Question 11
The length, breadth and height of a room are 8 m 25 cm, 6 m 75 cm and 4 m 50 cm,
respectively. Determine the longest rod which can measure the three dimensions of the room
exactly.
Solution
We are given the length, breadth and height of a room as 8m 25cm, 6m 75cm and 4m
50cm, respectively. We need to determine the largest room which can measure the
three dimensions of the room exactly.
First, we have to convert each dimension in cm
Length of room = 8m 25cm = 825cm
Breadth of room = 6m 75cm = 675cm
Height of room = 4m 50cm = 450cm.
Therefore, the required longest rod = H.C.F. of 825, 675 and 450.
First we consider 675 and 450.
By applying Euclid’s division lemma,
675= 450 X 1 + 225
450 = 225 X 2 + 0
Therefore, H.C.F. of 675 and 450 = 225
Now, we consider 225 and 825.
By applying Euclid’s division lemma
825 = 225 X 3 + 150
225 = 150 X 1 + 75
150 = 75 X 2 + 0
Therefore, H.C.F. of 825, 675 and 450 = 75
Hence, the length of required longest rod is 75 cm
Question 12
A hall has a certain number of chairs. Guests want to sit in different groups like in pairs, triplets, quadruplets, fives and sixes etc. When organiser arranges chairs in such pattern like 2’s, 3’s ,4’s.5’s and 6’s then 1,2,3,4 and 5 chairs are left respectively. But when he arranges in 11’s no chair will be left
A)In the hall how many least number of chairs are available?
B) If one chair is added to the total number of chairs, how many chairs will be left when
arranged in 7̕s
Solution
A)When chairs are arranged in pair , 1 chair left
When chairs are arranged in 3,s , 1 chair left
When chairs are arranged in 4,s, 1 chair left
And so on
So,we find (2-1)=(3-2)=(4-3)= (5-4)= (6-5)=1(k)
LCM of 2,3,4,5,6= 2x2x3x5=60
Required number= (60n-1)/11=remainder 0 (n is a natural number)
Taking n=9
(60×9-1)/11=539/11=remainder 0
Therefore least possible number is 539
B) 539+1=540
540/7 = remainder 1
1 chair left.
Question 13
A right angle triangle has base length √2 cm and height 1 cm . Find the hypotenuse and prove its length is a irrational number.
Solution
From Pythagoras theorem, Hypotenuse= √(Base^2 +Perpendicular ^2)
Hence, Hypotenuse= √3.
Now, to prove : √3 is irrational.
Let sqrt(3) is rotational number
sqrt(3) = p/q
[plq are co-prime integers & q = 0]
3= n ^ 2 = 3a ^ 2….(1)
3 is a factor of p²
→ 3 is a factor of p
.(2)
So p = 3m m is any integer
from (1)
9m ^ 2 = 3q ^ 2
= q ^ 2 = 3m ^ 2
3 is a factor of q
→ 3 is a factor of q
From (2) & (3)
…(3)
3 is a common factor of both p & q
It contradicts our supposition that p & q are co-prime integers.
Hence are supposition is wrong
sqrt(3) is irrational
Hence proved
Question 14
The traffic lights at three different road crossings change after 48 seconds, 72
seconds and 108 seconds respectively. If they change simultaneously at 7 am,
at what time will they change simultaneously again?
Solution
Let the lights change simultaneously exactly x seconds after 7 am.
Since the first light changes every 48 seconds, we have 48|x.
Similarly, 72|x and 108|x.
Hence we have l|x, where l is the Lowest common multiple of 48,72 and
108.
Also, precisely after l seconds, the lights will change simultaneously.
Hence the smallest value of x is l.
Hence the lights will change simultaneously, l seconds after 7 am.
Now we have
48=2×2×2×2×3,72=2×2×2×3×3,108=2×2×3×3×3
Hence LCM (48,72,108) =432
Hence we have l = 432.
Hence the lights will change simultaneously after 432 seconds = 7
minutes and 12 seconds after 7 am.
Hence the lights will change simultaneously at 07:07:12 am
Question – 15
Shree Sundar Kamla Nagar Society decided to give some food packets and
grocery packets to the needy on Independence Day. They have 36 food
packets and 72 grocery packets to distribute. They do not want to
discriminate among the needy, so they decided to distribute packets equally
among all.
(i) To how many maximum people can they distribute the packets?
(ii) How many food and grocery packets will each one get?
(iii) Now, they decide to add 54 packets of vegetables also. In this case to
how many maximum people can they distribute the packets?
(iv) How many total packets will each person get?
(v) If they decide to add 18 more vegetable packets and remove 36 grocery
packets, in this case to how many maximum people can they distribute the
packets?
Solution
(i) Maximum people to whom the packets can
be distributed = HCF(36,72)
36 = 18 x 2, 72 = 18 x 4
Ans – 18
(ii) Each one will get 2 food packets and 4
grocery packets
(iii) Maximum people to whom the packets can
be distributed = HCF(36,54,72)
36 = 18 x 2, 54 = 18 x 3, 72 = 18 x 4
Ans – 18
(iv) Each one will get 2food packets and 3
grocery packets and 4 vegetable packets
(v) No. of vegetable packets = 54 + 18 = 72
No. of grocery packets = 72 – 36 = 36
Maximum people to whom the packets can be
distributed = HCF(36,36,72)
36 = 36 x 1, 36 = 36 x 1, 72 = 36 x 2,
Ans – 36
Question -6
Students of class VI in their free period started playing a game. One of the students announced a prime number 3 in the class and the next students multiplied it by a prime number and passed it to the next student. Then again it
was multiplied by a prime number and passed on to the next student. In this way by multiplying to a prime number the last student got 90090 which the student announced in class.
Answer the following questions
(i) How many students are in the class?
(ii) Which is the highest prime number used by the student?
(iii) What is the least prime number used by the student?
(iv) Which prime number has been used maximum number of times?
Solution
90090 = 3 x 30030
30030 = 3 x 10010
10010 = 5 x 2002
2002 = 2 x 1001
1001 = 11 x 91
91 = 7 x 13
13 = 13 x 1
90090 = 2 x 32 x 5 x 7 x 11 x 13
(i) Total number of students in the class = No of prime numbers used to
multiply and reach 90090 = 7
(ii) Highest prime number used by the students = 13
(iii) Least prime number used by the student = 2
(iv) 2, 5, 7, 11 and 13 have been used minimum number of times
(v) Which prime number has been used minimum number of times?
Question -7
To enhance the reading skill of grade X students, the schools nominate you and
two of your friends to set up a class library. There are two sections- section A
and section B of grade X. There are 32 students in section A and 36 students in
section B.
Based on above information answer the following:
(i) What is the minimum number of books you will acquire for the class
library, so that they can be distributed equally among students of
section A or section B?
(ii) If the product of two positive integers is equal to the product of their
HCF and LCM is true, then HCF(32, 36) is
(iii) 36 can be expressed as a product of its primes as
(iv) 7× 11 × 13 × 15 + 15 is a
(v) If p and q are positive integers such that 𝑝 = 𝑎𝑏2 and 𝑞 = 𝑎2𝑏, where a and b are prime numbers then the LCM(p,q)
Solution
i. LCM(32, 36) = 288
ii. 32 X 36 = HCF(32, 36) X LCM(32, 36)
HCF(32, 36) = 4
iii. 22 X 32
iv. composite number
v. LCM(p, q) = 𝑎2 × 𝑏2
Question : 8
Given that √2 is an irrational number, prove that √8 is also irrational using a proof by
contradiction.
Solution
Let √8 be a rational number which can be expressed in the form of p/q where p and q
have no other common factor than 1.
√8= p/q
2√2=p/q
√2= p/2q
p/2q is rational but √2 is Irrational.
Thus our assumption is wrong.
√8 is an Irrational number.
Question – 19
A, B and C starts cycling around a circular path in the same direction at the same time.
Circumference of the path is 1980 m. If speed of A is 330 m/min, speed of B is 198 m/min
and that of C is 220 m/min and they start from the same point, then after what time will they
be together at the starting point?
Solution
As
TIME= Distance / Speed
Time taken by A to complete one round
= 1980/330 = 6 min
Time taken by B to complete one round
= 1980/198 = 10 min
Time taken by C to complete one round
= 1980/220 = 9 min
The three cyclists will be together after LCM (6, 10, 9)
6 = 2 × 3
10 = 2 × 5
9 = 32
LCM (6, 10, 9) = 21 × 32 × 5 = 90 min
Question 20
The LCM of two numbers is 14 times their HCF. The sum of LCM and HCF is 600. If one of
number is 280. Find the other number.
Solution
Given that the LCM of two numbers is 14 times their HCF and the sum of LCM and
HCF is 600, let’s denote the HCF as “h” and the two numbers as “a” and “b”.
Given:
LCM = 14h
LCM + HCF = 600
Substitute LCM from the first equation:
14h + h = 600
15h = 600
h = 40
Since one of the numbers is 280 and the HCF is 40, we can find the other number
using the formula:
LCM = (a x b) / HCF
14 x 40 = (280 x b) / 40
560 = 280b
b = 2
So, the other number is 2
SELF ASSESSMENT
1..A, B and C starts cycling around a circular path in the same direction at the same time.
Circumference of the path is 1980 m. If speed of A is 330 m/min, speed of B is 198 m/min
and that of C is 220 m/min and they start from the same point, then after what time will they
be together at the starting point?
2. Prove that √5 is an irrational number. Hence show that 3 + 2 √5 is also an irrational number.
3. Write a rational number between √2 and √3 . 5
4.Let d be the HCF of 25 and 50 . Find two numbers a and b , such that d= 25a + 50 b 5
5. Prove that one of every three consecutive integers is divisible by 3. 5
6.Prove that 3 + 2√5 is irrational. 5
7. Amita, Sneha, and Raghav start preparing cards for all persons of an old age home. In order
to complete one card, they take 10, 16 and 20 minutes respectively. If all of them started
together, after what time will they start preparing a new card together?
