ID : eu-6-LCM-and-HCF [1]
Grade 6
LCM and HCF
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Answer t he quest ions
(1)
What is the largest number that divides 1112 and 272 leaving remainder 8?
(2)
Which is the biggest number that divides 247, 184 and 529 leaving remainder 7, 8 and 9
respectively.
(3)
T wo tankers contain 497 litres and 203 litres of petrol respectively. What is the capacity of the
largest measuring container which can measure the petrol of either tanker exactly?
(4) Determine the two numbers nearest to 60000 which are exactly divisible by 7, 9, 8, 3, 2, 5, 6.
(5)
T he length, breadth and height of a room are 11 m 6 cm, 9 m 10 cm and 7 m 56 cm respectively.
What is the length of the longest rod which can measure the dimensions of the room exactly?
(6) Find the LCM (least common multiple) of the f ollowing numbers
52, 84, 48
Choose correct answer(s) f rom given choice
(7) A lighthouse has two lights-one that f lashes every 3 minutes, and another that f lashes every
1
1
minutes. Suppose the lights f lash together at noon. What is the f irst time af ter 3 pm that
2
they will f lash together again?
(8)
a. 3:04
b. 3:02
c. 3:03
d. 3:00
If a number is divisible by 42 and 24, it will be necessarily divisible by
a. 166
b. 174
c. 175
d. 168
(9) Viktoria and Veronika are f riends and f ootball coach too. Viktoria goes to Amity school every 4
days and Veronika goes to Amity every 6 days to deliver coaching classes. If they both delivered
the coaching sessions today, how many days in the next 48 days they both would take the
session on the same day ?
a. 6
b. 3
c. 4
d. 5
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ID : eu-6-LCM-and-HCF [2]
(10) Alina has two sheets of cloth. One sheet is 276 cms wide and the other sheet is 115 cms wide.
She wants to divide the sheets into strips of equal width that are as wide as possible without
wasting any cloth. How wide should she cut the strips?
a. 29 cms
b. 27 cms
c. 23 cms
d. 22 cms
Fill in t he blanks
(11) Find the Least Common Multiple (LCM) of f ollowing:
A)
LCM of 14, 147, 70 is
B)
.
D)
LCM of 72, 324, 27 is
LCM of 18, 70, 18, 56, 10 is
.
E)
.
LCM of 48, 108, 48, 12
.
(12)
C)
LCM of 4, 36, 4, 24 is
is
.
F)
LCM of 36, 450, 90, 25, 27
is
T he least common multiple (LCM) of 14 and 18 is
.
.
(13) 3 bells ring at intervals of 18, 144, 48 seconds respectively. If the bells ring together at 5
O'clock, they will ring together again at time
ring together in 132 minutes will be
:
:
. T he number of times they will
.
(14) Lukas saves €27.35 every day. T he minimum number of days in which he will be able to save an
exact amount of Rupees is
(15)
.
T he two numbers nearest to 8000 which are exactly divisible by 5, 2, 3, 9, 8, 6 are
and
.
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ID : eu-6-LCM-and-HCF [3]
Answers
(1)
24
Step 1
We have to f ind the largest number that divides 1112 and 272 leaving remainder 8.
Step 2
In other words, we have to f ind the largest number that divides (1112-8) and (272-8) leaving
no remainder.
Such number is the Highest Common Factor (HCF) of :
1104 [i.e., 1112 - 8] and 264 [i.e., 272 - 8].
Step 3
Let us now f ind the HCF of 1104 and 264.
All prime f actors of 1104:
2
| 1104
2 is a factor of 1104
2
| 552
2 is a factor of 552
2
| 276
2 is a factor of 276
2
| 138
2 is a factor of 138
3
| 69
3 is a factor of 69
23 | 23
23 is a factor of 23
|1
T heref ore,
1104 = 2 × 2 × 2 × 2 × 3 × 23.
Step 4
All prime f actors of 264:
2
| 264
2 is a factor of 264
2
| 132
2 is a factor of 132
2
| 66
2 is a factor of 66
3
| 33
3 is a factor of 33
11 | 11
11 is a factor of 11
|1
T heref ore, 264 = 2 × 2 × 2 × 3 × 11.
Step 5
T he HCF of 1104 and 264 is = 2 × 2 × 2 × 3 = 24.
Step 6
T hus, the largest number which divides 1112 and 272 leaving remainder 8 is 24.
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ID : eu-6-LCM-and-HCF [4]
(2)
8
Step 1
T he biggest number that divides 247, 184 and 529 leaving remainder 7, 8 and 9 respectively
is the same number that divides the numbers (247-7), (184-8) and (529-9) leaving no
remainder.
Step 2
In other words, we need to f ind the HCF of the f ollowing three numbers:
240 [Simplify 247-7],
, 176 [Simplify 184-8],
, 520 [Simplify 529-9].
Step 3
All prime f actors of 240:
2 | 240
2 is a factor of 240
2 | 120
2 is a factor of 120
2 | 60
2 is a factor of 60
2 | 30
2 is a factor of 30
3 | 15
3 is a factor of 15
5 |5
5 is a factor of 5
|1
T hus,
240 = 2 × 2 × 2 × 2 × 3 × 5.
Step 4
All prime f actors of 176:
2
| 176
2 is a factor of 176
2
| 88
2 is a factor of 88
2
| 44
2 is a factor of 44
2
| 22
2 is a factor of 22
11 | 11
11 is a factor of 11
|1
T hus,
176 = 2 × 2 × 2 × 2 × 11.
Step 5
All prime f actors of 520:
2
| 520
2 is a factor of 520
2
| 260
2 is a factor of 260
2
| 130
2 is a factor of 130
5
| 65
5 is a factor of 65
13 | 13
13 is a factor of 13
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ID : eu-6-LCM-and-HCF [5]
|1
T hus,
520 = 2 × 2 × 2 × 5 × 13.
Step 6
T he HCF of 240, 176 and 520 is = 2 × 2 × 2
= 8.
Step 7
T heref ore, the greatest number which divides 247, 184 and 529 leaving remainder 7, 8 and
9 respectively is 8.
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ID : eu-6-LCM-and-HCF [6]
(4) 57960, 60480
Step 1
Let's f irst f ind the smallest number that is exactly divisible by the numbers 7, 9, 8, 3, 2, 5, 6.
Such number will be the LCM of the numbers 7, 9, 8, 3, 2, 5, 6.
Step 2
Let us f ind the LCM of 7, 9, 8, 3, 2, 5, 6:
2 | 9, 8, 7, 6, 5, 3, 2
2 | 9, 4, 7, 3, 5, 3, 1
2 | 9, 2, 7, 3, 5, 3, 1
3 | 9, 1, 7, 3, 5, 3, 1
3 | 3, 1, 7, 1, 5, 1, 1
5 | 1, 1, 7, 1, 5, 1, 1
7 | 1, 1, 7, 1, 1, 1, 1
| 1, 1, 1, 1, 1, 1, 1
T he LCM is = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520
Step 3
Now, the other numbers that are exactly divisible by 7, 9, 8, 3, 2, 5, 6 will have to be the
multiples of their LCM. So, we will have to f ind the multiples of 2520 that are nearest to
60000.
Step 4
On dividing 60000 by 2520, we get a remainder of 2040. Hence 60000 - 2040 = 57960 and
60000 - 2040 + 2520 = 60480, both will be divisible by 2520.
Step 5
T hus the required numbers are 57960 and 60480.
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ID : eu-6-LCM-and-HCF [7]
(5)
14 cm
Step 1
According to the question, the length, breadth and height of the room are 11 m 6 cm, 9 m
10 cm and 7 m 56 cm respectively.
Now you have to convert each dimension into same unit. Since you know that 1m = 100 cm,
the length, breadth and height of the room in centimeters is 1106 cm, 910 cm and 756 cm
respectively.
Step 2
T he length of the rod which can be used to exactly measure all three dimensions of a room
should be a f actor of all three dimensions, that is, a f actor common to all three dimensions.
Since we have been asked to f ind the length of longest such rod, the length should be
equal to HCF of all three dimensions of the room.
Step 3
Let us now f ind the HCF of 1106, 910 and 756.
Step 4
All prime f actors of 1106:
2
| 1106
2 is a factor of 1106
7
| 553
7 is a factor of 553
79 | 79
79 is a factor of 79
|1
T heref ore,
1106 = 2 × 7 × 79.
Step 5
All prime f actors of 910:
2
| 910
2 is a factor of 910
5
| 455
5 is a factor of 455
7
| 91
7 is a factor of 91
13 | 13
13 is a factor of 13
|1
T heref ore,
910 = 2 × 5 × 7 × 13.
Step 6
All prime f actors of 756:
2 | 756
2 is a factor of 756
2 | 378
2 is a factor of 378
3 | 189
3 is a factor of 189
3 | 63
3 is a factor of 63
3 | 21
3 is a factor of 21
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ID : eu-6-LCM-and-HCF [8]
7 |7
7 is a factor of 7
|1
T heref ore,
756 = 2 × 2 × 3 × 3 × 3 × 7.
Step 7
T he HCF of 1106, 910 and 756 is = 2 × 7
= 14.
Step 8
Hence, the length of the longest rod which can measure the dimensions of the room
exactly is 14 cm.
(6) 4368
Let us f ind the LCM of 52, 84, 48:
2
| 84, 52, 48
2
| 42, 26, 24
2
| 21, 13, 12
2
| 21, 13, 6
3
| 21, 13, 3
7
| 7,
13, 1
13 | 1,
13, 1
| 1,
1,
1
T he LCM is = 2 × 2 × 2 × 2 × 3 × 7 × 13 = 4368
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ID : eu-6-LCM-and-HCF [9]
(7) c. 3:03
Step 1
T he f irst light f lashes every 3 minutes and the other light f lashes every 1
1
minutes.
2
Once the two lights f lash together, the amount of time af ter which the two lights will f lash
together again is equal to the LCM of 3 minutes and 1
1
minutes.
2
Step 2
Bef ore we calculate the LCM of 3 minutes and 1
1
minutes, let us translate both time
2
periods f rom minutes to seconds.
Since 1 minute = 60 seconds,
3 minutes = 3 × 60 = 180 seconds,
and 1
1
minutes =
2
3
× 60 = 90 seconds.
2
Step 3
Let us now calculate the LCM of 180 and 90.
All prime f actors of 180:
2 | 180
2 is a factor of 180
2 | 90
2 is a factor of 90
3 | 45
3 is a factor of 45
3 | 15
3 is a factor of 15
5 |5
5 is a factor of 5
|1
T hus,
180 = 2 × 2 × 3 × 3 × 5.
Step 4
All prime f actors of 90:
2 | 90
2 is a factor of 90
3 | 45
3 is a factor of 45
3 | 15
3 is a factor of 15
5 |5
5 is a factor of 5
|1
T hus,
90 = 2 × 3 × 3 × 5.
Step 5
T he LCM of 180 and 90 = 2 × 3 × 3 × 5 × 2 = 180.
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ID : eu-6-LCM-and-HCF [10]
Step 6
180 seconds in minutes =
180
minutes = 3 minutes.
60
Step 7
Now, we know that the two lights f lash together every 3 minutes. We have been told that
the two lights f lashes together at noon.
T his means, that times when they f lash together again are 12:3, 12:6, 12:9, 12:12, ...... 3:03,
...
T heref ore, the time af ter 3 PM that the two lights will f lash together again = 3:03 PM.
(8)
d. 168
Step 1
If a number is divisible by two dif f erent numbers, it is necessarily divisible by their L.C.M.
Step 2
L.C.M of 42 and 24 = 168.
Step 3
T hus, if a number is divisible by 42 and 24, it will be necessarily divisible by 168.
(9) c. 4
Step 1
Viktoria goes to Amity every 4 th days and Veronika goes to Amity every 6th days. So like this
they will both will take session on every 12 days.
Step 2
In order to f ind the number of days af ter which both of them goes together, we need to
f ind a number which is as small as possible and divisible by both 4 and 6.
T heref ore we need to f ind LCM of 4 and 6 which is 12 days.
Step 3
Now since both of them goes to school every 12th days. T heref ore number of session in
next 48 days they will take = 48 / 12 = 4
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ID : eu-6-LCM-and-HCF [11]
(10) c. 23 cms
Step 1
Let the width of strip be x cms
Step 2
Since no cloth should be wasted, 276 cms should be divisible by x cms
Similarly, 115 cms should be divisible by x cms
Step 3
Also x has to be as large as possible, theref ore x should be HCF of 276 and 115
Step 4
x = HCF(276, 115) = 23 cms
(11) A)
B)
C)
D)
E)
F)
(12)
1470
72
2520
648
432
2700
126
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ID : eu-6-LCM-and-HCF [12]
(13)
5
02
24
11
Step 1
T he 3 bells ring af ter intervals of 18, 144, 48 seconds respectively.
If the f irst bell rings just now, it will ring again af ter:
18 × 1 = 18 seconds
18 × 2 = 36 seconds
18 × 3 = 54 seconds
... and so on.
T his means, the number of seconds af ter which the 1st bell rings will be a multiple of 18.
Similarly, the number of seconds af ter which the 2nd and 3rd bells ring will be a multiple of
144 and 48 respectively.
Step 2
T his means, the number of seconds af ter which all 3 bells ring together must be a
(common) multiple of all 3: 18, 144, 48.
Step 3
We need to f ind the very f irst time af ter 5 O'clock, when the 3 bells will ring together. T his
will happen af ter a number of seconds af ter 5 O'clock, which is the least common multiple
(LCM) of 18, 144, 48.
Step 4
T he LCM of 18, 144, 48 is 144 seconds.
144 seconds in minutes =
144
60
=
12
minutes.
5
Step 5
T heref ore, all 3 bells will ring together
12
minutes past 5 O'clock, that is, at 5:02:24.
5
Step 6
We just f ound out that the 3 bells ring together every
12
minutes. T heref ore, the number
5
132
of times they will ring together in 132 minutes =
12
5
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=
132
× 5 = 11 × 5 = 55 times.
12
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ID : eu-6-LCM-and-HCF [13]
(14)
20
Step 1
We need to f ind the number of days af ter which he will have exact amount (no decimal)
If we observer €27.35, it has two parts,
€27 and €0.35
Step 2
T he f irst part €27 multiplied by any number of days will always result in f ull amount (no
decimal), so we need not worry about this part
Step 3
Now we need to f ind the least number of days, which when multiplied to €0.35, results in a
f ull amount. We know that this amount will be a multiple of 1 and will also be a multiple of
0.35. So smallest number which is divisible by both 1 and 0.35 will be HCF of 1 and 0.35.
Step 4
Since we are not taught calculating HCF of non-natural numbers, we can f irst calculate HCF
of 100 and 35, and then divide the result by 100.
HCF of 100 and 35 = 700
⇒ HCF of 1 and 0.35 = 700/100 = 7
Step 5
T heref ore number of days required to save €7, by saving €0.35 daily,
= 7 / 0.35 = 20 days
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ID : eu-6-LCM-and-HCF [14]
(15)
7920
8280
Step 1
Let's f irst f ind the smallest number that is exactly divisible by the numbers 5, 2, 3, 9, 8, 6.
T his number will be the LCM of the numbers 5, 2, 3, 9, 8, 6.
Step 2
Let us f ind the LCM of 5, 2, 3, 9, 8, 6:
2 | 9, 8, 6, 5, 3, 2
2 | 9, 4, 3, 5, 3, 1
2 | 9, 2, 3, 5, 3, 1
3 | 9, 1, 3, 5, 3, 1
3 | 3, 1, 1, 5, 1, 1
5 | 1, 1, 1, 5, 1, 1
| 1, 1, 1, 1, 1, 1
T he LCM is = 2 × 2 × 2 × 3 × 3 × 5 = 360
Step 3
Now, the other numbers that are exactly divisible by 5, 2, 3, 9, 8, 6 will have to be the
multiples of their LCM. So, we will have to f ind the multiples of 360 that are nearest to 8000.
Step 4
On dividing 8000 by 360, we get a remainder of 80. Hence either 8000 - 80 = 7920 and 8000
- 80 + 360 = 8280, both will be divisible by 360.
Step 5
T hus the required numbers are 7920 and 8280.
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