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Maths-

General

Easy

Question

What is the LCM of 6 and 16 :

48
6
16
12

The correct answer is: 48

LCM of 6 and 16 : 2 × 2 × 2 × 2 × 3 = 48

Related Questions to study

Find the Area of triangle , whose Base is 4/5 m and height is 15/8.

Find the Area of triangle , whose Base is 4/5 m and height is 15/8.

Convert 5 5/6 into improper fraction :

Convert 5 5/6 into improper fraction :

A cloth is 50.92m long , Raymond cut a 20.31m long part Find the length of remaining part

A cloth is 50.92m long , Raymond cut a 20.31m long part Find the length of remaining part

parallel

What is the LCM of 12 and 22 :

What is the LCM of 12 and 22 :

What is the LCM of 11 and 12 :

What is the LCM of 11 and 12 :

Find the LCM of 11 and 6 :

Find the LCM of 11 and 6 :

parallel

Convert 15 % into fraction :

Convert 15 % into fraction :

What is the LCM of 8 and 15 :

What is the LCM of 8 and 15 :

Find angle x and y :

Find angle x and y :

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What is the LC M of 9 and 18:

What is the LC M of 9 and 18:

Find the GCF of 9 and 16 :

Find the GCF of 9 and 16 :

The correct statement about the following disaccharide is:
I)
II)

The correct statement about the following disaccharide is:
I)
II)

Chemistry-General

parallel

A carbonyl compound P, which gives positive iodoform test, undergoes reaction with MeMgBr followed by dehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoes intramolecular aldol reaction to give predominantly S.
$table row cell P fraction numerator 1. M e M g B r over denominator 2. H to the power of plus end exponent comma H subscript 2 end subscript O end fraction Q fraction numerator 1. O subscript 3 end subscript over denominator 2. Z n comma H subscript 2 end subscript O end fraction R fraction numerator 1. O H to the power of minus end exponent over denominator 2. capital delta end fraction S end cell row cell 3. H subscript 2 end subscript S O subscript 4 end subscript comma capital delta end cell end table$
The structures of the products Q and R, respectively, are:

A carbonyl compound P, which gives positive iodoform test, undergoes reaction with MeMgBr followed by dehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoes intramolecular aldol reaction to give predominantly S.
$table row cell P fraction numerator 1. M e M g B r over denominator 2. H to the power of plus end exponent comma H subscript 2 end subscript O end fraction Q fraction numerator 1. O subscript 3 end subscript over denominator 2. Z n comma H subscript 2 end subscript O end fraction R fraction numerator 1. O H to the power of minus end exponent over denominator 2. capital delta end fraction S end cell row cell 3. H subscript 2 end subscript S O subscript 4 end subscript comma capital delta end cell end table$
The structures of the products Q and R, respectively, are:

Chemistry-General

Consider the following reactions I and II,
I)
II)
Products 'X' and 'Y' are respectively:

Consider the following reactions I and II,
I)
II)
Products 'X' and 'Y' are respectively:

Chemistry-General

In which of the following cases, the nitration will take place at meta-position?
I)
II)
III)
IV)

In which of the following cases, the nitration will take place at meta-position?
I)
II)
III)
IV)

Chemistry-General

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