In a town consisting of p persons x% can read and write. Of the males alone y% and of the females alone z% can read and write. Find the number of males in the village in terms of p, x, y and z if z p(x-z)/(y+x-z)
p(x-z)/(y+x-2z)
p(y-x)/(x-z)
p(x-z)/(y-z)
plz give d answer
Solution:
Let the number of males be 100m and the number of females be 100f.
Then, since the total number of persons in the town is p, so we have:
100m + 100f = p
or, 100(m + f) = p ------------- (1)
Since x% of the population can read and write, we have:
xp/100 = (x/100)(100m + 100f) = x(m + f) people who can read and write.
Of the males alone, y% can read and write and of the females alone, z% can read and write.
So, we now know that:
(y/100)(100m) = (m)(y) males can read and write and
(z/100)(100f) = (f)(z) females can read and write.
So, altogether (my + fz) people can read and write which is also equal to (mx + fx)
my + fz = mx + fx
or, m(y - x) = f(x - z)
or, f = m(y - x)/(x - z)
Putting the value of f in equation (1) and simplifying, we get:
I thought that when I calculate x, y and z percent of the respective quantities, I might not have to deal with fractions and so I assumed it to be 100m and 100f, but then I realized that it was all the same but still I did not edit/change my post. Anyways, Thanks for that.
there are three galleries in a coal mine.On the first day ,two galleries are operative and after some time ,the third gallery is made operative.With this,the output of the mine became half as large again . what is the capacity of the second gallery as a percentage of the first;if it is given that a four-month output of the first and the third galleries was the same as the annual output of the second gallery ?
70 %
64 %
60 %
None Of These
Capacity of 1st gallery................x Capacity of 2nd gallery................y Capacity of 3rd gallery................(x+y)/2 Therefore, from the given condition of the problem, we get
I have doubt in one of the prob. of the number system of Arun Sharma.
Here goes the prob. :
Find the max. power of 360 in 570! ??
Please explain the approach to be followed, when the same prime factor appears more than once in the number i.e. in 360.. :-|
360=2*2*2*3*3*5 = 2^3 * 3^2 * 5 So to make one 360 we need 3 2's 2 3's and one 5. Now we need to see how many of 2 3 and 5 are there in 570! umber 2's = 280, so combination of 3 2's ~ 280/3=93
number of 3's = 180 so combination of 2 3's ~ 180/2 =90
number of 5's =88. So min of (93, 90, 88 ) = 88 is the answer.
To get maximum power of 360, we have to divide the 570! by 360 i.e. => 2^565 * 3^283*5^140 / 2^3 * 3^2 * 5^1 => (2^3)^188* (3^2)^141 *(5^1) ^140 / 2^3 * 3^2 * 5^1 so, clearly at most the power could be ...140
Please lemme know if there is any shortcut method for the following question :
Find the remainder when 54^124 is divided by 17 ?
I know the way to determine it using the "mod" approach.... but you have to do it in steps.... Once you get the hang of the method, it's pretty simple and quick to use....
The answer is coming out to be 4 in this case... I can explain the method if you're interested in step-wise solution
if there is any other approach or shortcut method .. plz do lemme know..
This is exactly how I solve these types of problems as well..... but it's quick enough for me as I'd practiced a lot of these type of questions earlier.....
It may be solved using other methods as well.... but I don't use them very often.... maybe using Binomial theorem.... I'm not the right person to ask for this though 😃
a) (2^2) X (4^4) X (6^6) X (8^8 b) (17 X 23 X 51 X 32) + (15 X 17 X 16 X 22)
I am following the below mentioned approach :
a) (2^2) X (4^4) X (6^6) X (8^8 (mod 10) = (2^40 X 3^6) (mod 2*5) i solve (2^40 X 3^6) (mod 5) and get the rem as 4 and then divide by 2. Answer comes out to be 2 (which is correct) .
b) If i follow the same approach i am getting ans as 1 but the ans is 2
please lemme know where am i going wrong .....:dontgeti:
i.e. 2^40 * 3^6 => ( 2^10)^4 * 729 = (1024)^4 *729 = last digit is 4.... so, answer defintely not 2..Hence, your ans is Perfect....approach too gud.
2ndQ) Answer is 2...:) here u miss the buss...i.e. (17 X 23 X 51 X 32) -> last digit is 2 and (15 X 17 X 16 X 22) -> last digit is 0 soo, sum of them will be 2.... ofcourse, to exact (17 X 23 X 51 X 32) + (15 X 17 X 16 X 22) = 638112 +89760 which clearly says.... Last digit is 2.
One more question :
Find the digit in unit's place in the following :
a) (2^2) X (4^4) X (6^6) X (8^8 b) (17 X 23 X 51 X 32) + (15 X 17 X 16 X 22)
I am following the below mentioned approach :
a) (2^2) X (4^4) X (6^6) X (8^8 (mod 10) = (2^40 X 3^6) (mod 2*5) i solve (2^40 X 3^6) (mod 5) and get the rem as 4 and then divide by 2. Answer comes out to be 2 (which is correct) .
b) If i follow the same approach i am getting ans as 1 but the ans is 2
please lemme know where am i going wrong .....:dontgeti:
a) (2^2) X (4^4) X (6^6) X (8^8 b) (17 X 23 X 51 X 32) + (15 X 17 X 16 X 22)
I am following the below mentioned approach :
a) (2^2) X (4^4) X (6^6) X (8^8 (mod 10) = (2^40 X 3^6) (mod 2*5) i solve (2^40 X 3^6) (mod 5) and get the rem as 4 and then divide by 2. Answer comes out to be 2 (which is correct) .
b) If i follow the same approach i am getting ans as 1 but the ans is 2
please lemme know where am i going wrong .....:dontgeti:
@poonam for the second one b) (17 X 23 X 51 X 32) + (15 X 17 X 16 X 22) u can just multiply manually and add.. unit's digit of (17 X 23 X 51 X 32) is 2 and unit's digit of (15 X 17 X 16 X 22) is 0..adding this 2+0 we get 2 right? hope this is simple..
a) (2^2) X (4^4) X (6^6) X (8^8 ) first one is units digit will be 4x6x6x6..so units digit will be 4.. you can use cyclicity for this..will be much simpler..
i.e. 2^40 * 3^6 => ( 2^10)^4 * 729 = (1024)^4 *729 = last digit is 4.... so, answer defintely not 2..Hence, your ans is Perfect....approach too gud.
2ndQ) Answer is 2...:) here u miss the buss...i.e. (17 X 23 X 51 X 32) -> last digit is 2 and (15 X 17 X 16 X 22) -> last digit is 0 soo, sum of them will be 2.... ofcourse, to exact (17 X 23 X 51 X 32) + (15 X 17 X 16 X 22) = 638112 +89760 which clearly says.... Last digit is 2.
hey Naresh,
I am also getting the same answers after taking (mod 5) i.e. 4 and 2 respectively , but after that i divide it by 2 which is left out in the denominator and then i am getting answers as 2 and 1 respectively ..... :dontgeti:
