State whether the following sentences are right or wrong
The sex ratio is high in Brazil.
The sex ratio is high in Brazil.
Show Hint
Solution and Explanation
The sex ratio is high in Brazil.
Correct Answer: False
Step 1: Sex Ratio in Brazil
Brazil’s sex ratio is roughly balanced. At birth, there are about 1.05 males per female, but overall, females slightly outnumber males because of higher female life expectancy.
Step 2: Comparison with Other Countries
Some countries, like Qatar or India, show more skewed sex ratios. Brazil, however, follows global norms and does not have an unusually high ratio.
\[ \textbf{Brazil has a balanced sex ratio, not a high one.} \]
Top Questions on Practical Geography
- Observe the map and answer the questions given below:
- Maharashtra Class X Board - 2025
- Geography
- Practical Geography
- Give geographical reasons for the following
The waterways are not developed in Brazil.- Maharashtra Class X Board - 2025
- Geography
- Practical Geography
- Give geographical reasons for the following
Settlements are sparse in North-Eastern Brazil.- Maharashtra Class X Board - 2025
- Geography
- Practical Geography
- Give geographical reasons for the following
Tropical cyclones occur rarely in Brazil.- Maharashtra Class X Board - 2025
- Geography
- Practical Geography
- State whether the following sentences are right or wrong
Brazil is the only country in the world where both tigers and lions are found.- Maharashtra Class X Board - 2025
- Geography
- Practical Geography
Questions Asked in Maharashtra Class X Board exam
Complete the following activity to prove that the sum of squares of diagonals of a rhombus is equal to the sum of the squares of the sides.
Given: PQRS is a rhombus. Diagonals PR and SQ intersect each other at point T.
To prove: PS\(^2\) + SR\(^2\) + QR\(^2\) + PQ\(^2\) = PR\(^2\) + QS\(^2\)
Activity: Diagonals of a rhombus bisect each other.
In \(\triangle\)PQS, PT is the median and in \(\triangle\)QRS, RT is the median.
\(\therefore\) by Apollonius theorem,
\[\begin{aligned} PQ^2 + PS^2 &= \boxed{\phantom{X}} + 2QT^2 \quad \dots \text{(I)} \\ QR^2 + SR^2 &= \boxed{\phantom{X}} + 2QT^2 \quad \dots \text{(II)} \\ \text{Adding (I) and (II),} \quad PQ^2 + PS^2 + QR^2 + SR^2 &= 2(PT^2 + \boxed{\phantom{X}}) + 4QT^2 \\ &= 2(PT^2 + \boxed{\phantom{X}}) + 4QT^2 \quad (\text{RT = PT}) \\ &= 4PT^2 + 4QT^2 \\ &= (\boxed{\phantom{X}})^2 + (2QT)^2 \\ \therefore \quad PQ^2 + PS^2 + QR^2 + SR^2 &= PR^2 + \boxed{\phantom{X}} \\ \end{aligned}\]
- Maharashtra Class X Board - 2025
- Quadrilaterals
- Prove that tangent segments drawn from an external point to a circle are congruent.
- Draw a circle with radius 4.1 cm. Construct tangents to the circle from a point at a distance 7.3 cm from the centre.
- Maharashtra Class X Board - 2025
- Constructions
- \(\triangle\)ABC \(\sim\) \(\triangle\)PQR. In \(\triangle\)ABC, AB = 3.6 cm, BC = 4 cm and AC = 4.2 cm. The corresponding sides of \(\triangle\)ABC and \(\triangle\)PQR are in the ratio 2 : 3, construct \(\triangle\)ABC and \(\triangle\)PQR.
- Maharashtra Class X Board - 2025
- Constructions
- \(\square\)ABCD is a rectangle. Taking AD as a diameter, a semicircle AXD is drawn which intersects the diagonal BD at X. If AB = 12 cm, AD = 9 cm, then find the values of BD and BX.