Aptitude - Average - Discussion
Discussion Forum : Average - General Questions (Q.No. 9)
9.
A car owner buys petrol at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends Rs. 4000 each year?
Answer: Option
Explanation:
| Total quantity of petrol consumed in 3 years |
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Total amount spent = Rs. (3 x 4000) = Rs. 12000.
 Average cost = Rs. |
 |
12000 x 51 |  |
= Rs. | 6120 | = Rs. 7.98 |
| 76700 | 767 |
Discussion:
92 comments Page 2 of 10.
MONISH S L said:
4 years ago
Thanks @Krishna.
(3)
MONISH S L said:
4 years ago
Thanks @Krishna.
(2)
Arvind Sahu said:
9 years ago
As we know,
Average = (tatal cost) / (total quantity).
So according to this question, he is spending 4000/- per yr for 3 corresponding yr.
So total cost become:-
Total cost=4000*3=> 12000/-
And to find total quantity we have to find seperatly first and then add them-
1st yr (4000/7.5) = 533.3,
2nd yr (4000/8) = 500,
3rd yr (4000/8.50) = 470.5.
Total quantity = 533.3 + 500 + 470.5 = 1503.8.
Now we can easily divide and find the average like:-
Required Average = 12000/1503.8
= 7.98 answer.
Average = (tatal cost) / (total quantity).
So according to this question, he is spending 4000/- per yr for 3 corresponding yr.
So total cost become:-
Total cost=4000*3=> 12000/-
And to find total quantity we have to find seperatly first and then add them-
1st yr (4000/7.5) = 533.3,
2nd yr (4000/8) = 500,
3rd yr (4000/8.50) = 470.5.
Total quantity = 533.3 + 500 + 470.5 = 1503.8.
Now we can easily divide and find the average like:-
Required Average = 12000/1503.8
= 7.98 answer.
(2)
Sneha said:
4 years ago
= 4000( 100/750) + (1/8) + (100/850).
= 4000(2/15+1/8+2/17) cross multiplication,
= 4000(31/120+2/17) Cross mul,
= 4000(767/2040),
= 3068000/2040 decided by 4 on both side,
= 76700/51.
= 4000(2/15+1/8+2/17) cross multiplication,
= 4000(31/120+2/17) Cross mul,
= 4000(767/2040),
= 3068000/2040 decided by 4 on both side,
= 76700/51.
(2)
Mansingh budha said:
4 years ago
I do not understand this problem. How can I solve it? please solve it again.
(1)
Shivna said:
4 years ago
Thank you @M.V Krishna.
(1)
Venkat said:
9 years ago
Please solve this equation 4000(2/15 + 1/8 + 2/17) = ?
(1)
Ashwini madne said:
9 years ago
@Santhosh Sandy.
= 4000/7.50 + 4000/8 + 4000/8.50.
= 4000(1/7.50 + 1/8 + 1/8.50) ...... 1/7.50.
It can be written as 100/750 which is equal to 2/15 same as 1/8.50 can be written as 100/850 which is equal to 2/17.
= 4000(2/15 + 1/8 + 2/17).
= 4000[{(2 * 8) + (15 * 1)/15 * 8} + 2/17].
= 4000[{31/120 + 2/17}].
= 4000[767/2040].
= 76700/51.
= 4000/7.50 + 4000/8 + 4000/8.50.
= 4000(1/7.50 + 1/8 + 1/8.50) ...... 1/7.50.
It can be written as 100/750 which is equal to 2/15 same as 1/8.50 can be written as 100/850 which is equal to 2/17.
= 4000(2/15 + 1/8 + 2/17).
= 4000[{(2 * 8) + (15 * 1)/15 * 8} + 2/17].
= 4000[{31/120 + 2/17}].
= 4000[767/2040].
= 76700/51.
(1)
Jobin said:
1 decade ago
Normally average will be Rs. 8 if we won't take the factor of Rs. 4000 per year. When we consider the amount per year, number of liter consumed in the 1st year will be more than second and third.
So approximately the average will be less than 8 and greater than 7.5. And more nearer to 8. So the answer is 7.98.
So approximately the average will be less than 8 and greater than 7.5. And more nearer to 8. So the answer is 7.98.
(1)
Nitesh bhati said:
1 decade ago
1st year 4000/7.5 = 533.3.
2nd year 4000/8 = 500.
3rd year 4000/8.50 = 470.5.
= 533.3+500+470.5 = 1503.8.
= 1503.8*7.98 = 4000*3.
= 12000 = 12000.
LHS = RHS.
2nd year 4000/8 = 500.
3rd year 4000/8.50 = 470.5.
= 533.3+500+470.5 = 1503.8.
= 1503.8*7.98 = 4000*3.
= 12000 = 12000.
LHS = RHS.
(1)
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