Hi ,
I guess the answer is 36.3 miles/gallon....Answer shud be D...
Correct me if i was wrong...
Regards
Madhav
Yeah..the OA is Option (D)--> 36
Did you try to find the no. of gallons for 1 mile on the city / highway to solve the problem?
Yeah..the OA is Option (D)--> 36
Did you try to find the no. of gallons for 1 mile on the city / highway to solve the problem?
Hi Arch,
Basically the question is asking the combined average =Distance travelled/time taken
Total Distance travelled = 60 Miles
Total Time Taken = 10/25 + 50/40
Avg. = TDT/TTT = 36.3 miles/gallon
A certain portfolio consisted of 5 stocks, priced at $20, $35, $40, $45 and $70, respectively. On a given day, the price of one stock increased by 15%, while the price of another decreased by 35% and the prices of the remaining three remained constant. If the average price of a stock in the portfolio rose by approximately 2%, which of the following could be the prices of the shares that remained constant?
A) 20, 35, 70
B) 20, 45, 70
C) 20, 35, 40
D) 35, 40, 70
E) 35, 40, 45
Is the answer E??
Please ignore...
A certain portfolio consisted of 5 stocks, priced at $20, $35, $40, $45 and $70, respectively. On a given day, the price of one stock increased by 15%, while the price of another decreased by 35% and the prices of the remaining three remained constant. If the average price of a stock in the portfolio rose by approximately 2%, which of the following could be the prices of the shares that remained constant?
A) 20, 35, 70
B) 20, 45, 70
C) 20, 35, 40
D) 35, 40, 70
E) 35, 40, 45:new_ukliam2:
cost + 15% - 35%
20.00 23.00 14.95
35.00 40.25 26.16
40.00 46.00 29.90
45.00 51.75 33.64
70.00 80.50 52.33
The only thing that comes close to answer is 35 40 and 45 remaining constant.
cost + 15% - 35%
20.00 23.00 14.95
35.00 40.25 26.16
40.00 46.00 29.90
45.00 51.75 33.64
70.00 80.50 52.33
The only thing that comes close to answer is 35 40 and 45 remaining constant.
I solved it in a different way. I hope it's right.
Average= (20+35+40+45+70)/5
= 210/5
=42
Approx. 2% increase in average= 42*102/100
Total overall increase = (42*102/100)*5
= 214.2
So the approx. difference = 214.2 210
= 4.2
15% increase in one stock - 70*115/100 = 80.5
Rise of $10.5
35% decrease in another stock 20*65/100 = 13
Fall of $7
Difference > 10.5-7 = 3.5
This is closest to approx. increase of 4.2
So constant stocks will be 35, 40, 45
and hence answer will be E
I solved it in a different way. I hope it's right.
Average= (20+35+40+45+70)/5
= 210/5
=42
Approx. 2% increase in average= 42*102/100
Total overall increase = (42*102/100)*5
= 214.2
So the approx. difference = 214.2 210
= 4.2
15% increase in one stock - 70*115/100 = 80.5
Rise of $10.5
35% decrease in another stock 20*65/100 = 13
Fall of $7
Difference > 10.5-7 = 3.5
This is closest to approx. increase of 4.2
So constant stocks will be 35, 40, 45
and hence answer will be E
Well I did the same thing - but only had it in columns and had to all because it is mentioned approximately... so the chances of another possibility (say 3.9) is not there 😃
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then
put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the
sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is
numbered 5 ?
A. 1/6
B. 1/5
C. 1/3
D. 2/5
E. 2/3
Please post explanation also. Thanks.
put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the
sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is
numbered 5 ?
A. 1/6
B. 1/5
C. 1/3
D. 2/5
E. 2/3
Please post explanation also. Thanks.
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then
put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the
sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is
numbered 5 ?
A. 1/6
B. 1/5
C. 1/3
D. 2/5
E. 2/3
Please post explanation also. Thanks.
I think its D........
the possible outcomes are (5,3),(3,5)(4,4)(2,6)(6,2)
favourable outcomes are (5,3),(3,5)
so probability =2/5
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then
put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the
sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is
numbered 5 ?
A. 1/6
B. 1/5
C. 1/3
D. 2/5
E. 2/3
Please post explanation also. Thanks.
The answer is D - 2/5
To get a sum of 8 the possible outcomes are:
(2,6) (3,5) (4,4) (5,3) and (6,2)
Out of this two outcomes have 5... hence required probability is 2/5
In a sample of college students 40% are third yr students, 70% are not 2nd yr students. what fraction of students who are not 3rd yr are 2nd yr students?
A) 3/4
B)2/3
C)4/7
D)1/2
E)3/7
In a sample of college students 40% are third yr students, 70% are not 2nd yr students. what fraction of students who are not 3rd yr are 2nd yr students?
A) 3/4
B)2/3
C)4/7
D)1/2
E)3/7
My Answer would be D.
Consider - a total of 100 students..
3rd year = 40
not 2nd year = 70
rest of the students(not 2nd or 3rd years) = not second year - 3rd year
= 70 - 40 = 30
what fraction of students who are not 3rd yr are 2nd yr students =
who are 2nd yr/ who are not 3rd yr
who are 2nd yr = 100 - 70 = 30
who are not 3rd yr = who are 2nd yr + rest of the students(not 2nd or 3rd years) = 30 + 30 = 60
So, answer would be 30/ 60 = 1/2.
What is the OA?
In a sample of college students 40% are third yr students, 70% are not 2nd yr students. what fraction of students who are not 3rd yr are 2nd yr students?
A) 3/4
B)2/3
C)4/7
D)1/2
E)3/7
Suppose total no. of students are 100. And the 2 sets of 2nd and 3rd year students should be disjoint i.e. no one is both 2nd and 3rd year student.
40 are 3rd year students.
70 are not 2nd year student i.e. 30 are 2nd year students.
students who are not 3rd yr are 60.
And ratio of 2nd year students to the students who are not 3rd yr will be 30/60 = 1/2.
Answer should be D.
Whats the OA ?
One-fifth of the light switches produce by a cer足tain factory are defective. Four-fifths of the defective switches are rejected and 1/20 of the nondefective switches are rejected by mistake. If all the switches not rejected are sold, what percent of the switches sold by the factory are defective?
[LEFT](A) 4%
(B) 5%
(C) 6.25%
(D) 11%
(E) 16%[/LEFT]
In a sample of college students 40% are third yr students, 70% are not 2nd yr students. what fraction of students who are not 3rd yr are 2nd yr students?
A) 3/4
B)2/3
C)4/7
D)1/2
E)3/7
the answer should be D)1/2
what is OA?
the answer should be D)1/2
what is OA?
It just means original answer..
jumped the gun .. sorry
One-fifth of the light switches produce by a cer足tain factory are defective. Four-fifths of the defective switches are rejected and 1/20 of the nondefective switches are rejected by mistake. If all the switches not rejected are sold, what percent of the switches sold by the factory are defective?
[LEFT](A) 4%
(B) 5%
(C) 6.25%
(D) 11%
(E) 16%[/LEFT]
All the numbers can be calculated easily by assuming that total number of switches are 100.
Answer should be B.
Whats the OA ?
One-fifth of the light switches produce by a cer足tain factory are defective. Four-fifths of the defective switches are rejected and 1/20 of the nondefective switches are rejected by mistake. If all the switches not rejected are sold, what percent of the switches sold by the factory are defective?
[LEFT](A) 4%
(B) 5%
(C) 6.25%
(D) 11%
(E) 16%[/LEFT]
Answer should be
[LEFT](B) 5%
what is OA?[/LEFT]
One-fifth of the light switches produce by a cer足tain factory are defective. Four-fifths of the defective switches are rejected and 1/20 of the nondefective switches are rejected by mistake. If all the switches not rejected are sold, what percent of the switches sold by the factory are defective?
[LEFT](A) 4%
(B) 5%
(C) 6.25%
(D) 11%
(E) 16%[/LEFT]
Is the Answer B : 5%?
Let total no. of switches is 100
Defective: 20 (1/5 of the light switches are defective.)
Rejected: 16 (Four-fifths of the defective switches are rejected)
defective yet not rejected = 4
non defective = 100 - 20 = 80
non defective yet rejected by error = 4 (1/20 of the nondefective switches are rejected by mistake)
non defective and not rejected = 80 - 4 = 76
total non rejected = 76 + 4 = 80
defective but not rejected = 20 -16 = 4
solving => 80 * (x/100) = 4
=> x = 5%