Subject

Quantitative Aptitude

Class

SSCCGL Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

55.

To cover a distance with a speed of 60 km/hr, a train takes 15 hours. If it covers the same distance in 12 hours, what will be its speed?

  • 65 km/hr

  • 70 km/hr

  • 75 km/hr

  • 80 km/hr


C.

75 km/hr

Distance  = Speed x Time
              = 60 x 15 = 900 km
therefore space space Required space speed space equals space Distance over Time space equals space 900 over 12 space equals space 75 space km divided by hr

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57.

If straight x space plus space 1 over straight x space equals space 5 comma then the value of fraction numerator 5 straight x over denominator straight x squared space plus space 5 straight x space plus space 1 end fraction is:

  • 1 third

  • 1 fourth

  • 1 half

  • 1 fifth


C.

1 half

rightwards double arrow space space space straight x space plus space 1 over straight x space equals space 5 space space left parenthesis given right parenthesis
rightwards double arrow space space space fraction numerator 5 straight x over denominator straight x squared plus 5 straight x plus 1 end fraction space equals space fraction numerator 5 straight x over denominator straight x squared plus 1 plus 5 straight x end fraction
rightwards double arrow space space space fraction numerator 5 straight x over denominator straight x open parentheses straight x space plus space begin display style 1 over straight x end style close parentheses space plus space 5 straight x end fraction
rightwards double arrow space space space space fraction numerator 5 straight x over denominator straight x open square brackets open parentheses straight x space plus space begin display style 1 over straight x end style close parentheses space plus space 5 close square brackets end fraction
rightwards double arrow space space space space space space fraction numerator 5 over denominator open parentheses straight x space plus space begin display style 1 over straight x end style close parentheses plus 5 end fraction
equals space fraction numerator 5 over denominator 5 plus 5 end fraction space equals 5 over 10 space equals space 1 half

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54.

A man spends 15% of his income. If his expenditure is â‚¹ 75, his income in (₹ 75) is:

  • ₹ 400/-

  • ₹ 300/-

  • ₹ 750/-

  • ₹ 500/-


D.

₹ 500/-

Let the income of a man  = â‚¹ x.
According to the question,
  straight x space space cross times space 15 over 100 space equals space 75
straight x space equals space ₹ space 500 space divided by negative

 

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51.

The radius of a sphere and right circular cylinder is 'r'. Their volumes are equal. The ratio of the height and radius of the cylinder is

  • 3 : 1

  • 2 : 1

  • 3 : 2

  • 4 : 3


D.

4 : 3

Volume of sphere = Volume of sphere
   4 over 3 πr cubed space equals space πr squared straight h
  straight h over straight r space equals space 4 over 3
  rightwards double arrow space space space straight h space colon space straight r space equals space 4 space colon space space 3

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53.

A merchant buys 25 litre of milk daily at the rate of â‚¹ 12 per litre. He mixes 5 litre of water in it and sells at the rate of â‚¹ 10.40 per litre. His gain/loss is

  • 8% profit

  • 2% profit

  • 4% profit

  • 6% profit


C.

4% profit

Cost price of 25 litre of milk = 25 x 12 = â‚¹ 300/-
Selling price of 30 litre of mixture  = 30 x 10.40 = â‚¹ 312 /-
∴  Required profit percentage:
equals space fraction numerator 312 space minus space 300 over denominator 300 end fraction space cross times space 100
equals space 12 over 3 space equals space 4 percent sign


                

 

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58.

The point of intersection of all the three medians of a triangle is called its

  • Orthocentre

  • Incentre

  • Centroid

  • Circumcentre


C.

Centroid

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59.

The area of a triangle with vertices A(0, 8), O(0, 0) and B (5, 0) is

  • 8 sq units

  • 13 sq units

  • 20 sq units

  • 40 sq units


C.

20 sq units

x1y1 = (0, 8)
x2y2 = (0, 0)
x3y3 = (5, 0)
∴   Area of triangle:
     equals space 1 half open square brackets straight x subscript 1 left parenthesis straight y subscript 2 minus straight y subscript 3 right parenthesis space plus space straight x subscript 2 left parenthesis straight y subscript 3 space minus straight y subscript 1 right parenthesis space plus space straight x subscript 3 left parenthesis straight y subscript 1 minus straight y subscript 2 right parenthesis close square brackets
equals space 1 half open square brackets 0 left parenthesis 0 minus 0 right parenthesis space plus space 0 left parenthesis 0 minus 8 right parenthesis space plus space 5 left parenthesis 8 minus 0 right parenthesis close square brackets
equals space 1 half space cross times space open square brackets 0 plus 0 space plus space 40 close square brackets
equals space space 20 space sq space unit
   

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52.

When a discount of â‚¹ 42/- is allowed on the marked price of an article, the new reduced price becomes 86% of the original price. Find the marked price.

  • ₹ 250/-

  • ₹ 300/-

  • ₹ 350/-

  • ₹ 400/-


B.

₹ 300/-

Let the marked price be â‚¹ x.
According to the question,
 straight x minus 42 space equals space straight x space cross times space 86 over 100
100 straight x space minus space 4200 space equals space 86 straight x
14 straight x space equals space 4200
straight x space equals space ₹ space 300 divided by negative

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60.

The least number that must be subtracted from 1294 so that the remainder when divided by 9, 11, 13 will leave in each case the same remainder is 

  • 2

  • 3

  • 1

  • 4


C.

1

LCM of 9,11,13 is 1287.
On Dividing 1294 with 1287, we get remainder as 7,
Now to get remainder 6, 1 is to be deducted from 1294 so that 1293 when divided by 9,11,13 leaves 6 as remainder.

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56.

Sum of three consecutive integers is 51. The middle one is

  • 14

  • 15

  • 16

  • 17


D.

17

Let the numbers be x, x + 1 + x + 2
According to the question,
x + x +1 + x + 2  = 51
3x  = 51 - 3
3x = 48
x = 16
∴   Middle number = x + 1 = 16 + 1  = 17

 

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