Test: Class 10 Mathematics: CBSE Sample Question Paper- Term I (2021-22) - Class 10 MCQ

Here, divisor = 71, quotient = 37 and remainder = 42.
∴ Dividend = Divisor × Quotient + Remainder
= (71 × 37) + 42
= 2669.

Let x =0.121212.... ...(i)
100x =12.1212.... ...(ii)
On subtracting eqn. (i) from (ii)

Let the zeroes of the given polynomial be a and b, then sum of the zeroes = 4
i.e  p/5  = 4
⇒  p = 20

We have,
p(x) = x³ – 3x² + x + 1
Then, p(– 3) = (-3)³ – 3(– 3)² + (– 3)
+ 1
= – 27 – 27 – 3 + 1
= – 56.

Let the polynomial be f(x), then

Here, two points (–4, 3) and
(0, 0) are given, then
x₁ = – 4, y₁ = 3 and x₂ = 0, y₂ = 0
∴ Distance between them

Then, comparing on both sides, we get
a/3 = – 5
⇒ a = – 15.

Let the required ratio be k : 1, then the coordinates of P are

⇒ –15k + 15 = 3k + 3 and –10k + 25 = 11k + 11
⇒ –18k = –12 and –21k = –14
⇒ k = 2/3  and k = 2/3
Hence the point P divides AB in the ratio
2 : 3.

Distance between 

Let AE = x cm,
then EC = (5.6 – x) cm
and DE || BC   (given)

⇒ 3(5.6 – x) = 5x
⇒ 16.8 – 3x = 5x
⇒ 8x = 16.8
⇒ x = 2.1

∵ sin A = 1/√2,
∴ sin A = sin 45°
⇒ A = 45°
Now, tan A + cot A = tan 45° + cot 45°
= 1 + 1
= 2.

We have,
x = 0
or cos x = cos 0
or cos x = 1

= - 2/4
= - 1/2.

Given, radius (r) = 4 cm
∴ Circumference, (C) = 2πr
= 2 × 3.14 × 4 cm
= 25. 12 cm.

Area of a circle = 38.5 cm²

πr² = 38.5

Length of arc = 30 cm
So, 2 x 5 – 5p + 2 = 0
⇒ 10 + 2 – 5p = 0
⇒ 5p = 12
⇒ p = 2.4

Probability of an event can not be more than one or negative 9/8 > 1.

Total number outcomes = 80
i.e., n(S) = 80
Prime number less than 30 = 2, 3, 5, 7, 11,
13, 17, 19, 23, 29
i.e., n(E) = 10
∴ Required probability,

Let x be added to numerator and denominator of 2/7, so

28 = 2² × 7
and 42 = 2 x 3 x 7
∴ HCF (28, 42) = 2 x 7 = 14
and LCM (28, 42) = 2² x 3 x 7 = 84
Now, difference = 84 – 14 = 70.

Let father's age be x years and son's age be y, then
x + y = 40 ...(i)
and x – y = 20
On adding eq. (i) and (ii), we get
2x = 60
⇒ x = 30.

∵ The system of equations has unique
solution

i.e., all real values except - 14/5

Here,
r = 3.5 cm, θ = 90°

∴ Area of minor segment

Here, r = 18 cm and area of sector = 72 π
cm²

Given, 3x² + 2x – 1 = 0
⇒ 3x² + 3x – x – 1 = 0
⇒ 3x(x + 1) – 1 (x + 1) = 0
⇒ (x + 1) (3x – 1) = 0
⇒ x = – 1 and 1/3

Since ∠P in a semicircle, so
∠P = 90°
∴ In DRPQ, RQ² = PR² + PQ²
(Using Pythagoras theorem)
= (6)² + (8)²
= 100
∴ RQ = 10 cm
∴ Radius = 1/2RQ
= 5 cm

∵ DE || BC (given)
∴ ∠ADE = ∠ABC (corresponding angles)
and ∠AED = ∠ACB (corresponding angles)
∴ DADE ~ DABC (By AA similarity)