grade x mathematics (standard) #growwithgreen mock test · 2019-12-31 · of 1 mark each, two...

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Grade X

Mathematics (Standard) Mock Test

Math Mock Test(Standard) Grade - X

Time allowed: 3 Hours Maximum Marks: 80 General Instructions: (i) All 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. (vi) Use of calculators is not permitted.

Section A

Q 1- 10 are multiple choice questions. Select the most appropriate answer from the given options.

1. The smallest prime factor of two positive integers x and y are 5 and 7 respectively. What is the smallest prime factor of (x + y)?

A. 2 B. 3 C. 5 D. 7

2. If the zeroes of the quadratic polynomial, , are positive, then A. c and a have the same sign B. b and c have opposite signs C. both A and B D. none of the above

3. If the graphs of the equations intersect at exactly one point, then k cannot have its value as

A. B. C. 4 D. 8

4. If the points (2, 1), (0, 0) and (y, x) are collinear, then the value of (y − 2x)2 is A. 0 B. 1 C. 4 D. 9

5. The 5th term of the AP is A. B. C. D.

6. The areas of two similar triangles ∆ABC and ∆DEF are 81 cm2 and 144 cm2 respectively. If the longest side of ∆ABC be 6cm and the shortest side of ∆ABC be 2 cm, then the ratio between the longest side to the smallest side of the ∆DEF is

A. 3 B. 4 C. 5 D. 6

7. If , then the value of is

A.

B.

C.

D.

8. A semicircular sheet of radius 6 cm is rolled to form a right circular hollow cone. The radius of the cone is

A. 2 cm B. 3 cm C. 4 cm D. 5 cm

9. If has a common root then the value of ab, where b is not equal to zero is

A. 0 B. 1 C. 2 D. can not be determined

10. A number x is chosen at random from the numbers , −3, −2, −1, 0, 1, 2, 3, 4 the probability that | x | 2 is

A.

B.

C.

D.

(11 – 15) Fill in the blanks:

11. For any prime number p greater than 3, p2 upon division by 12, gives the remainder as x, the value of x2 is _____.

12. If the point (a, b) is equidistant from the points (4, 6) and (3, ), then the value of is _____.

13. The value of is _____.

OR

If then is equal to _____.

14. A line intersects a circle at two distinct points, then the distance between these two points is always _____ than or equal to the diameter of the circle.

15. The term _____ is inserted in between , such that the resulting sequence is an AP with common difference y.

(16 – 20) Answer the following:

16. Without performing actual division, find out whether the number will have a terminating or non-terminating decimal.

17. If ΔABC is right angled at 'C', then find the values of cos (A + B) and sin (A + B).

18. If the points (8, a), (a, −1), and (0, −6) are collinear, then find the value of a.

OR

Find the distance between A(a + b, b – a) and B(a – b, a + b).

19. In the given figure, AB and CD intersect at O. If AO = p, then find the length of OD in terms of p.

20. The mean of a data is If and h = 10, then find the assumed mean (A).

Section B

21. A farmer Babu has 10 cows and tied them at an equal distance of 4 m from each other in

a straight line along with his haystack as shown in the figure. Babu feeds all the cows separately starting from the haystack and he returns to the haystack after feeding each cow to get hay for the next. After feeding the last cow he rests there for a while and doesn't get back to the haystack. If the distance of the nearest cow from the haystack is 8 m, find (i) The total distance Babu will cover in order to feed all the cows. (ii) The time taken by Babu to feed all the cows, if Babu moves at a speed of 4 m/s.

22. The given figure shows three congruent triangles ΔABC, ΔBDE and ΔDFG. Prove that CE = EG.

OR

In the given figure, ∠ACB = ∠CDB

What is the length of AD?

23. From the top of a 10 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower.

OR

A straight highway leads to the foot of a tower. From the top of the tower, the angles of depression of two cars standing on the highway are observed as 30° and 45°. If the

distance between these two cars is 50 m, then find the height of the tower. [Use = 1.73]

24. A tangent AB at a point A of a circle of radius 7 cm meets a line through the centre O at a point B so that OB is 25 cm. Find the length of the tangent AB.

25. How many solid cones each of radius 2 cm and height 3 cm should be melted completely to form a solid sphere of radius 6 cm?

26. Anmol plays an online battle royale game. There is a wide range of weapons available in the game, but Anmol always uses an assault rifle for close-range and a sniper for long-range. The probability that Anmol misses a shot of the sniper is 0.3, and the probability that he successfully knocks down the opponent using the assault rifle is 0.4. If he took 20 shots of the sniper and 35 sprays of the assault rifle in a single match. Find (i) The sprays missed by Anmol in a close-range fight. (ii) The shots hit by Anmol on target in a long-range fight.

Section C

27. Prove that is irrational.

28. A polynomial 20x5 − 12x3 − 40x2 + 31, on division by 5x2 − 3, gives R as the remainder. Find the zeroes of the polynomial 2x2 + Rx − 15.

OR

The remainder obtained on dividing 6x3 + kx + 14x2 − 8 by 3x + 7 is 13.

Find the quotient and the value of k.

29. Find the solution of the pair of linear equations and .

OR

Solve the following pair of linear equations by cross-multiplication method.

4x − 3y = 5xy

2x − y = − 2xy

30. If the first term of an A.P. is 20 and the sum of its first 24 terms is −348, then find the first four terms of the A.P.

31. Rahul, a resident of Chandigarh got admission to a college situated in New Delhi. His college, the bus stand, and the metro station are situated in a straight line, such that the metro station is in between the bus stand and the college, and the distance between the metro station and his college is thrice the distance between the bus stand and the metro station. If the coordinates of the bus stand are (3, 4) and the coordinates of his college are ( ), (i)What are the coordinates of the metro station? (ii)What are the coordinates of Manu's place, if Rahul's friend Manu lives halfway between the college and the metro station? (iii)What is the distance between Manu's place and the bus stand?

32. If cosec A + cot A = x, then find the values of tan A, sin A, and cos A in terms of x.

OR

If then what is the value of cos2θ?

33. The given figure shows two concentric circles with centre O. The radii of the outer and inner circles are respectively 12 cm and 9 cm.

If the area of the shaded region is 55 cm2, then find the measure of ∠AOB.

34. If the mode of the following distribution is 68, then find the missing frequency corresponding to the class interval 60 − 80.

Class interval 0 − 20 20 − 40 40 − 60 60 − 80 80 − 100

Frequency 7 14 8 - 3

Section D

35. Find the roots of the equation pqx2 − (4qr − p2) x − 4pr = 0 by using the quadratic formula.

OR

The compound interest on a sum of Rs 80,000 at a certain rate of interest compounded annually for 2 years is Rs 13,312. By factorising quadratic equation, find the rate of interest.

36. The given figure shows an equilateral triangle ABC drawn on side BC of square BCED and another equilateral triangle FDC drawn on diagonal DC of the square BCED.

What is the ratio of the area of ΔABC to the area of ΔFDC?

OR

In the given figure, DE || BC and AM ⊥ BC.

If DE = 8 cm, BC = 10.4 cm, and ar (BCED) = 16.56 cm2, then find the length of AN.

37. The angle of elevation of top of the first building from the top and bottom of the second building are 30° and 45° respectively. Find the ratio of the height of second building to that of the first building.

OR

Arjun standing on a horizontal plane observes that a bird at a distance of 100 m from him is at an elevation of 60°. Aditi standing on a roof of a 30 m tall building observes that the same bird is making an angle of elevation of 45°. Both Arjun and Aditi are on the opposite sides of the bird. Find the distance of the girl from the bird.

38. Draw a circle of radius 4 cm. From a point 6 cm away from its centre, construct a pair of tangents to the circle and measure their lengths.

39. A solid iron pillar is in the shape of a cylinder and is surmounted by a cone of height 9 cm. The cylindrical portion is of height 100 cm and has a base diameter 14 cm. Find the mass of the pillar, if the mass of 1 cm3 of iron is 8 g.

40. The given data shows the marks obtained by students in a test.

Marks obtained Less than 10

Less than 20

Less than 30

Less than 40

Less than 50

Number of students 3 11 24 40 50

Find the median of the data graphically by drawing both the ogives.

Maths Mock Test Solutions

1. We have,

Since 2 is neither a factor of x nor a factor of y

Therefore both x and y are odd

(x + y) is even

Thus, the smallest prime factor of (x + y) is 2.

Hence, the correct answer is option A.

2. Let the given quadratic polynomial be f(x) = .

Suppose and be the zeroes of the given polynomial.

Thus, ,

But, ,

∴ , which is possible only when a and c both have the same sign and b and c both have the opposite sign.

Hence, the correct answer is option C.

3. We have,

The given system of equations has a unique solution and

Thus,


4. We have,

Now, the given points are collinear


5. We have,

The given AP can be rewritten as

Now, the common difference of this AP =

Thus, 5th term =

Hence, the correct answer is option C.

6. We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Similarly, the shortest side of DEF = cm

Thus, the ratio between the longest side to the smallest side of the ∆DEF =


7. Given: secx =

Now,


8. If a semicircular sheet of radius 6 cm is rolled to form a right circular hollow cone, then the radius of semicircular sheet = slant height of the cone.

And, the total area of the semicircular sheet = curved surface area of the cone

Hence, the correct answer is option B.

9.

Since is a root of also, then


10. The total numbers are 9.

Number x such that | x | 2 are −2, −1, 0, 1, 2

Total numbers x such that | x | 2 are 5

We know that probability =

Thus, the probability of getting a number x such that | x | 2 is equal to

Hence, the correct answer is option D.

11. We have,

Any prime number p greater than 3 can be represented in the form of , where n is a natural number.

So, square of any prime number p greater than 3 can be represented as,

p2 upon division by 12 gives the remainder as 1

Thus, For any prime number p greater than 3, p2 upon division by 12, gives the remainder as x, the value of x2 is .

12. We have,

Since the point (a, b) is equidistant from the points (4, 6) and (3, )

Thus, if the point (a, b) is equidistant from the points (4, 6) and (3, ), then the value of is .

13.

Hence, The value of is 1.

OR

Comparing with ,

a = cosx, b = sinx

Thus,

Hence, If then is equal to 1.

14. We have,

A line intersects a circle at two distinct points

Thus, the line segment joining these two points is a chord of the circle and length of the chord is always less than or equal to the diameter.

Hence, a line intersects a circle at two distinct points, then the distance between these two points is always less than or equal to the diameter of the circle.

15. We have,

The term to be inserted in between , such that the resulting sequence is an AP is the arithmetic mean of

Now, AM of =

So, the resulting sequence is: and the common difference is

Thus, the term x is inserted in between , such that the resulting sequence is an AP with common difference y.

16.

It can be observed that denominator of the simplified fraction is of the form 2n 5m, where n and m are non-negative integers.

Thus, will have a terminating decimal expansion.

17. In ΔABC,

A + B + C = 180º

Also,

18. The given points are collinear. Therefore, the area of triangle formed by these points is zero. That is,

∴ Required value of a is 4 or −10.

OR

The given points are A(a + b, b – a) and B(a – b, a + b).

19. In ΔAOC and ΔDOB:

∠A = ∠D = 110°

∠AOC = ∠DOB (Vertically opposite angles)

ΔAOC ~ ΔDOB (AA similarity criterion)

In similar triangles, corresponding sides are proportional.

Thus, the length of OD is .

20. Given: , and c = 10

We know

21. We have,

(i) Distance travelled by Babu to feed the first cow = 8 m

Distance travelled by Babu to feed the next, i.e. second cow = (8 + 8 + 4) m = 20 m

Distance travelled by Babu to feed the next, i.e. third cow = (4 + 8 + 8 + 4 + 4) m = 28 m

Distance travelled by Babu to feed the next, i.e. fourth cow = (4 + 4 + 8 + 8 + 4 + 4 + 4) m = 36 m

Distance travelled by Babu to feed the next, i.e. fifth cow = (4 + 4 + 4 + 8 + 8 + 4 + 4 + 4 + 4) = 44 m

We can clearly observe that the distance travelled by Babu separately from feeding the second cow till feeding the tenth cow forms an AP with the first term, a = 20 and, common difference, d = 8 and has a total of 9 terms

Thus, distance travelled by Babu to feed the tenth cow = 84 m

Now, the sum of this AP,

Thus, the total distance covered by Babu in order to feed all the cows = 8 + 468 = 476 m

(ii) The time taken by Babu to feed all the cows =

22. It is given that ΔABC ≅ ΔBDE ≅ ΔDFG

∴ AB = BD = DF ... (1)

BC = DE = FG ... (2)

AC = BE = DG ... (3)

∠BAC = ∠DBE = ∠FDG ... (4)

Now, ∠ABC + ∠EBC + ∠DBE = 180°

⇒ ∠EBC = 180° − 90° − ∠DBE = 90° − ∠DBE

Similarly, ∠GDE = 90° − ∠FDG

Using equation (4): ∠EBC = ∠GDE ... (5)

In ΔBCE and ΔDEG:

BC = DE {From (2)}

BE = DG {From (3)}

∠EBC = ∠GDE {From (5)}

∴ ΔBCE ≅ ΔDEG (By SAS congruency criterion)

It is known that the corresponding parts of congruent triangles are equal.

∴ CE = EG

OR

In ΔABC and ΔCBD:

∠ABC = ∠CBD [Common]

∠ACB = ∠CDB [Given]

ΔABC ∼ ΔCBD [By AA similarity criterion]

∴ AD = AB − BD = (7.2 − 3.2) cm = 4 cm

Thus, the length of AD is 4 cm.

23.

Let AB be the building of height 10 m and CD be the tower. Let the height of the tower be h and the distance between the building and the tower be x.

In right ΔABD,

∴ x = 10 m

In ΔAEC, we obtain

⇒ h = 10 × (1.732 + 1) m = 10 × 2.732 m = 27.32 m

Thus, the height of the tower is 27.32 m.

OR

The given information can be represented diagrammatically as:

Here, AB represents the tower. The points C and D represent the position of two cars on the highway.

According to the given information:

CD = 50 m, ∠ACB = 45°, and ∠ADB = 30°

Let AB = x

In ΔABD:

tan 30°

⇒ BD = x

In ΔABC:

tan 45°

1

⇒ BC = x

It is given that: CD = 50 m

⇒ BD − BC = 50 m

⇒ x − x = 50 m

⇒ x = 50 m

= 25 (1.73 + 1) m

= 25 × 2.73 m = 68.25 m

Thus, the height of the tower is 68.25 m.

24. The figure representing the given information can be drawn as follows:

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OA ⊥AB

Applying Pythagoras Theorem in right ΔOAB, we obtain

OB2= OA2+ AB2

⇒ (25 cm)2 = (7 cm)2+ AB2

⇒ AB2= 625 cm2− 49 cm2= 576 cm2

Thus, the length of the tangent AB is 24 cm.

25. Let n be the required number of solid cones.

It is known that the volume remains the same in conversion of solids from one shape to another.

∴Volume of n cones = Volume of sphere

⇒ n × Volume of a cone = Volume of sphere

It is known that:

Volume of cone

Volume of sphere =

Thus, the required number of cones is 72.

26. We have,

(i) Total number of sprays of the assault rifle by Anmol in the game = 35

Now, since Anmol always uses an assault rifle in a close-range fight

P(sprays on target in a close-range fight) = 0.4

Thus, P(sprays not on target in a close-range fight) =

Hence, sprays missed by Anmol in a close-range fight =

(ii) Total number of shots of the sniper by Anmol in the game = 20

Now, since Anmol always uses a sniper in a long-range fight

P(shots not on target in a long-range fight) = 0.3

Thus, P(shots on target in a long-range fight) =

Hence, shots hit by Anmol on target in a long-range fight =

27. Let us assume that is rational number.

So, we can find integers and ( 0) such that .

Suppose and have a common factor other than 1. Then, we divide by that common

factor to get , where and are co-prime.

Squaring both sides, we get :

3 divides 5 2

But 3 does not divide 5

3 divides 2

3 divides .

So, we can write , for some integer .

Put this value of in (1), we obtain :

15 divides 2 and so 15 divides .

3 and 5 both divides .

and both have at least 3 as a common factor. But this contradicts the fact that and have no common factors.

Our assumption was wrong.

So, we conclude that is irrational.

28. By using long division method, the polynomial 20x5 − 12x3 − 40x2 + 31 can be divided by 5x2 − 3 as:

Therefore, the given quadratic polynomial becomes 2x2 + 7x − 15.

This polynomial can be factored as:

= 2x2 + 10x − 3x − 15

= 2x (x + 5) − 3 (x + 5)

= (2x − 3) (x + 5)

Thus, the zeroes of the quadratic polynomial are and −5.

OR

Let p(x) = 6x3 + kx + 14x2 − 8

It is given that when p(x) is divided by 3x + 7, it gives 13 as the remainder.

Therefore, by remainder theorem:

According to division algorithm:

Dividend = Divisor × Quotient + Remainder

Let q(x) be the quotient.

∴ 6x3 + 14x2 − 9x − 8 = (3x + 7) × q(x) + 13

(3x + 7) × q(x) = 6x3 + 14x2 − 9x − 8 − 13

(3x + 7) × q(x) = 6x3 + 14x2 − 9x − 21

Thus, the required quotient is 2x2 − 3.

29. The pair of linear equations is given as:

Multiplying (1) by and (2) by :

Subtracting (4) from (3):

Substituting in (1):

Thus, the required solution is .

OR

4x − 3y = 5xy

2x − y = −2xy

On dividing the given equations by xy, we obtain

Taking , the given equations can be written in the form

4v − 3u − 5 = 0 … (1)

2v − u + 2 = 0 … (2)

The given equations can be solved by cross-multiplication method as

Therefore, is the required solution of the given pair of equations.

30. First term of the A.P. (a) = 20

Sum of first 24 terms (S24) = −348

We know that the sum of first n terms is given by

⇒ −29 = 40 + 23 d

⇒ −29 − 40 = 23d

⇒ −69 = 23d

First term (a) is given as20.

Second term (a2) = 20 − 3 = 17

Third term (a3) = a2 + d = 17 − 3 = 14

Fourth term (a4) = a3 + d = 14 − 3 = 11

Thus, the first four terms of the A.P. are 20, 17, 14, and 11.

31. We have,

Let the college is located at C( ), the metro station is located at M(x, y), and the bus stand is located at B(3, 4).

(i) Now, M divides CB in a ratio 3 : 1, such that

Using section formula, we get

Thus, the coordinates of the metro station are .

(ii) Since, Manu lives halfway between the college and the metro station

Thus, the coordinates of Manu's place are

(iii) Using Distance formula, we get

The distance between Manu's place and the bus stand = =

32. It is given that,

cosec A + cot A = x (1)

We know that, cosec2A − cot2A = 1

∴ (cosecA + cotA) (cosecA − cotA) = 1

⇒ x(cosecA − cotA) = 1

(2)

Adding equations (1) and (2), we obtain

OR

Thus, the value of cos2θ is .

33.

Let ∠AOB = θ

External radius, r1 = 12 cm

Internal radius, r2 = 9 cm

Area of the shaded region

It is given that area of the shaded region = 55 cm2

Thus, ∠AOB = 100°

34. Let the missing frequency corresponding to the class interval 60 − 80 be f.

Since the mode is 68, which lies in the class interval 60 − 80, the modal class of the given data is 60 − 80.

From the table, we can find that:

Lower class limit, l= 60

Class size, h= 20

Frequency (f1) of modal class = f

Frequency (f0) of class preceding the modal class = 8

Frequency (f2) of class succeeding the modal class = 3

Thus, the missing frequency of the class interval 60 − 80 is 18.

35. The given equation is: pqx2 − (4qr − p2) x − 4pr = 0

Here, a = pq, b = −(4qr − p2), c = − 4pr

Thus, the roots are and .

OR

Let the rate of interest R be x% per annum.

Amount, A received after 2 years = P + C.I. = Rs 80,000 + Rs 13,312 = Rs 93,312

Time period, n = 2 years

A = P

Since the rate of interest cannot be negative, x = 8

Thus, the rate of interest is 8% per annum.

36. BCED is a square and DC is its diagonal which divides it into two congruent right triangles.

Applying Pythagoras theorem in right ΔDBC, we obtain

DC2 = BC2 + BD2

= BC2 + BC2 (BD = BC)

DC2 = 2BC2 ...(1)

Since ΔABC and ΔFDC are equilateral triangles, each of their angles measures 60°.

∴ ∠A = ∠F = 60°

∠B = ∠D = 60°

∴ ΔABC ∼ ΔFDC (By AA similarity)

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

[From (1)]

Thus, the ratio of the area of ΔABC to the area of ΔFDC is 1 : 2.

OR

In ΔADE and ΔABC:

∠DAE = ∠BAC [Common]

∠ADE = ∠ABC [Corresponding angles]

∴ΔADE ∼ ΔABC [AA similarity criterion]

Area of trapezium is given by: × Sum of parallel sides × Height

∴ Area of BCDE = × (DE + BC) × MN

⇒ 16.56 cm2 = × (8 + 10.4) cm × MN

In ΔABM, DN is parallel to BM. Therefore, by B.P.T:

⇒ 13 AN = 10 AN + 18 cm

⇒ AN = 6 cm

Thus, the length of AN is 6 cm.

37. The given information can be represented diagrammatically as:

Here, AB and CD represent the first and second building.

According to the given information:

∠ACE = 30° and ∠ADB = 45°, where E is a point on AB such that CE || BD

Let AB = y

In ΔABD:

tan 45°

1

⇒ BD = y

∴ CE = y

In ΔAEC:

tan 30°

∴ DC = BE = AB − AE

Thus, required ratio

OR

In the figure, C denotes the position of the bird, A denotes the position of Arjun, and F denotes the position of Aditi. It is given that the bird is at a distance of 100 m from Arjun.

∴AC = 100 m

It is given that Aditi is standing on a roof of a 30 m tall building.

∴FD = EB = 30 m

In ΔACB,

In ΔCEF,

Thus, Aditi is at a distance of 79.9 m from the bird.

38.

Step of construction

Step: I- First of all we draw a circle of radius AB = 4 cm.

Step: II- Mark a point P from the centre at a distance of 6 cm from the point O.

Step: III -Draw the perpendicular bisector of OP, intersecting OP at Q.

Step: IV- Taking Q as centre and radius OQ = PQ, draw a circle to intersect the given circle at T and T’.

Step: V- Join PT and PT’ to obtain the required tangents.

Thus, PT and PT' are the required tangents.

The length of PT = PT' 4.5 cm

39. The iron pillar can be represented by a figure as:

Radius of the cylindrical portion, r =

Height of the cylindrical portion, h = 100 cm

Height of the conical portion, H = 9 cm

Radius of the conical portion, r = 7 cm

∴Volume of the pole

= Volume of the cylindrical portion + Volume of the conical portion

It is given that the mass of 1 cm3 of iron is 8 g.

∴Mass of 15,862 cm3 of iron = 15,862 × 8 g = 1,26,896 g

Thus, the mass of the pole is 1,26,896 g.

40. Firstly, the frequency distribution table for the given cumulative distribution table can be drawn as:

Marks obtained Frequency

0 − 10 3

10 − 20 8

20 − 30 13

30 − 40 16

40 − 50 10

A more than type frequency distribution table can be drawn as:

Marks obtained Cumulative Frequency

More than 0 50

More than 10 47

More than 20 39

More than 30 26

More than 40 10

Plotting the points on the graph and joining them to obtain the ogives:

A perpendicular to the x-axis is drawn from the point where both the ogives intersect. It can be observed that the point is 37.5.

Thus, the median of the data is 37.5.

grade x mathematics (standard) #growwithgreen mock test · 2019-12-31 · of 1 mark each, two...

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