Let the original number be x.According to the question we can write as (2/3)x+20x
Let the original number be x.According to the question we can write as (2/3)x+20xOn rearranging x−(2/3)x=20
Let the original number be x.According to the question we can write as (2/3)x+20xOn rearranging x−(2/3)x=20Now taking the L.C.M od 1 and 3 is 3
Let the original number be x.According to the question we can write as (2/3)x+20xOn rearranging x−(2/3)x=20Now taking the L.C.M od 1 and 3 is 3(3x−2x)/3=20
Let the original number be x.According to the question we can write as (2/3)x+20xOn rearranging x−(2/3)x=20Now taking the L.C.M od 1 and
Let the original number be x.According to the question we can write as (2/3)x+20xOn rearranging x−(2/3)x=20Now taking the L.C.M od 1 and 3 is 3(3x−2x)/3=20x/3=20Again by transposing x=60So the original number is 60
Answer:
60
Step-by-step explanation:
1-2/3=20
1/3=20
1= 1/3*3=20*3=60
Pls mark me brainliest
✿*゚¨゚✎・ ✿.。.:* *.:。✿*゚¨゚✎・✿.。.:* ANSWER♡LOVE♡ ✿*゚¨゚✎・ ✿.。.:*❄
Step-by-step explanation: