Balloon Recalculation going ON! - Page 173

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Kruts thumbnail
20th Anniversary Thumbnail Voyager Thumbnail
Posted: 19 years ago
Anna's term mark was 80%. Her exam mark was 90%. In calculating her final mark, the term mark was given a weight of 70% and the exam mark a weight of 30%. What was her final mark?

83%

OK...here goes the palindrome answer - 36!!
*dolly* thumbnail
20th Anniversary Thumbnail Sparkler Thumbnail Engager Level 1 Thumbnail
Posted: 19 years ago
pam yes dear it was correct abhi wrote that on last page

thank uuuu
Kruts thumbnail
20th Anniversary Thumbnail Voyager Thumbnail
Posted: 19 years ago
pams:[quote]if they dont want us to google, logic sums r welcome...y pure math?[/quote]

these are logic problems with math involved
😆
let me know if you want to understand the palindrome
bystander thumbnail
21st Anniversary Thumbnail Voyager Thumbnail + 2
Posted: 19 years ago
not sure if someone answered and i missed

Which Indian musician first conducted the London Philharmonic Orchestra?

Ravi Shankar
Naushad
Jaidev
S D Burman

is it the first one?
*dolly* thumbnail
20th Anniversary Thumbnail Sparkler Thumbnail Engager Level 1 Thumbnail
Posted: 19 years ago
yehhhh
thank you krutiiii 👏
Starttofinish thumbnail
20th Anniversary Thumbnail Rocker Thumbnail + 2
Posted: 19 years ago
Yes that is correct. These are mostly math qs. with logic and are more fun - according to me 😊
Starttofinish thumbnail
20th Anniversary Thumbnail Rocker Thumbnail + 2
Posted: 19 years ago
Math qs. are definitely better then the IF qs. Wish they had removed all IF qs.
princess belle thumbnail
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Posted: 19 years ago
hey kruti the mathmatecian help me with the pyramid question pls😉
Starttofinish thumbnail
20th Anniversary Thumbnail Rocker Thumbnail + 2
Posted: 19 years ago
I gtg. Bye guys. Catch you all later. Good night to all in India. Go get some sleep.
Kruts thumbnail
20th Anniversary Thumbnail Voyager Thumbnail
Posted: 19 years ago
two four digit palindromes that sum up to a five digit palindrome (it has to have 1 as first and last digits for this to even work):
a b b a
c d d c
-------
1 e f e 1

now a+c has to be less than 20 but greater than 10 to obtain a 1 at the end.

Thus a+c=11.

There are a total of eight possibilities but out of these, 4 are duplicates because we have not defined a and c yet...Thus the 4 possibilities are:
a 2 3 4 5
c 9 8 7 6

Let us take a=2 and c=9 for example:
Now we have 2 b b 2 and 9 d d 9 as our numbers.

Thus e is either equal to 1 or equal to 2 based on if we need to carry over a digit or not (11 -> 10 + 1).
If e=1 => b+d has no carry over digit => b+d=0
If e=2 => b+d has carry over digit...similar logic as earlier, b+d=11 is the only possibility.

Now, we know that b+d=0 or b+d=11:
If b+d=0...only once choice for b and d (both equal to zero individually) --> 4x1 = 4 possibilities.
If b+d=11...8 ways to choose b and d and 4 choices for a&c ==> 4x8 =32.

Therefore total possibilities = 4+32=36!!

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