Answer: B
Explanation:
Cylinder painted area = 2 \( \pi rh + \pi rÂ² \)
Cone painted area = \( \pi rl \)
\( \frac {2h + r}{\sqrt {(rÂ² + h1Â² )}} \) = 3 : 1
h = 6
Q2. The diameter of Road Roller is 84 cm and its length is 150 cm. It takes 600 revolutions to level once on a particular road. Then what is the area of that road in mÂ²?
Answer: A
Explanation:
Area
\(600 \times 2 \times \frac {22}{7} \times \frac {42}{100} \times \frac {150}{100} \) = 2376
Q3. A smaller triangle is having three sides. Another big triangle is having sides exactly double the sides of the smaller triangle. Then what is the ratio of Area of a Smaller triangle to Area of the Bigger triangle?
Answer: C
Explanation:
Smaller triangle sides = a, b, c
Area = \( \sqrt {s (s – a) (s – b) (s – c)} \)
s = \( \frac {a + b + c}{2} \)
= \( \sqrt \frac {(a + b + c)(b + c – a) (a + c – b) (a + b – c)}{4} \)
Bigger triangle = 2a, 2b, 2c
Area = \( \sqrt {(a + b + c)(b + c – a)(a + c – b)(a + b – c)} \)
Ratio = 1 : 4
Q4. 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?
Answer: C
Explanation:
It touches on midpoints on the sides of the square ABCD
Side = \( \sqrt {(10Â² +10Â²)} = \sqrt {200} \)
Area = 200 mÂ²
Q5. A hemispherical bowl of diameter 16cm is full of ice cream. Each student in a class is served exactly 4 scoops of ice cream. If the hemispherical scoop is having a radius of 2cm, then ice cream is served to how many students?
Answer: A
Explanation:
\( \frac {2}{3} \times \pi \times 8Â³ = n \times 4 \times \frac {2}{3} \pi 2Â³ \)
n = 16
Answer: B
Explanation:
R = 9, r =5
V = \( \frac {22}{7} \times 59(9 2 -5 2) \) = 10384
Q2. What is the radius of the circle whose area is equal to the sum of the areas of two circles whose radii are 20 cm and 21 cm?
Answer: C
Explanation:
\( \pi RÂ² = \pi r1Â² + \pi r2Â² \)
\( \pi RÂ² = \pi (r1Â² + r2Â²) \)
RÂ² = (400 + 441)
R = 29
Q3. 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?
Answer: D
Explanation:
\( \frac {22}{7} \times 7Â² \times 49 = n \times (\frac {7}{2})Â³ \)
n = 176
Q4. If the ratio of radius two Cylinders A and B are in the ratio of 2:1 and their heights are in the ratio of 2:1 respectively. The ratio of their total surface areas of Cylinder A to B is?
Answer:
Explanation:
Cylinder A: \( 2 \pi r1 (r1 + h1) \)
Cylinder B: 2 \( \pi r2 (r2 + h2) \)
r \( \frac {1}{r} 2 = 2 :1 ; h \frac {1} {h 2} = 2:1 \)
\( {T A }{T B} = 2 \pi r1 \frac {(r1 + h1)}{2} \pi r2 (r2 + h2) \)
Q5. The area of the Circular garden is 88704 mÂ². Outside the garden a road of 7m width laid around it. What would be the cost of laying road at Rs. 2/mÂ².
Answer: D
Explanation:
88704 = \( \frac {22}{7} \times r \)
r = 168
Outer radius = 168 + 7 = 175
Outer area = \( \frac {22}{7} \times 175 2 \) = 96250
Road area = 96250 â€“ 88704 = 7546
Cost = \( 7546 \times 2 \)= 15092
Answer: B
Explanation:
Volume of the Rectangular Block = \( 22 \times 18 \times 7 \)
Radius of the Cistern = 7 cm
Area of the Cylinder = \( \pi \times rÂ² \times h = \frac {22}{7} \times 7 \times 7 \times h \)
\( \frac {22}{7} \times 7 \times 7 \times h = 22 \times 18 \times 7; h \)= 18 cm
Q2. If the radius of the cylinder is doubled, but height is reduced by 50%. What is the percentage change in volume?
Answer: A
Explanation:
\( \frac {R1}{R2} = \frac {R}{2R} and \frac {H1}{H2} = \frac {H}{\frac {H}{2}} \)
Original Volume = \( \pi rÂ²h \)
New Volume = \( \pi (2r)Â² \frac {h}{2} \)
Change in Volume = \( \frac {(2 â€“ 1)}{1} \times 100 \) = 100
Q3. The perimeter of a rectangle and a square is 80 cm each. If the difference between their areas is 100 cm. Find the sides of the rectangle.
Answer: A
Explanation:
2(l + b) = 4a = 80
l + b = 40; a = 20 \( \Rightarrow \) aÂ² = 400
aÂ² â€“ lb = 100; 400 â€“ lb = 100; lb = 300
(l â€“ b)Â² = (l + b)Â² â€“ 4lb
(l â€“ b)Â² = 1600 â€“ 1200; l â€“ b = 20;
Q4. The length of a wall is 5/4 times of its height. If the area of the wall will be 500mÂ². What is the sum of the length and height of the wall?
Answer: B
Explanation:
\( l = 5x; h = 4x ; l \times h \)= 500
\( 20 xÂ² = 500 ; x = 5 \)
l + h = 25 + 20 = 45 m
Q5. A circular path runs around a circular garden. If the difference between the circumference of the outer circle and the inner circle is 88m. Find the width of Path?
Answer: A
Explanation:
Width of the Road = R â€“ r
2\( \pi R â€“ 2 \pi r \)= 88
R â€“ r = 14 m
Competitive Exams – Study Guide | ||
---|---|---|
Category | ||
Quantitative Aptitude |
Reasoning Ability |
General Awareness |
Computer Awareness |
English Knowledge |
Banking Awareness |
General Science |
World of Words |
Descriptive Test |
Competitive Exams – College Entrance Exams | |||
---|---|---|---|
Category | Notification | ||
Diploma | NITC New Delhi | ||
PG | GATE 2020 | ||
^{} Click Here For â€“ All India Entrance Exam Notifications |
Competitive Exams – Practice Sets | ||
---|---|---|
Category | Quiz | |
Quant Aptitude | Permutation and Combination | |
Spotting Errors | ||
Mensuration | ||
Reasoning Ability | Puzzles | |
Insurance Awareness | Insurance Awareness |
General Knowledge for Competitive Examinations | |
---|---|
Topic | Name of the Article |
GK – World | International Stadiums |
ICC Cricket World Cup 2019 | |
GK – India | Unlawful Activities Prevention Act |
SHREYAS | |
GK – Abbreviations | Sports Abbreviations |
Computer Abbreviations | |
GK – Banking & Insurance | SEBI ACT 1992 |
Role Of SEBI As A Regulator | |
GK – Science & Technology | Father of Different Fields â€“ Science & Technology |
Important Scientific Instruments |