IPMAT 2019 Question Paper IPM Indore Quantitative Ability. Solve questions from IPMAT 2019 Question Paper from IPM Indore and check the solutions to get adequate practice. The best way to ace IPMAT is by solving IPMAT Question Paper. To solve other IPMAT Sample papers, go here:**IPM Sample Paper**

Question 33: On a circular path of radius 6 m a boy starts from a point A on the circumference and walks along a chord AB of length 3 m. He then walks along another chord BC of length 2 m to reach point C. The point B lies on the minor arc AC. The distance between point C from point A is

- \\frac{\sqrt{15} + \sqrt{35}}{2}\\) m
- 8 m
- √13 m
- 6 m

\\frac{\sqrt{15} + \sqrt{35}}{2}\\) m

Try upto 40 hours for free

Learn from the best!

Limited Seats Available - Register Now!

We are given that,

OA = OB = OC = 6m.

AB = 3m.

BC = 2m.

Let’s say angle AOB = ∅ and angle BOC = ∝.

Cos ∅ = \\frac{AO^{2} + BO^{2} - AB^{2}}{2.𝐴𝑂.𝐵𝑂}\\) = \\frac{6^{2} + 6^{2} - 3^{2}}{2.6.6}\\) = \\frac{63}{72}\\) = \\frac{7}{8}\\)

Sin ∅ = \\sqrt{1 − cos^{2} ∅}\\) = \\sqrt{1 − (\frac{7}{8})^{2}}\\) = \\sqrt{\frac{64 - 49}{64}}\\) = \\sqrt{\frac{15}{64}}\\) = \\frac{\sqrt{15}}{8}\\)

Similarly Cos ∝ = \\frac{CO^{2} + BO^{2} - AC^{2}}{2.𝐴𝑂.C𝑂}\\)= \\frac{6^{2} + 6^{2} - 2^{2}}{2.6.6}\\)= \\frac{68}{72}\\) = \\frac{17}{18}\\)

Sin ∝ = \\sqrt{1 − cos^{2} ∝}\\) = \\sqrt{1 − (\frac{17}{18})^{2}}\\) = \\sqrt{\frac{35}{324}}\\) = \\frac{\sqrt{35}}{18}\\)

Cos (∅ + ∝) = Cos ∅ Cos ∝ - Sin ∅ Sin ∝

= \\frac{7}{8}\\) . \\frac{17}{18}\\) - \\frac{\sqrt{15}}{8}\\) . \\frac{\sqrt{35}}{18}\\) = \\frac{119 - 5\sqrt{21}}{8.18}\\)

Also,

Cos (∅ + ∝) = \\frac{AO^{2} + CO^{2} - AC^{2}}{2.𝐴𝑂.C𝑂}\\) =\\frac{6^{2} + 6^{2} - AC^{2}}{2.6.6}\\) = \\frac{72 - AC^{2}}{72}\\)

\\frac{72 - AC^{2}}{72}\\) = \\frac{119 - 5\sqrt{21}}{8*18}\\)

AC^{2}= 72 - \\frac{72(119 - 5\sqrt{21})}{8*18}\\)

= \\frac{72*8*18 - 72*119 + 72*5\sqrt{21}}{8*18}\\)

= \\frac{25 + 5\sqrt{21}}{2}\\)

Going by the options if AC = \\frac{\sqrt{15} + \sqrt{35}}{2}\\)

AC^{2}= \\frac{(\sqrt{15} + \sqrt{35})^{2}}{4}\\) = \\frac{15 + 35 + 2\sqrt{15*35}}{4}\\)

AC^{2}= \\frac{50 + 10\sqrt{21}}{4}\\)

AC^{2}= \\frac{25 + 5\sqrt{21}}{2}\\), which concurs with our calculated value of AC^{2}.

Therefore, The distance between C and A is \\frac{\sqrt{15} + \sqrt{35}}{2}\\) m.

The question is**"On a circular path of radius 6 m a boy starts from a point A on the circumference and walks along a chord AB of length 3 m. He then walks along another chord BC of length 2 m to reach point C. The point B lies on the minor arc AC. The distance between point C from point A is"**

Choice A is the correct answer

Copyrights © All Rights Reserved by 2IIM.com - A Fermat Education Initiative.

Privacy Policy | Terms & Conditions

CAT^{®} (Common Admission Test) is a registered trademark of the Indian Institutes of Management. This website is not endorsed or approved by IIMs.

2IIM Online CAT Coaching

A Fermat Education Initiative,

58/16, Indira Gandhi Street,

Kaveri Rangan Nagar, Saligramam, Chennai 600 093

**Mobile:** (91) 99626 48484 / 94459 38484

**WhatsApp:** WhatsApp Now

**Email: **info@2iim.com