Given Alphabets represents rings $E = \frac{9}{2} F = \frac{9}{4} G = \frac{9}{8} H$ and $D = 2 E$ . If parents want to buy two rings with the condition that the price difference between two should be minimal, then which two rings quality for this condition?

E & F

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G & H

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H & E

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D & E

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Solution

The correct option is B H & E
$E = \frac{9}{2} F = \frac{9}{4} G = \frac{9}{8} H$ and $D = 2 E$
Let price of $H = 80$ . Then
Price of $G = 40$ ; Price of $F = 20$ ;
Price of $E = 90$ ; and price of $D = 160$
Therefore price difference is minimum between H and E.

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Given Alphabets represents rings E=92F=94G=98H and D=2E. If parents want to buy two rings with the condition that the price difference between two should be minimal, then which two rings quality for this condition?

The correct option is B H & EE=92F=94G=98H and D=2ELet price of H=80. ThenPrice of G=40; Price of F=20;Price of E=90; and price of D=160Therefore price difference is minimum between H and E.

Given Alphabets represents rings $E = \frac{9}{2} F = \frac{9}{4} G = \frac{9}{8} H$ and $D = 2 E$ . If parents want to buy two rings with the condition that the price difference between two should be minimal, then which two rings quality for this condition?

The correct option is B H & E
$E = \frac{9}{2} F = \frac{9}{4} G = \frac{9}{8} H$ and $D = 2 E$
Let price of $H = 80$ . Then
Price of $G = 40$ ; Price of $F = 20$ ;
Price of $E = 90$ ; and price of $D = 160$
Therefore price difference is minimum between H and E.