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SSCCGL Quantitative Aptitude Solved Question Paper 2013

Multiple Choice Questions

101.

Three sides of a triangular field are of length 15 m, 20 m and 25 m long, respectively.

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102.

The value of (tan 1° tan 2° tan 3°.... tan 89°) is

undefined
0
1
89

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103.

A, B, C and D purchase a gift worth ₹ 60. A pays 1/2 of what others are paying, B pays 1/3rd of what others are paying and C pays 1/4 of what others are paying. What is the amount paid by D?

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104.

Mohan gets 3 marks for each correct sum and loses 2 marks for each wrong sum. He attempts 30 sums and obtains 40 marks. The number of sums solved correctly is

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105.

The greatest common divisor of
$3 to the power of 3 to the power of 333 end exponent space plus space 1$ and $3 to the power of 3 to the power of 334 end exponent space plus space 1$ is

$3 to the power of 3 to the power of 333 end exponent space plus space 1$
20
2
1

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106.

If the length of the sides of a triangle are in the ratio 4 : 5 : 6 and the inradius of the triangle is 3 cm, then the altitude of the triangle corresponding to the largest side as base is

10 cm
8 cm
7.5 cm
6 cm

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107.

The cost of manufacture of an article was ₹ 900. The trader wants to gain 25% after giving a discount of 10%. The marked price should be

₹ 1000
₹ 1500
₹ 1250
₹ 1200

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108.
If a² + 1 = a, then the value of a¹² + a⁶ + 1 is

2

3

-3

1

a² + 1 = a
$rightwards double arrow space space space space straight a space plus space 1 over straight a space equals space 1$
On Squaring both sides, we get,
$straight a squared space plus space 1 over straight a squared space plus space 2 space equals space 1 rightwards double arrow space space space straight a squared space plus space 1 over straight a squared space equals space minus 1$
On Cubing both sides, we get,
$space space space open parentheses straight a squared space plus space 1 over straight a squared close parentheses cubed space equals space left parenthesis negative 1 right parenthesis cubed rightwards double arrow space space straight a to the power of 6 space plus space 1 over straight a to the power of 6 space plus space 3 straight a squared space cross times space 1 over straight a squared open parentheses straight a squared space plus space 1 over straight a squared close parentheses space equals space minus 1 space space rightwards double arrow space space straight a to the power of 6 space plus space 1 over straight a to the power of 6 space plus space 3 space cross times space minus 1 space equals space minus 1 rightwards double arrow space space space space straight a to the power of 6 space plus space 1 over straight a to the power of 6 space plus space 1 space equals space 3 space space space space space space space space space space space space space... left parenthesis straight i right parenthesis$
If we put a = 1 in equation (i), it satisfies the equation. i.e. a = 1.
Now, put a = 1. in a¹² + a⁶ + 1 i.e (1)¹² + (1)⁶ + 1 = 3

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109.

If $straight x space equals space straight a space sec space straight theta space cos space straight ϕ comma space space straight y space equals space straight b space sec space straight theta space sin space straight ϕ$ and $straight z space equals space straight c space tan space straight theta$ . then the value of $straight x squared over straight a squared space plus space straight y squared over straight b squared space equals space straight z squared over straight c squared$

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110.

A person distributes his pens among four friends A, B, C and D in the ratio $1 third space colon space 1 fourth space colon space 1 fifth space colon space 1 over 6$ . What is the minimum number of pens that the person should have?