The document Class 10 Notes | EduRev is a part of the Class 10 Course Sample Papers For Class 10.

All you need of Class 10 at this link: Class 10

**Class X****Mathematics â€“ Standard (041)****Sample Question Paper 2019-20**

**Max. Marks: 80****Duration : 3 hrs****General Instructions:**

(i) All the questions are compulsory.

(ii) The question paper consists of 40 questions divided into 4 sections A, B, C, and D.

(iii) Section A comprises of 20 questions of 1 mark each. Section B comprises of 6 questions of 2 marks each. Section C comprises of 8 questions of 3 marks each. Section D comprises of 6 questions of 4 marks each.

(iv) There is no overall choice. However, an internal choice has been provided in two questions of 1 mark each, two questions of 2 marks each, three questions of 3 marks each, and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.

(v) Use of calculators is not permitted.

**Section A**

**Q.1. If the sum of the roots of the quadratic equation ax ^{2} - 5x + 11 = 0 is 15, then the value of a is : (1 Mark)(a) 11/15(b) 3(c) 15/11**

Ans:

Given, sum of the roots = 15

â‡’ - (-5)/a = 15 [âˆµ Sum of roots = - Coeff. of x / Coeff. of x

â‡’ a = 5/15 = 1/3

Ans:

â‡’ 7y (3y - 2) + (3y - 2) = 0 â‡’ (3y - 2) (7y + 1) = 0

â‡’ y = 2/3 or y = -1/7

[Here, the smallest two digit composite number =10 The smallest composite number = 4 Now, LCM of 4 and 10 = 20]

â‡’ d = t

t

â‡’ f

Given, DE||BC, so that Î”ABC ~ Î”ADE

Given, AD = 1 cm and BD = 2 cm, so that AB = AD + BD = 1 cm + 2 cm = 3 cm.

The ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

âˆ´ ar (Î”ABC)/ar (Î”ADE) = AB

(a) 1/4

(b) 1/3

(c) 3/8

(d) 1/2

Ans:

= (H, H), (H, T), (T, H), (T,T)

E = Event of getting tails on both the coins.

= (T, T)

âˆ´ P(E) = 1/4

[sin Î± = âˆš3/2 â‡’ Î± = 60

Now, Î² - Î± = 90

cos (Î± + Î²) = 0

cos(Î± + Î²) = cos 90Â°

Î± + Î² = 90Â°

Î± = 90Â° - Î²

sin (Î± - Î²) = sin (90Â° - Î² - Î²)

= sin (90Â° - 2Î²)

= cos 2Î²

Given tan 2A = cot (A - 24Â°)

â‡’ cot (90Â° - 2A) = cot (A - 24Â°) [âˆµ tan Î¸ = cot (90Â° - Î¸)]

â‡’ 90Â° - 24 = A - 24Â°

â‡’ 90Â° + 24Â° = A + 2A

â‡’ 3A - 114Â°

â‡’ A = 38Â°

[i.e., surface area of base + C.S.A. of cylinder + C.S.A. of cone]

Ans:

As 2, and (-3) are zeroes of polynomial

f(x) = x

f(2) = 0

(2)

4 + 2a + 2 + b = 0

2a + b = -6 ......(i)

And f(-3) = 0

(3)

9 - 3a - 3 + b = 0

-3a + b = -6

3a-b = 6 ..... (ii)

5a = 0 [Adding (i) and (ii)]

a = 0

But 2a + b = - 6 [From (i)]

2(0) + b = - 6

â‡’ b = - 6

Hence, the value of a = 0 and b = - 6.

Area of the sector = Î¸/360Âº x Ï€r

**Q.14. The line joining the observer's eye and the object is called the _______. (1 Mark)****Ans: **Line of sight**Q.15. A number is chosen at random from the numbers - 5, - 4, - 3, - 2, - 1, 0, 1 , 2, 3, 4, 5. Then the probability that square of this number is less than or equal to 1 is (1 Mark)****Ans:** 3/11 [âˆµ The required numbers are - 1, 0,1. Hence, the probability = 3/11]**Q.16. In the given figure,, if âˆ A = 90Â°, âˆ B = 90Â°, OB = 4.5 cm, OA = 6 cm and AP = 4 cm, then find QB. (1 Mark)****Ans: **In Î”PAO and Î”QBO,

âˆ A = âˆ B = 90Â° (Given)

âˆ POA = âˆ QOB

(Vertically Opposite Angle)

Since, Î”PAO ~ Î”QBO,

(by AA)

Then, OA/OB = PA/QB

Or 6/4.5 = 4/QB

or QB = 4 x 4.5 / 6

âˆ´ QB = 3 cm**Q.17. ****Write the condition for unique solution of a system of linear equations a _{1}x + b_{1}y + c_{1} = 0 Q and a_{2}x + b_{2}y + c_{2} = 0. (1 Mark)**

a

Hence, number of zeroes of the polynomial y = p(x) are three.

âˆ´ c - b = 2b - c

â‡’ 3b = 2c

â‡’ b/c = 2/3

Hence, b : c = 2: 3.

n(S) = 6

A = {1,2}

n(A) = 2

âˆ´ P(A) = n(A)/n(S) = 2/6 = 1/3

**Section B **

**Q.21.**** Find the HCF of 1260 and 7344 using Euclidâ€™s algorithm.****Or ****Show that every positive odd integer is of the form (4q + 1) or (4q + 3), where q is some integer. (2 Mark)****Ans:** Given integers are 1260 and 7344. Clearly 7344 > 1260.

Applying Euclidâ€™s division algorithm, we get

7344 = 1260 x 5 + 1044

1260 = 1044 x 1 + 216

1044 = 216 x 4 + 180

216 = 180 x 1 + 36

180 = 36 x 5 + 0

In the last equation, the remainder is zero and divisor in the last equation is 36.

Hence, HCF of 1260 and 7344 = 36.

Or

Let a be any odd positive integer.

When a is divided by 4, remainder will be one of 0, 1, 2, 3.

âˆ´ a = 4q + 0 i.e., 4q or 4q + 1 or 4q + 2 or 4q + 3

When a = 4q, which is even positive integer as it is divisible by 2.

When a = 4q + 2 = 2(2q + 1), which is even positive integer as it is divisible by 2.

Hence, every odd positive integer is of the form 4q + 1 or 4q + 3.**Q.22. Read the following passage and answer the questions that follows: ****A teacher of Xth class draw a figure on blackboard and asked one question about figure from students that, if Î”ACB ~ Î”APQ, AB = 6 cm, BC = 8 cm and PQ = 4 cm, then find the length of AQ. (2 Mark)****Ans:** As Î”ACB ~ Î”APQ

So, AB/BC = AQ/QP â‡’ 6/8 = AQ/4

[âˆµ AB = 6 cm, BC = 8 cm and PQ = 4 cm, given]

â‡’6 x 4 / 8 = AQ = 3cm**Q.23. In the given figure, DEFG is a square and âˆ BAC = 90Â°. Show that FG ^{2} = BG x FC. (2 Mark)**

âˆ A = âˆ G = 90Â°

âˆ ADE = âˆ DBG

[corresponding âˆ s]

âˆ´ By using AA similarity rule, we have

Î”ADE âˆ¼ Î”GBD

Similarly, Î”ADE âˆ¼ Î”FEC

â‡’ Î”GBD âˆ¼ Î”FEC

[AA similarity criterion]

âˆ´ GD/FC = GB/FE

â‡’ GD x FE = GB x FC

â‡’ FG x FG = BG x FC

[âˆµ GD = FE = DE = FG, as DEFG is a square]

â‡’ FG

Or

In an equilateral Î”ABC, draw AD âŠ¥ BC.

âˆ´ AD is also the median

i.e., BD = DC = BC/2 .

Now, in rt. âˆ ed Î”ABD

AB

= AD

4AB

4AB

4AB

3AB

[âˆµ AB = BC = CA]

|PQ| = |PR|

Squaring, we get

[x - (a + b)]^{2} + [y - (b - a)]^{2}

= (x - (a - b)]^{2 }+ [y - (a + b)]^{2}

or (x - a - b)^{2} [x - a + b]^{2}

= (y - a - b)^{2} - (y - b + a)^{2}

or, (x - a - b + x - a + b) (x - a - b - x + a - b)

= (y - a - b + y - b + a) (y - a - b - y + b - a)

or (2x - 2a) (-2b) = (2y - 2b) (-2a)

or, (x - a) = (y - b)a

or, bx = ay**Q.25. ****A die is thrown once. Find the probability of getting a number which (i) is a prime number (ii) lies between 2 and 6. (2 Mark)****Ans: **When a die is thrown once, the possible outcomes are 1, 2, 3, 4, 5 and 6.

Total number of all possible outcomes = 6

Let A and B be events of getting a prime number and a number lying between 2 and 6.

(i) Prime numbers are 2, 3 and 5, so number of cases in favour of event A = 3

Probability of getting a prime number = P(A)

= Number of cases in favour of event A/Total number of cases = 3/6 = 1/2

(ii) Numbers lying between 2 and 6 are 3, 4 and 5.

âˆ´ Number of cases in favour of event B = 3

Probability of getting a number that lies between 2 and 6

= P(B) = Number of cases in favour of event B/Total number of cases = 3/6 = 1/2**Q.26. ****Express 234 as the product of its prime factors.Or Show that any positive odd integer is of the form 4q +1 or 4q + 3, where q is a positive integer. (2 Mark)**

âˆ´ 234 = 2 x 3 x 3 x 13

Or

Let a be any positive odd integer and b = 4.

By Euclidâ€™s division algorithm, there exists integers^ and r such that

a = 4q + r, where 0

a = 4q or 4g + 1 or 4q + 2 or 4g + 3

[âˆµ r = 0,1,2, 3]

â‡’ a = 4g + 1 or a = 4q + 3

[âˆµ a is an odd integer â‡’ and a â‰ 4q + 2]

Hence, any odd integer is of the form 4q +1 or 4q + 3.

**Section C**

**Q.27. Prove that âˆš2 is an irrational number. ****(3 Mark)Ans:** Let us assume to the contrary that âˆš2 is a rational number. Then, there exists positive integers a and b such that

âˆš2 = a/b, where a and b are coprime i.e., their HCF is 1 and b â‰ 0.

â‡’âˆš2b = a

Squaring both sides, we get

2b

â‡’ 2 divides a

â‡’ 2 divides a ......(1)

[If a prime number r divides a

âˆ´ a = 2m, for some integer m

Substituting a = 2m in 2b

2b

â‡’ 2 divides b

â‡’ 2 divides b

From (1) and (2), we find that a and b have at least 2 as a common factor. But, this contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that âˆš2 is a rational number.

Age (in yr) | 5-14 | 15-24 | 25-34 | 35-44 | 45-54 | 55-64 |

Number of cases | 6 | 11 | 21 | 23 | 14 | 5 |

**Find the average age for which maximum cases occurred. (3 Mark)****Ans: **

Here, class intervals are not in inclusive form.

So, we first convert them in inclusive form by subtracting h/2 from the lower limit and adding h/2 to the upper limit of each class, where h is the difference between the lower limit of a class and the upper limit of the preceding class.

The given frequency distribution in inclusive form is as follows:

Age (in yr) | 4.5 - 14.5 | 14.5 - 24.5 | 24.5 - 34.5 | 34.5 - 44.5 | 44.5 - 54.5 | 54.5 - 64.5 |

Number of cases | 6 | 11 | 21 | 23 | 14 | 5 |

We observe that the class 34.5-44.5 has the maximum frequency.

So, it is the modal class such that

I = 34.5, h = 10, f = 23, f_{1} = 21 and f_{2} = 14**Q.29. In the figure, ABCDE is a pentagon with BE||CD and BC||DE. BC is perpendicular to CD. AB = 5 cm, AE = 5 cm, BE = 7 cm, BC = x - y and CD = x + y. If the perimeter of ABCDE is 27 cm, find the value of x and y, given x, y â‰ 0. (3 Mark)****Or****Solve the following system of equations:****21/x + 47/y = 100****47/x + 21/y = 162, x, y â‰ 0****Sol. **Here, AB = AE = 5 cm, BE = 7 cm,

BC = x - y and CD = x + y.

âˆ´ x + y = 7 ...(i)

And BC + CD + DE + AE + AB = 27

âˆ´ x - y + x + y + x - y + 5 + 5 = 27

âˆ´ 3x - y = 17 ...(ii)

Adding (i) and (ii), we obtain

4x = 24

â‡’ x = 6

From (i), we obtain 6 + y = 7 â‡’ y = 1

Hence, x = 6 and y = 1.

Or

Given equations are

21/x + 47/y = 110

47/x + 21/y = 162

Put 1/x = a and 1/y = b.

21a + 47b = 110 ......(i)

47a + 21b = 162 ......(ii)

Adding (i) and (ii), we obtain

68a + 68b = 272

a + b = 4 ......(iii)

Subtracting (i) and (ii), we obtain

26a - 26b = 52

a - b = 2 ...........(iv)

Adding (iii) and (iv) we have

2a = 6 â‡’ a = 3

From (iii), we have

3 + b = 4 b = 1

Now, a =1/x = 3 â‡’ x = 1/3

and, b = 1/y = 1 â‡’ y = 1**Q.30. Prove that:****cot Î¸ - tan Î¸ = 2cos ^{2}Î¸ - 1/sinÎ¸cosÎ¸ (3 Mark)**

= cosÎ¸/sinÎ¸ - sinÎ¸/cosÎ¸

= cos

= cos

= 2cos

Or

Prove that:

Ans:

L.H.S. = (sin Î¸ + cosec Î¸)

= sin

= (sin

= (sin

= 1 + (1 + cot

= 7 + tan

= R.H.S.

Or

We have

L.H.S = (1 + cot A - cosec A) (1 + tan A + sec A)

= (1 + cos A/ sin A - 1 / sin A)(1 + sin A/cos A + 1/cos A)

= (sin A + cos A - 1 / sin A) (cos A + sin A + 1 / cos A)

= (sin A + cos A)

= (sin

= 2 sin A cos A / sin A cos A = 2

= R. H. S.

One of the members of association, Sarika suggested that the paths should be constructed represented by the two linear equations x - 3y - 2 and - 2 x + 6y = 5.

x - 3y = 2

â‡’ x - 3y - 2 = 0 ...(i)

and -2x + 6y = 5

â‡’ -2x + 6y - 5 = 0 ...(ii)

Here, a

and a

Now, a

and c

Here, a

So, the paths represented by the equations are parallel, i.e. not intersect each other.

Here, ar(Î”ABC) = 1/2 x 24 x 7

= 84 m

Area of three sectors = Ï€r

= 1/2 x 22/7 x 3.5 x 3.5

= 77/4 = 19.25 m

Required area = Area of Î”ABC - Area of three sectors

= 84 - 19.25

= 64.75 m

275 = 176 x 1 + 99

176 = 99 x 1 + 77

99 + 77 x 1 + 22

77 = 22 x 3 + 11

22 = 11 x 2 + 0

â‡’ This HCF = 11

**Section D**

**Q.35. Two water taps together can fill a tank in 15/8 hours. The tap with longer diameter takes 2 hours less than the tap with smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.****Or****A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water (4 Mark)****Ans: **Let the time taken by the tap of smaller diameter = x hours

âˆ´ Time taken by the tap of larger diameter

= (x - 2) hours

âˆ´ Portion of the tank filled by the tap of smaller diameter in one hour = 1/x

and

Portion of the tank filled by the tap of larger diameter in one hour = 1/x-2

Thus, the portion of the tank filled by the two taps in one hour = 1/x + 1/x-2 ......(1)

It is given that both taps can fill the tank in = 15/8 hours

âˆ´ The portion of the tank filled by the two taps in one hour = 8/15 .....(2)

From (1) and (2), we get

1/x - 1/x - 2 = 8/15

â‡’ x - 2 + x / x(x - 2) = 8/15

â‡’ 15(2x - 2) 8(x^{2}-2x)

â‡’ 8x^{2} - 16x - 30x + 30 = 0

â‡’ 8x^{2} - 46x + 30 = 0

â‡’ 4x^{2} - 23x + 15 = 0

â‡’ 4x^{2} - 20x - 3x + 15 = 0

â‡’ 4x(x - 5) - 3 (x - 5) = 0

â‡’ (x - 5) (4x - 3) = 0

â‡’ x - 5 = 0 or x = 3/4

When, x = 3/4, x - 2 = 3/4 - 2 = 3-8/4 = - 5/4, which is impossible as time to fill cannot be negative.

âˆ´ When x = 5, x - 2 = 3

Hence, the larger pipe can fill the tank in 3 hours and the smaller pipe can fill the tank in 5 hour

Or

Let the speed of the boat in still water = x km/h

and the speed of the stream = y km/h

âˆ´ Speed of the boat going upstream

= (x - y) km/h

and

Speed of the boat going downstream

= (x + y) km/h

We know that Time = Distance/Speed

âˆ´ Time taken in going 30 km upstream

=(30/(x-y))h

and

Time taken in going 44 km downstream

= (44/(x+y))h

But, the total time of the journey is 10 hours.

âˆ´ 30/x - y + 44/x+ y = 10 ....(1)

Again, Time taken in going 40 km upstream = (40/(x-y))h

and Time taken in going 55 km downstream = (55/(x+y))h

In this case, total time of journey is given to be 13 hours.

âˆ´ 40/ x-y + 55/ x+y = 13 ...(2)

Multiplying (1) and (2) by 4 and 3 respectively, we pet

120/ x - y + 176 / x + y = 40 ...(3)

120/ x-y + 165/x+ y = 39 ......(4)

Subtracting (4) from (3), we get

11/ x + y = 1 â‡’ x + y = 11 ...(5)

Putting the value of x + y from (5) in (1), we have

30/x-y + 44/11 = 10 â‡’ 30/x - y = 6 â‡’ x - y = 5 ...(6)

Adding and subtracting (5) and (6), we get

2x = 16 and 2y = 6

â‡’ a = 8 and y = 3

âˆ´ Speed of the boat in still water = 8 km/h and speed of water stream = 3 km/h.**Q.36. The angles of a triangle are in AP. If the greatest angle equals to the sum of the other two, then find the angles. Also, find these angles are multiple of which angle.****Or****A sum of Rs. 1000 is invested at 8% simple interest per annum. Calculate the interest at the end of 1, 2, 3, ... years. Is the sequence of interests an AP? Find the interest at the end of 30 years. (4 Mark)****Ans: **Let the angles of a triangle, are (a - d), a and (a + d).

Since, sum of the three angles of a triangle = 180Â°

âˆ´ a - d + a + a + d = 180Â°

â‡’ 3a = 180Â° â‡’ a = 60Â°

So, the angles are 60Â° - d , 60Â° and 60Â° + d .

According to the question,

Greatest angle = Sum of two smaller angles

âˆ´ 60Â° + d = 60Â° - d + 60Â°

â‡’ 2d = 60Â°

â‡’ d = 30Â°

Hence, the required angles of a triangle are 60Â° - 30Â°, 60Â° and 60Â° + 30Â°, i.e. 30Â°, 60Â° and 90Â°.

Here, we see that above angles are multiple of 30Â°

Or

Sum of Rs. 1000 is invested at 8 % simple interest.

Interest at the end of 1st year

= 1000 x (8/100) x 1 = â‚¹ 80

Interest at the end of 2nd year

= 1000 x (8/100) x 2 = â‚¹ 160

Interest at the end of 3rd year

= 1000 x (8/100) x 3 = â‚¹ 240

Since, each interest is Rs. 80 more than the previous one.

Sequence of interest 80,160,240,... isanAR

Here, a = Rs. 80, n = 30

and d = a_{2} - a_{1} = Rs. (160 - 80) = Rs. 80

Now using a_{n} = a + (n - 1)d,

âˆ´ a_{30} = 80 + (30 -1) (80)

â‡’ a_{30} = 80 + (29) (80) = 80 + 2320 = 2400

Hence, interest at the end of 30 years is Rs. 2400**Q.37. A train covers a distance of 360 km at a uniform speed. Had the speed been 5 km/hour more, it would have taken 48 minutes less for the journey. Find the original speed of the train.****Or****Solve the following equation : (4 Mark)****Ans: **Let the original speed of the train be x km/h. According to the statement of the question, we have

360/x - 360/x+5 = 48/60

360(x+5-x/x(x+5)) = 4/5 (taking LCM of LHS)

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

âˆ´ x^{2} + 5x - 2250 = 0

â‡’ (x+50)(x-45) = 0

âˆ´ x = -50 or x = 45

Rejecting - ve value of x, because speed cannâ€™t be negative, we have

x = 45

Hence, the original speed of the train is 45 km/h.

Or

1/x - 1/x-2 = 3

x-2-x/x(x-2) = 3

-2 = 3x^{2} - 6x

â‡’ 3x2 - 6x + 2 = 0**Q.38. ****Two poles of equal heights arc 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 60Â° and 30Â° respectively. Find the height of the poles and the distances of the point from the poles. ORThe angle of elevation of the top of a hill from the foot of a tower is 60Â° and the angle of depression from the top of the tower of the foot of the hill is 30Â°. If tower is 50 meter high, find the height of the hill. (4 Mark)**

In Î”ABC, h/x = tan 60

â‡’ h = xâˆš3 ...(1)

In Î”EDC, h/80-x = tan30

â‡’ hâˆš3 = 80 - x

From (1), xâˆš3 x âˆš3 = 80 - x

â‡’ x = 20 m

âˆ´ h = 20âˆš3 m

âˆ´ The height of poles = 20âˆš3 m

Hence, the distance of poles from the point are 20 m and 60 m.

To prove : (Hypotenuse)

i.e., AC

Construction : Draw BD âŠ¥ AC.

Proof : Î”ADB ~ Î”ABC.

In Î”ADB and Î”ABC

âˆ ADB = âˆ ABC

and âˆ A = âˆ A

âˆ´ By A A-similarity criterion, we have

Î”ADB ~ Î”ABC

So, AD/AB = AB/AC [In similar triangles, corresponding sides are proportional]

â‡’ AD.AC = AB

In Î”BDC and Î”ASC

âˆ CDB = âˆ ABC [Each= 90Â°]

and âˆ C = âˆ C

âˆ´ Î”BDC ~ Î”ABC

So, CD/BC = BC/AC [In similar triangles, corresponding sides are proportional]

â‡’ CD.AC = BC

Adding (1) and (2), we have

AD.AC + CD.A C = AB

â‡’ (AD + CD) AC =AB

â‡’ ACAC = AB

Hence, AC

We know that any angle made by the diameter RQ in the semi-circle is 90Â°.

âˆ´ âˆ RPQ = 90Â°

In right angled Î”RPQ,

RQ

â‡’ RQ

â‡’ RQ

â‡’ RQ = âˆš625 [âˆµ side cannot be negative]

â‡’ RQ = 25 cm ...(1)

âˆ´ Area of right angled Î”RPQ = 1/2 x RP x PQ

[âˆµ area of triangle - 1/2 x base x height]

= 1/2 x 7 x 24 = 84cm

Area of semi-circle

= Ï€r2/2 = (22/7 x 2)(25/2)

= 11 x 625 / 28 = 6875/28 cm

Hence, area of the shaded region = Area of the semi-circle - Area of right angled Î”RPQ

= (6875/28) - (84)

= (6875 - 2352) / 28 (taking LCM)

= 4523 / 28 cm

85 docs

### Sample Question Paper (2019-20) - 2

- Doc | 15 pages
### Sample Question Paper (2019-20) - 3

- Doc | 23 pages
### Sample Question Paper (2019-20) - 4

- Doc | 17 pages
### Sample Question Paper (2019-20) - 5

- Doc | 19 pages
### Sample Question Paper (2019-20) - 6

- Doc | 17 pages
### Sample Question Paper (2019-20) - 7

- Doc | 17 pages