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Maths-

General

Easy

Question

Find the LCM of 20 and 16 :

40
20
80
60

The correct answer is: 80

LCM 69 of 20 and 16 : 22225 = 80

Related Questions to study

Find the surface area of Cube whose side is 20cm

Find the surface area of Cube whose side is 20cm

What is the LCM of 6 and 16 :

What is the LCM of 6 and 16 :

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.

parallel

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

What is the LCM of 12 and 22 :

What is the LCM of 12 and 22 :

parallel

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 :

Convert 15 % into fraction :

Convert 15 % into fraction :

parallel

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 :

What is the LC M of 9 and 18:

What is the LC M of 9 and 18:

parallel

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

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

parallel

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