Aptitude and Reasoning Questions

Let 138.009 + 341.981 - 146.305 = 123.6 + z

Then, z = (138.09+341.981) - (146.305+123.6)= 479.99 - 269.905 = 210.085

Let the total number of votes polled in the election be 100k.
Number of valid votes = 100k - 20% (100k) = 80k
Let the number of votes polled in favour of K and L be a and b respectively.
a - b = 15% (100k) => a = b + 15k
=> a + b = b + 15k + b
Now, 2b + 15k = 80k and hence b = 32.5k
It is given that 100k = 8720
32.5k = 32.5k/100k x 8720 = 2834

The number of valid votes polled in favour of L is 2834.

7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.

1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc

This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)

You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.

In each of these numbers, 7 is written once. Therefore, 243 times.

2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77

In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).

There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.

In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.

3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.

Therefore, the total number of times the digit 7 is written between 1 and 999 is

243 + 54 + 3 = 300

Let A be the event of first person hitting the target,

P(A) =33+5=38 (odd in favour)

Let B be the event of Second person hitting a target.

P(B)=23+2=25 (odd against)

Since both events are independent and both will hit the target so,

P(A∩B)= P(A)P(B)=38×25=320

The given series follows the rule,
+(5x1), +(5x2), +(5x3), +(5x4), +(5x5),....

So starting from 37, 37 + 5 = 42, 42 + 10 = 52, 52 + 15 = 67, 67 + 20 = 87,...
So wrong number is 47 which is to be 42.

As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply the 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

A loss of Rs.37