Quantitative Aptitude - Arithmetic Ability Questions

Total number of events=52

Number of aces =4

So,required probability = 4/52 =1/13

Two horses A and B, in a race of 6 horses... A has to finish before B

if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4!

if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4!

if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!

if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4! 

if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..

A cannot finish 6th, since he has to be ahead of B

Therefore total number of ways = 5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360

From the given data,

3500x7xt/100 = 500

=> t = 100/49 years

Now, in the second case

The interest per year = 49/100 x 800 = 392

 => 4900 x 1 x r/100 = 392

=> r = 8%

In case B, odd times even will give even

74.99% of 1299 + 8.98% of 1899 = ? 

$$\begin{gathered} ? = 75 x 1400100 + 9 x 1800100 \\ ? = 1050 + 162 \\ ? = 1212. \end{gathered}$$

  LCD is nothing but Lowest or Least Common Denominator

Here LCD of 12 and 18 means LCD of two fractions with denominators 12 and 18 respectively.

Therefore, LCM of 12 & 18 = 6 x 3 x 2 = 36 

•  How to calculate LCD ::

The lowest common denominator or least common denominator (LCD) is the least common multiple (LCM) of the denominators of a set of fractions.

Let the amount invested by saketh = RS. p

Now, that of sandeep = 20,000 x 6

saketh = 12 x p

Ratio of their earnings = 120000 : 12p = 6000 : (9000 - 6000)

=> $$\begin{gathered} 12000012p = 60003000 \\ => p= Rs. 5000 \end{gathered}$$

Hence, the amount investe by saketh  = Rs. p  = Rs. 5000.

Let the place be at a distance of 'd' kms

From the given data,

$$\begin{gathered} d5 - 1 + d5 + 1 = 7560 kms \\ d4 + d6 = 54 \\ \end{gathered}$$

5d/12 = 5/4 => d = 3 kms.

Hence, the place is 3 kms far.