## Solve this simple puzzle and put your answer in comments

Check your answer:- Click here to See Solution

Grape = 5 + 7 = 12

3Bananas = 7 -1 = 6

Apple + Grape + 1Banana = 7 + 12 + (6/3) = 21

One guy had a habit of spending the money as per the day of the month, for example he will spend $18 on 18th of the month.

Once he spent, $63 in continuous 5 days, **What could be the dates** ?

**What was the first day ?**

lets assume all the days were of the same month, in that case

x + (x+1) + (x+2) + (x+3) + (x+4) = 63

5x = 53

x is coming out to be in decimal, which can not be

**Case 2)**

now the other possibility is such that month changes after some days

thus there can be few possibilities

**A) **x + (x+1) + (x+2) + (x+3) + 1 = 63

4x = 56 => x = 14, but 1st cant come after 17th, so it is ruled out

**B)** x + (x+1) + (x+2) + 1 + 2 = 63

3x = 57 => x = 19, but again 1st cant come after 21, so it is ruled out as well

**C)** x + (x+1) + 1 + 2 + 3 = 63

2x = 56 => x = 28, this is possible for the month with 29 days, i.e. February of leap year

**so thee dates were 28,29,1,2,3**

1 9 _ 24 36 _ _ 56 _ _

Click here to See AnswersHide

For first(diff of 6): 0 6 12 **18** 24 30 **36** **42** 48 **54**

For second: put your answers in comments

For second: put your answers in comments

?One grand father

?One father

?One Grand Son

Sum of their ages is 140 years

Grand son’s age in month

is equal to grand fathers age in years

Grand son’s age in days

is equal to Father’s age in weeks

Now, Find out the age of all three.

Check your answer:-

Grand son’s age in years

Father’s age in years

Grand father’s age in years

Click here to See AnswersHide
Lets say G, F and S are ages of grand father, father and son in years

than

G+F+S = 140

S*12 = G

S*365 = (F*365)/7 => F = 7S

Than 12S + 7S + s = 140 => 20S = 140 => S = 7 Years

F = 49 Years

G = 84 Years

than

G+F+S = 140

S*12 = G

S*365 = (F*365)/7 => F = 7S

Than 12S + 7S + s = 140 => 20S = 140 => S = 7 Years

F = 49 Years

G = 84 Years

A lady starts from her house with some money to visit four Churches.

1. As soon as she enters Church1 her money gets doubled. She donates $100 and moves.

2. At Church2 again her money is doubled and she donates $100.

3. Her money is doubled again at Church3 and she donates $100.

4. At Church4 again her money gets doubled. She donates $100 again and returns home empty handed…

How much money she had when she started from her house???

Check your answer:-

Click here to See SolutionHide
Money before church4: $50

Money before church 3: 150/2 = $75

Money before church 2: 175/2 = $87.5

Money before Church 1: 187.5/2 = $93.75

Money before church 3: 150/2 = $75

Money before church 2: 175/2 = $87.5

Money before Church 1: 187.5/2 = $93.75

Puzzle Time challenge?

Chalo solve karo…

**Its a 5 digit number where…**

**1st digit denotes how many zeroes are there in the number**

**2nd digit denotes how many ones are there in the number**

**3rd digit denotes how many twos are there in the number**

**4th digit denotes how many threes are there in the number**

**5th digit denotes how many fours are there in the number**

**Can you guess the number?**

Check your answer:-

Lets say there are 4 0’s, so number should be 40000, but this violates that 5th digit tells how many 4’s are there.

Now lets say there 3 0’s then number can be 30010, but thus also violates, 2nd digit tells number of 1’s

Now if there are 2 0’s then 21200 satisfies all the conditions.

A man buy 3 accessories with 100 rupees. first one 5 rupees, second one 3 rupees, third one 0.5 rupees each. he bought total 100 pieces. How many pieces he bought for each accessory ?

First Accessory

Second Accessory

Third Accessory

Click here to See SolutionHide
lets say he bought x,y and z pieces.

Then we can make these two equations

Eq1: x + y + z = 100

Eq2: 5x + 3y + 0.5z = 100

Then we can make these two equations

Eq1: x + y + z = 100

Eq2: 5x + 3y + 0.5z = 100

2*Eq2 => 10x + 6y + z = 200

2*Eq2-Eq1 => 9x + 5y = 100

from this equation x should be a multiple of 5, for x = 5 => y = 11

for x = 10 => y = 2

for x = 15 or more y becomes -ve

So two possible answers

x = 5, y = 11, z = 84

x = 10, y = 2, z = 88

A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

A. 1/3

B. 1/4

C. 1/5

D. 1/7

Check your answer:-

Click here to See SolutionHide
Lets assume that we take x part of the mixture out, and assuming uniform distribution of water and syrup, we can say out of the x part, 3x/8 was water and 5x/8 was syrup, now on replacing this x part by water, both water and syrup part should become equal, we can form the below equation

Water Part = Syrup part

3 – 3x/8 + x = 5 – 5x/8

x +2x/8 = 2

5x/4 = 2

x = 8/5

initially total quantity was 8(i.e. 3+5), and we should take out 1/5th part out and replace it with water to make both equal.

There are some children on a school ground, a square has been drawn and all children are standing on a squares four lines, they are standing with same distance, four children are standing on a four corners. No 16 Is exactly opposite of No 6. How many children are there?

Check your answer:-

Assuming there are x children in every side of the square, thus a total of 4x -4 children will be there(not counting corner child twice).

Now we know there are atleast 16 children, thus 4x-4 >= 16. x >= 5

Lets assume first child is sitting on bottom left corner, than assume three cases

**Case 1: **x = 5

then 6th child will be standing on second position in second line of square, thus the exactly opposite child will be sitting on x + x-2 + x + x-1 = 4x-3th position, i.e. 17th position.

Which is not the case as per the question, thus x can’t be 5

**Case 2: **x = 6

then 6th child is sitting on the corner(point B), thus exactly opposite child is on x + x-2 + x= 3x-2th position, i.e. 16th.

Thus it satisfies all conditions of the question.

**Thus there are a total of 4x-4, i.e. 20 children in this case.**

**Case 3: **x > 6

then 6th child is standing somewhere in the bottom line (point A), as per this exactly opposite child will be standing at position x + x-2 + x -6 = 3x – 8

for 3x-8 = 16 => x = 24/3 = 8

**Thus there are a total of 4x-4 = 28 children in this case.**