Answer
Verified
91.5k+ views
Hint: It is known that the product of the HCF of the numbers and the LCM of the numbers is equal to the product of those numbers. So, use this formula to find the product in such a way that one of them satisfied the given condition of lying in between 200 and 300.
Complete step-by-step solution
Let us take the two numbers to be x and y.
Consider the HCF and LCM of the two numbers, which are 21 and 4641. We use the following formula to find the product of the two numbers.
$
{\text{HCF}} \times {\text{LCM}} = x \times y \\
\Rightarrow 21 \times 4641 = x \times y \\
$
Now, we look at the given options, in each of the given options one number is 273, we make a manipulation in such a way that we get a product of 273 and the second number. Let us factor both 21 and 4641.
$ \Rightarrow \left( {3 \times 7} \right) \times \left( {3 \times 7 \times 13 \times 17} \right) = x \times y$
Now we combine and multiply the factor that gives the product as 273 and combine and multiply the remaining to get the other number.
$
\Rightarrow \left( {21 \times 13} \right) \times \left( {21 \times 17} \right) = x \times y \\
\Rightarrow 273 \times 357 = x \times y \\
$
Thus, we may conclude that one number is 273 and the other is 357.
Hence, option (D) is the correct option.
Note: In the question having the options and a condition where many answers are possible, we begin by looking at the options first and analysing if they have any value in common. If yes, we try to fix that value and solve the question with respect to that value, as in this case it was 273. Even more than one answer would be valid for this question, but in order to match the options, we fixed ourselves to the number 273.
Complete step-by-step solution
Let us take the two numbers to be x and y.
Consider the HCF and LCM of the two numbers, which are 21 and 4641. We use the following formula to find the product of the two numbers.
$
{\text{HCF}} \times {\text{LCM}} = x \times y \\
\Rightarrow 21 \times 4641 = x \times y \\
$
Now, we look at the given options, in each of the given options one number is 273, we make a manipulation in such a way that we get a product of 273 and the second number. Let us factor both 21 and 4641.
$ \Rightarrow \left( {3 \times 7} \right) \times \left( {3 \times 7 \times 13 \times 17} \right) = x \times y$
Now we combine and multiply the factor that gives the product as 273 and combine and multiply the remaining to get the other number.
$
\Rightarrow \left( {21 \times 13} \right) \times \left( {21 \times 17} \right) = x \times y \\
\Rightarrow 273 \times 357 = x \times y \\
$
Thus, we may conclude that one number is 273 and the other is 357.
Hence, option (D) is the correct option.
Note: In the question having the options and a condition where many answers are possible, we begin by looking at the options first and analysing if they have any value in common. If yes, we try to fix that value and solve the question with respect to that value, as in this case it was 273. Even more than one answer would be valid for this question, but in order to match the options, we fixed ourselves to the number 273.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
Electric field due to uniformly charged sphere class 12 physics JEE_Main
If the distance between 1st crest and the third crest class 11 physics JEE_Main
Derive an expression for maximum speed of a car on class 11 physics JEE_Main
3 mole of gas X and 2 moles of gas Y enters from the class 11 physics JEE_Main
The vapour pressure of pure A is 10 torr and at the class 12 chemistry JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main