Kenneth had some money. He spent 1/7 of his money on food and $180 on
transport. He spent 1/5 of the remaining money on a new table. He also spent $1000 on a new television screen. He saved the remaining $1400 in the bank. How much money did Kenneth have at first?
Let X be the amount of money Kenneth had at first.
Kenneth spent X/7 on food + $180 + X/5 on a new table + $1000 on a new television screen.
Kenneth saved X - (X/7 + $180 + X/5 + $1000) = $1400 in the bank.
Multiplying through the parentheses, we get X - X/7 - $180 - X/5 - $1000 = $1400
Combining like terms, we get X - X/7 - X/5 = $3580
Combining the X terms, we get 35X/35 - 5X/35 - 7X/35 = $3580
Combining like terms, we get 23X/35 = $3580
Multiplying both sides by 35, we get 23X = $125300
Dividing both sides by 23, we get X = $<<5450=5450>>5450. Answer: \boxed{5450}.
Kenneth's total money = Food + Transport + Table + Television + Bank Savings
= (1/7 * Money) + 180 + (1/5 * (Money - 180)) + 1000 + 1400
= (1/7 * Money) + 180 + (1/5 * Money) - (1/5 * 180) + 1000 + 1400
= (1/7 * Money) + (2/5 * Money) + 1000 + 1400
= (7/7 * Money) + 1000 + 1400
= Money + 1000 + 1400
= Money + 2400
= 3710
Therefore, Kenneth had $3710 at first.
Apologies for the mistake in my previous response.
Your solution is correct. Kenneth had $3710 at first.
Let's solve this step-by-step.
Let's suppose Kenneth had x amount of money at first.
Step 1: He spent 1/7 of his money on food.
So, the money remaining after spending on food = x - (1/7)x = (6/7)x
Step 2: He spent $180 on transport.
So, the money remaining after spending on transport = (6/7)x - $180
Step 3: He spent 1/5 of the remaining money on a new table.
So, the money remaining after spending on the table = (1 - 1/5) * [(6/7)x - $180] = (4/5) * [(6/7)x - $180]
Step 4: He spent $1000 on a new television screen.
So, the money remaining after spending on the television screen = (4/5) * [(6/7)x - $180] - $1000
Step 5: He saved the remaining $1400 in the bank.
So, the money remaining after saving = [(4/5) * [(6/7)x - $180] - $1000] - $1400
According to the problem, the money remaining after saving is zero. So, we can now set up an equation and solve for x.
[(4/5) * [(6/7)x - $180] - $1000] - $1400 = 0
Now, let's solve for x.
[(4/5) * [(6/7)x - $180] - $1000] - $1400 = 0
[(4/5) * [(6/7)x - $180] - $1000] = $1400
[(4/5) * [(6/7)x - $180]] = $2400
[(4/5) * [(6/7)x]] - (4/5) * $180] = $2400
[(4/5) * [6/7] * x] - [(4/5) * $180] = $2400
(24/35) * x - (4/5) * $180 = $2400
Now, let's solve for x.
(24/35) * x - (4/5) * $180 = $2400
(24/35) * x = (4/5) * $180 + $2400
(24/35) * x = ($720/5) + $2400
(24/35) * x = $144 + $2400
(24/35) * x = $2544
Now, let's solve for x.
x = ($2544 * 35) / 24
x = $3710
Therefore, Kenneth had $3710 at first.