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Considering the alarming importance of CSAT in UPSC CSE Prelims exam and with enormous requests we received recently, InsightsIAS has started Daily CSAT Test to ensure students practice CSAT Questions on a daily basis. Regular Practice would help one overcome the fear of CSAT too.
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Looking forward to your enthusiastic participation (both in sending us questions and solving them on daily basis on this portal).
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Question 1 of 5
1. Question
ABCD is a square of 20 m. What is the area of the least-sized square that can be inscribed in it with its vertices on the sides of ABCD?
Correct
Answer: C. 200 m²
Explanation:
It touches on midpoints on the sides of the square ABCD
Side = √ (10² +10²) = √200
Area = 200 m²Incorrect
Answer: C. 200 m²
Explanation:
It touches on midpoints on the sides of the square ABCD
Side = √ (10² +10²) = √200
Area = 200 m² -
Question 2 of 5
2. Question
A well with 14 m diameter is dug up to 49 m deep. Now the soil taken out during dug is made into cubical blocks of 3.5m side each. Then how many such blocks were made?
Correct
Answer: D. 176
Explanation:
22/7*7²*49 = n*(7/2)³
n = 176Incorrect
Answer: D. 176
Explanation:
22/7*7²*49 = n*(7/2)³
n = 176 -
Question 3 of 5
3. Question
The area of the Circular garden is 88704 m². Outside the garden a road of 7m width laid around it. What would be the cost of laying road at Rs. 2/m².
Correct
Answer: D. Rs. 15,092
Explanation:
88704 = 22/7*r2
r = 168
Outer radius = 168+7 = 175
Outer area = 22/7*1752 = 96250
Road area = 96250 – 88704 = 7546
Cost = 7546*2 = 15092Incorrect
Answer: D. Rs. 15,092
Explanation:
88704 = 22/7*r2
r = 168
Outer radius = 168+7 = 175
Outer area = 22/7*1752 = 96250
Road area = 96250 – 88704 = 7546
Cost = 7546*2 = 15092 -
Question 4 of 5
4. Question
The length of a rectangle is reduced by 30%. By what percent would the width have to be increased to maintain the original area?
Correct
Correct Ans: B. 42.86%
Explanation:
Explanation :
—->Width=30*(100/100)-30
—-> = (3000/70) = 42.86%Incorrect
Correct Ans: B. 42.86%
Explanation:
Explanation :
—->Width=30*(100/100)-30
—-> = (3000/70) = 42.86% -
Question 5 of 5
5. Question
Six spherical cannon balls are tightly packed into a rectangular box in one layer. Each row has two cannon balls and each column has three. What part of the box is empty?
Correct
Correct Ans:(10/21)
Explanation:
Let-the diameter of each ball be 2r.
Lengthof the box=3*2r =6r
Breadth=2*2r=4r
Height=2r
Volume=6r*4r*2r=48r3
Volumeof6balls=6*(4/3)*(22/7)*r3 =(176r 3 /7)
Theareaofemptyspace=48r3 -(176r 3/7)
=(160r3/7)
Therequiredfraction=((160r3/7)/48r3)
= (10/21)Incorrect
Correct Ans:(10/21)
Explanation:
Let-the diameter of each ball be 2r.
Lengthof the box=3*2r =6r
Breadth=2*2r=4r
Height=2r
Volume=6r*4r*2r=48r3
Volumeof6balls=6*(4/3)*(22/7)*r3 =(176r 3 /7)
Theareaofemptyspace=48r3 -(176r 3/7)
=(160r3/7)
Therequiredfraction=((160r3/7)/48r3)
= (10/21)
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