let us assume the ten's digit number be x and unit's place digit be y.
Then two digit Original Number = 10x + y
According to question ? y = 2x - 1 ................(1)
When digits are interchanged then new number = 10y + x
If the digits at unit's and ten's place are interchanged, the difference between the new and the original number is less than the original number by 20,
New Number - Original Number = Original number - 20
? 10y + x - (10x + y) = 10x + y - 20
? 10y + x - 10x - y - 10x - y = - 20
? 8y - 19x = - 20
? 19x - 8y = 20 ...................(2)
Put the value of y from equation (1) in above equation (2).
? 19x - 8(2x - 1) = 20
? 19x - 16x - 8 = 20
? 7x = 20 + 8
? 7x = 28
? x = 28/7
? x = 4
Put the value of y in equation (1)
From (i) y = 2 x 4 -1 = y = 7
? original number = 10x + y =10 x 4 + 7 = 47.
Given that, 30 % of A = 20 % of B
? A/B = 20/30 = 2/3
? A : B = 2 : 3
Let initial quantity be Q, and final quantity be F
F = Q(1 - 8/Q)
=> Q = 20
? log5[(x2 + x ) / x] = 2
? log10(x + 1) = 2
? x + 1 = 25
? x = 24
∴ Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).
We know that speed is inversely proportional to time.
Given that, (Speed of A ) : (speed of B ) = 2 : 7
?(Time taken by A ) : (Time taken by B ) = 1/2 : 1/7 = 7 : 2
Let th distance = D
and usual speed = V
According to the question,
D/(3V/4) - D/V = 2
? D/V = 3 x 2 = 6
Time taken to cover the distance with usual speed = 6 h
?2n = 64
? 2n/2 = 26
? n/2 = 6
? n = 12
Distance = 160 km
Relative Speed = 8 + 2 = 10
Time = Distance/Relative speed = 160/10 = 16 h
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