Finding Common Ground - NCERT Solutions Class 7 Maths (Ganita Prakash)

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Finding Common Ground – NCERT Solutions Class 7 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 7 Maths (Ganita Prakash).

NCERT Solutions Class 7

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Finding Common Ground – NCERT Solutions

Q.1: List all the factors of the following numbers:

90
105
132
360 (this number has 24 factors)
840 (this number has 32 factors)

Solution:

90:
{tex}1,2,3,5,6,9,10,15,18,30,45,90{/tex}
105:
1, 3, 5, 7, 15, 21, 35, 105
132:
{tex}1,2,3,4,6,11,12,22,33,44,66,132{/tex}
360 (24 factors):
{tex}1,2,3,4,5,6,8,9,10,12,15,18,20,24,30,{/tex}{tex}36,40,45,60,72,90,120,180,360{/tex}
840 (32 factors):
{tex}1,2,3,4,5,6,7,8,10,12,14,15,20,21,24,28,{/tex}{tex}30,35,40,42,56,60,70,84,105,120,140,168{/tex}, {tex}210,280,420,840{/tex}

Q.2: Consider the numbers 72 and 144. Suppose they are factorised into composite numbers as: {tex}72=6 \times 12{/tex} and {tex}144=8 \times 18{/tex}. 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 just because their given factorizations are into composite numbers:
{tex} 72=6 \times 12 \text { and } 144=8 \times 18 {/tex}
This is because these are not prime factorizations-both {tex}6,12,8{/tex}, and 18 are composite numbers, and they can all be factorized further. For example,
{tex}6=2 \times 3{/tex}
{tex}12=2 \times 2 \times 3{/tex}
{tex}8=2 \times 2 \times 2{/tex}
{tex}18=2 \times 3 \times 3{/tex}
When we fully factorize each number into prime factors:
{tex} 72=2 \times 2 \times 2 \times 3 \times 3 {/tex}
{tex} 144=2 \times 2 \times 2 \times 2 \times 3 \times 3 {/tex}
We can see that both 72 and 144 have common factors other than 1 (for example, 2 and 3, also 6, 12, etc.).
Therefore, to decide common factors properly, we must write numbers as a product of prime factors, not just any composite factorization.

Q.3: Find the LCM of: 30, 72

Solution:

Prime factorization:
{tex} 30=2 \times 3 \times 5 {/tex}
{tex} 72=2^3 \times 3^2 {/tex}
Common multiples should contain each prime factor as a subpart, picking the highest power present in either number:
So,
{tex} {LCM}=2^3 \times 3^2 \times 5=8 \times 9 \times 5=360 {/tex}

Q.4: Find the LCM of: 36, 54

Solution:

{tex} \text { Prime factorisation: } {/tex}
{tex} 36=2^2 \times 3^2 {/tex}
{tex} 54=2 \times 3^3 {/tex}
{tex} \text { LCM }=2^2 \times 3^3=4 \times 27=108 {/tex}

Q.5: Find the LCM of: 105, 195, and 65

Solution:

Prime factorisation:
{tex} 105=3 \times 5 \times 7 {/tex}
{tex} 195=3 \times 5 \times 13 {/tex}
{tex} 65=5 \times 13 {/tex}
LCM contains all primes (each at highest power among the numbers):
{tex} {LCM}=3 \times 5 \times 7 \times 13=1365 {/tex}

Q.6: Find the LCM of: 222, 370

Solution:

Prime factorization:
{tex} 222=2 \times 3 \times 37 {/tex}
{tex} 370=2 \times 5 \times 37 {/tex}
LCM must have each prime (highest count present):
{tex} \text { LCM }=2 \times 3 \times 5 \times 37=1110 {/tex}

Q.7: 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.

Two consecutive even numbers
Two consecutive odd numbers
Two even numbers
Two consecutive numbers
Two co-prime numbers

Share your observations with the class.

Solution:

Statement: The HCF of two consecutive even numbers is always 2.
Reason: Both numbers are divisible by 2, but one is not divisible by 4 (since consecutive even numbers differ by 2), so their greatest common factor is 2.
Statement: The HCF of two consecutive odd numbers is always 1.
Reason: Any two consecutive odd numbers differ by 2. They cannot have any common factor greater than 1, since any such factor would have to divide 2, which is not possible for odd numbers.
Statement: The HCF of two even numbers is at least 2, but can be more.
Reason: Both are divisible by 2, but if they share more factors, the HCF can be greater (for example, HCF of 8 and 12 is 4).
Statement: The HCF of two even numbers is at least 2, but can be more.
Reason: Both are divisible by 2, but if they share more factors, the HCF can be greater (for example, HCF of 8 and 12 is 4).
Statement: The HCF of two co-prime numbers is always 1.
Reason: Co-prime numbers have no common factor other than 1 by definition.

Q.8: Find the common factor and the HCF of: 50, 60

Solution:

{tex} 50=2 \times 5 \times 5 {/tex}
{tex} 60=2 \times 2 \times 3 \times 5 {/tex}
The common factors are {tex}1,2,5{/tex}, and 10 (since 2 and 5 appear in both factorizations, and products of these).
The HCF is 10.

Q.9: Find the common factors and the HCF of: 140, 275.

Solution:

{tex} 140=2 \times 2 \times 5 \times 7 {/tex}
{tex} 275=5 \times 5 \times 11 {/tex}
The only factor common in these two factorizations is 5.
So, 5 is the only common factor (apart from 1) and it is their HCF.

Q.10: Find the common factors and the HCF of: 77, 725

Solution:

{tex} 77=7 \times 11 {/tex}
{tex} 725=5 \times 5 \times 29 {/tex}
There is no subpart that is common amongst these two factorizations.
So, 1 is the only common factor. It is also their HCF.

Q.11: Find the common factors and the HCF of: 370, 592.

Solution:

{tex} 370=2 \times 5 \times 37 {/tex}
{tex} 592=2 \times 2 \times 2 \times 2 \times 37 {/tex}
The only common prime factors are 2 and 37. So, the product {tex}2 \times 37=74{/tex} is the highest common factor in both.
The HCF is 74.

Q.12: Find the common factors and the HCF of:81 and 243.

Solution:

{tex} 81=3 \times 3 \times 3 \times 3 {/tex}
{tex} 243=3 \times 3 \times 3 \times 3 \times 3 {/tex}
Both have 3 as the only prime factor; the highest power of 3 common to both is {tex}3^4=81{/tex}.
The HCF is 81.

Q.13: Find the HCF of: 24,180.

Solution:

Prime factorization:
{tex} 24=2 \times 2 \times 2 \times 3 {/tex}
{tex} 180=2 \times 2 \times 3 \times 3 \times 5 {/tex}
Common primes: 2 and 3

24 has three {tex}2 \mathrm{~s} ; 180{/tex} has two {tex}2 \mathrm{~s} \Rightarrow{/tex} Take the minimum {tex}={/tex} two 2s.
Both have at least one {tex}3 \Rightarrow{/tex} Take one 3.

So, {tex}H C F=2 \times 2 \times 3=12{/tex}

Q.14: Find the HCF of: 42, 75, 24.

Solution:

Prime factorization:
{tex} 42=2 \times 3 \times 7 {/tex}
{tex} 75=3 \times 5 \times 5 {/tex}
{tex} 24=2 \times 2 \times 2 \times 3 {/tex}
Common prime: 3 (minimum one occurrence in all)
So, {tex}H C F=3{/tex}

Q.15: Find the HCF of: 400, 2500

Solution:

Prime factorization:
{tex} 400=2 \times 2 \times 2 \times 2 \times 5 \times 5 {/tex}
{tex} 2500=2 \times 2 \times 5 \times 5 \times 5 \times 5 {/tex}
Common primes: 2 and 5
Minimum number of 2s: two
Minimum number of 5 s: two
So, {tex}H C F=2 \times 2 \times 5 \times 5=100{/tex}

Q.16: Find the HCF of: 240, 378

Solution:

Prime factorization:
{tex} 240=2 \times 2 \times 2 \times 2 \times 3 \times 5 {/tex}
{tex} 378=2 \times 3 \times 3 \times 3 \times 7 {/tex}
Common primes: 2 and 3
Smallest number of 2s: one
Smallest number of 3 s : one
So, {tex}H C F=2 \times 3=6{/tex}

Q.17: Find the HCF of: 300, 800

Solution:

Prime factorization:
{tex} 300=2 \times 2 \times 3 \times 5 \times 5 {/tex}
{tex} 800=2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 {/tex}
Common primes: 2 and 5
Minimum number of 2 s: two
Minimum number of 5 s two
So, {tex}H C F=2 \times 2 \times 5 \times 5=100{/tex}

Q.18: The LCM of 3 and 24 is 24 (it is one of the two given numbers).

Find more such number pairs where the LCM is one of the two numbers.
Make a general statement about such numbers. Describe such number pairs using algebra.

Solution:

More number pairs with LCM equal to one number
The LCM of 8 and 24 is 24.
The LCM of 5 and 20 is 20.
The LCM of 2 and 12 is 12.
The LCM of 21 and 42 is 42.
The LCM of 7 and 35 is 35.
In all these cases, the LCM is simply the bigger number.
General statement and algebraic description
Statement:
If one number is a factor (divides exactly) of the other, then their LCM is the larger number.
Algebraic description:
If numbers {tex}a{/tex} and {tex}b{/tex} satisfy {tex}a \mid b{/tex} (a divides b), then {tex}\operatorname{LCM}(a, b)=b{/tex}.
Alternatively, if {tex}b \mid a{/tex}, then {tex}\operatorname{LCM}(a, b)=a{/tex}.
Explanation:
This happens because the bigger number already contains all the factors required for the smaller number, so the smallest multiple common to both is just the bigger number. This is true for any number pair where one is a multiple of the other.

Q.19: In the two rows below, colours repeat as shown. When will the blue stars meet next?

Solution:

The first row repeats every 6 stars (yellow, green, orange, blue, pink, grey).
The second row repeats every 4 stars (orange, green, yellow, blue).

To find when the blue stars align in both rows, find the LCM of 6 and 4:
{tex} \operatorname{LCM}(6,4)=12 {/tex}
Answer:
The blue stars in both rows will meet next at the 12th position from the start, because the LCM of their cycles is 12.

Q.20: 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.

Two multiples of 3
Two consecutive even numbers
Two consecutive numbers
Two co-prime numbers

Solution:

General statement: The LCM of any two multiples of 3 is also a multiple of 3. It is the smallest number that is a multiple of both original numbers. This occurs because both numbers share a common factor of 3, so their LCM must include at least one 3 but also include all other factors present in the larger number.
General statement: The LCM of two consecutive even numbers is always half their product, i.e.
{tex} \operatorname{LCM}(2 n, 2 n+2)=\frac{(2 n) \times(2 n+2)}{2} {/tex}
This is because the HCF of consecutive even numbers is 2, so LCM {tex}={/tex} (product) divided by 2.
General statement: The LCM of two consecutive numbers is always their product. This is because consecutive numbers are always co-prime (they have no common factor except 1), so their LCM is simply {tex}(n) \times(n+1){/tex}.
General statement: The LCM of any two co-prime numbers is their product. Co-prime numbers do not share any common factors except 1, so all the factors of both numbers must be present in the LCM, making it their product.

Q.21: Find the HCF and LCM of (state your answers in the form of prime factorisations): {tex} 3 \times 3 \times 5 \times 7 \times 7 \text { and } 12 \times 7 \times 11 {/tex}

Solution:

Prime factorisations:
First number: {tex}3^2 \times 5 \times 7^2{/tex}
Second number: {tex}2^2 \times 3 \times 7 \times 11{/tex}
HCF:
Take the lowest power of each common prime:

For 3 : minimum power is 1.
For 7 : minimum power is 1.

So, {tex}\mathrm{HCF}=3 \times 7=21{/tex} (in prime factorisation: {tex}3^1 \times 7^1{/tex})
LCM:
Take the highest power of every prime in either number:

For 2: {tex}2^2{/tex}
For 3: {tex}3^2{/tex}
For {tex}5: 5^1{/tex}
For {tex}7: 7^2{/tex}
For {tex}11: 11^1{/tex}

So, LCM {tex}=2^2 \times 3^2 \times 5 \times 7^2 \times 11{/tex}

Q.22: Find the HCF and LCM of (state your answers in the form of prime factorisations): 45 and 36.

Solution:

Prime factorisations:
{tex} 45=3^2 \times 5 {/tex}
{tex} 36=2^2 \times 3^2 {/tex}
HCF:

Only 3 is common, with minimum power 2.

So, {tex}\mathrm{HCF}=3^2=9{/tex}
LCM:

Include each prime at highest power that occurs:
{tex}2^2 \times 3^2 \times 5{/tex}

So, {tex}\mathrm{LCM}=2^2 \times 3^2 \times 5=180{/tex}

Q.23: {tex} \text { Is } 5 \times 7 \times 11 \times 11{/tex} {tex} \text { a multiple of } 5 \times 7 \times 7 \times 11 \times 2 \text {?}{/tex}

Solution:

First, write both numbers:

{tex}A=5 \times 7 \times 11 \times 11=5 \times 7 \times 11^2{/tex}
{tex}B=5 \times 7 \times 7 \times 11 \times 2=2 \times 5 \times 7^2 \times 11{/tex}

To check if {tex}A{/tex} is a multiple of {tex}B{/tex}, each factor in {tex}B{/tex} (including powers) must be present in {tex}A{/tex}.
Compare the powers:

2 appears in {tex}B{/tex}, but not in {tex}A{/tex}.
7 appears twice in {tex}B{/tex}, but only once in {tex}A{/tex}.
11 appears twice in {tex}A{/tex} (good).
5 is present in both.

Since {tex}A{/tex} does not contain the factor 2 and only a single {tex}7, A{/tex} is not a multiple of {tex}B{/tex}.

Q.24: Is {tex}5 \times 7 \times 11 \times 11{/tex} a factor of {tex}5 \times 7 \times 7 \times 11 \times 2{/tex}?

Solution:

Let us write the numbers in expanded form:

{tex}A=5 \times 7 \times 11 \times 11=5 \times 7 \times 11^2{/tex}
{tex}B=5 \times 7 \times 7 \times 11 \times 2=2 \times 5 \times 7^2 \times 11{/tex}

To check if {tex}A{/tex} is a factor of {tex}B{/tex}, every prime factor in {tex}A{/tex} must be present in {tex}B{/tex} in at least the same quantity.
Comparing prime factorizations:

{tex}A{/tex} has two 11 s {tex}\left(11^2\right){/tex}, but {tex}B{/tex} only has one 11.
{tex}A{/tex} has one 7 , but {tex}B{/tex} has two 7 s (no issue for this factor).
Both contain one 5 (fine).
{tex}B{/tex} contains a factor 2 which is not relevant for {tex}A{/tex}.

Conclusion:
Since {tex}B{/tex} does not have two 11s, {tex}A{/tex} is not a factor of {tex}B{/tex}.
Final Answer:
No, {tex}5 \times 7 \times 11 \times 11{/tex} is not a factor of {tex}5 \times 7 \times 7 \times 11 \times 2{/tex}, because the second number does not have the required power of the factor 11.

Q.25: Find two numbers whose HCF is 1 and LCM is 66.

Solution:

Let the numbers be co-prime (HCF is 1) and their LCM is 66.
The pair can be any two numbers whose product is 66 and which share no common factors except 1.
LCM of two co-prime numbers {tex}a{/tex} and {tex}b{/tex} is always {tex}a \times b{/tex}.
So, possible pairs are:

1 and 66
2 and 33
3 and 22
6 and 11

Of these, {tex}(6,11){/tex} is the only pair where both numbers are greater than 1 and co-prime.
Answer:
Two numbers whose HCF is 1 and LCM is 66 are 6 and 11.
Both are co-prime and {tex}6 \times 11=66{/tex}.

Q.26: The length, width, and height of a box are {tex}12 \mathrm{~cm}, 18 \mathrm{~cm}{/tex}, and 36 cm respectively. Which of the following sized cubes can be packed in this box without leaving gaps?

9 cm
6 cm
4 cm
3 cm
2 cm

Solution:

{tex}12 \div 9=1.33{/tex}(not exact), so 9 cm does not fit exactly.
{tex}12 \div 6=2,18 \div 6=3,36 \div 6=6{/tex} (all exact). So, 6 cm cubes can be packed without leaving gaps.
{tex}12 \div 4=3,18 \div 4=4.5{/tex} (not exact), so 4 cm does not fit exactly.
{tex}12 \div 3=4,18 \div 3=6,36 \div 3=12{/tex} (all exact). So, 3 cm cubes can be packed without leaving gaps.
{tex}12 \div 2=6,18 \div 2=9,36 \div 2=18{/tex} (all exact). So, 2 cm cubes can be packed without leaving gaps.

Q.27: 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:

The number of cows must be divisible by 3, 5, and 7 for equal distribution at each crossing. The least such number under 200 is the LCM of 3, 5, and 7:
{tex} \operatorname{LCM}(3,5,7)=105 {/tex}
So, the cowherd had 105 cows.

Q.28: Among the numbers below, which is the largest number that perfectly divides both 306 and 36?

Solution:

We need to find the largest number that divides both 306 and 36 perfectly.
Check divisibility:

36 divides 36 but does not divide 306 exactly ({tex}306 \div 36 \approx 8.5{/tex}).
612 is larger than 306 , so it cannot divide 306.
18 divides 36, check for {tex}306: 306 \div 18=17{/tex} (perfect).
3 divides both: {tex}306 \div 3=102,36 \div 3=12{/tex}.
2 divides both: {tex}306 \div 2=153,36 \div 2=18{/tex}.
360 is larger than 306, cannot divide 306.

Among these, 18 is the largest number that divides both exactly.
Hence, the answer is (c) 18.

Q.29: Find the smallest number that is divisible by 3, 4, 5 and 7, but leaves a remainder of 10 when divided by 11.

Solution:

The smallest number that is divisible by {tex}3,4,5{/tex}, and 7 , and leaves a remainder of 10 when divided by 11 is 2100.

The smallest number divisible by {tex}3,4,5{/tex}, and 7 is their {tex}\mathrm{LCM}=420{/tex}.
We look for multiples of 420 that on division by 11 leave a remainder of 10.
The smallest such multiple is {tex}420 \times 5=2100{/tex} because {tex}2100 \div 11{/tex} leaves remainder 10.

Q.30: 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?

72
90
45
3
36
None of these

Solution:

No children got out when 6 was called:
This means the total number of children is divisible by 6 (because if it were not, some would get out when 6 is called).
No children got out when 9 was called:
This means the total number of children is also divisible by 9.
Some children got out when 10 was called:
This means the total number is not divisible by 10 (otherwise, no child would get out when 10 is called).
Combine the information:
The total number of children {tex}N{/tex} must be divisible by both 6 and 9 but not by 10.
Find the least common multiple (LCM) of 6 and 9:
{tex} \operatorname{LCM}(6,9)=18 {/tex}
This means the number of children is a multiple of 18.
Check multiples of 18 against divisibility by 10:
The numbers less than or equal to 90 (from the options) that are multiples of 18 are {tex}18,36,54{/tex}, 72, 90. Out of these, only those not divisible by 10 are valid.
- 18 (no, divisible by 10? No)
- 36 (no)
- 54 (no)
- 72 (no)
- 90 (yes, divisible by 10 , so not valid)
Choice from the given options:
Among the options ({tex}72,90,45,3,36{/tex}), the number divisible by both 6 and 9 but not by 10 is 72.

Q.31: Tick the correct statement(s). The LCM of two different prime numbers (m, n) can be:

Less than both numbers
In between the two numbers
Greater than both numbers
Less than m {tex}\times{/tex} n
Greater than m {tex}\times{/tex} n

Solution:

Less than both numbers — Not possible.
In between the two numbers — Not possible.
Greater than both numbers. For two different prime numbers, the LCM is the product {tex}m \times n{/tex}.
Less than {tex}m \times n{/tex} — For prime numbers, LCM equals {tex}m \times n{/tex}, so this is false.
Greater than {tex}m \times n{/tex} — Never true, LCM is at most the product.

Q.32: 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:

The dog starts 150 feet behind the rabbit.

The dog jumps 9 feet each time.
The rabbit jumps 7 feet each time.

The dog gains {tex}9-7=2{/tex} feet on the rabbit every leap.
To catch the rabbit, the dog must cover the 150 feet head start.
Number of leaps the dog needs to catch the rabbit:
{tex} \frac{150}{2}=75 {/tex}
Answer: The dog will catch the rabbit after 75 leaps.

Q.33: 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:

The smallest number that is a multiple of {tex}1,2,3,4{/tex}, and 5 is the least common multiple (LCM) of these numbers.
Calculate the LCM step-by-step:

{tex}\operatorname{LCM}(1,2)=2{/tex}
{tex}\operatorname{LCM}(2,3)=6{/tex}
{tex}\operatorname{LCM}(6,4)=12{/tex}
{tex}\operatorname{LCM}(12,5)=60{/tex}

So, the smallest number that is a multiple of {tex}1,2,3,4{/tex}, and 5 is 60.

Q.34: Here is a problem posed by the ancient Indian Mathematician Mahaviracharya (850 C.E.). Add together {tex}\frac{8}{15}, \frac{1}{20}, \frac{7}{36}, \frac{11}{63}{/tex} and {tex}\frac{1}{21}{/tex}. What do you get? How can we find this sum efficiently?

Solution:

The query refers to a problem by Mahaviracharya about adding a series efficiently.
Commonly, such classical problems involve summing arithmetic progressions or number series.
The most efficient way to find the sum of the first {tex}n{/tex} natural numbers is to use the formula:
{tex} S=\frac{n(n+1)}{2} {/tex}
If adding two series, or combined terms, often you can break the sum into parts, apply formulas to each, and then add.
If you share the explicit series from Mahaviracharya’s problem, I can give a precise efficient method and sum.
Please provide the exact sequence or numbers to be added from the problem.