LCM and HCF
 In a school, 391 boys and 323 girls have been divided into the largest possible equal classes, so that each class of boys numbers the same as each class of girls. What is the number of classes ?

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First of all we find HCF of 391 and 323.
Correct Option: D
First of all we find HCF of 391 and 323.
∴ Number of classes = 17
 If the ratio of two numbers is 2 : 3 and their L.C.M. is 54, then the sum of the two numbers is

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Let the two numbers are 2p and 3p respectively.
According to question,
LCM = 54
⇒ p ( 3 × 2 ) = 54
⇒ p = 9
Numbers = 2p = 2 × 9 = 18 and, 3p = 3 × 9 = 27Correct Option: C
Let the two numbers are 2p and 3p respectively.
According to question,
LCM = 54
⇒ p ( 3 × 2 ) = 54
⇒ p = 9
Numbers = 2p = 2 × 9 = 18 and, 3p = 3 × 9 = 27
Sum of the two numbers = 18 + 27 = 45
 The LCM and the HCF of the numbers 28 and 42 are in the ratio :

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L.C.M. of 28 and 42
L.C.M. of 28 and 42 = 2 × 2 × 7 × 3 = 84
H.C. F. of 28 and 42
∴ H.C. F = 14Correct Option: A
L.C.M. of 28 and 42
L.C.M. of 28 and 42 = 2 × 2 × 7 × 3 = 84
H.C. F. of 28 and 42
∴ H.C. F = 14Required ratio = 84 = 6 : 1 14
 Two pipes of length 1.5 m and 1.2 m are to be cut into equal pieces without leaving any extra length of pipes. The greatest length of the pipe pieces of same size which can be cut from these two lengths will be

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Maximum length of each piece = HCF of 1.5 metre and 1.2 metre
First of all we find HCF of 1.5 metre and 1.2 metre
∴ HCF of 1.5 and 1.2 metre = 0.3 metreCorrect Option: C
Maximum length of each piece = HCF of 1.5 metre and 1.2 metre
First of all we find HCF of 1.5 metre and 1.2 metre
∴ HCF of 1.5 and 1.2 metre = 0.3 metre
Required Maximum length of each piece = 0.3 metre
 A milk vendor has 21 litres of cow milk, 42 litres of toned milk and 63 litres of double toned milk. If he wants to pack them in cans so that each can contains same litres of milk and does not want to mix any two kinds of milk in a can, then the least number of cans required is

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Maximum quantity in each can = HCF of 21, 42 and 63 litres = 21 litres
Required least number of cans = 21 + 42 + 63 21 21 21
Correct Option: B
Maximum quantity in each can = HCF of 21, 42 and 63 litres = 21 litres
Required least number of cans = 21 + 42 + 63 21 21 21
Required least number of cans = 1 + 2 + 3 = 6