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Rep Power: 17 | 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?
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Rep Power: 4 | 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) |
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| (e)x-member Join Date: Jul 2006 Location: India
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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 ![]() | |
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Rep Power: 4 | Re: Puzzles ocntest -114 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 |
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