1. When four dice are thrown randomly, what is the probability that the number attained in each of them is the same?
1) 1/1296 b) 1/36 c) 1/216 d) 1/18 e) None of these
Ans. C
Same number = [1,1,1,1] or [2,2,2,2] or . . . . [6,6,6,6]
Total favorable cases = 6
Total cases = 6^4
Probability = 6/6^4 = 1/6^3 = 1/216
2. A can do a piece of work in 10 days and B can do the same piece of work in 20 days. They start the work together, but after 5 days A leaves B will do the remaining piece of work in :
1) 6 days 2) 5 days 3) 4 days 4) 10 days
Ans : A can do a piece of work in 10 days
Hence A’s work of one day = 1/10
B can do a piece of work in 20 days
Hence B’s work of one day = 1/20
A and B started work together and worked for 5 days
Hence their combined work of 5 days = 5/10 + 5/20 = ½ + ¼ = ¾
Remaining work after 5 days = 1 – ¾ = ¼
Now B alone work hence number of days that B will finish this ¼ piece of work = numbers of days to complete the work / part of work
= 20× ¼ = 5 days
3. A dealer offered a machine for sale
₹27500, but even if had charged 10% less, he would have made a profit of 10%. The actual cost of the machine is
1) 1/1296 b) 1/36 c) 1/216 d) 1/18 e) None of these
Ans. C
Same number = [1,1,1,1] or [2,2,2,2] or . . . . [6,6,6,6]
Total favorable cases = 6
Total cases = 6^4
Probability = 6/6^4 = 1/6^3 = 1/216
2. A can do a piece of work in 10 days and B can do the same piece of work in 20 days. They start the work together, but after 5 days A leaves B will do the remaining piece of work in :
1) 6 days 2) 5 days 3) 4 days 4) 10 days
Ans : A can do a piece of work in 10 days
Hence A’s work of one day = 1/10
B can do a piece of work in 20 days
Hence B’s work of one day = 1/20
A and B started work together and worked for 5 days
Hence their combined work of 5 days = 5/10 + 5/20 = ½ + ¼ = ¾
Remaining work after 5 days = 1 – ¾ = ¼
Now B alone work hence number of days that B will finish this ¼ piece of work = numbers of days to complete the work / part of work
= 20× ¼ = 5 days
3. A dealer offered a machine for sale
₹27500, but even if had charged 10% less, he would have made a profit of 10%. The actual cost of the machine is
1) 24250 2) 22500 3) 22275 4) 22000
Ans : 10% of 27500 = Rs. 2750
S.P. = 27500 – 2750
= Rs. 24750
C.P.
= Rs. 22500
S.P. = 27500 – 2750 = Rs. 24750
C.P. = Rs. 22500
4. In an examination, 1100 boys and 900 girls appeared. 50% of the boys and 40% of the girls passed the examination. The percentage of candidates who failed ?
1) 45 2) 45.5 3) 50 4) 54.5
Directions:Study the graphs carefully to answer the questions that follow:
Total number of children in 6 different schools and the percentage of boys in them
Total number of children in 6 different schools and the percentage of boys in them
What is the total percentage of girls in schools R and U together (rounded off to two digits after decimal)
Girls in R = 2000 x 72.5/100 = 1450
Girls in U = 1000 x 82.5/100 = 825
Total no. of girls = (1450+825) = 2275
Required Percentage of girls = 2275/3000 x 100 = 75.83%
Girls in U = 1000 x 82.5/100 = 825
Total no. of girls = (1450+825) = 2275
Required Percentage of girls = 2275/3000 x 100 = 75.83%
What is the total number of girls in school T
Total no of girls in T = 1250 x (100-40)/100 = 750
The total number of students in school R, is approximately what percent of the total number of students in school S?
Required Percentage = 2000/2250 x 100 = 88.8 % = 89%
What is the average number of girls in schools P and Q together?
Average no. of girls = (2500 x (100-40)/100 + 3000 x (100-45)/100 )/2
= (1500+1650)/2 = 1575
= (1500+1650)/2 = 1575
What is the respective ratio of the number of boys in the school P to the number of boys in school Q?
Required Ratio = (2500 x 40/100)/(3000 x 45/100) = 1000/1350 = 20:27
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