## LCM Questions​

### To find the LCM Questions and Answers using the prime factorization method, do the following:

• Step 1: Display each integer as the sum of its prime elements.
• Step 2: LCM will be the product of all prime factors’ highest powers.

To find the LCM, multiply all prime factors. But the common factors are included only once.

### To find the LCM Questions and Answers using the division method, do the following:

• Step 1: Begin by writing the numbers in a horizontal line separated by commas.

• Step 2: Next, divide all of the numbers given by the smallest prime number.

• Step 3: Below the previous line, write the quotients and undivided numbers in a new line.

• Step 4: Continue this process until we reach a point where there are no prime factors in common.

• Step 5: The product of all the divisors and the numbers in the final line is LCM.

### For any two numbers, use the L.C.M formula:

• If we know the greatest common divisor (GCD) of two numbers, we can simply calculate LCM using the following formula:
LCM = $\frac {a*b}{GCF of (a,b)}$

• To find the LCM of two fractions, we must first find the LCM of the numerators and the HCF of the denominators.
Additionally, both of these outcomes will be reported as a fraction.
Thus, LCM =$\frac { L.C.M of Numerator }{H.C.F of Denominator}$

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1. The ratio between two numbers is 13:15 and their LCM is 39780. The numbers are:

2652,3060

2652,3060

79.56%

884, 1020

884, 1020

10.87%

884, 1040

884, 1040

5.52%

670, 1340

670, 1340

4.05%

Let the numbers be 13X and 15X

Their LCM will be (13*5)X

Now (13*15)X= 39780

Therefore X= 204

Hence the numbers are (13*204, 15*204) or 2652 and 3060

2. Find the lowest number which when divided by 15, 27, 35, and 42 leaves a remainder of 9 in each case.

1890

1890

20.11%

1899

1899

64.37%

1880

1880

6.13%

1881

1881

9.39%

The lowest number which is exactly divisible by numbers 15 27 35 and 42 will be the LCM of these numbers which are mentioned below:

 3 15   27  35  42 5 5   9   35   14 7 1   9   7   14 9 1   9   1   2 2 1   1   1   1

= 3*5*7*9*2= 1890

Therefore the required number is 1890+9= 1899

3. Find the smallest number which when increased by 8 is perfectly divisible 30  45  65  78?

1160

1160

7.64%

1150

1150

7.85%

1162

1162

73.97%

1168

1168

10.54%

The required number will be=

= {LCM of (30   45   65   78) – 8}

LCM of 30   45   65   78 =

5 30    45    65  78
3 6   9   13    78
13  2   3   13   26
2 2   3   1   2

5*3*13*2*3 = 1170

Hence the required number = 1170-8= 1162

4. Find the lowest number, which when decreased by five, is exactly divided by 20   28   35   105?

420

420

9.28%

425

425

82.7%

430

430

2.95%

440

440

5.06%

Clearly the required which when reduced by 5 will be  LCM of 20   28   35   105

Hence the required number = {LCM of 20   28   35   105) + 5}

LCM of 20   28   35   105 =

 5 20   28   35 105 7 4   28   7   21 4 4   4   1   3 1   1   1   3

= 5*7*4*3 = 420

420+5= 425

5. Find the lowest number which when divided by 35   45   55 leaves the remainders 18   28   38   respectively?

3440

3440

7.58%

3448

3448

77.78%

3445

3445

9.6%

3450

3450

5.05%

Here the difference between the divisor and its corresponding remainder = 17 (35-18= 17 similarly 45-28= 17 and 55-38= 17)

Therefore the required number will be = { (LCM of 35   45   55) – 17}

 5 35   45   55 7   9   11

= (5*7*9*11) – 17= 3465-17

= 3448

Therefore option B is the correct one.

6. Five bells begin to toll together and toll respectively at intervals of 6, 7, 8, 9, and 12 seconds. How many seconds will they toll together again?

72 sec

72 sec

7.39%

612 sec

612 sec

9.24%

504 sec

504 sec

78.98%

318 sec

318 sec

4.39%

The number of times the bells will toll together = LCM of 6, 7, 8, 9, and 12 = 504

7. If the two numbers are in the given ratio, which is 4:5 and their H.C.F is 40, find their L.C.M.

37

37

3.94%

45

45

10.1%

40

40

64.78%

None of the above

None of the above

21.18%

Let the number be 4y and 5y

L.C.M = 20 y

20y = 40

Y = 2

So the numbers will be 4*2 = 8

5*2 = 10

L.C.M of 8 and 10 is 40

8. The given ratio of two numbers is 3:2. If the L.C.M of them is 30, then calculate their sum.

34

34

5.24%

50

50

15.26%

55

55

14.35%

None of the above

None of the above

65.15%

Let the given numbers be = 3y and 2y

LCM = 6Y

6Y = 30

Y = 5

The two numbers will be 3*5 = 15 and 2*5 = 10

Their sum will be = 10+15 = 25

9. Find the LCM of two numbers whose product is 2025 and their HCF is 15?

30375

30375

5.21%

2040

2040

4.98%

135

135

86.26%

2010

2010

3.55%

LCM=$\frac{product of two numbers}{HCF} = \frac{2025}{15}=135$

10. The product of two-digit numbers is 2160, and their GCM is 12. The numbers are:

12,180

12,180

36.8%

72, 30

72, 30

10.67%

36, 60

36, 60

50.13%

96, 25

96, 25

2.4%

Let the numbers be 12a and 12b

Then 12a*12b= 2160 or ab=15

Therefore the values of co-primes a and b are (1,15) and (3,5)

Hence the two-digit numbers are (3*12, 5*12), i.e. 36, 60

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