10

Present weight = 40 kg + m kg= (40+ m) kg

Perimeter of triangle = Sum of its all sides

Total number of students = p
Money collect = Rs 50 per student
Total money collect =Rs 50 P
Advance money paid for transport = Rs 1800
Money left with them = Rs(50 P - 1800)

By inspection of the above table, we find that t =12 satisfies the equation \(\frac{z}{3}\)=4.
Hence, z = 12 is its solution.

p = 10 is a solution of equation p - 5 = 5.

6x

Multiplication and addition

8x

r

Perimeter (P) of the regular pentagon with side l = Sum of the lengths of all sides of the regular pentagon

= AB+ BC + CD +DE+EA
=l+l+l+l+l
\(=5\times l\)
∴ Perimeter = 5l
Hence, perimeter of regular pentagon is 5l

2n

In figure,
(a) Number of squares = 1 and number of matchsticks = 4 = 3\(\times\)1+ 1 = 3 x Number of square + 1
(b) Number of squares = 2 and number of matchsticks = 7 = 3\(\times\)2 + 1 = 3 x Number of squares + 1
(c) Number of squares = 3 and number of matchsticks = 10 = 3\(\times\)3 + 1 =3 x Number of squares + 1
(d) Number of squares = 4 and number of matchsticks = 13 =3\(\times\)4 +1 = 3 x Number of squares + 1
Thus, if number of squares = x
Then, number of matchsticks = 3\(\times\)Number of squares + 1 =3x+1
Hence, the required rule that gives the number of
matchsticks is 3x + 1,where x is the number of squares.

Let number of students = 5
Given, number of pencils distributed to each student = 5
When there is one student i.e. 5 = 1, then the number of pencils = 5 \(\times\) 1 i.e. 5
When there are two students i.e. 5 = 2, then the number of pencils = 5 \(\times\) 2 i.e. 10
When there are three students i.e. 5 = 3, then the number of pencils = 5 \(\times\) 3 i.e. 15
When there are 5 students i.e. 5 = 5, then the number of pencils = 5 \(\times\) 5. i.e. 55
Hence, 55 pencils are needed for 5 students.

Rules for the pattern of letters L, C and F are as follow:

In question 1, letter T and V havesame rule as that givenbyL. This happens because the number of matchsticks required in each of them is 2.