CBSE Class 10 Maths Question Paper 2018 (Delhi & Outside Delhi) with Solutions

Students can use CBSE Previous Year Question Papers Class 10 Maths with Solutions and CBSE Class 10 Maths Question Paper 2018 (Delhi & Outside Delhi) to familiarize themselves with the exam format and marking scheme.

CBSE Class 10 Maths Question Paper 2018 (Delhi & Outside Delhi) with Solutions

Time allowed: 3 hours
Maximum marks: 80

General Instructions:
Read the following instructions carefully and follow them:

1. All questions are compulsory.
2. This question paper consists of 30 questions divided into four sections – A, B, C and D.
3. Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions and 4 marks each.
4. There is no overall choice. However, an internal choice has been provided You have to attempt only one of the alternatives in all such questions.
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Section A

Questions number 1 to 6 carry 1 mark each.

Question 1.
If x = 3 is one root of the quadratic equation x ² – 2kx -6 = 0, then find the value of k.
Answer:
We have, x ² – 2kx – 6 = 0, where x = 3
⇒ (3) ² – 2k(3) – 6 = 0
⇒ 9 – 6k – 6 = 0 ⇒ 3 – 6k = 0
⇒ -6k = -3 ⇒ k = \(\frac{1}{2}\)

Question 2.
What is the HCF of smallest prime number and the smallest composite number?
Answer:
Smallest prime number = 2
Smallest composite number = 4 or 2 ²
∴ Their HCF = 2

Question 3.
Find the distance of a point F(x, y) from the origin.
Answer:
P(x, y), O(0,0)
∴ OP = \(\sqrt{(x-0)^2+(y-0)^2}\) = \(\sqrt{x^2+y^2}\) units

Question 4.
In an AP, if the common difference (d) = -4, and the seventh term (a ₇ ) is 4, then find the first term.
Answer:
Given: a ₇ = 4, d = -4
⇒ a + 6d = 4 …..[∵ a _n = a + (n – 1)d
⇒ a + 6(-4) = 4 ⇒ a – 24 = 4
⇒ a = 24 + 4 ⇒ a = 28

Question 5.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.

Question 6.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.

Section B

Questions number 7 to 12 carry 2 marks each.

Question 7.
Given that √2 is irrational, prove that (5 + 3√2) is an 1rrational number.
Answer:
Let us assume, to the contrary, that 5 + 3√2 is rational
That is, we can find co-prime a and b (b ≠ 0) such that 5 + 3√2 = \(\frac{a}{b}\)
⇒ 3√2 = \(\frac{a}{b}\) – 5 ⇒ √2 \(\frac{a}{3 b}\) – \(\frac{5}{3}\) \(\frac{a-5 b}{3 b}\)
Since a and b are integers, we get rational, and so \(\frac{a}{3 b}\) – \(\frac{5}{3}\) is rational.
But this contradicts the fact that √2 is irrational.
This contradiction has arisen because of our incorrect assumption that 5 + 3√2 is rational.
So, we conclude that 5 + 3√2 is irrational.

Question 8.
In Fig., ABCD is a rectangle. Find the values of x and y.

Answer:
Given: ABCD is a rectangle.
DC = AB, BC = AD. …[∵ Opposite sides of a rectangle are equal
x + y = 30
14 + y + y = 30 …[From (i)
2y = 30 – 14
y = \(\frac{16}{2}\) = 8
x – y = 14
x = 14 + y ……….. (i)
From (i), x = 14 + 8 = 22
∴ x = 22 cm, y = 8 cm

Question 9.
Find the sum of first 8 muLtiples of 3.
Answer:
To find: 3 + 6 + 9 + 8 terms
Here 1 ^st term, a = 3,
common difference, d = 3,n = 8
As we know, S _n = \(\frac{n}{2}\) [2a + (n – 1 )d]
S ₈ = \(\frac{8}{2}\)[6 + (7)3]
= 4(6 + 21) = 4(27) = 108

Question 10.
Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3). Hence find ni.
Answer:

Let AP : PB = K : 1
Coordinates of P = Coordinates of P
(\(\frac{6 K+2}{K+1}, \frac{-3 K+3}{K+1}\)) = (4, m) ….[Using section formula

\(\frac{6 \mathrm{~K}+2}{\mathrm{~K}+1}\) = 4
⇒ 6K + 2 = 4K + 4
⇒ 6K – 4K = 4 – 2
⇒ 2K = 2
⇒ K = 1 ………. (i)

\(\frac{-3 \mathrm{~K}+3}{\mathrm{~K}+1}\) = m
⇒ \(\frac{-3+3}{1+1}\) = m …[From (i)
∴ m = 0

Question 11.
Two different dice are tossed together. Find the probability:
(i) of getting a doublet
(ii) of getting a sum 10, of the numbers on the two dice.
Answer:
Two dice can be thrown as 6 × 6 = 36 ways
(i) ‘a doublet’ can be obtained as (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), i.e., 6 ways
P(a doublet) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

(ii) “Sum 10” can be obtained as (4, 6), (6, 4), (5, 5), i.e., 3 ways
P(sum 10) = \(\frac{3}{36}\) = \(\frac{1}{12}\)

Question 12.
An integer is chosen at random between I and 100. Find the probability that it is:
(i) divisible by 8. (ii) not divisible by 8.
Answer:
Integers between 1 and 100 are
2, 3, 4, 5, 6,…. 99, i.e., 98
(i) Integers divisible by 8 are
8, 16, 24,…. 96, i.e., 12
P(divisible by 8) = \(\frac{12}{98}\) = \(\frac{6}{49}\)

(ii) Integers not divisible by 8 are 98 – 12 = 86
P(not divisible by 8) = \(\frac{86}{98}\) = \(\frac{43}{49}\)

Section C

Questions number 13 to 22 carry 3 marks each.

Question 13.
Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given numbers.
Answer:

404 = 2 ² × 101
96 = 2 ⁵ × 3
HCF = 2 ² = 4
LCM = 2 ⁵ × 101 × 3
= 32 × 303
= 9696
LHS = HCF × LCM
= 4 × 9696
= 38,784 ………. (i)
RHS = Product of two given nos.
= 404 × 96
= 38,784 ………….(ii)
From (i) and (ii), LHS = RHS …(Hence proved)

Question 14.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.

Question 15.
If A(-2, 1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides.
Answer:

Question 16.
A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.
Answer:
Let the usual speed of the plane = x km/hr
Then the increased speed of a plane = (x + 100) km/hr
Distance = 1500 km
According to the Question,
\(\frac{1500}{x}-\frac{1500}{x+100}\) = \(\frac{30}{60}\)
…..[∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\) = 30 mins. = \(\frac{30}{60}\) = \(\frac{1}{2}\) hr.
⇒ \(\frac{1500(x+100-x)}{x(x+100)}\) = \(\frac{1}{2}\)
⇒ x(x + 100) = 2 × 1,50,000
⇒ x ² + 100x – 3,00,000 = 0
⇒ x ² + 600x – 500x – 3,00,000 = 0
⇒ x(x + 600) – 500(x + 600) = 0
⇒ (x + 600) (x – 500) = 0
⇒ x + 600 = 0 or x – 500 = 0
⇒ x = -600 (rejected) or x = 500
∵ Speed of a plane cannot be negative.
∴ Usual speed of a plane = 500 km/hr

Question 17.
Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonals.
Answer:
Part I.
Theorem: State and prove Thales Theorem.
Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given: In ΔABC, DE || BC

Part II.
Given: ΔACP and ΔBCQ are two similar Δs, drawn on the diagonal and side of a square ABCD.
To Prove: 2ΔBCQ = ΔACP
Proof: ΔACP ~ ΔBCQ

Question 18.
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Answer:
Given: PT and PS are tangents from an external point P to the circle with centre O.

To prove: PT = PS
Const.: Join O to P, T and S.
Proof: In ΔOTP and ΔOSP
OT = OS … [radii of same circle
OP = OP …..[common
∠OTP = ∠OSP…[each 90°
ΔOTP ΔOSP …[R.H.S.
∴ PT = PS …[c.p.c.t.

Question 19.
If 4 tan θ = 3, evaluate ([late]\frac{4 \sin \theta-\cos \theta+1}{4 \sin \theta+\cos \theta-1}[/latex])
Answer:

4 tan θ = 3
tan θ = \(\frac{3}{4}\)
Let BC = 3K, AB = 4K
In rt. ΔABC,
AC ² = AB ² + BC ² …[Pythagoras’ theorem
AC ² = (4K) ² + (3K) ²
= 16K ² + 9K ² = 25K ²
AC = + 5K …[∵ AC cannot be negative

Question 20.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.

Question 21.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Figure. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm. Find the total surface area of the article.

Or
Question 21.
A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?
Answer:
Total Surface Area of the article
= C.S. area of cyl. + 2(C.S. area of hemisphere)
= 2πrh + 2(2πr ² ) = 2πr(h + 2 r)
= 2 × \(\frac{22}{7}\) × \(\frac{35}{10}\) (10 + 7)
= 22 × 17 = 374 cm ²
Or
Here r = \(\frac{24{2}\) = 12m, h = 3.5 m or \(\frac{7}{2}\) m
Volume of rice = \(\frac{1}{3}\)πr ² h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 12 × 12 × \(\frac{7}{2}\) = 528 m ³
Slant height, l = \(\sqrt{r^2+h^2}\) …[Using phythagoras’ theorem
= \(\sqrt{(12)^2+\left(\frac{7}{2}\right)^2}\) = \(\sqrt{144+\frac{49}{4}}\)
= \(\sqrt{\frac{576+49}{4}}\) = \(\sqrt{\frac{625}{4}}\) = \(\frac{25}{2}\) m
C.S. area of rice = πrl
= \(\frac{22}{7}\) × 12 × \(\frac{25}{2}\) = \(\frac{3300}{7}\) m ²
= 471.43 m ² (approx.)
Therefore, canvas cloth is required to cover the heap of rice = 471.43 m ² (approx.).

Question 22.
The table below shows the salaries of 280 persons:

Calculate the median salary of the data.
Answer:

Here n = 280, \(\frac{n}{2}\) = \(\frac{280}{2}\) = 140
Median class is 10 – 15.
Median = l + \(\frac{\frac{n}{2}-c . f .}{f}\) h …[where l = 10, f = 133, c.f. = 49, h = 5
= 10 + \(\frac{140-49}{133}\) × 5
= 10 + \(\frac{91 \times 5}{133}\) = 10 + \(\frac{455}{133}\)
= 10 + 3.42
= ₹13.421 (in thousand) (approx.)
= ₹13,421

Section D

Questions number 23 to 30 carry 4 marks each.

Question 23.
A motor boat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Or
Question 23.
A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed?
Answer:
Let speed of the stream be x km/ hr,
Speed of the boat upstream = (18 – x) km/hr
and Speed of the boat downstream = (18 + x) km/hr
Distance = 24 km
According to the Question
\(\frac{24}{18-x}\) = \(\frac{24}{18+x}\) = 1 ………. [∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)
\(\frac{24[18+x-(18-x)]}{(18-x)(18+x)}\) = 1
\(\frac{24(2 x)}{324-x^2}\) = 1 ⇒ 48x = 324 – x ²
⇒ x ² + 48x – 324 = 0
⇒ x ² + 54x – 6x – 324 = 0
⇒ x(x + 54) – 6(x + 54) = 0
⇒(x – 6) (x + 54) = 0
x – 6 = 0 or x + 54 = 0
x = 6 or x = -54 (rejected)
Since speed cannot be negative.
∴ Speed of stream, x = 6 km/hr
Or
Let the original average speed of the train = x km/hr.
Then the increased average speed of the train = (x + 6) km/hr.
As per the question,
\(\frac{63}{x}+\frac{72}{x+6}\) = 3 …[∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)
⇒ \(\frac{63(x+6)+72 x}{x(x+6)}\) = 3
⇒ 3x(x + 6) = 63x + 378 + 72x
⇒ 3x(x + 6) = 135x + 378
⇒ 3x ² + 18x – 135x – 378 = 0
⇒ 3x ² – 117x – 378 = 0
⇒ x ² – 39x – 126 = 0 …[Dividing both sides by 3
⇒ x ² – 42x + 3x – 126 = 0
⇒ x(x – 42) + 3(x – 42) = 0
⇒ (x – 42)(x + 3) = 0
⇒ x – 42 = 0 or x + 3 = 0
⇒ x = 42 or x = -3 (rejected)
∵ Speed cannot be negative.
The original average speed of the train = 42 km/hr.

Question 24.
The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7 : 15. Find the numbers.
Answer:
Let 4 consecutive numbers in AP be
a – 3d, a – d, a + d, a + 3d
As per the question,
a – 3d + a- d + a + d + a + 3d = 32
⇒ 4a = 32 ⇒ a = \(\frac{32}{4}\) = 8
Now, \(\frac{(a-3 d)(a+3 d)}{(a-d)(a+d)}\) = \(\frac{7}{15}\)
⇒ \(\frac{a^2-9 d^2}{a^2-d^2}\) = \(\frac{7}{15}\)
⇒ \(\frac{64-9 d^2}{64-d^2}\) = \(\frac{7}{15}\) ….[∵a = 8
⇒ 448 – 7d ² = 960 – 135d ²
⇒ 135d ² – 7d ² = 960 – 448
⇒ 128d ² = 512
⇒ d ² = \(\frac{512}{128}\)
⇒ d ² = 4 ⇒ d = ±2

Question 25.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.

Question 26.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.

Question 27.
Prove that : \(\frac{\sin A-2 \sin ^3 A}{2 \cos ^3 A-\cos A}\) = tan A
Answer:

Question 28.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.

Question 29.
As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use √3 = 1.732]
Answer:

Let AB be the light house, C and D the ships.
In rt. ΔABC, tan 45° = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
⇒ 1 = \(\frac{100}{B C}\)
⇒ BC = 100

In rt. ΔABD,
tan 30° = \(\frac{\mathrm{AB}}{\mathrm{BD}}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{100}{B D}\) ⇒ BD = 100√3m
∴ The distance between the two ships,
CD = BD – BC
= 100√3 – 100 = 100(√3 – 1)
= 100(1.732 – 1)
= 100(0.732) = 73.2 m

Question 30.
The mean of the following distribution is 18. Find the frequency of the class 19 – 21:

Answer:

Class	Frequency(f i )	x i	f i x i
11 – 13	3	12	36
13 – 15	6	14	84
15 – 17	9	26	144
17 – 19	13	18	234
19 – 21	f	20	20f
21 – 23	5	22	110
23 – 25	4	24	96
	Σf i = 40 + f		Σf i x i = 70 + 20f

∴ Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) …[Given Mean = 18
⇒ \(\frac{18}{1}\) = \(\frac{704+20 f}{40+f}\)
⇒ 704 + 20f = 720 + 18f
⇒ 20f – 18f = 720 – 704
⇒ 2f = 16 ∴ f = 8