Exercise 3.7
1. Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.
We need the largest weight that divides both 75 and 69 exactly → this is the HCF.
Prime factorisation:
\(75 = 3 \times 5 \times 5\)
\(69 = 3 \times 23\)
Common factor: 3
\(75 = 3 \times 5 \times 5\)
\(69 = 3 \times 23\)
Common factor: 3
Maximum weight = 3 kg
2. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?
We need the smallest distance that is a multiple of 63, 70, and 77 → this is the LCM.
LCM = \(2 \times 3 \times 3 \times 5 \times 7 \times 11 = 6930\)
Minimum distance walked = 6930 cm
3. The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.
We need the longest tape that divides all three dimensions → HCF of 825, 675, 450.
\(825 = 3 \times 5^2 \times 11\)
\(675 = 3^3 \times 5^2\)
\(450 = 2 \times 3^2 \times 5^2\)
Common factors: \(3 \times 5^2 = 75\)
\(675 = 3^3 \times 5^2\)
\(450 = 2 \times 3^2 \times 5^2\)
Common factors: \(3 \times 5^2 = 75\)
Longest tape = 75 cm
4. Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.
To find the number we need to first find LCM of 6, 8, 12.
LCM = \(2 \times 2 \times 2 \times 3 = 24\)
So the smallest 3-digit number which is a multiple of 24 is:
24 × ? = smallest 3-digit number
24 × 5 = 120 (3-digit number)
So the smallest 3-digit number which is a multiple of 24 is:
24 × ? = smallest 3-digit number
24 × 5 = 120 (3-digit number)
Smallest 3-digit number = 120
5. Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.
Find LCM of 8, 10, 12.
LCM = \(2 \times 2 \times 2 \times 3 \times 5 = 120\)
120 is the smallest 3-digit number exactly divisible by 8, 10 and 12.
The greatest 3-digit number is 999.
Divide 999 by 120
\(999 \div 120 ≈ 8.325\)
So,
we will multiply 120 by 8 to get the greatest 3-digit number
\(8 \times 120 = 960\)
120 is the smallest 3-digit number exactly divisible by 8, 10 and 12.
The greatest 3-digit number is 999.
Divide 999 by 120
\(999 \div 120 ≈ 8.325\)
So,
we will multiply 120 by 8 to get the greatest 3-digit number
\(8 \times 120 = 960\)
Greatest 3-digit number = 960
6. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again?
Find LCM of 48, 72, 108 → this gives the time interval (in seconds) after which they sync again.
LCM = \(2^4 \times 3^3 = 432\) seconds
Convert 432 seconds to minutes and seconds:
\(432 \div 60 = 7\) minutes and \(12\) seconds
So, they will change again after 7 minutes and 12 seconds.
If they changed at 7:00:00 a.m. initially.Convert 432 seconds to minutes and seconds:
So, they will change again after 7 minutes and 12 seconds.
Add 7 minutes and 12 seconds:
\(7:00:00 + 0:07:12 = 7:07:12\)
They will change again at 7:07:12 a.m.
7. Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.
We need the **largest container** that divides all three → HCF of 403, 434, 465.
\(403 = 13 \times 31\)
\(434 = 2 \times 7 \times 31\)
\(465 = 3 \times 5 \times 31\)
Common factor (HCF)= 31
\(434 = 2 \times 7 \times 31\)
\(465 = 3 \times 5 \times 31\)
Common factor (HCF)= 31
Maximum capacity = 31 litres
8. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.
Let the number be \(N\).
Then \(N - 5\) is divisible by ( 6, 15, and 18.)
So, \(N - 5 = \text{LCM of } (6,15,18)\)
LCM = \(2 \times 3^2 \times 5 = 90\)
So, \(N - 5 = 90\)
\(N = 90 + 5 = 95\)
So, \(N - 5 = 90\)
\(N = 90 + 5 = 95\)
Least number = 95
9. Find the smallest 4-digit number which is divisible by 18, 24 and 32.
Find LCM of 18, 24, 32.
LCM = \(2^5 \times 3^2 = 32 \times 9 = 288\)
∴ 288 is the smallest number divisible by 18, 24 and 32.
We need the smallest 4-digit number which is a multiple of 288.
Smallest 4-digit number = 1000
Divide 1000 by 288:
\(1000 \div 288 ≈ 3.47\)
If we multiply (\(288 \times 3 = 864\)) we get a 3-digit number.
So, we will multiply (\(288 \times 4 = 1152\)) to get the smallest 4-digit number.
∴ 288 is the smallest number divisible by 18, 24 and 32.
We need the smallest 4-digit number which is a multiple of 288.
Smallest 4-digit number = 1000
Divide 1000 by 288:
\(1000 \div 288 ≈ 3.47\)
If we multiply (\(288 \times 3 = 864\)) we get a 3-digit number.
So, we will multiply (\(288 \times 4 = 1152\)) to get the smallest 4-digit number.
∴ Smallest 4-digit number = 1152
10. Find the LCM of the following numbers: (a) 9 and 4 (b) 12 and 5 (c) 6 and 5 (d) 15 and 4. Observe a common property. Is LCM the product of two numbers in each case?
(a): LCM of 9 and 4
Yes LCM = product of 9 and 4 = 36
(b): LCM of 12 and 5
Yes LCM = product of 12 and 5 = 60
(c): LCM of 6 and 5
Yes LCM = product of 6 and 5 = 30
(d): LCM of 15 and 4
Yes LCM = product of 15 and 4 = 60
Observation: In all cases, the two numbers are Co-Prime (no common prime factors).
So, LCM = product of the numbers.
11. Find the LCM of the following numbers in which one number is the factor of the other:
(a) 5, 20
(b) 6, 18
(c) 12, 48
(d) 9, 45.
What do you observe?
(a) 5, 20
(b) 6, 18
(c) 12, 48
(d) 9, 45.
What do you observe?
(a): LCM of 5 and 20
(b): LCM of 6 and 18
(c): LCM of 12 and 48
(d): LCM of 9 and 45
Observation: In each case, Larger number is the LCM.