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Daily CSAT Practice Test
Everyday 5 Questions from Aptitude, Logical Reasoning, and Reading Comprehension will be covered from Monday to Saturday.
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Question 1 of 5
1. Question
8 men and 4 women together can complete a piece of work in 6 days. The work done by a man in one day is double the work done by a woman in one day. If 8 men and 4 women started working and after 2 days 4 men left and 4 new women joined, in how many more days will the work be completed?
Correct
Solution (a)
Let the work done by a woman in one day = x units
The work done by a man in one day = 2x units
8 men and 4 women together can complete a piece of work in 6 days.
Total work = (8*2x+4*x)6 = 120x units
Amount of work done by 8 men and 4 women in 2 days = (8*2x+ 4*x)2 = 40x units
The remaining (120x40x) 80x units will be done by 8 women and 4 men
Number of days taken by 8 women and 4 men to complete 80x units of work = 80x/16x =5 days.
Incorrect
Solution (a)
Let the work done by a woman in one day = x units
The work done by a man in one day = 2x units
8 men and 4 women together can complete a piece of work in 6 days.
Total work = (8*2x+4*x)6 = 120x units
Amount of work done by 8 men and 4 women in 2 days = (8*2x+ 4*x)2 = 40x units
The remaining (120x40x) 80x units will be done by 8 women and 4 men
Number of days taken by 8 women and 4 men to complete 80x units of work = 80x/16x =5 days.

Question 2 of 5
2. Question
One side of an equilateral triangle is 24 cm. The midpoints of its sides are joined to form another triangle whose midpoints are in turn joined to form still another triangle. This process continues indefinitely. Find the sum of the perimeters of all the triangles.
Correct
Solution (a)
Side of an equilateral triangle = 24 cm
The midpoints of its sides are joined to form another triangle whose midpoints are joined to form another triangle.
This process continues indefinitely.
Side of 2nd equilateral triangle=12 cm [Line segment joining midpoint of two sides of a triangle is parallel to the third side and half of it]
Similarly, Side of third equilateral triangle=6 cm
Sides are 3, 1.5, 3/4….
Perimeter of 1st triangle = 24+24+24 = 72 cm
Perimeter of 2nd triangle = 12+12+12 = 36 cm
Perimeter of 3rd triangle = 6+6+6 = 18 cm
……….
The sum of the perimeter of all the triangles = 72+36+18+9+9/2+9/4+……..
This is a geometric progression having a common ratio = 72/36 = 2
Sum of an infinite G.P = first term *common ratio
72 × 2 = 144, is the sum of the perimeter of all triangles.
Incorrect
Solution (a)
Side of an equilateral triangle = 24 cm
The midpoints of its sides are joined to form another triangle whose midpoints are joined to form another triangle.
This process continues indefinitely.
Side of 2nd equilateral triangle=12 cm [Line segment joining midpoint of two sides of a triangle is parallel to the third side and half of it]
Similarly, Side of third equilateral triangle=6 cm
Sides are 3, 1.5, 3/4….
Perimeter of 1st triangle = 24+24+24 = 72 cm
Perimeter of 2nd triangle = 12+12+12 = 36 cm
Perimeter of 3rd triangle = 6+6+6 = 18 cm
……….
The sum of the perimeter of all the triangles = 72+36+18+9+9/2+9/4+……..
This is a geometric progression having a common ratio = 72/36 = 2
Sum of an infinite G.P = first term *common ratio
72 × 2 = 144, is the sum of the perimeter of all triangles.

Question 3 of 5
3. Question
If nCx = 56 and nPx = 336, then Find n and x?
Correct
Solution (c)
We know that
nPr = nCr*r!
Given, nPx = 336
nCx*x! = 336
x! = 336/56
x! = 6
x = 3
nCx = 56
n!/(nx)!*x! = 56
n!/(n3)!*3! = 56
n!/(n3)! = 56*6 = 336
If n = 8, 8!/(83)! = 8*7*6 = 336
Therefore, n = 8, and x = 3
Incorrect
Solution (c)
We know that
nPr = nCr*r!
Given, nPx = 336
nCx*x! = 336
x! = 336/56
x! = 6
x = 3
nCx = 56
n!/(nx)!*x! = 56
n!/(n3)!*3! = 56
n!/(n3)! = 56*6 = 336
If n = 8, 8!/(83)! = 8*7*6 = 336
Therefore, n = 8, and x = 3

Question 4 of 5
4. Question
A trader makes a profit equal to the selling price of 75 articles when he sold 100 of the articles. What % profit did he make in the transaction?
Correct
Solution (c)
It is given that, the profit on 100 articles = SP of 75 articles
100(SPCP) = 75*SP
4(SPCP) = 3SP
SP = 4CP
Profit % = [(SP – CP)/CP]*100
= [4CP – CP/CP]*100 = 300%
Incorrect
Solution (c)
It is given that, the profit on 100 articles = SP of 75 articles
100(SPCP) = 75*SP
4(SPCP) = 3SP
SP = 4CP
Profit % = [(SP – CP)/CP]*100
= [4CP – CP/CP]*100 = 300%

Question 5 of 5
5. Question
During a Cricket tournament organised by College, 4/5 of the seats in the lower deck of the stadium were sold. If 1/4 of all the seating in the stadium is located in the lower deck, and if 2/3 of all the seats in the stadium were sold, then what fraction of the unsold seats in the stadium was in the lower deck?
Correct
Solution (c)
Let the total number of seats in the stadium be x
The number of seats in the lower deck be ‘a’ and the number of seats in the upper deck be ‘b’
x = a+b
a = x/4 and b = 3x/4
Now, in the lower deck 4a/5 seats were sold and a/5 seats were unsold.
Number of total seats sold in the stadium = 2x/3
Number of unsold seats in the lower deck = a/5 = x/20
Number of unsold seats in the stadium = x/3
Required fraction = (x/20)/(x/3) = 3/20
Therefore, option c is the correct.
Incorrect
Solution (c)
Let the total number of seats in the stadium be x
The number of seats in the lower deck be ‘a’ and the number of seats in the upper deck be ‘b’
x = a+b
a = x/4 and b = 3x/4
Now, in the lower deck 4a/5 seats were sold and a/5 seats were unsold.
Number of total seats sold in the stadium = 2x/3
Number of unsold seats in the lower deck = a/5 = x/20
Number of unsold seats in the stadium = x/3
Required fraction = (x/20)/(x/3) = 3/20
Therefore, option c is the correct.
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