Question 1
Location of B is North of A and location of C is East of A. The distances AB and AC are 10 km and 24 km. To go from B to C, how far and in what general direction does a person need to go?
(A) 26 Km; North-West
(B) 26 Km; South-East
(C) 26 Km; South-West
(D) 26 Km; North-East
Answer the questions that follow based on the passage given below:
Ten people namely Litesh, Pawan, Neetu, Parul, Dinesh, Sheema, Anil, Dharam, Dolly and Sheela go for a magic show and sit on seats numbered 1 to 10. There are three couples amongst them. Each couple has only one child, who always sits next to its mother. A family always sits together. Sheela, Dolly, Parul and Sheema are females while the others are males.
– Dinesh sits on seat number 6 which is immediately next to Nitu’s mother’s seat.
– Sheela sits on the seat whose number is both a perfect square and a perfect cube.
– Dharam is Dolly’s father, and they both sit in prime numbered seats.
– Litesh, the bachelor sits next to Pawan.
– The children are Neetu, Dolly and Dinesh.
Question 2
Who is Dinesh’s mother, if the person sitting two places away from the person sitting immediately next to Neetu is Parul?
(A) Sheema
(B) Sheela
(C) Parul
(D) Cannot be determined
Question 3
Who is siting five seats away from the person who is sitting two places away from Dharam’s wife?
(A) Litesh
(B) Pawan
(C) Neetu
(D) Cannot be determined
Question 4
What is Anil’s seat number?
(A) 4
(B) 5
(C) 9
(D) 10
Question 5
Apart from Pawan, who else is sitting next to Litesh?
(A) Dolly
(B) Dharam
(C) Pawan
(D) Nobody
N.B. Please note that the official timings for publication of these questions is 11 am daily.
SOLUTIONS TO DAILY CSAT MISSION # 77
1. (B) 2. (B) 3. (A) 4. (B) 5. (D)
Explanations
1. Distance covered by the man in four minutes = 3 x (1000/60) x 4 = 200 metres.
So, distance travelled by the carriage in four minutes = 200 m + 100 m = 300 m.
Therefore the speed of the carriage = 300/4 m/min = 300/4 x 60/1000 kmph = 4.5 kmph.
2. Let him take ‘t’ minutes to cover 40% of the journey. So time taken to cover 60% of the journey = t + 10.
So, (t + 10)/t = 60/40, which gives t = 20.
Total time for the journey = t + (t + 10) = 50 minutes = 5/6 hours. Speed = 48 km/hr. So, the distance = 48 x 5/6 = 40 km.
3. Let the total journey be 120 km long.
So, with stoppages, it takes 3 hours to complete the journey (because avg speed is 40).
Without stoppages, it would take 2 hours to complete the journey (because avg speed is 60).
From this, we can conclude that in the three hour journey, it would have spent 1 hour in halts. So, time per hour spent in halts = 1/3 of an hour = 20 minutes.
4. Let ‘t’ minutes be the time he takes if he arrives at exactly the correct time for the class.
At 5 kmph, he arrives 3 minutes later i.e. time taken = (t + 3) minutes.
At 6 kmph, he arrives 7 minutes earlier i.e. time taken = (t – 7) minutes.
Since the distance remains the same, 5(t + 3) = 6 (t – 7), which gives t = 57 minutes.
At, 5 kmph, he takes 60 minutes (1 hour) to reach the class. So, distance = 5 km.
5. Let the time Rahul takes in catching up to Aryan be ‘t’ minutes.
In this time, at 50 m/min, Rahul has covered 50t metres.
And, Aryan has ran for five more minutes. So Aryan has covered 40(t+5) metres.
But since Rahul has caught up, both must have covered the same distance from the starting point.
So, 50t = 40(t+5) which gives t = 20 minutes.
In 20 minutes at 60 m/min, the dog must have covered 1200 metres.
Comments
6 responses to “CSAT DAILY MISSION #76”
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1.B
2.D
3.C
4.A
5.D -
B..C..C..A..D
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bacad
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plz pls plz sir/madam provide with paper 1 gs questions now Its most imp so my humble request u start GS questions that we can improve
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BCBAD
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b a c a d
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