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REAL NUMBERS
Exercise 3.1
1)
Use Euclid’s division algorithm to find the HCF of the
following numbers.
i)
65 and 117
Let a=117 & b=65
65)117(1
65
a=(b*q)+r
117=(65*1)+52
------------52
Let a=65 & b=52
52)65(1
52
a=(b*q)+r
65=(52*1)+13
------------13
Let a=52 & b=13
13)52(4
52
-------------
a=(b*q)+r
52=(13*4)+0
00
Since the remainder is zero.
So the HCF of 65 and 117 is 13.
ii)
237 and 81
Let a=237 & b=81
81)237(2
162
a=(b*q)+r
237=(81*2)+75
------------75
Let a=81 & b=75
75)81(1
75
a=(b*q)+r
81=(75*1)+6
------------6
Let a=75 & b=6
6)75(12
a=(b*q)+r
6
75=(6*12)+3
------------15
12
------------03
Let a=6 & b=3
03)06(2
06
a=(b*q)+r
06=(03*2)+0
------------00
The HCF of 237 & 81 is 3.
iii)
55 and 210
Let a=210 & b=55
55)210(3
165
------------45
a=(b*q)+r
210=(55*3)+45
Let a=55 & b=45
45)55(1
45
a=(b*q)+r
55=(45*1)+10
------------10
Let a=45 & b=10
10)45(4
40
a=(b*q)+r
45=(10*4)+5
------------05
Let a=10 & b=5
05)10(2
10
a=(b*q)+r
10=(5*2)+0
------------00
The HCF of 55 & 210 is 5.
iv)
305 and 793
Let a=793 & b=305
305)793(2
610
a=(b*q)+r
793=(305*2)+183
------------183
Let a=305 & b=183
183)305(1
183
a=(b*q)+r
305=(183*1)+122
------------122
Let a=183 & b=122
122)183(1
122
------------61
Let a=122 & b=61
a=(b*q)+r
183=(122*1)+61
61)122(2
122
a=(b*q)+r
122=(61*2)+0
------------00
The HCF of 305 & 793 is 61.
2)
Show that any positive even integer is of the form 4q or
4q+2, where q is a whole number.
Soln: Let a be any even integer & b=4.
By division Lemma there exists integers q & r.
Such that
a=4q+r, where 0 ≤ r < 4 and q is a whole number.
0≤r<4

r=0,1,2,3.

a= 4q or a= 4q+1 or a=4q+2 or a=4q+3
but a is an even integer

a = 4q or a = 4q+2
but a ≠ 4q + 1 or a ≠ 4q+3
hence, any even integer is of the form 4q or 4q+2 where q
is a whole number.
3)
Use Euclid’s division Lemma to show that the squares of
any positive integer are either of the form 3m or 3m+1
for some integer but not of the form 3m+2.
Soln: Let a be any positive integer.
Then it is of the form 3q or 3q+1 or 3q+2. So we have the
following cases.
Case 1 : When a=3q
a2 = (3q)2 = 9q2 = 3q(3q) = 3m where m = 3q2
case 2 : When a = 3q+1
a2 = (3q+1)2 = (3q)2 + 2(3q)(1) + (1)2
= 3q(3q+2) + 1
= 3m+1 where m= q(3q+2)
Case 3: When a= 3q+2
a2 = (3q+2)2 = (3q)2 + 2(3q)(2) + (2)2
= 9q2 + 12q + 4
= 9q2 + 12q + 3 + 1
= 3(3q2+4q+1) + 1
= 3m+1 where m= 3q 2+4q+1
Hence a is of the form 3m or 3m+1.
4)
Prove that the product of three consecutive positive
integers is divisible by 6.
Soln: Let n, n+1, n+2 be three consecutive positive
integers.
If n= 2 then
Product of three consecutive numbers is
n (n+1) (n+2) = 2 (2+1) (2+2)
= 2 (3) (4)
= 24
Thus 24 is divisible by 6.

The product of three consecutive positive integers n,n+1,
n+2 is divisible by 6.
5)
There are 75 roses and 45 lily flowers. These are to be
made into bouquets containing both the flowers. All the
bouquets should contain the same number of flowers.
Find the number of bouquets that can be formed and the
number of flowers in them.
Soln: Resolving 75 and 45 into prime factors, we get.
5 75
5 15
3
75 = 52 X 3
5
75
5
15
3
5
45
3
9
3
45 = 5 X 32
HCF (75, 45) = 5 X 3 = 15

The number of bouquets that can be formed = 15

6)
The number of flowers in them = 5+3 =8
The length and breadth of a rectangular field is 110m and
30m respectively. Calculate the length of the longest rod
which can measure the length and breadth of the field
exactly.
Soln: Resolving 110 and 30 into prime factors, we get.
5 75
5 15
3
110 = 2 X 5 X 11
30 = 2 X 3 X 5
HCF(110, 30) = 2 X 5 = 10
Therefore the length of the longest rod which can measure
length & breadth is 10m.
Exercise 3.2
1)
Express each number as a product of prime factors.
i)
120
120 = 23 X 3 X 5
2 120
3 60
5 20
2 4
2
ii)
3825
3825 = 32 X 52 X 17
5 3825
5 765
3 153
3 51
17
iii) 6762
6762 = 2 X 3 X 72 X 23
2 6762
3 3381
7 1127
7 161
23
Iv ) 32844
32844 = 22 x 3 X 7 X 17 X 23
2
32844
2
16422
3
8211
7
2737
17 391
23
2)
If 25025 = P1x1, P2x2 , P3x3 , P4x4 find the value of P1, P2, P3, P4,
and x1, x2, x3, x4.
Soln. 25025 = 52 X 71 X 111 X 131
25025 = P1x1, P2x2 , P3x3 , P4x4
P1 = 5, P2 = 7, P3 = 11, P4 = 13
And
x1 = 2 , x2 = 1 , x3 = 1 , x4 = 1
3)
Find the LCM and HCF of the following integers by
expressing them as product of primes.
i)
12, 15 & 30
12 = 22 X 3
15 = 3 X 5
30 = 2 X 3 X 5
Therefore LCM= 60 & HCF = 3
ii)
18, 81 & 108
18 = 32 X 2
81 = 34
108 = 22 X 33
Therefore HCF = 32 = 9 & LCM = 34 X 22 = 324
4)
Find the HCF & LCM of the pairs of integers & verify that
LCM (a,b) X HCF (a,b) = a X b.
i)
16 & 80
16 = 24
80 = 24 X 5
HCF = 16
LCM = 80
LXH=aXb
80 X 16 = 16 X 80
1280 = 1280
ii)
125 & 55
125 = 52
55 = 5 X 11
HCF = 5
LCM = 1375
LXH=aXb
1375 X 5 = 125 X 55
6875 = 6875
5)
If HCF of 52 & 182 is 26, find their LCM.
H = 26, L =? A = 52, B = 182
HXL=AXB
26 X L = 52 X 182
L = = 364.
6)
Find the HCF of 105 and 1515 by prime factorization method
and hence find its LCM.
105 = 3X5X7
5
105
3
21
7
5
1515
3
303
101
7)
Find the smallest number which when increased by 17 is
exactly divisible by both 520 and 468
520 = 52X10
468 = 52X9
LCM = 52X10X9 = 4680
The smallest number which when increased by 17 which is
exactly divisible by 520 & 468 is 4680 – 17 = 4663.
8)
A rectangular hall is 18m 72cm long & 13m 20cm broad. It is
to be paved with square tiles of the same size. Find the least
possible number of such tiles.
Area of rectangle = lXb
= 1872 X 1320
= 24X32X13X23X3X5X11
HCF = 23X3 = 24
The greatest possible length of square tiles = 24
The least possible lengths of such tiles = 78 X 55 = 4290
Exercise 3.3
1)
Prove that is an irrational number
Let us assume as a rational number.
Therefore it is in the form of p/q,
Then, p & q are integers having no common factor other
than 1.
Now = p/q
By cross multiplication
q=p
Squaring on the both sides
( q)2 =p2
5q2 = p2 , q2 = p2/ 5
5 divides p2
5 divides a----------(1)
p/5 = r for some positive integer r
p = 5r
squaring on both sides
p2 = 25 r2
but p2 = 5q2
5q2 = 25r2
q2 = 5r2
5 divides q2
5 divides q ------------(2)
From (1) & (2) we find p & q have at least 5 as a
common factor, this contradicts the fact that p & q are coprime.
Hence is an irrational number.
2)
Prove that the following are irrational numbers.
1.
2
- Proof: let us assume on the contradictory that 2 is
rational number and its simplest form be a/b.
2 = a/b where a & b are +ve integers & co prime.
a/2b =
since a and 2b are non zero integers so a/2b is rational.
It follows that is rational.
This contradicts the fact that is irrational.
2.
- Proof: let us assume that is a rational number.
Hence = p/q
= 4 p/q
This follows that is a rational number.
But this contradicts the fact that is irrational number.
Hence /4 is an irrational number.
3.
3+
- Let us assume that 3+ is a rational number.
Hence 3+ = p/q
= p/q – 3
This is in the form of p/q.
Hence is a rational number.
But this contradicts to the fact that
number.
Hence 3+ is an irrational number.
is an irrational