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Question of the week  28 (In the figure above, O is the center .....)
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Updated on: 27 Feb 2019, 05:36
Question Stats:
25% (02:37) correct 75% (03:24) wrong based on 71 sessions
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eGMAT Question of the Week #28In the figure above, O is the center of the circle, with radius 10 units and PA is a tangent, whose length is equal to 24 units. If the lengths of PB and PC are distinct positive integers, and PB < PC, then find the sum of all possible lengths of PB? A. 16 B. 18 C. 34 D. 58 E. 126 Attachment:
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Re: Question of the week  28 (In the figure above, O is the center .....)
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21 Dec 2018, 20:55
EgmatQuantExpert wrote: In the figure above, O is the center of the circle, with radius 10 units and PA is a tangent, whose length is equal to 24 units. If the lengths of PB and PC are distinct positive integers, and PB < PC, then find the sum of all possible lengths of PB? A. 16 B. 18 C. 34 D. 58 E. 126 It will be maximum when Pb~PC, that is PB is the tangent, and at that time it will be 24, so PB<24.. Now for Min, when PC lies on the hypotenuse itself... So PO = √(24^2+10^2)=√676= 26, PB will be 2610=16.. So min value is 16 and max is less than 24.. Now the property that will be useful here is (PA)^2=PB*PC... So PB will be a factor of 24^2 but less than 24 and greater than or equal to 16.. So PB can be only 16 and 18.. Sum is 16+18=34 C
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Re: Question of the week  28 (In the figure above, O is the center .....)
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25 Dec 2018, 03:30
EgmatQuantExpertPlease provide detail solution... EgmatQuantExpert wrote: In the figure above, O is the center of the circle, with radius 10 units and PA is a tangent, whose length is equal to 24 units. If the lengths of PB and PC are distinct positive integers, and PB < PC, then find the sum of all possible lengths of PB? A. 16 B. 18 C. 34 D. 58 E. 126



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Re: Question of the week  28 (In the figure above, O is the center .....)
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28 Dec 2018, 03:39
Solution Given:• O is the center of the circle • BC is a chord • Radius of the circle = 10 • PA is the tangent, whose length = 24 • Lengths of PB and PC are integers, and PB < PC To find:• The sum of all possible lengths of PB Approach and Working: • Let us draw lines from A to B and from A to C Now, if we observe in triangles PAB and PAC, • ∠APB = ∠APC (common angle) • ∠PAB = ∠PCA (angles in alternate segments) Therefore, we can say that the two triangles are similar, • Thus, \(\frac{PA}{PB} = \frac{PC}{PA}\) • \(PA^2 = PB * PC\) Implies, PB * PC = \(24^2\) (substituting the value of PA) • Since, PB < PC, PB must be less than 24, and the minimum value that PB can take is when B is the intersection point of line PO and the circle. • In such case, \(PB = PO – OB = √(PA^2 + OA^2) – 10 = √676 – 10 = 26 – 10 = 16\) • Thus, 16 ≤ PB < 24, and PB must contain only 2 or 3 as its prime factors, since PB is a factor of \(24^2\) and \(24^2 = 2^6 * 3^2\) Therefore, the only possible values of PB that satisfies the above two conditions are PB = 16 or 18. So, the sum of all possible lengths of PB = 16 + 18 = 34 Hence the correct answer is Option C. Answer: C
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Re: Question of the week  28 (In the figure above, O is the center .....)
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29 Dec 2018, 09:16
chetan2u wrote: EgmatQuantExpert wrote: In the figure above, O is the center of the circle, with radius 10 units and PA is a tangent, whose length is equal to 24 units. If the lengths of PB and PC are distinct positive integers, and PB < PC, then find the sum of all possible lengths of PB? A. 16 B. 18 C. 34 D. 58 E. 126 It will be maximum when Pb~PC, that is PB is the tangent, and at that time it will be 24, so PB<24.. Now for Min, when PC lies on the hypotenuse itself... So PO = √(24^2+10^2)=√676= 26, PB will be 2610=16.. So min value is 16 and max is less than 24.. Now the property that will be useful here is (PA)^2=PB*PC... So PB will be a factor of 24^2 but less than 24 and greater than or equal to 16.. So PB can be only 16 and 18.. Sum is 16+18=34 C That was a very comprehensive approach Chetan Sir.
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Re: Question of the week  28 (In the figure above, O is the center .....)
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30 Dec 2018, 08:22
EgmatQuantExpert wrote: Solution Given:• O is the center of the circle • BC is a chord • Radius of the circle = 10 • PA is the tangent, whose length = 24 • Lengths of PB and PC are integers, and PB < PC To find:• The sum of all possible lengths of PB Approach and Working: • Let us draw lines from A to B and from A to C Now, if we observe in triangles PAB and PAC, • ∠APB = ∠APC (common angle) • ∠PAB = ∠PCA (angles in alternate segments) Therefore, we can say that the two triangles are similar, • Thus, \(\frac{PA}{PB} = \frac{PC}{PA}\) • \(PA^2 = PB * PC\) Implies, PB * PC = \(24^2\) (substituting the value of PA) • Since, PB < PC, PB must be less than 24, and the minimum value that PB can take is when B is the intersection point of line PO and the circle. • In such case, \(PB = PO – OB = √(PA^2 + OA^2) – 10 = √676 – 10 = 26 – 10 = 16\) • Thus, 16 ≤ PB < 24, and PB must contain only 2 or 3 as its prime factors, since PB is a factor of \(24^2\) and \(24^2 = 2^6 * 3^2\) Therefore, the only possible values of PB that satisfies the above two conditions are PB = 16 or 18. So, the sum of all possible lengths of PB = 16 + 18 = 34 Hence the correct answer is Option C. Answer: CI have a doubt In the figure we know: Triangle PAO = Right Angled Triangle PA = 24 AO = 10 => PO = 26 Now lets look at Triangle POC: PO = 26 OC = 10 => PC<PO + OC => PC < 26+10 => PC < 36 Since PA*PA = PB*PC For PC = 36 => 24*24= 36*PB => PB = 16 Hence PB != 16 That leaves us with only 1 value i.e. PB = 18 Please let me know where I went wrong. Thanks in advance.



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Re: Question of the week  28 (In the figure above, O is the center .....)
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04 Jan 2019, 05:08
eabhgoy wrote: I have a doubt In the figure we know: Triangle PAO = Right Angled Triangle PA = 24 AO = 10 => PO = 26 Now lets look at Triangle POC: PO = 26 OC = 10 => PC<PO + OC => PC < 26+10 => PC < 36 Since PA*PA = PB*PC For PC = 36 => 24*24= 36*PB => PB = 16 Hence PB != 16 That leaves us with only 1 value i.e. PB = 18 Please let me know where I went wrong. Thanks in advance. Hi eabhgoy, PC can be equal to OC + OP, when C is one of the end points of the diameter. In that case PC = 36 and PB will be 36  20 = 16. Regards,
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Re: Question of the week  28 (In the figure above, O is the center .....)
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06 Jan 2019, 00:23
EgmatQuantExpert wrote: eabhgoy wrote: I have a doubt In the figure we know: Triangle PAO = Right Angled Triangle PA = 24 AO = 10 => PO = 26 Now lets look at Triangle POC: PO = 26 OC = 10 => PC<PO + OC => PC < 26+10 => PC < 36 Since PA*PA = PB*PC For PC = 36 => 24*24= 36*PB => PB = 16 Hence PB != 16 That leaves us with only 1 value i.e. PB = 18 Please let me know where I went wrong. Thanks in advance. Hi eabhgoy, PC can be equal to OC + OP, when C is one of the end points of the diameter. In that case PC = 36 and PB will be 36  20 = 16. Regards, Thank you So since OC is radius then it is always a part of a diameter chord.




Re: Question of the week  28 (In the figure above, O is the center .....)
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