Question

An English word consists of $9$ alphabets. The sum of twice the number of vowels and three times the number of consonants present in the word is equal to four more that four times the total numbers of vowels in the English alphabets. The product of the number of vowels and consonants present in the word is

• A. $9$
• B. $20$
• C. $18$
• D. $14$

Answer 1

Answer: (C)

$18$

Let no of vowels present in the word $=x$

Let no of consonants present in the word $=y$

According to problem

$x+y=9\Rightarrow y=9-x.....\left(i\right)$

$2x+3y=4+4\left(5\right)$

$2x+3y=4+20$

$2x+3y=24......\left(ii\right)$

Put $\left(i\right)$ in $\left(ii\right)$

$2x+3y(9-x)=24$

$2x+27-3x=24$

$-x=24-27$

$x=3$

$y=6$

$\therefore$ No of vowels $=3$

No of consonants $=6$

Product of no. of vowels and consonants present $=3\times 6$

$=18$

$\therefore$ Ans $=18$