Question
‘M’ has 15% less balls than ‘N’. ‘N’ has 10%
more balls than ‘O’. By how much percent the number of balls with ‘M’ are less than those with ‘O’?Solution
ATQ,
Let, number of balls with ‘N’ = 11z
Number of balls with ‘M’ = 0.85 × 11z = 9.35z
Number of balls with ‘O’ = (11z/1.1) = 10z
So, required percentage = {(10z – 9.35z)/10z} × 100 = 6.5%
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