Subject

Quantitative Aptitude

Class

SSCCGL Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

32.

At an election, a candidate secures 40% of the votes, but is defeated by the other candidate by a majority of 298 votes. Find the total number of votes recorded.

  • 1580

  • 1490

  • 1470

  • 1530


B.

1490

Let total number of votes recorded  = x
  Now according to the question,
               (60% of x) - (40% of x) = 298
                               = 20% of x = 298
rightwards double arrow space space space space straight x space cross times space space 20 over 100 space equals space 289 space space rightwards double arrow space space straight x over 5 space equals space 298
rightwards double arrow space space space space space straight x space equals space 5 space cross times space 298 space rightwards double arrow space space straight x space equals space 1490

606 Views

27.

Find the wrong number in the given series
3, 7, 16, 35, 70, 153

  • 70

  • 16

  • 153

  • 35


A.

70

The sequence of given series is as follows:
3 rightwards arrow with cross times 2 plus 1 on top 7 rightwards arrow with cross times 2 plus 2 on top 16 rightwards arrow with cross times 2 plus 3 on top 35 rightwards arrow with cross times 2 plus 4 on top stack box enclose 70 with 74 on top rightwards arrow with cross times 2 plus 5 on top 153

331 Views

26.

A number when divided by 6 leaves remainder 3. When the square of the same number is divided by 6, the remainder is

  • 0

  • 2

  • 1

  • 3


D.

3

As we know that,
        If a number or square of a number is divided by the same number, a remainder is same in both the conditions.
Hence, square of a number is divided by 6, we get 3 as remainder.

1312 Views

30.

Two persons ride towards each other from two places 55km apart, one riding at 12 km/h and the other at 10 km/h. When will they be 11 km apart?

  • 2h and 30 min

  • 1h and 30 min

  • 2h

  • 2h and 45 min


C.

2h

Relative speed of both persons  = 12 + 10 = 22 km/h
 Now, distance between both of them = 55 -  11 = 44 km/h
  Time, when distance is 11 km between both of them
                = fraction numerator Total space distance over denominator Relative space Speed end fraction equals 44 over 22 equals 2 straight h
Hence, they will be apart 11 km after 2 hour.

484 Views

31.

A train 150 m long passes a telegraphic post in 12 seconds. Find the speed of the train.

  • 50 km/h

  • 12.5 km/h

  • 25 km/h

  • 45 km/h


D.

45 km/h

Length of train  = 150 m
                       = 150 over 1000 km space equals space 3 over 20 km
Time  = 12 s = fraction numerator 12 over denominator 60 space cross times 60 end fraction straight h space equals space 1 over 300 straight h
∴  Speed of train  = fraction numerator 3 divided by 20 over denominator 1 divided by 300 end fraction space equals space fraction numerator 3 space cross times space 300 over denominator 20 space cross times space 1 end fraction space equals space 45 space km divided by straight h

316 Views

29.

In a parade of school students, the number of boys and girls are in the ratio of  9:7 respectively and the total number of boys and girls are 256. Find the number of girls.

  • 102

  • 112

  • 118

  • 128


B.

112

Let, number of boys and girls are 9x and 7x, respectively.
  Given, total number of boys and girls  = 256
∴   Number of girls  = fraction numerator 7 straight x over denominator left parenthesis 9 straight x plus 7 straight x right parenthesis end fraction cross times 256 space equals space fraction numerator 7 straight x space cross times space 256 over denominator 16 straight x end fraction space equals space fraction numerator 7 cross times 256 over denominator 16 end fraction
                      = 7 x 16 = 112



459 Views

35.

Successive discount of 20% and 10% is given on an item of ₹700, find the selling price.

  • 504

  • 196

  • 582

  • 601


A.

504

Here, r1 = 20%, r2 = 10% and marked price  = ₹700
∴   SP of an item = 700 open parentheses fraction numerator 100 minus 20 over denominator 100 end fraction close parentheses open parentheses fraction numerator 100 minus 10 over denominator 100 end fraction close parentheses
                         = 700 space cross times space 80 over 100 cross times 90 over 100
                         = 7 x 8 x 9 = ₹ 504

284 Views

33.

If straight x minus 1 over straight x space equals 2, then what is the value of straight x squared plus 1 over straight x squared ?

  • 4

  • 5

  • 3

  • 6


D.

6

Given straight x minus 1 over straight x equals 2,
Now, squaring both the sides,
space rightwards double arrow space space straight x squared plus 1 over straight x squared minus 2 space equals space 4 space rightwards double arrow space space straight x squared plus 1 over straight x squared space equals space 4 plus 2
rightwards double arrow space space space straight x squared plus 1 over straight x squared space equals space 6

263 Views

34.

The ratio between the height of tower and the point at some distance is 5 square root of 3 colon 5. What will be the angle of elevation?

  • 30 degree
  • 60 degree
  • 90 degree
  • 45 degree

B.

60 degree

Let, height of tower  = 5 square root of 3 straight x
and distance of a point = 5x
Let, angle of elevation  = straight theta
Then,  in  increment ABC,
                    tanθ space equals AC over BC space
rightwards double arrow space space space space space space tanθ space equals space fraction numerator 5 square root of 3 over denominator 5 straight x end fraction straight x

              equals space square root of 3
rightwards double arrow space space space tanθ space equals space square root of 3 space equals space tan 60 degree
rightwards double arrow space space space space space straight theta space equals space 60 degree

564 Views

28.

The averages of runs scored by a player in 11 innings is 63 and the average of his first innings is 60 and the average of last six innings is 65. Find the runs scored in the sixth inning. 

  • 60

  • 54

  • 67

  • 57


D.

57

Average score of runs of 11 innings = 63
    Total score of runs of 11 innings  = 11 x 63 = 693
Average score of runs of first 6 innings  = 60
  Total score of runs of 6 innings  = 6 x 60 = 360
      Now, average score of runs of last 6 innings  = 65
Total score of runs of last 6 innings = 6 x 65 = 390
  Hence, Total runs scored in sixth innings
                               = 360 + 390 - 693 = 750 - 693 = 57

541 Views