Given interchanges- Sign- displaystyle div quad andquad - and numbers 4 and 2displaystyle 4-2div 1-frac - 3 - 2 - displaystyle 2-4div 3-6displaystyle 4-2div 3-3displaystyle 2-4div 5-8

Question

Toppr Admin · Accepted Answer

On interchanging-xA0-xF7-and- and 4 and 2 in -a- we get the equation as2-xF7-4-1-32or24-1-32or32-32 which is true-

Given interchanges: Sign: $\div a n d +$ and numbers 4 and 2
$4 + 2 \div 1 = \frac{3}{2}$
$2 + 4 \div 3 = 6$
$4 + 2 \div 3 = 3$
$2 + 4 \div 5 = 8$

On interchanging $\div a n d +$ and 4 and 2 in (a), we get the equation as
$2 \div 4 + 1 = \frac{3}{2} o r \frac{2}{4} + 1 = \frac{3}{2} o r \frac{3}{2} = \frac{3}{2}$ which is true.

Given interchanges: Sign: ÷and+ and numbers 4 and 24+2÷1=322+4÷3=64+2÷3=32+4÷5=8

On interchanging ÷and+ and 4 and 2 in (a), we get the equation as2÷4+1=32or24+1=32or32=32 which is true.

Given interchanges: Sign: $\div a n d +$ and numbers 4 and 2
$4 + 2 \div 1 = \frac{3}{2}$
$2 + 4 \div 3 = 6$
$4 + 2 \div 3 = 3$
$2 + 4 \div 5 = 8$

On interchanging $\div a n d +$ and 4 and 2 in (a), we get the equation as
$2 \div 4 + 1 = \frac{3}{2} o r \frac{2}{4} + 1 = \frac{3}{2} o r \frac{3}{2} = \frac{3}{2}$ which is true.