💡First, calculate the area of a single strip:[text{Area of one strip} = text{Length} times text{Width} = 15 times 1 = 15]There are four strips, so the total area without any overlaps would be:[4 times 15 = 60]Next, analyze the overlaps and layout:- Two strips overlap perpendicularly at their centers. Since they overlap completely in this area, they contribute only 15 units of area instead of 30.- The other two strips are placed end to end without overlapping, contributing a total of 30 units of area.Calculate the total area covered:- Area covered by two perpendicular overlapping strips: (15)- Area covered by two end-to-end strips: (15 + 15 = 30)Thus, the total area covered is:[15 + 30 = 45]The final answer is:[boxed{C) 45}]

💡First, I need to determine the number of paintings Yanni has in total.I know that 3 paintings are 5 feet by 5 feet each. Calculating their total area:3 paintings × (5 ft × 5 ft) = 3 × 25 sq ft = 75 sq ft.There is 1 painting that is 10 feet by 8 feet:1 painting × (10 ft × 8 ft) = 80 sq ft.The final painting is 5 feet tall and 9 feet wide:1 painting × (5 ft × 9 ft) = 45 sq ft.Adding up the areas of all the paintings:75 sq ft + 80 sq ft + 45 sq ft = 200 sq ft.Since the combined area matches the total area Yanni has, the number of paintings is:3 + 1 + 1 = 5 paintings.

💡Okay, so I have this geometry problem about a parallelogram ABCD with a point M inside it. Through M, they draw two lines: PR parallel to BC and QS parallel to AB. Points P, Q, R, and S are on the sides AB, BC, CD, and DA respectively. I need to prove that lines BS, PD, and MC all intersect at one point.Hmm, let me try to visualize this. Parallelogram ABCD means opposite sides are equal and parallel. So AB is parallel to CD, and BC is parallel to AD. Point M is somewhere inside this figure. Then, from M, they draw PR parallel to BC, so PR must be a horizontal line if BC is horizontal. Similarly, QS is drawn parallel to AB, so QS would be a vertical line if AB is vertical.Wait, actually, in a general parallelogram, the sides aren't necessarily horizontal or vertical, but they are parallel. So PR is parallel to BC, which is one pair of sides, and QS is parallel to AB, the other pair. So PR and QS are like grid lines inside the parallelogram, intersecting at point M.Points P, Q, R, S are on AB, BC, CD, DA respectively. So P is on AB, Q is on BC, R is on CD, and S is on DA. So PR connects AB to CD, and QS connects BC to DA.Now, I need to show that lines BS, PD, and MC intersect at one point. So BS is a line from B to S, PD is a line from P to D, and MC is a line from M to C. These three lines should meet at a single point.I remember that in projective geometry, if certain lines are drawn in a parallelogram, they can intersect at a common point due to the properties of parallel lines and the ratios they create. Maybe I can use similar triangles or some proportionality here.Let me try to assign coordinates to the points to make this more concrete. Let's place the parallelogram ABCD in a coordinate system. Let me assume point A is at (0,0). Since it's a parallelogram, I can let point B be at (a,0), point D at (0,b), and then point C would be at (a,b). So sides AB are from (0,0) to (a,0), BC from (a,0) to (a,b), CD from (a,b) to (0,b), and DA from (0,b) to (0,0).Point M is inside the parallelogram. Let me assign coordinates to M as (m,n), where 0 < m < a and 0 < n < b.Now, line PR is drawn through M parallel to BC. Since BC is vertical (from (a,0) to (a,b)), a line parallel to BC would also be vertical. So PR is a vertical line passing through M. Therefore, point P is the intersection of this vertical line with AB, which is the line y=0. So P would be at (m,0). Similarly, point R is the intersection of this vertical line with CD, which is the line y=b. So R is at (m,b).Similarly, line QS is drawn through M parallel to AB. Since AB is horizontal (from (0,0) to (a,0)), a line parallel to AB would also be horizontal. So QS is a horizontal line passing through M. Therefore, point Q is the intersection of this horizontal line with BC, which is the line x=a. So Q is at (a,n). Similarly, point S is the intersection of this horizontal line with DA, which is the line x=0. So S is at (0,n).Okay, so now I have coordinates for all these points:- P is (m,0)- Q is (a,n)- R is (m,b)- S is (0,n)- M is (m,n)Now, I need to find the equations of lines BS, PD, and MC and show that they intersect at a single point.First, let's find the equation of line BS. Points B and S are at (a,0) and (0,n) respectively.The slope of BS is (n - 0)/(0 - a) = -n/a. So the equation is y = (-n/a)x + n.Wait, let me check that. When x = a, y should be 0: y = (-n/a)*a + n = -n + n = 0. Correct. When x = 0, y = n. Correct. So equation of BS is y = (-n/a)x + n.Next, equation of PD. Points P and D are at (m,0) and (0,b) respectively.Slope of PD is (b - 0)/(0 - m) = -b/m. So equation is y = (-b/m)x + b.Check: When x = m, y = (-b/m)*m + b = -b + b = 0. Correct. When x = 0, y = b. Correct. So equation of PD is y = (-b/m)x + b.Now, equation of MC. Points M and C are at (m,n) and (a,b) respectively.Slope of MC is (b - n)/(a - m). So equation is y - n = [(b - n)/(a - m)](x - m).Simplify: y = [(b - n)/(a - m)](x - m) + n.Let me write that as y = [(b - n)/(a - m)]x - [(b - n)/(a - m)]m + n.Simplify the constants: - [(b - n)m/(a - m)] + n = [ - (b - n)m + n(a - m) ] / (a - m) = [ -bm + mn + an - mn ] / (a - m) = (an - bm)/(a - m).So equation of MC is y = [(b - n)/(a - m)]x + (an - bm)/(a - m).Now, I need to find the intersection point of BS and PD, and then check if it lies on MC.First, find intersection of BS and PD.Set their equations equal:(-n/a)x + n = (-b/m)x + b.Bring all terms to one side:[(-n/a) + (b/m)]x + (n - b) = 0.Factor x:x*(b/m - n/a) = b - n.So x = (b - n)/(b/m - n/a) = (b - n)/( (ab - mn)/(am) ) = (b - n)*(am)/(ab - mn).Simplify numerator and denominator:x = am(b - n)/(ab - mn).Similarly, y can be found by plugging back into one of the equations, say BS:y = (-n/a)x + n.So y = (-n/a)*(am(b - n)/(ab - mn)) + n = (-n m (b - n))/(ab - mn) + n.Combine terms:y = [ -n m (b - n) + n(ab - mn) ] / (ab - mn) = [ -n m b + n² m + a b n - m n² ] / (ab - mn).Simplify numerator:- n m b + n² m + a b n - m n² = (-n m b + a b n) + (n² m - m n²) = b n (a - m) + 0 = b n (a - m).So y = [b n (a - m)] / (ab - mn).Therefore, the intersection point of BS and PD is at:x = am(b - n)/(ab - mn),y = b n (a - m)/(ab - mn).Now, let's see if this point lies on MC.The equation of MC is y = [(b - n)/(a - m)]x + (an - bm)/(a - m).Let me plug x into this equation and see if y matches.Compute y from MC:y = [(b - n)/(a - m)] * [am(b - n)/(ab - mn)] + (an - bm)/(a - m).Simplify:First term: [(b - n)/(a - m)] * [am(b - n)/(ab - mn)] = [am(b - n)^2]/[(a - m)(ab - mn)].Second term: (an - bm)/(a - m).Combine the two terms:[am(b - n)^2 + (an - bm)(ab - mn)] / [(a - m)(ab - mn)].Let me expand the numerator:am(b - n)^2 + (an - bm)(ab - mn).First, expand am(b - n)^2:am(b² - 2b n + n²).Second, expand (an - bm)(ab - mn):= an*ab - an*mn - bm*ab + bm*mn= a² n b - a m n² - a b² m + b m² n.So total numerator:am(b² - 2b n + n²) + a² n b - a m n² - a b² m + b m² n.Let me distribute am:= a m b² - 2 a m b n + a m n² + a² n b - a m n² - a b² m + b m² n.Now, let's combine like terms:1. a m b² - a b² m = 0.2. -2 a m b n.3. a m n² - a m n² = 0.4. a² n b.5. + b m² n.So numerator simplifies to:-2 a m b n + a² n b + b m² n.Factor out b n:= b n ( -2 a m + a² + m² ).= b n (a² - 2 a m + m²).= b n (a - m)^2.So numerator is b n (a - m)^2.Denominator is (a - m)(ab - mn).So y = [b n (a - m)^2] / [(a - m)(ab - mn)] = [b n (a - m)] / (ab - mn).Which matches the y-coordinate we found earlier.Therefore, the intersection point of BS and PD lies on MC, meaning that all three lines intersect at the same point.So, we've shown that the three lines BS, PD, and MC intersect at a single point.**Final Answer**boxed{text{The lines } BS, PD, text{ and } MC text{ intersect at one point.}}

💡Okay, so I have this problem: I need to find all positive integers M less than 10 such that 5 divides 1989^M + M^1989. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, I remember that if 5 divides a number, then that number modulo 5 should be 0. So, I can rewrite the condition as:1989^M + M^1989 ≡ 0 mod 5Alright, so I need to compute 1989^M mod 5 and M^1989 mod 5, then add them together and see if the result is 0 mod 5.Let me start by simplifying 1989 mod 5. To find 1989 mod 5, I can divide 1989 by 5 and find the remainder. Let's see, 5 times 397 is 1985, so 1989 minus 1985 is 4. Wait, no, that would make 1989 ≡ 4 mod 5. But wait, 4 is equivalent to -1 mod 5 because 4 + 1 = 5, which is 0 mod 5. So, 1989 ≡ -1 mod 5. That might be useful later.So, 1989 ≡ -1 mod 5. Therefore, 1989^M ≡ (-1)^M mod 5. That simplifies things a bit. So, depending on whether M is odd or even, (-1)^M will be -1 or 1 mod 5.Now, let's think about M^1989 mod 5. Since M is less than 10, I can consider each M from 1 to 9 and compute M^1989 mod 5 individually. But that might take some time. Maybe there's a pattern or a theorem I can use to simplify this.I remember Euler's theorem, which says that if M and 5 are coprime, then M^φ(5) ≡ 1 mod 5, where φ(5) is Euler's totient function. Since 5 is prime, φ(5) = 4. So, for M not divisible by 5, M^4 ≡ 1 mod 5. That means M^1989 can be simplified by expressing 1989 in terms of multiples of 4.Let me compute 1989 divided by 4. 4 times 497 is 1988, so 1989 = 4*497 + 1. Therefore, M^1989 = M^(4*497 + 1) = (M^4)^497 * M^1. Since M^4 ≡ 1 mod 5, this simplifies to 1^497 * M ≡ M mod 5. So, for M not divisible by 5, M^1989 ≡ M mod 5.But wait, what if M is divisible by 5? If M is 5, then M^1989 is 5^1989, which is obviously 0 mod 5. So, for M = 5, M^1989 ≡ 0 mod 5.So, summarizing:- If M is not divisible by 5, then M^1989 ≡ M mod 5.- If M is divisible by 5, then M^1989 ≡ 0 mod 5.But since M is less than 10, the only M divisible by 5 is M = 5.Now, let's go back to our original equation:1989^M + M^1989 ≡ (-1)^M + M^1989 ≡ 0 mod 5We need to consider two cases: M is even or odd.Case 1: M is even.If M is even, then (-1)^M = 1. So, the equation becomes:1 + M^1989 ≡ 0 mod 5Which implies:M^1989 ≡ -1 mod 5But earlier, we saw that for M not divisible by 5, M^1989 ≡ M mod 5. So, if M is even and not 5, then:M ≡ -1 mod 5Which means M ≡ 4 mod 5.Since M is less than 10, the possible values are M = 4 and M = 9. But wait, M = 9 is odd, so we need to check if M = 4 satisfies the condition.Let me check M = 4:1989^4 + 4^1989 ≡ (-1)^4 + 4^1989 ≡ 1 + 4 mod 5 (since 4^1989 ≡ 4 mod 5)1 + 4 = 5 ≡ 0 mod 5. So, M = 4 works.Case 2: M is odd.If M is odd, then (-1)^M = -1. So, the equation becomes:-1 + M^1989 ≡ 0 mod 5Which implies:M^1989 ≡ 1 mod 5Again, for M not divisible by 5, M^1989 ≡ M mod 5. So, if M is odd and not 5, then:M ≡ 1 mod 5Which means M ≡ 1 mod 5.Since M is less than 10, the possible values are M = 1 and M = 6. But wait, M = 6 is even, so we need to check M = 1.Let me check M = 1:1989^1 + 1^1989 ≡ (-1)^1 + 1 ≡ -1 + 1 ≡ 0 mod 5. So, M = 1 works.Now, what about M = 5? Let's check it separately.M = 5:1989^5 + 5^1989 ≡ (-1)^5 + 0 ≡ -1 + 0 ≡ -1 ≡ 4 mod 5 ≠ 0. So, M = 5 doesn't work.So, from the above cases, the only M values that satisfy the condition are M = 1 and M = 4.Wait, let me double-check M = 9, which is odd. Earlier, I thought M = 9 is odd, so let's see:M = 9:1989^9 + 9^1989 ≡ (-1)^9 + 9^1989 ≡ -1 + (9 mod 5)^1989 ≡ -1 + 4^1989But 4^1989 ≡ 4 mod 5, so:-1 + 4 ≡ 3 mod 5 ≠ 0. So, M = 9 doesn't work.Similarly, M = 6:1989^6 + 6^1989 ≡ (-1)^6 + 6^1989 ≡ 1 + (6 mod 5)^1989 ≡ 1 + 1^1989 ≡ 1 + 1 ≡ 2 mod 5 ≠ 0. So, M = 6 doesn't work.So, indeed, only M = 1 and M = 4 satisfy the condition.I think that's it. I've checked all possible M from 1 to 9, and only M = 1 and M = 4 work.

💡Okay, so I have this problem where I need to find the minimum value of the expression (frac{1}{sin^{2}{theta}} + frac{4}{cos^{2}{theta}}) where (theta) is between 0 and (frac{pi}{2}). Hmm, that seems a bit tricky, but let me try to break it down step by step.First, I remember that in trigonometry, the Pythagorean identity says that (sin^{2}{theta} + cos^{2}{theta} = 1). Maybe I can use that somehow. The expression I have involves reciprocals of (sin^{2}{theta}) and (cos^{2}{theta}), so perhaps I can relate them through this identity.Wait, another thought: sometimes when dealing with expressions that have reciprocals like this, it's helpful to use substitution. Maybe I can let (x = sin^{2}{theta}) and (y = cos^{2}{theta}). Then, since (x + y = 1), I can express the problem in terms of (x) and (y). Let me try that.So, substituting, the expression becomes (frac{1}{x} + frac{4}{y}). And since (x + y = 1), I can write (y = 1 - x). So now, the expression is (frac{1}{x} + frac{4}{1 - x}). Now, I need to find the minimum value of this function for (x) in the interval (0,1) because (theta) is between 0 and (frac{pi}{2}), so both (sin{theta}) and (cos{theta}) are positive, and their squares are between 0 and 1.Alright, so now I have a function (f(x) = frac{1}{x} + frac{4}{1 - x}) defined on (0,1). To find its minimum, I can use calculus. I'll take the derivative of (f(x)) with respect to (x), set it equal to zero, and solve for (x). That should give me the critical points, and then I can check if it's a minimum.Let's compute the derivative:(f'(x) = -frac{1}{x^{2}} + frac{4}{(1 - x)^{2}})Set this equal to zero:(-frac{1}{x^{2}} + frac{4}{(1 - x)^{2}} = 0)Moving one term to the other side:(frac{4}{(1 - x)^{2}} = frac{1}{x^{2}})Cross-multiplying:(4x^{2} = (1 - x)^{2})Expanding the right-hand side:(4x^{2} = 1 - 2x + x^{2})Bring all terms to one side:(4x^{2} - x^{2} + 2x - 1 = 0)Simplify:(3x^{2} + 2x - 1 = 0)Now, I need to solve this quadratic equation. Let's use the quadratic formula:(x = frac{-b pm sqrt{b^{2} - 4ac}}{2a})Here, (a = 3), (b = 2), and (c = -1). Plugging these in:(x = frac{-2 pm sqrt{(2)^{2} - 4(3)(-1)}}{2(3)})Calculate the discriminant:(4 + 12 = 16)So,(x = frac{-2 pm 4}{6})This gives two solutions:1. (x = frac{-2 + 4}{6} = frac{2}{6} = frac{1}{3})2. (x = frac{-2 - 4}{6} = frac{-6}{6} = -1)But (x) must be between 0 and 1, so we discard (x = -1). Thus, (x = frac{1}{3}).Now, let's verify if this is a minimum. We can use the second derivative test. First, compute the second derivative of (f(x)):(f''(x) = frac{2}{x^{3}} + frac{8}{(1 - x)^{3}})Since (x) is in (0,1), both terms are positive, so (f''(x) > 0). Therefore, the function is concave upward at (x = frac{1}{3}), which means it's a local minimum. Since the function tends to infinity as (x) approaches 0 or 1, this local minimum is indeed the global minimum on the interval.Now, let's compute the minimum value by plugging (x = frac{1}{3}) back into the function:(fleft(frac{1}{3}right) = frac{1}{frac{1}{3}} + frac{4}{1 - frac{1}{3}} = 3 + frac{4}{frac{2}{3}} = 3 + 6 = 9)So, the minimum value is 9.Wait a second, I remember that sometimes with these kinds of optimization problems, using the AM-GM inequality can be a quicker method without calculus. Let me try that approach to verify.The expression is (frac{1}{sin^{2}{theta}} + frac{4}{cos^{2}{theta}}). Let me denote (a = frac{1}{sin{theta}}) and (b = frac{2}{cos{theta}}). Then, the expression becomes (a^{2} + b^{2}). But I also know that (a sin{theta} = 1) and (b cos{theta} = 2). Hmm, maybe that's not directly helpful. Alternatively, perhaps I can use the Cauchy-Schwarz inequality.Wait, another idea: using the Cauchy-Schwarz inequality on the terms. Let me think.Alternatively, since I have two terms, maybe I can use the method of Lagrange multipliers, but that might be overkill. Alternatively, maybe I can set up a substitution where I let (t = tan{theta}), since (tan{theta} = frac{sin{theta}}{cos{theta}}), which might simplify the expression.Let me try that substitution. Let (t = tan{theta}), so (sin{theta} = frac{t}{sqrt{1 + t^{2}}}) and (cos{theta} = frac{1}{sqrt{1 + t^{2}}}). Then, (sin^{2}{theta} = frac{t^{2}}{1 + t^{2}}) and (cos^{2}{theta} = frac{1}{1 + t^{2}}).Substituting into the expression:(frac{1}{sin^{2}{theta}} + frac{4}{cos^{2}{theta}} = frac{1 + t^{2}}{t^{2}} + 4(1 + t^{2}))Simplify:(frac{1}{t^{2}} + 1 + 4 + 4t^{2} = frac{1}{t^{2}} + 4t^{2} + 5)So, the expression reduces to (frac{1}{t^{2}} + 4t^{2} + 5). Now, I can let (u = t^{2}), so (u > 0), and the expression becomes (frac{1}{u} + 4u + 5).Now, to find the minimum of (g(u) = frac{1}{u} + 4u + 5) for (u > 0). Again, I can use calculus here. Take the derivative:(g'(u) = -frac{1}{u^{2}} + 4)Set equal to zero:(-frac{1}{u^{2}} + 4 = 0)So,(frac{1}{u^{2}} = 4)Which gives (u^{2} = frac{1}{4}), so (u = frac{1}{2}) since (u > 0).Now, check the second derivative to confirm it's a minimum:(g''(u) = frac{2}{u^{3}}), which is positive for (u > 0), so it's a minimum.Compute the minimum value:(gleft(frac{1}{2}right) = frac{1}{frac{1}{2}} + 4 times frac{1}{2} + 5 = 2 + 2 + 5 = 9)Same result as before. So, whether I use substitution and calculus or trigonometric substitution and calculus, I get the minimum value as 9.Alternatively, maybe I can use the AM-GM inequality on the terms (frac{1}{sin^{2}{theta}}) and (frac{4}{cos^{2}{theta}}). Let me try that.The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. So, for two terms (a) and (b), we have:(frac{a + b}{2} geq sqrt{ab})But in this case, I have two terms, but they are not symmetric. Maybe I can adjust the weights. Alternatively, perhaps I can consider multiple variables.Wait, another approach: Let me consider the expression (frac{1}{sin^{2}{theta}} + frac{4}{cos^{2}{theta}}) and try to write it as a sum where I can apply AM-GM.Let me denote (a = frac{1}{sin{theta}}) and (b = frac{2}{cos{theta}}). Then, the expression becomes (a^{2} + b^{2}). But I also know that (a sin{theta} = 1) and (b cos{theta} = 2). Maybe I can relate these.Alternatively, perhaps I can use the Cauchy-Schwarz inequality. The Cauchy-Schwarz inequality states that for vectors (u) and (v), ((u cdot v)^2 leq (u cdot u)(v cdot v)). Maybe I can set up vectors such that their dot product relates to my expression.Let me consider vectors (u = (frac{1}{sin{theta}}, frac{2}{cos{theta}})) and (v = (sin{theta}, cos{theta})). Then, (u cdot v = frac{1}{sin{theta}} cdot sin{theta} + frac{2}{cos{theta}} cdot cos{theta} = 1 + 2 = 3).On the other hand, by Cauchy-Schwarz,((u cdot v)^2 leq (u cdot u)(v cdot v))So,(3^2 leq left(frac{1}{sin^{2}{theta}} + frac{4}{cos^{2}{theta}}right)(sin^{2}{theta} + cos^{2}{theta}))But (sin^{2}{theta} + cos^{2}{theta} = 1), so,(9 leq left(frac{1}{sin^{2}{theta}} + frac{4}{cos^{2}{theta}}right) times 1)Therefore,(frac{1}{sin^{2}{theta}} + frac{4}{cos^{2}{theta}} geq 9)So, the minimum value is 9, achieved when the vectors (u) and (v) are proportional, i.e., when (frac{frac{1}{sin{theta}}}{sin{theta}} = frac{frac{2}{cos{theta}}}{cos{theta}}), which simplifies to (frac{1}{sin^{2}{theta}} = frac{2}{cos^{2}{theta}}). Solving this gives (tan^{2}{theta} = frac{1}{2}), so (tan{theta} = frac{1}{sqrt{2}}), which means (theta = arctan{left(frac{1}{sqrt{2}}right)}). Plugging this back, we can find (sin{theta}) and (cos{theta}), but since we already found the minimum value using calculus and substitution, this confirms that the minimum is indeed 9.So, regardless of the method I use—whether substitution with calculus, trigonometric substitution with calculus, or applying the Cauchy-Schwarz inequality—I consistently arrive at the conclusion that the minimum value of the given expression is 9.I think I've explored multiple methods to solve this problem, and each approach leads me to the same answer. It's reassuring to see that different techniques converge to the same result. I don't see any mistakes in my reasoning, so I feel confident that the minimum value is indeed 9.**Final Answer**The minimum value is (boxed{9}).

💡First, I need to calculate the cost of each furniture piece after applying the discounts and taxes, but before the service charge.For the Sofa:- The original price is 600 with a 20% discount.- Discount amount: 20% of 600 = 120.- Discounted price: 600 - 120 = 480.- Sales tax: 5% of 480 = 24.- Price after tax: 480 + 24 = 504.For the Dining Table:- The original price is 400 with a 15% discount.- Discount amount: 15% of 400 = 60.- Discounted price: 400 - 60 = 340.- Sales tax: 6% of 340 = 20.40.- Price after tax: 340 + 20.40 = 360.40.For the Chair Set:- The original price is 300 with a 25% discount.- Discount amount: 25% of 300 = 75.- Discounted price: 300 - 75 = 225.- Sales tax: 4% of 225 = 9.- Price after tax: 225 + 9 = 234.For the Coffee Table:- The original price is 100 with a 10% discount.- Discount amount: 10% of 100 = 10.- Discounted price: 100 - 10 = 90.- Sales tax: 3% of 90 = 2.70.- Price after tax: 90 + 2.70 = 92.70.Next, I'll sum up the costs of all the furniture pieces after tax to find the total cost before the service charge:Total cost before service charge = 504 (Sofa) + 360.40 (Dining Table) + 234 (Chair Set) + 92.70 (Coffee Table) = 1191.10.Now, applying the 2% service charge on the total cost:Service charge = 2% of 1191.10 = 23.822.Adding the service charge to the total cost gives:Total cost with service charge = 1191.10 + 23.822 = 1214.922.Since Linda spent 3/4 of her savings on furniture, the remaining 1/4 was spent on the TV, which cost 200. Therefore, her original savings can be calculated as:200 = 1/4 of original savings.Original savings = 200 * 4 = 800.Thus, Linda's original savings were boxed{800}.