Question

BRAIN TEASER!
Proper explanation needed.

There is a straight highway. Four different villages lie on that highway. The distance between them is different. The third village is 60km away from the first village; the fourth is 40 km away from the second; the third is 10 km near to the fourth that it is to the second. Can you calculate the distance between the fourth and the first village?

Answer 1

Let A,B,C and D be the four villages

Distance between A and B = 60 km (Given)
Distance between B and D = 40 km (Given)
Distance between B and C is greater than C and D by 10 km (Given)

Distance between B and C + Distance between C and D = Distance between B and D

BC + CD = BD
BC + (BC -10) = 40
BC + BC - 10 = 40
2BC = 40+10
2BC = 50
BC = 50/2 = 25

Therefore, CD = BC -10
                  CD = 25 -10 = 15

Distance between A and D = AC + CD
                                            = 60 + 15
                                            = 75 (Ans)



                 

Answer 2

Answer:

Concept: using the logical reasoning to solve this question.

Given: distance between third village and first village is 60km, fourth and second is 40km, distance of third and fourth  is 10km near  than that to the second and third.

To find: distance from first and fourth .

Step-by-step explanation:

Let A,B,C and D be the four villages.

Distance between A and B = $60$km .

Distance between B and D = $40$ km .

Distance between B and C is greater than C and D by $10$ km.  

Now, Distance between B and C + Distance between C and D = Distance between B and D.

$BC + CD = BD$

$BC + (BC -10) = 40$

$2BC = 40+10$

$2BC = 50$

$BC = 50/2 = 25$.

Therefore, $CD = BC -10$

                 $CD = 25 -10 = 15$.

Distance between A and D = $AC + CD$

                                           = $60 + 15$

                                           = $75$.

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