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 hrs

Max. 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 Case Based integrated units of assessment questions (4 marks each).
- 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 question of Section E.
- Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.

Section – A

Section A consists of 20 Questions of 1 marks each

Question 1.

If two positive integers a and b are written as a = x^{3}y^{2} and b = xy^{3} where x and yare prime numbers, then HCF (a, b) is

(a) xy

(b) xy^{2}

(c) x^{3}y^{3}

(d) x^{2}y^{2}

Answer:

(b) xy^{2}

Here, a = x^{3}y^{2} = xy^{2} (x^{2})

and b = xy^{3}

= xy^{2} (y)

∴ HCF (a, b) = xy^{2}

Question 2.

The LCM of smallest two-digit composite number and smallest composite number is

(a) 12

(b) 4

(c) 20

(d) 44

Answer:

(c) 20

Smallest composite number = 4 = 2 × 2 = 22

Smallest two-digit composite number = 10 = 2 × 5

∴ LCM (4, 10) = 2 × 2 × 5 = 20

Question 3.

If x = 3 is one of the roots of the quadratic equation x^{2} – 2kx – 6 = 0, then the value of k is

(a) – \(\frac{1}{2}\)

(b) \(\frac{1}{2}\)

(c) 3

(d) 2

Answer:

(b) \(\frac{1}{2}\)

Given quadratic equation is x^{2} – 2kx – 6 = 0

∵ x = 3 is a root of the equation.

(3)^{2} – 2k (3) – 6 = 0

⇒ 9 – 6k – 6 = 0

⇒ 3 = 6k

⇒ k = \(\frac{1}{2}\)

Question 4.

The pair of equations y = 0 and y = – 7 has

(a) one solution

(b) two solutions

(c) infinitely many solutions

(d) no solution

Answer:

(d) no solution

The pair of equations are y = 0 and y = – 7

The graph of the equations are parallel to each other, therefore, the given pair has no solution.

Question 5.

Value(s) of k for which the quadratic equation 2x^{2} – kx + k = 0 has equal roots is

(a) only 0

(b) 4

(c) only 8

(d) 0, 8

Answer:

(d) 0, 8

Since, the given quadratic equation 2x^{2} – kx + k = 0 has equal roots.

Discriminant, (-k)^{2} – 4(2)(k) = 0

⇒ k^{2}-8k = 0

⇒ k(k – 8) = 0

⇒ k = 0 or k = 8

Question 6.

The distance of the point (3, 5) from X-axis is k units, then k equals

(a) 3

(b) – 3

(c) 5

(d) – 5

Answer:

(c) 5

From the graph, we can find, k = 5 units.

Question 7.

If in ΔABC and ΔPQR, we have \(\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}\), then

(a) ΔPQR ~ ΔCAB

(b) ΔPQR ~ ΔABC

(c) ΔCBA ~ ΔPQR

(d) ΔBCA ~ ΔPQR

Answer:

(a) ΔPQR ~ ΔCAB

Given, in ΔABC and ΔPQR,

\(\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}\)

Corresponding sides of ACAB and A PQR are in the same ratio.

∴ Δ PQR ~ Δ CAB

Question 8.

Which of the following is NOT a similarity criterion?

(a) AA

(b) SAS

(c) AAA

(d) RHS

Answer:

(d) RHS

RHS is not a similarity criterion.

Question 9.

In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ZPOQ = 110°, then ZPTQ is equal to

(a) 60°

(b) 70°

(c) 80°

(d) 90°

Answer:

(b) 70°

PT ⊥ OP and QT ⊥ OQ

[∵ tangent to a circle is perpendicular to the radius through the point of contact]

⇒ ∠OPT = 90°

and ∠OQT = 90°

Now, in quadrilateral OQTP,

∠O + ∠Q + ∠T + ∠P = 360°

[angle sum property of quadrilateral]

⇒ 110° + 90°+ ∠T + 90° = 360°

⇒ ∠T = 70°

or ∠PTQ = 70°

Question 10.

If cos A = \(\frac{4}{5}\), then the value of tan A is

(a) \(\frac{3}{5}\)

(b) \(\frac{3}{4}\)

(c) \(\frac{4}{3}\)

(d) \(\frac{1}{8}\)

Answer:

(b) \(\frac{3}{4}\)

We know that in a right ∆ABC (say)

cos A = \(\frac{\text { Base }}{\text { Hypotenuse }}\)

= \(\frac{A B}{A C}=\frac{4}{5}\)

⇒ AB = 4x

and AC = 5x

In right angled ∆ABC,

AC^{2} = AB^{2} + BC^{2}

(5x)^{2} = (4x)^{2} + BC^{2}

⇒ BC^{2} = 9x^{2}

⇒ BC = 3x

We know,

tan A = \(\frac{\text { Perpendicular }}{\text { Base }}\)

= \(\frac{B C}{A B}\)

= \(\frac{3 x}{4 x}=\frac{3}{4}\)

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 ……………

(a) 30°

(b) 45°

(c) 60°

(d) 90°

Answer:

(b) 45°

Let AB be the tower and BC be its shadow.

Given, AB = BC

In right angled ∆ABC,

tan C = \(\frac{\text { Perpendicuar }}{\text { Base }}\)

= \(\frac{A B}{B C}\) = 1

⇒ tan C = tan 45° [∵ tan 45° = 1]

⇒ C = 45°

Question 12.

1 – cos^{2} A is equal to

(a) sin^{2} A

(b) tan^{2} A

(c) 1 – sin^{2} A

(d) sec^{2} A

Answer:

(a) sin^{2} A

We know trigonometric identity,

sin^{2} A + cos^{2} A = 1

⇒ sin^{2} A = 1 – cos^{2} A

Question 13.

The radius of a circle is same as the side of a square. Their perimeters are in the ratio

(a) 1 : 1

(b) 2 : π

(c) π : 2

(d) √π : 2

Answer:

(c) π : 2

Let the radius of a circle and side of a square be x unit.

Then, perimeter of circle = 2πx

and perimeter of square = 4x

∴ \(\frac{\text { Perimeter of circle }}{\text { Perimeter of square }}=\frac{2 \pi x}{4 x}\)

= \(\frac{\pi}{2}\)

Question 14.

The area of the circle is 154 cm^{2}. The radius of the circle is

(a) 7 cm

(b) 14 cm

(c) 3.5 cm

(d) 17.5 cm

Answer:

(a) 7 cm

Let the radius of the circle be r cm.

Then, area of circle = πr² = 154

⇒ \(\frac{22}{7}\) × r^{2} = 154

⇒ r^{2} = 49

r = 7 cm

Question 15.

When a dice is thrown once, the probability of getting an even number less than 4 is

(a) 1/4

(b) 0

(c) 1/2

(d) 1/6

Answer:

(d) 1/6

Let E be the event of getting an even number less than 4.

Outcomes favourable to E : {2}

Possible outcomes: {1, 2, 3, 4, 5, 6}

∵ P(E) = Number of outcomes favourable to E / Total number of possible outcomes

= \(\frac{1}{6}\)

Question 16.

For the following distribution

The lower limit of modal class is

(a) 15

(b) 25

(c) 30

(d) 35

Answer:

(a) 15

Modal class is the class with highest frequency.

Here, modal class = 15 – 20

∴ Lower limit of modal class = 15

Question 17.

A rectangular sheet of paper 40 cm × 22 cm, is rolled to form a hollow cylinder of height 40cm. The radius of the cylinder (in cm) is

(a) 3.5

(b) 7

(c) \(\frac{80}{7}\)

(d) 5

Answer:

(a) 3.5

Let r be the radius of cylinder.

After the sheet is folded, breadth of the rectangular sheet will become the perimeter of the base of the cylinder.

∴ 22 = 2πr

r = \(\frac{22 \times 7}{22 \times 2}\)

r = \(\frac{7}{2}\) = 3.5 cm

Question 18.

Consider the following frequency distribution

The median class is

(a) 6 – 12

(b) 12 – 18

(c) 18 – 24

(d) 24 – 30

Answer:

(b) 12 – 18

Class | Frequency (f) | Cumulative Frequency (cf) |

0 – 6 | 12 | 12 |

6 – 12 | 10 | 22 |

12 – 18 | 15 | 37 |

18 – 24 | 8 | 45 |

24 – 30 | 11 | 56 |

Total | n = Xf = 56 |

\(\frac{n}{2}=\frac{56}{2}\) = 28

Median class is the class whose cumulative frequency is greater than (nearest to) \(\frac{n}{2}\) i.e. 28.

Median class is 12 – 18.

Question 19.

Assertion (A) The point (0, 5) lies on F-axis.

Reason (R) The x-coordinate of the point on F-axis is zero.

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

(b) Both Assertion and Reason are tme but Reason is not the correct explanation of Assertion.

(c) Assertion is true but Reason is false

(d) Assertion is false but Reason is true

Answer:

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Given point is (0, 5), abscissa (x – coordinate) is 0 and ordinate (y – coordinate) is 5.

Question 20.

Assertion (A) : The HCF of two numbers is 5 and their product is 150. Then, their LCM is 40.

Reason (R) : For any two positive integers a and b, HCF (a,b) xLCM (a,b) = axb.

(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

(c) Assertion is true but Reason is false

(d) Assertion is false but Reason is true

Answer:

(d) Assertion is false but Reason is true

We know that HCF and LCM of two numbers (a, b) is given by

HCF (a, b) × LCM (a,b) = a × b

⇒ 5 × LCM(a, b) = 150

⇒ LCM (a, b) = \(\frac{150}{5}\) = 30

Section – B

Section B consists of 5 Questions of 2 marks each.

Question 21.

Find whether the following pair of linear equations is consistent or inconsistent 3x + 2y = 8, 6x – 4y = 9.

Answer:

Given, pair of linear equation is 3x + 2y = 8

⇒ 3x + 2y – 8 = 0

and 6x – 4y = 9

⇒ 6x – 4y – 9 = 0

Here, \(\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}\)

and \(\frac{b_1}{b_2}=\frac{2}{-4}=\frac{-1}{2}\)

[where a_{1}, a_{2} are coefficients of x and b_{1}, b_{2} are coefficients of y]

∵ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

The pair of linear equations is consistent.

Question 22.

In the given figure, if ABCD is a trapezium in which AB || CD || EF, then prove that \(\frac{A E}{E D}=\frac{B F}{F C}\).

Answer:

Given, ABCD is a trapezium.

Join DB.

Let DB intersects EF and G.

In ∆DAS,

using basic proportionality theorem,

\(\frac{D E}{E A}=\frac{D G}{G B}\)

⇒ \(\frac{E A}{D E}=\frac{G B}{D G}\) …………..(i)

In ABCD,

using basic proportionality theorem.

\(\frac{B F}{F C}=\frac{B G}{G D}\) ………….(ii)

Using Eqs. (i) and (ii), we get

\(\frac{E A}{D E}=\frac{B F}{F C}\)

\(\frac{A E}{E D}=\frac{B F}{F C}\)

Or

In the given figure, if AD = 6 cm, DB = 9 cm, AE = 8 cm and EC = 12 cm and ∠ADE = 48°. Find ∠ABC.

Answer:

In ∆ADE and ∆ABC,

\(\frac{A D}{D B}=\frac{6}{9}=\frac{2}{3}\)

and \(\frac{A E}{E C}=\frac{8}{12}=\frac{2}{3}\)

⇒ \(\frac{A D}{D B}=\frac{A E}{E C}=\frac{2}{3}\)

∴ DE || BC [by converse of BPT]

Since DE || BC and AB is the transversal.

∴ ∠ADE = ∠ABC [corresponding angles]

⇒ ∠ABC = 48°

Question 23.

The length of a tangent from a point A at distance 5cm from the centre of the circle is 4cm. Find the radius of the circle.

Answer:

Let O be the center of circle

and let B be the point of contact of tangent and circle

OB ⊥ AB

[∵ tangent to a circle is perpendicular to the radius through the point of contact]

In right angled ∆OBA,

OA^{2} = OB^{2} + AB^{2}

5^{2} = OB^{2} + 4^{2}

OB^{2} = 25 – 16

OB^{2} = 9

⇒ OB = 3 cm

Question 24.

Evaluate sin^{2} 60°+ 2 tan 45°- cos^{2} 30°.

Answer:

sin^{2} 60°+ 2 tan 45°- cos^{2} 30°.

= \(\left(\frac{\sqrt{3}}{2}\right)^2\) + 2 (1) + \(\left(\frac{1}{2}\right)^2\)

= \(\frac{3}{4}+2-\frac{1}{4}\)

= \(\frac{2}{4}\) + 2

= \(\frac{1}{2}\) + 2

= \(\frac{5}{2}\)

Question 25.

What is the diameter of a circle whose area is equal to the sum of the areas of two circles of radius 40 cm and 9 cm?

Answer:

Let radius of two circles be r_{1} and r_{2}

and let r be the radius of big circle.

Given, r_{1} = 40 cm

r_{2} = 9 cm

According to the question,

πr² = πr_{1}² + πr_{2}²

r^{2} = r_{1}^{2} + r_{2}^{2}

= 40^{2} + 9^{2}

= 1600 + 81 = 1681

r = \(\sqrt{1681}\) = 41 cm

:. Diameter of the required circle = 2r

= 2 × 41 = 82 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)

Answer:

Given, radius of a circle is 10 cm.

∴ Area of minor segment ACB = Area of sector OACB – Area of tOAB

Area of sector OACB = \(\frac{90}{360}\) π × (radius)^{2}

= \(\frac{1}{4}\) × 314 × (10)^{2}

= 78.5 cm^{2}

∴ Area of ∆OAB, = \(\frac{1}{2}\) × Base × Height

= \(\frac{1}{2}\) × QA × QB

= \(\frac{10 \times 10}{2}\) = 50 cm

∴ Required area = 78.5 – 50 = 28.5 cm^{2}

Section – C

Section C consists of 6 Questions of 3 marks each

Question 26.

Prove that √3 is an irrational number.

Answer:

Let us assume that √3 is a rational number.

∴ √3 = \(\frac{p}{q}\)

[where p and q are co prime and q ≠ 0]

On squaring both sides, we get

3 = \(\frac{p^2}{q^2}\)

⇒ p^{2} = 3q^{2}

∵ p^{2} is a multiple of 3.

⇒ 3 divides p^{2}.

∴ 3 divides p.

∴ p = 3a for some integer a.

Substituting for p, we get

(3a)^{2} = 3q^{2}

⇒ q^{2} = 3a^{2}

⇒ q^{2} is a multiple of 3.

⇒ 3 divides q^{2}.

∴ 3 divides q.

⇒ p and q have atleast 3 as a common factor.

But this contradicts the statement that p and q are co-primes.

This contradiction has arisen because of our incorrect assumption that T3 is rational.

Hence, √3 is an irrational number.

Question 27.

Find the zeroes of the quadratic polynomial 4s^{2} – 4s + 1 and verify the relationship between the zeroes and the coefficients.

Answer:

Let f(s) = 4s^{2} – 4s + 1

= 4s^{2} – 2s – 2s + 1

= 2s (2s – 1) – 1 (2s – 1)

For zeroes of f(s), put f(s) = 0

⇒ (2s – 1) (2s – 1) = 0

⇒ s = \(\frac{1}{2}\), \(\frac{1}{2}\)

Now, sum of zeroes = \(\frac{1}{2}+\frac{1}{2}\) = 1

and product of zeroes = \(\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)

We know,

sum of zeroes = \(\frac{- \text { Coefficient of } s}{\text { Coefficient of } s^2}\)

= – \(\frac{(-4)}{4}\) = 1

and product of zeroes = \(\frac{\text { Constant term }}{\text { Coefficient of } s^2}\)

= \(\frac{1}{4}\)

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.

Answer:

Let the cost of each bat be ₹ x

and the cost of each ball be ₹ y,

According to the question,

4x + y = 2050 …………..(i)

3x + 2y = 1600 …………(ii)

On multiplying Eq. (i) by 2, we get

8x + 2y = 4100 …………(iii)

On subtracting Eq (ii) from Eq. (iii), we get

5x = 2500

⇒ x = 500

Substituting x = 500 in Eq. (ii), we get

3(500) + 2y = 1600

⇒ 2y = 1600 – 1l500

⇒ y = 50

Cost of each bat and ball is ₹ 500 and ₹ 50, respectively.

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.

Answer:

Let the fixed charge be ₹ x

and the charge for each extra day be ₹ y.

Total amount paid by Saritha for 7 days = ₹ 27

⇒ x + 4y = 27 …………..(i)

Total amount paid by Susy for 5 days = ₹ 21

⇒ x + 2y = 21 ……………(ii)

On subtracting Eq. (ii) from Eq. (i), we get

2y = 6

⇒ y = 3

From Eq. (ii),

x + 2 (3) = 21

⇒ x = 15

Hence, the fixed charge is 15 and charge for each extra day is ₹ 3.

Question 29.

A circle touches all the four sides of quadrilateral ABCD. Prove that AB + CD = AD + BC.

Answer:

Let the circle intersects AB, BC, CD and DA at point L,M, N, and O respectively.

Now, AL = LO …………(i)

[∵ tangents from an external point to circle are equal in Length]

Similarly, DN = DO ………….(ii)

CN = CM ………..(iii)

BL = BM …………(iv)

On adding Eqs. (i), (ii), (iii) and (iv), we get

AL + DN + CN + BL = AO + DO + CM + BM

⇒ (AL + BL) + (DN + CN) = (AO + DO) + (CM + BM)

⇒ AB + CD = AD + BC

[∵ AL + BL = AB,

DN + CN = CD,

AO + DO = AD

CM + BM = BC]

Question 30.

Prove that (cosec θ – cot θ)^{2} = \(\frac{1-\cos \theta}{1+\cos \theta}\)

Answer:

L.H.S. = (cosec θ – cot θ)^{2}

Hence proved.

Or

Prove that sec A (1 – sin A) (sec A + tan A) = 1

Answer:

L.H.S = sec A (1 – sin A) (sec A + tan A)

Question 31.

A bag contains 6 red, 4 black and some white balls.

(i) Find the number of white balls in the bag, if the probability of drawing a white ball is 1/3.

Answer:

Let number of white balls be n.

Total number of balls = 6 + 4+ n = 10 + n

Let E be the event of drawing a white ball.

Given, P(E) = \(\frac{1}{3}\)

⇒ \(\frac{\text { Number of outcomes favourable to } E}{\text { Total number of outcomes }}=\frac{1}{3}\)

⇒ \(\frac{n}{10+n}=\frac{1}{3}\)

⇒ 3n = 10 + n

⇒ 2n = 10

⇒ n = 5

(ii) How many red balls should be removed from the bag for the probability of drawing a white ball to be \(\frac{1}{2}\)?

Answer:

Now, total number of balls = 10 + n

= 10 + 5 = 15

Let x red balls are removed from the bag.

∴ Total number of balls in the the bag = 15 – x

Now, P(E) = \(\frac{1}{2}\)

⇒ \(\frac{5}{15-x}=\frac{1}{2}\)

⇒ 10 = 15 – x

⇒ x = 5

Section – D

Section D consists of 4 Questions of 5 marks each

Question 32.

A train travels 360 km at a uniform speed. If the speed had been 5 km / h more, it would have taken 1 h less for the same journey, then find the speed of the train.

Answer:

Let the speed of the train be x km/h.

According to the question,

\(\frac{360}{x}-\frac{360}{x+5}\) = 1

⇒ 360 (x + 5) – 360x = x (x + 5)

⇒ 360x + 1800 – 360x = x^{2} + 5x

⇒ x^{2} + x – 1800 = 0

⇒ x^{2} + 45x – 40x – 1800 = 0

⇒ x (x + 45) – 40 (x + 45) = 0

⇒ (x + 45) (x – 40) = 0

x = – 45, x =40

x = 40 [speed can’t be negative]

∴ Speed of the train is 40 km/h.

Or

A motor boat whose speed is 18 km/h in still water takes 1 h more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

Answer:

Let the speed of the stream be x krrIhr.

Speed of the boat in downstream = (x + 18) km/h

Speed of the boat in upstream = (18 – x) km/h

According to the question,

\(\frac{24}{18-x}-\frac{24}{x+18}\) = 1

⇒ 24 (x + 18) – 24 (18 – x) = (18 – x) (18 + x)

⇒ 24x + 24x = 324 – x^{2}

⇒ x^{2} + 48x – 324 = 0

⇒ x^{2} + 54x – 6x – 324 = 0

⇒ x (x + 54) – 6 (x + 54) = 0

⇒ (x + 54) (x – 6) = 0

⇒ x = – 54 or x = 6

∴ x = 6 [∵ speed can’t be negative]

Question 33.

(i) 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.

Answer:

Let ABC be a triangle in which a line parallel to side BC intersects the other two skies AB and AC at D and E, respectively

Join BD, CE and draw EF ⊥ AC and DG ⊥ AB

Area of ∆AED = \(\frac{1}{2}\) × Base × Height

= \(\frac{1}{2}\) × AE × GD

ar (∆ AED) = \(\frac{1}{2}\) × AE × GD

Area of ∆BED = ar (∆ BED)

= \(\frac{1}{2}\) × BE × DG

Also, ar (∆ AED) = \(\frac{1}{2}\) × AD × EF

ar (∆ CED) = \(\frac{1}{2}\) × CD × EF

∵ \(\frac{\ {ar}(\triangle A E D)}{\ {ar}(\triangle B E D)}=\frac{\frac{1}{2} \times A E \times G D}{\frac{1}{2} \times B E \times G D}=\frac{A E}{B E}\) ………..(i)

and \(\frac{\ {ar}(\triangle A E D)}{\ {ar}(\triangle C E D)}=\frac{\frac{1}{2} \times A D \times E F}{\frac{1}{2} \times C D \times E F}=\frac{A D}{C D}\) ………….(ii)

since, ∆ BED and ∆ CED are on the same base and between the same parallels.

∴ ar (∆ BED) = ar (∆ CED) ………………(iii)

From Eqs. (i), (ii) and (iii), we have

\(\frac{A E}{E B}=\frac{A D}{D C}\)

Hence proved.

(ii) In ∆ PQR, S and T are points on PQ and PR, respectively. If \(\frac{P S}{S Q}=\frac{P T}{T R}\) and ∠PST = ∠PRQ, then prove that PQR is an isosceles triangle.

Answer:

Given, \(\frac{P S}{S Q}=\frac{P T}{T R}\)

By converse of BPT,

ST || QR

⇒ PO is the transversal.

∴ ∠PST = ∠POR [corresponding angles]

Given, ∠PST = ∠PRQ

∠PQR = ∠PRQ

⇒ PQ = PR

[sides opposite to equal angles are equal]

∴ ∆ PQR is an isosceles triangle.

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.

Answer:

Radius of the cylinder = Radius of hemisphere ends (r)

= \(\frac{5}{2}\) = 2.5 cm

Height of the cylindrical part (h) = (14 – 2 × 2.5) mm

= 14 – 5 = 9 mm

∴ Surface area of the capsule = CSA of cylindrical part + CSA of hemispherical ends

= 2πrh + 2 × (2πr^{2})

= 2πr (h + 2r)

= 2 × \(\frac{22}{7}\) × 2.5 × (9 + 2 × 2.5)

= 2 × \(\frac{22}{7}\) × 2.5 × 14

= 220 mm^{2}

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:

Radius of hemispherical ends and cylindrical part (r) = \(\frac{2.8}{2}\) = 14 cm

Height of hemispherical ends (h) = 5 – 2 × 1.4

= 5 – 2.8 = 2.2 cm

∴ Volume of one gulab jamun = 2 × Volume of hemispherical ends + Volume of cylinder

= 2 × \(\frac{2}{3}\) πr^{3} + πr^{2}h

= πr^{2} (\(\frac{4 r}{3}\) + h)

= \(\frac{22}{7}\) × 1.4 × 1.4 (\(\frac{4}{3}\) × 1.4 + 2.2)

= \(\frac{22}{7}\) × 1.4 × 1.4 × 4.602

= 25.02 cm^{3}

Volume of 45 such gulab jamuns = 45 × 25.02

= 1125.9 cm^{3}

and volume of sugar syrup in 45 gulab jamuns = \(\frac{30}{100}\) × 1125.9 = 337.77 cm^{3}

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.

Answer:

Assumed mean, a = 3250

∴ \(\bar{x}=a+\frac{\Sigma f_i d_i}{\Sigma f_i}\)

= 3250 + \(\frac{64000}{400}\)

= 3250 + 160 = 3410

Section – E

Section E consists of 3 Questions of 2 marks each

Question 36.

**Case Study 1**

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 yr and 22600 in 9th yr.

(i) In which year, the production is ₹ 29,200?

Answer:

Let the production increases by d every year

and the production in the 1st yr be a.

Given, production in 6th yr = 16000

⇒ a_{6} = 16000

a + (6 – 1) d = 16000

a+ 5d = 16000 ………….(i)

Also, production in 9th yr = 22600

⇒ a_{9} = 22600

a + (9 – 1) d = 22600

a + 8d = 22600 ……………..(ii)

On subtracting Eq. (i) from Eq. (ii), we get

3d = 6600

⇒ d = 2200

a = 5000 [using Eq. (i)]

Let on nth yr. production = 29200

⇒ a_{n} = 29200

∴ a + (n – 1) d = 29200

⇒ 5000 + (n – 1) d = 29200

⇒ (n – 1) 2200 = 24200

⇒ n – 1 = 11

⇒ n = 12

In 12th year, the production is 29200.

(ii) Find the production during 8th yr.

Answer:

Production during 8th yr,

a_{8} = a + 7d

= 5000 + 7 (2200)

= 5000 + 15400

= 20400

Or

Find the production during first 3 yrs.

Answer:

Production during first 3 yr = Production in 1st yr + Production in 2nd yr + Production in 3rd yr

= a_{1} + a_{2} + a_{3}

S_{3} = \(\frac{3}{2}\) [2a + (3 – 1) d]

[∵ S_{n} = \(\frac{n}{2}\) [2a + (n – 1) d]

= \(\frac{3}{2}\) [2 (5000) + 2 (2200)]

= 3 [5000 + 2200]

= 7200 × 3

= 21600

(iii) Find the difference of the production during 7th yr and 4th yr.

Answer:

a_{7} – a_{4} = a- + 6d – (a + 3d)

= 6d – 3d

= 3d

= 3 (2200)

= 6600

Question 37.

**Case Study 2**

Alia and Shagun 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 far from Shagun’s house. Suppose the school is situated at the point O i.e. 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.

(i) How far is Alia’s house from Shagun’s house?

Answer:

Distance between Alias house and Shaguns house

AB = \(\sqrt{(2-2)^2+(1-3)^2}\)

= \(\sqrt{(-2)^2}\) = 2 units

[∵ distance between two points P(x_{1}, y_{1}) and Q (x_{2}, y_{2})

PQ = \(\left.\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\right]\)

(ii) How far is the library from Shagun’s house?

Answer:

Distance between library and shagun’s house

BC = \(\sqrt{(2-4)^2+(1-1)^2}\)

= \(\sqrt{(-2)^2+0}\)

= √4

= 2 units

(iii) Show that for Shagun’s house, school is farther as compared to Alia’s house and library.

Answer:

Distance between Alia’s house and library,

AC = \(\sqrt{(4-2)^2+(1-3)^2}=\sqrt{2^2+(-2)^2}\)

= 2√2 units

Distance between Shagun’s school and Shaguns house = OB

= \(\sqrt{(2-0)^2+(1-0)^2}\)

= \(\sqrt{4+1}=\sqrt{5}\) units

∵ OB > AC

For Shaguns house, school is farther compared to Alia’s house and library.

Or

Show that Alia’s house, Shagun’s house and library form an isosceles right triangle.

Answer:

Distance between Shaguns house and library = \(\sqrt{(4-2)^2+(1-1)^2}\)

= 2 units

Now, AC^{2} = 8 units

and AB^{2} + BC^{2} = 2^{2} + 2^{2} = 8 units

∴ AC^{2} = AB^{2} + BC^{2}

⇒ ABC is a right angled triangle, right angled at B.

Also, AB = BC = 2 units

∴ ABC is an isosceles right angled triangle.

Question 38.

**Case Study 3**

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.

(i) What is the measure of ∠APD?

Answer:

XY || PQ

∴ ∠XAP = ∠APQ = 45° = ∠APD [alternate angles]

(ii) If ∠YAQ = 30°, then ∠AQD is also 30°. Why?

Answer:

∠YAQ = ∠AQD = 30° [alternate angles]

(iii) How far is boat P from the light house?

Answer:

In right angled ∆ADP

tan 45° = \(\frac{AD}{PD}\)

⇒ 1 = \(\frac{100}{PD}\)

PD = 100 m

Hence, the boat P is at a distance of 100 m from the light house.

Or

How far is boat Q from the light house?

Answer:

In right angled ∆ADQ,

tan 30° = \(\frac{AD}{DQ}\)

⇒ \(\frac{1}{\sqrt{3}}=\frac{A D}{D Q}\)

⇒ DQ = 100 √3 m

Hence, the boat Q is at a distance of 100√3 from the light house.