CBSE Sample Papers For Class 10 Maths Mock Paper 1

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CBSE Class10 Maths Sample Question Paper -1 2021-22

Mathematics is one of the most challenging and important subjects. This retains the highest weightage irrespective of the stream one will choose in higher education. It also helps in developing some crucial skills like logical reasoning and calculation skills. This is the reason Mathematics is counted as one of the main subjects in Class 10 in CBSE and almost every board.

Though some find solving Mathematics problems an uphill task, this can become one of the easiest scoring subjects, if given proper attention and dedication. There is a very bright scope to get full marks in this subject provided calculations and all the steps are mentioned in an accurate method. Thus it can also be called a mark-boosting subject. Perform well in this exam and one’s overall grade in board exams can be improved. For revision, students can use Extramarks cbse Sample Papers For Class 10 Maths Mock Paper 1.

Maths Mock Paper-1 for CBSE Class 10 Board Exams

Class 10 plays a crucial role in a student’s life. This is the class where students appear for the board exams for the very first time. The mark sheet of this Class has its significance. Thus students of Class 10 must chalk out a prime and proper schedule to study for board exams.

The first thing they should do is to cover up NCERT books. These books are produced and published by the National Council Of Education Research and Training to impart vital and comprehensive knowledge to students. A major portion of the question paper is set from these books. Thus having deep knowledge and understanding of these books is important.

Apart from this, the role of CBSE Sample Papers cannot be neglected. Practising sample papers is a crucial step in preparation. For this, students may refer to CBSE Sample Papers For Class 10 Maths Mock Paper 1 available on Extramarks. Solving a few sample papers helps in polishing and brushing up the concepts in a substantial manner.

Download CBSE Sample Papers Class 10 Maths

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Another advantage of opting for Extramarks is that once done with the exam, students can evaluate their answer sheets with the help of CBSE Sample Papers For Class 10 Maths Mock Paper 1 with solutions provided by Extramarks. With their mistakes noted, they can directly go back to the topics they are still weak at. Hence gradually and steadily their performance gets improved and they face the exams with stability and confidence.

But where to find the resources?

Finding resources is quite easy in today’s online world. But the authenticity of these resources should be confirmed. The resources provided by Extramarks are trustworthy, scrutinised and authentic. Thus students of Class 10 can refer to Extramarks to prepare for Mathematics. Here all the required sources are provided. Students of Class 10 can find the latest CBSE syllabus for all the subjects along with Mathematics. The different topics are explained and elaborated under CBSE revision notes. Not only this, Extramarks provides the questions which are most expected under the link CBSE important questions and CBSE extra questions. All this material is prepared by the seasoned and highly qualified subject experts hired by Extramarks.

Why practice with the Extramarks mock tests-01 for Class 10 CBSE?

It is recommended that students of Extramarks should practice CBSE Sample Papers For Class 10 Maths Mock Paper 1 to grab the concepts of Mathematics in a better way. However, this recommendation is not void and is based upon a couple of logic. Some of the reasons for practising with CBSE Sample Papers For Class 10 Maths Mock Paper 1 are discussed below:

The more practice one does, the more confident one feels. These CBSE Sample Papers For Class 10 Maths Mock Paper 1 are created in a pattern which is close to the pattern of the final examination. These CBSE Sample Papers For Class 10 Maths Mock Paper 1 help students understand difficult sort of questions which can be expected in exams and multiple ways in which questions can be approached.
These CBSE Sample Papers For Class 10 Maths Mock Paper 1 have their answers uploaded on the Extramarks website too. The answers are given in a well-represented, systematic and step-by-step manner. Thus it becomes easy for students to comprehend any concept and to track the mistakes committed by them while solving CBSE Sample Papers For Class 10 Maths Mock Paper 1. This self-evaluation helps in improving the performance of students in the long run.
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Q1. x and y are two digit numbers. If their HCF is 7 and LCM is 105, then x and y are

Opt:

21 and 35

28 and 35

25 and 35

25 and 45

Ans:

21 and 35

Q2. If the difference between the roots of the equation x² + px + 8 is 2, then the value of p is

Opt:

±2

±4

±6

±8

Ans: ±6

Q3. If one root of x² – x – m = 0 is square of the other, then m =

Opt:

2 \pm \sqrt{3}

3 \pm \sqrt{2}

2 \pm \sqrt{5}

5 \pm \sqrt{2}

Ans:

2 \pm \sqrt{5}

Q4. If 3^{(x + y)} = 81 and 81^{(x – y)} = 3, then the values of x and y are

Opt:

17/8, 9/8.

17/8, 15/8.

17/8, 11/8.

5/8, 11/8.

Ans:

17/8, 15/8.

Q5.

\begin{matrix} If α and β are the zeroes of a quadratic polynomial \\ such that α + β = 15 and α - β = 5, then the quadratic \\ polynomial is \end{matrix}

Opt:

3x² – 25x – 11

x² – 15x + 50

3x² + 25x – 11

x² + 15x + 50

Ans: x² – 15x + 50

Q6. The angle of a triangle are in AP and the least being half the greatest. Then, what are the measures of the angles?

Opt:

30°, 60°, 90°

40°, 50°, 90°

20°, 60°, 100°

40°, 60°, 80°

Ans: 40°, 60°, 80°

Q7. Three fair dice are thrown together then the probability that the sum of the numbers is more than or equal to 17 will be

Opt:

2/27.

1/18.

1/27.

1/54.

Ans: 1/54.

Q8. If the distance between points A(3, m) and B(4, 1) is 10, then the value(s) of m are:

Opt:

–2, 4.

–2, –4.

– 4, 2.

4, 2.

Ans: –2, 4.

Q9. Three cubes of a metal are of edges 3 cm, 4 cm and 5 cm. These are melted together and from the melted material another cube is formed. What is the edge of this cube?

Opt:

10 cm

9 cm

8 cm

6 cm

Ans: 6 cm

Q10. AB and CD are two chords of a circle intersecting at the point P outside the circle. If PA = 12 cm, PC = 15, CD = 7 cm, then AB is equal to

Opt: 2 cm

4 cm

8 cm

10 cm

Ans: 2 cm

Q11. If p and q are the roots of the quadratic equation x²+ mx + m² + a = 0, then the value of p²+ q² + pq is

Opt:

–a.

± m².

Ans: –a.

Q12. There are 40 students in class of whom 25 are female and 15 are male. A teacher has to select one student out of male and female as a class representative. Teacher writes the name of each student on a separate paper, the papers being identical. Then teacher puts the papers in a box and stirs them thoroughly. Teacher then draws one paper from the box. The probability that the name written on the paper is the name of a female student is

Opt:

1/8.

3/8.

5/8.

7/8.

Ans: 5/8.

Q13. If the vertices of a triangle are (1, k), B(4, –3), (–9, 7) and its area is 15 square units, then the value of k is

Opt:

-7 units.

-3 units.

3 units.

7 units.

Ans: -3 units.

Q14.

\begin{matrix} If α, β are the zeroes of the polynomial f (x) = x^{2} + x + 1, then \\ What is the value \frac{1}{α} + \frac{1}{β} ? \end{matrix}

Opt:

–1

-1/7

1/7

Ans: –1

Q15. The tenth term from the end of the AP 8, 10, 12, …, 126 is

Opt:

106.

108.

110.

112.

Ans: 108.

Q16.

Opt:

542/979

316/979

301/979

271/979

Ans: 271/979

Q17. Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

Assertion: When 32.345675 is expressed in the form

, q ≠ 0, then the denominator is of the form 2^m × 5ⁿ, where m and n are non-negative integers.

Reason: 32.345675 is a terminating decimal fraction.

Opt: Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

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

Assertion (A) is true but Reason (R) is false.

Assertion (A) is false but Reason (R) is true.

Ans: Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Q18.

Assertion: The midpoint of line segment joining the points (2, 3) and (4, –3) is (3, 0).

Reason: The midpoint of line segment joining the points (x₁, y₁) and (x₂, y₂) is

(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) .

(2x1+x2,2y1+y2).

Opt:

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

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

Assertion (A) is true but Reason (R) is false.

Assertion (A) is false but Reason (R) is true.

Ans:

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

Q19. The wheel of a motor cycle is 70 cm in diameter, making 40 revolutions in every 10 seconds. The speed of the motorcycle in km/hour is

Opt:

22.32.

27.68.

31.68.

36.24.

Ans: 31.68.

Q20.

Opt: RTS. TPQ. PTS. PRT.

Ans: RTS.
Q21. Find the values of k for which the system of linear equations has no solution.
2x + 3y = 7, kx + 9y = 28.

Ans:

\begin{matrix} The given pair of equations are \\ 2 x + 3 y = 7 \\ kx + 9 y = 28 \\ For no solutions we have, \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ \Rightarrow \frac{2}{k} = \frac{3}{9} \neq \frac{7}{28} \\ \Rightarrow \frac{2}{k} = \frac{1}{3} \neq \frac{1}{4} \\ From first and second ratio \\ we have,k = 6. \\ So, there is no solution for k = 6. \end{matrix}

Q22. In the given figure, ∠ABC = 90° and BD ⊥ AC. If BD = 8 cm, AD = 4 cm, then find the value of CD.

Ans:

\begin{matrix} Since ∠ ABD + ∠ DBC = 90 ° . .. (1) \\ and ∠ DBC + ∠ DCB = 90 ° . .. (2) Remaining angles of traingle \\ So from equation (1) and (2), we get \\ ∠ ABD = ∠ DCB \\ In ΔDBA and ΔDCB, \\ ∠ ABD = ∠ DCB Proved above \\ and ∠ D = ∠ D Common \\ Therefore, ΔDBA ~ ΔDCB By AA - Similarity \\ \Rightarrow \frac{BD}{CD} = \frac{AD}{BD} \\ \Rightarrow CD = \frac{{BD}^{2}}{AD} \\ CD = \frac{{(8)}^{2}}{4} \\ = 16 cm \end{matrix}

Q23. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the distance of chord PQ from centre.

Ans:

Since OT is perpendicular bisector of PQ, therefore
But PQ = 8 cm
PR + RQ = 8
PR + PR = 8
PR = 8/2 = 4cm
In right triangle ORP, we have
OP² = OR² + PR²
OR²= OP² – PR²
= 25 – 16
= 9
OR = 3cm.
Thus, the distance of chord from centre is 3 cm.
PR = RQ

Q24. Simplify: sec²x(1 – sin²x) + secB (sinB/tanB)

Ans:

We have,
sec²x(1 – sin²x) + secB(sinB/tanB)
= (1/cos²x)cos²x + (1/cosB)(sinB.cosB/sinB)
= 1+1
= 2.

Q25. If 2x = secA and 2/y = tanA, then find the value of 2(x² – 1/y²).

Ans:

Since 2x = secA ⇒ x = (1/2)secA
and 2/y = tanA⇒1/y=(1/2)tanA
So, 2(x² – 1/y²) = 2{(1/4)sec²A – (1/4)tan²A}
= (2/4){ sec²A – tan²A}
= 1/2(1)
= 1/2.

Q26. Show that the points (1,-1),(5,2) and (9,5) are collinear.

Ans:

\begin{matrix} Let the points are A (1, - 1), B (5, 2) and C (9, 5) \\ AB = \sqrt{{(5 - 1)}^{2} + {(2 + 1)}^{2}} = \sqrt{16 + 9} = 5 \\ BC = \sqrt{{(5 - 9)}^{2} + {(2 - 5)}^{2}} = \sqrt{16 + 9} = 5 \\ AC = \sqrt{{(1 - 9)}^{2} + {(- 1 - 5)}^{2}} = \sqrt{64 + 36} = 10 \\ Clearly, AB + BC = AC \\ Thus, points A, B, C are collinear . \end{matrix}

Q27. Find the distance between the points (acosA+bsinA,0) and Q(0,asinA-BcosA).

Ans:

\begin{matrix} The distance \\ PQ = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} \\ PQ = \sqrt{{(acosA + bsinA)}^{2} + {(bcosA - asinA)}^{2}} \\ PQ = \sqrt{a^{2} (\cos^{2} A + \sin^{2} A) + b^{2} (\cos^{2} A + \sin^{2} A)} \\ PQ = \sqrt{a^{2} + b^{2}} \end{matrix}

Q28. Show that 5 – √3 is irrational.

Ans:

Let us assume, to the contrary, that 5 – √3 is rational.
That is, we can find co-prime a and b (b ≠ 0) such that 5 – √3 = a/b
Therefore, 5 – a/b = √3
Rearranging this equation, we get √3 = 5 – (a/b) = (5b – a)/b
Since a and b are integers, we get 5 – (a/b) is rational, and so √3 is rational.
But this contradicts the fact that √3 is irrational.
Therefore, our assumption that 5 – √3 is rational is incorrect.
So, we conclude that 5 – √3 is irrational.

Q29. Divide the polynomial P(t) = 2t³– 11t²+ 16t – 4 by the polynomial g(t) = t²– 2t +1 and verify the division algorithm.

Ans:

\begin{matrix} 2 t - 7 \\ t^{2} - 2 t + 1 2 t^{3} - 11 t^{2} + 16 t - 4 \\ 2 t^{3} - 4 t^{2} + 2 t \\ - + - \\ \bar{\begin{matrix} - 7 t^{2} + 14 t - 4 \\ - 7 t^{2} + 14 t - 7 \\ + - + \end{matrix}} \\ \bar{3} \\ We have, quotient q (t) = 2 t - 7 and remainder r (t) = 3 \\ Now, Quotient ×Divisor+Remainder \\ = (2 t - 7) (t^{2} - 2 + 1) + 3 \\ = 2 t^{3} - 4 t^{2} + 2 t - 7 t^{2} + 14 t - 7 + 3 \\ = 2 t^{3} - 11 t^{2} + 16 t - 4 = Dividend \\ Hence, the division algorithm is verified. \end{matrix}

Q30. The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 75 and for a journey of 15 km, the charge paid is ₹ 110. How much does a person have to pay for travelling a distance of 25 km?

Ans:

Let the fixed charge of taxi be ₹ x per km
The running charges be ₹ y km/hr
As per the given conditions
x + 10y = 75 …(1)
x + 15y = 110 …(2)
Subtracting (2) from (1), we get
5y = 35
y = 7
Substituting the value of y in (1), we get
x + 10(7) = 75
x + 70 = 75
x = 5
So, the total charges for travelling 25 km
= x + 25y
= 5 + 25(7)
= ₹ 180

Q31. In triangle ABC, ∠C = 3∠B = 2(∠A + ∠B). Find the three angles.

Ans:

\begin{matrix} Let ∠ A = x°, ∠ B = y °, ∠ C = 3 ∠ B, i.e., ∠ C = 3 y ° \\ 3 ∠ B = 2 (∠ A + ∠ B) \\ 3 y = 2 (x + y) \Rightarrow y = 2 x . \dots (1) \\ Since ∠ A, ∠ B and ∠ C are angles of a traingle \\ So ∠ A+ ∠ B+ ∠ C=180° \\ x + y + 3y = 180 \\ x + 4y = 180 . \dots (2) \\ Putting y = 2x in eqn (2) \\ x + 8 x = 180 \\ ∠ A = x = 20 °, ∠ B = y = 2 x = 40 ° ∠ C = 3 y = 120 ° \end{matrix}

Q32. If 7 cosecA – 3cotA = 7, prove that 7cotA – 3cosecA = 3.

Ans:

7cosecA – 3cotA = 7
7cosecA – 7 = 3cotA
7(cosecA – 1) = 3cotA
Multiplying by (cosecA + 1) both sides, we get-
7(cosecA – 1) (cosecA + 1) =3cotA (cosecA + 1)
7(cosec²A – 1) =3CotA (cosecA + 1)
7cot²A = 3 cotA (cosecA + 1)
7cotA = 3(cosecA + 1)
7cotA – 3 cosecA = 3

Q33. In the given fig. AB || MN, if PA = x – 2, PM = x; PB = x – 1 and PN = x + 2, find the value of ‘x’.

Ans:

\begin{matrix} Since, AB ∥ MN \\ ∴ \frac{PA}{AM} = \frac{PB}{BN} \\ \Rightarrow \frac{PA}{AM} + 1 = \frac{PB}{BN} + 1 Adding 1 on both sides \\ \frac{PA + AM}{AM} = \frac{PB + BN}{BN} \\ \frac{x}{x - x + 2} = \frac{x + 2}{x + 2 - x + 1} \\ \Rightarrow \frac{x}{2} = \frac{x + 2}{1} \\ \Rightarrow 3 x = 2 x + 4 \\ \Rightarrow x = 4 . \end{matrix}

Q34. A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical part are 5 cm and 13 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Find the surface area of the toy if the total height of the cone is 30 cm.

Ans:

\begin{matrix} Total surface area of Toy = Surface area of conical part \\ + surface area of cylindrical part + surface area of hemisphere \\ Total height of toy = 30 cm \\ Vertical height of conical part = 30 - (5 + 13) = 12 cm \\ Applying Pythagoras Theorem, we get \\ l = 13 cm \\ Now, the surface area of conical part = πrl = π \times 5 \times 13 = 65 π \\ Surface area of cylindrical part = 2 πrh = 2 π \times 5 \times 30 = 130 π \\ Surface area of hemispherical part = 2 {πr}^{2} = 2 π \times 5 \times 5 = 50 π \\ Total surface area = 65 π + 130 π + 50 π = 245 π \\ = 245 \times \frac{22}{7} = 35 \times 22 = 770 {cm}^{2} \end{matrix}

Q35. A circus tent is cylindrical upto a height of 3 m and conical above it. If the diameter of the base is 105 m and the vertical height of the conical part is 7.26 m, find the total canvas used in making the tent.

Ans:

\begin{matrix} Total canvas used = Total curved surface area of tent \\ Total curved surface area of the tent \\ = Curved surface area of the cylindrical portion + Curved surface area of the conical portion \\ Radius of the cylindrical portion = \frac{D}{2} = \frac{105}{2} = 52 .5 m \\ Height of the cylindrical portion = 3 m \\ Curved surface ​​ area of the cylindrical portion = 2 πrH = 2 \times \frac{22}{7} \times 52 .5 \times 3 m^{2} \\ For the conical part : \\ l = slant height = \sqrt{h^{2} + r^{2}} = \sqrt{{(7 .26)}^{2} + {(52 .5)}^{2}} = 52 .99 \approx 53 m \\ Curved surface area of the conical part = πrl = \frac{22}{7} \times 52 .5 \times 53 \\ Total curved surface area = 2 \times 22 \times 7 .5 \times 3 + 22 \times 7 .5 \times 53 \\ = 22 \times 7 .5 \times (6 + 53) \\ = 22 \times 7 .5 \times 59 = 9735 m^{2} \end{matrix}

Q36. Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Use the above theorem to prove that PR² = PQ² + QR² – 2QM.QR

Ans:

Given: A right triangle ABC, right angled at B.
To prove: (Hypotenuse)² = (Base)²+ (Perpendicular)²
i.e., AC² = AB² + BC²
Construction: Draw BD ⊥ AC

\begin{matrix} Proof : ∠ A = ∠ A common \\ ∠ ADB = ∠ ABC each 90 ° \\ ∴ ΔADB ~ ΔABC AA - criterion if similarity \\ ∴ \frac{AD}{AB} = \frac{AB}{AC} sides are proportional \\ \Rightarrow {AB}^{2} = AD \times AC . .. (1) \\ Again in ΔCDB and ΔCBA \\ ∠ C = ∠ C \\ ∠ BDC = ∠ ABC \\ ∴ ΔCDB ~ ΔCBA \\ ∴ \frac{CD}{BC} = \frac{BC}{CA} \\ \Rightarrow {BC}^{2} = CD \times AC . .. (2) \\ Adding (1) and (2), we get \\ {AB}^{2} + {BC}^{2} = AD \times AC + CD \times AC \\ = AC AD + CD \\ = AC \times AC \\ = {AC}^{2} \\ Hence, {AC}^{2} = {AB}^{2} + {BC}^{2} \\ Now we prove the second part of the question . \end{matrix}

\begin{matrix} A triangle PQR in which ∠ Q is acute and PM ⊥ QR . \\ We have to prove that {PR}^{2} = {PQ}^{2} + {QR}^{2} - 2 QM . QR \\ In right traingle PMQ, we have \\ {PQ}^{2} = {PM}^{2} + {QM}^{2} Using the theorem proved above \\ \Rightarrow {PM}^{2} = {PQ}^{2} - {QM}^{2} . .. (1) \\ In right traingle PMR, we have \\ {PR}^{2} = {PM}^{2} + {MR}^{2} Using the above result \\ = {PM}^{2} + {(QR - QM)}^{2} ∵ QR = QM + MR \Rightarrow MR = QR - QM \\ = {PM}^{2} + {QR}^{2} + {QM}^{2} - 2 QM . QR \\ = ({PQ}^{2} - {QM}^{2}) + ({QR}^{2} + {QM}^{2} - 2 QM . QR) Using (1) \\ = {PQ}^{2} + {QR}^{2} - 2 QM . QR \end{matrix}

Q37. Prove that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Ans:

Given: ΔABC and ΔPQR such that ΔABC ~ ΔPQR

To prove : \frac{area (ΔABC)}{area (ΔPQR)} = \frac{{AB}^{2}}{{PQ}^{2}} = \frac{{BC}^{2}}{{QR}^{2}} = \frac{{CA}^{2}}{{RP}^{2}}

Construction: Draw AD ⊥ BC and PS ⊥ QR

\begin{matrix} Proof : \frac{area (ΔABC)}{area (ΔPQR)} = \frac{\frac{1}{2} \times BC \times AD}{\frac{1}{2} \times QR \times PS} Area of = \frac{1}{2} \times base \times height \\ \Rightarrow \frac{area (ΔABC)}{area (ΔPQR)} = \frac{BC \times AD}{QR \times PS} . .. (1) \\ Now, in ΔADB and ΔPQS, we have \\ ∠ B = ∠ Q As ΔABC ~ ΔPQR \\ ∠ ADB = ∠ PSQ each 90 ° \\ ∠ BAD = ∠ QPS \\ Thus, ΔADB and ΔPSQ are equiangular and hence they are similar . \\ Consequently, \frac{AD}{PS} = \frac{AB}{PQ} . .. (2) \\ If Δ ‘ s are similar, the ratio of their corresponding sides is same \\ But \frac{AB}{PQ} = \frac{BC}{QR} \\ \Rightarrow \frac{AD}{PS} = \frac{BC}{QR} using (2) . .. (3) \\ Now, from (1) and (3), we get \\ \frac{area (ΔABC)}{area (ΔPQR)} = \frac{BC}{QR} \times \frac{AD}{PS} \\ \Rightarrow \frac{area (ΔABC)}{area (ΔPQR)} = \frac{BC}{QR} \times \frac{BC}{QR} using (3) \\ \Rightarrow \frac{area (ΔABC)}{area (ΔPQR)} = \frac{{BC}^{2}}{{QR}^{2}} . .. (4) \\ As ΔABC ~ ΔPQR, therefore \\ \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} \\ Hence, \frac{area (ΔABC)}{area (ΔPQR)} = \frac{{AB}^{2}}{{PQ}^{2}} = \frac{{BC}^{2}}{{QR}^{2}} = \frac{{CA}^{2}}{{RP}^{2}} From (4) and (5) \end{matrix}

Q38.

The median of the following data is 28.5. Find the values of x and y, if the total frequency is 60.

Class Interval:	0-10	10-20	20-30	30-40	40-50	50-60
No. of Students:	5	x	20	15	y	5

Ans:

Class Intervals	Frequency	Cumulative Frequency (cf)
0-10	5	5
10-20	x	5 + x
20-30	20	25 + x
30-40	15	40 + x
40-50	y	40 +x + y
50-60	5	45 + x + y
	∑f = 60

\begin{matrix} We have, \\ Median = 28 .5 \\ It lies in the class interval 20 - 30 \\ So, 20 - 30 is the median class . \\ ∴ l = 20, h = 10, f = 20, F = 5 + x and N = 60 \\ Median = l + \frac{\frac{N}{2} - F}{f} \times h \\ \Rightarrow 28 .5 = 20 + \frac{30 - 5 - x}{20} \times 10 \\ \Rightarrow 28 .5 = 20 + \frac{25 - x}{2} \\ \Rightarrow 8 .5 = \frac{25 - x}{2} \\ \Rightarrow 25 - x = 17 \\ \Rightarrow x = 8 \\ We have, N = 60 \\ ∴ 45 + x + y = 60 \Rightarrow x + y = 15 \\ Putting x = 8 in x + y = 15, we get y = 7 \\ Putting x = 8 in x + y = 15, we get y = 7 \\ Hence, x = 8 and y = 7 \end{matrix}

Q39. Case Study – 2

Aman wants to buy a car and plans to take a loan from a bank for his car. He repays his total loan of ₹2,50,000 by paying an instalment at the start of every month of ₹1000. If the instalment is increased by ₹100 every month, answer the following:

(a)
What is the amount paid by Aman in 25^th instalment?

(b)
What is the total amount of loan paid in 25 instalments?

(c)
Find the total number of instalments in which the loan is repaid. Also, find the number of instalments in which Aman pays half of the loan.

What is the amount paid by Aman in 30^th instalment. Also, find the total amount of loan paid by him in 30 instalments.

Ans:

According to question, the A.P. formed is as follows:

1000, 1100, 1200, 1300, …., 250000

Here, a = 1000, d = 100, a_n = 250000

(a)
We know: a_n = a + (n – 1)d

For n = 25, we get:

a₂₅ = 1000 + (25 – 1)(100)

= 1000 + 2400

= 3400

⸫ Amount paid in 25^th instalment = ₹3400

\begin{matrix} (b) \\ W e k n o w : S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ F o r n = 25, w e g e t : \\ S_{25} = \frac{25}{2} [2 (1000) + (25 - 1) (100)] \\ = \frac{25}{2} [2000 + 2400] = \frac{25}{2} (2600) \\ = 25 (1300) \\ ∴ T o t a l a m o u n t p a i d i n 25 i n s t a l m e n t s = ₹ 32500 \end{matrix}

(b)We know : Sn=2n[2a+(n−1)d]Forn=25,weget:S25=225[2(1000)+(25−1)(100)]=225[2000+2400]=225(2600)=25(1300)∴Totalamountpaidin25instalments=₹32500

a_n = a + (n – 1)d

250000 = 1000 + (n – 1)(100)

250000 – 1000 = (n – 1)(100)

n – 1 = 2490

n = 2450

So, total number of instalments = 2450

Now, amount of half of the loan = ₹125000

We know:

a_n = a + (n – 1)d

125000 = 1000 + (n – 1)(100)

125000 – 1000 = (n – 1)(100)

n – 1 = 1240

n = 1241

⸫ Aman pays half of the loan in 1241 instalments.

We know: a_n = a + (n – 1)d

For n = 30, we get:

a₃₀ = 1000 + (30 – 1)(100)

= 1000 + 2900

= 3900

⸫ Amount paid in 25^th instalment = ₹3900

\begin{matrix} N o w, \\ W e k n o w : S_{n} = \frac{n}{2} [2 a + [n - 1] d] \\ F o r n = 30, w e g e t : \\ S_{30} = \frac{30}{2} [2 (1000) + 30 - 1 (100)] \\ = 15 [2000 + 2900] = 15 (3100) \\ = 46500 \\ ∴ T o t a l a m o u n t p a i d i n 25 i n s t a l m e n t s = ₹ 46500 \end{matrix}

Now, We know :Sn=2n[2a+[n−1]d]Forn=30,weget:S30=230[2(1000)+30−1(100)]=15[2000+2900]=15(3100)=46500∴Totalamountpaidin25instalments=₹46500

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