Q.1. Based on Euclid’s algorithm: a = bq + r ; Using Euclid’s algorithm: Find the HCF of 825 and 175.
Ans:Since 825>175, we apply division lemma to 825 and 175 to get825 = 175 x 4 + 125. Since r ≠ 0, we apply division lemma to 175 and 125 to get 175 = 125 x 1 + 50 Again applying division lemma to 125 and 50 we get, 125 = 50 x 2 + 25. Once again applying division lemma to 50 and 25 we get. 50 = 25 x 2 + 0. Since remainder has now become 0, this implies that HCF of 825 and 125 is 25. Q.2. Based on Showing that every positive integer is either of the given forms: Solved example: Prove that every odd positive integer is either of the form 4q + 1 or 4q + 3 for some integer q. Ans: Let a be any odd positive integer (first line of problem) and let b = 4. Using division Lemma we can write a = bq + r, for some integer q, where 0≤r<4. So a can be 4q, 4q + 1, 4q + 2 or 4q + 3. But since a is odd, a cannot be 4q or 4q + 2. Therefore any odd integer is of the form 4q + 1 or 4q + 3. Q.Find H C F (26,91) if LCM(26,91) is 182 Sol: We know that LCM x HCF = Product of numbers. or 182 x HCF = 26 x 91 or HCF = 26 x 91 = 13 Hence HCF (26, 91) = 13. Q. prove that √5 is irrational. Solution: let us assume on the contrary that √5 is rational. That is we can find co-primes a and b b (≠0) such that √5 = a/b. Or √5b = a. Squaring both sides we get 5b2 = a2. This means 5 divides a2. Hence it follows that 5 divides a. So we can write a = 5c for some integer c. Putting this value of a we get 5b2 = (5c)2 Or 5b2 = 25c2 Or b2 = 5b2. It follows that 5 divides b2. Hence 5 divides b. Now a and b have at least 5 as a common factor. But this contradicts the fact that a and b are co-primes. This contradiction has arisen because of our incorrect assumption that √5 is rational. Hence it follows that √5 is irrational. Q. prove that product of three consecutive positive integers is divisible by 6? Ans: Let three consecutive positive integers be, n, n + 1 and n + 2. Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2. ∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer. If n = 3p, then n is divisible by 3. If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3. If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3. So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3. ⇒ n (n + 1) (n + 2) is divisible by 3. Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1. ∴ n = 2q or 2q + 1, where q is some integer. If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2. If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2. So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2. ⇒ n (n + 1) (n + 2) is divisible by 2. Since, n (n + 1) (n + 2) is divisible by 2 and 3. ∴ n (n + 1) (n + 2) is divisible by 6. Q. Express HCF of F 65 and 117 in the form of 65m +117n Ans: Among 65 and 117; 117 > 65 Since 117 > 65, we apply the division lemma to 117 and 65 to obtain 117 = 65 x 1 + 52 … Step 1 Since remainder 52 ≠ 0, we apply the division lemma to 65 and 52 to obtain 65 = 52 x 1 + 13 … Step 2 Since remainder 13 ≠ 0, we apply the division lemma to 52 and 13 to obtain 52 = 4 x 13 + 0 … Step 3 In this step the remainder is zero. Thus, the divisor i.e. 13 in this step is the H.C.F. of the given numbers The H.C.F. of 65 and 117 is 13 From Step 2: 13 = 65 – 52 x 1 … Step 4 From Step 1: 52 = 117 – 65 x 1 Thus, from Step 4, it is obtained 13 = 65 – (117 – 65 x 1) x 1 ⇒13 = 65 x 2 – 117 ⇒13 = 65 x 2 + 117 x (–1) In the above relationship the H.C.F. of 65 and 117 is of the form 65m + 117 n, where m = 2 and n = –1 1oth maths term-1
44 Comments
Veena
13/8/2013 04:33:42 pm
Really I also need an answer
Reply
Tom
14/5/2012 05:48:11 am
Opss.......its great
Reply
Jb
19/7/2016 07:04:49 pm
Reply
Nimra
29/5/2012 09:19:21 am
Hy! It's easy to prepare frm these questions.
Reply
Jackline
16/6/2012 05:54:05 am
really..superb questions
Reply
Mahima
17/6/2012 02:41:54 pm
Superb
Reply
surabhi
4/7/2012 09:25:51 am
it so really for helpful
Reply
meet
13/7/2012 01:28:43 pm
its just really helpful :)
Reply
12/8/2012 11:55:14 pm
love it! very interesting topics, I hope the incoming comments and suggestion are equally positive. Thanks for sharing information that is actually helpful.
Reply
poonam
29/8/2012 06:14:06 am
not helpful
Reply
summi
3/4/2013 01:27:50 pm
Fore more solved problems
Reply
kartikey
7/4/2013 11:03:46 am
kuch samajh hi nahi aaya
Reply
harini
1/5/2013 10:06:58 am
hi give me for all chapter solutions so, i can get some good marks
Reply
rAjEsH../
8/5/2013 01:22:18 pm
it gud is .i love it just soouestions nice i go home and give ut my tmmrow. n score gud marks I.
Reply
varun
28/6/2013 05:59:56 am
almost helpful .. only easy questions i recommend rd sharma
Reply
arpit
30/6/2013 02:51:04 am
thanks to one who has written these these help me in my holiday homework
Reply
ramita
30/6/2013 02:51:57 am
ya its true
Reply
khushpriya
17/9/2013 04:14:57 am
its really helpful...thnx
Reply
nivetha
17/3/2014 10:58:39 am
now only i completed std 9 so i could not understand any of these things but i already worked out these sums so it is very easy
Reply
nithish balaji
17/3/2014 10:59:23 am
so boaring
Reply
parinnethi
29/3/2014 08:44:44 am
this topics are seeming to b interesting.....i'm fond of euclid division lemma....
Reply
Aaisha Thomas
10/4/2014 12:30:50 am
Its very useful
Reply
bhooom
18/4/2014 05:10:02 am
plz more questions
Reply
Rahul
21/4/2014 11:04:12 am
hi , i didnt quite understand the 5th one though.. others were great
Reply
katheeja rocks
15/6/2014 08:29:20 am
very simple
Reply
vs meghana
6/8/2014 10:33:35 am
its really useful helpful for other classes
Reply
kumud
10/9/2014 02:55:58 pm
its really helpful
Reply
Arjun
7/4/2015 02:25:59 pm
This is bomb
Reply
Janaharshini R
18/6/2015 01:00:36 pm
It is so utile while exercising the sums in the chapter "Real Numbers" for exams.Thanks for dedicating these kind of websites.
Reply
Nameera
21/6/2015 06:01:43 am
NOT HELPFUL TO 10 CLASS STUDENTS. WASTE. NOT ABLE TO UNDERSTAND
Reply
anuja
14/5/2016 06:56:11 pm
nice but can be expressed in an easy way.
Reply
sp
19/5/2016 02:26:16 pm
so booooring
Reply
abi
31/5/2016 06:55:15 pm
waste
Reply
Leave a Reply. |
Blog SeaRCH Link
All
Join Us For Update |