Subject

Quantitative Aptitude

Class

SSCCGL Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

61.

The length of the base of an isosceles triangle is 2x - 2y + 4z, and its perimeter is 4x - 2y + 6z. Then the length of each of the equal sides is:

  • x + y

  • x + y + z

  • 2(x + y)

  • x + z

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62.

If θ > 0, be acute angle, the value of θ in degrees satisfying
fraction numerator cos squared straight theta space minus space 3 space cosθ space plus space 2 over denominator sin squared straight theta end fraction space equals space 1 is:

  • 90°

  • 30°

  • 45°

  • 60°

60 Views

63.

If the numbers cube root of 9 space comma space fourth root of 20 space comma space space root index 6 of 25 are arranged in ascending order, then the right arrangement is:

  • root index 6 of 25 space less than space fourth root of 20 space less than space cube root of 9

  • cube root of 9 space less than space fourth root of 20 space less than space root index 6 of 25

  • fourth root of 20 space less than space root index 6 of 25 space less than space cube root of 9

  • root index 6 of 25 space less than space cube root of 9 space less than space fourth root of 20

54 Views

64.

Visitors to a show were charged â‚¹ 15 each on the first day, â‚¹ 7.50 on the second day, â‚¹ 2.50 on the third day and total attendance on three days were in the ratio 2 : 5 : 13  respectively. The average charge per person for the entire three day is

  • ₹ 5

  • ₹ 5.50

  • ₹ 6

  • ₹ 7

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65.

If fraction numerator 2 plus straight a over denominator straight a end fraction space plus space fraction numerator 2 plus straight b over denominator straight b end fraction space plus space fraction numerator 2 plus straight c over denominator straight c end fraction space equals space 4 comma then the value of fraction numerator ab plus bc plus ca over denominator abc end fraction is:

  • 2

  • 1

  • 0

  • 1 half


D.

1 half

fraction numerator 2 plus straight a over denominator straight a end fraction space plus space fraction numerator 2 plus straight b over denominator straight b end fraction space plus space fraction numerator 2 plus straight c over denominator straight c end fraction space equals space 4
fraction numerator 2 plus straight a over denominator straight a end fraction minus 1 space plus space fraction numerator 2 space plus space straight b over denominator space straight b end fraction space minus space 1 space plus space fraction numerator 2 space plus space straight c over denominator straight c end fraction minus 1 space equals space 4 minus 3
fraction numerator 2 plus straight a minus straight a over denominator straight a end fraction space plus space fraction numerator 2 plus straight b minus straight b over denominator straight b end fraction space plus space fraction numerator 2 plus straight c minus straight c over denominator straight c end fraction space equals space 1
2 over straight a space plus space 2 over straight b space plus space 2 over straight c space equals space 1
2 open square brackets 1 over straight a space plus space 1 over straight b space plus space 1 over straight c close square brackets space equals space 1
fraction numerator bc space plus space ac space plus space ab over denominator abc end fraction space equals space 1 half
therefore space space space fraction numerator ab space plus space bc space plus space ca over denominator abc end fraction space equals space 1 half

 

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66.

If straight x space plus space straight y plus space straight z space equals space 1 comma space space  1 over straight x plus 1 over straight y plus 1 over straight z space equals space 1 and xyz = -1, then x3 + y3 + z3 is equal to:

  • -1

  • 1

  • -2

  • 2

51 Views

67.

In Δ PQR, L and M are two points on the sides PQ and PR respectively such that LM || QR. If PL = 2 cm, LQ = 6 cm and PM = 1.5 cm,  then MR in cm is

  • 0.5

  • 4.5

  • 9

  • 8

74 Views

68.

The length of the radius of a circle with centre O is 5 cm and the length of the chord AB is 8 cm. The distance of the chord AB from the point O is

  • 2 cm

  • 3 cm

  • 4 cm

  • 15 cm

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69.

The upper part of a tree broke at a certain height makes an angle of 60° with the ground at a distance of 10 m from its feet. The original height of the tree was

  • 20√3 m

  • 10√3 m

  • 10(2 + √3) m

  • 10(2 - √3) m

51 Views

70.

What would be the compound interest of â‚¹ 25,000/- for 2 years at 5% per annum?

  • ₹ 2,500

  • ₹ 2,562.5

  • ₹ 2,425.25

  • ₹ 5,512.5

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