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Budhan covers a distance of $19$ km in $2$ hours by cycling one fourth of the time and walking the rest. The next day he cycles (at the same speed as before) for half the time and walks the rest (at the same speed as before) and covers $26$ km in $2$ hours. The speed in km/h at which Budhan walk is

  1. $1$
  2. $4$
  3. $5$
  4. $6$
in Numerical Ability by Veteran (60.6k points)
edited by | 244 views
Migrated from GO Civil 8 months ago by Arjun

4 Answers

+2 votes
Best answer
Let the speed by cycling be $X$ km/h and speed of walking be $Y$ km/h

So, $(1/2)X + (3/2)Y = 19\qquad \to (1)$

And, $X + Y = 26\qquad \to (2)$

Solving (i) & (ii) we get, $X = 20$ and $Y = 6$

Thus, the speed in km/h at which Budhan walk is $6.$

Correct choice: option "D."
by Boss (15.5k points)
selected by
+2 votes

Let us suppose speed of cycling is a km/h and walking is b km/h.

so, 

19 = 0.5a + 1.5 b

26 = a + b

on solving these two equation we get,

b=6 km/h

so answer is option D.

by (153 points)
0 votes

Distance=speed * Time

Let cycle speed is X  and walking speed is Y

19=(x/4+3y/4)

x+3y-38=0...........{1}

26=(x/2+y/2)

x+y+26=0..............{2}

On solving 1&2

Y=6

by (93 points)
0 votes
Let cycling speed be $c$ and walking speed be $w$.

First equation would be : $c \times \frac{1}{4}\times 2$ + $w \times \frac{3}{4}\times2$=$19$

i.e $c \times \frac{1}{2}$ + $w \times \frac{3}{2}$=$19$

i.e $c+3\times w=38$

Second equation would be $c \times \frac{1}{2}\times 2$ + $w \times \frac{1}{2} \times 2 $=$26$

i.e $c+w=26$

Therefore from both eqns. we have :

$26-w+3w=38$ $\implies 26+2w=38 \implies 2w=12 \implies w=6$

Therefore walking speed is 6.
by Active (2.4k points)
0
Is it told $2$ hours for cycling and $2$ hours for walking? It is told only $2$ hours by cycling.

right?
0
Total time is 2 hours.
Answer:

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