Let us assume the one adult fare be a and one child fare be c.
According to question,
Bus fare of one Adult is six times the fare of one child.
a = 6c ........................(1)
If an adult's bus fare is 114.
a = 114
put the value of a in equation (1). We will get
6c = 114
? c = 114/6
? c = 19
Now we have to calculate amount paid by 4 adults and 5 children for Bus fare
Bus fare of 4 adults and 5 child = 4a + 5c
Put the value of a and c in above equation. We will get
? Bus fare of 4 adults and 5 child = 4 x 114 + 5 x 19
? Bus fare of 4 adults and 5 child = 456 + 95
? Bus fare of 4 adults and 5 child =?/- 551
47/10000 = .0047
.02 =(2/100) x 100% =2%
Let N / 11 = 233
Then, N = 233 x 11 = 2563
? Missing digit is 5.
121012 = 12 x 10084 + 4
? remainder = 4
We have the important relation, More work, More time (days)
? A piece of work can be done in 6 days.
? Three times of work of same type can be done in 6 x 3
= 18 days
? = 750.0003 ÷ 19.999
? ? ? 750 ÷ 20
? ? ? 375 ? 38
Subtract 20, 25, 30, 35, 40, 45 from successive numbers. So 0 is wrong.
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