PASASAGE:
An airline has a certain free luggage allowance & charges for excess luggage at a fixed rate per kg. Two passengers Raja & Praja have 60 kg of luggage between them, and are charged Rs.1200 & Rs.2400 for excess luggage. Had the entire luggage belonged to one of them , the excess luggage charge would have been Rs. 5400.
Q.1 What is the weight of the Praja’s luggage ?
A) 20 kg
B) 35 kg
C) 40 kg
D) 30 kg
Ans. B
Let free luggage allowed be x kg
Raja & Praja are allowed free luggage of 2xkg
Let charge for excess luggage= Rs. Y /kg
Raja & Praja pay Rs. 1200 + Rs. 2400 = 3600 excess luggage
Y(60-2x) = 3600–i
If entire luggage had belonged to one of them , excess luggage charges would have been Rs.5400
Y (60-x) = 5400—ii
From these two equations:
60y- 2xy = 3600
60y – xy =5400
Now. Solve these two equations;
Y = 120
X = 15 kg
Since Praja is charged Rs.2400 for excess luggage is 2400/120 = 20kg
Hence weight of Praja’s luggage + her excess luggage
= 15 + 20kg
=35 kg
Q.2 What is the free luggage allowance per person ?
A) 10 kg
B) 15 kg
C) 20 kg
D) 30 kg
Ans. B
Q.3 Two boys begin together to write out a booklet containing 8190 lines. The first boy starts with the first line , writing at the rate of 200 lines an hour & the second boy starts with the last line, then writes 8189th line & so on, proceeding backward at the rate of 150 lines an hour. At what line will they meet ?
A) 4910
B) 4680
C) 4280
D) 5200
Ans. B
Relative speed = 200 + 150 = 350 lines/hr
So, they will meet after =8190 / 350 = 117/5 hrs
Therefore, they will meet at 117/5*200 = 4680th line from the beginning
Q.4 If 20% of P = 30% of Q = 1/6 of R, then P:Q:R is:
A) 3 : 2: 16
B) 2 :3 : 16
C) 15 : 10 : 18
D) 10 : 15 : 18
Ans. C
PQR=1/5:3/10:1/6
=15:10:18
Q.5 A local train is going from Delhi to Saharanpur & it has 4 stoppages in between. If two passengers enter in the train during the journey , how many different sets of tickets they may had ?
A) 90
B) 120
C) 45
D) 180
Ans. C
1st station = 4tickets
2nd station = 3 tickets
3rd station = 2 tickets
4th station = 1 ticket
Total no. of tickets = 4+3+2+1 = 10
Therefore, sets of ticket they may had = 10C2 = 45
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