Subject

Quantitative Aptitude

Class

SSCCGL Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

10.

If the ratio of the angles of a quadrilateral is 2 : 7 : 2 : 7, then it is a

  • trapezium

  • parallelogram

  • square

  • rhombus


B.

parallelogram

Clearly, from the ratios, it can be drawn that opposite angles are equal. And the opposite angles of the parallelogram are always equal.
 rightwards double arrow space space space space 2 straight x plus 7 straight x plus 2 straight x plus 7 straight x space equals 360 degree
rightwards double arrow space space space space 18 straight x space equals space 360 degree
rightwards double arrow space space space straight x space equals space 20 degree
∴  First angle  = 2x = 2 x 20 = 40°
   Second angle  = 7x = 7 x 20 = 140°
    Third angle  = 2x = 2 x 20 = 40°
   Fourth angle  = 7x = 7 x 20 = 140°

 

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7.

One of the angles of a right-angled is 15°, and the hypotenuse is 1 metre. The area of the triangle (in square cm) is

  • 1220

  • 1200

  • 1250

  • 1215


C.

1250



Area of triangle  =  1 half cross times straight B cross times straight P
Base space of space triangle space space equals space cos space 15 space degree space equals space straight B over straight H
cos space 15 degree space equals space cos thin space left parenthesis 45 degree space minus space 30 degree right parenthesis
       equals space cos space 45 degree. space cos space 30 degree space plus space sin space 45 degree. space sin space 30 degree
equals space fraction numerator 1 over denominator square root of 2 end fraction cross times space fraction numerator square root of 3 over denominator 2 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space cross times space 1 half space equals space fraction numerator square root of 3 plus 1 over denominator 2 square root of 2 end fraction
 
Perpendicular of triangle  = sin space 15 degree space equals space straight P over straight H
   sin space 15 degree space equals space sin space left parenthesis 45 degree space minus space 30 degree right parenthesis
space equals space sin space 45 degree space cross times space cos space 30 degree space minus space cos space 45 degree space cross times space sin space 30 degree
equals space fraction numerator 1 over denominator square root of 2 end fraction cross times fraction numerator square root of 3 over denominator 2 end fraction space minus space fraction numerator 1 over denominator square root of 2 end fraction cross times space 1 half space equals space fraction numerator square root of 3 minus 1 over denominator 2 square root of 2 end fraction
Now, Area of triangle  equals space 1 half cross times open parentheses fraction numerator square root of 3 begin display style minus end style begin display style 1 end style over denominator 2 square root of 2 end fraction cross times fraction numerator square root of 3 begin display style minus end style begin display style 1 end style over denominator 2 square root of 2 end fraction close parentheses
                        equals space 1 half space cross times space fraction numerator left parenthesis 3 minus 1 right parenthesis over denominator 8 end fraction space equals space 1 over 8 space cm squared
equals space 1 over 8 space cross times space 10000 space equals space 1250 space cm squared



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3.

The sum of money which becomes ₹ 2420 at 10% rate of compound interest after two years is

  • ₹ 2000

  • ₹ 1000

  • ₹ 2500

  • ₹ 1500


A.

₹ 2000

Required solution: Use shortcut method:
    rightwards double arrow space space fraction numerator 2420 space cross times space 100 space cross times space 100 over denominator 110 space cross times space 110 end fraction space equals space 2000



368 Views

8.

Assume that a drop of water is spherical and its diameter is one-tenth of a cm. A conical glass has a height equal to the diameter of its rim. If 32000 drops of water fill the glass completely, then the height of the glass (in cm) is

  • 3

  • 1

  • 4

  • 2


C.

4

Volume of one drop of water equals space 4 over 3 πr cubed
equals space 4 over 3 cross times straight pi cross times space open parentheses 1 over 20 close parentheses cubed space cm cubed
∴   Volume of 32000 drops of water
     equals space fraction numerator 4 straight pi over denominator 3 end fraction space cross times space fraction numerator 32000 over denominator 20 space cross times space 20 space cross times space 20 end fraction space cm cubed
equals space fraction numerator 16 straight pi over denominator 3 end fraction cm cubed

∴  Volume of glass
               equals space 1 third πR squared straight H
Here,   Here comma space space space straight R space equals space straight H over 2

therefore space space 1 third straight pi open parentheses straight H over 2 close parentheses squared. space straight H space equals space fraction numerator 16 straight pi over denominator 3 end fraction
rightwards double arrow space space straight H cubed over 4 space equals space 16
rightwards double arrow space space space straight H cubed space equals space 64
therefore space space space straight H space equals space cube root of 64 space equals space 4 space cm.

315 Views

2.

When Principal = ₹ S, rate of interest  = 2r % p.a., then a person will get after 3 years at compound interest

  • ₹ space fraction numerator 6 Sr over denominator 100 end fraction
  • ₹ space straight S open parentheses 1 plus straight r over 100 close parentheses cubed
  • ₹ space straight S open parentheses 1 plus straight r over 50 close parentheses cubed
  • ₹ space 3 straight S open parentheses 1 plus straight r over 100 close parentheses cubed

C.

₹ space straight S open parentheses 1 plus straight r over 50 close parentheses cubed

Principal (P) = Rs. S
Rate (r) = 2r% p.a.
therefore space space straight A space equals space straight P open parentheses 1 plus straight r over 100 close parentheses to the power of straight n
equals straight S open parentheses 1 plus fraction numerator 2 straight r over denominator 100 end fraction close parentheses cubed space equals space straight S open parentheses 1 plus straight r over 50 close parentheses cubed

282 Views

6.

The in-radius of a triangle is 6 cm, and the sum of the lengths of its sides is 50 cm. The area of the triangle (in square cm.)

  • 150

  • 50

  • 300

  • 56


A.

150

OD = OE = OF = 6 cm.
Area of triangle ABC = Area of (Δ AOB + Δ BOC + Δ AOC)

equals space 1 half cross times space AB space cross times space OF space plus space 1 half space cross times space BC space cross times space OD space plus space 1 half space cross times space AC space cross times space DE
equals space 1 half cross times space 6 space cross times space left parenthesis AB space plus space BC space plus space CA right parenthesis
equals space 1 half space cross times space 6 space cross times space 50 space equals space 150 space space cm squared.

440 Views

5.

A spherical ball of radius 1 cm is dropped into a conical vessel of radius 3 cm and slant height 6 cm. The volume of water (in cm3), that can just immerse the ball, is

  • fraction numerator 5 straight pi over denominator 3 end fraction
  • straight pi over 3
  • 3 straight pi
  • fraction numerator 4 straight pi over denominator 3 end fraction

A.

fraction numerator 5 straight pi over denominator 3 end fraction

Shortcut method:
equals space 4 over 3 πr cubed space plus space 1 fourth cross times 4 over 3 πr cubed
equals space 5 over 3 πr cubed space equals space 5 over 3 straight pi space cm cubed


1295 Views

9.

If the height of a cylinder is 4 times its circumference, the volume of the cylinder in terms of its circumference c, is

  • fraction numerator 2 straight c cubed over denominator straight pi end fraction
  • 4 πc cubed
  • straight c cubed over straight pi
  • 2 πc cubed

C.

straight c cubed over straight pi

If the radius of base of cylinder be r units, then,
Height  =  4 cross times 2 πr space equals space 8 πr
therefore space space space 2 πr space equals space straight c
therefore space space space straight r space equals space fraction numerator straight c over denominator 2 straight pi end fraction space space and space space space straight h space equals space 4 straight c
therefore space space Volume space of space cylinder space space equals space πr squared straight h space equals space fraction numerator πc squared over denominator 4 straight pi squared end fraction cross times space 4 straight c space equals space straight c cubed over straight pi space cubic space units.

455 Views

4.

From a solid right circular cylinder of length 4 cm and diameter 6 cm, a conical cavity of the same height and base is hollowed out. The whole surface area of the remaining solid (in square cm) is

  • 48 space straight pi
  • 15 space straight pi
  • 63 space straight pi
  • 24 space straight pi

A.

48 space straight pi

The remaining solid is shown as below:

Now, required surface area = 2 πrh space plus space πr squared space plus space πrl
                          therefore space space space straight l space equals space square root of straight r squared plus straight h squared end root space equals space square root of 3 squared plus 4 squared end root space equals space square root of 25 space equals space 5 space cm                             
 space equals straight pi space left parenthesis 2 space cross times space 3 space cross times space 4 space plus space 3 space cross times space 5 space plus space 3 squared right parenthesis 
 equals space straight pi left parenthesis 24 plus 15 plus 9 right parenthesis space equals space 48 space straight pi

538 Views

1.

The principal which gives ₹ 1 interest per day at a rate of 5% simple interest per annum is

  • ₹ 5000

  • ₹ 7300 

  • ₹ 36500

  • ₹ 3650


B.

₹ 7300 

100 SI = P x R x T
100 x 1 = P x 5 x (1/365)
rightwards double arrow space space fraction numerator 100 space cross times space 1 space cross times space 365 over denominator 5 end fraction space equals space Rs space 7300










581 Views