R U Genius in Mathematics?????

[deepika]

oho sabse haath milana tha....i got it wrong...anywhich ways this thread is not meant for me ha ha ha ha

5 handshakes

 

[deepika]

Akshay ji i give up u mere bas ka thota toh engineering na karti mai

Try this Deepika...

 

[akshaymalik84]

Akshay ji i give up u mere bas ka thota toh engineering na karti mai

hahahaha.....chalo theek se Deepika ji

 

[anil_rathee]

i got it

bhai Akshay i got the answer but its only in case () are allowed...Tell me if () not allowed then i will think of another option

8/(3-(8/3))

= 8/(1/3)
= 24

Get 24 from 8,8,3,3 by applying addition,subtraction,multiply and
division

 

[akshaymalik84]

bhai Akshay i got the answer but its only in case () are allowed...Tell me if () not allowed then i will think of another option

8/(3-(8/3))

= 8/(1/3)
= 24

Sahi answer bhai Anil,.....there are still one or two ways to solve it.

 

[anil_rathee]

Ramanujam nahi hai bhai hum

bhai ek neii hi bahut dimaag chuss liya
itna dimaag bhi naa sei

Naya question de de ibb
usmein dimmag laga lenge

Sahi answer bhai Anil,.....there are still one or two ways to solve it.

 

[akshaymalik84]

One more

A B C
......D
.......E F G
.............H
..............I

Each of the letter represent different digit 1,2,3,4,5,6,7,8 and 9.

And A+B+C = C+D+E = E+F+G = G+H+I = 13

what is the value of E?

 

[kapdal]

One more

A B C
......D
.......E F G
.............H
..............I

Each of the letter represent different digit 1,2,3,4,5,6,7,8 and 9.

And A+B+C = C+D+E = E+F+G = G+H+I = 13

what is the value of E?

E= 4

Prima facie, looks like no solution as no. of variables are greater than no. of equations. However, you can split 13 into likely combinations:
1 3 9
1 4 8
1 5 7
2 3 8
2 4 7
2 5 6
3 4 6

Now, you have to pick 4 combos from above, such that there are only 3 numbers repeated only twice (C, E and G) and all other numbers not being repeated. You can see that those combos are:
1 3 9
1 4 8
2 4 7
2 5 6

Now arrange them in such that C E and G are common in above equation:
9 3 1 = 1 8 4 = 4 7 2 = 2 5 6

Hence, E= 4
From above, clearly, C = 1 and G = 2. However, other alphabets that can be solved are the middle ones in the middle equations, that is D=8 and F=7. A, B, H and I can't be solved uniquely as they have 2 solutions each.

 

[akshaymalik84]

E= 4

A, B, H and I can't be solved uniquely as they have 2 solutions each.

thats why i ask value of E only

 

[kapdal]

thats why i ask value of E only

I know, but I was just giving the whole solution with rationale. Guess there is nothing wrong with that

 

[akshaymalik84]

I know, but I was just giving the whole solution with rationale. Guess there is nothing wrong with that

well done kapil

 

[srgorchhiya]

45 if i am not wrong

9+8+7+6+5+4+3+2+1=45

 

[akshaymalik84]

A man wanted to get into his work building, but he had forgotten his code. However, he did remember five clues. These are what those clues were:
The fifth number plus the third number equals fourteen.
The fourth number is one more than the second number.
The first number is one less than twice the second number.
The second number plus the third number equals ten.
The sum of all five numbers is 30.
What were the five numbers and in what order?

 

[rakeshsehrawat]

Code is 74658

Let code be ABCDE
CLUE A+B+C+D+E =30, E+C=14, D=B+1, A=2B-1
BY USING CLUES WE GET A+B+D=16=2B-1+B+B+1
B=4
A=7
D=5
C=6
E=8