@vijay_chandola could you please explain a bit more, just didn't get the question at all. i mean how to proceed.
@Faruq said:Guppy has a watch that shows the date without the month and the year. By default, the watch displays 31 days in each month. Therefore, at the end of all the months with less than 31 days the date on the watch needs to be readjusted. On 10th March 2001 it showed the right date as โฌห10 โฌโข. What date would it show on 15th May 2002, if it is known that Guppy never readjusted his watch during this period?(a) 23 (b) 7 (c) 8 (d) 22
15-8=7
months with 30 days april, june,sept, nov,april (5 months, so running 5 days backward)
months with 28 days feb. (3 days backward)
so 15-5-3=7
OA=7?
months with 30 days april, june,sept, nov,april (5 months, so running 5 days backward)
months with 28 days feb. (3 days backward)
so 15-5-3=7
OA=7?
@somnathbhatta said:@vijay_chandola could you please explain a bit more, just didn't get the question at all. i mean how to proceed.
divide the whole circle in 18 equal parts (19*18 =342)
Rest part wil be off angle 360-342=18.
Difference = 19-18=1.
==> this one degree can divide the whole circle in 360 parts :)
@ravi.theja said:actual no.of day 365 + 21 + 15 = 401.but the watch shows 31 * 12 + 21 + 15 = 408.watch shows 7 days more soon 15th ==> 22 will be shown??
ravi ,i think actual no of days should be 365 + 21 + 30 + 15 = 431
we need to take the month april
we need to take the month april
@The_Loser said:BW 100 to 199 --- we have 19 numbers having 2 as a digit in them. 200 to 299 --- 119 numbers having 2 as a digit in them. 300 to 499 --- 19 . . . 700 to 799 --- 1919 * 6 + 119 = 233??
100 to 199 --- we have 20 numbers having 2 as a digit in them.
200 to 299 --- 120 numbers having 2 as a digit in them.
300 to 499 --- 20
. . . 700 to 799 --- 20
20 * 6 + 120 = 240
200 to 299 --- 120 numbers having 2 as a digit in them.
300 to 499 --- 20
. . . 700 to 799 --- 20
20 * 6 + 120 = 240
@vijay_chandola said:51 P.S. Sir kahan ho ajkl?
@pirateiim478 said:51??
@vijay_chandola Bhai Ab se CAT13 k liye yahi hu bas
If i take 11 to 20
41 to 50
71 to 80
91 to 100
Now if i will take any one another No. then it will form at least one pair of difference 10
So iss tarah se toh 40+1=41 number hue.
Please Explain Bhai
If i take 11 to 20
41 to 50
71 to 80
91 to 100
Now if i will take any one another No. then it will form at least one pair of difference 10
So iss tarah se toh 40+1=41 number hue.
Please Explain Bhai
@Faruq said:@vijay_chandola Bhai Ab se CAT13 k liye yahi hu bas
If i take 11 to 20
41 to 50
71 to 80
91 to 100
Now if i will take any one another No. then it will form at least one pair of difference 10
So iss tarah se toh 40+1=41 number hue.
Please Explain Bhai
I took 50 numbers out of first 100 such that none of the pairs has numbers would differ by ten. take 5 sets from 1 to 100 { 1-20, 20-40, 40-60.... 80-100}
From each set, we can take half of the numbers such that gap would not be 10.
So, 5*10 = 50 numbers
Now, we have to pick any one number to ensure that we have picked the required pair.
@Faruq said:@vijay_chandola Agree with your method
But what is wrong with my method.???
I'm not sure 
, but might be missing some of the pairs by starting it from 11-20
, but might be missing some of the pairs by starting it from 11-20If we take 1-10, 20-31, 31-40, it gives 51 answer. :)
Consider the expression 1รท2รท3รท5รท7รท11รท13รท17รท19 . If you are allowed to put as many parentheses as you want, how many different values can be obtained for this expression?
a. 256
b. 512
c. 64
d. 128
e. 1024
a. 256
b. 512
c. 64
d. 128
e. 1024
@Faruq said:Consider the expression 1รท2รท3รท5รท7รท11รท13รท17รท19 . If you are allowed to put as many parentheses as you want, how many different values can be obtained for this expression? a. 256 b. 512 c. 64 d. 128 e. 1024
We can put the parenthsis at 9 places.
So, 2^9 = 512? :neutral:
@Faruq
hmmm made a general expression
for
1) (1/2)= only 1 possible
2) 1/2/3 = (1/2)/3 &
1/(2/3)
2 possible cases
Similarly this was the genreal term is 2^(n-2)(n being the no of numerals involved)
Hence 2^(9-2)=2^7
for
1) (1/2)= only 1 possible
2) 1/2/3 = (1/2)/3 &
1/(2/3)
2 possible cases
Similarly this was the genreal term is 2^(n-2)(n being the no of numerals involved)
Hence 2^(9-2)=2^7
@vijay_chandola @ani4588
We would like to form sets of 3 words from the letters contained in the set S={A, B, C, D, E, F, G, H, I} such that each word has 3 distinct letters and no two words share the same letter. How many different such sets of words can we have?
a.60479
b.60478
c.60483
d.60480
e.60477
Yeh toh question hi ni samajh aa raha
We would like to form sets of 3 words from the letters contained in the set S={A, B, C, D, E, F, G, H, I} such that each word has 3 distinct letters and no two words share the same letter. How many different such sets of words can we have?
a.60479
b.60478
c.60483
d.60480
e.60477
Yeh toh question hi ni samajh aa raha
7^13+1 /6 ? Remainder