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There is a number which is greater than the aggregate of its third. tenth and the twelfth parts by 58.Can u guess the number?
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- Let the number be x
It is given that number is greater than the aggregate of its third, tenth and twelvth parts by 58
X-(X/3+ X/10 + X/12) =58
take LCM
X-(20X + 6X + 5X)/60 = 58
(60X- 31X)/60 = 58
29X/60 =58
Now, X = (58 * 60)/29 = 60 *2 = 120.
Thus, the number is 120.
- 12 years agoHelpfull: Yes(18) No(2)
- let the no. is x
X-(X/3-X/10 - X/12) =58
X*(1-1/3-1/10-1/12)=58
x*(60-20-6-5)/60 = 58
x*(60-31)/60=58
x*29/60=58
x=58*29/60
x=2*60
x=120 - 6 years agoHelpfull: Yes(2) No(0)
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