Quantitative Aptitude - Arithmetic Ability Questions

Let the number be 'x'. Then, from given data

x/2 + x/3 + x/4 = x+22
13x/12 = x+22
x = 264

Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F of 551 and 1073 = 29;
First number = 551/29 = 19
Third number = 1073/29 = 37.
Required sum = 19 + 29 + 37 = 85.

Given exp = 18225102+18225104+18225106+18225108

               = 1822510+18225102+18225103+18225104

               = 13510+135100+1351000+13510000

               = 13.5+1.35+0.135+0.0135=14.9985

Let first part is 'K' and second part is 'L'
then K + L = 50000---------eq1
Total profit = profit on x + profit on y
7000 = (K x 10 x 1)/100 + (L x 20 x 1)/100
70000 = K + 2L-----------------------------------eq2
70000 = 50000 + L
so L = 20000 then K = 50000 - 20000 = 30000
first part on 10% is = Rs.30000

Let the numbers be a and b.
We know that product of two numbers = Product of their HCF and LCM
Then, a + b = 55 and ab = 5 x 120 = 600.
=> The required sum = (1/a) + (1/b) = (a+b)/ab
=55/600 = 11/120

let original radius = r and  new radius = (50/100) r = r/2

original area = πr2 and new area = πr/22

decrease in area =  3πr2/4*1πr2 *100 = 75%

Given number is 987 = 3 x 7 x 47.

So, required number must be divisible by each one of 3, 7, 47.

None of the numbers in 553681 and 555181 are divisible by 3. While 556581 is not divisible by 7.
Correct answer is 555681.

     D _  C  _ A          --------(1)

     D > B > C           --------(2)

from (1) and (2)

     D > B > C > A   ---------(3)

Again      E _ B _ A

But B > A, from (3)

So          E > D > B > C > A      [ Since B is the average of E and A so it is eqidistant from both E and A]