Question
56. (3) ( 1.49=frac{14 cdot 9}{10}, 159=frac{5.9}{10} )
यदि ( 14.9=a ) एवं ( 5.1=b )
तब,
व्यंजक ( =frac{1}{10}left(frac{a^{2}-b^{2}}{a-b}right)=frac{1}{10}(a+b) )
( =frac{1}{10}(14.9+5.1) )
( =frac{1}{10} times 20=2 )

1:49x14.9-0.51x5.1 50. 14.9-5.1 - बराबर है। । (1) 0-20 (2) 20-00 (3) 2:00 (4) 22-00
Solution
