EXERCISE 3.7
1). Renu purchases two bags of fertilisers of weight 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.
Ans. weight of two bags of fertiliser = 75 kg and 69 kg
Maximum weight = HCF of both bags weight
75 = 3 x 5 x 5
69 = 3 x 23
HCF 3
∴3 kg is the maximum value of weight which can measure the weight of the fertiliser exact number of times.
2). Three boys step off together from the same spot. Their steps measure 63cm, 70cm and 77cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?
Ans. First boy steps measure = 63 cm
Second boy steps measure = 70 cm
Third boy steps measure = 77 cm
Minimum distance = LCM of all 3 measures
LCM of 63cm, 70cm, 77cm is
LCM = 2 x 3 x 3 x 5 x 7 x 11
=6930
∴Each should cover a minimum distance of 6930cm to complete the steps.
3). The length and breadth and height of a room are 825cm, 675cm and 450cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.
Ans. Length of the room = 825 cm
Breadth of the room = 675 cm
Height of the room = 450 cm
Longest tape required = HCF
825 = 3 x 5 x 5 x 11
675 = 3 x 3 x 3 x 5 x 5
450 = 2 x 3 x 3 x 5 x 5
HCF = 3 x 5 x 5
= 75 cm.
Hence the longest tape is 75cm which can measure the three dimension of the room.
4). Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.
Ans. LCM of 6, 8, 12 = smallest number
LCM of 6, 8, 12 = 2 x 2 x 2 x 3
= 24
We need to find the smallest 3-digit multiple of 24
24 x 4 = 96 and 24 x 5 = 120
Hence 120 is the smallest 3-digit number which is exactly divisible by 6, 8, 12.
5). Determine the largest number exactly divisible by 8, 10 and 12.
Ans. LCM of 8, 10, 12
= 2 x 2 x 2 x 3 x 5
= 120
We need to find the largest 3-digit multiple of 120.
120 x 8 = 960 and 120 x 9 = 1080
Hence 960 is the greatest 3-digit number exactly divisible by 8, 10 and 12.
6). The traffic light at three different road crossings changes after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7:am, at what time will they change simultaneously again?
Ans. LCM of 48, 72, 108 = time period after which these light change.
LCM = 2 x 2 x 2 x 2 x 3 x 3 x 3
=432
Hence the lights will change together after every 432 seconds. Which is equal to 7 min 12 seconds past 7:am.
7). Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find
the maximum capacity of the container that can measure the diesel of the three
containers exact number of times.
Ans. HCF of three tankers = maximum capacity of a container required
HCF of 403 = 13 x 31
443 = 2 x 7 x 31
465 = 3 x 5 x 31
HCF = 31
Hence a container of 31 litres can measure the diesel of the three containers exact number of times.
8). Find the least number which when divided by 6, 15 and 18 leaves remainder 5 in each case.
Ans. LCM of 6, 15, 18
LCM = 2 x 3 x 3 x 5 = 90
Required number = 90 + 5 = 95
Hence 95 is the required least number.
9). Find the smallest four digit number which is divisible by 18, 24 and 32.
Ans. LCM of 18, 24 and 32
LCM = 2 x 2 x 2 x 2 x 2 x 3 x 3
= 288
We need the smallest 4-digit number which has to be the multiple of 288
288 x 3 = 864, 288 x 4 = 1152
1152 is the smallest 4-digit number which is divisible by 18, 24 and 32.
9). Find the LCM of the following numbers in which one number is always a multiple of 3
9 and 4
Ans. LCM of 9, 4 = 2 x 2 x 3 x 3
= 36
12 and 5
Ans. LCM of 12, 5 = 2 x 2 x 3 x 5
= 60
6 and 5
Ans. LCM of 6, 5 = 2 x 3 x 5
=30
15 and 4
Ans. LCM of 15, 4 = 2 x 2 x 3 x 5
= 60
Yes in each case LCM is the product of the two given numbers.
10). Find the LCM of the following numbers in which one number is the factor of the other
a) 5, 20
Ans. LCM = 2 x 2 x 5 = 20
b) 6, 18
Ans. LCM = 2 x 3 x 3 = 18
c) 12, 48
Ans. LCM = 2 x 2 x 2 x 2 x 3 = 48
d) 9, 45
Ans. LCM = 3 x 3 x 5 =45
∴The LCM of the given numbers in each case is the larger than the two numbers.