Puzzles ocntest -114

[HELP]

1)[/b].It was a beautiful sunny morning. The air was fresh and a mild wind was blowing against my wind screen. I was driving from Bangalore to Brindavan Gardens. It took me 1 hour and 30 minutes to complete the journey. .After lunch I returned to Bangalore. I drove for 90 minutes. How do you explain it?


2)[/b]. Can you name the smallest integer that can be written with two digits?


3)[/b].There are thirty-four lines that an~ tangent to a circle, and these line$ create regions in the plane. Can you tell how many. of these regions are not enclosed?


4)[/b]. Here is a multiplication:


159 x 48 = 7632


Can you see something peculiar in this multiplication? Yes, all the nine digits are different. How many other similar numbers can you think of?5)[/b]. What is the sum of all numbers between 100 and 1000 which are divisible by 14?

[srividya_tkr]

Answers:

1. 1 Hr and 30 min is same as that of 90 minutes

2. I think its 1/1,2/2,.....9/9(check with Shakuntala devi puzzles book)

3. 32 * 2 =68(since every tangent provides 2 point of intersection on the circle boundaries)

4.Check shakuntala devi book

5. 14(8 + 9 +.......+71) sice 14 * 8 =112(above 100 and multiple of 14) and 14*71 =994(below 1000)

Now

use sum of n no.s formula i.e. n(n+1)

----------

= 14( from sum of 1 to 71 subtractthe sum of 1 to 7,since we need the sum from 8 to 71 which is a multiple of 14)

=14 [71(72) - 7(8)]

-------- ------

22

= 14 [ 2556 - 28 ]

=14*2528

=35392(check the values,but I think this method will work out)

[WISDOM]

Originally Posted by srividya_tkr

Answers:

1. 1 Hr and 30 min is same as that of 90 minutes

2. I think its 1/1,2/2,.....9/9(check with Shakuntala devi puzzles book)

3. 32 * 2 =68(since every tangent provides 2 point of intersection on the circle boundaries)

4.Check shakuntala devi book

5. 14(8 + 9 +.......+71) sice 14 * 8 =112(above 100 and multiple of 14) and 14*71 =994(below 1000)

Now

use sum of n no.s formula i.e. n(n+1)

----------

= 14( from sum of 1 to 71 subtractthe sum of 1 to 7,since we need the sum from 8 to 71 which is a multiple of 14)

=14 [71(72) - 7(8)]

-------- ------

22

= 14 [ 2556 - 28 ]

=14*2528

=35392(check the values,but I think this method will work out)

just wondering ans to 2nd question can be 0 to the power 1 that wud be 0. don't know this question isin shakuntla aunty!

liked u'r way of sloving 5th question. easy one but many can mess up in this

[diya_11]

5)Find the sum of all numbers between 100 and 1000 divisible by 7.

Ans)The numbers divisble by 7 above 100 and below 1000 are
112,126, ...... 994.
Thse r in A.p so a=112 and d=14
Assume it contains n terms, then
112+(n-1)14=994
14(n-1)=994-112
14(n-1)=882
n=64
Sum=(n/2)*(first term+last term)
sum=(64/2)*(112+994)
=32*1106
=35392