indices

45.
The answer is : A. 121/1197

[5•49^n - 3•7^(2n -2)] / [7^(2n+2) - 7^(2n-1)]
= [5•7^2•7^(2n-2) - 3•7^(2n -2)] / [7^4•7^(2n-2) - 7•7^(2n-2)]
= [245•7^(2n-2) - 3•7^(2n -2)] / [2401•7^(2n-2) - 7•7^(2n-2)]
= (245 - 3)•7^(2n-2) / (2401 - 7)•7^(2n-2)
= 242/2394
= 121/1197


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44.
The answer is : D. 1/30

[25^(n+1) - 5^(2n+1)] / [25^(n + 2) - 5^(2n+2)]
= [5^2(n+1) - 5^(2n+1)] / [5^2(n + 2) - 5^(2n+2)]
= [5^(2n+2) - 5^(2n+1)] / [5^(2n + 4) - 5^(2n+2)]
= [5•5^(2n+1) - 5^(2n+1)] / [5^3•5^(2n + 1) - 5•5^(2n+1)]
= (5 - 1)•5^(2n+1) / (125 - 5)•5^(2n+1)
= 4/120
= 1/30


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53.
The answer is : D. I, II and III

I. 6/√3 = 6√3/(√3)² = 6√3/3 = 2√3
II. √432 - √300 = 12√3 - 10√3 = 2√3
III. √12 = 2√3


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54.
The answer is : A. 21310

A816 = 10x16 + 8 = 16810
1011012 = 1x2^5 + 1x2^3 + 1x2^2 + 1 = 4510

A816 + 1011012 = 16810 + 4510 = 21310


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56.
The answer is : C. 1E016

10(a² + 3ab + b²)
= 10[(a² + 2ab + b²) + ab]
= 10[(a + b)² + ab]
= 10{[(√11 + √7) + (√11 - √7)]² + (√11 + √7)(√11 - √7)}
= 10[(2√11)² + (11 - 7)]
= 10(44 + 4)
= 48010
= 1E016

16 |­ 480
16 | _30 ...... 0
_____ 1 ...... 14


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57.
The answer is : C. I and II only

I. 12 = 110
Hence, I is true.

II. 916 x 910 = 910 x 910 = 8110
Hence, II is true.

III. 562910 is an odd number, but 10101111111002 is aneven number.
Hence, III is false.