4 votes 4 votes For all integers $y>1, \: \langle y \rangle = 2y+(2y-1)+(2y-2) + \dots +1$. What is the value of $\langle 3 \rangle \times \langle 2 \rangle $? Where $\times$ is a multiplication operator? $116$ $210$ $263$ $478$ Quantitative Aptitude ugcnetcse-june2019-paper1 general-aptitude quantitative-aptitude + – Arjun asked Jul 2, 2019 • edited Jun 17, 2020 by soujanyareddy13 Arjun 1.9k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
4 votes 4 votes Given Function $\langle y \rangle=2y+(2y-1)+(2y-2)+ \dots+1\to\qquad(i)$ put $y=2$ at equation $(i)$ $<2> = 4+3+2+1= 10$ & Put $y=3$ equation $(i)$ $ <3> = 6+5+4+3+2+1=21$ So $\langle3\rangle\times\langle2\rangle=21*10=210$ Option $(B)$ is correct. Hira Thakur answered Jul 3, 2019 • edited May 16, 2021 by Hira Thakur Hira Thakur comment Share Follow See all 0 reply Please log in or register to add a comment.
3 votes 3 votes $ \langle y \rangle = 2y+(2y-1)+(2y-2) + \dots +1$ we can write like this $ \langle y \rangle = 1+2+3 +\dots +(2y-2) +(2y-1) +2y $ It is arithmetic progression whose first term is $a = 1$ and difference $d =1$ and last terms $l = 2y$ Sum of the AP series $S = \dfrac{n}{2}\left(a+ l \right)$ Now, find the sum $ \langle y \rangle = \dfrac{2y}{2}(1+2y)$ $ \implies \langle y \rangle = y(1+2y) $ $ \implies \langle y \rangle = y + 2y^{2}\rightarrow(1)$ Now find the value of $ \langle 3 \rangle = 3 + 18 = 21$ and $ \langle 2 \rangle = 2 + 8 = 10$ The value of $ \langle 3 \rangle \times \langle 2 \rangle = 21 \times 10 = 210$ Option $(B)$ is the correct answer. Lakshman Bhaiya answered Jul 4, 2019 Lakshman Bhaiya comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Option B is the correct answer <y> is simply the sum of the natural numbers till 2y, starting from 1 i.e. <y> = 1+2+3+4+..........+2y so <3> = 1+2+3+........+2*3 = 1+2+3+4+5+6 = 21 and <2> = 1+2+3+4 = 10 hence <3>*<2> = 21*10 = 210 Rudr Pawan answered Sep 14, 2019 Rudr Pawan comment Share Follow See all 0 reply Please log in or register to add a comment.