Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Basic with Solutions Set 1 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Basic Set 1 with Solutions
Time: 3 Hours
Maximum Marks: 80
General Instructions:
- This Question Paper has 5 Sections A, B, C, D and E.
- Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
- Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
- Section C has 6 Short Answer-II (SA-II) type questions carrying 3 marks each.
- Section D has 4 Long Answer (LA) type questions carrying 5 marks each.
- Section E has 3 sourced based/Case Based/passage based/integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
- All Questions are compulsory. However, an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2 Questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E.
- Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.
Section – A (20 Marks)
Question 1.
If two positive integers a and b are written as a = x3y2 and b = xy3, where x, y are prime number, then HCF (a, b) is: 1
(a) xy
(b) xy2
(c) x3y3
(d) x2y2
Answer:
(b) xy2
Explanation: Given that
a = x3y2
=x × x × x × y × y
and
b =xy3
= x × y × y × y
∴ HCF of a and b = HCF (x3y2, xy3)
= x × y × y
= xy2
Question 2.
The LCM of smallest two digit composite number and smallest composite number is: 1
(a) 12
(b) 4
(c) 20
(d) 44
Answer:
(c) 20
Explanation: As we know, the smallest two digit composite number is 10 and the smalLest composite number is 4.
By prime factorisation, we get
4 = 1 × 2 × 2
10 = 1 × 2 × 5
Now, LCM of (4, 10) = 2 × 2 × 5 = 20
Question 3.
If x = 3 is one of the roots of the quadratic equation x2 – 2kx – 6 = 0, then the value of k is: 1
(a) –\(\frac{1}{2}\)
(b) \(\frac{1}{2}\)
(c) 3
(d) 2
Answer:
(b) \(\frac{1}{2}\)
Explanation: Given one root of x2 – 2kx – 6 = 0
is x = 3
Putting x = 3 in above quadratic eqn., we get
(3)2 – 2k × 3 – 6 = 0
9 – 6k – 6 = 0
-6k = 6 – 9
-6k = – 3
k = \(\frac{-3}{-6}=\frac{1}{2}\)
Hence, the value of k = \(\frac{1}{2}\)
Question 4.
The pair of equations y = 0 and y = -7 has: 1
(a) one solution
(b) two solution
(c) infinitely many solutions
(d) no solution
Answer:
(d) no solution
Explanation: Given, the pair of equations are y = 0 and y = -7.
Hence b1 = 1, c1 = 0
b2 = 1, c2 = 7
So \(\frac{b_1}{b_2}=\frac{1}{1}\) = 1
So \(\frac{c_1}{c_2}=\frac{0}{-7}\) = 0
∴ \(\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Therefore, the pair of linear equations has no solution.
Question 5.
Value(s) of k for which the quadratic equation 2x2 – kx + k = 0 has equal roots is: 1
(a) 0 only
(b) 4
(c) 8 only
(d) 0,8
Answer:
(d) 0, 8
Explanation: Given, equation is
2x2 – kx + k = 0
Here, a = 2, b = – k and c = k
So, D = b2 – 4ac
= (-k)2 – 4 x 2 x k
= k2 – 8k
Since, roots are equal, then
D = 0
k2 – 8k = 0
k (k – 8) = 0
k = 0, 8
Question 6.
The distance of the point (3, 5) from x-axis is k units, then k equals: 1
(a) 3
(b) 4
(c) 5
(d) 8
Answer:
(c) 5
Explanation: The distance of point (3, 5) from the x-axis is equal to the ordinate of the given coordinates. So, the distance from x-axis is 5 units.
Question 7.
If in ΔABC and ΔPQR, \(\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{\mathrm{CA}}{\mathrm{PQ}}\) then: 1
(a) ΔPQR ~ ΔCAB
(b) ΔPQR ~ ΔABC
(c) ΔCBA ~ ΔPQR
(d) ΔBCA ~ ΔPQR
Answer:
(a) ΔPQR ~ ΔCAB
ExpLanation: In ΔABC and ΔPQR
\(\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{\mathrm{CA}}{\mathrm{PQ}}\) (Given)
By SSS criterion, the corresponding vertices are as follows:
A ↔ Q
B ↔ R
C ↔ P
Therefore, we have ΔPQR ~ ΔCAB
Question 8.
Which of the following is NOT a similarity criterion of triangles? 1
(a) AA
(b) SAS
(c) AAA
(d) RHS
Answer:
(d) RHS
Explanation: RHS is not a similarity criterion, it is a congruence criterion.
Question 9.
In figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to: 1
(a) 60°
(b) 70°
(c) 80°
(d) 90°
Answer:
(b) 70°
Explanation: Given that ∠POQ = 110°
∠OQT = ∠OPT = 90°
[∵ A tangent at any point of a cricle is perpendicular to the radius at any point of contact]
Also,
∠TQO + ∠QOP + ∠OPT + ∠PTQ = 360°
[Sum of all angles of a quadriateral is 360°]
∠PTQ = 360° – 90° – 90° – 110°
∠PTQ = 70°
Question 10.
If cos A = g , then tan A is: 1
(a) \(\frac{3}{5}\)
(b) \(\frac{3}{4}\)
(c) \(\frac{4}{3}\)
(d) \(\frac{1}{8}\)
Answer:
(b) \(\frac{3}{4}\)
Explanation: Given, cos A = \(\frac{4}{5}\)
Question 11.
If the height of the tower is equal to the length of its shadow, then the angle of elevation of the sun is …………….. . 1
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(b) 45°
Explanation:
tanθ = tan45°
θ = 45°
Question 12.
(1 – cos2 A) is equal to: 1
(a) sin2 A
(b) tan2 A
(c) 1 – sin2 A
(d) sec2 A
Answer:
(a) sin2 A
Explanation: We know that,
sin2A + cos2A = 1
1 – cos2A = sin2A
Question 13.
The radius of a circle is same as the side of a square. Their perimeters are in the ratio: 1
(a) 1 : 1
(b) 2 : π
(c) π : 2
(d) \(\sqrt{\pi}\) : 2
Answer:
(c) π : 2
Explanation: Given that
Radius of circle = side of square
Let radius of circle be x
Radius of circle = Side of square = x
Ratio of perimeter = \(\frac{\text { Perimeter of circle }}{\text { Perimeter of square }}\)
= \(\frac{2 \pi x}{4 x}\)
= \(\frac{\pi}{2}\)
Question 14.
The area of the circle is 154 cm2. The radius of the circle is: 1
(a) 7 cm
(b) 14 cm
(c) 3.5 cm
(d) 17.5 cm
Answer:
(a) 7 cm
Explanation: Given, area of circle = 154 cm2
∵ Area of circle = πr2
πr2 = 154
r2 = 154 × \(\frac{7}{22}\)
r2 = 7 × 7
r = 7 cm
Hence, the radius of circle is 7 cm.
Question 15.
When a dice is thrown once, the probability of getting an even number less than 4 is: 1
(a) \(\frac{1}{4}\)
(b) 0
(c) \(\frac{1}{2}\)
(d) \(\frac{1}{6}\)
Answer:
(d) \(\frac{1}{6}\)
Explanation: When a die is thrown the possible outcome are 1, 2, 3, 4, 5 and 6.
favourable outcomes (even number Less than 4) is 2.
So, total number of favourable outcomes is 1.
P (even number less than 4) = \(\frac{1}{6}\)
Question 16.
For the following distribution: 1
Class | Frequency |
0 – 5 | 10 |
5 – 10 | 15 |
10 – 15 | 12 |
15 – 20 | 20 |
20 – 25 | 9 |
The Lower limit of modal class is:
(a) 15
(b) 20
(c) 10
(d) 5
Answer:
(a) 15
Explanation:
Class | Frequency | Cumulative Frequency |
0 – 5 | 10 | 10 |
5 – 10 | 15 | 25 |
10 – 15 | 12 | 37 |
15 – 20 | 20 | 57 |
20 – 25 | 9 | 66 |
The highest frequency is 20, which lies in the interval 15 – 20.
So, the lower limit of modal class is 15.
Question 17.
A rectangular sheet of paper 40 cm × 22 cm, is rolled to form a hollow cylinder of height 40 cm. The radius of the cylinder (in cm) is: 1
(a) 3.5
(b) 7
(c) \(\frac{80}{7}\)
(d) 5
Answer:
(a) 3.5
Explanation: Given, rectangular sheet of paper 40 cm × 22 cm is rolled to form a hollow cylinder of height 40 cm. When the sheet is rolled 22 cm becomes the circumference of the base of circle.
Circumference = 22 cm
2πr = 22 cm
2 × \(\frac{22}{7}\) × r =22 cm
r = \(\frac{7}{2}\)
r = 3.5 cm
Question 18.
Consider the following frequency distribution: 1
Class | Frequency |
0 – 6 | 12 |
6 – 12 | 10 |
12 – 18 | 15 |
18 – 24 | 8 |
24 – 30 | 11 |
The median class is:
(a) 6 – 12
(b) 12 – 18
(c) 18 – 24
(d) 24 – 30
Answer:
(b) 12 – 18
Explanation:
Class | Frequency | Cumulative Frequency |
0 – 6 | 12 | 12 |
6 – 12 | 10 | 22 |
12 – 18 | 15 | 37 |
18 – 24 | 8 | 45 |
24 – 30 | 11 | 56 |
Here, N = 56
∴ \(\frac{N}{2}=\frac{56}{2}\) = 28, which Lies in the interval 12 – 18.
Question 19.
Assertion (A): The point (0, 4) lies on y-axis.
Reason (R): The x-coordinate of a point on y-axis is zero. 1
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
Explanation:
We can see that the point (0, 4) lies on y – axis and the x – coordinate of a point on y – axis is zero.
Question 20.
Assertion (A): The HCF of two numbers is 5 and their product is 150. Than their LCM is 40.
Reason (R): For any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b. 1
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer:
(d) Assertion (A) is false but reason (R) is true.
Explanation: We know that
HCF (a, b) × LCM (a, b) = a × b
Given that HCF (a, b) = 5
a × b = 150
5 × LCM (a, b) = 150
LCM (a, b) = \(\frac{150}{5}\)
LCM (a, b) = 30
Section – B (10 Marks)
Question 21.
Find whether the following pair of linear equations is consistent or inconsistent: 2
3x + 2y = 8
6x – 4y = 9
Answer:
3x + 2y = 8
6x – 4y = 9
a1 = 3, b1 = 2, c1 = 8
a2 = 6, b2 = -4, c2 = 9
\(\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2} \quad \frac{b_1}{b_2}=\frac{2}{-4}=\frac{-1}{2}\) and \(\frac{c_1}{c_2}=\frac{8}{9}\)
\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
The given pair of linear equations are consistent
Question 22.
In the given figure, if ABCD is a trapezium in which AB || CD || EF, then prove that \(\frac{A E}{E D}=\frac{B F}{F C}\). 2
OR
In figure, if AD = 6 cm, DB = 9 cm, AE = 8 cm and EC = 12 cm and ∠ADE = 48°. Find ∠ABC.
Answer:
Given:-AB || CD || EF
To prove:- \(\frac{A E}{E D}=\frac{B F}{F C}\)
Construction:- Join BD to intersect EF at G.
OR
Given AD = 6 cm, DB = 9 cm AE = 8 cm, EC = 12 cm, ∠ADE = 48 To find:- ∠ABC=?
Proof:
In ΔABC
\(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{6}{9}=\frac{2}{3}\) ………. (i)
\(\frac{A E}{E C}=\frac{8}{12}=\frac{2}{3}\) ………. (ii)
From (i) and (ii)
\(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
DE || BC [Converse of BPT]
∠ADE = ∠ABC [Corresponding angles]
⇒ ∠ABC = 48 °
Question 23.
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle. 2
Answer:
In Δ OTA, ∠OTA = 90 °
By Pythagoras theorem
OA2 = OT2 + AT2
(5)2 = OT2 + (4)2
25 – 16 = OT2
9 = OT2
OT = 3cm
Radius of circle = 3cm.
Question 24.
Evaluate: sin2 60° + 2tan 45° – cos230°. 2
Answer:
sin2 60° + 2 tan 45° – cos2 30°
= (\(\frac{\sqrt{3}}{2}\))2 + 2(1) – (\(\frac{\sqrt{3}}{2}\))2
= \(\frac{3}{4}\) + 2 – \(\frac{3}{4}\)
= 2
Question 25.
Find the diameter of a circle whose area is equal to the sum of the areas of two circles of radii 40 cm and 9 cm.
OR
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of minor segment. (Use π = 3.14) 2
Answer:
Area of the circle = sum of areas of 2 circLes
πR2 = π(40)2 + π(9)2
πR2 = π × (402 + 92)
R2 = 1600 + 81
R2 = 1681
R = 41 cm.
Diameter of given circle
= 41 × 2
= 82 cm
Radius of circle = 10 cm, θ = 90°
Area of minor segment = \(\frac{\theta}{360^{\circ}}\)πr2 – Area of Δ
= \(\frac{\theta}{360^{\circ}}\) × πr2 – \(\frac{1}{2}\) × b × h
= \(\frac{90^{\circ}}{360^{\circ}}\) × 3.14 × 10 × 10 – \(\frac{1}{2}\) × 10 × 10
= \(\frac{314}{4}\) – 50
= 78.5 – 50 = 28.5 cm2
Area of minor segment = 28.5 cm2
Section – C (18 Marks)
Question 26.
Prove that \(\sqrt{3}\) is an irrational number. 3
Answer:
Let us assume that \(\sqrt{3}\) be a rational number \(\sqrt{3}=\frac{a}{b}\) where a and b are co-prime, squaring both the sides
(\(\sqrt{3}\))2 =(\(\frac{a}{b}\))2
3 = \(\frac{a^2}{b^2}\)
⇒ a2 = 3b2
a2 is divisible by 3, so a is also divisible by 3 …(i)
Let, a = 3c for any integer c.
(3c)2 = 3b2
9c2 =3b2
b2 = 3c2
Since, b2 is divisible by 3 so, b is also divisible by 3 …(ii)
From (i) & (ii) we can say that 3 is a factor of a and b which is contradicting the fact that a and b are co- prime.
Thus, our assumption that \(\sqrt{3}\) is a rational number is wrong.
Hence, \(\sqrt{3}\) is an irrational number.
Question 27.
Find the zeroes of the quadratic polynomial 4s2 – 4s + 1 and verify the relationship between the zeroes and the coefficients. 3
Answer:
P(s) = 4s2 + 4s + 1
4s2 – 2s – 2s + 1 = 0
2s (2s – 1) – 1 (2s – 1) = 0
Question 28.
The coach of a cricket team buys 4 bats and 1 ball for ₹ 2050. Later, she buys 3 bats and 2 balls for 1600. Find the cost of each bat and each ball.
OR
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days, while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. 3
Answer:
Let cost of one bat be ₹ x
Let cost of one ball be ₹ y
ATQ
4x + 1y = 2050 …….. (i)
3x + 2y = 1600 …….. (ii)
from (i) 4x + 1y = 2050
y = 2050 – 4x
Substitute value of y in (ii)
3x+ 2(2050 – 4x) = 1600
3x + 4100 – 8x =1600
-5x= -2500
x= 500
Substitute value of x in (i)
4x + 1y = 2050
4(500) + y = 2050
2000 + y = 2050
y = 50
Hence
Cost of one bat = ₹ 500
Cost of one ball = ₹ 50
OR
Let the fixed charge for first 3 days = ₹ x
And additional charge after 3 days = ₹ y
ATQ
x + 4y = 27 ……… (i)
x + 2y = 21 …….. (ii)
Subtract eqn (ii) from (i)
2y = 6
y = 3
Substitute value of y in (ii)
x + 2(3) = 21
x= 21 – 6
x = 15
Fixed charge = ₹ 15
Additional charge per day = ₹ 3
Question 29.
A circle touches all the four sides of quadrilateral ABCD. Prove that AB + CD = AD + BC. 3
Answer:
Given circle touching sides of ABCD at P, Q, R and S
To prove: AB + CD = AD + BC
Proof
AP = AS …….. (i)
[tangents from an external point to a circle are equal in length]
PB = BQ …… (ii)
DR = DS …… (iii)
CR = CQ …… (iv)
Adding eqn (i), (ii), (iii) & (iv)
AP + BP + DR + CR = AS + DS + BQ + CQ
AB + DC = AD + BC
Question 30.
Prove that:
(cosec θ – cot θ)2 = \(\frac{1-\cos \theta}{1+\cos \theta}\)
OR
Prove that sec A (1 – sin A) (sec A + tan A) = 1. 3
Answer:
LHS = RHS,
Hence Proved.
OR
sec A (1 – sin A) (sec A + tan A) = 1
= 1 = R.H.S
L.H.S = R.H.S
Hence Proved.
Question 31.
A bag contains 6 red, 4 black and some white balls.
(A) Find the number of white balls in the bag if the probability of drawing a white ball is \(\frac{1}{3}\).
(B) How many red balls should be removed from the bag for the probability of drawing a white ball to be \(\frac{1}{2}\)? 3
Answer:
(A) Red balls = 6, Black balls = 4, White balls = x
P(white ball) = [late]\frac{x}{10+x}=\frac{1}{3}[/latex]
⇒ 3x = 10 + x
⇒ x = 5 white balls
(B) Let y red balls be removed, black balls = 4,
white balls = 5
P(white balls) = \(\frac{5}{(6-y)+4+5}=\frac{1}{2}\)
⇒ \(\frac{5}{15-y}=\frac{1}{2}\)
⇒ 10 = 15 – y = 5
So, 5 balls should be removed.
Section – D (20 Marks)
Question 32.
A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.
OR
A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. 5
Answer:
Let the speed of train be × km/hr
distance = 360 km
Speed = \(\frac{\text { distance }}{\text { time }}\)
Time = \(\frac{360}{x}\)
New speed = (x + 5) km/hr
Time = \(\frac{D}{5}\)
x + 5 = \(\frac{360}{\left(\frac{360}{x}-1\right)}\)
(x + 5)(\(\frac{360}{x}\) – 1) = 360
(x + 5)(360 – x) = 360x
-x2 – 5x + 1800 = 0
x2 + 5x – 1800 = 0
x2 + 45x – 40x – 1800 = 0
x(x + 45) – 40(x + 45) = 0
(x + 45)(x – 40) = 0
x + 45 = 0, x – 40 = 0
x = -45, x = 40
Speed cannot be negative
Speed of train = 40km/hr
OR
Let the speed of the stream = xkm/hr
Speed of boat = 18 km/hr
Upstream speed = (18 – x)km/hr
Downstream speed =(18 + x) km/hr
Time taken (upstream) = \(\frac{24}{(18+x)}\)
Time taken (downstream) = \(\frac{24}{(18-x)}\)
ATQ
\(\frac{24}{(18-x)}=\frac{24}{(18+x)}\) + 1
\(\frac{24}{(18-x)}-\frac{24}{(18+x)}\) = 1
24(18 + x) – 24(18 – x)= (18 – x)(18 + x)
24(18 + x – 18 + x) = (18)2 – x2
24(2x) = 324 – x2
48x – 324 + x2 = 0
x2 + 48x – 324 = 0
x2 – 6x + 54x – 324 = 0
x(x – 6) + 54(x – 6) = 0
(x – 6) (x + 54) = 0
x – 6 = 0, x + 54 = 0
x = 6 , x = -54
Speed cannot be negative
Speed of stream=6 km/hr
Question 33.
Prove that 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.
In ΔPQR, S and T are points on PQ and PR respectively. \(\) and ∠PST = ∠PRQ.
Prove that PQR is an isosceles triangle. 5
Answer:
Given, in ΔABC, DE || BC
To prove \(\frac{A D}{D B}=\frac{A E}{E C}\)
Construction: join BE and CD
Draw DM ⊥ AC and EN ⊥ AB
Proof:
Area of ΔADE = \(\frac{1}{2}\) × b × h
= \(\frac{1}{2}\) × AD × EN ………… (i)
Area (ΔDBE) = \(\frac{1}{2}\) × DB × EN ……… (ii)
Divide eqn (i) by (ii)
area ΔADE = \(\frac{1}{2}\) × AE × DM …………. (iv)
area ΔDEC = \(\frac{1}{2}\) × EC × DM ……….. (v)
Divide eqn (iv) by (v)
ΔBDE and ΔDEC are on the same base DE and between same parallel lines BC and DE
∴ area (ΔDBE) = area (DEC)
Hence,
[LHS or (iii) = RHS of (vi)]
\(\frac{A D}{D B}=\frac{A E}{E C}\)
[RHS of (iii) RHS of (vi)]
Since \(\frac{P S}{S Q}=\frac{P T}{T R}\)
∴ ST || QR [by converse of BPT]
∠PST = ∠PQR [Corresponding ongles] [given]
But ∠PST = ∠PRQ
∠PQR= ∠PRQ
PR = PQ
[sides opposite to equal angles ore equal]
Hence, ΔPQR is isosceles.
Question 34.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck at each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
OR
A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like cylinder with two hemisphericaL ends with Length 5 cm and diameter 2.8 cm.
Answer:
Diameter of cylinder and hemisphere = 5 mm
Radius, (r) = \(\frac{5}{2}\)
Total length = 14 mm
Height of cylinder = 14 – 5 = 9 mm
CSA of cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × \(\frac{5}{2}\) × 9
= \(\frac{990}{7}\)mm2
CSA of hemispheres = 2πr2
= 2 × \(\frac{22}{7}\) × (\(\frac{5}{2}\))2
= \(\frac{275}{7}\) mm2
CSA of 2 hemispheres = 2 × \(\frac{275}{7}\)
= \(\frac{550}{7}\) mm2
Total area of capsule = \(\frac{990}{7}+\frac{550}{7}\)
= \(\frac{1540}{7}\)
= 220 mm2
OR
Diameter of cylinder = 2.8 cm
radius of cylinder = \(\frac{2.8}{2}\) = 1.4 cm
Radius of cylinder = Radius of hemisphere
= 1.4 cm
Height of cylinder = 5 – 2.8
= 2.2 cm
Volume of 1 Gulab jamun
= vol. of cylinder + 2 × vol. of hemisphere
= πr2h + 2 × \(\frac{2}{3}\)πr3
= \(\frac{22}{7}\) (1.4)2 × 2.2 + 2 × \(\frac{2}{3}\) × \(\frac{22}{7}\) × (1.4)3
= 13.55 + 11.50
= 25.05 cm3
Volume of 45 Gulabjamun = 45 × 25.05
Syrup in 45 Gulabjamun = 30% × 45 × 25.05
= \(\frac{30}{100}\) × 45 × 25.05 1
= 338.175 cm3
≈ 338 cm3
Question 35.
The following table gives the distribution of the life time of 400 neon lamps:
Life time (in hours) | Number of Lamps |
1500 – 2000 | 14 |
2000 – 2500 | 56 |
2500 – 3000 | 60 |
3000 – 3500 | 86 |
3500 – 4000 | 74 |
4000 – 4500 | 62 |
4500 – 5000 | 48 |
Find the average Life time of a Lamp. 5
Answer:
Mean = a + \(\frac{\Sigma f d}{\Sigma f}\)
a = 3250
Mean= 3250 + \(\frac{64000}{400}\)
= 3250 + 160
= 3410
Average life of lamp is 3410 hr
Section – E (12 Marks)
CASE STUDY 1
Question 36.
India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year.
Based on the above information, answer the following questions:
(A) In which year, the production is 29,200 sets? 1
(B) Find the production in the 8th year.
OR
Find the production in first 3 years. 2
(C) Find the difference of the production in 7th year and 4th year. 1
Answer:
a6 = 16000, a9= 22600
a + 5d = 16000 ………… (i)
a + 8d = 22600 ………….. (ii)
Substitute a = 1600 – 5d from (i)
16000 – 5d + 8d= 22600
3d = 22600 – 16000
3d = 6600
d = \(\frac{6600}{3}\) = 2200
a = 16000 – 5(2200)
a = 16000 – 11000
a = 5000
(A) an = 29200, a = 5000, d = 2200
= an + (n – 1)d
29200 = S000+(n – 1) 2200
29200 – 5000 = 2200n – 2200
24200 + 2200 = 2200n
26400 = 2200n
n = \(\frac{264}{22}\)
n = 12
In 12th year the production was ₹ 29200
(B) n = 8, a = 5000, d = 2200
an = a + (n – 1 )d
= 5000 + (8 – 1) 2200
= 5000 + 7 × 2200
= 5000 + 15400
= 20400
The production during 8th year is 20400
OR
n = 3, a = 5000,d = 2200
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
= \(\frac{3}{2}\)[2(5000) + (3 – 1)2200]
S3 = \(\frac{3}{2}\)(10000 + 2 × 2200)
= 3 × 7200
= 21600
(C) a4 = a + 3d
= 5000 + 3 (2200)
= 5000 + 6600
= 11600
a7 = a + 6d
= 5000 + 6 × 2200
= 5000 + 13200
= 18200
a7 – a4 = 18200 – 11600 = 6600
CASE STUDY 2
Question 37.
Alia and Sha gun are friends living on the same street in Patel Nagar. Shagun’s house is at the intersection of one Street with another Street on which there is a library. They both study in the same school and that is not for from
Sha gun’s house. Suppose the school is situated at the point O, Le., the origin, Alia’s house is at A. Shagun’s house is at B and library is at C.
Based on the above information, answer the folLowing questions:
(A) How far is ALia’s house from Shagun’s house? 1
(B) How for is the Library from Shagun’s house? 1
(C) Show that for Shagun, school Is farther compared to AL1a’s house and library.
OR
Show that Alia’s house, shagun’s house and Library for an isoscetes right triangLe. 2
Answer:
Coordinates of A (2, 3) Alia’s house
Coordinates of B (2, 1) Shagun’s house
Coordinates of C (4, 1) Library
Distance between Alias house and Shagun’s house, AB = 2 units
Distance between Library and Shagun’s house, CB = 2 units
OB is greter than AB and CB,
For Shagun, school [O] is farther than Alia’s house [A] and Library [C]
OR
C (4, 1), A (2, 3)
CA = \(\sqrt{(2-4)^2+(3-1)^2}\)
= \(\sqrt{(-2)^2+2^2}=\sqrt{4+4}=\sqrt{8}\)
= 2√2 units AC2 = 8
Distance between Alia’s house and Shagun’s house, AB = 2 units
Distance between Libaray and Shagun’s house, CB = 2 units
AB2 + BC2 = 22 + 22 = 4 + 4 = 8 = AC2
Therefore, A, B and C from an isosceles right triangle.
CASE STUDY 3
Question 38.
A boy is standing on the top of Light house. He observed that boat P and boat Q are approaching the Light house from opposite directions. He finds that angLe of depression of boat P is 45° and angLe of depression of boat Q is 30°. He aLso knows that height of the Light house is 100 m.
Based on the above information, answer the following questions:
(A) What is the measure of ∠APD?. 1
(B) If ∠YAQ = 30°, then ∠AQD is also 30°, Why? 1
(C) Find length of PD.
OR
Find length of DQ. 2
Answer:
(A) XY || PQ and AP is transversal.
∠APD = ∠PAX [Alternative interior angles]
∠APD = 45°
(B) Since, XY || PQ and AQ is a transversal So, alternate interior angles are equal Hence, ∠YAQ = ∠AQD = 30°
(C) In ΔADP, θ = 45°
tan θ = \(\frac{P}{B}\)
tan 45° = \(\frac{100}{P D}\)
PD = 100 m
Boat P is 100 m from the light house
OR
In ΔADQ, Φ = 30°
tan Φ = \(\frac{P}{B}\)
tan 30° = \(\frac{100}{\mathrm{DQ}}\)
\(\frac{1}{\sqrt{3}}=\frac{100}{\mathrm{DQ}}\)
DQ = 100√3 m