- Two persons appear in an examination for the same post. The probability of first person selected is 1/4 and the probability of second person selected is 1/2. What is the probability that only one of them is selected?
1/41/22/32/5NoneOption B
Solution:
P(1st person selected)=1/4, P(1st person not selected)=3/4
P(2nd person selected)=1/2, P(2nd person not selected)=1/2
Then probability of only one of them is selected=(1/4*1/2) + (3/4*1/2)
=1/8 +3/8
=4/8 ==>1/2. - A and B together can do a piece of work in 30 days. A having worked for 16 days, B finishes the remaining work alone in 44 days. In how many days shall B finish the whole work alone?
30days45days60days40daysNoneOption C
Solution:
Let A’s 1 day’s work = x and B’s 1 day’s work = y.
Then x+y =1/30 and 16x+44y=1
Solving both we get x=1//60, y=1/60.
So B complete the work alone=60days.
- A boat takes 9/4 times as long to row a distance upstream as to row the same distance downstream. If it takes 3 hours to cover a distance of 54 km downstream, what is the still speed of the boat?
11.5km/hr12km/hr13km/hr13.5km/hrNoneOption C
Solution:
Downstream speed=54/3=18km/hr
Upstream speed=54/(3*9/4)=8km/hr
Speed of the boat=(18+8)/2=26/2
=13km/hr.
- A bookseller buy 30 books for Rs. 2400 and sells them at a profit equal to the selling price of 6 books. What is the selling price of two dozen books, if the price of each book is same?
Rs2000Rs2400Rs2600Rs2250NoneOption B
Solution:
Cost price = 2400 /30 = Rs. 80.
Now,
Selling Price of 30 books = CP of 30 books + SP of 6 books.
SP of 24 books = Rs. 2400
SP of 1 book = 2400 /24 = Rs 100.
then Selling price of two dozen 24 book = 24*100 = Rs. 2400.
- The mean weight of 100 students in a class is 46 kg. The mean weight of boys is 50 and of girls is 40 kg. Therefore, the number of boys is:
45405560NoneOption D
Solution:
Let number of boys are x and then number of girls = (100-x).
Thus,
50x+(100-x)*40 = 46*100
==>x = 60.
Number of boys = 60.
- In Sham’s opinion , his weight is greater than 45kg but less than 52kg. His brother think that sham’s weight is greater than 40 kg but less than 50kg. His mother’s view is that his weight cannot be greater than 48kg. If all of their assumption is correct, then what is the average of different probable weight of sham?
42kg45kg47kg51kgNoneOption C
Solution:
Let Sham’s weight be xkg.
Sham’s assumption 45<x<52
His brother ‘s assumption 40<x<50
His mother’s assumption x<48
In all assumption’s common weight is 46, 47, 48.
Then its average=(46+47+48)/3=47kg.
- Machine A can print two lakh books in 8 hours. Machine B can print the same number of books in 10 hours while machine C can print the same in 12 hours. All the machines started printing at 7 A.M. Machine A is stopped at 9 A.M. and the remaining two machines complete work. Approximately at what time will the printing of two lakh books be completed?
12pm10am10.30am11amNoneOption D
Solution:
A’s work in 1 hr = 1/8
B’s work in 1 hr = 1/10
C’s work in 1 hr = 1/12
Work done by A,B and C in 1 hr = 1/8 + 1/10 + 1/1
= 37/120
Work done by A and C in 1 hrs = 1/10 + 1/12 = 22/120 =11/60.
From 7 am to 9 am
all machines worked for 2 hrs= 2 × (37/120) = 37/60
remaining work = 1- 37/60 = 23/60
then hrs taken by B and C to complete the remaining work = (23/60) / (11/60) = 23/11 =2.02
Hence the work will be completed approximately 2 hours ie around 11 am.
- A dishonest shopkeeper, at the time of selling and purchasing, weights 10% less and 20% more per kilogram respectively. Find the percentage profit earned by him. (Assuming he sells at Cost Price)
40%32.5%28.65%33.33%NoneOption D
Solution:
Let the CP be Rs100.
While purchasing he would take 1200 grams for the price of 1000 grams. While selling he would sell 900 grams for the price of 1000 grams.
Since, CP = SP
The profit earned by him,
= (300 *100)/900 = 33.33%.
- How many words can be formed by re-arranging the letters of the word CEMENT such that C and T occupy the first and last position respectively?
12162410NoneOption A
Solution:
As C and T should occupy the first and last position, the first and last position can be filled in only one following way.=1!
The remaining 4 positions can be filled in 4! Ways.
Hence by rearranging the letters of the word CEMENT we can form,
1×4!= 4! =24ways.
E is repeated =24/2!=12ways.
- Pipe A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened to fill a tank. The time when the tank be full, it was found that a leak pipe was also opened along with pipes A and B. It is turned off immediately and then after 10 minutes, the tank gets full. Find the time in which the leak pipe can empty the full tank?
360/49 mins320/41 mins380/49 mins2900/41 minsNoneOption A
Solution:
Pipes A and B can fill the tank = 15*20/(15+20)
= 60/7 mins
Let leak pipe can empty tank in x mins
A and B were opened for (60/7) + 10 = 130/7 mins
And the leak was opened for 60/7 mins
So (7/60) * 130/7 – (1/x) * 60/7 = 1
x = 360/49 mins.
For Q9 ans should be 12 na?
S rit letter E is repeated so we have to divide it by 2!.
Ok thnq :))
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