Download 10th CBSE {SA - 1} Revision Pack Booklet - 4

Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
Revision Question Bank
Real Numbers
1.
Find the HCF of 30 and 70 by using factor tree method.
Solution.
Now,
And
30 = 2 × 3 × 5 and 70 = 2 × 5 × 7
HCF (30, 70) = 2 × 5 = 10
2.
Prove that 5  2 3 is an irrational number.
Solution.
Let 5–2 3 is a rational number.
p
Then, 5 – 2 3 = , where p and q are co-prime integers and q  0.
q
p
5 p
 3
5– = 2 3  
q
2 2q
p
Here, is a rational number.
2q
5 p
 
is a rational number.
2 2q
 3 is a rational number.
But this contradicts the fact that 3 is an irrational number.
Hence, our assumption that (5–2 3 ) is a rational number is not correct.
3.
Hence, (5–2 3 ) is an rational number.
Find the HCF of 81 and 237 and express it as a linear combination of 81 and 237.
Solution.
Given, integers are 81 and 237 such that 81 < 237. Applying division lemma to 81 and
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721
3
Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
237, we get
 81 237  2


237 = 81 × 2 + 75
……(i) 
162 

75 
Since, the remainder 75  0. So, consider the divisor 81 and the remainder 75 and apply
division lemma, we get
 75 81 1


81 = 75 × 1+ 6
……..(ii) 
75 

6 
We consider the new divisor 75, the new remainder 6 and apply division lemma, we get
 6 75 12


75 = 6×12+3 ...(iii) 
72 

3 
We consider the new divisor 6, the new remainder 3 and apply division lemma, we get
 3 6  2


6=3×2 + 0 ...(iv) 
6 

0 
The remainder at this stage is zero. So, the divisor at this stage or the remainder at the
earlier stage i.e., 3 is the HCF of 81 and 237.
To represent the HCF as a linear combination of the given two numbers, we start from the
last but one step and successively eliminate the previous remainders as follows.
From Eq. (iii), we have
3 = 75 – 6 ×12
 = 75– (81– 75 ×1) ×12
on substituting, 6  81–75  1 


obtained
from
Eq.
ii




= 75–12×81+12×75
= 13×75–12×81
= 13×(237 –51× 2)–12×81
Ton substituting, 75  237 – 81  2


obtained from Eq.  i 

= 13 × 237–26×81–12×81
= 13× 237 –38×81 = 237x + 81y,
where x= 13 and y = –38
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721
4
Pioneer Education {The Best Way To Success}
4.
5.
6.
IIT – JEE /AIPMT/NTSE/Olympiads Classes
Find the HCF of 65 and 117 and express it in the form 65m +117n.
Solution.
Given, integers are 65 and 117 such that 117 >65. Applying division lemma to 65 and 117,
we get
 65 117 1


117 = 65×1 + 52 ...(i) 

65
 52

Since, the remainder 52  0. So, we apply the division lemma to the divisor 65 and the
remainder 52, we get
 52 65 1


65 = 52 ×1 +13
...(ii) 
52 

13 
We consider the new divisor 52, the new remainder 13 and apply division lemma, we get
52 = 13 ×4 + 0
At this stage the remainder is zero. So, the last divisor or the non-zero remainder at the
earlier stage Le., 13 is the HCF of 65 and 117. From Eq. (ii), we have 13 = 65 – 52 × 1
= 65–(117–65 ×1)
on substituting 52  117  65  1


obtained from Eq.  i 

= 65–117 + 65×1
= 65×2 + 117×(–1)
= 65m + H7n , where m = 2 and n = –1.
Prove that one of the even three consecutive positive integers is divisible by 3.
Solution.
Let n, n +1 and n + 2 be three consecutive positive integers.
We know that, n is of the form 3q, 3g+l or 3q+ 2. So, we have the following cases
Case I When n = 3q
In this case, n is divisible by 3 but n +1 and n + 2 are not divisible by 3.
Case II When n = 3q +1
In this case, n +2 = 3<y +1+2 = 3(9 +1) is divisible by 3 but n and n +1 are not divisible by
3.
Case III When n = 3g+2
In this case, n +1 = 3<j+l+2 = 3(q +l) is divisible by 3 but n and n +2 are not divisible by 3.
Hence, one of n, n+1 and n +2 is divisible by 3.
Prove that, if x and y are odd positive integers, then x2 + y2 is even but not divisible by 4.
Solution.
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721
5
Pioneer Education {The Best Way To Success}
7.
8.
IIT – JEE /AIPMT/NTSE/Olympiads Classes
We know that, any odd positive integer is of the form 2q+1 for some integer q.
So, let x = 2m +1 and y = 2n +1 for some integers m and n.
Now, x2 +y2 =(2m+1)2 +(2n+1)2
= (4m2 +1+4m)+(4n2 +1+4m)
=4(m2 +n2)+4(m + n)+2
= 4{(ma2+n2)+(m + n)}+2 = 4g+2,
where q= (m2 +n2)+(m + n)
x2 +y2 is an even and leaves remainder 2 when divided by 4.
x2 + y2 is an even but not divisible by 4.
Prove that the square of any positive integer is of the form 3m or 3m +1 but not of the
form 3m+2.
Solution.
Since, positive integer n is of the form 3q, 3q+1 or 3q+2.
If n=3q, then n2=9q2
n2 =3(3q2 ) =3m, where m = 3q2.
If n = 3q + 1, then
n2 = (3q + 1)2 =9q2 +6q+1 =3q(3q+ 2)+1
=3m + 1, where m = q(3q+2)
If n = 3q+2, then
n2 = 9q2 +12q+4=9q2 +12q +3+1
=3(3q2 +4g + 1)+1 = 3m+1 where m = 3q2+4q+ 1
Hence, n2 is of the form 3m or 3m+1 but not of the form 3m+2.
A sweet seller has 420 kaju barfis and 130 badam barfis, he wants to stack them in such a
way that each stack has the same number and then take up the least area of the tray.
What is the number of burfis that can be placed in each stack for this purpose?
Solution.
The area of the tray that is used up in stacking the burfis will be least, if the sweet seller
stacks maximum number of burfis in each stack. Since, each stack must have the same
number of burfis. Therefore, the number of stacks will be least, if the number of burfis in
each stack is equal to the HCF of 420 and 130.
In order to find the HCF of 420 and 130, let us apply Euclid's division lemma to 420 and
130, we get
420 = 130×3 + 30 ...(i)
Let us now consider the divisor 130, the remainder 30 and apply division lemma, we get
130 = 30×4+10
...(ii)
Considering now divisor 30, the remainder 10 and apply division lemma, we get
30 = 3×10 + 0
...(iii)
Since, the remainder at this stage is zero. Therefore, last divisor 10 is the HCF of 420 and
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721
6
Pioneer Education {The Best Way To Success}
9.
IIT – JEE /AIPMT/NTSE/Olympiads Classes
130. Hence, the sweet seller can make stacks of 10 burfis of each kind to cover the least
area of the tray.
Three sets of English, Hindi and Mathematics books have to be stacked in such a way that
all the books are stored topic-wise and the height of each stack is same. The number of
English books is 96, the number of Hindi books is 240 and the number of Mathematics
books is 336. Assuming that the books are of same thickness. Determine the number of
stacks of English, Hindi and Mathematics books.
Solution.
In order to arrange the books as required, we have to find the largest number that divides
96, 240 and 336 exactly. Clearly, such a number is their HCF
Now,
96 =2×2×2×2×2×3= 25 ×3
240=2×2×2×2×3×5 = 25 ×3×5
336 = 2×2×2×2×3×7 = 24 ×3×7
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721
7
Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
HCF (96,240,336) = 24 X 3 = 48
Hence, there must be 48 books in each stack.
1
10. State whether 1.456 + is a rational number.
9
Solution.
Here, 1.456 = 1.45656 ... is a non-terminating repeating decimal.
 1.456 is a rational number.
p
1
and is in the form of , where q  0.
q
6
1
 is a rational number. 9
9
We know that, the sum of two rational numbers is a rational number.
1
Hence, 1.456 + is a rational number.
9
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721
8
Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
Chapter Test {Real Numbers}
M: Marks: 40
M: Time: 40 Min.
1.
If the HCF of 85 and 153 is expressible in the form of 85m –153, then find the value of m.
Solution:
[4]
Now,
153 = 85 × 1 + 68
 85 = 68 × 1+17
and 68 = 17 × 4 + 0
 HCF of 85 and 153 is 17.
According to the question,
17 = 85m – 153  85m=170
170
 M
2
85
2.
If p is a prime number, then prove that
p is irrational.
[4]
Solution:
Let p be a prime number and if possible, let p be rational.
m
, where m and n are integers having no common factor other than 1 and n  0 .
n
m
Then, p 
n
On squaring both sides, we get
Let p 
2
( p)2  m 
 
1
n
p m2
  2
1 n
 pn2  m2
[since, p divides pn2]
 p divides m2
 p divides m.
[ p is prime and p divides m2  p divides m]
Let m = pq for some integer q.
On putting m = pq in Eq. (i), we get
pn2 = p2q2
 n2 = pq2
 p divides n2
 p divides n.
[ p is prime and p divides n2  p divides n]
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721
9
Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
Thus, p is a common factor of m and n but this contradicts the fact that m and n have no
common factor other than 1.
The contradiction arises by assuming that p is rational.
Hence, p is irrational.
3.
4.
Given, a real number in decimal expansion is 43.123456789 . Check whether they are
p
rational or not. If they are rational and of the form , then what can you say about the
q
prime factors of q?
[4]
Solution:
Given, decimal number is 43.123456789 , which is non-terminating repeating numbers.
Hence, it is a rational number.
p
Let
= x = 43.123456789
...(i)
q
On multiplying both sides by 1000000000, we get
1000000000 x = 43123456789.123456789
...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
1000000000x – x
= 43123456789123456789 …..– 43.123456789…..
 999999999x = 43.123456746
43123456746 479149194 p

 (say)
 x
999999999 111111111 q
where, p = 4791495194 and q = 111111111, which is not of the form 2n. 5n i.e., the
prime factors of q are not of the form 2n . 5n.
Find the greatest number that will divide 445, 572 and 699 leaving remainders 4, 5 and
6, respectively.
[4]
Solution:
Since, the remainders are 4,5 and 6, respectively.
 We have to find the HCF of 445 – 4,572 – 5 and 699–6
i.e., 441, 567 and 693.
For the HCF of 441, 567 and 693, we have
441 =3× 3 ×7 × 7=32 × 72
567 =3×3 ×3 × 3×7=34 ×7
693 = 3 × 3 × 7 × 11 = 32 × 7 × 11
HCF of 441, 567 and 693 = 32 × 7
= 9 × 7 = 63
Hence, required number = 63
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721
10
Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
5.
Show that every positive odd integer is of the form (6q + 1) or (6q + 3) or (6q + 5) for
some integer q.
[4]
Solution:
Let a be a given positive odd integer. On dividing a by 6, let q be the quotient and r be
the remainder. Then, by Euclid's algorithm, we have
a = 6q + r, where 0  r  6
 a = 6q + r, where r = 0,1, 2, 3, 4, 5
 a = 6q or a = 6q +1 or a = 6q + 2
or a = 6q + 3 or a = 6q + 4 or a = 6q + 5
But a = 6q, a = 6q + 2, a = 6q + 4, given even values of a. Thus, when a is odd, it is of the
form 6q +1 or 6q + 3 or 6q + 5 for some integer q.
6.
Prove that 2 3 – 7 is an irrational.
Solution:
[4]
Let 2 3  7 is rational.
p
Then, 2 3 – 7 = , where p and q are integers
q
p
 2 3  7
q
p  7q
 2 3
q
p  7q
 3
2q
p  7q
As, p and g are integers , then
is a rational number.
2q
Then
3 is a rational number.
we know that, 3 is an irrational number. Hence, our assumption that (2 3 – 7) is a
rational, is wrong.
7.
Hence, 2 3 – 7 is an irrational.
In a seminar, the number of participants in English, Hindi and Mathematics are 60, 84
and 108, respectively.
(i) Find the minimum number of rooms required, if in each room the same number of
participants are to be seated and all of them being in the same subject.
(ii) Which mathematical concept is used to solve the above question?
(iii) What values (quality) are hidden behind conducting the seminar?
[4]
Solution:
The number of room will be minimum, if each room accommodate maximum
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721
11
Pioneer Education {The Best Way To Success}
8.
9.
IIT – JEE /AIPMT/NTSE/Olympiads Classes
number of participants. Since, in each room of the same number of participants are
to be seated and all of them must be of the same subject. Therefore, the number of
participants in each room must be the HCF of 60,84 and 108. The prime
factorisation of 60, 84 and 108 are as under
60 = 22 × 3 × 5
84 = 22 × 3×7
and 108 = 22 × 33
HCF of 60, 84 and 108 = 2 2 × 3= 12
Therefore, in each room 12 participants can be seated.
Number of room required
Total number of participants
=
12
60  84  108
=
12
252
 21
=
12
(ii) Euclid's division algorithm.
(iii) Equality and honesty.
Find the largest number which divides 248 and 1032 leaving remainder 8 in each case.
Solution:
It is given that the required number when divides 248 and 1032, the remainder is 8 in
each case. This means,
248 – 8 = 240 and 1032 – 8 = 1024
It follows from this that the required number is a common factor of 240 and 1024.
Therefore, it is the HCF of 240 and 1024.
 By Euclid’s algorithm,
1024 = 240 × 4 + 64
240 = 64 × 3 + 48
64 = 48 × 1 = 48
48 = 16 × 3 + 0
Clearly, HCF of 240 and 1024 is the last divisor i.e., 16.
Hence, required number = 16.
Find the HCF of 231 and 396, using Euclid's division algorithm.
[4]
Solution:
Given integers are 231 and 396.
Clearly, 396 > 231. Applying Euclid’s division lemma to 396 and 231, we get
396 = 231 × 1 + 165.
Since the remainder 165  0. So, we apply the division lemma to the divisor 231 and
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721
12
Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
remainder 165 to get :
231 = 165 × 1 + 66.
Now, we apply division lemma to the new divisor 165 and new remainder 66 to get:
165 = 66 × 2 + 33
We now consider the new divisor 66 and the new remainder 33, and apply division
lemma to get :
66 = 33 × 2 = 0
The remainder at this stage is zero. So, the divisor at this stage or the remainder at the
previous stage i.e., 33 is the HCF of 396 and 231.
10. Use Euclid's division lemma to show that the square of any positive integer is either of
the form 3m or (3 m +1) for some integer m.
[4]
Solution:
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 I When a = 3q
In this case, we have
a2 = (3q)2 = 9q2 = 3q (3q) = 3m, where m = 3q
Case II When a = 3q + 1
In this case, we have
a2 = (3q + 1)2 = 9q2 + 6q + 1 = 3q (3q + 2) + 1 = 3m + 1, where m = q(3q + 2)
Case III When a = 3q + 2
In this case, we have
a2 = (3q + 2)2 = 9q2 + 12q + 4 = 9q2 + 12q + 3 + 1
= 3(3q2 + 4q + 1) + 1 = 3m + 1, where m = 3q2 + 4q + 1
Hence, a is of the form 3m or 3m + 1.
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721
13