Aptitude Questions and Answers

Here Number of odd days in 1600 years = 0

Number of odd days in 300 years from 1600 to 1900 = 5*3 = 2 week + 1 odd day= 1 odd day

Number of odd days in 84 years= 21 leap year + 63 days = 21*2 + 63*1 = 105 days = 0 odd days

Number of odd days in 20 may = 31 days of Jan. + 28 days of feb + 31 days of mar. + 30 days in april + 20 days in may = 140 days = 0 odd day

So total number of odd days = 0+1+0+0=1 = Monday

The clock loses 1% time during the first week.

In a day there are 24 hours and in a week there are 7 days. Therefore, there are 7 * 24 = 168 hours in a week.

If the clock loses 1% time during the first week, then it will show a time which is 1% of 168 hours less than 12 Noon at the end of the first week = 1.68 hours less.

Subsequently, the clock gains 2% during the next week. The second week has 168 hours and the clock gains 2% time = 2% of 168 hours = 3.36 hours more than the actual time.

As it lost 1.68 hours during the first week and then gained 3.36 hours during the next week, the net result will be a -1.68 + 3.36 = 1.68 hour net gain in time.

So the clock will show a time which is 1.68 hours more than 12 Noon two weeks from the time it was set right.

1.68 hours = 1 hour and 40.8 minutes = 1 hour + 40 minutes + 48 seconds.

i.e. 1 : 40 : 48 P.M.

A runs 25% as fast as B.
That is, if B runs 100m in a given time, then A will run 125m in the same time
In other words, if A runs 5m in a given time, then B will run 4m in the same time.

Therefore, if the length of a race is 5m, then A can give B a start of 1m so that they finish the race in a dead heat.

Start : length of race :: 1 : 5

In this question, we know that the start is 7m.

Hence, the length of the race will be 7 * 5 = 35m.

Let the speed of the bus be x m/s
speed of the car = 68km/hr (or) `170/9` m/s
in 10s the car covers a distance of 100m (40m + 60m = 100m) at a speed of (`170/9` – x)m/s
distance = speed*time
`100 = (170/9 -x) **10`
x = `80/9` m/s
Now convert it to km/hr which is 32 km/hr

Speed of the train relative to man = (68 - 8) kmph = (60* `5/18`) m/sec = (`50/3`)m/sec
Time taken by the train to cross the man I = Time taken by It to cover 150 m at `50/3` m / sec
= 150 *`3/ 50` sec = 9 sec

Let us assume the value of x to be 10%

Therefore, the number of sheep in the herd at the beginning of year 2001 (end of 2000) will be 1 million + 10% of 1 million = 1.1 million

In 2001, the numbers decrease by y% and at the end of the year the number sheep in the herd = 1 million.
i.e., 0.1 million sheep have died in 2001.

In terms of the percentage of the number of sheep alive at the beginning of 2001,

it will be `(0.1/1.1) ** 100% = 9.09%`

From the above illustration it is clear that x>yx

As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply initial amount by 1.5 thrice and by 0.5 thrice (in any order).

The overall resultant will remain same.

So final amount with the person will be (in all cases):

=64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs 27

Hence the final result is: 64-27 =37

The only way to roll higher on one die is if the magicians rolls between 2 and 5, inclusive, with two dice. Where he to roll a six with two dice than there is no way he could eclipse that number by rolling one die.

Below is the probability of rolling a certain number with two dice.

‘2’ – 1/36

‘3’ – 2/36

‘4’ – 3/36

‘5’-  4/36

Now the only numbers a magician can roll with the one die and win is between 3 and 6, inclusive. The chances of rolling any are always 1/6.

Next, we have to combine the probability distribution relating to two dice with that relating to the one.

The chances of a magician rolling any given number with one die are 1/6. So let’s start with the lowest number he can roll: a ‘3.’ To win with this roll, he will have to roll a ‘2’ with two dice, the odds of which are 1/36. So .

So the chances of him rolling a ‘3’ with one die and winning are 1/216.

Let’s repeat this logic for the next roll, ‘4.’ Chances of rolling are 1/6. Only way a ‘4’ wins is if he rolls a ‘2’ or a ‘3’ with two dice.

Odds of rolling a ‘2’ with two dice  + odds of rolling a ‘3’ with two dice = . Combine this with the odds of rolling a ‘4’ ( which is 1/6): .

If he rolls a ‘5’ with one die, he can win if he rolls a ‘4’ with two dice, the probability is 3/36. He can also win if he rolls a ‘2’ or a ‘3’ with two dice, the number outcomes we just found: 3/36. So we add

Next, if he rolls a ‘6’ on one die, he can beat ‘2’ through ‘5’ with the two dice. Number of ways to roll a ‘5’ = 4/36. Combining this with the odds of rolling a ‘6’ on one die with the odds or rolling ‘2’, ‘3’, ‘4’, or ‘5’ with the two dice we get: 10/216.

You may be wondering why I left the denominator as 216. Well, this allows us to add up all the instances he can possibly win:

The series is 1³ + 1², 2³ + 2², 3³ + 3²

`"Total age of all students" /22 = 21 `

Total age of all students= 22 *21

`"Total age of all students + age of the teacher"/(22+1) = 21+1`

Total age of all students + age of the teacher = 22 * 23

Age of the teacher = 23×22 – 22×21 = 22(23-21) = 22×2 = 44