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29 May 2020, 03:24
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C
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In the figure above, rectangle PQRS is a shaded region inside the square ABCD. What is the probability that a point chosen at random from the square ABCD will lie inside the shaded rectangle PQRS?
(1) The length of a diagonal of rectangle PQRS is 55% the length of a diagonal of square ABCD
(2) The length of side PQ is 20% greater than the length of side QR
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1.png [ 2.28 KiB | Viewed 5810 times ]
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
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29 May 2020, 03:52
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Suppose PQ=a; QR=b; AB=l
Probability = Area of rectangle/ Area of square \(= \frac{a*b}{l^2} = \frac{a}{l}*\frac{b}{l}\)
Statement 1
Length of diagonal of rectangle \(= \sqrt{a^2+b^2}\)
Length of the diagonal of square = \(\sqrt{2}l\)
\( \frac{\sqrt{a^2+b^2}}{\sqrt{2}l}\) = \(\frac{11}{20}\)
\(\frac{a^2}{l^2} +\frac{b^2}{l^2} = \frac{121}{200}\).....(1)
We can't find \(\frac{a}{l}\) and \(\frac{b}{l}\), using above equation
Insufficient
Statement 2- We know a/b. But there is no way to find the ratios of the area of rectangle and square.
Insufficient
Combining both statements
\(\frac{b}{l} = \frac{5}{6}(\frac{a}{l})\) (From statement 2). Also \(\frac{a}{l}, \frac{b}{l} >0\)
Hence, we can find the unique value of \(\frac{a}{l}\) and \(\frac{b}{l}\), using equation (1) (From statement 1)
Sufficient
Probability = Area of rectangle/ Area of square \(= \frac{a*b}{l^2} = \frac{a}{l}*\frac{b}{l}\)
Statement 1
Length of diagonal of rectangle \(= \sqrt{a^2+b^2}\)
Length of the diagonal of square = \(\sqrt{2}l\)
\( \frac{\sqrt{a^2+b^2}}{\sqrt{2}l}\) = \(\frac{11}{20}\)
\(\frac{a^2}{l^2} +\frac{b^2}{l^2} = \frac{121}{200}\).....(1)
We can't find \(\frac{a}{l}\) and \(\frac{b}{l}\), using above equation
Insufficient
Statement 2- We know a/b. But there is no way to find the ratios of the area of rectangle and square.
Insufficient
Combining both statements
\(\frac{b}{l} = \frac{5}{6}(\frac{a}{l})\) (From statement 2). Also \(\frac{a}{l}, \frac{b}{l} >0\)
Hence, we can find the unique value of \(\frac{a}{l}\) and \(\frac{b}{l}\), using equation (1) (From statement 1)
Sufficient
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29 May 2020, 03:55
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IMO C
Statement 1: The length of a diagonal of rectangle PQRS is 55% the length of a diagonal of square ABCD (but the diagonal could be anywhere moreover we need value of length and breadth of rectangle to determine area) (not sufficient)
Statement 2: The length of side PQ is 20% greater than the length of side QR
(But the rectangle can of any length and breadth, need some parameter to draw area with respect to square) (not sufficient)
Statement 1&2:
Let diagonal of square be 100
Therefore side S of square will be
100^2 = S^2+S^2
10000 = 2S^2
5000 = S^2
S= 50√2 and area of square is 5000 ------(1)
The length of rectangle diagonal is 55% of that of square= 100*55/100=55
Also L = 120B/100 -------(2)
55^2=L^2+B^2
3025 = (1.2b)^2+b^2
3025 = 1.44b^2+b^2
3025 = 2.44b^2
B^2 = 1239.75
B= 35.21 and L = 35.21*120/100=42.25
Area of rectangle (shaded region) = L*B = 1487.7 ------(3)
Probability of that point being in shaded region is = 1487.7/5000 = 0.2975 or 0.3 (approx)
Posted from my mobile device
Statement 1: The length of a diagonal of rectangle PQRS is 55% the length of a diagonal of square ABCD (but the diagonal could be anywhere moreover we need value of length and breadth of rectangle to determine area) (not sufficient)
Statement 2: The length of side PQ is 20% greater than the length of side QR
(But the rectangle can of any length and breadth, need some parameter to draw area with respect to square) (not sufficient)
Statement 1&2:
Let diagonal of square be 100
Therefore side S of square will be
100^2 = S^2+S^2
10000 = 2S^2
5000 = S^2
S= 50√2 and area of square is 5000 ------(1)
The length of rectangle diagonal is 55% of that of square= 100*55/100=55
Also L = 120B/100 -------(2)
55^2=L^2+B^2
3025 = (1.2b)^2+b^2
3025 = 1.44b^2+b^2
3025 = 2.44b^2
B^2 = 1239.75
B= 35.21 and L = 35.21*120/100=42.25
Area of rectangle (shaded region) = L*B = 1487.7 ------(3)
Probability of that point being in shaded region is = 1487.7/5000 = 0.2975 or 0.3 (approx)
Posted from my mobile device
29 May 2020, 11:13
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See the attachment. Answer C
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1.PNG [ 31.26 KiB | Viewed 4993 times ]
