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- Question
41, 43, 47, 53, 61, 71, 73, 81
Options- A. 61
- B. 71
- C. 73
- D. 81
- Correct Answer
- 81
ExplanationEach of the numbers except 81 is a prime number. - 1. A man goes to the fair in Funcity with his son and faithful dog. Unfortunately man misses his son which he realises 20 minutes later. The son comes back towards his home at the speed of 20 m/min and man follows him at 40 m/min. The dog runs to the son (child) and comes back to the man (father) to show him the direction of his son. It keeps moving to and fro at 60 m/min between son and father, till the man meets the son. what is the distance travelled by the dog in the direction of the son?
Options- A. 800 m
- B. 1675 m
- C. 848 m
- D. 1000 m Discuss
- 2. Den Bosh and Eastbourne are two famous cities 300 km apart. Maradona starts from Den Bosch at 8 : 24 am. An hour later Pele starts from Den Bosch. After travelling for 1 hour, Pele reaches Nottingham that Maradona had passed 40 minutes earlier. Nottingham falls on the way from Den Bosh to Eastbourne at the same time, what is the speeds of the time Maradona and Pele respectively?
Options- A. 100 km/h, 125 km/h
- B. 60 km/h, 80 km/h
- C. 60 km/h, 75 km/h
- D. 75 km/h, 100 km/h Discuss
- 3. A walks at a uniform rate of 2 km/h and 2 h after his start, B cycles after him at the uniform rate of 5 km/h. How far from the starting point will B catch A?
Options- A. 62/3 km
- B. 61/3 km
- C. 65/7 km
- D. 63/7 km Discuss
- 4. A man, on tour travel first 160 km. at 64 km/hr. and the next 160 km. at 80 km/hr. The average speed for first 320 km. of the tour is?
Options- A. 35.55 km/hr
- B. 71.11 km/hr
- C. 36 km/hr
- D. 72 km/hr Discuss
- 5. Two ladies simultaneousely leave cities A and B connected by a strainght road and travel towards each other. The first lady travels 2km/h faster than the second lady and reaches B one hour before the second lady reaches A. The two cities A and B are 24 km apart. How many kilometres does each lady travel in one hour?
Options- A. 5 km,3 km
- B. 7 km, 5 km
- C. 8 km, 6 km
- D. 16 km, 14 km Discuss
- 6. A boy walking at a speed of 15 km/h reaches his school 20 min late. Next time he increases his speed by 5 km/h but still he late by 5 min. Find the distance of the school from his home.?
Options- A. 5 km
- B. 10 km
- C. 15 km
- D. 20 km Discuss
- 7. A person covers half of his journey at 40 km/h, one-third at 60 km/h and rest of journey at 30 km/h. Find his average speed for the whole journey.?
Options- A. 45 km/h
- B. 42. 35 km/h
- C. 50 km/h
- D. 40 km/h Discuss
- 8. If Sohail walks from his home to office at 16 km/h, he is late by 5 min. If he walks at 20 km/h, he reaches 10 min before the office time. Find the distance of his office from his house.
Options- A. 22 km
- B. 20 km
- C. 18 km
- D. 16 km Discuss
- 9. A takes 4 h more than the time taken by B to walk D km. If A doubles his speed, he can make it in 2 h less than that of B. How much time does B require for walking D km?
Options- A. 8 h
- B. 4 h
- C. 6 h
- D. 9 h Discuss
- 10. Nilu covers a distance by walking for 6 h. While returning, his speed decreases by 2km/h and he takes 9 h to cover the same distance. What was her speed while returning?
Options- A. 2 km/h
- B. 5 km/h
- C. 4 km/h
- D. 7 km/h Discuss
More questions
Correct Answer: 1000 m
Explanation:
In 20 minutes the difference between man and his son = 20 x 20 = 400 m
Distance travelled by dog when he goes towards son = (400/40) x 60 = 600 m and time required is 10 minutes
In 10 minutes the remaining difference between man and son = 400 - (20 x 10) = 200 m
Note: Relative speed of dog with child is 40 km/h and the same with man is 100 km/h.
Time taken by dog to meet the man = 200/100 = 2 min
In 2 minute the remaining distance between child and man = 200 - ( 2 x 20) = 160 m
Now, the time taken by dog to meet the child again = 160/40 = 4 min
In 4 minutes he covers 4 x 60 = 240 m distance while going towards the son.
In 4 minute the remaining distance between man and child = 160 - (4 x 20) = 80 m
Time required by dog to meet man once again = 80/100 = 0.8 min
In 0.8 min remaining distance between man and child = 80 - (0.8 x 20) = 64 min
Now, time taken by dog to meet the child again = (64/40) x (8/5) min
? Distance travelled by dog = (8/5) x 60 = 96 m
Thus, we can observe that every next time dog just go 2/5th of previous distance to meet the child in the direction of child.
So, we can calculate the total distance covered by dog in the direction of child with the help of GP formula.
Here, first term (a) = 600 and common ratio (r) = 2/5
Sum of the infinite GP = a/(1 - r)
= 600/(1 - 2/5) = (600 x 5)/3 = 1000 m
Correct Answer: 75 km/h, 100 km/h
Explanation:
Let pele covers D km in 1 hours. So Maradona takes (2 h - 40 min) = 1 h 20 min to cover D km.
Let speed of Maradona and pele be M and P respectively than.
D = M x 4/3 and D = P x 1
M/P = 3/4
Again, 300/M - 300/P = 1
300/3k - 300/4k = 1
k = 25
M = 3k = 75 km/h and
P = 4k = 100 km/h
Correct Answer: 62/3 km
Explanation:
Let B catches A after T h.
Then, distance travelled by A in (T + 2) h = Distance travelled by B in T h
According to the question,
2 (T + 2 ) = 5T
? 2T + 4 = 5T
? 3T = 4
? T = 4/3 h
Distance travelled by B in 4/3 h
= (4/3) x 5 = 20/3 = 62/3 km
Correct Answer: 71.11 km/hr
Explanation:
Average speed = 2uv / ( u + v ) km/hr
=(2 x 64 x 80) / (64 + 80) km/hr
= (2 x 64 x 80) / 144 km/hr.
= 71.11 km/hr
Correct Answer: 8 km, 6 km
Explanation:
Let the speed of first lady be (v + 2)km/h and speed of second lady will be v km/h.
Total distance = 24 km
For first lady, Speed = Distance/Time
v + 2 = 24/t ...(i)
Let time taken by first lady to cover distance of 24 km be t h.
Then, time taken by second lady to cover same distance will be(t + 1)h.
For second lady, Speed =Distance/time
v = 24/(t + 1) ...(ii)
Form Eqs.(i) and (ii), we get
24/t - 2 = 24/(t + 1)
? (24 - 2t)/t = 24/(t + 1)
? (24 - 2t)(t + 1) = 24t
? 24t - 2t2 + 24 - 2t = 24t
? 2t2 + 2t - 24 = 0
? t2 + t - 12 = 0
? t2 + 4t - 3t - 12 = 0
? t(t + 4) -3(t + 4) = 0
? (t - 3)(t + 4) = 0
? t = 3, -4
? t = 3 (t ?- 4)
? The distance travelled by the first lady in one hour
= 24/t = 24/3 = 8 km
And distance travelled by the second lady in one hour
= 24/t + 1
= 24/3 + 1 = 6 km
Correct Answer: 15 km
Explanation:
Let the distance = x
Here, the difference in time
= 20 - 5 = 15 min
= 15/60 = 1/4h
Speed during next journey
= 15 + 5 = 20 km/h
According to the question,
x/15 - x/20 = 1/4
? 4x - 3x/60 = 1/4
? x = 60/4 = 15
? x = 15 km
Correct Answer: 42. 35 km/h
Explanation:
Let total distance be 100 km.
Then, Average = 100/[ (50/40) + (30/60) + (20/30)]
= 100/( 5/4 + 1/2 + 2/3) = 100/[(15 + 6 + 8)/12]
= (100 x 12)/29
= 42.35
Correct Answer: 20 km
Explanation:
Let required distance = L
According to the question,
L/16 - L/20 = 15/20
? (5L - 4L)/80 = 1/4
? L = (1/4) x 80 = 20 km
Correct Answer: 8 h
Explanation:
Let B takes x h to walk D km.
Then, A takes ( x + 4 ) h to walk D km.
With doube of the speed,
A will take ( x+ 4)/2 h.
According to the question,
x - (x + 4)/2 = 2
? 2x - (x + 4) = 4
? 2x - x - 4 = 4
? x = 4 + 4 = 8 h
Correct Answer: 4 km/h
Explanation:
Let the speed in return journey = x
According to the question,
6( x + 2 ) = 9x
? 6x + 12 = 9x
? 9x - 6x = 12
? 3x = 12
? x = 12/3 = 4km/h
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