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 HCF and LCM Questions

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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:

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 11296
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

 

Therefore, Required Number is of the form 840k+3.

 

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

 

Therefore, Required  Number = (840 x 2+3)=1683

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38 10835
Q:

A room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in Width. Find the least number of square tiles of equal size required to cover the entire floor of the room.

A) 107 B) 117
C) 127 D) 137
 
Answer & Explanation Answer: B) 117

Explanation:

Let us calculate both the length and width of the room in centimeters.

Length = 6 meters and 24 centimeters = 624 cm

width = 4 meters and 32 centimeters = 432 cm

As we want the least number of square tiles required, it means the length of each square tile should be as large as possible.Further,the length of each square tile should be a factor of both the length and width of the room.

Hence, the length of each square tile will be equal to the HCF of the length and width of the room = HCF of 624 and 432 = 48

Thus, the number of square tiles required = (624 x 432 ) / (48 x 48) = 13 x 9 = 117

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20 9945
Q:

A heap of coconuts is divided into groups of 2, 3 and 5 and each time one coconut is left over. The least number of Coconuts in the heap is  ?

A) 63 B) 31
C) 16 D) 27
 
Answer & Explanation Answer: B) 31

Explanation:

To get the least number of coconuts :
LCM = 30 => 30 + 1 = 31

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33 9631
Q:

Three numbers are in the ratio of 3:4:5 and their L.C.M is 3600.Their HCF is:

A) 40 B) 60
C) 100 D) 120
 
Answer & Explanation Answer: B) 60

Explanation:

Let the numbers be 3x, 4x, 5x.

 

Then, their L.C.M = 60x.

 

So, 60x=3600 or x=60.

 

Therefore,  The numbers are (3 x 60), (4 x 60), (5 x 60).

 

Hence,required H.C.F=60

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14 9032
Q:

If the HCF of 210 and 55 is expressible in the form of 210 x 5 + 55P, then value of P = ?

A) -23 B) 27
C) 16 D) -19
 
Answer & Explanation Answer: D) -19

Explanation:

HCF of 210 and 55 is 5

Now, 210x5 + 55P = 5

=> 1050 + 55P = 5

=> 55P = -1045

=> P = -1045/55

=> P = -19.

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28 8975
Q:

A drink vendor has 368 liters of Maaza, 80 liters of Pepsi and 144 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) 47 B) 46
C) 37 D) 35
 
Answer & Explanation Answer: C) 37

Explanation:

The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.
Number of cans of Maaza = 368/16 = 23
Number of cans of Pepsi = 80/16 = 5
Number of cans of Sprite = 144/16 = 9
The total number of cans required = 23 + 5 + 9 = 37 cans.

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31 8148
Q:

H.C.F of 4 x 27 x 3125, 8 x 9 x 25 x 7 and 16 x 81 x 5 x 11 x 49 is :

A) 360 B) 180
C) 90 D) 120
 
Answer & Explanation Answer: B) 180

Explanation:

4 x 27 x 3125 = 22 × 33 × 55 ;

 

8 x 9 x 25 x 7 = 23 × 32 × 52 × 7

 

16 x 81 x 5 x 11 x 49 = 24 × 34 × 5 × 11 × 72

 

H.C.F = 22 × 32 × 5 = 180.

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22 7860