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CBSE Class 10 Maths Question Paper 2018 Comptt (Delhi & Outside Delhi) with Solutions
Time allowed: 3 hours
Maximum marks: 80
General Instructions:
Read the following instructions carefully and follow them:
- All questions are compulsory.
- This question paper consists of 30 questions divided into four sections – A, B, C and D.
- Section A contains 6 questions ofl 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.
- There is no overall choice. However, an internal choice has been provided in jew questions. You have to attempt only one of the alternatives in all such questions.
- Use of calculator is not permitted.
Section A
Questions number 1 to 6 carry 1 mark each.
Question 1.
Write whether \(\frac{2 \sqrt{45}+3 \sqrt{20}}{2 \sqrt{5}}\) on simplification gives an irrational or a rational number.
Answer:
Question 2.
If x = a, y = b is the solution of the pair of equations x – y = 2 and x + y = 4, find the values of a and b.
Answer:
x – y = 2
a – b = 2 …[∵ x = a, y = b
a = 2 + b …….. (i)
x + y = 4
a + b = 4 [∵ x = a, y = b
2 + b + b = 4 …[From (i)
2b = 4 – 2 ⇒ 2b = 2
b = \(\frac{2}{2}\) = 1
From (i), a = 2 + b = 2 + 1 = 3
∴ a = 3, b = 1
Question 3.
If one root of 5x
2
+ 13x + k = 0 is the reciprocal of the other root, then find value of k.
Answer:
Let roots be α and \(\frac{1}{\alpha}\).
Given. 5x
2
+ 13x + k = 0
Here, a = 5, b = 13, c = k
Product of roots, α × \(\frac{1}{\alpha}\) = \(\frac{c}{a}\) = \(\frac{k}{5}\)
⇒ 1 = \(\frac{k}{5}\) ∴ k = 5
Question 4.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.
Question 5.
Two cubes have their volumes in the ratio 1 : 27. Find the ratio of their surface areas.
Answer:
Let side of I
st
cube be x units,
and side of II
nd
cube be y units.
∴ Required Ratio = 1 : 9
Question 6.
A(5, 1); B(1, 5) and C(-3, -1) are the vertices of ΔABC. Find the length of median AD.
Answer:
Section B
Questions number 7 to 12 carry 2 marks each.
Question 7.
Given that √3 is an irrational number, prove that (2 + √3) is an irrational number.
Answer:
Let us assume, to the contrary, that 2 + √3 is rational.
That is, we can find co-prime a and b(b ≠ 0)
such that 2 + √3 =
⇒ √3 = \(\frac{a}{b}\) – 2 ⇒ √3 = \(\frac{a-2 b}{b}\)
Since, a and b are integers, we get (\(\frac{a}{b}\) – 2)
is rational, and so √3 is rational.
But this contradicts the fact that √3 is irrational. This contradiction has arisen because of our incorrect assumption that 2 + √3 is rational.
So, we conclude that 2 + √3 is irrational.
Question 8.
X is a point on the side BC of ΔABC. XM and XN are drawn parallel to AB and AC respectively meeting AB in N and AC in M. MN produced meets CB produced at T. Prove that TX
2
= TB × TC.
Answer:
Given. XM || AB and XN || AC
To Prove. TX
2
= TB × TC
Proof. In ΔTXM, BN || XM …[∵ XM || AB
\(\frac{\mathrm{TB}}{\mathrm{TX}}\) = \(\frac{\mathrm{TN}}{\mathrm{TM}}\) ………. (i) ….[Thales theorem
In Δ TMC, XN || CM …[∵ XN || AC
\(\frac{\mathrm{TX}}{\mathrm{TC}}\) = \(\frac{\mathrm{TN}}{\mathrm{TM}}\) ………. (ii) ….[Thales theorem
From (i) & (ii), \(\frac{\mathrm{TB}}{\mathrm{TX}}\) = \(\frac{\mathrm{TX}}{\mathrm{TC}}\)
⇒ TX
2
= TB × TC (Hence Proved)
Question 9.
In Fig., ABC is a triangle in which ∠B = 90°, BC = 48 cm and AB = 14 cm.
A circle is inscribed in the triangle, whose centre is O. Find radius r of in circle.
Answer:
Construction. Join OA,
OB and OC.
Area of ΔABC
= \(\frac{1}{2}\) × AB × BC
= \(\frac{1}{2}\) × 14 × 48
= 7 × 48 = 336 cm
2
Now, ar(ΔABC)
= ar(ΔAOB) + ar(ΔBOC) + ar(ΔAOC)
⇒ 336 = \(\frac{\mathrm{AB} \times r}{2}+\frac{\mathrm{BC} \times r}{2}+\frac{\mathrm{AC} \times r}{2}\) …[∵ Area of Δ = \(\frac{1}{2}\) × Base × Height
⇒ 336 = \(\frac{r}{2}\) (AB + BC + AC)
⇒ 336 × 2 = r(14 + 48 + 50)
…[AC = \(\sqrt{48^2+14^2}\) = \(\sqrt{2500}\) = 50 cm
⇒ 336 × 2 = r × 112
⇒ r = \(\frac{336 \times 2}{112}\) = 6
∴ Radius of in-circle, r = 6 cm
Question 10.
Find the linear relation between x and y such that P(x, y) is equidistant from the points A(1, 4) and B(-1, 2).
Answer:
PA = PB …[Given
PA
2
= PB
2
…[squaring both sides
⇒ (x – 1)
2
+ (y – 4)
2
= (x + 1)
2
+ (y – 2)
2
⇒ x
2
– 2x + 1 + y
2
– 8y + 16 = x
2
+ 2x + 1 + y
2
– 4y + 4
⇒ – 2x – 8y + 16 – 2x + 4y – 4 = 0
⇒ – 4x – 4y + 12 = 0 …[Dividing both sides by -4
⇒ x + y – 3 = 0
Therefore, x + y – 3 = 0is the linear relation between x and y.
Question 11.
A, B, C are interior angles of ΔABC. Prove that cosec(\(\frac{A+B}{2}\)) = sec\(\frac{C}{2}\).
Answer:
In ΔABC,
∠A + ∠B + ∠C = 180° …[Angle-Sum-Property of a Δ
⇒ A + B = 180° – C ……..(i)
LHS.: cosec(\(\frac{A+B}{2}\))
= cosec(\(\frac{180^{\circ}-\mathrm{C}}{2}\)) ….. [From (i)
= cosec(\(\frac{180^{\circ}-\mathrm{C}}{2}\))
= cosec(90° – \(\frac{C}{2}\) = sec \(\frac{C}{2}\) …[∵ cosec(90° – θ) = sec θ
= R.H.S. (Hence Proved)
Question 12.
A right circular cylinder and a cone have equal bases and equal heights. If their curved surface areas are in the ratio 8 : 5, show that the ratio between radius of their bases to their height is 3 : 4.
Answer:
Let r be the radii of cylinder & cone, and h be the heights of cylinder & cone.
⇒ 16 (r
2
+ h
2
) = 25h
2
…[Squaring both sides
⇒ 16r
2
+ 16h
2
= 25h
2
⇒ 16r
2
= 25h
2
– 16h
2
⇒ 16r
2
= 9h
2
⇒ \(\frac{r^2}{h^2}=\frac{9}{16}\) ⇒ \(\frac{r}{h}=\frac{3}{4}\) …[Taking square-roots on both sides
Hence, the ratio between radius and height is 3 : 4. (Hence Proved)
Section C
Questions number 13 to 22 carry 3 marks each.
Question 13.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.
Question 14.
Divide 27 into two parts such that the sum of their reciprocals is \(\frac{3}{20}\) .
Answer:
Let the two parts be x and 27 – x.
According to question, \(\frac{1}{x}+\frac{1}{27-x}=\frac{3}{20}\)
⇒ \(\frac{27-x+x}{x(27-x)}=\frac{3}{20}\)
⇒ 27 × 20 = 3x(27 – x)
⇒ 180 = 27x – x
2
⇒ x
2
– 27x + 180 = 0
⇒ x
2
– 15x – 12x + 180 = 0
⇒ x(x – 15) – 12(x – 15) = 0
⇒ (x – 12)(x – 15) = 0
Either x – 12 = 0 or x – 15 = 0
x = 12 or x = 15
When x = 12, two parts are 12,15.
When x = 15, two parts are 15,12.
Question 15.
In an A.P. if sum of its first n terms is 3n
2
+ 5n and its kth term is 164, find the value of k.
Answer:
S
n
= 3n
2
+ 5n …[Given
When n = 1, S
1
= 3(1)
2
+ 5(1) = 8
When n = 2, S
2
= 3(2)
2
+ 5(2) = 12 + 10 = 22
a
1
= S
1
= 8; a
2
= S
2
– S
1
= 22 – 8 = 14
d = a
2
– a
1
= 14 – 8 = 6
We have, k
th
term = 164
⇒ a + (k – 1)d = 164 ⇒ 8 + (k – 1)6 = 164
⇒ (k – 1)6 = 164 – 8 ⇒ (k – 1)6 = 156
⇒ (k – 1) = \(\frac{156}{6}\) ⇒ k = 26 + 1
∴ k = 27
Question 16.
If coordinates of two adjacent vertices of a parallelogram are (3, 2), (1, 0) and diagonals bisect each other at (2, -5), find coordinates of the other two vertices.
Answer:
Let C(a, b) and D(c, d).
Mid-point of AC = Mid-point of BD …[Diagonals of a ||gm bisect each other.
Here, Mid-point of AC = Coordinates of P a + 3 b + 2
(\(\frac{a+3}{2}\), \(\frac{b+2}{2}\)) = (2, -5)
⇒ \(\frac{a+3}{2}\) = 2
a + 3 = 4
⇒ a = 1
⇒ \(\frac{b+2}{2}\) = -5
b + 2 =-10
⇒ b = -12
Pt. C(1, -12)
Now, Mid-point of BD = Coordinates of P
(\(\frac{c+1}{2}\), \(\frac{d+0}{2}\)) = (2, -5)
⇒ \(\frac{c+1}{2}\) = 2
c + 1 =4
c = 3
\(\frac{d+0}{2}\) = -5
d = -10
Pt. D(3, -10).
Question 17.
In Fig., AB is a chord of length 8 cm of a circle of radius 5 cm. The tangents to the circle at A and B intersect at P. Find the length of AP.
Or
Question 17.
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Answer:
Construction. Join OP.
Let OP intersect AB at a point M. Then ΔAPB is isosceles and OP is the angle bisector of ∠APB.
So, OP⊥AB
AM = MB = 4 cm …[∵⊥ from the centre bisects the chord
In rt. ΔOMA,
OM = \(\sqrt{\mathrm{OA}^2-\mathrm{AM}^2}\) …[Pythagoras’ theorem
OM = \(\sqrt{5^2-4^2}\) = \(\sqrt{25-16}\) = √9 = 3cm
Let AP = x cm and PM = y cm
In rt. ΔAMP,
PA
2
= AM
2
+ PM
2
x
2
= 4
2
+ y
2
⇒ x
2
– y
2
= 16
In rt. AOAP,
OA
2
+ PA
2
= OP
2
…[Pythagoras’ theorem
(5)
2
+ x
2
= (y + 3)
2
25 + x
2
= y
2
+ 6y + 9
x
2
– y
2
= 6y + 9 – 25
⇒ x
2
– y
2
= 6y – 16
From solving (i) & (ii),
6y – 16 = 16
⇒ 6y = 32 ⇒ y = \(\frac{32}{6}\) = \(\frac{16}{3}\)
From (i), x
2
– (\(\frac{16}{3}\))
2
= 16
⇒ x
2
= 16 + \(\frac{256}{3}\) = \(\frac{144+256}{9}\) = \(\frac{400}{9}\)
⇒ x = \(\frac{20}{3\) or \(6 . \overline{6}\) cm
Therefore, length of AP = x = \(6 . \overline{6}\) cm.
Or
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 18.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.
Question 19.
Prove that: (\(\frac{1+\tan ^2 \mathrm{~A}}{1+\cot ^2 \mathrm{~A}}\)) = (\(\frac{1-\tan A}{1-\cot A}\)) = tan
2
A
Answer:
Question 20.
The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of distances travelled by their tips in 48 hours.
Or
Question 20.
The side of a square is 10 cm. Find the area between inscribed and circumscribed circles of the square.
Answer:
Length of short hand = 4 cm
Distance travelled by short hand in 12 hrs.
= 2πr = 2(\(\frac{22}{7}\)) × 4 = \(\frac{176}{7}\) cm
Distance travelled by short hand in two days or 48 hrs.
\(\frac{176}{7}\) × 4= \(\frac{704}{7}\) cm ……… (i)
Length of long hand = 6 cm
Distance travelled by long hand in 1 hr.
= 2 × \(\frac{22}{7}\) × 6 \(\frac{264}{7}\) cm
Distance travelled by long hand in 2 days i.e., 48 hrs.
= \(\frac{264}{7}\) × 48 = \(\frac{12672}{7}\) cm …….. (ii)
∴ Total distance travelled by both hands
= \(\frac{704}{7}\) + \(\frac{12672}{7}\) = \(\frac{13376}{7}\) = 1910.86 cm
Or
Given.ABCD is a square i.e., AB = BC
In rt. ΔABC
AC
2
= AB
2
+ BC
2
= 10
2
+ 10
2
…(using Pythagoras’ Theorem
= 100 + 100 = 200
AC = \(\sqrt{2 \times 100}\) = 10√2 cm
Radius (outer circle),
OA = R = \(\frac{\mathrm{AC}}{2}\) = \(\frac{10 \sqrt{2}}{2}\) = 5√2
Radius (inner circle),
OE = r = \(\frac{\mathrm{EF}}{2}\) = \(\frac{\mathrm{AB}}{2}\) = \(\frac{10}{2}\) = 5 cm
∴ Area between inscribed and circumscribed circle = ar. of outer circle – ar. of inner circle
= πR
2
– πr
2
= π(R
2
– r
2
)
= \(\frac{22{7}\)[(5√2)
2
– (5)
2
]
= \(\frac{22{7}\) (50 – 25) = \(\frac{22 \times 25}{7}\) = \(\frac{550}{7}\)
= 78.57 cm
2
(approx.)
Question 21.
If sin (A + 2B) = \(\frac{\sqrt{3}}{2}\) and cos (A + 4B) = 0, A > B, and A + 4B ≤ 90°, then find A and B.
Answer:
sin(A + 2B) = \(\frac{\sqrt{3}}{2}\)
⇒ sin(A + 2B) = sin 60°
⇒ A + 2B – 60°
A=60° – 2B …….. (i)
A = 60° – 2(15°) …[From (ii)
∴ A = 60° – 30° – 30°
cos(A + 4B) = 0
cos(A + 4B) = cos 90°
⇒ A + 4B = 90°
⇒ 60° – 2B + 4B = 90° …[From (i)
⇒ 2B = 90° – 60°
∴ B = 15° ………… (ii)
Question 22.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.
Section D
Questions number 23 to 30 carry 4 marks each.
Question 23.
For what values of m and n the following system of linear equations has infinitely many solutions.
3x + 4y = 12; (m + n)x + 2(m – n)y = 5m – 1
Answer:
Here, \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\) ……[∵ Infinite solutions
\(\frac{3}{m+n}=\frac{4}{2(m-n)}=\frac{12}{5 m-1}\)
Taking first two
\(\frac{3}{m+n}=\frac{2}{m-n}\)
⇒ 3m – 3n = 2m + 2n
⇒ 3m – 2m = 2n + 3n
⇒ m = 5n …(i)
At n = 1, …[From (ii)
∴ m = 5(1) = 5
Taking last two
\(\frac{2}{m-n}=\frac{12}{5 m-1}\)
⇒ 6m – 6n = 5m – 1
⇒ 6m – 5m – 6n = – 1
⇒ m – 6n = -1
⇒ 5n – 6n = -1 …..[Frorn (i)
⇒ – n = – 1 ⇒ n = 1
Question 24.
Not in Current Syllabus.
Answer:
Not in Current Syllabus.
Question 25.
A faster train takes one hour less than a slower train for a journey of 200 km. If the speed of slower train is 10 km/hr less than that of faster train, find the speeds of two trains.
Or
Question 25.
Solve for x: \(\frac{1}{a+b+x}\) = \(\frac{1}{a}\) + \(\frac{1}{b}\) + \(\frac{1}{x}\), a ≠ 0,b ≠ 0, x ≠ 0
Answer:
Let speed of fast train be x km/hr.
and speed of slow train be (x -10) km/hr
Given. Distance travelled = 200 km
According to Question,
\(\frac{200}{x-10}-\frac{200}{x}\) = 1 (∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)
⇒ 200(\(\frac{1}{x-10}-\frac{1}{x}\)) = 1
⇒ 200(\(\frac{x-(x-10)}{x(x-10)}\)) = 1
⇒ x(x – 10) = 2000
⇒ x
2
– 10x – 2000 = 0
⇒ x
2
– 50x + 40x – 2000 = 0
⇒ x(x – 50) + 40(x – 50) = 0
⇒ (x – 50) (x + 40) = 0
⇒ x – 50 = 0 or x + 40 = 0
⇒ x = 50 or x = -40 (rejected)
As speed of train cannot be negative.
∴ Speed of fast train = 50 km/hr
Speed of slow train = (x – 10)
= 40 km/hr
Or
⇒ \(\frac{-1}{x(a+b+x)}=\frac{1}{a b}\)
⇒ x(a + b + x) = -ab
⇒ x
2
+ ax + bx + ab = 0
⇒ x(x + a) + b(x + a) = 0
⇒ (x + o)(x + b) = 0
⇒ x + a = 0 or x + b = 0
⇒ x = -a or x = -b
∴ x = -a, -b
Question 26.
Not in Current Syllabus
Answer:
Not in Current Syllabus
Question 27.
The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of depression from the top of tower to the foot of hill is 30°. If tower is 50 metre high, find the height of the hill.
Or
Question 27.
Two poles of equal height are standing opposite to each other on either side of the road which is 80 m wide. From a point in between them on the road, the angles of elevation of the top of poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.
Answer:
Let CD, the height of the hill = H, Tower be AB and distance between them
BC = x m
In rt. ΔABC,
tan 30° = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
\(\frac{1}{\sqrt{3}}=\frac{50}{x}\)
x = 50√3 ……(i)
In rt. ΔBCD,
tan 60° = \(\frac{\mathrm{CD}}{\mathrm{BC}}\) ⇒ √3 = \(\frac{\mathrm{H}}{x}\)
H = x√3
Substituting the value of x from (i), we get
H = 50 √3 × √3 = 50 × 3 = 150 m
∴ Height of the hill = 150 m
Or
Let AB = DE = y m be the two poles such that BD = 80m, CD = x m and BC = 80 – x m
In rt. ΔCDE,
tan 30° = \(\frac{\mathrm{DE}}{\mathrm{CD}}\)
\(\frac{1}{\sqrt{3}}=\frac{y}{x}\) ⇒ x = √3y …….(i)
In rt. ΔABC,
tan 60° = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
\(\frac{\sqrt{3}}{1}=\frac{y}{80-x}\) ⇒ y = √3 (80 – x)
⇒ y = √3 (80 – √3y) …[From (i)
⇒ y = 80 √3 – 3y ⇒ 4y = 80√3
⇒ y = \(\frac{80 \sqrt{3}}{4}\) = 20√3m = 20(1.73) = 34.6 m
Putting the value of y in (i),
x = √3 (20√3 ) ⇒ x = 60 m
∴ CD, x = 60 m
BC = 80 – x = 80 – 60 = 20 m
Therefore, Height of poles = 20√3 m and Distance of poles from the point are 20 m and 60 m.
Question 28.
A man donates 10 aluminum buckets to an orphanage. A bucket made of aluminum is of height 20 cm and has its upper and lowest ends of radius 36 cm and 21 cm respectively. Find the cost of preparing 10 buckets if the cost of aluminum sheet is ₹42 per 100 cm
2
.
Answer:
Upper radius of the bucket, R = 36 cm
Lower radius of the bucket, r = 21 cm
Height of the bucket, h = 20 cm
Slant height of the bucket,
l = \(\sqrt{(R-r)^2+h^2}\) = \(\sqrt{(36-21)^2+20^2}\)
= \(\sqrt{15^2+20^2}\) = \(\sqrt{225+400}\) = \(\sqrt{625}\) = 25cm
T.S.A. of 1 bucket = π(R + r)l + πr
2
= π[(R + r)l + r
2
] = \(\frac{22}{7}\) [(36 + 21)25 + (21)
2
]
= \(\frac{22}{7}\) (1425 + 441) = \(\frac{22}{7}\) × 1866 cm
2
Cost of 1 buckets @ of ₹42 per 100 cm
2
= \(\frac{42}{100}\) × \(\frac{22}{7}\) × 1866 = ₹2463.12
∴ Cost of 10 buckets = 10 × ₹2463.12 = ₹24631.20
Question 29.
Find the mean and mode for the following data:
Answer:
| Classes | Frequency (f i ) | x i | f i x i |
| 10 – 20 | 4 | 15 | 60 |
| 20 – 30 | 8 | 25 | 200 |
| 30 – 40 | 10f 0 | 35 | 350 |
| 40 – 50 | 12f 1 | 45 | 540 |
| 50 – 60 | 10f 2 | 55 | 550 |
| 60 – 70 | 4 | 65 | 260 |
| 70 – 80 | 2 | 75 | 150 |
| Σf i = 50 | Σf i x i = 2110 |
∴ Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{2110}{50}\) = 42.2
Maximum frequency is 12.
∴ Modal class is 40 – 50
∴ Mode = l + \(\frac{f_1-f_0}{2 f_1-f_0-f_2}\) × h ….[l = 40, h = 10, f
0
= 10, f
1
= 12, f
2
= 10
= 40 + \(\frac{12-10}{24-10-10}\) × 10
= 40 + (\(\frac{2}{4}\) × 10) = 40 + 5 = 45
Question 30.
A box contains cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that number on the drawn card is:
(i) a prime number (ii) a composite number (iii) a number divisible by 3
Or
Question 30.
The King, Queen and Jack of clubs are removed from a pack of 52 cards and then the remaining cards are well shuffled. A card is selected from the remaining cards. Find the probability of getting a card:
(i) of spades (ii) of black king (iii) of clubs (iv) of jacks.
Answer:
(i) Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19 i.e., 8
P(a prime number) = \(\frac{8}{20}\) = \(\frac{2}{5}\)
(ii) Composite numbers from 1 to 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, i.e., 11
P(a composite number) = \(\frac{11}{20}\)
(iii) Number divisible by 3 from 1 to 20 are 3, 6, 9,12,15,18, i.e., 6
P (number divisible by 3)
= \(\frac{6}{20}\) = \(\frac{3}{10}\)
Or
Total number of cards in a deck = 52
∴ Remaining cards = 52 – 3 = 49
(i) P(a spade card) = \(\frac{13}{49}\)
(ii) P(a black king) = \(\frac{2-1}{49}\) = \(\frac{1}{49}\)
(iii) P(a club card) = \(\frac{13-3}{49}\) = \(\frac{10}{49}\)
(iv) P(a jack card) = \(\frac{4-1}{49}\) = \(\frac{3}{49}\)
