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Approximation & Line Graph Test 235
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Approximation & Line Graph Test 235

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  • Question 1/10
    1 / -0.25

    Directions For Questions

    Direction: What approximate value will come in place of the question mark (?) in the following question? (You are not expected to calculate the exact value)

    ...view full instructions

    219.9% of 8498 + √1370 + ³√2200=?
    Solutions
    219.9% of 8498 + √1370 + ³√2200=?

    220% of 8500 + √1369 + ³√2197=?
    18700 + 37 + 13 = 18750
  • Question 2/10
    1 / -0.25

    Directions For Questions

    Direction: What approximate value will come in place of the question mark (?) in the following question? (You are not expected to calculate the exact value)

    ...view full instructions

    518.004 × 3.96 + 40.02% of 759.86 +? = 9150.03
    Solutions
    518.004 × 3.96 + 40.02% of 759.86 +? = 9150.03
    On approximating
    518 × 4 + 40 × 760 ÷ 100 + ? = 9150
    or, 2072 + 304 + ? = 9150 
    or, ? = 6774
  • Question 3/10
    1 / -0.25

    Directions For Questions

    Direction: What approximate value will come in place of the question mark (?) in the following question? (You are not expected to calculate the exact value.

    ...view full instructions

    124.96% of 1200.13 + 55% of 14.98 + 105% of 20.03 = ?
    Solutions
    124.96% of 1200.13 + 55% of 14.98 + 105% of 20.03 = ?
    125% of 1200 + 55% of 15 +105% of 20


  • Question 4/10
    1 / -0.25

    Directions For Questions

    Direction: What approximate value should come at the place of question mark (?) in the following question (You are not expected to calculate the exact value)?

    ...view full instructions

    6894.986 ÷23+ 502.005 + 598.020 = ? - 23.91× 6.806
    Solutions
    6894.986 ÷23+ 502.005 + 598.020 = ? - 23.91× 6.806 
    6900÷23 +500 + 600 = ? -24×7
    300+500+600 = ?- 168
    1400+168= ?
    ?= 1568 =1570 (approx)
  • Question 5/10
    1 / -0.25

    Directions For Questions

    Direction: What approximate value will come in place of the question mark (?) in the following question? (You are not expected to calculate the exact value)

    ...view full instructions

    (30.9)2 – (29.02)2 – {634.3 – (400.7 + 194.62)} = 3?
    Solutions

    (30.9)2 – (29.02)2 – {634.3 – (400.7 + 194.62)} = 3?

    (31)2 – (29)2 – {634 – (400 + 195)} = 3?

    961 – 841 – {634 – 595} = 3?

    120 – 39 = 3?

    81 = 3?

    ? = 4

  • Question 6/10
    1 / -0.25

    Directions For Questions

    Direction: Study the following graph carefully and answer the questions that follow.

    The following line graph gives the percentage of the number of candidates who qualified an examination out of the total number of candidates who appeared for the examination over a period of seven years from 1994 to 2000.

    ...view full instructions

    The difference between the percentage points of candidates qualified out of those that appeared was maximum for which of the following pair of years?
    Solutions
    The differences between the percentage points of candidates that qualified out of those that appeared for the given(in options) pair of years are:
    For 1997 and 1998 = 80 - 50 = 30 
    For 1996 and 1997 = 60 - 50 = 10
    For 1994 and 1995 = 50 - 30 = 20
    For 1995 and 1996 = 60 - 50 = 10
    For 1999 and 2000 = 80 - 60 = 20 
    Thus, the maximum difference is between the years 1997 and 1998
  • Question 7/10
    1 / -0.25

    Directions For Questions

    Direction: Study the following graph carefully and answer the questions that follow.

    The following line graph gives the percentage of the number of candidates who qualified an examination out of the total number of candidates who appeared for the examination over a period of seven years from 1994 to 2000.

    ...view full instructions

    In which pair of years was the number of candidates qualified the same?
    Solutions
    The graph gives the data for the percentage of candidates qualified to appear and unless the absolute values of number of candidates qualified or candidates appeared, so we cannot compare the absolute values for any two years. Hence, the data is inadequate to solve this question.
  • Question 8/10
    1 / -0.25

    Directions For Questions

    Direction: Study the following graph carefully and answer the questions that follow.
    The following line graph gives the percentage of the number of candidates who qualified an examination out of the total number of candidates who appeared for the examination over a period of seven years from 1994 to 2000.

    ...view full instructions

    If the number of candidates qualified in 1998 was 21200, what was the number of candidates appeared in 1998?
    Solutions
    Let the number of candidates that appeared in 1998 be X.

    80% of X = 21200

    ⇒ X = = 26500
  • Question 9/10
    1 / -0.25

    Directions For Questions

    Direction: Study the following graph carefully and answer the questions that follow.
    The following line graph gives the percentage of the number of candidates who qualified an examination out of the total number of candidates who appeared for the examination over a period of seven years from 1994 to 2000.

    ...view full instructions

    If the total number of candidates appeared in 1998 and 1999 together was 47400, then what is the total number of candidates qualified in these two years together?
    Solutions
    Since both years percentages are equal
    Number of qualified candidates = 80% of total = 47400 = 37920
  • Question 10/10
    1 / -0.25

    Directions For Questions

    Direction: Study the following graph carefully and answer the questions that follow.
    The following line graph gives the percentage of the number of candidates who qualified an examination out of the total number of candidates who appeared for the examination over a period of seven years from 1994 to 2000.

    ...view full instructions

    If there was an increase in the qualifying percentage by 5 points in the year of 1995, how many students would have qualified if there were 20000 students?
    Solutions
    Required answer = (50 + 5)% × 20000 = 11000
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