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If a man saves ` 4 more each year than he did the year before and if he saves ` 20 in the first year,after how many years will his savings be more than ` 1000 altogether?
- a)19 years
- b)20 years
- c)21years
- d)18 years
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
If a man saves ` 4 more each year than he did the year before and if h...
We need the sum of the series 20 + 24 + 28 to cross 1000. Trying out the options, we can see that
in 20 years the sum of his savings would be: 20 + 24 + 28 + … + 96. The sum of this series would
be 20 × 58 =1160. If we remove the 20th year we will get the series for savings for 19 years. The
series would be 20 + 24 + 28 + …. 92. Sum of the series would be 1160 – 96 = 1064. If we
remove the 19th year’s savings the savings would be 1064 – 92 which would go below 1000.
Thus, after 19 years his savings would cross 1000. Option (a) is correct.
in 20 years the sum of his savings would be: 20 + 24 + 28 + … + 96. The sum of this series would
be 20 × 58 =1160. If we remove the 20th year we will get the series for savings for 19 years. The
series would be 20 + 24 + 28 + …. 92. Sum of the series would be 1160 – 96 = 1064. If we
remove the 19th year’s savings the savings would be 1064 – 92 which would go below 1000.
Thus, after 19 years his savings would cross 1000. Option (a) is correct.
Most Upvoted Answer
If a man saves ` 4 more each year than he did the year before and if h...
Solution:
To solve this problem, we can use the formula for the sum of an arithmetic progression:
Sn = n/2[2a + (n-1)d]
where Sn is the sum of the first n terms, a is the first term, d is the common difference, and n is the number of terms.
We know that the man saves 4 more each year than he did the year before, so the common difference is d = 4.
We also know that he saves `20 in the first year, so the first term is a = 20.
We want to find out after how many years his savings will be more than `1000 altogether, so we want to find the value of n that satisfies the inequality Sn > 1000.
Substituting the values that we know into the formula, we get:
Sn = n/2[2a + (n-1)d]
Sn = n/2[2(20) + (n-1)(4)]
Sn = n/2[40 + 4n - 4]
Sn = n/2[36 + 4n]
Sn = 18n + 2n^2
Now we can set up the inequality:
Sn > 1000
18n + 2n^2 > 1000
2n^2 + 18n - 1000 > 0
We can solve this quadratic inequality by factoring or using the quadratic formula, but we can also estimate the value of n by looking at the options given.
Option A is 19 years, which means that the man's savings would be:
S19 = 19/2[2(20) + (19-1)(4)]
S19 = 19/2[40 + 72]
S19 = 19/2[112]
S19 = 1064
This is more than `1000, so option A is the correct answer.
We can also check the other options:
Option B: S20 = 20/2[2(20) + (20-1)(4)] = 1120
Option C: S21 = 21/2[2(20) + (21-1)(4)] = 1204
Option D: S18 = 18/2[2(20) + (18-1)(4)] = 988
Therefore, the correct answer is option A, 19 years.
To solve this problem, we can use the formula for the sum of an arithmetic progression:
Sn = n/2[2a + (n-1)d]
where Sn is the sum of the first n terms, a is the first term, d is the common difference, and n is the number of terms.
We know that the man saves 4 more each year than he did the year before, so the common difference is d = 4.
We also know that he saves `20 in the first year, so the first term is a = 20.
We want to find out after how many years his savings will be more than `1000 altogether, so we want to find the value of n that satisfies the inequality Sn > 1000.
Substituting the values that we know into the formula, we get:
Sn = n/2[2a + (n-1)d]
Sn = n/2[2(20) + (n-1)(4)]
Sn = n/2[40 + 4n - 4]
Sn = n/2[36 + 4n]
Sn = 18n + 2n^2
Now we can set up the inequality:
Sn > 1000
18n + 2n^2 > 1000
2n^2 + 18n - 1000 > 0
We can solve this quadratic inequality by factoring or using the quadratic formula, but we can also estimate the value of n by looking at the options given.
Option A is 19 years, which means that the man's savings would be:
S19 = 19/2[2(20) + (19-1)(4)]
S19 = 19/2[40 + 72]
S19 = 19/2[112]
S19 = 1064
This is more than `1000, so option A is the correct answer.
We can also check the other options:
Option B: S20 = 20/2[2(20) + (20-1)(4)] = 1120
Option C: S21 = 21/2[2(20) + (21-1)(4)] = 1204
Option D: S18 = 18/2[2(20) + (18-1)(4)] = 988
Therefore, the correct answer is option A, 19 years.
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