find the multiples of numerator and denominator....
then divide the numerator until a prime no is left in denominator....
after that the method will become very simple ....
for 2nd question also there are various values coming..
196, 441 and other also..
so hw can we determine a single value..
:shocked:
How did u do the 1st question?what is the logic?
A three digit no. 'mnp' is a perfect square and the no. of factors it has is also a perfect square.It is also known that the digits m,n,p are distinct.
1)if (m+n+p) is also a perfect square,what is the no. of factors of the six digit no. mnpmnp?
a)32
b)72
c)48
d)cannot be determined
my take for both the questions is
CBD option d..
if correct will tell u the approach..
209598 SaysThe ans for ist question is (d) and for the 2nd question is (c)
for 2nd question also there are various values coming..
196, 441 and other also..
so hw can we determine a single value..
:shocked:
Given In the Question = a,b,c are distinct....
so mnp = 196...(such that m+n+p = 16)
But Only 196 has Perfect Square Factors which are 9
Now mnpmnp = 196 * 1001 = 2^2 * 7^3 * 11 * 13
Total Factors = 48
A three digit no. 'mnp' is a perfect square and the no. of factors it has is also a perfect square.It is also known that the digits m,n,p are distinct.
2)if the fourth power of the product of the digits of the no. mnp is not divisible by 5, what is the no. of factors of the 9 digit no,mnpmnpmnp?
a)32
b)72
c)48
d)cannot be determined
mnpmnpmnp = A Prime Number * mnp * Another Single Digit
Prime Number = 333667..For Length Of 9 digits...We have to multiply with this atleast...
Please Watch This Link..How I Got 333667
mnp = 196,256,
So mnpmnpmnp = 196196196 = 333667 * 2^2 * 3 * 7^2
Number Of Factors = 2*3*2*3 = 36
Or mnpmnpmnp = 256256256 = 333667 * 2^7 * 3
Number Of Factors = 2*8*2 = 32
Answer is a) 32...Please Check
Am I missing Something
PS: I've not checked for other squares...
So I'll go for CBD...Please Tell.What's the OA??
2)How many integer values of x and y satisty the exp 4x+3y=3 where xa)284
b)285
c)286
d)none
-1000
Put x=0,y=1
Next x=-3,y=5
So y = 4k+1..K from 0 to 249..Number of Integral Solutions = 250
4x+3y = 3
x=3,y=-3
x=6,y=-7
So y = -(4k+3)...From 0 to 249...Number of Integral Solutions = 250
Total Solutions = 500
None Of These
209598 SaysHow did u do the 1st question?what is the logic?
i just checked for the no..which are perfect squares and also have factors as perfect square..
like all perfect cubes have 4 factors..
and after finding that u will see that there are various such no.s so answer will become CBD.
and there is no need of finding all no.if u are able to find any two no. ans is CBD..and we all knw there are more than 1 three digit cube...
mnpmnpmnp = A Prime Number * mnp * Another Single Digit
Prime Number = 333667..For Length Of 9 digits...We have to multiply with this atleast...
Please Watch This Link..How I Got 333667
mnp = 196
So mnpmnpmnp = 196196196 = 333667 * 2^2 * 3 * 7^2
Number Of Factors = 2*3*2*3 = 36
Am I missing Something
PS: I've not checked for other squares...
So I'll go for CBD...Please Tell.What's the OA??
brother just tell me 1 thing hw did u came to know that mnp the no. is 196??
i think there are various other no.s also which are perfect squares and have the no. of factors as perfect squares ..
i am also a bit cnfused..after knwing the answer for 2nd question..
mnpmnpmnp = A Prime Number * mnp * Another Single Digit
Prime Number = 333667..For Length Of 9 digits...We have to multiply with this atleast...
Please Watch This Link..How I Got 333667
mnp = 196
So mnpmnpmnp = 196196196 = 333667 * 2^2 * 3 * 7^2
Number Of Factors = 2*3*2*3 = 36
Am I missing Something
PS: I've not checked for other squares...
So I'll go for CBD...Please Tell.What's the OA??
i have also gone for CBD but the oa is option c..
brother just tell me 1 thing hw did u came to know that mnp the no. is 196??
i think there are various other no.s also which are perfect squares and have the no. of factors as perfect squares ..
i am also a bit cnfused..after knwing the answer for 2nd question..
Oh.. Sorry..Here I thought mnp = 196..coz i was in dilemma ki M+n+p = perfect square...but it was for previous question...editing my post...ASAP....Thanks for pointing out.....
PS: Chek Out ...Post Edited
Take x = 1,y=1/2..Value = x+y+xy = 1+1/2 + 1/2 = 2
Take x=2 and y=1/3...Value = x+y+xy = 2+1/3 + 2/3 = 3
Take x=3 and y=1/4..value = 3+1/4 + 3/4 = 4
...Similarly last term would be 1971 + 1/1972 + 1971/1972 = 1972
superb!!!!!
-10004x+3y = 3
Put x=0,y=1
Next x=-3,y=5
So y = 4k+1..K from 0 to 249..Number of Integral Solutions = 250
4x+3y = 3
x=3,y=-3
x=6,y=-7
So y = -(4k+3)...From 0 to 249...Number of Integral Solutions = 250
Total Solutions = 500
None Of These
The answer is a)284.Even i dont know how.Tats y i posted this question
There were 7 bowlers in a team of 16 players shortlisted to play the next world cup.Statisticians discovered that if you looked at the no. of wickets taken by any of the 7 bowlers of the current team,the no. of wickets taken by them had a strange property.The no's were such that for any team selection of 11 players(having 1 to 7 bowlers) by using the no. of wickets taken by each bowler and attaching a coefficients of +1,0 or -1 to each value available and adding the resultant values, any no. from 1 to 1093,both included could be formed.If we denote W1,W2,W3,W4,W5,W6 and W7 as the 7 values in ascending order what could be the ans to these questions:
1)find the value of W1+2W2+3W3+4W4+5W5+6W6
a)2005
b)1995
c)1985
d)none
2)Find the index of the largest power of 3 contaiened in W1,W2,W3,W4,W5,W6 and W7
a)15
b)10
c)21
d)6
3)if the sum of the coefficients is 0,find the smallest no. that can be obtained
a)-1067
b)-729
c)-1041
d)-1054
There were 7 bowlers in a team of 16 players shortlisted to play the next world cup.Statisticians discovered that if you looked at the no. of wickets taken by any of the 7 bowlers of the current team,the no. of wickets taken by them had a strange property.The no's were such that for any team selection of 11 players(having 1 to 7 bowlers) by using the no. of wickets taken by each bowler and attaching a coefficients of +1,0 or -1 to each value available and adding the resultant values, any no. from 1 to 1093,both included could be formed.If we denote W1,W2,W3,W4,W5,W6 and W7 as the 7 values in ascending order what could be the ans to these questions:
1)find the value of W1+2W2+3W3+4W4+5W5+6W6
a)2005
b)1995
c)1985
d)none
2)Find the index of the largest power of 3 contaiened in W1,W2,W3,W4,W5,W6 and W7
a)15
b)10
c)21
d)6
3)if the sum of the coefficients is 0,find the smallest no. that can be obtained
a)-1067
b)-729
c)-1041
d)-1054
My Take In Bold...Whats the OA?
There were 7 bowlers in a team of 16 players shortlisted to play the next world cup.Statisticians discovered that if you looked at the no. of wickets taken by any of the 7 bowlers of the current team,the no. of wickets taken by them had a strange property.The no's were such that for any team selection of 11 players(having 1 to 7 bowlers) by using the no. of wickets taken by each bowler and attaching a coefficients of +1,0 or -1 to each value available and adding the resultant values, any no. from 1 to 1093,both included could be formed.If we denote W1,W2,W3,W4,W5,W6 and W7 as the 7 values in ascending order what could be the ans to these questions:
1)find the value of W1+2W2+3W3+4W4+5W5+6W6
a)2005
b)1995
c)1985
d)none
2)Find the index of the largest power of 3 contaiened in W1,W2,W3,W4,W5,W6 and W7
a)15
b)10
c)21
d)6
3)if the sum of the coefficients is 0,find the smallest no. that can be obtained
a)-1067
b)-729
c)-1041
d)-1054
W1 = 3^0 = 1
W2 = 3^1 = 3
W3 = 3^2 = 9
W4 = 3^3 = 27
W5 = 3^4 = 81
W6 = 3^5 = 243
W7 = 3^6 = 729
w1+w2+w3+w4+w5+w6+w7 = 1093
Proceed Further....
1. 2005
2. 6
3. -1067
W1 = 3^0 = 1
W2 = 3^1 = 3
W3 = 3^2 = 9
W4 = 3^3 = 27
W5 = 3^4 = 81
W6 = 3^5 = 243
W7 = 3^6 = 729
w1+w2+w3+w4+w5+w6+w7 = 1093
Proceed Further....
1. 2005
2. 6
3. -1067
How did u the values of W1 to W7?Please explain.
Can anyone tell me the approach to this problem
In an examination,the maximum marks for each of the three papers is 50 each.THe maximum marks for the fourth paper is 100.Find the number of ways with which a student can score 60% marks in aggregate.
pls it would be very helpful if someone tells me the solution quickly
209598 SaysHow did u the values of W1 to W7?Please explain.
Well...Sum Can be obtained by .... Coffecients..-1,0,1...
So All Values must be in Base 3....
From 3^0 to 3^6......So I got them.....
Also Remember In such type of questions..You should be aware of such shortcuts....
Happy Learning
Hey puys, this is my first post here. have been following the site for quite some time though. awesome site i must say..
Had a few queries with arun sharma from the chapter percentages.
Pg 158 Q.44.
A salesman is appointed with a basic salary of 1200/month and the condition that for every sales of 10000 and above he will get 50% of basic and 10% of sales. this scheme does not operate for first 10000 of sales. what would be the value of sales if he wants to earn 7600 in a particular month?
Pg 158 Q42
Aus scored X runs in 50 overs. india tied the score in 20% less overs. if indias avg run rate had been 33.33% higher the scores would have been tied 10 overs earlier. find how many runs were scored by australia.
A.250
B.240
C.200
D. Cannot be determined.
My Sol. I started solving the sum through options.
Aus----240----50overs----4.8
Ind-----240----40overs----6
ind-----240---30overs-----8 (6*4/3)
hence all conditions are fullfilled
Apparently as per the book the ans cannot be determined which i couldnt understand why??
Thanks.
Hey puys, this is my first post here. have been following the site for quite some time though. awesome site i must say..
Had a few queries with arun sharma from the chapter percentages.
Pg 158 Q.44.
A salesman is appointed with a basic salary of 1200/month and the condition that for every sales of 10000 and above he will get 50% of basic and 10% of sales. this scheme does not operate for first 10000 of sales. what would be the value of sales if he wants to earn 7600 in a particular month?
Pg 158 Q42
Aus scored X runs in 50 overs. india tied the score in 10% less overs. if indias avg run rate had been 33.33% higher the scores would have been tied 10 overs earlier. find how many runs were scored by australia.
A.250
B.240
C.200
D. Cannot be determined.
My Sol. I started solving the sum through options.
Aus----240----50overs----4.8
Ind-----240----40overs----6
ind-----240---30overs-----8 (6*4/3)
hence all conditions are fullfilled
Apparently as per the book the ans cannot be determined which i couldnt understand why??
Thanks.
Q 44) Is the answer 78,000 .. ?
Q 42) If the Aus scored, 'y' runs then,
40*r = y
And, 30*r*4/3 = y => 40*r = 30*r*4/3
Here, r could be any real number.
And, as you have more than one solutions to the problem the answer is 'cannot be determined'
..
..
please help me with the solution ...
q. what is the sum of factors of 18000 which are divisible by 8 but not by 25?