Wrong Looking Correct Mathematical Equation!

The following question it puts forth you:

25 - 55 + (85 + 65) = ?

Then, you are told that even though you might think its wrong, the correct answer is actually 5!

Whats your reaction to it? How can this be true? 

 That's how it's perfectly correct!

That's How Equation is Correct!

Why it was looking wrong? 

If you read the data carefully then you will notice '!' attached to number 5 which is being claimed answer. Actually claimed answer is 5! not 5 Read it again...

"Then, you are told that even though you might think its wrong, the correct answer is actually 5!."

Now use of '!' is not limited to the sentences only. In mathematics it's a 'factorial'.

So 5! = 5 x 4 x 3 x 2 x 1 = 120 and 25 - 55 + (85 + 65) = 120 and hence,

25 - 55 + (85 + 65) = 5! 

Now doesn't it look the correct equation? 

How Accurate You are?

In a competitive exam, each correct answer could win you 10 points and each wrong answer could lose you 5 points. You sat in the exam and answered all the 20 questions, which were given in the exam.

When you checked the result, you had scored 125 marks in the test.
Can you calculate how many answers given by you were correct and wrong ?

These should be those numbers! 

Source 

  

Analysis Of Your Result

What was the test ?

Let C be the number of correct answers and W be the number of wrong answers.

Since there are 20 question in total,

C + W = 20    .....(1)

and the score 125 must be subtraction of marks obtained for correct answer and marks due to wrong answers.

10C - 5W = 125   .....(2)

Multiplying (1) by 5 and then adding it to (2),

5C + 5W + 10C - 5W = 100 + 125

15C = 225

C = 15.

From (1), W = 20 - C = 20 - 5 = 5.

Hence, your 15 answers are correct while 5 answers are wrong.

Who Will Be Not Out?

It is a strange cricket match in which batsman is getting bowled in the very first ball he faced. That means on ten consecutive balls ten players get out.

Assuming no extras in the match, which batsman will be not out at the end of the innings?  

Know that lucky player!

Source 

"He Will Be Not Out!"

What happened in the match?

First let's number all the players from 1 to 11 as Batsman 1, Batsman 2, Batsman 3 & so on with last player as Batsman 11. 

Now let's take a look at what must have happened during 1st over.

1st Ball : Batsman-1 got out
2nd Ball : Batsman-3 got out
3rd Ball : Batsman-4 got out
4th Ball : Batsman-5 got out
5th Ball : Batsman-6 got out
6th Ball : Batsman-7 got out

Batsman 8 comes in 

Batsman 2 is still standing at non-striker end watching fall of wickets. Remember in the match all batsman are bowled out so no change in strike because of run out or before catch is taken.

At the end of first over, the strike is rotated and Batsman 2 comes on strike while Batsman 8 at the non striker end.

Now here is what happens in second over.

1st Ball : Batsman-2 got out
2nd Ball : Batsman-9 got out
3rd Ball : Batsman-10 got out
4th Ball : Batsman-11 got out

So the only batsman left NOT OUT is Batsman 8 standing at the non-striker end.
 

What's Next in the Series?

Can you find the next number in series?
5, 8, 17, 47, 242, ? 

And this should be the next number!

Source 

The Next Number In The Series

What was the series?

Let's take a look at the series once again.

5, 8, 17, 47, 242, ? 

Here if we observe carefully, we can notice that,

5^2 - 8 = 17
8^2 - 17 = 47
17^2 - 47 = 242


In short, (n the number) = [Square of (n-2) th number] - [(n-1) th number].

Hence,

47^2 - 242 = 1967.  

I Found! Did you?

A Mathematical Buy!

In a classic wine shop in Flobecq, Belgium, list of three most popular wines are:

- The cost of 1 French wine bottle: 500$
- The cost of 1 German wine bottle: 100$
- The cost of 20 Dutch wine bottles: 100$

Homer Simpson entered the wine shop and he needs to buy

- All three types of wine shop.
- Needs to buy Dutch wine bottles in multiple of 20.
- Need to buy 100 wine bottles

Simpson has only 10000$. How many wine bottle(s) of each type, Simpson must buy? 

Simpson must buy.......Read More.... 

Source 
 

A Buy To Be Mathematical...

What was needed to buy?

Let's recollect the data where cost of each kind of wine bottle is listed.

- The cost of 1 French wine bottle: 500$
- The cost of 1 German wine bottle: 100$
- The cost of 20 Dutch wine bottles: 100$ (Cost of 1 bottle = 5$)

 

Now let F be the number of French bottle, G be the number of German bottles and D be the number of Dutch bottle that Simpson should buy.

F + G + D = 100

G = 100 - F - D   .....(1)

Total cost of all bottle must be $10000.

500F + 100G + 5D = 10000

Substituting (1) in above,

500F + 100(100 - F - D) + 5D = 10000

500F + 10000 - 100F - 100D + 5D = 10000

400F - 95D = 0

400F = 95D 
 
80F = 19D

D/F = 80/19

Possible values of D and F are 80 and 19 respectively.

From (1),

G = 100 - 19 - 80 =  1.

Let's verify if all these fits in his budget or not.  

19 French wine bottles would cost 19 x 500 = 9500, 1 German wine would cost = 1 x 100 = 100 and 80 Dutch wine bottles would cost 80 x 5 = 400. Remember we have got number of Dutch bottles in multiple of 20. Hence total cost = 9500 + 100 + 400 = 10000.

Hence with $10000, Simpson should buy 19 French, 1 German and 80 Dutch wine bottles if conditions of buying 100 bottles & Dutch bottles in multiple of 20 are applied.   

 

Just Try To Crack It!

Can you tell the correct key?

Here is the step-by-step process!

Source 

Cracking of The Code in Steps...

What was the challenge?

Let's number the clues as 1, 2, 3.

Now following step by step process here onward. 

1. The numbers 3 & 1 are common in first & third combinations. Now both must not be the part of original number as in that case Clue 1 will be invalid. 

2. If numbers 3 & 1 are not correct in third combination then the correct 2 numbers must be among 5,7,9.

3. But it can't be both 7 and 9 as again that would make clue 1 invalid! Hence, the 5 is part of the original key & in correct position as in third combination. So we have got first digit of key as a 5.

4. The 1 correct number in clue 2 is 5 & that's in wrong position. If other is assumed to be 9 & to be in right position then it contradict the clue 3. So the second digit must be 7.

5.The only correct number in clue 1 is 7 & that's in wrong position. That means numbers 1,3,4,9 must not be the part of the code.

6. Since 3,4,9 eliminated in previous steps, the only number that is correct and in right position must be 6 in suggestion made by clue 2. So far we have got 3 digits of the code as 576XX.

7. Last 2 digits can be any combination from 0,2,5,7,6,8. Now addition of all digits is equal to the number formed by last 2 digits. It's impossible that the addition of all digits exceeds 50. Hence, the second last digit must be 2.

8. Now both 57620 or 57628 are perfectly valid where sum of all digits equals to number formed by last 2 digits. 

  

Fill In Next Three Blanks

Can you find the next three numbers in the given series?

4, 6, 12, 18, 30, 42, 60, 72, 102, 108,_,_,_

Escape to the answer!

Source 

Numbers Deserving Blanks

How was series looking?

Let's take a look at the series once again.

4, 6, 12, 18, 30, 42, 60, 72, 102, 108,_,_,_ 

If we carefully observe the series, we can find that every number is in between 2 PRIME numbers.

4 is in between 3 and 5. 

6 is in between 5 and 7.

12 is in between 11 and 13.

So on........

108 is in between 107 and 109.

Hence next numbers should be 138, 150 and 180!

 

Mixed up Apples at the Farm

Someone has mixed up the apples. Read carefully what has happened. Can you find a way to help solve the problem.

• There are 10 baskets containing apples.

• There are various amounts of apples in each basket ranging from 10 to 20.

• 9 of the baskets contain apples weighing 4 ounces each.

• 1 of the baskets contains apples weighing 5 ounces each.

• All the apples look the same.

• The equipment you have is a set of scales and an empty basket.

• It is late and the truck is waiting to take the apples to market. You only have time to make one measurement using the scales. 

Take out the basket contains apples weighing 5 ounces each.

Here is how you can identify that basket! 

Source 
  

Sorting of Mixed Up Apples

How they were mixed?

Let's number all the baskets from 1 to 10. Just take out 1 apple from first basket, 2 from second, 3 from third & so on. We would have 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 apples now. If each of them was of 4 oz, then these 55 apples together would have weighed as 55 x 4 = 220 oz. Since 1 basket have apples each weighing 5 oz, these 55 apples would weigh more than 220 oz.

Let's say it weighs 222 oz, then the 2 apples taken by second basket must be of 5 oz each. And if it weighs 228 then there are 8 apples weighing more than 4 oz (i.e. 5 oz each). Hence that eight basket must have all apples weighing.

So depending on how much weight of 55 apples exceeds 220 oz, we can identify the basket with apples weighing 5 oz each.  

 

The CryptArithmetic Problem

Can you solve the below alphametic riddle by replacing letters of words by a numbers so that the below equation holds true?

BASE +
BALL
---------
GAMES
----------

 
Find numbers replaced letters here! 

Source 

The CryptArithmetic Problem's Solution

What was the problem?

Let's first recall the given equation.

  BASE +
  BALL
---------
GAMES
----------

We are assuming repeating the numbers are not allowed. 

Let's first take last 2 digits operation into consideration i.e. SE + LL = ES or 1ES (carry in 2 digit operation can't exceed 1). For a moment, let's assume no carry generated.

10S + E + 10L + L = 10E + S .....(1)

9 (E - S) = 11L

To satisfy this equation L must be 9 and (E - S) must be equal to 11. But difference between 2 digits can't exceed 9. Hence, SE + LL must have generated carry.So rewriting (1),

10S + E + 10L + L = 100 + 10E + S

9 (E - S) + 100 = 11L

Now if [9 (E - S)] exceeds 99 then L must be greater than 9. But L must be digit from 0 to 9. Hence, [9 (E - S)] must be negative bringing down LHS below 100. Only value of E - S to satisfy the given condition is -5 with L = 5. Or we can say, S - E = 5.

Now, possible pairs for SE are (9,4), (8,3), (7,2), (6,1), (5,0). Out of these only (8,3) is pair that makes equation SE + LL = SE + 55 = 1ES i.e. 83 + 55 = 138. Hence, S = 8 and E = 3.

Replacing letters with numbers that we have got so far.

    1---------
  BA83 +
  BA55
---------
GAM38
----------

Now, M = 2A + 1. Hence, M must be odd number that could be any one among 1,7,9 (since 3 and 5 already used for E and L respectively).

If M = 1, then A = 0 and B must be 5. But L = 5 hence M can't be 1

If M = 7, then A = 3 or A = 8. If A = 3 then B = 1.5 and that's not valid digit. And if A = 8 then it generates carry 1 and 2B + 1 = 8 again leaves B = 3.5 - not a perfect digit.

If M = 9, then A = 4 (A = 9 not possible as M = 9) and B must be 7 with carry G = 1.Hence for first 2 digits we have 74 + 74 + 1 = 149.

Finally, rewriting the entire equation with numbers replacing digits as -

    1
---------
  7483 +
  7455
---------
14938
----------

So numbers for letters are S = 8, E = 3, L = 5, A = 4, B = 7, M = 9 and G = 1.