STUDENT CONTEST #5 – WINNERS
The list of winners and solutions for the 5th contest have been shared below.
(This contest ended on 31ST May, 2019)
42 min read
STUDENT CONTEST #5 – WINNERS
The list of winners and solutions for the 5th contest have been shared below.
(This contest ended on 31ST May, 2019)
LIST OF WINNERS OF ROUND #5
(There was an error in the original list posted. The updated list as of 30 June, 2019, is shared below.)
| Name (First) | Name (Last) | School | City | Class/Grade |
| VIDAAL | KAMPANI | DELHI PRIVATE SCHOOL | DUBAI | 4 |
| BHAVYA | YADAV | DELHI PUBLIC SCHOOL | FEROZEPUR | 5 |
| Hiyanshi | Mehta | Bombay Scottish School, Powai | Mumbai | 5 |
| ASHMIT | TIWARI | B. K. Birla Public School ,Kalyan (W) | Kalyan (W) ,Thane | 4 |
| Vaibhavi | Venkateswaran | Bombay Scottish School,Powai | Mumbai | 5 |
| Prachi | Kothari | Children’s Academy, Bachani Nagar, Malad East, Mumbai | Mumbai | 5 |
| Name (First) | Name (Last) | School | City | Class/Grade |
| Aaryaveer | Adyanthaya | Children’s Academy, Ashok Nagar | Mumbai | 6 |
| HRISHITA | PATNAIK | DELHI PRIVATE SCHOOL, SHARJAH | SHARJAH | 7 |
| Dhruv | gupta | Salwan public school | New delhi | 7 |
| Jai | Bellare | Bombay Scottish School Mahim | Mumbai | 8 |
| Jayden | Antony Thomas | Peepal Prodigy | Coimbatore | 6 |
| BHAVI | NAIK | Children’s Academy | Mumbai, India | 6 |
| AASHKA | SHAH | Children’s Academy (Ashok Nagar, Kandivali East) | MUMBAI | 6 |
| PINAKI | MAHESHWARI | DELHI PRIVATE SCHOOL DUBAI | DUBAI | 7 |
| REAGAN | PEREIRA | BOMBAY SCOTTISH SCHOOL,MAHIM | Mumbai, Maharashtra | 8 |
| Krishanth sorna Aravintha | Losanan | DPS Dubai | Dubai | 6 |
| Prasiddhi | Borgaonkar | B.K.Birla Public School, Kalyan | Dombivali East | 6 |
| Harsh | Fozdar | Children’s Academy, Ashok Nagar | Mumbai | 6 |
| Damandeep | Singh | Delhi public school ferozepur | Ferozepur city | 8 |
| Sanskriti | Verma | B K Birla Public School | Kalyan west | 7 |
| Name (First) | Name (Last) | School | City | Class/Grade |
| Vedant | Lal | Delhi Private School, Dubai | Dubai | 10 |
| Manya | Narula | Delhi Private School | Sharjah | 9 |
CONTEST BEGINS
SOLUTIONS
CLASSES 3-5
Below figures are of different phases of the moon observed by a person on the earth.
In which phase of the moon, is its surface receiving the maximum light of the sun? Explain your answer.
Solution:
| During its revolution around the earth, the Moon goes through all its phases as shown in question. Just like the Earth, half of the Moon is always lit by the Sun while the other half is in darkness. The phases we see, result from the angle, the Moon makes with the Sun as viewed from Earth. Figure I, is the diagram which generally is given in books. This image shows how the Moon is seen from the Earth at given points in its orbit. It does not show which side of the Moon is lit by the Sun. The side lit by the Sun is always the side that is pointed toward the Sun, as seen in the figure II. |
We only see the Moon because sunlight gets reflected back to us from its surface. If we could magically look down on our solar system, we would see that the half of the Moon facing the Sun is always lit. But the lit side does not always face the Earth! As the Moon circles the Earth, the part of the lit side we see changes. These changes are known as the phases of the Moon. So, the half of the Moon surface is always lit by the sun irrespective of its phase we see from the Earth.
Q1. Which of the following figures has a bigger perimeter? Explain your answer.
(Hint: scale or thread can be used to compare perimeter)
Solution:
Perimeter is the length of the path or boundary that surrounds a shape. Take a thread, choose a starting point and place is around the boundary of a shape till you reach back to the starting point. Break the thread at this point. Now, measure the length of thread using a scale. It will give you the perimeter of the choose shape.
In figure I, length of thread AB will give the perimeter. We can also calculate the approximate perimeter by measuring the lengths of line segments surrounding the shape.
Perimeter of figure I = 10 + 4 + 4 + [(3 + 3)Total number of teeth in the comb]
= 18 + (6 20) = 18 + 120 = 138 cm
Perimeter of figure II = 10 + 8 + 10 + 8 = 36 cm
Hence, perimeter of shape shown in figure I is way bigger than that of figure II.
Q2. Which of the following figures has bigger area? Explain your answer.
CLASSES 6-8
In the given figure, a lighted candle and its shadow formed on the table inside a closed room is shown. Can this shadow be formed by the light of the candle? Explain your answer.
Solution:
Light of the candle forms its own shadow but it will be very small as shown in fig 1. To form a shadow shown in question, there has to be another light source kept at an elevation opposite to the candle as shown in fig II.
Figure I
Figure II
Q1. Height of a father and his son in the mobile picture shown is proportional to their actual heights. If actual height of the son is 120 cm, then what is the actual height of the father?
Solve this question, both using proportion and unitary method.
Solution:
Suppose actual height of the father = x cm
From the given question, Ratio of height of father to height of son in mobile = Ratio of actual height of father to actual height of son
=> 12:8 = x:120
=> 12/8 = x/120
=> X = 180 cm
Let us solve this question using unitary method.
If height of son would be 8 cm, height of father = 12 cm
If height of son would be 1 cm, height of father = 12/8 cm
So, when height of son is 120 cm, height of father = 12/8 * 120 = 180 cm
CLASSES 9-10
Q: Consider two objects kept close to each other. You observe that both of them start moving towards each other on their own. Do you think one of them must be a magnet? Give reason to support your answer.
Solution:
Two objects can attract each other in the following conditions.
- If one of them is magnet and the other object is made of magnetic materials like iron and nickel. Again, two magnets kept in proper orientation can also attract each other.
- If the two bodies have electrostatic charges of opposite nature, then they will attract each other.
III. Gravitational force which is attractive by nature, always acts between two bodies. Thus, both the bodies attract each other. But as this force is very small in magnitude, it can’t be visualized easily. But if one of the objects is very large as compared to the other, then gravitational force can be visualized.
Q1. A hexagonal prism and a cylinder of same height is shown in the figure. Perimeter of the circular face of the cylinder is equal to the perimeter of the hexagonal face of the prism. Which shape will have bigger curved (or lateral) surface area, prism or cylinder? Explain your answer.
Solution:
The nets of cylinder and hexagonal prism is shown below. Rectangular region in each case represents curved surface area corresponding to shapes.
So, curved surface area of cylinder = Perimeter of circular face h
and, curved surface area of hexagonal prism = Perimeter of hexagonal face h
Since, Perimeter of circular face = Perimeter of hexagonal face
=> curved surface area of cylinder = curved surface area of hexagonal prism
In case of any query contact us at: +91-9741877802