# Tutor profile: Praveen V.

## Questions

### Subject: Pre-Algebra

4 friends ordered an extra large pizza, and cut it into slices of same size. Monu had $$\frac{1}{5} th$$ of the pizza, Sonu had $$\frac{1}{3} rd$$, Tonu had $$\frac{1}{6}th$$ and Ronu had $$\frac{3}{10}th$$ of the pizza. If there were 60 slices totally, how many slices did each have. Who had the largest share ?

This is a simple word problem in fractions, where we determine "what portion of the food" did each of the four friends eat. Total number of slices = $$60$$ Monu had $$\frac{1}{5} th$$ of pizza = $$\frac{1}{5} x 60 slices$$ = $$12$$ slices of pizza Sonu had $$\frac{1}{3} th$$ of pizza = $$\frac{1}{3} x 60 slices$$ = $$20$$ slices of pizza Tonu had $$\frac{1}{6} th$$ of pizza = $$\frac{1}{6} x 60 slices$$ = $$10$$ slices of pizza Ronu had $$\frac{3}{10} th$$ of pizza = $$\frac{3}{10} x 60 slices$$ = $$18$$ slices of pizza Among the 4 friends, Sonu had the maximum share with 20 slices of pizza.

### Subject: Computer Science (General)

A system in an computer network uses the IP address $$205.123.4.192$$. (a) What is the network and host id of the computer? (b) Is it a Class A, B or C type network?

IP address of the computer system = $$205.123.4.192$$ (a) Network Id refers to the ID common to all computers on a network, while host ID is the unique ID of a particular computer on a network The first three 8 bit addresses refer to the Network ID, while the last 8 bit address refers to the host ID. In this case, Network ID = $$205.123.4$$ Host ID = $$192$$ (b) Most IP addresses fall into one of three classes: • If the 32-bit address starts with a 0 bit, the address is a Class A address. • If the 32-bit address starts with the bits 10, the address is a Class B address. • If the 32-bit address starts with bits 110, the address is a Class C address. The first 8 bits of the IP address = $$205$$ Represented as binary, it is equal to $$11001101$$ As it starts with 110, it is a Class C address.

### Subject: Algebra

The perimeter of a rectangular field can be found using the formula $$2l + 2w$$, where $$l$$ represents the length and $$w$$ represents the width of the field. Find the length and width of the field if Ram runs around the field in $$60 minutes$$ at $$10 Km/h$$ and the length of the field is twice its width.

This problem involves two different concepts - perimeter and speed. In word problems, identifying such constructs and separating them makes our work easier and the problem far less confusing. $$\underline{Perimeter}$$ Perimeter of the rectangular field = $$2l + 2w$$ ( $$l \rightarrow length, w \rightarrow width$$) $$\hspace{1cm}\longrightarrow ➀$$ Length of field = 2 x Width of field i.e. $$l = 2w$$ Substituting in ➀, we get Perimeter of field = $$2x(2w) + 2w$$ = $$4w+2w$$ = $$6w$$ $$\hspace{2.5cm}\longrightarrow ➁$$ $$\underline{Speed}$$ As we know, Speed is defined as distance travelled per unit time. $$Speed = \frac{Distance}{Time}$$ $$\implies Distance = Speed x Time$$ Speed at which Ram was walking = $$10Km/h$$ Time within which Ram walks the field = $$60 minutes$$ = $$\frac{60}{60} hours$$ = $$1 hour$$ $$Distance = Speed x Time$$ $$= 10 x 1 = 10Km$$ $$\hspace{8cm}\longrightarrow ➂$$ The distance travelled by Ram is also the perimeter of the field. Substituting value of ➂ in ➁, we get $$Perimeter of field = 6w$$ $$ 10Km = 6w$$ $$Width of field w = \frac{10}{6} = 1.66Km$$ [Answer] $$Length of field = 2w$$ $$= 2 x 1.667 = 3.34Km$$ [Answer]

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