Quantitative Aptitude - Arithmetic Ability Questions

Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then 

 Required probability = PA∩B∩C 

= PA PBA PCA∩B

 Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9

When  a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white

 ∴PBA=317 

Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.

 ∴PCA∩B =216=18

 Hence the required probability = 29×317×18=1204

At 5% more rate, the increase in S.I for 10 years = Rs.600  (given)

So, at 5% more rate, the increase in SI for 1 year = 600/10 = Rs.60/-

i.e. Rs.60 is 5% of the invested sum

So, 1% of the invested sum = 60/5

Therefore, the invested sum = 60 × 100/5 = Rs.1200

Simple interest is given by the formula SI = (pnr/100), where p is the principal, n is the numberof years for which it is invested, r is the rate of interest per annum

In this case, Rs. 1250 has become Rs.10,000.

Therefore, the interest earned = 10,000 – 1250 = 8750.

8750 = [(1250 x n x 12.5)/100]

=> n = 700 / 12.5 = 56 years.

From the directions given : 

From the fig. it is clear that 

1) neighbours of B are G , H.

2) E is facing H.

Let p% of a day is 3 hrs.

We know that a day has 24 hrs.

=> p x 24/100 = 3

=> p = 300/24

=> p = 12.5%

Hence, 3 hrs is 12.5% of a day.

Let their prices be 3x,  5x and 7x.

Then, 3x + 5x + 7x = (15000 * 3) or x = 3000.

Cost of cheapest item = 3x = Rs. 9000.

NOTE :

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2:  Add +6 to the Given Year.

Solution : 

Given Year is 2024, Which is a leap year.

So, Add +28 to the given year (i.e 2024 + 28) = 2052

Therfore, The calendar of the year 2024 can be used again in the year 2052.