Subject

Quantitative Aptitude

Class

SSCCGL Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The compound interest on a certain sum of money for 2 years at 5% is ₹328, then the sum is

  • ₹ 3000

  • ₹ 3600

  • ₹ 3200

  • ₹ 3400

1217 Views

2.

The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be 1/27 th of the volume of the given cone, at what height above the base is the section made?

  • 19 cm

  • 20 cm

  • 12 cm

  • 15 cm

241 Views

3.

ABCD is a trapezium with AD and BC parallel sides. E is a point on BC. The ratio of the area of ABCD to that of AED is 

593 Views

4.

If the surface area of sphere is 346.5 cm2, then its radius [ taking π = 22/7 ] is

  • 7 cm

  • 3.25 cm

  • 5.25 cm

  • 9 cm

262 Views

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5.

An interior angle of a regular polygon is 5 times its exterior angle. Then the number of sides of the polygon is

  • 14

  • 16

  • 12

  • 18

332 Views

6.

The height of the right pyramid whose area of the base is 30m2 and volume is 500 m3, is

  • 50 m

  • 60 m

  • 40 m

  • 20 m

238 Views

7.

The base of a right prism is a right angle triangle with two sides 5cm and 12 cm. The height of the prism is 10 cm. The height of the prism is 10 cm. The total surface area of the prism is

  • 360 sq cm

  • 300 sq cm

  • 330 sq cm

  • 325 sq cm

510 Views

8.

In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides. The area of the remaining portion of the triangle is ( √3 = 1.732 )

  • 98.55 sq cm

  • 100 sq cm

  • 101 sq cm

  • 95 sq cm

1098 Views

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9.

The base of a right prism is an equilateral triangle. If the lateral surface area and volume is 120 cm2, 40√3 cm3 respectively then the side of base of the prism is

  • 4 cm

  • 5 cm

  • 7 cm

  • 40 cm

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10.

Perimeter of a rhombus is 2p unit and sum of length of diagonals is m unit, then area of the rhombus is


C.

Let d1, d2 be the diagonals of the rhombus
Perimeter of rhombus (P) = 2 square root of straight d subscript 1 squared plus straight d subscript 2 squared end root
                         Perimeter = 2p(given)

∴       2 straight p space equals space 2 square root of straight d subscript 1 squared space plus space straight d subscript 2 squared end root
Squaring both the sides, we get
         straight p squared space equals space straight d subscript 1 squared space plus space straight d subscript 2 squared       ...(i)
Now, sum of diagonals (d1 + d2) =  m (given)
  Again, Squaring both the sides, we get
               left parenthesis straight d subscript 1 space plus space straight d subscript 2 right parenthesis squared space equals space straight m squared
         rightwards double arrow space space straight d subscript 1 squared space plus space straight d subscript 2 squared space plus space 2 space cross times space straight d subscript 1 space cross times space straight d subscript 2 space equals space straight m squared
         rightwards double arrow space space space straight p squared space plus space 2. straight d subscript 1. straight d subscript 2 space equals space straight m squared space space space space left square bracket using space left parenthesis straight i right parenthesis right square bracket
rightwards double arrow space space space space space space space space space space 2. space straight d subscript 1. space straight d subscript 2 space equals space straight m squared space minus space straight p squared
rightwards double arrow space space space space space space space space space space space straight d subscript 1. space straight d subscript 2 space equals space fraction numerator straight m squared minus straight p squared over denominator 2 end fraction space.... left parenthesis ii right parenthesis

Area of rhombus = 1 half cross times space straight d subscript 1 space cross times space straight d subscript 2
                        equals space 1 half space cross times space fraction numerator straight m squared minus straight p squared over denominator 2 end fraction space space space left square bracket using space left parenthesis ii right parenthesis right square bracket
equals space 1 fourth left parenthesis straight m squared minus straight p squared right parenthesis
                          
 
         

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