Let A, B, C and D pay ?x, ?y, ?z and ?a
According to the question,
x = 1/2(y + z + a)
2x = (y + z + a) ..........(i)
y = 1/3(x + z + a)
3y = (x + z + a) ...........(ii)
z = 1/4(x + y + a)
4z = (x + y + a) ..........(iii)
Also, x + y + z + a = 60 ...............(iv)
Now, put the value of x + y + a = 4z from (iii) in (iv)
Then, 4z + z = 60 ? 5z = 60
? z = 12
Similarly, on putting the value of x + z + a = 3y from (ii) in (iv), we get
3y + y = 60 = ? 4y = 60
? y = 15
Again, on putting the value of (y + z + a) = 2x from (i) in (iv), we get
2x + x = 60 ? 3x = 60
&ther4; x = 20
Now, x + y + z + a = 60
On putting the value of x, y and z, we get
12 + 15 + 20 + a = 60
? a = 60 - 47 = ? 13
.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.
NA
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