
Find the first four common multiples of 3 and 4.
a) 24, 28, 32, 36
b) 24, 27, 33, 36
c) 12, 24, 36, 48
d) 12, 15, 20, 24
Answer
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Hint: In the above type of question we need to find the correct option of the given four. The correct option in these four given options will be the one which will have terms divisible by both 3 and 4. So to find this we will divide each term by 3 and 4 and if the result of all the mentioned terms in the option given is an integer then we are going to choose that option as the correct option.
Complete step by step solution:
In the above given question we need to find the correct option out of the given four options in the question.
The correct option out of the four is the one which have the terms all divisible by 3 and 4 i.e. all the terms in the option when divided by 3 and 4 should leave remainder as zero or the result when divided by 3 and 4 will be equivalent to an integer.
So to proceed with this we are going to solve it by going through the options.
Let us take option a), we have 24, 28, 32, 36. Out of these, only 24 and 36 are divisible by 3. So, we can rule out this option.
Now, coming to option b), we have 24, 27, 33, 36. If we notice, we find that all terms are divisible by 3. But, coming to terms like 27, 33, we find that these are not divisible by 4. So, we can rule out this option too.
Now let us see option c) it has terms 12, 24, 36, 48. Now when we divide these terms by 3 all the terms are divisible by 3 and give integer result and the same thing when divided by 4 i.e. give integer results so this is the correct option.
Let us take option d) it has terms 12, 15, 20, 24. When divided by 3 we will be able to see that 12, 15 and 24 is divisible but 20 is not, this option is wrong.
Therefore, we have the correct option as option c) 12, 24, 36, 48.
So, the correct answer is “Option C”.
Note: In the above mentioned question make use of the options provided by the question. It is the easiest way and the fastest way to solve a question, whenever we have a question and there are options, always check whether or not this question can be solved through it. Alternatively we can also write down multiple of 3 and 4 separately and then take out the first four common multiples.
Complete step by step solution:
In the above given question we need to find the correct option out of the given four options in the question.
The correct option out of the four is the one which have the terms all divisible by 3 and 4 i.e. all the terms in the option when divided by 3 and 4 should leave remainder as zero or the result when divided by 3 and 4 will be equivalent to an integer.
So to proceed with this we are going to solve it by going through the options.
Let us take option a), we have 24, 28, 32, 36. Out of these, only 24 and 36 are divisible by 3. So, we can rule out this option.
Now, coming to option b), we have 24, 27, 33, 36. If we notice, we find that all terms are divisible by 3. But, coming to terms like 27, 33, we find that these are not divisible by 4. So, we can rule out this option too.
Now let us see option c) it has terms 12, 24, 36, 48. Now when we divide these terms by 3 all the terms are divisible by 3 and give integer result and the same thing when divided by 4 i.e. give integer results so this is the correct option.
Let us take option d) it has terms 12, 15, 20, 24. When divided by 3 we will be able to see that 12, 15 and 24 is divisible but 20 is not, this option is wrong.
Therefore, we have the correct option as option c) 12, 24, 36, 48.
So, the correct answer is “Option C”.
Note: In the above mentioned question make use of the options provided by the question. It is the easiest way and the fastest way to solve a question, whenever we have a question and there are options, always check whether or not this question can be solved through it. Alternatively we can also write down multiple of 3 and 4 separately and then take out the first four common multiples.
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