Sum of the ratios = 3 + 2 = 5
Semi-perimeter = Length + Breadth = 50 m
\(\therefore\) Length = \(\frac { 3 }{ 5 } \times 50\) = 30 m
Breadth = \(\frac { 2 }{ 5 } \times 50\) = 20 m

Length of each part = 5.5 cm.

For finding the HCF of 16 and 20, we use division method.

Hence, HCF of 16 and 20 is 4.

1.54 + 1.80
ones           tenths             Hundredths
1                   5                       4
1                   8                       0
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3                  3                           4
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Number of hundredths in 1.54 = 4
Number of hundredths in 1.80 = 0
\(\therefore \)Number of hundredths in the sum of 1.54
and 1.80=4+0=4
Number of tenths in 1.54 = 5
Number of tenths in 1.80 = 8
5 tenths + 8 tenths = (5 + 8) tenths = 13 tenths
= 10 tenths + 3 tenths
= lone + 3 tenths
\(\therefore \) Number of tenths in the sum of 1.54 and
1.80 = 3
Number of ones in 1.54 = 1
Number of ones in 1.80 = 1
Number of ones as carry = 1
\(\therefore \) Number of ones in the sum of 1.54 and 1.80
=1+1+1=3
Therefore, 1.54 + 1.80 = 3.34

Given, equation 2x - 5 = 30 ⇒  LHS = 2x - 5
On putting x = 4 in LHS ⇒  2\(\times\) 4 - 5 = 8 - 5 = 3 and RHS = 30
So, LHS≠RHS
Hence, x = 4 is not a solution of 2x - 5 = 30.

2 x Radius = Diameter

In 4235, hundred digit is 2 < 5.
∴ The required rounded number 4235 = 4000

To find 6 - 2
Let us start from 6 and make 2 equal jumps to the left of 6. Each jump is equal to 1 unit. Now, we reach at 4.

\(\therefore\)  6 - 2 = 4.

Like fractions: Fractions having same denominators

Given, merchant have 120 L of oil one kind, 180 L of another kind and 240 L of a third kind.
The required capacity of a tin will be HCF of 120, 180 and 240

HCF of 120 and 180 is 60.
Now, we find the HCF of 60 and 240

The HCF of 60 and 240 is 60.
Hence, HCF of 120, 180 and 240 is 60.
\(\therefore\) The required capacity of a tin = 60 L

16 cm²

Let Sarita's present age = y yr
(i) Sarita's age after 5 yr from now = (y + 5) yr
(ii) Sarita's age 3 yr back = (y - 3) yr
(iii) Given, Sarita's grandfather is 6 times of her age So,
Sarita's grandfather age = 6\(\times\)Sarita's present age =6\(\times\)y=6yyr
(iv) Given, grandmother is 2 yr younger than grandfather. Age of grandfather = 6y yr [from (iii)]
∴  Age of grandmother = [Grandfather's age - 2] yr =(6y-2)yr
(v) Given, Sarita's father's age is 5 yr more than 3 times Sarita's age.
Sarita's present age =y yr
3 times of Sarita's age = 3y yr
∴  Sarita's father age = 3\(\times\)Sarita's age + 5 = (3y + 5) yr

According to the question,
Digit at ten's place = 4
Digit at unit's place = \(\frac{1}{4}\) of ten's place digit =\(\frac{1}{4}\times4=1\)
Digit at hundred's place = 0
Digit at thousand's place = 5 x Digit of unit's place
=5\(\times\)1=5
Digit at ten thousand's place = 2 x Digit of ten's place
=2\(\times\)4=8
∴ Required number = 85041

Rs 5000

(a) The following figure shows a closed curve, but not a polygon, because a polygon is made by line segments only.

(b) The following figure shows an open curve made up entirely of line segments.

(c) It is not possible, because a polygon of two sides cannot be drawn.

We know that, tetrahedron is a regular pyramid made up of 4 equilateral triangles.

30^o

In figures (a) and (c), there ate only two lines of symmetry i. e. horizontal line of symmetry and vertical line of symmetry.

But for figure (b), we can draw more than two lines of symmetry as shown below.

The given fractions are unlike, we should first get their equivalent fractions with a denominator, which is common multiple of 6 and 15 i.e 30.
now \(\frac { 5\times 5 }{ 6\times 5 } =\frac { 25 }{ 30 } ,\frac { 13\times 2 }{ 15\times 2 } =\frac { 26 }{ 30 } \)
Since\(\frac { 26 }{ 30 } >\frac { 25 }{ 30 } \)so \(\frac { 13 }{ 15 } >\frac { 5 }{ 6 } \)

Rs. 27666050, ~Rs. 48622530

We have, 30 and 415
30 = 1 x 30; 30 = 2 x 15; 30 = 3 x 10; 30 = 5 x 6
Factors of 30 are 1,2,3,5,6,10,15 and 30 and 415 = 1 x 415; 415 = 5 x 83
Factors of 415 are 1, 5, 83 and 415.
\(\because\)Common factors of 30 and 415 are 1 and 5.
So, 30 and 415 are not co-prime numbers.

The joins of the vertices, which are not adjacent i.e. opposite to each other, are called diagonals of quadrilateral. A rough sketch of a quadrilateral PQRS with its diagonals and their names are given below:

Here, PR and QS are its diagonals. They intersect each other at point O. So, point O is in the interior of the quadrilateral PQRS.

We know that, one less than given number gives a predecessor. So, the preceding numbers of the given numbers are given as below:

Whole numbers between 32 and 53 are as follows:
33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49,  50, 51 and 52.
Hence, total number of whole numbers between 32 and 53 is 20.
Alternate Method
We can find the total numbers by using (b - a) - 1.
Where, a = First number and b = Last number
Here, a = 32, b = 53
Whole numbers between 32 and 53 = (b - a) - 1
                                                    = (53 - 32) -1 [putting the values of a and b)
                                                    = 21 - 1 = 20.

Number of tiles =\(\frac { Area\quad of\quad the\quad floor }{ Area\quad of\quad one\quad tile } =\frac { 40\times 30 }{ 2\times 2 } =\frac { 1200 }{ 4 } \)
=300

Given population is 2538473.
Rounded off to nearest thousands = 2538000

According to the associative law under addition,
125 + (68 + 32) = (125 + 68) + 32.

Perimeter = 5 x 1 = 5 m