Question

How many such $4^{\prime }s$ are there in the given below arrangement which are immediately preceded by a pair of numbers whose product is more than the product of a pair of numbers immediately following $4$?

$4378445643241744521468948947864254462$

• A. $3$
• B. $4$
• C. $5$
• D. $7$

Answer 1

The correct option is D $7$

Total number of 4's = $13$

The 4's in the arrangement are:

$437$ : There is no pair of numbers preceding $\Rightarrow$ not eligible

$78445$ : $7\times 8>4\times 5\Rightarrow$ eligible

$84456$ : $8\times 4>5\times 6\Rightarrow$ eligible

$56432$ : $5\times 6>3\times 2\Rightarrow$ eligible

$32417$ : $3\times 2<1\times 7\Rightarrow$ not eligible

$17445$ : $1\times 7<4\times 5\Rightarrow$ not eligible

$74452$ : $7\times 4>5\times 2\Rightarrow$ eligible

$21468$ : $2\times 1<6\times 8\Rightarrow$ not eligible

$89489$ : $8\times 9=8\times 9\Rightarrow$ not eligible

$89478$ : $8\times 9>7\times 8\Rightarrow$ eligible

$86425$ : $8\times 6>2\times 5\Rightarrow$ eligible

$25446$ : $2\times 5<4\times 6\Rightarrow$ not eligible

$54462$ : $5\times 4>6\times 2\Rightarrow$ eligible

$\therefore$ Number of 4's satisfying the given criteria = $7$. $\left[D\right]$