Two numbers are less than a third number by 35% and 48% respectively. Find the percent by which the second number is less than the first.
Let the third number be 100. Then, first number will be = 100 – 35 = 65 And the second number will be = 100 – 48 = 52 Required % = [(65 – 52)/65]×100 = 20%
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int main(){
int i;
for(i=0;i
int i;
i=20;
print(“%d”, i);
}
return 0;
}
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