Playing With Numbers
EXERCISE 3.7
CBSE Class 6 Maths
Q1: 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 maximum weight = HCF of 75 and 69.
Factors of 75 = 3 ✕ 5 ✕ 5
Factors of 69 = 3 ✕ 69
HCF = 3
∴ the required weight is 3 kg.
Q2: 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:
To find minimum distance = LCM of 63, 70 and 77
7 │ 63, 70, 77
───┼────────────
9 │ 9, 10, 11
───┼────────────
10 │ 1, 10, 11
───┼────────────
11 │ 1, 1, 11
───┼────────────
│ 1, 1, 1
L.C.M. of 63, 70 and 77 = 7 ✕ 9 ✕ 10 ✕ 11 = 6930 cm.
∴ the minimum distance is 6930 cm.
Q3: 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 measurement of longest tape = HCF of 825 cm, 675 cm and 450 cm.
Factors of 825 = 3 ✕ 5 ✕ 5 ✕ 11
Factors of 675 = 3 ✕ 5 ✕ 5 ✕ 3 ✕ 3
Factors of 450 = 2 ✕ 3 ✕ 3 ✕ 5 ✕ 5
HCF = 3 ✕ 5 ✕ 5 = 75 cm
∴ the longest tape is 75 cm.
Q4: Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.
Answer:
LCM of 6, 8 and 12 is:
2 │ 6, 8, 12
───┼────────────
2 │ 3, 4, 6
───┼────────────
2 │ 3, 2, 3
───┼────────────
3 │ 3, 1, 3
───┼────────────
│ 1, 1, 1
LCM = 2 ✕ 2 ✕ 2 ✕ 3 = 24
The smallest 3-digit number = 100
To find the number, we have to divide 100 by 24
100 = 24 x 4 + 4
∴ the required number = 24 ✕ 5 = 120.
Q5: Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.
Answer:
LCM of 8, 10, 12 =
2 │ 8, 10, 12
───┼────────────
2 │ 4, 5, 6
───┼────────────
2 │ 2, 5, 3
───┼────────────
3 │ 1, 5, 3
───┼────────────
5 │ 1, 5, 1
───┼────────────
│ 1, 1, 1
LCM = 2 ✕ 2 ✕ 2 ✕ 3 ✕ 5 = 120
The largest three digit number = 999
120 ) 999 (8
-960
──────
39
Thus the required number = 999 – 39 = 960
Q6: 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:
LCM of 48, 72, 108 =
2 │ 48, 72, 108
───┼────────────
2 │ 24, 36, 54
───┼────────────
2 │ 12, 18, 27
───┼────────────
2 │ 6, 9, 27
───┼────────────
3 │ 3, 9, 27
───┼────────────
3 │ 1, 3, 9
───┼────────────
3 │ 1, 1, 3
───┼────────────
│ 1, 1, 1
LCM = 2 ✕ 2 ✕ 2 ✕ 2 ✕ 3 ✕ 3 ✕ 3 = 432 seconds
The lights will change simultaneously after every 432 seconds.
432 second = 7 minutes 12 seconds
∴ Time = 7 am + 7 minutes 12 seconds = 7:07:12 a.m.
Q7: 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:
Maximum capacity = HCF (403, 434, 46)
Factors of 403 = 13 ✕ 31
Factors of 434 = 2 ✕ 7 ✕ 31
Factors of 465 = 3 ✕ 5 ✕ 31
HCF = 31
∴ 31 litres of container is required to measure the quantity.
Q8: Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.
Answer:
LCM of 6, 15 and 18 =
2 │ 6, 15, 18
───┼────────────
3 │ 3, 5, 9
───┼────────────
3 │ 1, 5, 3
───┼────────────
5 │ 1, 5, 1
───┼────────────
│ 1, 1, 1
LCM = 2 ✕ 3 ✕ 3 ✕ 5 = 90
∴ the required number = 90 + 5 = 95
Q9: Find the smallest 4-digit number which is divisible by 18, 24 and 32.
Answer:
LCM of 18, 24 and 32 =
2 │ 18, 24, 32
───┼────────────
2 │ 9, 6, 16
───┼────────────
2 │ 9, 3, 8
───┼────────────
2 │ 9, 3, 4
───┼────────────
2 │ 9, 3, 2
───┼────────────
3 │ 9, 3, 1
───┼────────────
3 │ 3, 1, 1
───┼────────────
│ 1, 1, 1
= 2 ✕ 2 ✕ 2 ✕ 2 ✕ 2 ✕ 3 ✕ 3 = 288
The smallest four-digit number = 1000
────────
288 ) 1000 (3
-864
──────
136
∴ the required number = 1000 + (288 – 136) = 1152.
or = 288 ✕ 4 = 1152
Q10: 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) LCM of 9 and 4
2 │ 9, 4
───┼────────────
2 │ 9, 2
───┼────────────
3 │ 9, 1
───┼────────────
3 │ 3, 1
───┼────────────
│ 1, 1
= 2 ✕ 2 ✕ 3 ✕ 3 = 36
(b) 12 and 5
2 │ 12, 5
───┼────────────
2 │ 6, 5
───┼────────────
3 │ 3, 5
───┼────────────
5 │ 1, 5
───┼────────────
│ 1, 1
= 2 ✕ 2 ✕ 3 ✕ 5 = 60
(c) 6 and 5
2 │ 6, 5
───┼────────────
3 │ 3, 5
───┼────────────
5 │ 1, 5
───┼────────────
│ 1, 1
= 2 ✕ 3 ✕ 5 = 30
(d) 15 and 4
2 │ 15, 4
───┼────────────
2 │ 15, 2
───┼────────────
3 │ 15, 1
───┼────────────
5 │ 5, 1
───┼────────────
│ 1, 1
= 2 ✕ 2 ✕ 3 ✕ 5 = 60
In all these cases we find, Yes, the LCM is equal to the product of two numbers in each case.
Q11: 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) 5, 20
2 │ 5, 20
───┼────────────
2 │ 5, 10
───┼────────────
5 │ 5, 5
───┼────────────
│ 1, 1
= 2 ✕ 2 ✕ 5 = 20
(b) 6, 18
2 │ 6, 18
───┼────────────
3 │ 3, 9
───┼────────────
3 │ 1, 3
───┼────────────
│ 1, 1
= 2 ✕ 3 ✕ 3 = 18
(c) 12, 48
2 │ 12, 48
───┼────────────
2 │ 6, 24
───┼────────────
2 │ 3, 12
───┼────────────
2 │ 3, 6
───┼────────────
3 │ 3, 3
───┼────────────
│ 1, 1
= 2 ✕ 2 ✕ 2 ✕ 2 ✕ 3 = 48
(d) 9, 45
3 │ 9, 45
───┼────────────
3 │ 3, 15
───┼────────────
5 │ 1, 5
───┼────────────
│ 1, 1
= 3 ✕ 3 ✕ 5 = 45
From the above cases, we find that if the smaller number is a factor of larger number, the LCM of these two numbers is equal to the larger number.
No comments:
Post a Comment
We love to hear your thoughts about this post!
Note: only a member of this blog may post a comment.