1.
If the cost of a diamond varies directly with the square of its weight and if the weight of the
diamond is 5

2 thoughts on “<br /> 1.<br /> If the cost of a diamond varies directly with the square of its weight and if the weight of the<br /> diamond is 5”

1. Answer: Rs. 8640

   Given,

   ▶ The cost of diamond varies directly with the square of its weight. Also,

   • When the weight of diamond is 5 g, then the cost of the diamond is Rs. 1500.

   We have to find the cost of the diamond of 12 grams.

   According to the given statement,

   ⇒ Cost ∝ Weight²

   ⇒ Cost = k . Weight²

   [ k = proportionality constant ]

   ⇒ 1500 = k . (5)²

   ⇒ k = 1500/25

   ⇒ k = 60

   Now, We have

   • k = 60, Weight = 12 g

   Let’s find the cost using the above formula,

   ⇒ Cost = k . Weight²

   ⇒ Cost = 60 . (12)²

   ⇒ Cost = 60 × 144

   ⇒ Cost = Rs. 8640

   Hence, The cost of 12g diamond is Rs. 8640

   Note:

   You may think the question can also be solved by just getting the cost of 1g diamond and multiply by 12 to get the cost of 12g of diamond, In that case, Cost will be Rs. 3600.

   But, you should notice that the cost of diamond is not dependent on the weight linearly and that’s why you would be wrong if you did that way.

   Reply

2. Given : The cost of a diamond varies directly with the square of its weight & if the weight of the

   diamond is 5 grams, then the cost is Rs.1500 .

   Exigency To Find : The cost of 12 grams diamond .

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   ❍ The Given Statement :

   • The cost of a diamond varies directly with the square of its weight .

   Or ,

   • Cost of Diamond [tex]\propto[/tex] square of Weight

   Or ,

   • Cost of Diamond [tex]\propto[/tex] Weight ²

   As , We now that ,

   • k is the Proportionality Constant .

   • [tex]\dag\:\:\boxed {\sf{ Cost \:of\:Diamond = k \times (Weight) ^2 }}\\[/tex]

   Where ,

   • Cost of Diamond = ₹ 1500
   • Weight of Diamond = 5 gm

   ⠀⠀⠀⠀⠀⠀[tex]\underline {\frak{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}\\[/tex]

   [tex]\qquad :\implies \sf{ 1500 = k ( 5)^2 }\\[/tex]

   [tex]\qquad :\implies \sf{ 1500 = k ( 25 )}\\[/tex]

   [tex]\qquad :\implies \sf{ \cancel {\dfrac{1500}{25}} = k }\\[/tex]

   ⠀⠀⠀⠀⠀[tex]\underline {\boxed{\pink{ \mathrm { k = \: 60}}}}\:\bf{\bigstar}\\[/tex]

   ⠀⠀⠀⠀⠀Finding Cost of 12 gm of Diamond :

   As, We already know that ,

   • [tex]\dag\:\:\boxed {\sf{ Cost \:of\:Diamond = k \times (Weight )^2 }}\\[/tex]

   Where,

   • k = 60 .
   • Weight of Diamond = 12 gm
   • Cost of Diamond = ??

   ⠀⠀⠀⠀⠀⠀[tex]\underline {\frak{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}\\[/tex]

   [tex]\qquad :\implies \sf{ Cost \:of\:Diamond\:= 60 \times ( 12)^2 }\\[/tex]

   [tex]\qquad :\implies \sf{ Cost \:of\:Diamond\:= 60 \times ( 144 )}\\[/tex]

   ⠀⠀⠀⠀⠀[tex]\underline {\boxed{\pink{ \mathrm { Cost \:of\:Diamond\:= Rs.8640\: }}}}\:\bf{\bigstar}\\[/tex]

   Therefore,

   ⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \mathrm { Hence, \:The\:cost\:of\:12\:gm \:of\:Diamond \:is\:\bf{Rs.8640\: }}}}\\[/tex]

   ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

   Reply