find distance only Practice Questions Answers Test with Solutions & More Shortcuts

Question : 1 [P.N.B. (PO) 2010]

Mohan walked 30 m towards South, took a left turn and walked 15 m. He, then took a right turn and walked 20 m. He again took a right turn and walked 15 m. How far is the from the starting point?

a) 95 m

b) 50 m

c)  70 m

d) Cannot be determined

e) None of these

Answer: (b)

According to the question, the direction diagram will be as follows.

Therefore, Required distance, PT = PQ + QT = 30 + 20 = 50 m.

Question : 2

Ankit started walking towards North. After walking 30 m, he turned towards left and walked 40 m. He, then turned left and walked 30 m. He again turned left and walked 50 m. How far is the from his original position?

a) 40 m

b) 50 m

c) 30 m

d) 10 m

e) 12 m

Answer: (d)

According to the question, the direction diagram will be as follows. Therefore, Required distance, PT = ST - SP = 50 - 40 = 10 m.

Question : 3 [M.P. (PCS) 2008]

Molly travelled from point A to point B which is 5 ft. He, then travelled 6 ft to his right and then turned to left and went 4 ft. Finally, he again went 6 ft to his left. How far is he from the point B now?

a) 4 ft

b) 10 ft

c) 6 ft

d) 5 ft

Answer: (a)

According to the question, the direction diagram will be as follows. Therefore, Required distance, BE = 4 ft.

Question : 4

Amol travels a distance of 10 ft from A to B. Then, he turns to left and travels 5 ft and again turns to left and travels 2 ft. Finally, he turns to left and moves 5 ft. How far is he now from his original position?

a) 5 ft

b) 6 ft

c) 8 ft

d) 12 ft

e) 10 ft

Answer: (b)

According to the question, the direction diagram is as follows. A = Original position,

E = Final position AB = 10 ft, BC = DE = 5 ft CD = EB = 2 ft

Therefore, Required distance, AE = AB - EB = 10 -2 = 8 ft.

Question : 5 [S.B.I. (Clerk) 2011]

Shyam walks 6 m towards East, then turns right and walks 9 m. Again, he turns to his left and walks 6 m. At what distance is he now from his original point?

a) 15 m

b) 21 m

c) 18 m

d) Cannot be determined

e) None of these

Answer: (a)

According to the question, the direction diagram is as follows. O is the mid - point of BC and

AD OB = OC = $9/2$ = 4.5 m

OA = $√{(AB)^2 + (OB)^2}= √{(6)^2+ (4.5)^2}$

= $√{(36 + 20.25)} = √{(56.25)}$ = 7.5 m

OD = OA = 7.5

Therefore, Required distance, AD = AO + OD = 7.5 + 7.5 = 15 m.

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