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  1. C > B > A
  2. C > A > B
  3. B > C > A
  4. B > A > C
  5. A > B < C
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1 Answers

Option 2 : C > A > B

Quantity A: The ratio of children in three classes X, Y, Z is 39 : 64 : 58.

Let the number of children in each class be 39a, 64a, 58a. Then, (39 + 64 + 58)a = 1288

⇒ a = 1288/161 ⇒ a = 8

∴ Number of children in class X, Y and Z respectively = 39 × 8, 64 × 8, 58 × 8 = 312, 512 and 464

In class X number of boys is 26 less than girl ⇒ b = g – 26      ----(1)

Also, b + g = 312      ----(2)

Solving (1), and (2) ----> b(boys) = 143, g(girls) = 169

Thus, B = 169

Now, Number of boys in class Y is 67.83% more than the number of boys in class X.

Thus, number of boys in class Y=167.83% of 143 = (167.83 × 143)/100 = 239.99 = 240 (rounded off to nearest whole number)

Now, total number of children in class Y = 512

Number of girls in class Y = total children – Number of boys ⇒512 – 240 = 272

Number of girls in class Y = 272

∴ C = 272

Girls in class Z are 25.36% less than the number of girls in class Y.

Thus, number of girls in class Z = 74.64% of 272 = (74.64 × 272)/100 = 203.03 = 203 (rounded off to nearest whole number)

Now, total number of children in class z = 464

Number of boys in class Z = total children – Number of girls ⇒464 – 203 = 261

Number of boys in class Z = 261

⇒ A = 261

Hence, C > A > B.

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