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Let us suppose the word ‘MEETBOOK’ has been concatenated with the inverse of the word ‘APTITUDE’. Then find the letter which appears 7th to the right of the vowel which appears nearest to the letter which is 12th to the left of the vowel nearest to the right end.

Let us suppose the word ‘MEETBOOK’ has been concatenated with the inverse of the word ‘APTITUDE’. Then find the letter which appears 7th to the right of the vowel which appears nearest to the letter which is 12th to the left of the vowel nearest to the right end. Correct Answer D

1) The arrangement of word after concatenation are:

Left End M E E T B O O K E D U T I T P A Right End

2) Vowel nearest to the right end

M E E T B O O K E D U T I T P A

3) 12th letter left to A is T.

M E E T B O O K E D U T I T P A

4) Vowel nearest T is E.

M E E T B O O K E D U T I T P A

5) Letter 7th to the right of E is D.

M E E T B O O K E D U T I T P A

Hence, D is the correct answer.

Related Questions

If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
If the word ‘SPRITE’ is concatenated with the word ‘IMMEDIATELY’. Then find the vowel which is nearest to the letter which is 2nd to the left of the letter which is 7th from right end.
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Let the function
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Study the following information to answer the questions given below: A company wants to hire young professionals where students of any graduate stream can participate in the drive. But company put some eligibility criteria. There are two rounds, round one consist of aptitude test and 2nd round is personal interview. 1) Student having no backlogs is eligible to participate in the drive. 2) Student must have at least 68 percentages in 10th and 12th. 3) Student having greater than 85 percentages in graduation without any backlogs are eligible for direct interview round. If the 1st or 2nd criteria not fulfilled then, 1) If a candidate has a work experience of 1 year in any field he is eligible to appear in aptitude round. But if the candidate has not clear his backlogs he is not eligible to participate in the drive. Amit a commerce student has a score of 69 percentages in 10th and 12th. He is currently working in an IT company for 15 months. In graduation he scored only 65 percentages and also clears his backlogs.
The value of the expression $$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$   $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$    $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$   = ?
The Hamiltonian of a particle is given by $$H = \frac{{{p^2}}}{{2m}} + V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right) + \phi \left( { + \left| {\overrightarrow {\bf{r}} } \right|} \right)\overrightarrow {\bf{L}} .\overrightarrow {\bf{S}} ,$$        where $$\overrightarrow {\bf{S}} $$ is the spin, $$V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  and $$\phi \left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  are potential functions and $$\overrightarrow {\bf{L}} \left( { = \overrightarrow {\bf{r}} \times \overrightarrow {\bf{p}} } \right)$$   is the angular momentum. The Hamiltonian does not commute with
$$\frac{{{{\left( {4.53 - 3.07} \right)}^2}}}{{\left( {3.07 - 2.15} \right)\left( {2.15 - 4.53} \right)}} + \, $$     $$\frac{{{{\left( {3.07 - 2.15} \right)}^2}}}{{\left( {2.15 - 4.53} \right)\left( {4.53 - 3.07} \right)}} + \,\, $$     $$\frac{{{{\left( {2.15 - 4.53} \right)}^2}}}{{\left( {4.53 - 3.07} \right)\left( {3.07 - 2.15} \right)}}$$     is simplified to :
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