Lakshya Dunivedi
mula Academy, Dewas
Subject: Maths
Class: 10
Worksheet
24
245
At
Q.7
Chapter 1
Real Numbers
isit
Q. 1
The decimal expansion of the rational no. 43 will terminate after how many places of
decimal? Write the decimal expansion
Q. 2
(4,0.0215]
Show that 5 x 7 x 11 + 11 is a composite number.
Q. 3 Using euclid's division algorithm, find the H.C.F. of 272 & 1040.
(16)
Q. 4 Can 12" ends with the digit o for any natural number n; check.
Q. 5 The H.C.F of two no. is 23 & their L.C.M. is 1449. If one no. is 207,find the other number. [161]
Q. 6
Using euclid's division lemma, show that the cube of any +ve integer is of the form 9 m or
9m + 1 or 9m + 8 for some integer m.
Prove that V2 is irrational.
Q.8 The length, breadth &height of a room are 8 m 25m, 6m 75 cm & 4m 50 cm respectively.
Determine,
The length of the longest rod that can be placed in this room (in cm)
2. The length of the longest rod which can measure the three dimensions of the room exactly. (75cm]
There are 156, 208 & 260 students in groups A,B, & Cresp. buses are field trip, herd, if the same no. of
students should be accommodated in each bus & reparate bus for separate bus for separate group is
needed. Find the no. of buses arequired,
[3+4+5=12]
Q. 10 Using euclid's division lemma, show that the square of any +ve integer is either of the form 3m or
31n + 1 for some integer m.
Q. 11 Show that 3+3+5 is an irrational number.
"
1.
Answers
Step-by-step explanation:
2.Since product of two odd number is odd
therefore, 5×7×11 is odd number
similarly 13×17 is an odd number
We know, sum of two odd number is even.
Therefore,(5×7×11)+(13×17)= even number
we know, 2 is only even prime number and remaining all even number are composite.
Hence,(5×7×11) + (13×17) is a composite number.
3. here ,420>272
therefore 420= 272 × 1 + 148
272= 148 × 1 + 124
148= 124 × 1 + 24
124= 24 × 5 + 4
24= 4 × 6 + 0
since remainder is 0, thus the HCF is 4.
5.Let the two numbers be a and b.
Let the value of a be 207.
Given: HCF = 23 and LCM = 1449
We know, a × b = HCF × LCM
⇒ 207× b = 23 × 1449
⇒ ∴ b = (23 ×1449)/207
= 33327/207
= 161
Hence, the other number b is 161.
6.According to Euclid’s Division Lemma
Let take a as any positive integer and b = 9.
Then using Euclid’s algorithm we get a = 9q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2, 3, 4, 5 , 6 , 7 , 8, because 0 ≤r < b and the value of b is 9
Sp possible forms will 9q, 9q+1, 9q+2,9q+3,9q+4,9q+5,9q+6,9q+7 and 9q+8
to get the cube of these values use the formula
(a+b)³ = a³ + 3a²b+ 3ab² + b³
In this formula value of a is always 9q
So plug the value we get
(9q+b)³ = 729q³ + 243q²b + 27qb² + b³
Now divide by 9 we get quotient = 81q³ + 27q²b + 3qb² and remainder is b³
So we have to consider the value of b³
b = 0 we get 9m+0 = 9m
b = 1 then 1³ = 1 so we get 9m +1
b = 2 then 2³ = 8 so we get 9m + 8
b = 3 then 3³ = 27 and it is divisible by 9 so we get 9m
b = 4 then 4³ =64 divide by 9 we get 1 as remainder so we get 9m +1
b=5 then 5³=125 divide by 9 we get 8 as remainder so we get 9m+8
b=6 then 6³=216 divide by 9 no remainder there so we get 9m
b=7 then 7³ = 343 divide by 9 we get 1 as remainder so we get 9m+1
b=8 then 8³ = 512 divide by 9 we get 8 as remainder so we get 9m+8
So all values are in form of 9m , 9m+1 or 9m+8.
6.Let's suppose √2 is a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero.
We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further.
From the equality √2 = a/b it follows that 2 = a2/b2, or a2 = 2 · b2. So the square of a is an even number since it is two times something.
From this we know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a · a would be odd too. Odd number times odd number is always odd. Check it if you don't believe me!
Okay, if a itself is an even number, then a is 2 times some other whole number. In symbols, a = 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction.
If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:
2 = (2k)2/b2
2 = 4k2/b2
2*b2 = 4k2
b2 = 2k2
This means that b2 is even, from which follows again that b itself is even. And that is a contradiction!!!
WHY is that a contradiction? Because we started the whole process assuming that a/b was simplified to lowest terms, and now it turns out that a and b both would be even. We ended at a contradiction; thus our original assumption (that √2 is rational) is not correct. Therefore √2 cannot be rational.
10.Let take a as any positive integer and b = 3.
Then using Euclid’s algorithm we get a = 3q + r
here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2 because 0 < r < b and the value of b is 3 So our possible values will 3q+0 , 3q+1 and 3q+2
Now find the square of values
Use the formula (a+b)² = a² + 2ab +b² to open the square bracket
(3q)² = 9q² if we divide by 3 we get no remainder
we can write it as 3*(3q²) so it is in form of 3m here m = 3q²
(3q+1)² = (3q)² + 2*3q*1 + 1²
=9q² + 6q +1 now divide by 3 we get 1 remainder
so we can write it as 3(3q² + 2q) +1 so we can write it in form of 3m+1 and value of m is 3q² + 2q here
(3q+2)² = (3q)² + 2*3q*2 + 2²
=9q² + 12q +4 now divide by 3 we get 1 remainder
so we can write it as 3(3q² + 4q +1) +1 so we can write it in form of 3m +1 and value of m will 3q² + 4q +1
Square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
or
let ' a' be any positive integer and b = 3.
we know, a = bq + r , 0 < r< b.
now, a = 3q + r , 0<r < 3.
the possibilities of remainder = 0,1 or 2
Case I - a = 3q
a2 = 9q2
= 3 x ( 3q2)
= 3m (where m = 3q2)
Case II - a = 3q +1
a2 = ( 3q +1 )2
= 9q2 + 6q +1
= 3 (3q2 +2q ) + 1
= 3m +1 (where m = 3q2 + 2q )
Case III - a = 3q + 2
a2 = (3q +2 )2
= 9q2 + 12q + 4
= 9q2 +12q + 3 + 1
= 3 (3q2 + 4q + 1 ) + 1
= 3m + 1 where m = 3q2 + 4q + 1)
From all the above cases it is clear that square of any positive integer ( as in this case a2 ) is either of the form 3m or 3m +1.
Hope it helps,