Finding Common Ground Class 7 Solutions RBSE Maths Ganita Prakash Part 2 Chapter 3
PracticingGanita Prakash Class 7 Solutionsand RBSE Class 7 Maths Part 2 Chapter 3 Finding Common Ground Solutions Question Answer helps develop logical thinking and accuracy.
Ganita Prakash Class 7 Part 2 Chapter 3 Solutions
Class 7 Ganita Prakash Part 2 Chapter 3 Solutions
RBSE Class 7 Maths Ganita Prakash Part 2 Chapter 3 Solutions
In-text Questions
Page 48
Question 1.
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) 14 and 30
Factors of 14 : 1, 2, 7, 14
Factors of 30 : 1, 2, 3, 5, 6, 10, 15, 30
Common factors : 1, 2
HCF = 2
(b) 7 and 11
Both are prime numbers.
Common factor = 1
HCF = 1
(c) 30 and 50
Factors of 30 : 1, 2, 3, 5, 6, 10, 15, 30
Factors of 50 : 1,2, 5, 10, 25, 50
Common factors: 1, 2, 5, 10
HCF = 10
(d) 28 and 42
Factors of 28 : 1, 2, 4, 7, 14, 28
Factors of 42 : 1, 2, 3, 6, 7, 14, 21, 42
Common factors : 1, 2, 7, 14
HCF = 14
Question 2.
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 same as the HCF of the two numbers. Reason : To collect both the treasures, using the jump of longest size.
Page 51
Question 1.
Similarly, is 2 × 7 = 14 a factor of 840? Why or why not? Is 2 × 2 × 2 a factor of 840? Why or why not? Is 3 × 3 × 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,
Both 2 and 7 appear in the prime factorisation of 840.
Since 14 can be formed using the primes of 840, it divides 840 exactly.
Yes,
Three 2s occur in the prime factorisation, so 8 (which uses all three 2s) is also a factor of 840.
No,
The number 840 has only one factor of 3. Since 27 needs three 3 s, it cannot be formed from the prime factors of 840 and therefore does not divide 840.
Yes, we can use this idea to list down all the possible factors of a number using just its prime factors.
Page 55
Question 1.
What about the largest common multiple? Does such a number exist?
Solution:
For any common multiple of two numbers, we can always create a larger one by multiplying it by 2, 3, or any other whole number. This process can continue forever. For example, if 24 is a common multiple of two numbers, then :
48, 72, 96, 120, … are also common multiples.
Since this list never ends, there is no greatest (largest) common multiple – common multiples are unlimited.
Only the least common multiple (LCM) exists, not a largest one.
Page 56
Question 1.
Do you remember the ‘Idli-Vada’ game from Grade 6 (see chapter ‘Prime Time’)? Two numbers are chosen and whenever players come to their multiples, ‘idli’ or ‘vada’ should be called out depending on whose multiple the number is. If the number happens to be a common multiple, then ‘idli-vada’ should be called out. In each problem below, the two numbers corresponding to ‘idli’ and ‘vada’ are given. 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 Idli-Vada game, ‘idli-vada’ is called when a number is a common multiple of both given numbers.
The first such number is the LCM (Least Common Multiple) of the two numbers.
(a) 4 and 6
LCM of 4 and 6 = 12 First ‘idli-vada’ at 12
(b) 7 and 11
Since both are primes,
LCM = 7 × 11 = 77 First ‘idli-vada’ at 77
(c) 14 and 30
14 = 2 × 7
30 = 2 × 3 × 5
LCM = 2 × 7 × 3 × 5 = 210
First ‘idli-vada’ at 210
(d) 15 and 55
15 = 3 × 5
55 = 5 × 11
LCM = 3 × 5 × 11 = 165
First ‘idli-vada’ at 165
Yes, the answer is always the LCM.
Reason:
Players call ‘idli-vada’ at numbers that are multiples of both chosen numbers (common multiples).
The least such number is exactly the LCM, so it is always the first number where ‘idli-vada’ is called.
Page 59
Question 1.
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 a number. The other number can be any multiple of m, i.e., m and km.
Here, k is any positive integer. m = 6⇒ pairs : (6, 12), (6, 18), (6, 24), …
In all cases, HCF = m.
(b) The other number could be any factor of 7k such that the smaller number divides the larger one.
So a suitable pair is : 7k, k.
If 7k = 21 (so k = 3), the pair is (21, 3) and HCF = 3.
Page 59
Question 1.
What happens to the HCF of two numbers if both numbers are doubled?
Take some pairs of numbers and explore. Are you able to see why the HCF will also double?
Solution:
If both numbers are doubled, then both numbers get an extra factor of 2 in their prime factors. This 2 will be included as a factor in the largest common subpart, and so the HCF will double. For example, consider the numbers 270 and 50.
HCF = 2 × 5 = 10
Let us double these numbers to get 540 and 100.
HCF = 2 × 2 × 5 = 20.
Page 60
Question 1.
Consider the following two multiples of 14 – 14 × 6, 14 × 9. What is their HCF?
Solution:
Clearly, 14 is a common factor. Let us calculate the prime factorisations.
HCF = 14 × 3 = 42.
Question 2.
Here are some more numbers where both numbers are multiples of the same number. Find their HCF:
(a) 18 × 10, 18 × 15
(b) 10 × 38, 10 × 21
(c) 5 × 13, 5 × 20
(d) 12 × 16, 12 × 20
Solution:
(a) HCF = 18 × HCF (10, 15)
HCF (10, 15) = 5
HCF = 18 × 5 = 90
(b) HCF = 10 × HCF (38, 21)
HCF (38, 21) = 1
HCF = 10
(c) HCF = 5 × HCF (13, 20)
HCF (13, 20) = 1
HCF = 5
(d) HCF = 12 × HCF (16, 20)
HCF (16, 20) = 4
HCF – 12 × 4 = 48
Question 3.
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:
Part (c) in above question also have the same thing.
Other examples are :
7 × 4 and 7 × 9
HCF (4, 9) = 1
HCF = 7
12 × 5 and 12 × 11
HCF (5, 11) = 1
HCF = 12
Page 62
Question 1.
You can try this method (division method) for these pairs of numbers.
(a) 90 and 150
Solution:
(a)
LCM = 2 × 3 × 3 × 5 × 1 × 5 = 450
(b)
LCM = 2 × 2 × 3 × 7 × 11 = 924
Page 62
Question 1.
Which is greater-the LCM of two numbers or their product?
Solution:
The product of two numbers is always greater than or equal to their LCM
Question 2.
You could analyse the above statement using examples. Then try to reason or prove, why the LCM is never greater than the product of the numbers.
Solution:
Example 1 (numbers with a common factor):
Take 12 and 18.
Prime factorisations : 12 = 22× 3, 18 = 2 × 32.
LCM = 22× 32= 36. Product = 12 × 18 = 216. So LCM (36) is much less than the product (216).
Example 2 (coprime numbers) :
Take 8 and 15.
They have no common prime factors, so LCM = 8 × 15 = 120. Here LCM equals the product because the numbers are coprime.
The product a × b contains all prime factors of both a and b, but with multiplicities multiplied. The LCM only needs each prime to appear with the highest power that appears in either number, not multiplied. So the LCM can’t be larger than the raw product – at worst (when there are no common primes) it equals the product.
The LCM of two positive integers is always less than or equal to their product; equality holds precisely when the two numbers are coprime.
Question 3.
Consider the numbers 105 and 95. Find their LCM.
Solution:
Factorising them into their primes :
105 = 3 × 5 × 7
95 = 5 × 19
LCM = 3 × 5 × 7 × 19.
Let us consider the product in the factorised form:
105 × 95 = 3 × 5 × 5 × 7 × 19
Question 4.
Is the LCM a factor of the product? If yes, what should it be multiplied with to get the product?
Solution:
Yes, the LCM is a factor of the product.
To get the product from the LCM, multiply by the remaining factor 5 :
LCM × 5 = 105 × 95
So, the LCM must be multiplied by 5 to obtain the product.
Question 5.
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) Prime factorisation
45 = 32× 5
105 = 3 × 5 × 7
LCM = 32 × 5 × 7 = 315
45 × 105 = 4725
So, 4725 ÷ 315 = 15
HCF = 3 × 5 = 15
So, LCM need to be multiplied by HCF to get the product of numbers.
Same pattern is observed in all.
Page 63
Question 1.
Do you see that, in each case, the number by which the LCM is multiplied to get the product is actually the HCF?
Solution:
Yes.
Our observations seem to suggest the following:
HCF × LCM = Product of the two numbers.
Question 2.
Why does this happen? Can you give an explanation or proof?
Solution:
This happens because of how the prime factors of the two numbers are shared between the HCF and the LCM.
The common prime factors of both numbers make up the HCF.
The highest powers of all prime factors (common and non-common) make up the LCM.
When we multiply HCF and LCM, every prime factor from both numbers is included exactly as it appears in the product of the two numbers.
So, the prime factors of the two numbers are correctly “divided” between the HCF and the LCM, and together they recreate the full product of the numbers :
HCF × LCM = Product of the two numbers
Question 3.
Explore whether this property holds when 3 numbers are considered.
Solution:
No, this property does not hold for three numbers.
Example : 2, 3, 4
HCF (2, 3, 4) = 1
LCM (2, 3, 4) = 12
Product = 2 × 3 × 4 = 24
HCF × LCM = 1 × 12 = 12 × 24
Class 7 Maths Finding Common Ground Solutions
Figure it Out (Page 51)
Question 1.
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) Factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.
(b) Factors of 105 are 1, 3, 5, 7, 15, 21, 35, 105.
(c) Factors of 132 are 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132.
(d) Factors of 360 are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360.
(e) Factors of 840 are 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, 840.
Figure it Out (Page 53)
Question 1.
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
How do we directly find the HCF without listing all the factors?
Solution:
(a) Factors of 50 : 1,2, 5, 10, 25,
Factors of 60 : 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Common factors : 1, 2, 5, 10 HCF = 10
(b) Factors of 140 : 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70, 140
Factors of 275 : 1, 5, 11, 25, 55, 275 Common factors : 1, 5 HCF = 5
(c) Factors of 77 : 1, 7, 11, 77
Factors of 725 : 1, 5, 25, 29, 145, 725
Common factor: 1
HCF = 1
(d) Factors of 370 : 1, 2, 5, 10, 37, 74, 185, 370
Factors of 592 : 1, 2, 4, 8, 16, 37, 74, 148, 296, 592
Common factors : 1, 2, 37, 74
HCF = 74
(e) Factors of 81 : 1, 3, 9, 27, 81
Factors of 243 : 1, 3, 9, 27, 81, 243
Common factors : 1, 3, 9, 27, 81
HCF = 81
We can directly find the HCF without listing prime factors by finding the prime factorisation of the given numbers simultaneously and identifying the common prime factors by which both numbers can be divided. The product of these common prime factors is their HCF.
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) Prime factorisation :
24 = 2 × 2 × 2 × 3 = 23× 3
180 = 2 × 2 × 3 × 3 × 5 = 22× 32× 5
Common prime factors : 22× 3
HCF = 4 × 3 = 12
(b) Prime factorisation:
42 = 2 × 3 × 7
75 = 3 × 5 × 5 24 = 2 × 2 × 2 × 3 = 23× 3
Common prime factor to all : 3
HCF = 3
(c) Prime factorisation:
240 = 24× 3 × 5
378 = 2 × 33× 7
Common prime factors : 2 × 3
HCF = 6
(d) Prime factorisation:
400 = 24× 52
2500 = 22× 54
Common prime factors : 22× 52
HCF = 4 × 25 = 100
(e) Prime factorisation:
300 = 22× 3 × 52
800 = 25× 52
Common prime factors : 22× 52
HCF = 4 × 25 = 100
Question 2.
Consider the numbers 72 and 144. Suppose they are factorised into composite numbers as : 72 = 6 × 12 and 144 = 8 × 18. Seeing this, can one say that these two numbers have no common factor other than 1? Why not?
Solution:
No, we cannot say that 72 and 144 have no common factor other than 1.
Even though the composite factors shown (6, 12, 8, 18) do not look the same, this does not mean the numbers have no common factors. Composite factors can be further broken into prime factors.
72 = 23× 32
144 = 24× 32
They clearly share common prime factors 2 and 3, so they have several common factors such as 2, 3, 4, 6, 9, 12, 18, 36 and more.
Their HCF is 72.
Thus, we must always use prime factorisation to correctly find common factors, not just compare composite factors.
Figure it Out (Page 58)
Question 1.
Find the LCM of the following numbers :
(a) 30, 72
(b) 36, 54
(c) 105, 195, 65
(d) 222, 370
Solution:
(a) Prime factorisation :
30 = 2 × 3 × 5
72 = 23× 32
Take highest powers of all primes :
LCM = 23× 32× 5
LCM = 360
(b) Prime factorisation:
36 = 22× 32
54 = 2 × 33
Highest powers:
LCM = 22× 33
LCM = 108
(c) Prime factorisation:
105 = 3 × 5 × 7
195 = 3 × 5 × 13
65 = 5 × 13
Highest powers :
LCM = 3 × 5 × 7 × 13
LCM = 1365
(d) Prime factorisation :
222 = 2 × 3 × 37
370 = 2 × 5 × 37
Highest powers :
LCM = 2 × 3 × 5 × 37
LCM = 1110
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) They always have 2 as a common factor, so their HCF is 2.
(b) They have no common factor other than 1, so their HCF is 1.
(c) They are always divisible by 2, so their HCF is at least 2 (it may be greater depending on the numbers).
(d) They always have no common factor except 1.
(e) They have no common factor 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) The LCM of two numbers is one of the numbers when one number is a multiple of the other.
Examples :
- LCM (4, 20) = 20
- LCM (6, 18) = 18
- LCM (5, 25) = 25
- LCM (7, 35) = 35
- LCM (8, 32) = 32
(b) Let the two numbers be n and kn, where A: is a positive integer.
Since kn is already a multiple of n, the smallest number divisible by both is kn.
LCM (n, kn) = kn
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) If a = 3m and b = 3n, then both numbers contain the factor 3.
Therefore, LCM (a, b) = 3 × LCM (m, n)
The LCM will always be a multiple of 3.
(b) Let the numbers be 2k and 2 (k + 1).
They share only the common factor 2, so :
LCM 2k, (2k + 2) =
The LCM is always greater than the larger number.
(c) Let the numbers be n and n + 1.
Since consecutive numbers are always co¬prime :
gcd (n, n + 1) = 1
LCM (n, n + 1) = n(n + 1)
The LCM equals the product of the numbers.
(d) Let the numbers be a and b such that :
gcd (a, b) = 1
LCM (a, b) = a × b
The LCM is simply the product of the two numbers.
Figure it Out (Pages 63-64)
Question 1.
In the two rows below, colours repeat as shown. When will the blue stars meet next?
Solution:
In the first row blue star is repeated at the position 4, 10..
So, next blue star will be at 16. (at the interval of 6)
In the second row, blue stars are at the position 4, 8, 12..
So, next blue star will be at 16. (at the interval of 4)
Question 2.
(a) Is 5 × 7 × 11 × 11 a multiple of 5 × 7 × 7 × 11 × 2?
(b) Is 5 × 7 × 11 × 11 a factor of 5 × 7 × 7 × 11 × 2?
Solution:
(a) To be a multiple, the first number must contain all prime factors of the second number in equal or greater powers.
First number : 51× 71× 112
Second number : 2 × 5 × 72× 11
The first number is missing the factor 2 and has only one 7 instead of two.
No, it is not a multiple.
(b) To be a factor, every prime factor of the smaller number must appear in the larger number with equal or greater power.
The first number needs 112, but the second number has only one 11.
No, it is not a factor.
Question 3.
Find the HCF and LCM of the following (state your answers in the form of prime factorisations) :
(a) 3 3x5x7x7 and 12x7x11
(b) 45 and 36
Solution: (a) Numbers :
1st = 3 × 3 × 5 × 7 × 7 = 32× 5 × 72
2nd = 12 × 7 × 11 = (22× 3) × 7 × 11 = 22× 3 × 7 × 11
HCF = 3 × 7 = 21
LCM = 22× 32× 5 × 72× 11
(b) Numbers :
45 = 32× 5
36 = 22× 32
HCF : Common prime factors with lowest powers.
Then, HCF = 32
LCM : All prime factors with highest powers.
Then, LCM = 22× 32× 5
Question 4.
Find two numbers whose HCF is 1 and LCM is 66.
Solution:
Since, HCF × LCM = Product of the two numbers
and the HCF = 1, the two numbers must be co-prime and their product must equal the LCM:
1 × 66 = 66
So we need two co-prime factors of 66.
Prime factorisation : 66 = 2 × 3 × 11
Split the primes into two groups with no common factor :
6 = 2 × 3 and 11 These are co-prime.
The two numbers are 6 and 11.
HCF (6, 11) = 1
LCM (6, 11) = 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:
We need a number of cows that is divisible by 3, 5, and 7, and is less than 200.
The LCM of 3, 5 and 7 is LCM (3, 5, 7) = 3 × 5 × 7 = 105
Any number that works must be a multiple of 105. The multiples are 105, 210,only 105 is less than 200.
So the cowherd had 105 cows.
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:
To pack cubes into the box without leaving gaps, the edge length of each cube must exactly divide all three dimensions of the box : 12 cm, 18 cm, and 36 cm.
Find the common factors of 12, 18, and 36.
Prime factorisations :
12 = 22× 3
18 = 2 × 32
36 = 22× 32
Common factors of 12, 18 and 36 are : 1, 2, 3, 6
Out of the given options, the common factor condition is satisfied by 4, 3 and 2.
Cubes that can be packed without gaps :
6 cm, 3 cm, and 2 cm Correct options : (b), (d), and (e)
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:
Prime factorisation :
306 = 2 × 3 × 3 × 17 = 2 × 32× 17
36 = 22× 32
HCF = 2 × 32= 2 × 9 = 18
The largest number that perfectly divides both 306 and 36 is : (c) 18 (their HCF).
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:
Prime factorisation :
3 = 3
4 = 22
5 = 5
7 = 7
LCM = 22× 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420
So, the number must be a multiple of420. It should leave a remainder of 10 when divided by 11.
We need : N ≡ 10 (mod 11) ⇒ 420k ≡ 10 (mod 11)
420 mod 11 = ?
11 × 38 = 418 → 420 – 418 = 2
420 ≡ 2 (mod 11)
So : 2k ≡ 10 (mod 11)
The modular inverse of 2 modulo 11 is 6 (since 2 × 6 = 12 ≡ 1 mod 11).
k ≡ 10 × 6 ≡ 60 ≡ 5 (mod 11)
Smallest positive k = 5k = 5k = 5
N = 420 × 5 = 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:
When 6 was called, no one got out -»total number of children is divisible by 6 When 9 was called, no one got out → total number of children is divisible by 9 When 10 was called, some got out → total number of children is not divisible by 10 Let the total number of children be N.
Find the LCM of 6 and 9
Prime factorisation:
6 = 2 × 3
9 = 32
LCM (6, 9) = 2 × 3 × 3 = 18 So number of children must be multiple of 18.
Here 72, 90 and 36 are multiple of 18 in which only 72 and 36 are not multiple of 10.
Hence, number of children could have been played initially = 72 or 36.
So option (a) and (e) are correct.
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 × n
(e) Greater than m × n
Solution:
Two different prime numbers have no common factor other than 1.
So, HCF (m, n) = 1
LCM (m, n) = m × n/HCF (m, n) = m × n
Analyze the options
(a) Less than both numbers → (LCM = m × n > m and > n), so it is wrong statement.
(b) In between the two numbers → so it is wrong statement.
(c) Greater than both numbers → (LCM = m × n > m and > n) so it is a correct statement.
(d) Less than m × n → (LCM = m × n) so it is wrong statement.
(e) Greater than m × n → so it is wrong statement.
So, correct statement is (c).
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:
Rabbit moves 7 feet per leap → total distance = 7n
Dog moves 9 feet per leap → total distance = 9n
The rabbit has a head start of 150 feet, so the dog catches the rabbit when :
9 n = 7n + 150
9n – 7n = 150
2n = 150
n = 75
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:
Number | Prime Factorisation |
1 | – |
2 | 2 |
3 | 3 |
4 | 22 |
5 | 5 |
6 | 2 × 3 |
8 | 23 |
9 | 32 |
10 | 2 × 5 |
Prime Factorisation
Take the highest powers of all primes
- 2 → highest power = 23(from 8)
- 3 → highest power = 32(from 9)
- 5 → highest power = 51(from 5 or 10)
LCM = 23× 32× 5 = 360
The smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, 10 is 360
Question 13.
Here is a problem posed by the ancient Indian Mathematician Mahaviracharya (850 C.E.). Add together . What do you get? How can we find this sum efficiently?
Solution:
We need to add :
Find the LCM of all denominators
Prime factorisations :
15 = 3 × 5
20 = 22× 5
36 = 22× 32
63 = 32× 7
21 = 3 × 7
LCM = 22× 32× 5 × 7 = 4 × 9 × 5 × 7 × 1260
Convert each fraction to denominator 1260
Add the numerators : 672 + 63 + 245 + 220 + 60 = 1260
Then, final answer is : = 1
Using LCM avoids repeated simplifications and gives a clean, fast solution – just as used by ancient mathematicians like Mahaviracharya.
It’s Puzzle Time!
Mystery Colours!
You might have noticed and wondered about these different circle designs around the page numbers on each page!
The picture below shows all the designs for the numbers from 1 to 100.
Try to decode the colour scheme for each number.
There are several interesting patterns here.
Share your observations with your classmates.
Extending this scheme, colour the page numbers from 101-110.
Solution:
The colour scheme is according to number of factors a particular number. The number of colours around the number in a circle is equal to the number of factors that number had.
The colouring work will be done by students themself