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61. A student has to secure 40% marks to pass. If he gets 40 marks and fails by 40 marks, find the maximum marks set for the examination.
  A.  200
  B.  500
  C.  700
  D.  150
     
   
View Answer

Shortcut:
The pass marks in an examination is a%. If a candidate who secures 'b' marks fails by 'c' marks, then the maximum marks are given by
100(b+c) / a
).
Here, a = 40, b = 40, c = 40
Using these values in the shortcut, we get:
Maximum Marks =
100(40+40) / 40
=
100 x 80 / 40
= 200
Hence, maximum marks are 200.

62. If an examination a candidate must get 80% marks to pass. If a candidate who gets 210 marks, fails by 50 marks, find the maximum marks.
  A.  255
  B.  325
  C.  295
  D.  360
     
   
View Answer

Shortcut:
The pass marks in an examination is a%. If a candidate who secures 'b' marks fails by 'c' marks, then the maximum marks are given by
100(b+c) / a
).
Here, a = 80, b = 210, c = 50
Using these values in the shortcut, we get:
Maximum Marks =
100(210+50) / 80
=
100 x 260 / 80
= 325
Hence, maximum marks are 325.

63. A candidate scores 25% and fails by 30 marks, while another candidate who scores 50% marks, gets 20 marks more than the minimum required marks to pass the examination. Find the maximum marks for the examination.
  A.  410
  B.  200
  C.  120
  D.  230
     
   
View Answer

Shortcut:
A candidate scoring a% in an examination fails by 'b' marks, while another candidate who scores c% marks get 'd' marks more than the minimum required pass marks. Then the maximum marks for that examination are given by
100(b + d) / c − a
.
Here, a = 25, b = 30, c = 50, d = 20
Using these values in the shortcut, we get:
Maximum Marks =
100(30 + 20) / 50 − 25
=
100 x 50 / 25
= 200
Hence, maximum marks are 200.

64. In measuring the sides of a rectangle, one side is taken 5% in excess and the other 4% in deficit. Find the error per cent in area calculated from the measurement.
  A.  +
4 / 5
%
  B.  +
3 / 5
%
  C.  +
3 / 2
%
  D.  +
5 / 4
%
     
   
View Answer

Shortcut:
In measuring the sides of a rectangle, one side is taken a% in excess and the other b% in deficit. The error per cent in area calculated from the measurement is given by
(
a − b −
ab / 100
)
% excess or deficit will be according to +ve or -ve signs respectively.
Here, a = 5, b = 4
Using these values in the shortcut, we get:
Net change =
(
5 − 4 −
5x4 / 100
)
% =
(
1 −
20 / 100
)
% =
(
1 −
1 / 5
)
% = +
4 / 5
%
Hence, net excess of
4 / 5
% in the measurement.

65. If one of the sides of a rectangle is increased by 20% and the other is increased by 5%. Find the per cent value by which the area changes.
  A.  +15%
  B.  +26%
  C.  +29%
  D.  +34%
     
   
View Answer

Shortcut:
If one of the sides of a rectangle is increased by a% and the other is increased by b% then the per cent value by which area changes is given by
(
a + b +
ab / 100
)
% increase.
Here, a = 20, b = 5
Using these values in the shortcut, we get:
Net change =
(
20 + 5 +
20x5 / 100
)
% =
(
25 +
100 / 100
)
% = (25 + 1)% = 26%
Hence, net excess of 26% in the area.

66. In measuring sides of a rectangle, one side is taken 5% in deficit and the other 4% in deficit. Find the error per cent in area calulated from the measurement.
  A.  -8.8%
  B.  -5.8%
  C.  +4.2%
  D.  +7%
     
   
View Answer

Shortcut:
If one of the sides of a rectangle is decreased by a% and the other is decreased by b% then the per cent value by which area changes is given by
(
a + b −
ab / 100
)
% decrease.
Here, a = 5, b = 4
Using these values in the shortcut, we get:
Net change =
(
5 + 4 −
5x4 / 100
)
% =
(
9 −
20 / 100
)
% = (9 − 0.2)% = 8.8%
Hence, net decrease of 8.8% in the area.

67. In an examination, 40% of the students failed in Maths, 30% failed in English and 10% failed in both. Find the percentage of students who passed in both the subjects.
  A.  15%
  B.  30%
  C.  20%
  D.  40%
     
   
View Answer

Shortcut:
In an examination a% failed in English and b% failed in maths. If c% of students failed in both the subjects the percentage of students who passed in both subjects is given by 100 − (a + b − c)%
Here, a = 40, b = 30, c = 10
Using these values in the shortcut, we get:
Net change = 100 − (40 + 30 − 10)% = [100 − (70 − 10)]% = [100 − 60]% = 40%
Hence, 40% passed in both subjects.

68. A man spends 75% of his income. His income increases by 20% and his expenditure also increases by 10%. Find the percentage increase in his savings.
  A.  42%
  B.  20%
  C.  50%
  D.  55%
     
   
View Answer

Shortcut:
A man spends a% of his income. His income is increased by b% and his expenditure also increases by c%, then the percentage increase in his savings is given by
(
100b − ac / 100 − a
)
%
Here, a = 75, b = 20, c = 10
Using these values in the shortcut, we get:
Percentage increase in saving =
(
100 x 20 − 75 x 10 / 100 − 75
)
% =
(
2000 − 750 / 25
)
% =
(
1250 / 25
)
% = 50%
Hence, percentage increase in saving is 50%.

69. A solution of slat and water contains 15% salt by weight. Of it 30kg water evaporates and the solution now contains 20% of salt. Find the original quantitiy of solution.
  A.  130 kg
  B.  120 kg
  C.  90 kg
  D.  100 kg
     
   
View Answer

Shortcut:
A solution of salt and water contains a% salt by weight. Out of it 'Z' kg water evaporates and the solution now contains b% of salt. The original quantity of solution is given by Z
(
b / b - a
)
kg.
Here, Z = 30, a = 15, b = 20
Using these values in the shortcut, we get:
Original Quantity = 30
(
20 / 20 - 15
)
kg = 30
(
20 / 5
)
kg = 30 x 4 = 120 kg
Hence, original quantity of solution is 120 kg.

70. In a library, 20% of the books are in Hindi, 50% of the remaining are in English and 30% of the remaining are in French. The remaining 6300 books are in regional languages. What is the total number of books in the library?
  A.  22,500
  B.  18,500
  C.  22,400
  D.  21,500
     
   
View Answer

Shortcut:
If a% of a thing is one type, b% of the remaining thing is of second type. c% of the remaining thing is of third type and the value of remaining things is given as 'Z'. Then, the total number of things is obtained by the Z
(
100 / 100 - a
)
(
100 / 100 - b
)
(
100 / 100 - c
)
.
Here, Z = 6300, a = 20, b = 50, c = 30
Using these values in the shortcut, we get:
Total number of things = 6300
(
100 / 100 - 20
)
(
100 / 100 - 50
)
(
100 / 100 - 30
)
= 6300
(
100 / 80
)
(
100 / 50
)
(
100 / 70
)
=
6300 x 100 x 100 x 100 / 80 x 50 x 70
= 90 x 25 x 10 = 22500
Hence, total number of books are 22500.

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