Quantitative Aptitude - Arithmetic Ability Questions

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 number of 20 ps coins = x and

number of 25 ps coins = y

Given total coins in the bag = 50

x + y = 50.......(1)

But the total money in the bag = Rs. 10.25

0.20x + 0.25y = 10.25

20x + 25y = 1025.........(2)

Now multiplying (1) by 25 we get

25x+25y=1250.............(3)

By solving (2) and (3)

20x + 25y = 1025;

=> x = 45;

Then, the no. of 20 ps coins are 45.

Total number of persons = 9

Out of 9 persons 4 persons can be selected in 9C4 ways =126

1 man,1 woman and 2 children can be selected in 3C1*2C1*4C1 ways =36

So,required probability = 36/126 =2/7

Given exp = 18225102+18225104+18225106+18225108

               = 1822510+18225102+18225103+18225104

               = 13510+135100+1351000+13510000

               = 13.5+1.35+0.135+0.0135=14.9985

Initial quantity of copper =80100 x 50 = 40 g 

And that of Bronze = 50 - 40 = 10 g

Let 'p' gm of copper is added to the mixture

=> 50 + p x 90100 = 40 + p

=> 45 + 0.9p = 40 + p

=> p = 50 g

Hence, 50 gms of copper is added to the mixture, so that the copper is increased to 90%.

Converting each of the given fractions into decimal form, we get

2/3 = 0.66, 3/4 = 0.75, 4/5 = 0.8, 5/6 = 0.833 

Since 0.833>0.8>0.75>0.66 

So, 5/6 > 4/5 > 3/4 > 2/3 

Therefore,  Required Difference = 5/6 - 2/3 = 1/6

At first glance it might seem that this problem cannot be solved because we do not have enough
information. It can be solved as long as you double whatever amount you start with. If we start with
$100, then P = $100 and FV = $200.

FV=P(1+r/n)^nt

a^3 + b^3 = (a+ b) * (a^2 + b^2 - ab)

[ (10.3)^3 + (1)^3 ] / [(10.3)^2  + (1)^2 - (1 * 10.3)]

 => a + b = 10.3 + 1 = 11.3