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**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. For some integer q, every odd integer is of the form : (1 Mark)****(a) q ****(b) q + 1 ****(c) 2q ****(d) 2q + 1****Ans. **(d)

According to Euclid's division

lemma, a = bq + r, where o __<__ r < b.

If b = 2, then a = 2q or a = 2q + 1

Since a = 2q + 1 is not divisible by 2, then 2q + 1 is an odd integer.**Q.2. The product of the HCF and LCM of the smallest prime number and the smallest composite number is : (1 Mark)**

[Smallest prime number = 2 Smallest composite number = 4 Product of their HCF and LCM = 2x4 = 8]

In a right angled Î”ABC, sin A = sin(90Â°- C)= cosC [âˆµ A + C = 90Â°]

(a) 1

(b) 0

(e) - 1

(d) 2

We have,

In Î”ABD and Î”CAD, âˆ ADB = âˆ ADC = 90Â° and âˆ ABD = âˆ CAD = Î¸. By AA Similarity, we get, Î”ABD ~ Î”CAD

Ans.

We have,

sec A = cosec B

â‡’ sec A = sec (90Â° - B)

â‡’ A = 99Â° - B

â‡’ A+B = 90Â°

The sharpened part of the pencil is cone and unsharpened part is cylinder.

We have, tan 48Â° tan 13Â° tan 42Â° tan 77Â°

= tan (90Â° - 42Â°) tan 13Â° tan 42Â° tan (90Â° - 13Â°)

= cot 42Â° tan 13Â° tan 42Â° cot 13Â° [âˆµ tan (90Â° - A) = cot A]

= (tan 42Â° cot 42Â°)(tan 13Â° cot 13Â°)

= 1 x 1 = 1 [âˆµ tan A cot A = 1]

= 21 + 11

= 32 cm

OR

Let the distance between the foot of the ladder and the wall, AB be x.

then AC, the length of the ladder = 2x

â‡’

Ans.

[Here, a = 53 and a3 = a + 2d = 23

â‡’ 2d = 23 - 53 â‡’ d = - 15

Now, a

and a

= 53 - 60 = - 7]

Or

The value of sin Î¸ lies between ..........

Ans.

Putting in above equation.

âˆ´ cos72Â° - cos72Â° = 0

Or

-1

â‡’ x

â‡’ x

â‡’ x

â‡’

â‡’ Hence, the values of x are 3 and - 3.

âˆ´

Ans.

âˆ´ AP = AR = 4 cm

BP = BQ = 3 cm

CQ = CR

CR = AC - AR

= 11 - 4

= 7 cm

Hence, BC = BQ + CQ = 3 + 7 = 10 cm

Or

Chord AB of larger circle is tangent to smaller circle.

âˆ´ OP is perpendicular bisector of AB.

âˆ´ In right Î”OBP, we have

OB

13

â‡’ BP

= 144

â‡’ BP = 12 cm

Now, AB = 2 x BP

= 2 x 12 = 24 cm

Given, 3 is a zero of the polynomial f(x). Therefore, put x = 3 in Eq. (i), and consider f(3) = 0

âˆ´ f(3) = 2(3)

â‡’ 2 x 9 + 3 + k = 0

â‡’ 18 + 3 + k = 0

â‡’ k = - 21

Since - 4 is a zero of the polynomial f(x) = x

âˆ´ f(-4) = 0

â‡’ (- 4)

â‡’ 16 + 4 - 3k - 2 = 0

â‡’ 18 = 3k

k = 6

Hence, for k = 6, - 4 is a zero of the polynomial x

**Section B**

**Q.21. Using Euclid's algorithm, find the HCF of 240 and 228. (2 Mark)**

and 228 = 12 x 19 + 0

Hence, HCF of 240 and 228 = 12

Join OQ and OR

Now, in Î”PQO and Î”PRO, we have

PO = PQ [common]

OQ = OR [radii of same circle]

PQ = PR

[tangents from an external point]

[by SSS congruence rule]

â‡’ âˆ OPQ = âˆ OPR

â‡’ O lies on the bisector of the lines l

So, it is the modal class such that

l = 1500, h = 500, f

âˆ´ Total number of all possible outcomes = 6

Let A and B be events of getting a composite number and a prime number respectively.

(i) The composite numbers from 1 to 6 are 4 and 6,

âˆ´ Number of cases in favour of event A = 2

Probability of getting a composite number

= 2/6 = 1/3.

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

Probability of getting a prime number

= 3/6 = 1/2

Adding equations (i) and (ii),

2A = 90Âº

âˆ´ A = 90Âº/2 = 45

Putting this value of A in equation (i),

B = 60Â°- A = 60Â°- 45Â° = 15Â°

Hence, A = 45Â° and B = 15Â°.

OR

Each one of the circular plate is also a cylinder. Its volume is,

The volume of right circular cylinder

Also, volume of sphere

**Section C**

**Q.27. Three different coins are tossed together. Find the probability of getting (3 Mark)**

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

â‡’ n(S) = 8

(ii) At least two heads {HHT, HTH, THH, HHH}

(iii) At least two tails {TTH, THT, HTT, TTT}

Let the value of first prize be Rs. x

By the given condition, we have

x, x - 20, x - 40, ...

Now, a

a

Since a

Therefore, it is an A.P. with common difference (d) = - 20

Let S = x + (x - 20) + (x - 40) + ... upto 7 terms

Here, a = x, d = - 20 and n = 7

Hence, amount of each prize (in Rs.) respectively is

160, 160 - 20, 160 - 40, 160 - 60, 160 - 80, 160 - 100, 160 - 120 i.e., Rs. 160, Rs. 140, Rs. 120, Rs. 100, Rs. 80, Rs. 60 and Rs. 40.

Clearly, a

a

and a

(i) âˆµ a

âˆ´ a + 5d = 0

â‡’ a = - 5d [using Eq. (i)]

(ii) On substituting a = - 5 d in Eqs, (ii) and (iii), we get

a

and a

From Eqs. (iv) and (v), we get

a

Then âˆš3 = a/b, where a and b are co -prime i.e., their HCF = 1 and b â‰ 0.

âˆ´ â‡’

Squaring both sides, we get

3b

â‡’ a

â‡’ a is divisible by 3. [If r(prime) divides a

Let a = 3m, where m is an integer.

Substituting a = 3m in (1) , we get

3b

This means that b

So, we conclude that âˆš3 is an irrational number.

Or

The required number is the HCF of the numbers

1251 - 1 = 1250, 9377 - 2 = 9375 and 15628 - 3 = 15625

First we find HCF of 1250 and 9375 by Euclid's algorithm as given below :

9375 = 1250 x 7 + 625

1250 = 625 x 2 + 0

625 is the HCF of 1250 and 9375.

Now, we find the HCF of 625 and 15625 by Euclid's algorithm.

15625 = 625 x 25 + 0

Therefore, the HCF of 1250, 9375 and 15625 is 625.

Hence, the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively is 625.

By eqs. (i) and (ii),

OR

Area of quad. ABGD = arÎ”ABD | arÎ”BCD

= 53. sq. units

Area of ar ABCD = | (-4) (- 6 - 5) + (-1) (5 + 5) + 4 (- 5 + 6)|

19 sq. units.

Hence, area of quad. ABCD = 53 + 19 = 72 sq. units**Q.32. If tan A = n tan B and sin A = m sin B. (3 Mark)**

Or

Let E be a point on the road such that BE = xm,

DE = (100 - x) m, âˆ AEB = 60Â° and âˆ CED = 30Â°

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

â‡’ 3x = 100 - x â‡’ 4x = 100

âˆ´ Area cleaned at each sweep of a blade = Area of sector with r - 21 cm and Î¸ = 120Â°.

Since there are two wipers, therefore, total area cleaned at each sweep of the blades

= (2 x 462) cm

**Section D**

**Q.35. Solve the following pair of linear equations graphically:x + 3y = 6 and 2x - 3y = 12.**

OR

The denominator of a fraction is one more than twice its numerator. If the sum of the fraction and its reciprocal is find the fraction.

and 2x - 3y = 12 .....(ii)

â‡’

Plotting the above points and drawing lines joining them, we get the graphs of the equations x + 3y = 6 and 2x - 3y = 12.

The two lines intersect each other at point B (6, 0).

Hence, x - 6 and y = 0.

Again, Î”OAB is the region bounded by the line 2x - 3y = 12 and both the co-ordinate axes.

OR

Let numerator be x.

Then, die fraction =

â‡’ 11x

11x

Hence, fraction = 3/7

So, by AA criterion of similarity, we have

Thus, in Î”AXB and Î”PYQ, we have

So, by SAS similarity, we have

From (iv) and (v), we have

From (iii) and (vii), we have

Then, âˆ SDAB = âˆ OCD [alternate interior angles]

âˆ AOB = âˆ DOC

[vertically opposite angles]

and âˆ ABO = âˆ CDO [alternate interior angles]

âˆ´ Î”OAB ~ Î”OCD [by AAA similarity criterion]

[if two triangles are similar, then their corresponding sides are proportional] Hence proved.

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by the another cylinder of height 60 cm and radius 8 cm. Find mass of the pole, given that 1 cm

Total surface area of the cube (S

= 6 x (6)

= 216 cm

Diameter of the hemisphere = 4.2 cm

âˆ´ Radius of the hemisphere = 2,1 cm

Area of the base of the hemisphere (S

Curved surface area of the hemisphere (S3) = 2Ï€r

âˆ´ Total surface area of the block (S) = (S

= (216 - 13.86 + 27.72) cm

= 229.86 cm

Volume of the cube (V

âˆ´ Volume of the block (V) = V

= (216 + 19.404) cm

= 235.404 cm

Or

Let R and H denote radius of base and height of the bigger cylinder and r and h denote radius of base and height of the smaller cylinder surmounted on bigger cylinder; V be the volume of the solid iron pole.

Diameter of bigger cylinder = 24 cm; radius of the bigger cylinder, R - 12 cm and height of the bigger cylinder, H = 220 cm and radius of the smaller cylinder, r = 8 cm and height of the smaller cylinder, h = 60 cm.

Volume of the solid iron pole formed of two cylinders,

V = Ï€R

â‡’ V = Ï€(12

â‡’ V = Ï€(144 x 220 + 64 x 60) cm

â‡’ V = Ï€(31680 + 3840) cm

â‡’ V = 35520Ï€ cm

= 35520 x 3.14 cm

= 111632.8 cm

It is given that mass of 1 cm

Let the speed of the boat be x m/min.

âˆ´ Distance covered in 2 minutes = 2x

âˆ´ CD = 2x

Let BC be y m.

In Î”ABC, AB/BC = tan 60Âº

â‡’ y + 2x = 150 ...(ii)

Substituting the value of y from (i) in (ii),

Speed of the boat = 25(3-âˆš3) m/min.

Mode : We have Modal class = 22 - 30

[âˆµ Max. frequency 36 belongs to 20 - 30]

l = 20, h = 10, f

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