UPSC Insta–DART (Daily Aptitude and Reasoning Test) 27 Dec 2024
Kartavya Desk Staff
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.We are naming this initiative as Insta– DART – Daily Aptitude and Reasoning Test. We hope you will be able to use DART to hit bull’s eye in CSAT paper and comfortably score 100+ even in the most difficult question paper that UPSC can give you in CSP-2021. Your peace of mind after every step of this exam is very important for us.
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 The ratio of the length and the perimeter of a rectangle is 2:7. What is the ratio of the length and breadth of the rectangle? a) 4:3 b) 4:5 c) 5:4 d) 5:3 Correct Answer: A Explanation Length: Perimeter = l : 2(l+b) =>2x:7x Perimeter 2(l+b) = 2 (2x+b) =7x 4x+2b=7x; 2b=3x; b=3/2; Length : Breadth = 2 :3/2 =>4:3 Incorrect Answer: A Explanation Length: Perimeter = l : 2(l+b) =>2x:7x Perimeter 2(l+b) = 2 (2x+b) =7x 4x+2b=7x; 2b=3x; b=3/2; Length : Breadth = 2 :3/2 =>4:3
#### 1. Question
The ratio of the length and the perimeter of a rectangle is 2:7. What is the ratio of the length and breadth of the rectangle?
Explanation
Length: Perimeter = l : 2(l+b) =>2x:7x
Perimeter 2(l+b) = 2 (2x+b) =7x
2b=3x; b=3/2;
Length : Breadth = 2 :3/2 =>4:3
Explanation
Length: Perimeter = l : 2(l+b) =>2x:7x
Perimeter 2(l+b) = 2 (2x+b) =7x
2b=3x; b=3/2;
Length : Breadth = 2 :3/2 =>4:3
• Question 2 of 5 2. Question The perimeter of an isosceles triangle is 125 cm. If the base is 33 cm, find the length of the equal sides. a) 32 cm b) 46 cm c) 34 cm d) 42 cm Correct Answer: B Explanation Perimeter of isosceles triangle = 125 cm Base = 33 cm Let each equal side = x cm x + 33 + x = 125 cm 2x = 125 – 33 = 92 Length of each equal side, x = 46 cm Incorrect Answer: B Explanation Perimeter of isosceles triangle = 125 cm Base = 33 cm Let each equal side = x cm x + 33 + x = 125 cm 2x = 125 – 33 = 92 Length of each equal side, x = 46 cm
#### 2. Question
The perimeter of an isosceles triangle is 125 cm. If the base is 33 cm, find the length of the equal sides.
Explanation
Perimeter of isosceles triangle = 125 cm
Base = 33 cm
Let each equal side = x cm
x + 33 + x = 125 cm
2x = 125 – 33 = 92
Length of each equal side, x = 46 cm
Explanation
Perimeter of isosceles triangle = 125 cm
Base = 33 cm
Let each equal side = x cm
x + 33 + x = 125 cm
2x = 125 – 33 = 92
Length of each equal side, x = 46 cm
• Question 3 of 5 3. Question If the base radius of 2 cylinders are in the ratio 3:4 and their heights are in the ratio of 4:9, then the ratio of their volumes is: a) 2:1 b) 1:4 c) 1:2 d) 4:1 Correct Answer: B Explanation It is given that; base radius of 2 cylinders are in the ratio 3:4 and their heights are in the ratio of 4:9. We know that, Volume of cylinder = πr2h Volume ∝ radius2 Volume ∝ height Hence ratio of their volumes is => 334 : 449 = 1:4 Incorrect Answer: B Explanation It is given that; base radius of 2 cylinders are in the ratio 3:4 and their heights are in the ratio of 4:9. We know that, Volume of cylinder = πr2h Volume ∝ radius2 Volume ∝ height Hence ratio of their volumes is => 334 : 449 = 1:4
#### 3. Question
If the base radius of 2 cylinders are in the ratio 3:4 and their heights are in the ratio of 4:9, then the ratio of their volumes is:
Explanation
It is given that; base radius of 2 cylinders are in the ratio 3:4 and their heights are in the ratio of 4:9.
We know that, Volume of cylinder = πr2h
Volume ∝ radius2
Volume ∝ height
Hence ratio of their volumes is => 334 : 449 = 1:4
Explanation
It is given that; base radius of 2 cylinders are in the ratio 3:4 and their heights are in the ratio of 4:9.
We know that, Volume of cylinder = πr2h
Volume ∝ radius2
Volume ∝ height
Hence ratio of their volumes is => 334 : 449 = 1:4
• Question 4 of 5 4. Question If the length of a rectangle is increased by 40% and the breadth is decreased by 20%, then the area of the rectangle is increased by x%. The value of x is: a) 20 b) 12 c) 16 d) 8 Correct Answer: B Explanation Area of rectangle = length Breadth = lb Length=1.4l; Breadth =0.8b; New Area = 1.4 l 0.8b =1.12lb Thus 12% increase in area of rectangle Incorrect Answer: B Explanation Area of rectangle = length Breadth = lb Length=1.4l; Breadth =0.8b; New Area = 1.4 l 0.8b =1.12lb Thus 12% increase in area of rectangle
#### 4. Question
If the length of a rectangle is increased by 40% and the breadth is decreased by 20%, then the area of the rectangle is increased by x%. The value of x is:
Explanation
Area of rectangle = length * Breadth = lb
Length=1.4l; Breadth =0.8b;
New Area = 1.4 l *0.8b =1.12lb
Thus 12% increase in area of rectangle
Explanation
Area of rectangle = length * Breadth = lb
Length=1.4l; Breadth =0.8b;
New Area = 1.4 l *0.8b =1.12lb
Thus 12% increase in area of rectangle
• Question 5 of 5 5. Question A metallic sphere of diameter 40 cm is melted into a smaller sphere of radius 0.5 cm. How many such small balls can be made? a) 64,000 b) 32,000 c) 3200 d) 6400 Correct Answer: A Explanation Let x spheres be made out of one big sphere Volume of big sphere = Volume of x small spheres 4/3 π R3 = x 4/3 π r3 203 =x 0.53 X = (20/0.5)3 =64000 Incorrect Answer: A Explanation Let x spheres be made out of one big sphere Volume of big sphere = Volume of x small spheres 4/3 π R3 = x 4/3 π r3 203 =x 0.53 X = (20/0.5)3 =64000
#### 5. Question
A metallic sphere of diameter 40 cm is melted into a smaller sphere of radius 0.5 cm. How many such small balls can be made?
Explanation
Let x spheres be made out of one big sphere
Volume of big sphere = Volume of x small spheres
4/3 π R3 = x 4/3 π *r3
203 =x *0.53
X = (20/0.5)3 =64000
Explanation
Let x spheres be made out of one big sphere
Volume of big sphere = Volume of x small spheres
4/3 π R3 = x 4/3 π *r3
203 =x *0.53
X = (20/0.5)3 =64000
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