1. Two cars started simultaneously toward each other from town A and B, that are 480 km apart. It took the first car traveling from A to B, 8 hours to cover the distance and the second car traveling from B to A, 12 hours. Determine at what distance from A the two cars meet.
A.288 km
B.200 km
C.300 km
D.196 km
E.None of these
Sol. Speed of 1st Car= (480 / 8) km/hr = 60 km/hr
Speed of 2nd Car= (480 / 12) km/hr = 40 km/hr
Time taken to meet= (480 / 100) hours = 4.8 hours
Distance from A where the two cars will meet = (4.8 × 60)km = 288km
2. How many motor vehicle registration number plates can be formed with the digits 1, 2, 3, 4, 5 (no digits being repeated) if it is given that registration number can have 1 to 5 digits?
A.100
B.120
C.205
D.325
E.None of these
Sol.If number is 1 digit number, Number of ways= 5.
If number is 2 digit number, Number of ways=5×4=20
If 3 digit number, Number of ways= 5×4×3= 60
If 4 digit number, Number of ways=5×4×3×2=120
If 5 digit number, Number of ways=5×4×3×2×1=120.
By adding these we get the sum as 325.
3. A shopkeeper buys an article for Rs. 400 and marks it for sale at a price that gives him 80% profit on his cost. He, however, gives a 15% discount on the marked price to his customer. Calculate the actual percentage profit made by the shopkeeper.
A.62%
B.64%
C.53%
D.54%
E.None of these
Sol. Cost Price = 400.
In oredr to make a profit of 80%, the Mark Price should be 720.
But on giving a discount of 15%, the selling price became 612.
Hence profit = ( 612 - 400 ) = 212 and profit % = ( 212 / 400 ) x 100 = 53%.
4. How many new words can be formed with the word MANAGEMENT all ending in G?
A.10!/ [(2!)^4 – 1]
B.9!/(2!)^4
C.10!/(2!)^4
D.9!/[(2!)^4 – 1]
E.None of these
Sol. If the last place is filled with G, we are left with 9 letters which can be permuted in 9! ways.
But in the remaining 9 letters M, A, N and E repeat 2 times each.
Therefore, number of all words ending in G = 9! / ( 2! )^4.
5. If a team of four persons is to be selected from 8 males and 8 females, then in how many ways can the selections be made to include at least one female.
A.3500
B.875
C.1200
D.1750
E.None of these
Sol. If we subtract the cases in which all are males, we will be left with the cases where atleast one female is selected.
Therefore, the total ways in which 4 members can be selected from 16 is 16C4 and total ways in which 4 males can be selected is 8C4. So required ways to select at least one female is 16C4 - 8C4 = 1750.
6. A bag contains 3 red, 6 white and 7 black balls. Two balls are drawn at random. What is the probability that both are black?
A. 1/8
B. 7/40
C. 12/40
D. 13/40
E. None of these
Sol. The reqd. probability is 7C2 / 16C2 = 7 / 40.
7. A 65 feet long ladder is leaning against a straight wall. Its lower end is 25 feet from the bottom of the wall. If the upper end is moved down by 8 feet, how much further away will the lower end of the ladder be from the base of the wall?
A.8 feet
B.10 feet
C.14 feet
D.52 feet
E.None of these
Sol. Height =[(65)^2-(25)^2]^0.5 = 60;
Now when it moves down 8feet.
distance from lower end of wall to ladder =[(65)^2-(52)^2]^0.5 = 39;
Therefore the lower end moves furthur ( 39 - 25 ) = 14 meters.
8. Three groups of children contain 3 girls and 1 boy, 2 girls and 2 boys, and 1 girl and 2 boys respectively. One child is selected at random from each group. The probability that all the three selected are girls is
A.3/8
B.1/8
C.5/8
D.2/3
E.None of these
Sol. The only case wil be = (3/4)*(2/4)*(1/3) = 1/8
9. X can do a piece of work in 10 days, Y in 12 days and Z in 15 days working independently. They all start the work together, but X leaves after 2 days and Y leaves 3 days before the work is completed. In how many days is the work completed?
A.5 days
B.6 days
C.7 days
D.8 days
E.None of these
Sol. Work done per day when all three work together = (1 / 10) + (1 / 12) + (1 / 15) = (6 + 5 + 4) /60 = 15 / 60 = 1 / 4.
In 2 days total work done by them= 2 / 4 = 1 / 2,
In last 3 days work is only done by Z. Therefore, piece of work done = 3 / 15 = 1/5.
Remaining work = 1 -(1 / 2) - (1 / 5) = 3 / 10 which is done by Y and Z in 2 days.
Total number of days in which work is completed = (2 + 3 + 2)= 7days.
10. Two pipes can fill a cistern in 4 and 6 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom of the cistern, it takes 36 minutes extra for the cistern to be filled up. When the cistern is full, in what time will the leak empty it?
A. 12
B. 10
C. 15
D. 16
E. None of these
Sol. Work done by both pipes in a hour = (1/4)+(1/6) = 5/12 . So both pipes can fill the tank in 12/5 hours.
But because of the leak time taken by them is (12/5) + (36/60)=3 hours
Considering the leak takes x hours to empty the tank, we get;
(1/4) + (1/6) – (1/X) = 1/3.
On solving we get X = 12 hours .
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