GSEB Class 6 Maths Solutions Chapter 3 Playing With Numbers Exercise 3.7

 

Question 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.
Answer: The greatest value of weight that can exactly measure the given weights will be their HCF.
Now, factors of 75 are: 1, 3, 5, 15, 25, and 75.
Factors of 69 are: 1, 3, 23, and 69.
Common factors are: 1 and 3.
The HCF of 75 and 69 is 3.
Hence, the needed maximum weight value is 3 kg.
In simple words: To find the largest weight that can measure both bags perfectly, we calculate the HCF of their weights. The common factors are 1 and 3, so the highest common factor is 3. This means 3 kg is the biggest weight that works.

Exam Tip: When asked to find the "maximum value" or "greatest measure" that can exactly divide given numbers, always think of calculating the Highest Common Factor (HCF).

 

Question 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?
Answer: The smallest distance each boy should walk must be the least common multiple of their step measurements.
To find the LCM of 63, 70, and 77, we have:

The LCM of 63, 70, and 77 is \( 2 \times 3 \times 3 \times 5 \times 7 \times 11 = 6,930 \).
So, the needed minimum distance is 6,930 cm.
In simple words: We need to find the smallest distance that is a multiple of all three step lengths. We do this by calculating the LCM of 63, 70, and 77, which comes out to be 6,930 cm.

Exam Tip: For problems asking for the "minimum distance" or "when they will meet again" where quantities start at different intervals, calculate the Least Common Multiple (LCM).

 

Question 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.
Answer: The longest tape will be the H.C.F of 825, 675, and 450.
We have:

Prime factors are:
\( 825 = 3 \times 5 \times 5 \times 11 \)
\( 675 = 3 \times 3 \times 3 \times 5 \times 5 \)
\( 450 = 2 \times 3 \times 3 \times 5 \times 5 \)
The HCF of 825, 675, and 450 is \( 3 \times 5 \times 5 = 75 \).
So, the needed length for the longest tape is 75 cm.
In simple words: To find the longest tape that can measure all three dimensions exactly, we need to calculate the HCF of the given lengths. After finding the prime factors for 825, 675, and 450, the common factors give us an HCF of 75, which is the tape's length.

Exam Tip: Remember that "longest tape" or "maximum capacity" problems usually require finding the Highest Common Factor (HCF).

 

Question 4. Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.
Answer: The LCM of 6, 8, and 12 is exactly divisible by them.

The LCM of 6, 8, and 12 is \( 2 \times 2 \times 2 \times 3 = 24 \).
All other multiples of 24 will also be divisible by 6, 8, and 12. But we need the smallest multiple of 24 that has 3-digits.
The smallest 3-digit number is 100.
Now, we divide 100 by 24:
\( 100 \div 24 = 4 \) with a remainder of \( 4 \).
\( (24 \times 4) = 96 \)
So, the multiple of 24 just above 100 is \( (100 - 4) + 24 \) or 120.
Therefore, the needed number is 120.
In simple words: First, find the LCM of 6, 8, and 12, which is 24. We need the smallest 3-digit number that 24 divides evenly. Since 100 is the smallest 3-digit number, we divide 100 by 24. We get 4 with a remainder of 4. To get the next multiple of 24, we add \( 24 - 4 = 20 \) to 100, which gives us 120.

Exam Tip: To find the smallest number divisible by a set of numbers, first find their LCM. Then, if a specific digit count (e.g., 3-digit) is required, adjust the LCM by finding its smallest multiple that meets that condition.

 

Question 5. Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.
Answer: To find the LCM of 8, 10, and 12, we have:

The LCM is \( 2 \times 2 \times 2 \times 3 \times 5 = 120 \).
Since all other multiples of 120 will also be divisible by 8, 10, and 12, we need the greatest 3-digit number that is a multiple of 120.
The largest 3-digit number is 999.
We divide 999 by 120:
\( 999 \div 120 = 8 \) with a remainder of \( 39 \).
\( (120 \times 8) = 960 \)
So, \( 999 - 39 \), which is 960, is a multiple of 120.
Thus, the needed number is 960.
In simple words: First, find the LCM of 8, 10, and 12, which is 120. Then, find the largest 3-digit number that 120 can divide evenly. We divide 999 (the largest 3-digit number) by 120, which gives us 8 with a remainder of 39. To find the largest multiple, we subtract the remainder from 999, resulting in 960.

Exam Tip: When finding the greatest number within a range divisible by multiple numbers, calculate the LCM, then divide the largest number in the range by the LCM and subtract the remainder from the range's largest number.

 

Question 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?
Answer: The needed time will be the LCM of 48, 72, and 108.
To find their LCM, we have:

The LCM is \( 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 432 \).
The needed time is 432 seconds, which means 7 minutes and 12 seconds.
\( [432 \text{ seconds} = 7 \text{ minutes } 12 \text{ seconds}] \)
Therefore, the traffic lights will change simultaneously again at 7 minutes and 12 seconds past 7 a.m.
In simple words: To find when the lights will change together again, we need the LCM of 48, 72, and 108 seconds. The LCM is 432 seconds. We convert 432 seconds into minutes and seconds, which is 7 minutes and 12 seconds. So, the lights will change together at 7:07:12 a.m.

Exam Tip: Problems involving cycles repeating at different intervals (like traffic lights or bells ringing) always require finding the Least Common Multiple (LCM) to determine when they will coincide again.

 

Question 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.
Answer: The maximum capacity of the needed measure will be equal to the HCF of 403, 434, and 465 litres.

Factors of 403 are 1, 13, 31, and 403.
Factors of 434 are 1, 2, 7, 31, and 434.
Factors of 465 are 1, 3, 5, 31, and 465.
Their common factors are: 1 and 31.
The HCF is 31.
Thus, the maximum capacity of the needed container is 31 litres.
In simple words: To find the largest container that can measure the diesel from all three tankers perfectly, we need to find the HCF of 403, 434, and 465. By breaking down each number into its factors, we find that 31 is the highest common factor. So, a 31-litre container is the answer.

Exam Tip: Always look for keywords like "maximum capacity," "largest measure," or "greatest number" which indicate you need to find the HCF (Highest Common Factor).

 

Question 8. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.
Answer: We first find the LCM of 6, 15, and 18.

The LCM of 6, 15, and 18 is \( 2 \times 3 \times 3 \times 5 = 90 \).
Since 90 is the smallest number that can be divided exactly by 6, 15, and 18. To get a remainder of 5, the least number will be \( 90 + 5 \), which is 95.
Therefore, the needed number is 95.
In simple words: First, find the LCM of 6, 15, and 18, which is 90. This means 90 is the smallest number that divides evenly by all three. Since we want a remainder of 5 each time, we just add 5 to the LCM, making the answer 95.

Exam Tip: If a problem asks for a number that leaves a specific remainder when divided by several numbers, first find the LCM of those divisors, then add the remainder to the LCM.

 

Question 9. Find the smallest 4-digit number which is divisible by 18, 24 and 32.
Answer: We first find the LCM of 18, 24, and 32.

The LCM is \( 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 288 \).
Since 288 is the smallest number that is divisible by 18, 24, and 32, but it is not a 4-digit number.
The smallest 4-digit number is 1000.
Now, we divide 1000 by 288:
\( 1000 \div 288 = 3 \) with a remainder of \( 136 \).
\( (288 \times 3) = 864 \)
The multiple of 288 that is just above 1000 is \( 1000 - 136 + 288 \), which equals 1152.
Therefore, the needed number is 1152.
In simple words: First, calculate the LCM of 18, 24, and 32, which is 288. We need the smallest 4-digit number divisible by 288. The smallest 4-digit number is 1000. When we divide 1000 by 288, we get 3 with a remainder of 136. To find the next multiple of 288 (the first 4-digit one), we subtract the remainder from 1000 and then add 288, giving us 1152.

Exam Tip: When finding the smallest number with a certain number of digits divisible by multiple numbers, first find their LCM. Then, determine how many times the LCM fits into the smallest number of that digit count and adjust to find the next complete multiple.

 

Question 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 in the obtained LCMs. Is LCM the product of two numbers in each case?
Answer:
(a) For 9 and 4:

The product of 9 and 4 is \( 9 \times 4 = 36 \).
The LCM of 9 and 4 is \( 2 \times 2 \times 3 \times 3 = 36 \).
Since the product of 9 and 4 equals their LCM, the LCM is the product of 9 and 4.

(b) For 12 and 5:

The product of 12 and 5 is \( 12 \times 5 = 60 \).
The LCM of 12 and 5 is \( 2 \times 2 \times 3 \times 5 = 60 \).
Since the product of 12 and 5 equals their LCM, the LCM is the product of 12 and 5.

(c) For 6 and 5:

The product of 6 and 5 is \( 6 \times 5 = 30 \).
The LCM of 6 and 5 is \( 2 \times 3 \times 5 = 30 \).
Since the product of 6 and 5 equals their LCM, the LCM is the product of 6 and 5.

(d) For 15 and 4:

The product of 15 and 4 is \( 15 \times 4 = 60 \).
The LCM of 15 and 4 is \( 2 \times 2 \times 3 \times 5 = 60 \).
Since the product of 15 and 4 equals their LCM, the LCM is the product of 15 and 4.

We observe that in all these cases, the LCM of the two numbers is equal to the product of the two numbers. This happens because the pairs of numbers are coprime (they share no common factors other than 1).
In simple words: For each pair of numbers, we found their LCM. We noticed that in every case, the LCM was the same as multiplying the two numbers together. This pattern occurs when the numbers don't share any common factors except 1.

Exam Tip: Remember a key property: if two numbers are coprime (their HCF is 1), then their LCM is always equal to their product.

 

Question 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 in the results obtained?
Answer:
(a) For 5 and 20:

The LCM of 5 and 20 is \( 2 \times 2 \times 5 = 20 \).

(b) For 6 and 18:

The LCM of 6 and 18 is \( 2 \times 3 \times 3 = 18 \).

(c) For 12 and 48:

The LCM of 12 and 48 is \( 2 \times 2 \times 2 \times 2 \times 3 = 48 \).

(d) For 9 and 45:

The LCM of 9 and 45 is \( 3 \times 3 \times 5 = 45 \).

We observe that when one number is a factor of the other, the LCM of the two numbers is always the greater number.
In simple words: When one number can divide the other number evenly, the LCM of these two numbers will simply be the larger of the two numbers. For instance, if 5 is a factor of 20, then the LCM is 20.

Exam Tip: A useful shortcut for LCM: if the smaller number is a factor of the larger number, then the LCM of the two numbers is always the larger number itself.