Edit MetaData
1 year, 6 months ago
1 year ago
Suppose that sunita travelled 'x' distance in 'y' time (with 450 speed) and 33.3-x distance in 60-y time
Now time=Distance/Speed
So,
Y=x/450 ( eqn 1) &
60-y= 33300-x/630 bcoz distance is in km which you have to convert to meters and speed is 1.4 times initial
Now replace the Y in eqn 2 and solve to get x=11250 m=11.25 km
1 year, 4 months ago
shivani jee, mujhe engilsh samjh me nhi aata hai hindi medium me sawal bhejiye
1 year, 6 months ago
I'm going to solve this question using the mixtures method.
First part speed=450m/min=27 km/hr
Second part speed = 37.8 km/hr
27 37.8
\ /
33.3
/ \
4.5 6.3
Ratio of distance travelled with speed 27kmph and 37.8 kmph is 4.5:6.3
4.5x+6.3x=33.3 km
x=33.3/10.8
Distance travelled at lower speed (27kmph or 450m/min) = 4.5x = 33.3*4.5/10.8 = 13.875 km
At higher speed = 6.3x = 6.3*33.3/10.8 = 19.425 km
Answer is 13.875 km.
Quick, Easy and Effective Revision
By proceeding you agree to create your account
Free CAT Formulae PDF will be sent to your email address soon !!!