Question: 41 -
In a factory there are three types of Machines M1 , M2 and M3 which produces 25%, 35% and 40% of the total products respectively. M1, M2 and M3 produces 2% , 4% and 5% defective products, respectively. What is the percentage of non-defective products?
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89%
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96.1%
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97%
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86.1%
Answer:
96.1%
Solution:
Non- defective products
{(25x0.98+35x0.96+40x0.95)/100} x 100 = 96.1
Non- defective products
{(25x0.98+35x0.96+40x0.95)/100} x 100 = 96.1
Question: 42 -
In measuring the sides of a rectangle errors of 5 % and 3 % in excess are made. The error percent in the calculates area is
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8.15%
-
8. 35 %
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7.15%
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6. 25%
Answer:
8.15%
Solution:
Let length = x units and breadth = Y units
Then actual area = xy sq.units
Length shown = (105/100 × x)units = 21x/20 units;
Breadth shown = (103/100 × Y)
Calculated area = (21x/20 × 103y/100)sq.units
= 2163xy/2000 sq.units
Error = (2163xy/2000 - xy)
= 163xy/2000
Error % = (163xy/2000 × 1/xy × 100)%
= 163/20 % = 8.15 %
Let length = x units and breadth = Y units
Then actual area = xy sq.units
Length shown = (105/100 × x)units = 21x/20 units;
Breadth shown = (103/100 × Y)
Calculated area = (21x/20 × 103y/100)sq.units
= 2163xy/2000 sq.units
Error = (2163xy/2000 - xy)
= 163xy/2000
Error % = (163xy/2000 × 1/xy × 100)%
= 163/20 % = 8.15 %
Question: 43 -
Fresh grapes contain 80 % water dry grapes contain 10 % water. If the weight of dry grapes is 250 kg. What was its total weight when it was fresh?
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1225kg
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1000kg
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1125kg
-
1100kg
Answer:
1125kg
Solution:
Let the weight of fresh grapes be x kg
Quantity of water in it = (80/100 × x)kg = 4x/5 kg
Quantity of pulp in it = (x – 4x/5)kg = x/5 kg
Quantity of water in 250 kg dry grapes
= (10/100 × 250)kg = 25kg
Quantity of pulp in it = (250 - 25)kg = 225 kg
Therefore, x/5 = 225
=> x = 1125
Let the weight of fresh grapes be x kg
Quantity of water in it = (80/100 × x)kg = 4x/5 kg
Quantity of pulp in it = (x – 4x/5)kg = x/5 kg
Quantity of water in 250 kg dry grapes
= (10/100 × 250)kg = 25kg
Quantity of pulp in it = (250 - 25)kg = 225 kg
Therefore, x/5 = 225
=> x = 1125
Question: 44 -
The average of a set of whole numbers is 27.2. when the 20% of the elements are eliminated from the set of numbers then the average become 34. The number of elements in the new set of numbers can be :
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27
-
63
-
35
-
52
Answer:
52
Solution:
only (c) is correct since it is divisible by 4.
Let the original number of element be x then the new no.of elements will be
4x/5 = K
So K must be divisible by 4
Since, x = Kx5/4
only (c) is correct since it is divisible by 4.
Let the original number of element be x then the new no.of elements will be
4x/5 = K
So K must be divisible by 4
Since, x = Kx5/4