FACTS  AND  FORMULAE  FOR  HCF  AND  LCM  QUESTIONS

 

 

I.Factors and Multiples : If a number 'a' divides another number 'b' exactly, we say that 'a' is a factor of 'b'. In this case, b is called a multiple of a.

 

II.Highest Common Factor (H.C.F) or Greatest Common Measure (G.C.M) or Greatest Common Divisor (G.C.D) : The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly. There are two methods of finding the H.C.F. of a given set of numbers :

1. Factorization Method : Express each one of the given numbers as the product of prime factors.The product of least powers of common prime factors gives H.C.F.

2. Division Method : Suppose we have to find the H.C.F. of two given numbers. Divide the larger number by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is the required H.C.F.

Finding the H.C.F. of more than two numbers : Suppose we have to find the H.C.F. of three numbers. Then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given numbers. Similarly, the H.C.F. of more than three numbers may be obtained. 

 

III.Least Common Multiple (L.C.M) : The least number which is exactly divisible by each one of the given numbers is called their L.C.M.

1. Factorization Method of Finding L.C.M: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors.

2. Common Division Method (Short-cut Method) of Finding L.C.M : Arrange the given numbers in a row in any order. Divide by a number which divides exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.

 

IV. Product of two numbers = Product of their H.C.F and L.C.M

 

V. Co-primes : Two numbers are said to be co-primes if their H.C.F. is 1.

 

VI. H.C.F and L.C.M of Fractions :

1. H.C.F = H.C.F. of Numerators / L.C.M of Numerators

2. L.C.M = L.C.M of Numerators / H.C.F of Denominators

 

VII. H.C.F and L.C.M of Decimal Fractions : In given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.

 

VIII. Comparison of Fractions : Find the L.C.M. of the denominators of the given fractions. Convert each of the fractions into an equivalent fraction with L.C.M. as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.

Q:

The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

A) 1677 B) 1683
C) 2523 D) 3363
 
Answer & Explanation Answer: B) 1683

Explanation:

L.C.M. of 5, 6, 7, 8 = 840.

 

 Required number is of the form 840k + 3

 

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

 

 Required number = (840 x 2 + 3) = 1683.

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25 20200
Q:

The maximum number of students  among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is:

A) 91 B) 910
C) 1001 D) 1911
 
Answer & Explanation Answer: A) 91

Explanation:

Required number of students = H.C.F  of 1001 and 910  = 91

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34 20160
Q:

Find the greatest number that will divide 964, 1238 and 1400 leaving remainder of 41,31 and 51 respectively ?

A) 64 B) 69
C) 71 D) 58
 
Answer & Explanation Answer: C) 71

Explanation:

To find the greatest number which divides the numbers 964, 1238 and 1400 leaving the remainders 41, 31 and 51 is nothing but the HCF of (964 - 41), (1238 - 31), (1400 - 51).

Therefore, HCF of 923, 1207 and 1349 is 71.

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20 14038
Q:

LCM of 80, 85, 90

A) 11440 B) 11998
C) 12240 D) 12880
 
Answer & Explanation Answer: C) 12240

Explanation:

LCM of (80, 85, 90) can be found by prime factorizing them.

 

80 → 2 × 2 × 2 × 2 × 5

85 → 17 × 5

90 → 2 × 3 × 3 × 5

L.C.M of (80,85,90) = 2 × 2 x 2 × 2 × 3 × 3 × 5 × 17

= 16 x 9 x 85

= 144 x 85

= 12240

 

L.C.M of (80,85,90) = 12240.

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97 13750
Q:

The greatest possible length which can be used to measure exactly the length 7m, 3m 85cm, 12 m 95 cm is

A) 15 cm B) 25 cm
C) 35 cm D) 42 cm
 
Answer & Explanation Answer: C) 35 cm

Explanation:

Required Length = H.C.F of 700 cm, 385 cm and 1295 c

                             = 35 cm.

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28 13715
Q:

A drink vendor has 80 liters of Mazza, 144 liters of Pepsi, and 368 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required?

A) 35 B) 36
C) 37 D) 38
 
Answer & Explanation Answer: C) 37

Explanation:

If we want to pack the drinks in the least number of cans possible, then each can should contain the maximum numbers of liters possible.As each can contains the same number liters of a drink, the number of liters in each can is a comman factor for 80,144 and 368; and it is also the highest such factor, as we need to store the maximum number of liters in each can.

So, the number of liters in each can  = HCF of 80,144 and 368 = 16 liters.

Now, number of cans of Maaza = 80/16 = 5

Number of cans of Pepsi = 144/16 = 9

Number of cans of Sprite = 368/16 = 23

Thus, the total number of cans required = 5 + 9 + 23 = 37 

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24 12964
Q:

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

A) 4 B) 10
C) 15 D) 16
 
Answer & Explanation Answer: D) 16

Explanation:

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

So, the bells will toll together after every 120 seconds(2 minutes).

In 30 minutes,they will together (30/2)+1=16 times

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20 12541
Q:

In a palace, three different types of coins are there namely gold, silver and bronze. The number of gold, silver and bronze coins is 18000, 9600 and 3600 respectively. Find the minimum number of rooms required if in each room should give the same number of coins of the same type ?

A) 26 B) 24
C) 18 D) 12
 
Answer & Explanation Answer: A) 26

Explanation:

Gold coins = 18000 , Silver coins = 9600 , Bronze coins = 3600

Find a number which exactly divide all these numbers 

That is HCF of 18000, 9600& 3600 

All the value has 00 at end so the factor will also have 00.

HCF for 180, 96 & 36.

 

Factors of  

180 = 3 x 3 x 5 x 2 x 2

96 = 2 x 2 x 2 x 2 x 2 x 3 

36 = 2 x 2 x 3 x 3 

Common factors are 2x2×3=12

 Therefore, Actual HCF is 1200

 

  

Gold Coins 18000/1200 will be in 15 rooms

Silver Coins 9600/1200 will be in 8 rooms

Bronze Coins 3600/1200 will be in 3 rooms

Total rooms will be (15+8+3)  =  26 rooms.

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13 12425