Test: Class 9 Math: CBSE Sample Question Paper Term I - 1 - Class 9 MCQ

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Infinitely many equations satisfy x = 2 and y = 3 as infinitely many lines pass through a single point.

In ΔAOB , using angle sum property of triangles, we have
40 + 60 + ∠AOB = 180
⇒ ∠AOB = 180 - 100 = 80
Also, we have
x + y + ∠AOB = 180 ...(Angles on a line)
⇒  x + y + 80 = 180
⇒  x + y = 100 

Area = 1/2 x Base x Height
= 1/2 x 8 x 10
= 40 cm² 

0 is an integer and all integers are rational numbers.

a point.
As, x = -9/2 is a point.

y : z = 3 : 7
Let common ratio be a
y = 3a
z = 7a
x = z (corresponding angle)
x = 7a
x + y =180° (interior angle)
7a + 3a =180°
10 a = 180°
a = 180/10
a = 18
x = 7a
x = 7x 18
x =126°

In ΔABC we have:
∠A + ∠B + ∠C = 180° [Sum of the angles of a trianlge]
⇒ 50° + ∠B + ∠C = 180°
⇒  ∠B + ∠C = 130°
⇒ 1/2∠B + 1/2∠C = 65°
In ΔOBC we have:
∠OBC + ∠OCB + ∠BOC = 180°
⇒ 1/2∠B + 1/2∠C + ∠BOC  = 180° [Using (i)]
⇒ 65° + ∠BOC = 180°
⇒ ∠BOC = 115°

10^x = 64
⇒ 
⇒ 
⇒ 
Now, 
= 8  x 10 = 80

We have:
3x + 72 = 180° [ AOB is a straight line]
⇒ 3x = 108
⇒ x = 36
Also,
∠AOC + ​∠COD + ​∠BOD = 180°[ AOB is a straight line]
⇒ 36° + 90° + y = 180°
⇒ y = 54°

= 
= 
= 
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= 1/9

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= 
= 
= 
= 
= 16 + 1/2
= 33/2

59^o, 60^o, 61^o
Clearly, x + (x + 1) + (x + 2) = 180
⇒ 3x + 3 = 180
⇒ 3(x + 1) = 180
⇒ x + 1 = 60
⇒ x = 59
Hence the angles are 59, 60, 61

(a, -2) lies on the graph of 3x - y = 10
3(a) - (-2) = 10
3a + 2 = 10
3a = 10-2
3a = 8
a = 8/3

Let the lower limit of a class = x
class size = 10
Upper limit = x + 10
Now, mid - value =  = x + 5 = 42 (given)
x = 37 = lower limit
x + 10 = 47 = upper limit
pper and lower limits are 47, 37

Since longest altitude is drawn opposite to the shortest side in a triangle.
Area of triangle = 1/2 x Base x Height
⇒ 420√5 = 1/2 x 35 x Height
⇒ Height = 
= 24√5 cm

If  be the mean of the n observations q X_i, ..., X_n then we have

⇒ 
Let  be the mean of n values X_i, ..., X_n. So, we have

⇒ 
The sum of the deviations of n values X_i, ..., X_n from their mean  is
= 
= 
= 
= 0

= 
= 
= 
= 5/5
= 1

Let,
AB, CD and EF intersect at O
∠COB = ∠AOD (Vertically opposite angle)
∠AOD = 3x + 10 ....(i)
∠AOE + ∠AOD + ∠DOF = 180o (Linear pair)
x + 3x + 10^o + 90^o = 180^o
4x + 100^o = 180^o
4x = 80^o
x = 20^o

There is only one pair i.e., (1, 1) which satisfies the given equation but in positive real numbers, real numbers and rational numbers there are many pairs to satisfy the given linear equation. Hence, unique solution is possible only in case of Natural numbers.

Let the sides of ΔABC be n, n + 1, n + 2
⇒ Perimeter = n + n + 1 + n + 2
⇒ (9 + 9 + 9) = 3n + 3
⇒ 27 = 3n + 3
⇒ 3n = 24
 ⇒ n = 8 cm
Thus, the shortest side is 8 cm

A linear equation in two variables is of the form ax + by + c = 0 as a and b are cofficient of x and y
so if a = 0 and b = 0 or either of one is zero in that  case the equation will be one variable or their will be no equation respectively. 
therefore when a ≠ 0 and b ≠ 0 then only the equation will be in two variable

It lies on y-axis so it's abscissa = 0 and it lies on y-axis at a distance of 4unit.
Thus point will be (0, 4).

10^2y = 25
10^2y =  25
(10^y)² = (5)² 
⇒ 10^y = 5
Now 10^-y
= 1/10^y
= 1/5

Area of equilateral triangle = 81√3cm²
⇒ 
⇒ (Side)²  = 81 x4
⇒ (Side)² = 324
Side = 18 cm
Now,
Height = 
= 9√3cm

In ΔAOB andΔDOC
∠OAB = ∠ODC (alternate interior angles)
∠OBAv=v∠OCD
OB = OC (given)
So, from ASA congruence ,we have 
ΔAOB ≅ ΔDOC 
Now, from CPCT ,we have
AB = CD
OA = OD which means O is the mid-point of AD.
Hence ,all the given statements are true.

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= 1/2

 In 3rd quadrant values of Abscissa and Ordinate both are —ve , i.e, (— , +)
Here,since value of both x and y-coordinate are -ve so it will lies in 3rd quadrant

Add the values corresponding to the height of the bar before 40.
6 + 3 + 4 + 2 = 15

Area of triangle with sides a, b, c (A) = 
New sides are 2a, 2b and 2c
Then 
New area =  
= = 
= 4A
Increased area = 4A - A = 3A
% of increased area = 3A/A x 100 = 300%