
=>remainder on [204*5 + 120] divided by 9
=>remainder on [ 6*5 + 3] divided by 9
=>remainder on 33 divide by 9
=>6
hope that helps ..... and if u find any flaw just let me know.
hi guys plz help me out in some of the problems of P&C; which is always my weak area.....
Q> the number of natural no.s' of 2 or more than 2 digits in which digits from left to right are in increasing order is?
(a) 127
(b) 128
(c) 502
(d) 501
(e) 512
Lets try to stop the motor of the World
Hey mate, can you please post the question 'exactly' as it is printed in the book?
The reason I am saying this is:
Natural numbers range from 0 to infinity.
So there are infinite numbers (of more than 2 digits) of the form that every next digit is greater than previous.
Please correct me if I am misinterpreting the problem.
Please provide me with the solution to the below mentioned problem:
CH: Profit n Loss
Q.NO. 21(LOD I)
Pg: 90
Q. How much % more than the CP should a shopkeeper mark his goods, so that after allowing a discount of 12.5% he should have a gain of 5% on his outlay?
Please provide me with the solution to the below mentioned problem:
CH: Profit n Loss
Q.NO. 21(LOD I)
Pg: 90
Q. How much % more than the CP should a shopkeeper mark his goods, so that after allowing a discount of 12.5% he should have a gain of 5% on his outlay?
x-12.5-(12.5x/100)=5
x=20
Please provide me with the solution to the below mentioned problem:
CH: Profit n Loss
Q.NO. 21(LOD I)
Pg: 90
Q. How much % more than the CP should a shopkeeper mark his goods, so that after allowing a discount of 12.5% he should have a gain of 5% on his outlay?
let the cost price be 'c' and the % more than c.p at which he should mark his goods be 'x'
so,
- (12.5/100) = c + 5c/100
solving this equation we get,
7x + 700 = 840
x=20
thus answer is 20%
If still you feel something is not clear, you can clarify....
The equation u hav used is basically: M.P-Discount=SP ?
Please provide me with the solution to the below mentioned problem:
CH: Profit n Loss
Q.NO. 21(LOD I)
Pg: 90
Q. How much % more than the CP should a shopkeeper mark his goods, so that after allowing a discount of 12.5% he should have a gain of 5% on his outlay?
assume c.p =100 ...let the % markup be x....
(100+x)-(100+x)*0.125 = 105
(cost price) - (discount) = 5% gain
solve to get 20%
Hey mate, can you please post the question 'exactly' as it is printed in the book?
The reason I am saying this is:
Natural numbers range from 0 to infinity.
So there are infinite numbers (of more than 2 digits) of the form that every next digit is greater than previous.
Please correct me if I am misinterpreting the problem.
i think you are getting it wrong buddy.....
since digits are higher than previous one so in natural no.s' 1 to 9 first digit will be 12 and last digit will be 123456789 in b/w these no.s we have to find no. of possible no. ......this wat i think and i m not gettin how to make such combinations.....
ques in exactly as per source.....
how to find the maximum value of n such that 77! is perfectly divisible by 720^n.
a) 35
b) 18
c) 17
d) 36
Puys,help reqd..
P-385,Q-20,24
how to calculate f^-1(x)
What will be the domain of the function f(x)=(8-x)C(5-x) for+ve values of x?
Puys,help reqd..
P-385,Q-20,24
how to calculate f^-1(x)
What will be the domain of the function f(x)=(8-x)C(5-x) for+ve values of x?
20.
f(x)= (1/x)+1, we need to find the inverse, to approach the problem in the simplest way, interchange x and f(x), we get x=(1/f(x))+1, rearranging it we get, f(x)=1/(x-1) , so answer is c
24.
This question is deceptive yet simple, they are just asking us the domain, and that too positive ones, since we cant exceed the value to be more than 5, as 5-x would turn negative, hence the soln is {1,2,3,4,5} i.e c
how to find the maximum value of n such that 77! is perfectly divisible by 720^n.
a) 35
b) 18
c) 17
d) 36
720=(2^4)*5*(3^2)
If we divide 77 by the factors of 720 i.e 2, 3, 5(and its powers) and in case of 2 and 3 divide it by 4 and 2 again at the end, we get around 18, 17 and 18 , so the answer has to be 17 i guess
(77/2)+(77/4)+(77/8 )+(77/16)+(77/32)+(77/64)= 38+19+9+4+2+1=73/4 = 18
Similarly with 3 and 5 we get 17 and 18, Hence 17 is the answer since its the lowest
Chap 6 Profit loss, LOD 2 , Q no 16
A shopkeeper makes a profit of Q% by selling an object for Rs 24 . Had the CP and SP be interchanged, It would have led to a loss of 62.5Q%. With the later cost price, what should be the new selling price to get profit of Q%?
a) 34.40 b)32.5 c)25.60 d)38.8 e)none of these
count me in from now on..