Subject

Quantitative Aptitude

Class

SSCCGL Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

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


A.

50 m

Volume of pyramid
    equals space 1 third space cross times space left parenthesis area space of space base right parenthesis space cross times space height
rightwards double arrow space 500 space equals space 1 third space cross times space 30 space cross times space straight h
rightwards double arrow space space 10 straight h space equals space 500
rightwards double arrow space space space space space straight h space equals space 500 over 10 space equals space 50 space metre

238 Views

9.

The base of a right prism is an equilateral triangle. If the lateral surface area and volume is 120 cm240 square root of 3 space cm cubed respectively then the side of base of the prism is

  • 4 cm

  • 5 cm

  • 7 cm

  • 40 cm


A.

4 cm

Lateral surface area of prism  = Base perimeter x Height
  Base Perimeter is of equilateral triangle  = 3a 
 Height (h)
∴     120 cm2 = 3a x h
  rightwards double arrow space space straight a space cross times space straight h space equals space 120 over 3 space equals space space 40 space cm squared ... (i)
Volume of prism  = Base Area x Height
rightwards double arrow space space 40 square root of 3 space equals space fraction numerator square root of 3 over denominator 4 end fraction space cross times space left parenthesis side right parenthesis squared space cross times space straight h
rightwards double arrow space space fraction numerator 40 square root of 3 space cross times space 4 over denominator square root of 3 end fraction space equals space straight a squared space cross times space straight h
rightwards double arrow space space space space space space 160 space equals space straight a squared space cross times space straight h space space space space space space space... left parenthesis ii right parenthesis
Dividing equation (ii) by (i),
Side(a) = 160 over 40 equals space space 4 space cm

849 Views

10.
Perimeter of a rhombus is 2p unit and sum of length of diagonals is m unit, then area of the rhombus is
  • 1 fourth straight m squared straight p space sq space unit
  • 1 fourth mp squared space sq space unit
  • 1 fourth left parenthesis straight m squared minus straight p squared right parenthesis space sq space unit
  • 1 fourth left parenthesis straight p squared minus straight m squared right parenthesis space sq space unit

C.

1 fourth left parenthesis straight m squared minus straight p squared right parenthesis space sq space unit

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
                          
 
         

786 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 left parenthesis square root of 3 space equals space 1.732 right parenthesis

  • 98.55 sq cm

  • 100 sq cm

  • 101 sq cm

  • 95 sq cm


A.

98.55 sq cm

Side (a) = 24 cm

Area space of space triangle space space equals space fraction numerator square root of 3 over denominator 4 end fraction space cross times space left parenthesis side right parenthesis squared
space equals space fraction numerator square root of 3 over denominator 4 end fraction space cross times space 24 space cross times space 24 space equals space 144 square root of 3 space sq. space cm space equals space 144 space cross times space 1.732 space equals space 249.408 space cm squared

In radius = fraction numerator straight a over denominator 2 square root of 3 end fraction space equals space fraction numerator 24 over denominator 2 square root of 3 end fraction space equals space 4 square root of 3 space cm
Area of circle = πr squared
space equals space 22 over 7 cross times space 4 square root of 3 space cross times space 4 square root of 3
equals space 1056 over 7 space equals space 150.86 space cm squared
Area of remaining part
= (249.408 - 150.86) cm2
equals space 98.548 space sq. cm space almost equal to space 98.55 space cm squared


1098 Views

4.

If the surface area of sphere is 346.5 cm2, then its radius open square brackets taking space straight pi space equals space 22 over 7 close square brackets is

  • 7 cm

  • 3.25 cm

  • 5.25 cm

  • 9 cm


C.

5.25 cm

Surface area of sphere:
      equals space space 4 πr squared
equals space space 4 space cross times space 22 over 7 space cross times space straight r squared space equals space 346.5
rightwards double arrow space space straight r squared space equals space fraction numerator 346.5 space cross times space 7 over denominator 4 space cross times space 22 end fraction space equals space 27.5625
rightwards double arrow space space straight r space equals space square root of 27.5625 end root space equals space 5.25 space cm

        

262 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 

  • fraction numerator AD with bar on top over denominator BC end fraction
  • fraction numerator BE with bar on top over denominator EC with bar on top end fraction
  • fraction numerator AD with bar on top space plus space BE with bar on top over denominator AD with bar on top space plus space CE with bar on top end fraction
  • fraction numerator AD with bar on top space plus space BC with bar on top over denominator AD with bar on top end fraction

D.

fraction numerator AD with bar on top space plus space BC with bar on top over denominator AD with bar on top end fraction

EF is perpendicular on side AD.
∴  Area of trapezium
    equals space 1 half left parenthesis AD space plus space BC right parenthesis space cross times space EF
Area of ΔAED = 1 half cross times AD cross times EF
∴  Required ratio
             equals space fraction numerator begin display style 1 half end style cross times left parenthesis AD space plus space BC right parenthesis space cross times space EF over denominator begin display style 1 half end style cross times space AD space cross times space EF end fraction
equals space fraction numerator AD space plus space BC over denominator AD end fraction
593 Views

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


C.

₹ 3200

Shortcut Method:
Go with option (3) and check it.
Principal is Rs3200
3200 rightwards arrow with 5 percent sign space equals space 160 on top left parenthesis 3200 plus 160 space equals space 3360 right parenthesis rightwards arrow with 5 percent sign space equals space 168 on top left parenthesis 160 plus 168 space equals space 328 right parenthesis

1217 Views

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


C.

12

Sum of interior and exterior angle  = 180°
  I + E = 180°
Let the exterior angle be x.
rightwards double arrow space space 5 straight x space plus space straight x space equals space 180 degree
rightwards double arrow space space space space space space 6 straight x space equals space 180 degree
rightwards double arrow space space space space space space straight x space space equals space 30 degree
Number of sides of polygon:
equals space fraction numerator Sum space of space exterior space angle over denominator Each space exterior space angle end fraction space equals space fraction numerator 360 degree over denominator 30 degree end fraction space equals space 12

332 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 over 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


B.

20 cm


Let H and R be the height and radius of bigger cone and h and r be that of smaller cone.
∠A is common and MN || OB.
therefore triangle AOB space space tilde space space triangle AMN
So,
    rightwards double arrow space space AO over AM space equals space BO over MN
rightwards double arrow space space space 30 over straight h space equals space straight R over straight r space space space space space space space space space space space space... left parenthesis straight i right parenthesis

Volume of bigger cone equals space 1 third πR squared straight H
Volume of smaller cone = 1 third πr squared straight h
According to the question,
   open parentheses 1 third πR squared straight H close parentheses cross times space 1 over 27 space equals space 1 third πr squared straight h
 rightwards double arrow space space fraction numerator straight R squared straight H over denominator 27 end fraction space equals space straight r squared straight h
rightwards double arrow space space 27 straight r squared straight h space equals space straight R squared straight H
rightwards double arrow space space fraction numerator 27 straight h over denominator straight H end fraction space equals space straight R squared over straight r squared

rightwards double arrow space space fraction numerator 27 straight H over denominator straight H end fraction space equals space open parentheses 30 over straight h close parentheses squared space space space space space... left square bracket From space left parenthesis straight i right parenthesis right square bracket
rightwards double arrow space space space fraction numerator 27 straight h over denominator straight H end fraction space equals space 900 over straight h squared

rightwards double arrow space space 27 straight h cubed space equals space 900 space straight H space equals space 900 space cross times space 30
rightwards double arrow space space straight h cubed space equals space fraction numerator 900 space cross times space 30 over denominator 27 end fraction space equals space 1000
rightwards double arrow space space straight h space equals space cube root of 1000 space equals space 10 space cm
∴  Required height  = 30 - 10 = 20 cm  
   
241 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


C.

330 sq cm

Perpendicular (P) = 5 , Base (B) = 12,  straight H space equals space square root of straight p squared plus straight b squared end root space equals space square root of 5 squared plus 12 squared end root space equals space 13cm
Curved Surface area  = Base Perimeter x Height
                              = (5 + 12 + 13) x 10 = 300 cm2
Base Area = 1 half cross times 5 cross times 12 = 30 cm2
Total Surface Area = CSA + 2Base Area
                           = 300 + 30 = 330 cm2.

510 Views