[NOTE : for example in 2^5 , here ^5 is the power of 2 or 2 rest to 5. Apply this for every found question.]
- Q1
Karan has 180 blue marbles and 150 red marbles. He wants to pack them into packets containing equal number of marbles of the same colour. What is the maximum number of marbles that each packet can hold?
Answer:
Given: Number of blue marbles = 180
Number of red marbles = 150
Each packet contains the same number of marbles.
Now- Calculate the number of packets and number of marbles in each packet
Let the maximum number of marbles that each packet can hold be ‘x’.
For Blue marbles: Let the 180 blue marbles be divided into 'a' packets containing 'x' marbles each.
So, 180 = ax.
Because number of marbles and number of packets are integers, x and a are factors of 180.
For Red marbles: Let the 150 red marbles be divided into 'b' packets containing 'x' marbles each.
So, 150 = bx.
Because number of marbles and number of packets are integers, x and b are factors of 150.
∴ x is a factor of 180 and 150. i.e., x is a common factor of 180 and 150.
Because ‘x’ is the maximum number of marbles that each packet contains, x is the HCF of 180 and 150
- Apply Euclid’s Lemma to find HCF
Apply Euclid’s Lemma on 180 and 150
180 = 150 × 1 + 30
The remainder is not ‘0’
Now, Apply Euclid’s Lemma on 150 and 30
Note: If the remainder is not zero in any step, the divisor of that step becomes the new dividend for the next iteration and the remainder of that step becomes the divisor for the next iteration.
150 = 30 × 5 + 0
The remainder is ‘0’.
If the remainder becomes zero in a step, the process is stopped.
The divisor of the step in which the remainder becomes zero is the HCF.
The divisor of step 2 is 30 and it is the HCF.
So, the maximum number of marbles in each packet is 30
- Q2 : The Muscle Gym has bought 63 treadmills and 108 elliptical machines. The gym divides them into several identical sets of treadmills and elliptical machines for its branches located throughout the city, with no exercise equipment left over. What is the greatest number of branches the gym can have in the city?
Answer:
Given: Number of Treadmills = 63
Number of Elliptical machines = 108
Now - What will represent the greatest number of branches?
Let the greatest number of branches in the city be 'n'.
Let the number of treadmills sent to each branch be 't' and the number of ellipticals sent to each branch be 'e'.
63 treadmills were distributed to 'n' branches with each branch getting 't' treadmills.
i.e., nt = 63
108 elliptical machines were distributed to 'n' branches with each branch getting 'e' elliptical machines.
i.e., ne = 108.
So, n is a factor of 63 and 108.
'n' is the greatest number of branches the gym can have in the city. So, 'n' is the HCF of 63 and 108.
- Find HCF of 63 and 108
Prime factorise 63 and 108.
63 = 3^2 × 7
108 = 2^2 × 3^3
Step 2: List down common primes.
3 is the only prime common to both the numbers.
Step 3: HCF is the product of lowest power of common primes in the 2 numbers.
Lowest power of 3 between 63 and 108 is 3^2.
Therefore, HCF = 9
The greatest number of branches the gym can have in the city is 9.
- Q3: Anish goes fishing every 5th day and Balaji goes fishing every 7th day. If Anish and Balaji both went fishing today, how many days until they will go fishing on the same day again?
Answer-
Given: Anish goes fishing every 5th day.
Balaji goes fishing every 7th day.
- Find the next instance when Anish and Balaji go fishing together
Let today be day 0.
Anish and Balaji went fishing today.
Anish will go fishing again in 5 days and again in another 5 days from then.
i.e., Anish will go fishing on the following days from today: 5, 10, 15, 20, ...
Anish will therefore, go fishing on days that are multiples of 5.
Balaji will go fishing again in 7 days and again in another 7 days from then.
i.e., Balaji will go fishing on the following days from today: 7, 14, 21, ....
Balaji will therefore, go fishing on days that are multiples of 7.
Anish and Balaji will go fishing on the same day again if that day is a multiple of 5 and 7.
i.e., on days that are common multiples of 5 and 7.
The first such common multiple is the lowest common multiple or LCM of 5 and 7.
- Find LCM of 5 and 7
Prime factorise 5 and 7
Both are prime numbers.
Step 2: LCM is the product of the largest power of all primes
= 5 × 7 = 35
They will go fishing on the same day again 35 days from today.
- Q4: Find the LCM and HCF of the following :
(i) 2^5× 5^4 × 7^2 × 13^6 and 2^3 × 5^6 × 7 × 17^3
(ii) a^5 × b^2 × c^2 × d^5 and a^7 × b^3 × e × f^3 where a, b, c, d, e, and f are prime.
Answer-
(i)LCM and HCF of 2^5 × 5^4 × 7^2 × 13^6 and 2^3 × 5^6 × 7 × 17^3
Notice that the two numbers are already prime factorized. That makes life really easy.
Let us compute HCF first.
The common primes in the two numbers are 2, 5, and 7.
The lowest power of 2, 5, and 7 are respectively 23, 54, and 71
∴ HCF of 2^5 × 5^4 × 7^2 × 13^6 and 2^3 × 5^6 × 7 × 17^3 is 2^3 × 5^4 × 7
Let us compute LCM next.
The primes found in the two numbers are 2, 5, 7, 13, and 17.
The highest power of 2, 5, 7, 13, and 17 are 25, 56, 72, 136, and 173 respectively.
∴ LCM of 2^5 × 5^4 × 7^2 × 13^6 and 2^3 × 5^6 × 7 × 17^3 is 2^5 × 5^6 × 7^2 × 13^6 × 17^3
(ii) Find the LCM and HCF of a^5 × b^2 × c^2 × d^5 and a^7 × b^3 × e × f^3
Given: a, b, c, d, e, and f are prime numbers.
Let us compute HCF first.
The common primes in the two numbers are a and b.
The lowest power of a and b in the two numbers are a5 and b2 respectively.
∴ HCF of a^5 × b^2 × c^2 × d^5 and a^7 × b^3 × e × f^3 is a^5 × b^2
Let us compute LCM next.
The primes found in the two numbers are a, b, c, d, e, and f.
The highest power of a, b, c, d, e, and f are a^7, b^3, c^2, d^5, e^1, and f^3 respectively.
∴ LCM of a^5 × b^2 × c^2 × d^5 and a^7 × b^3 × e × f^3 is a^7 × b^3 × c^2 × d^5 × e × f^3
- Q5: Tamanna is arranging black marbles in groups of 13 and purple marbles in groups of 25. If she has the same number of black and purple marbles, what is the smallest number of marbles of each colour that she could have?
Answer-
Given: Black marbles is arranged in group of 13.
Purple marbles is arranged in group of 25.
Step 1 - What represents the least number of marbles?
Tamanna is arranging black marbles in groups of 13.
So, the number of black marbles she has will be a multiple of 13.
She is arranging purple marbles in groups of 25.
So, the number of purple marbles she has will be a multiple of 25.
She has the same number of black and purple marbles.
i.e., the number of marbles of each colour that she has is a multiple of 13 and a multiple of 25.
Or the number of marbles of each colour that she has is a common multiple of 13 and 25.
Because we are finding the smallest number of marbles of each colour, we have to compute the lowest common multiple or the LCM of 13 and 25.
Step 2 - Find LCM of 13 and 25
Step 1: Prime factorise 13 and 25
13 is a prime number.
25 = 5^2
Step 2: List down all primes in the two numbers
13 and 5 are prime numbers in the two numbers.
Step 3: LCM is the product of the greatest power of all primes in the 2 numbers.
LCM = 13 × 5^2 = 13 × 25 = 325
The smallest number of marbles of each colour that she could have is 325.
- Q6: Katya has 49 paintings and 35 medals. She wants to display them in groups throughout her house, each with the same combination of paintings and medals, with none left over. What is the greatest number of groups Katya can display?
Answer-
Given: Total number of paintings = 49.
Total number of medals = 35.
Calculate the Maximum Number of Groups
Let the greatest number of groups that Katya can display be 'n'.
Let the number of paintings in each group be 'p' and the number of medals in each group be 'm'.
49 paintings have been divided into 'n' groups each containing 'p' paintings.
i.e., np = 49
35 medals have been divided into 'n' groups each containing 'm' medals.
i.e., nm = 35
So, 'n' is a factor of 49 and 35.
Because n is the greatest number of groups, n is the HCF of 49 and 35.
- Find HCF of 49 and 35
Prime factorise 49 and 35.
49 = 7 × 7 = 7^2
35 = 7 × 5
List down common primes.
The only common prime between the 49 and 35 is 7.
HCF is the product of the lowest power of common primes.
HCF = 7
The greatest number of groups in which Katya can display paintings and medals is 7.
- Q7: What is the largest number that divides 967 and 1767 leaving remainders of 71 and 103 respectively?
Answer -
Given: The two numbers are 967 and 1767.
Let x be the largest number that divides both 967 and 1767 and leaves remainders of 71 and 103 respectively.
- Decoding the information given in the question
Remainder of
967/x
967/x is 71.
So, 967 – 71 = 896 is divisible by x.
Remainder of
1767/x is 103.
So, 1767 - 103 = 1664 is divisible by x.
So, x divides both 896 and 1664.
In other words, x is a factor of both 896 and 1664. i.e., x is a common factor of 896 and 1664.
- Apply Euclid's Lemma to find the HCF
Apply Euclid's Lemma on 1664 with 896 as divisor.
1664 = 896 × 1 + 768
The remainder is not zero.
Step 2: Apply Euclid's Lemma on 896 with 768 as divisor.
896 = 768 × 1 + 128
The remainder is not zero.
Step 3: Apply Euclid's Lemma on 768 with 128 as divisor.
768 = 128 × 6 + 0
The remainder is zero.
∴ The divisor of this step, 128 is the HCF of 896 and 1664.
128 is the largest number that divides 967 and 1767 leaving remainders 71 and 103 respectively.
- Q8: Find the largest number that will divide 382 and 710 and leaves a remainder 13 in each case.
Answer-
Given: The two numbers are 382 and 710
Let 'x' be the largest number that divides 382 and 710, leaving 13 as remainder in both cases.
- Decoding the given information
Remainder of
382/x is 13
∴ 382 – 13 = 369 is divisible by x.
Similarly, remainder of
710/x is 13
∴ 710 – 13 = 697 is divisible by x.
So, x divides both 369 and 697.
Or, x is a factor common of both 369 and 697.
Because x is the largest such number, x is the HCF of 369 and 697.
- Use Euclid's Division Lemma to find HCF of 369 and 697
Step 1: Apply Euclid's Lemma on 697 with 369 as divisor
697 = 369 × 1 + 328
The remainder is not zero.
: Apply Euclid's Division Lemma on 369 with 328 as divisor
369 = 328 × 1 + 41
The remainder is not zero.
: Apply Euclid's Lemma on 328 with 41 as divisor
328 = 41 × 8 + 0
The remainder is zero.
∴ The divisor of this step, 41 is the HCF of 369 and 697
The largest number that divides 382 and 710 leaving remainder 13 in each case is 41.
- Q9: What is the largest number that divides 170, 220, and 420 leaving remainder 8, 4 and 15 respectively?
Answer-
Given Data: The three numbers are 170, 220, and 420.
- Decoding Given Information
Let x be the largest number that divides 170, 220, and 420 leaving reaminders of 8, 4, and 15 respectively.
Remainder of
170/x is 8.
=> 170 – 8 = 162 is divisible by x.
Remainder of
220/x is 4.
=> 220 – 4 = 216 is divisible by x.
Remainder of
420/x is 15.
=> 420 – 15 = 405 is divisible by x.
So, we can infer that x divides 162, 216, and 405 without leaving any remainder.
Or, x is a factor of 162, 216, and 405. i.e., x is a common factor of 162, 216, and 405.
Because x is the largest such number, x is the HCF OF 162, 216 and 405.
- Use Euclid's Division Lemma to find the HCF of the 3 Numbers
Apply Euclid's Division Lemma on 216 with 162 as divisor.
216 = 162 × 1 + 54
The remainder is not zero.
Step 2: Apply Euclid's Division Lemma on 162 with 54 as divisor.
162 = 54 × 3 + 0.
The remainder is 0.
∴ The divisor of this step, 54 is the HCF of 162 and 216.
Find HCF of 54 and 405
Apply Euclid's Division Lemma on 405 with 54 as divisor.
405 = 54 × 7 + 27
The remainder is not zero.
: Apply Euclid's Division Lemma on 54 with 27 as divisor.
54 = 27 × 2 + 0
The remainder is zero.
∴ The divisor of this step, 27 is the HCF of 54 and 405.
Hence, 27 is the HCF of 162, 216, and 405.
27 is the largest number that divides 170, 220, and 420 leaving remainders of 8, 4, and 15 respectively.
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