0 votes 0 votes $(BE)^{2} = MPB$, where $B, E, M$ & $P$ are distinct integers, then $M$? $2$ $3$ $9$ None of these Quantitative Aptitude cat1998 quantitative-aptitude + – makhdoom ghaya asked Apr 20, 2017 makhdoom ghaya 8.1k points 518 views answer comment Share See all 0 reply Please log in or register to add a comment.
1 votes 1 votes There is only one possibility. $19^{2}=361$ Here B=1 E=9 M=3 P=6 Here all numbers are distinct. Hence,Option(B) 3 should be the correct choice. LeenSharma answered May 4, 2017 LeenSharma 228 points comment Share See all 0 reply Please log in or register to add a comment.