a dishonest milkman purchased milk @ rs. 10/ litre and mixed 5 lt. of water in it .By selling i\the mixture @ rate of 10/lt he earns a profit of 25%. The quantity of the amount of the mixture that he had was ?? a. 15lt. b. 20 lt. c. 25lt. d. 30 lt.
help me out on alligations....
say x litres of milk are mixed with 5 litres.. now 10x will be the cost for x+5 litres
10-10x/x+5 =1/4(10x/x+5)
solving we get x=20
quantity of amount of the mixture that he had = 20+5 =25 litres..
What is the highest power of 54 that divides 31! exactly? a)0 b)1 c)4
What is the greatest power of 1000 which can divide 80! exactly? a)16 b)20 c)6
pls do show the approach of solving dis sort of prblms.
1) 54= 3^3*2
highest power of 3 =14 but we need highest power of 3^3 hence =4
highest power of 2 = 26
since the least power of 4 and 26 is 4 hence option c - 4
2)similar approach 1000 = 2^3*5^3
highest power of 2^3 in 80! = =29
highest power of 5^3 in 80! = =6 hence option c - 6..
profit 10% so actual mixture price 10*.75 =7.5 /lit
now using alligation 10 0(taking water as 0/lit) 7.5 7.5 2.5
so 3:1
here 1 ==5 3 ==15
total 20.
dude, they have asked to find the quantity of mixture and not milk!! so, mixture is 20+5=25 litres 😉 u check if its 25 litre, you will get profit as 25%
A pipe can fill a tank in x hours and another can empty it in y hours. If the tank is 1/3rd full, then the number of hours in which they will together fill it is.....
My answer is coming out to be 2xy/3(y-x) but it is wrong as per book.Please suggest
dude, you are right,even m getting same ans.book ans is wrong 😉
When 30! is computed,it ends in7 zeroes .What is the digit that immediately precedes the zeroes? a)2 b)4 c)6 d)8 e)7 pls do show d entire process.
ans of this ques is option c..... and reason for this is when we want to calc. last non zero digit.....in short v hv 2 remove all the zero digit this means we have to remove 2^7*5^7 from the 30!....then we will be haveing numbers like 2^19*3^...and so on....of which need 2 cal only unit digit bcoz that unit digit will only b the 1st non zero digit,i hope i hv xplained...bcoz i found it really hard 2 xplain this by writing...sry....no offence meant....but i must say a gud question
This is the 13th question from LOD III of the Permutation and Combination chapter.It goes like this:-
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 in which a student can score 60% of the aggregate. (a) 3,30,850 (b) 2,33,551 (c) 1,10,551 (d) 2,20,800 (e) None of these.
This is somewhat like the question w+x+y+z>= N where 0
When 30! is computed,it ends in7 zeroes .What is the digit that immediately precedes the zeroes? a)2 b)4 c)6 d)8 e)7 pls do show d entire process.
I think the answer is option (c). For removing the last 7 zeroes, we have to divide by 10^7. This can be achieved by 2^7 * 5^7. Hence we remove this from the standard form of the number. Now what we get as the standard form of the number is this:- 2^19 * 3^14 * 5^0 * 7^4 * 11^2 * 13^2 * 17 * 19 * 23 * 29
Now it's easy from here. Calculate the unit's digit of this, and that will the last digit. Comes out as 6.
I think the answer is option (c). For removing the last 7 zeroes, we have to divide by 10^7. This can be achieved by 2^7 * 5^7. Hence we remove this from the standard form of the number. Now what we get as the standard form of the number is this:- 2^19 * 3^14 * 5^0 * 7^4 * 11^2 * 13^2 * 17 * 19 * 23 * 29
Now it's easy from here. Calculate the unit's digit of this, and that will the last digit. Comes out as 6.
yaar just calc prop i hv done this this twice...thts why i m damm sure ki ans 8 hi h
