9:{"metadata":[["$","title","0",{"children":"iimkbmsat0025 - If one angle of a parallelogram is 24^ less than twice its adjacent angle, then the largest angle of the parallelogram is | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Let one ang<= of the paral<=logram be x. Since one ang<= is 24° <=ss than twice its adjacent ang<=, the adjacent ang<= is 2x - 24°.

In a paral<=logram, adjacent ang<=s are supp<=mentary (add up to 180°): x + (2x - 24°) = 180°

Simplifying the equation: 3x - 24° = 180° 3x = 204° x = 68°

Finding both ang<=s: First ang<= = 68° Second ang<= = 2(68°) - 24° = 136° - 24° = 112°

In a paral<=logram, opposite ang<=s are equal, so we have: Two ang<=s of 68° each Two ang<=s of 112° each Therefore, the lar>=st ang<= is 112°."}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"IIM-K (BMSAT),QA,Geometry,Quadrilaterals"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"property":"og:title","content":"iimkbmsat0025 - If one angle of a parallelogram is 24^ less than twice its adjacent angle, then the largest angle of the parallelogram is | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined Let one ang<= of the paral<=logram be x. Since one ang<= is 24° <=ss than twice its adjacent ang<=, the adjacent ang<= is 2x - 24°.

In a paral<=logram, adjacent ang<=s are supp<=mentary (add up to 180°): x + (2x - 24°) = 180°

Simplifying the equation: 3x - 24° = 180° 3x = 204° x = 68°

Finding both ang<=s: First ang<= = 68° Second ang<= = 2(68°) - 24° = 136° - 24° = 112°

In a paral<=logram, opposite ang<=s are equal, so we have: Two ang<=s of 68° each Two ang<=s of 112° each Therefore, the lar>=st ang<= is 112°."}],["$","meta","9",{"property":"og:url","content":"/iimk-bmsat-past-year-questions/2025/QA/40"}],["$","meta","10",{"property":"og:site_name","content":"AfterBoards | JIPMAT Preparation"}],["$","meta","11",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","12",{"property":"og:type","content":"website"}],["$","meta","13",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","14",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","15",{"name":"twitter:title","content":"iimkbmsat0025 - If one angle of a parallelogram is 24^ less than twice its adjacent angle, then the largest angle of the parallelogram is | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined Let one ang<= of the paral<=logram be x. Since one ang<= is 24° <=ss than twice its adjacent ang<=, the adjacent ang<= is 2x - 24°.

In a paral<=logram, adjacent ang<=s are supp<=mentary (add up to 180°): x + (2x - 24°) = 180°

Simplifying the equation: 3x - 24° = 180° 3x = 204° x = 68°

Finding both ang<=s: First ang<= = 68° Second ang<= = 2(68°) - 24° = 136° - 24° = 112°

In a paral<=logram, opposite ang<=s are equal, so we have: Two ang<=s of 68° each Two ang<=s of 112° each Therefore, the lar>=st ang<= is 112°."}],["$","meta","17",{"name":"twitter:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","link","18",{"rel":"shortcut icon","href":"https://www.afterboards.in/favicon.ico","type":"image/x-icon","sizes":"48x48"}],"$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f"],"error":null,"digest":"$undefined"}

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