10th class mathematics important questions
1.Use Euclid's calculation to track down the H.C.F of 900 and 270?
Euclid's calculation is a technique for tracking down the most noteworthy normal variable (HCF), otherwise called the best normal divisor (GCD), of two numbers. This is the way you can utilize Euclid's calculation to track.
down the HCF of 900 and 270:
Begin with the two numbers: 900 and 270.
Partition the bigger number (900) by the more modest number (270):
900 ÷ 270 = 3 with a rest of 90.
Presently, supplant the bigger number (900) with the more modest number (270), and supplant the more modest number (270) with the rest of).
In this way, presently we have: 270 and 90.
Rehash the division:
270 ÷ 90 = 3 with a rest of 0.
Since we have a rest of 0, the interaction stops.
The last non-zero remaining portion is 90. This remaining portion is the HCF of 900 and 270.
Answer: In this way, the HCF of 900 and 270 is 90.
2.Find A ⋂ B when A= {5,6,7,8} and B= {7,8,9,10}?
The image "⋂" addresses the convergence of sets A and B, which comprises of the components that are normal to the two sets. For this situation, An and B are as per the following:
A = {5, 6, 7, 8}
B = {7, 8, 9, 10}
To find the convergence A ⋂ B, you want to distinguish the components that show up in the two sets:
A ⋂ B = {7, 8}
In this way, the crossing point of sets A and B is {7, 8}. These are the components that are normal to the two sets.
3.Formulate a pair of linear equations in two variables 5 pencils and 7 pens together cost's 50 whereas 7 pencils and 5 pens together cost Rs 46?
Solution:
Let pencil to be x and allow pen to be y.
5x+7y=50
7x+5y=46
Increase the principal condition by 7.
35x+49y=350
Increase the second condition by 5.
35x+25y=230
solving them we get 24y=120 which gives y is 5 rupees
substituting in the second equation we get x is 3 rupees.
4.write the nature of the roots of the quadratic equation2x² - 3x - 5 = 0
Solution:
Given, the quadratic equation is 2x² - 3x - 5 = 0
We have to find the roots of the equation.
By using the quadratic formula,
x = [-b ± √b² - 4ac]/2a
Here, a = 2, b = -3 and c = -5
b² - 4ac = (-3) ² - 4(2) (-5)
= 9 + 40
= 49
x = [- (-3) ± √49]/2(2)
x = [3 ± 7]/4
Now, x = (3+7)/4 = 10/4 = 5/2
x = (3-7)/4 = -4/4 = -1
Therefore, the roots of the equation are 5/2 and -1.
5.find the centroid of the triangle whose vertices are (3, -5), (-7,4), and (10, -2)?
Solution:
Presently we will put the worth of x1, x2, x3.
also, y1, y2, y3
from the directions of the given vertices. Thus, we get.
⇒G= [3−7+103,4−2−53]
On tackling we get.
⇒G= [63 −33]
On separating we get.
⇒G= [2−1]
6.find the volume of the right circular cone with radius 6 cm and height 7 cm?
Solution:
Volume of a cone of base radius r, and height h, = 1/3πr²h
i) Radius of the cone, r = 6 cm
Height of the cone, h = 7 cm
Volume of the cone = 1/3πr²h
= 1/3 × 22/7 × 6 cm × 6 cm × 7 cm
= 264 cm³
7.A dice is thrown at once to find the probability of getting an even prime number on its face?
Solution:
We use the basic formula of probability to solve the problem.
Number of outcomes on throwing a die is (1, 2, 3, 4, 5, 6) = 6
Number of prime numbers on dice are 1, 3 and 5 = 3
(i) Probability of getting a prime number = Number of prime number/total number of outcomes
= 3/6 = 1/2
8.A joker’s cap is in the form of a right circular cone whose base radius is 7 cm and height is 24 cm. Find the area of the sheet required to make 10 such caps ?
Solution
Given: A right circular cone (joker’s cap) of base radius 7 cm and height 24 cm.
Since the cap is conical in shape, the area of the sheet required to make each cap will be equal to the curved surface area of the cone.
The curved surface area of a right circular cone of base radius(r) and slant height(l) is πrl
Slant height, l = √r² + h² where h is the height of the cone.
Radius, r = 7 cm Height, h = 24 cm
Slant height,
l = √r² + h²
= √(7)² + (24)²
= √49 + 576
= √625
= 25 cm
Area of the sheet required to make each cap = πrl
= 22/7 × 7 cm × 25 cm
= 550 cm²
Area of the sheet required to make 10 such caps = 10 × 550 cm2 = 5500 cm2
Thus, the area of the sheet required to make 10 such caps is 5500 cm2.
9.Check whether (5,-2), (6,4), and (7,-2) are the vertices of an isosceles triangle
An isosceles triangle is a triangle that has two sides of equal length.
To check whether the given points are vertices of an isosceles triangle, we need to check if the distance between any of the 2 points should be the same for two pairs of given points.
Let the points (5, - 2), (6, 4), and (7, - 2) represent the vertices A, B, and C of the given triangle
We know that the distance between the two points is given by the Distance Formula, Distance Formula = √[(x₂ - x₁)2 + (y₂ - y₁)2]
To find AB, that is distance between points A (5, - 2) and B (6, 4), let x₁ = 5, y₁ = -2, x₂ = 6, y₂ = 4
AB = √[( 6 - 5 )2 + (4 - (-2))2]
= √[(1)2 + (6)2]
= √1 + 36
= √37
To find BC, distance between Points B (6, 4) and C (7, - 2), let x₁ = 6, y₁ = 4, x₂ = 7, y₂ = - 2
BC = √ [( 7 - 6 )2 + (-2 - 4)2]
= √[(1)2 + (- 6)2]
= √1 + 36
= √37
To find AC, that is distance between Points A (5, - 2) and C (7, - 2), let x₁ = 5, y₁ = - 2, x₂ = 7, y₂ = - 2
AC = √ [( 7 - 5 )2 + (-2 - (- 2))2]
= √[(2)2 + (0)2]
= 2
From the above values of AB, BC and AC we can conclude that AB = BC. As the two sides are equal in length, therefore, ABC is an isosceles triangle.
10.Write the formula for finding the median of groups and explain each term in it ?
Median of grouped data is the median of the data that is continuous and in the form of frequency distribution. Median is the middlemost value of the given data that separates the lower half of the data from the higher half.
Before starting to use the median of the grouped data formula, find the median class by using the above procedure. Then the formula to calculate the median of grouped data is l + [(n/2−c)/f] × h, where:
l = lower limit of median class
n = total number of observations
c = cumulative frequency of the preceding (of median class) class
f = frequency of median class
h = size of each class
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