Students often refer to Class 7 Ganita Prakash Solutions and NCERT Class 7 Maths Part 2 Chapter 3 Finding Common Ground Question Answer Solutions to verify their answers.
Class 7 Maths Ganita Prakash Part 2 Chapter 3 Solutions
Ganita Prakash Class 7 Chapter 3 Solutions Finding Common Ground
Class 7 Maths Ganita Prakash Part 2 Chapter 3 Finding Common Ground Solutions Question Answer
3.1 The Greatest of All (Pages 47-48)
Sameeksha is building …. size 4 ft.
Question.
How many tiles of this size should she purchase?
Solution:
Area of the room = 12 x 16 = 192 square feet
Area of one tile = 4 x 4 =16 square feet
Area of the room 192
Number of tiles = \(\frac{\text { Area of the room }}{\text { Area of one tile }}=\frac{192}{16}\)=12
Area of one tile 16
Thus, Sameeksha should purchase 12 tiles of size 4 ft x 4 ft to cover the room completely.
Lekhana purchases …. minimise the number of bags.
Question.
Which weight should she choose to minimise the number of bags?
Solution:
To minimise the number of bags, Lekhana should maximise the weight of each bag. So she should choose
the highest common factor of 84 and 108 i.e., 12 . Therefore, the weight of each bag should be 12 kg.
Now, let us calculate how many bags she will need:
- From the first farm: \(\frac{84}{12}\)=7 bags
- From the second farm: \(\frac{108}{12}\)=9 bags
To meet her needs, Lekhana should use bags weighing 12 kg each. She will require 7 bags for the first farm and 9 bags for the second farm.
Question.
Do you remember the “Jump Jackpot” game from Grade 6 (see the chapter ‘Prime Time’)? Grumpy places a treasure on a number and Jumpy chooses a jump size and tries to collect the treasure. In each case below, the two numbers upon which treasures are kept are given.
Find the longest jump size (starting from 0) using which jumpy can land on both the numbers having the treasure.
(a) 14 and 30
(b) 7 and 11
(c) 30 and 50
(d) 28 and 42
Solution:
(a) The treasures are at 14 and 30.
We need to find the longest jump size (starting from 0) that can land exactly on both 14 and 30.
The factors of 14 are: 1, 2, 7, 14
The factors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30
Common factors of both 14 and 30 are: 1, 2
So, 2 is the longest possible jump size.
(b) The treasures are at 7 and 11.
The factors of 7 are: 1,7
The factors of 11 are: 1, 11
The common factor of both 7 and 11 is: 1
So, 1 is the longest possible jump size.
(c) The treasures are at 30 and 50.
The factors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30
The factors of 50 are: 1, 2, 5, 10, 25, 50
The common factors of both 30 and 50 are: 1,2, 5, 10
So, 10 is the longest possible jump size.
(d) The treasures are at 28 and 42.
The factors of 28 are: 1, 2, 4, 7, 14, 28
The factors of 42 are: 1, 2, 3, 6, 7, 14, 21, 42
The common factors of both 28 and 42 are: 1,2, 7, 14
So, 14 is the longest possible jump size
Question.
Is the longest jump size for the numbers the same as their HCF? Explain why it is so.
Solution:
Yes, the longest jump size is the Flighest Common Factor (HCF) of the given numbers because the HCF represents the greatest number that divides both numbers exactly, allowing Jumpy to use this step size to land on both treasure locations.
Question.
Can this process be simplified? Can it be made more reliable?
Solution:
Yes, the process of finding the Highest Common Factor (HCF) can be simplified and made more reliable using the Division Method.
- Divide the larger number by the smaller number and get the remainder.
- Replace the larger number with the smaller number, and the smaller number with the remainder.
- Repeat the process until the remainder is 0. The divisor at this step will be the HCF.
Factors of a Number Using Prime Factorisation (Page 51)
Question.
Similarly, is 2 x 7 = 14 a factor of 840? Why or why not?
Is 2 x 2 x 2 a factor of 840? Why or why not?
Is 3 x 3 x 3 a factor of 840? Why or why not?
Can we use this idea to list down all the possible factors of a number using just its prime factors?
Solution:
Yes, 2x2x7 = 28 is a factor of 840.
This is because the prime factorisation of 840 is 2x2x2x3x5x7 and 28
i.e. 2 x 2 x 7 is a part of this factorisation.
Yes, 2x2x2 = 8 is a factor of 840.
Again, considering the prime factorisation of 840, 8 = 2 x 2 x 2, and since 840 has three 2’s in its factorisation, 8 is a factor of 840.
No, 3 x 3 x 3 = 27 is not a factor of 840.
While 840 includes only one 3 in its prime factorisation, 27 is not a factor of 840.
Yes, we can list all the possible factors of a number using its prime factorisation. By taking all combinations of the prime factors in the correct powers, we can generate every factor of the number.
Question.
Check that all the factors of 225 occur in this list.
Solution:
We know that:
1 x 225 = 225, 3 x 75 = 225, 5 x 45 = 225, 9 x 25 = 225, 15 x 15 = 225
Thus, 1,3, 5, 9, 15, 25,45, 75, and 225 are the factors of 225.
They all occur in the above list.
Figure It Out (Page 51)
List all the factors of the following numbers:
(a) 90
(b) 105
(c) 132
(d) 360 (this number has 24 factors)
(e) 840 (this number has 32 factors).
Solution:
(a)
Prime Factors: 2, 3, 5
Combination of two prime factors:
2 x 5 = 10, 5 x 3 = 15, 2 x 3 = 6, 3 x 3 = 9
Combination of three prime factors:
2 x 3 x 5 = 30, 2 x 3 x 3 = 18, 3 x 3 x 5 = 45
Combination of four prime factors:
2 x 3 x 3 x 5 = 90
Adding 1 to this list of factors, we get the factors 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90.
(b)
Combination of two prime factors:
3 x 5 = 15, 5 x 7 = 35, 7 x 3 = 21
Combination of three prime factors: 3 x 5 x 7 = 105
Adding 1 to this list of factors 1,
we get the factors 1, 3, 5, 7, 15, 21, 35, 105.
(c)
Prime Factors: 2, 3, 11
Combination of two prime factors:
2 x 2 = 4, 2 x 3 = 6, 3 x 11=33, 2 x 11=22
Combination of three prime factors:
2 x 2 x 3 = 12,2 x 3 x 11 = 66, 2 x 2 x 11=44
Combination of four prime factors:
2 x 2 x 3 x 11 = 132
Adding 1 to this list of factors,
we get the factors 1, 2,3,4, 6, 11, 12, 22,33,44, 66, and 132.
(d) Do it yourself.
(e)
Prime Factors: 2, 3, 5, and 7
Combination of two prime factors:
2 x 2 = 4,2 x 3 = 6, 2 x 5 = 10,
3 x 5 = 15,5 x 7 = 35, 3 x 7 = 21, 7 x 2= 14
Combination of three prime factors:
2 x 2 x 2 = 8, 2 x 2 x 3 = 12, 2 x 3 x 5 = 30, 3 x 5 x 7 = 105,
2 x 5 x 7 = 70, 2 x 2 x 5 = 20, 2 x 2 x 7 = 28, 2 x 3 x 7 = 42
Combination of four prime factors:
2 x 2 x 2 x 3 = 24, 2 x 2 x 2 x 5 = 40, 2 x 2 x 2 x 7
= 56, 2 x 2 x 3 x 5 = 60, 2 x 2 x 3 x 7 = 84,
2 x 2 x 5 x 7=140, 2 x 3 x 5 x 7 = 210
Combination of five prime factors:
2 x 2 x 2 x 5 x 7 = 280, 2 x 2 x 2 x3 x 5 = 120,
2 x 2 x 2 x 3 x 7 = 168, 2 x 2 x 3 x 5 x 7 = 420
Combination of six prime factors: 2 x 2 x 2 x 3 x 5 x 7 = 840
Adding 1 to this list of factors,
we get factors 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, and 840.
Finding the HCF of Numbers Using Prime Factorisation
Figure It Out (Page 53)
Find the common factors and the HCF of the following numbers:
(a) 50, 60
(b) 140, 275
(c) 77, 725
(d) 370, 592
(e) 81, 243
Solution:
(a) Factorising the numbers, we get 50=2x 5 x5
The highest among the common factors, HCF is 2 x 5 = 10.
(b) Factorising the numbers, we get
The highest among the common factors, HCF is 5.
(c) Factorising the numbers, we get
77 = 7 x 11
725 = 5 x 5 x 29
There is no common factor between these two factorisations.
So, 1 is the only common factor, HCF is 1.
(d) Factorising the numbers, we get
The highest among the common factors, HCF is 74.
(e) Same as above.
Figure It Out (Page 54)
Question 1.
Find the HCF of the following numbers:
(a) 24, 180
(b) 42, 75, 24
(c) 240, 378
(d) 400, 2500
(e) 300, 800
Solution:
(a) 24 = 2 x 2 x 2 x 3, 180 = 2 x 2 x 3 x 3 x 5
Here common primes are 2 and 3.
24 contains the minimum number of 3s which occurs once.
So, the largest common sub-part should contain only one 3.
180 contains the minimum number of 2s which occurs two times.
So, the largest common subpart should contain only two 2s i.e., 2 x 2.
So, HCF = 2 x 2 x 3 = 12
(b) 42 = 2 x 3 x 7, 75 = 3 x 5 x 5, 24 = 2 x 2 x 2 x 3
Here common prime is 3.
All 42, 75, and 24 contain the minimum number of 3s which occurs once.
So, the largest common subpart should contain only one 3.
So, HCF = 3.
(c) Same as above. (Ans. 6)
(d) 400= 2 x 2 x 2 x 2 x 5 x 5
2500 = 2x 2 x 5 x 5 x 5 x 5
Here common primes are 2 and 5.
400 contains the minimum number of 5s which occurs two times.
So, the largest common subpart should contain only two 5s i.e., 5 x 5.
2500 contains the minimum number of 2s which occurs two times.
So, the largest common subpart should contain only two 2s i.e., 2 x 2.
So, HCF = 2 x 2 x 5 x 5 = 100
(e) 400= 2 x 2 x 2 x 2 x 5 x 5
2500 = 2x 2 x 5 x 5 x 5 x 5
Here common primes are 2 and 5.
400 contains the minimum number of 5s which occurs two times.
So, the largest common subpart should contain only two 5s i.e., 5 x 5.
2500 contains the minimum number of 2s which occurs two times.
So, the largest common subpart should contain only two 2s i.e., 2 x 2.
So, HCF = 2 x 2 x 5 x 5 = 100
Question 2.
Consider the numbers 72 and 144. Suppose they are factorised into composite numbers as: 72 = 6 x 12 and 144 = 8 x 18. Seeing this, can one say that these two numbers have no common factor other than 1 ? Why not?
Solution:
Given, 72 = 6 x 12 and 144 = 8×18
Now, 6, 12, 8, and 18 are composite numbers, so they can be factorised further.
Thus, no one can say that these two numbers have no common factor other than 1.
Further, the factorisations 72 = 6 x 12 and 144 = 8×18 do not show the complete list of prime factors of the numbers.
72 = 2 x 2 x 2 x 3 x 3 1
44 = 2 x 2 x 2 x 2 x 3 x 3
So, HCF = 2 x 2 x 2 x 3 x 3 = 72, which is other than 1.
3.2 Least, but not Last! (Page 56)
Do you remember … are given. it Find the first number for which ‘idli-vada’ will be called out:
(a) 4 and 6
(b) 7 and 11
(c) 14 and 30
(d) 15 and 55 Is the answer always the LCM of the two numbers? Explain.
Solution:
In the game:
We call “Idli” for multiples of the first number.
We call “Vada” for multiples of the second number.
We call “Idli-Vada” when it’s a common multiple -i.e. when the number is a multiple of both.
The first number for which this happens is the LCM of the two numbers.
(a) The first number for which the players should say, ‘idli-vada’ is 12, which is a multiple of 4 and 6.
(b) The first number for which the players should say, ‘idli-vada’ is 77, which is a multiple of 7 and 11.
(c) The first number for which the players should say, ‘idli-vada’ is 210, which is a multiple of 14 and 30.
(d) The first number for which the players should say, ‘idli-vada’ is 165, which is a multiple of 15 and 55.
Yes, the answer is always the LCM of the two numbers. This is because the game involves finding the smallest number that is a common multiple of both numbers.
Question:
How do we find the LCM of two numbers using their prime factors?
Solution:
To find the LCM (Lowest Common Multiple) of two numbers using their prime factors, follow these
Steps:
- Find the prime factorisation of each number.
- Identify all the prime factors that appear in either number.
- For each prime factor, take the highest power of that prime that appears in any of the numbers.
- Multiply all these highest powers together. The product will be the LCM of the two numbers.
Finding LCM through Prime Factorisation (Page 57)
Is 2 x 3 x 5 x 7 also a common multiple?
Solution:
2 x 3 x 5 x 7 = 210
2 x 3 x 5 x 7 is indeed a common multiple of 14 and 35, but not the least common multiple because it includes an extra factor 3.
Figure It Out (Page 58)
Find the LCM of the following numbers:
(a) 30, 72
(b) 36, 54
(c) 105, 195, 65
(d) 222,370
Solution:
(a) Factorising them into their primes, we get
30 = 2 x 3 x 5
72 = 2 x 2 x 2 x 3 x 3
So, LCM of 30 and 72 = 2 x 2 x 2 x 3 x 3 x 5 = 360
(b) Lactorising them into their primes, we get
36 = 2 x 2 x 3 x 3
54 = 2 x 3 x 3 x 3
So, LCM of 36 and 54 = 2 x 2 x 3 x 3 x 3 = 108
(c) Factorising them into their primes, we get
105 = 3 x 5 x 7
195 = 3 x 5 x 13
65 = 5 x 13
So, LCM of 105,195, and 65 = 3 x 5 x 7 x 13 = 1365
(d) Factorising them into their primes, we get
222 = 2 x 3 x 37 370 = 2 x 5 x 37
So, LCM of 222, and 370 = 2 x 3 x 5 x 37= 1110
Finding Common Ground’129
3.3 Patterns, Properties, and a Pretty Procedure! (Pages 58-59)
Question.
Find more such number pairs where the HCF is one of the two numbers. Flow can we describe such pairs of numbers?
Solution:
Consider some examples:
HCF (4, 12) = 4, HCF (5, 20) = 5, HCF (7,28) = 7, HCF (9, 27) = 9, HCF (10, 30)= 10
Such pairs of numbers are those where one number is a factor (or divisor) of the other
Question.
For number pairs satisfying this property (i.e., one of the numbers is the HCF)
(a) if m is a number, what could be the other number?
(b) if 7k is a number, what could be the other number?
Solution:
(a) If m is one of the numbers, then the other number must be a multiple of m, which can be written as:
p x m, where p is any positive integer.
For example: If m = 4 and p -3, the other number is 4 x 3=12.
(b) In this case, one number is Ik, then the other number must be a factor of Ik.
Since Ik is a multiple of 7, the other number could simply be 7.
For example: If k=4, the numbers are 7 x 4 = 28 and 7.
The HCF of 7 and 7k will always be 7.
Figure It Out (Page 59)
Question 1.
Make a general statement about the HCF for the following pairs of numbers. You could consider examples before coming up with general statements. Look for possible explanations of why they hold.
(a) Two consecutive even numbers
(b) Two consecutive odd numbers
(c) Two even numbers
(d) Two consecutive numbers
(e) Two co-prime numbers Share your observations with the class.
Solution:
(a) For example, consider the numbers 2 and 4, as well as 8 and 10. The highest common factor (HCF) of two consecutive even numbers is always 2, since 2 is their only common factor. These numbers are divisible by 2 but do not share any larger common factor.
(b) The HCF of two consecutive odd numbers is always 1 because odd numbers are not divisible by 2, and consecutive odd numbers share no common factors. They are co-prime.
(c) The HCF of two even numbers is the largest power of 2 that divides both numbers, which depends on the specific numbers, such as 8 and 12 or 16 and 24, the HCF are 4 and 8 respectively.
(d) Let us take pair of numbers, 8 and 9, 15 and 16. The highest common factor (HCF) of two consecutive numbers is always 1 because consecutive numbers share no common factors other than 1.
(e) Let us take pair of numbers, 8 and 15, and 9 and 10. The highest common factor (HCF) of two coprime numbers is always 1, because by definition, co-prime numbers have no common factors other than 1.
Question 2.
The LCM of 3 and 24 is 24 (it is one of the two given numbers).
(a) Find more such number pairs where the LCM is one of the two numbers.
(b) Make a general statement about such numbers. Describe such number pairs using algebra.
Solution:
(a) (5, 15): LCM = 15 (it is one of the numbers)
(6, 18): LCM = 18 (it is one of the numbers)
(7, 21): LCM = 21 (it is one of the numbers).
(b) The LCM of two numbers will be equal to one of the numbers if the smaller number is a factor of the larger number.
Let the two numbers be n and k x n (a multiple of n)
Then LCM (n, k x n) = k x n
Thus, if one number is a multiple of the other, then their LCM is the larger number.
Question 3.
Make a general statement about the LCM for the following pairs of numbers. You could consider examples before coming up with these general statements. Look for possible explanations of why they hold.
(a) Two multiples of 3
(b) Two consecutive even numbers
(c) Two consecutive numbers
(d) Two co-prime numbers
Solution:
(a) Let two pairs of multiples of 3 are 6 and 9, and 12 and 15, then, LCM of 6 and 9 is 18, and LCM of 12 and 15 is 60.
The LCM is also a multiple of 3. Thus the LCM of two multiples of 3 is also a multiple of 3.
(b) Since the LCM of 2 and 4 is 4, and LCM of 4 and 6 is 12.
The LCM of two consecutive even numbers is also even, and generally greater than both numbers, because they are not usually factors of each other.
(c) Since, LCM of 6 and 7 is 42, and LCM of 11 and 12 is 132.
The LCM of two consecutive numbers is always their product because their HCF is always 1.
(d) LCM of 4 and 9 is 36, and LCM of 7 and 10 is 70.
The LCM of two co-prime numbers is their product.
(Page 60)
Question.
Here are some more numbers where both numbers are multiples of the same number. Find their HCF:
(a) 18 x 10, 18 x 15
(b) 10 x 38, 10 x 21
(c) 5 x 13, 5 x 20
(d) 12 x 16, 12 x 20
Solution:
Question.
In which of these cases is the HCF the same as the common multiplier, like problem (b) where the HCF is 10? Explore a few more examples of this type to understand when this happens.
Solution:
In case (c) also the HCF is the same as the common multiplier. When the common prime factors are the only factors that divide both numbers, the product of these common factors is same as HCF.
Efficient Procedures for HCF and LCM (Pages 60-62)
Question.
How do we use this to find the HCF of 84 and 180? Explore.
[Hint: Observe that 84 = 2 x 2 x 3 x 7; and 180 = 2 x 2 x 3 x 15 similar to prime factorisation]
Solution:
So, HCF will be 2 x 2 x 3 = 12
Question.
You can try this method for these pairs of numbers.
(a) 90 and 150 (b) 84 and 132
Solution:
(a)
Here, HCF will be 10 x 3 = 30 and
LCM will be 10 x 3 x 3 x 5 = 450.
(b) 84 and 132
Here, HCF will be 6 x 2 = 12 and
LCM will be 6x2x7x11 = 924.
(Pages 62-63)
Question.
Which is greater – the LCM of two numbers or their product?
Solution:
The LCM of two numbers is never greater than their product. It is equal to their product only when the numbers are co-prime (i.e., their HCF is 1). If the numbers share common factors, the LCM will be less than their product.
Question.
Explore whether the LCM is a factor of the product in the following cases. If yes, identify the number that the LCM should be multiplied by to get the product. Do you see any pattern? Use these numbers:
(a) 45, 105
(b) 275, 352
(c) 222, 370
Solution:
(a) 45 = 3 x 3 x 5
105 = 3 x 5 x 7
LCM = 3 x 3 x 5 x 7 = 315
The product of numbers in factorised form is 45 x 105 = 3 x 3 x 3 x 5 x 5 x 7
Yes, LCM is a factor of the product.
It should be multiplied by 15 to get the product.
Thus. LCM x 15 = 45 x 105
(b) 275 = 5 x 5 x 11
352 = 2 x 2 x 2 x 2 x 2x 11
LCM = 2 x 2 x 2 x 2 x 2 x 5 x 5 x 11
The product of numbers in factorised form is
275 x 352 = 2 x 2 x 2 x 2 x 2 x 5 x 5 x 11 x 11
Yes, LCM is a factor of the product.
It should be multiplied by 11 to get the product.
Thus, LCM x 11 =275 x 352
(c) 222 = 2 x 3 x 37
370 = 2 x 5 x 37
LCM = 2 x 3 x 5 x 37= 1110
The product of numbers in factorised form is 222 x 370 = 3 x 2 x 37 x 5 x 2 x 37
Yes, LCM is a factor of the product.
So, it should be multiplied by 74 to get the product.
Thus, LCM x 74 = 222 x 370 it
Question.
Why does this happen? Can you give an explanation or proof?
Solution:
The relationship LCM x HCF = Product of two numbers holds due to the way the LCM and HCF are defined. When we multiply the LCM and HCF of two numbers, we obtain the same result as multiplying the two numbers directly. This occurs because the HCF accounts for the common factors, while the LCM considers all the factors from both numbers, counting them only once.
Question.
Explore whether this property holds when 3 numbers are considered.
Solution:
Do it yourself.
Figure It Out (Pages 63-64)
Question 1.
In the two rows below, colours repeat as shown. When will the blue stars meet next?
(Refer the original image from NCERT Textbook Page 63)
Solution:
The blue stars will meet again at position 16.
This is because the blue stars in the top row repeat itself at every 6th position and in the bottom row every 4th positions.
So, the next common position is after the LCM of 6 and 4, which is 12.
Since their starting points are different and they are meeting first at position 4, after that they will meet again at position 16 (i.e., 12th position after the first meeting point)
Question 2.
(a) Is 5 x 7 x 11 x 11 a multiple of 5 x 7 x 7 x 11 x 2?
(b) Is 5 x 7 x 11 x 11 a factor of 5 x 7 x 7 x 11 x 2?
Solution:
(a) 5 x 7 x 11 x 11 is not a multiple of 5 x 7 x 7 x 11 x 2, because the factors of the second number also include 2 and one more 7, which are not present in the first number.
(b) 5 x 7 x 11 x 11 is not a factor of 5 x 7 x 7 x 11 x 2, because in 5 x 7 x 11 x 11 one extra 11 is present, which is not present in 5 x 7 x 7 x 11 x 2.
Question 3.
Find the HCF and LCM of the following (state your answers in the form of prime factorisations):
(a) 3 x 3 x 5 x 7 x 7 and 12 x 7 x 11
(b) 45 and 36
Solution:
(a) Prime factorisations:
3 x 3 x 5 x 7 x 7 = 3 x 3 x 5 x 7 x 7
12 x 7 x 11 = 2 x 2 x 3 x 7 x 11
HCF = 3 x 7 = 21
and LCM = 2 x 2 x 3 x 3 x 5 x 7 x 7 x 11 = 97020
(b) Prime factorisations:
45 = 3 x 3 x 5
36 = 2 x 2 x 3 x 3
HCF = 3 x 3 = 9
and LCM = 2 x 2 x 3 x 3 x 5 = 180
Question 4.
Find two numbers whose HCF is 1 and LCM is 66.
Solution:
Prime factorisation of 66 = 2 x 3 x 11
We need two numbers such that their product
= HCF x LCM = 1 x 66 = 66
Possible pairs: 1 x 66,2 x 33, 3 x 22, 6 x 11
whose HCF is 1 and LCM is 66.
Question 5.
A cowherd took all his cows to graze in the fields. The cows came to a crossing with 3 gates. An equal number of cows passed through each gate. Later at another crossing with 5 gates again an equal number of cows passed through each gate. The same happened at the third crossing with 7 gates. If the cowherd had less than 200 cows, how many cows did he have? (Based on the folklore mathematics from Karnataka.)
Solution:
LCM of 3, 5, and 7 is 105.
The number of cows must be divisible by 3,5, and 7
(i.e., the LCM of these numbers).
The next multiple of 105 is 210 > 200.
But, it is given that the total number of cows must be less than 200.
So, the possible number of cows is 105.
Question 6.
The length, width, and height of a box are 12 cm, 18 cm, and 36 cm respectively. Which of the following sized cubes can be packed in this box without leaving gaps?
(a) 9 cm
(b) 6 cm
(c) 4 cm
(d) 3 cm
(e) 2 cm
Solution:
The length, width, and height of the box are 12 cm, 18 cm, and 36 cm.
12 = 2 x 2 x 3
18 = 2 x 3 x 3
36 = 2 x 2 x 3 x 3
Here, the common factors are 2, 3 and 6
So, cubes of side lengths 2 cm, 3 cm and 6 cm can be packed fully in this box (without leaving any gap).
Question 7.
Among the numbers below, which is the largest number that perfectly divides both 306 and 36?
(a) 36
(b) 612
(c) 18
(d) 3
(e) 2
(f) 360
Solution:
(c) Here, 306 = 2 x 3 x 3 x 17 and 36 = 2 x 3 x 2 x 3
HCF of 306 and 36 = 18
So, 18 is the largest number that perfectly divides both 306’and 36.
Question 8.
Find the smallest number that is divisible by 3,4, 5 and 7 but leaves a remainder of 10 when divided by 11.
Solution:
First, we will find the LCM of 3, 4, 5, and 7 by prime factorisation:
3 = 3, 4 = 2 x 2, 5 = 5, 7 = 7 LCM = 2 x 2 x 3 x 5 x 7 = 420
But we want a number when divided by 11, gives the remainder 10.
We need to check multiples of 420 to find one that gives a remainder of 10 when divided by 11.
Consider 420 x 5 = 2100
When we divide 2100 by 11, we get the remainder 10. Thus, the smallest number is 2100.
Question 9.
Children are playing ‘Fire in the Mountain’. When the number 6 was called out, no one got out. When the number 9 was called out, no one got out. But when the number 10 was called out, some people got out. How many children could have been playing initially?
(a) 72
(b) 90
(c) 45
(d) 3
(e) 36
(f) None of these
Solution:
(a and e): It is given that when the number 6 was called out, no one got out. When the number 9 was called out, no one got out. But when the number 10 was called out, some people got out.
Thus the number of children must be divisible by 6 and 9 but not by 10.
6 = 2 x 3 and 9 = 3 x 3. So, the LCM of 6 and 9 is 18.
The number of children playing initially must be a multiple of 18 but not divisible by 10.
By checking options, we get that 72 and 36 are two numbers that are both multiples of 6 and 9 but not divisible by 10.
Question 10.
Tick the correct statement(s): The LCM of two different prime numbers (m, n) can be:
(a) Less than both numbers
(b) In between the two numbers
(c) Greater than both numbers
(d) Less than m x n
(e) Greater than m x n
Solution:
(c) LCM of two prime numbers (a, b) = a x b, which is greater than both the numbers.
Question 11.
A dog is chasing a rabbit that has a head start of 150 feet. It jumps .9 feet every time the rabbit jumps 7 feet. In how many leaps does the dog catch up with the rabbit?
Solution:
In every leap, the dog jumps 9 ft, and the rabbit jumps 7 ft.
Distance covered per leap by the dog = 9-7 = 2 feet.
Rabbit has 150 feet head start.
Number of leaps required = \(\frac{150}{2}\) = 75
So, the dog catch up the rabbit in 75 leaps.
Question 12.
What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, 10? Do you remember the answer from Grade 6, Chapter 5?
Solution:
To find the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, and 10.
That means we need to find their Least Common Multiple (LCM).
LCM of 1, 2, 3, 4, 5, 6, 8, 9 and 10 = 2 x 2 x 2 x 3 x 3 x 5 = 8 x 9 x 5 = 360
The smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, and 10 is 360.
Question 13.
Here is a problem posed by the ancient Indian Mathematician Mahaviracharya (850 C.E.) Add together
\(\frac{8}{15}, \frac{1}{20}, \frac{7}{36}, \frac{11}{63} \text { and } \frac{1}{21}\). What do you get? How can we 15 20 36 63 21 find this sum efficiently?
Solution:
Prime factorisation of denominator of each fractions:
15 = 3 x 5
20 = 2 x 2 x 5
36 = 2 x 2 x 3 x 3
63 = 3 x 3 x 7
21 = 3 x 7
LCM = 2 x 2 x 3 x 3 x 5 x 7
= 4 x 9 x 5 x 7 = 1260
Convert each fraction to a fraction with denominator 1260.
