CBSE Sample Papers For Class 9 Maths Mock Paper 1

Q1. If x = 1 +

2

, then the value of x+(1/x) will be

Opt:

2

22

3

32

Ans:

22

Q2. If the volume of a cuboid is 4a² – 12a, then its possible dimensions are given by

Opt:

2, 2a and a – 3

2, 2a and a + 3

2, 2a and a – 4

2, 2a and a + 4

Ans: 2, 2a and a – 3

Q3. If ABCD is a square, X and Y are points on sides AD and BC respectively such that AY = BX, then

Opt: BY > AX

BY < AX

BY = BA

BY = AX

Ans: BY = AX

Q4. If PQRS is an isosceles trapezium, then the measure of R is

Opt:

68°

69°

70°

71°

Ans: 69°

Q5. At which point does the line 3x – 5y = –10 cuts y-axis?

Opt:

(0, 1)

(0, 2)

(0, 3)

(0, 4)

Ans:”

(0, 2)

Q6. PQRS is a parallelogram, whose one side is ‘a’ and the other side is ‘b’. If ABCD is a rectangle with same side ‘a’ and ‘b’, then

Opt:

area (ABCD) = (2/3) area (PQRS)

area (ABCD) ≤ area (PQRS)

area (ABCD) ≥ area (PQRS)

area (ABCD) = (3/4) area (PQRS)

Ans: area (ABCD) ≥ area (PQRS)

Q7. A point, whose x – coordinate is zero and y – coordinate is non-zero, will lie

Opt:

on the x-axis

on the y-axis

at the origin

in the first quadrant

Ans: on the y-axis

Q8. In fig. if AC = BD, then

Opt:

AB = CD

AC = CD

BC = CD

AB = BC

Ans: AB = CD

Q9.

In the figure, if AB║ CD, CD ║ EF and y : z = 3 : 7, then the value of x is

Opt:

126°

100°

90°

75°

Ans: 126°

Q10. A die is thrown 1000 times with the frequencies 183, 150, 153, 148, 176 and 190 for the outcomes 1, 2, 3, 4, 5 and 6 respectively. Now, the probability of getting 5 is

Opt:

0.824

0.810

0.190

0.176

Ans: 0.176

Q11. The ratio of volumes of two spheres are is 8 : 27. What is the ratio of their surface areas?

Opt:

4 : 9

5: 9

2 : 3

1 : 3

Ans: 4 : 9

Q12. If the base of an isosceles triangle is 12 cm and one of its equal sides is 14 cm, then the area of the triangle is

Opt:

2410 cm2. 1210 cm2. 24 cm2. 610 cm2.

Ans:

2410 cm2.

Q13. Two parallel chords of a circle whose diameter is 13 cm are respectively 5 cm and 12 cm in length. If both the chords lie in a semi-circle, then the distance between them is

Opt:

8.5 cm

5 cm

3.5 cm

3 cm

Ans: 3.5 cm

Q14. The mean score of 25 observations is 80 and the mean score of another 55 observations is 65, then the mean score of the whole data set is _______________.

Opt:

69.69

73.5

75

145

Ans: 69.69

Q15.

If x−1 is a factor of 2x2+kx+2,​ then the value of k is

Opt:

2+2 2−2 2−2 −2−2

Ans:

−2−2

Q16. A card is drawn from a packet of 100 cards numbered 1 to 100. The probability of drawing a number which is a perfect square is

Opt:

1/100

2/100

1/10

2/10

Ans: 1/10

Q17. If the point (2, 5) lies on the graph of the equation 5y = bx + 13, then the value of b will be

Opt:

5

6

7

8

Ans: 6

Q18. Two irrational numbers between 2 and 2.5 are

Opt:

5 and 2×5 5 and 2×5 5 and 2×5 2 and 5

Ans:

5 and 2×5

Q19. Direction for questions 19 & 20: In question numbers 19 and 20, a statement of

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

Assertion (A): In a cylinder, if radius is halved and height is doubled, the volume will remain same.

Reason (R):  Volume of cylinder is given by base area multiplied by its height.

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

Q20. Assertion(A): In rolling a die, the probability of getting a number 1 is

16\frac{1}{6}

61​​
.  Reason(R): When a die is rolled, the possible outcomes are the number 1, 2, 3, 4, 5 and 6.

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).

Q21. There is a slide in a park. One of its side walls has been painted in some colours with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15m, 11m and 6m, find the area painted in colour.

Ans:

Side walls is in the triangular form with sides a = 15 m, b = 6m and c = 11m.s = a+b+c2 =15+6+112 =322 = 16 mArea of side wall = ss-as-bs-c = 1616-1516-616-11 = 16×1×10×5 = 2×2×2×2×2×5×5 = 202 m2

 

Q22. A coin is tossed 100 times. It is observed that 60 times head comes up and 40 times tail comes up then find the probability that neither a head nor a tail comes up.

Ans: Total number of times in which either head
or tail comes up = 100
Number of times in which neither a head
nor a tail comes up = 0
Therefore,
The required probability = 0/100
= 0.

Q23. Find the zeroes of the quadratic polynomial x² – 4x +3.

Ans:

Let f(x) = x² – 4x +3 = x² – 3x – x +3
= x(x – 3) – (x – 3)
= (x – 1)(x – 3)
Zeroes of f(x) are given by f(x) = 0
Therefore, x – 1 = 0 or x – 3 = 0, i.e. x = 1 or x = 3.
Therefore, the zeroes of x² – 4x +3 are 1 and 3.”

Q24. A godown is in the form of a cuboid measuring 60 m x 40 m x 20 m. How many cuboidal boxes can be stored in it if the volume of one box 0.8 m³?

Ans:

Volume of one box = 0.8 m3Volume of one godown = 60×40×20 = 48000 m3Number of boxes that can ba stored in the godown = Volume of godownVolume of one box = 48000.8 = 60000 boxes.

Q25. The lateral surface of a cylinder is equal to the curved surface of a cone. If the radius is the same, find the ratio of the height of the cylinder and slant height of the cone.

Ans:

Let the radius of both = r, height of cylinder = h and slant height of cone = 1Lateral surface area of cylinder = 2πrhCorved surface area of cone = πrl2πrh = πrl Given 2h = 1h1 = 12 = 1:2

Q26. In the given figure, PQR and QST are two quadrilateral triangles such that S is the mid-point of QR.

Prove that arΔQST=12arΔPQT

Ans:

Since triangles PQR and QST are equilateral triangles.∴∠PRQ=∠RQT=60∘⇒QT∥PR⇒arΔPQT=arΔRQT...1Now,arΔQST=12arΔRQTST is a median of the ΔRQT⇒arΔRQT=2arΔQST...2Now,from1and2⇒arΔPQT=2arΔQSTorarΔQST=12arΔPQT

Q27. In the following figure, PQRS is a trapezium in which PQ || SR. Prove that ar(ΔQOR) = ar(ΔPOS).

Ans:

Since ΔPQR and ΔPQS are on the same base PQ and between the same parallels PQ and SR.
∴                    ar(ΔPQR) = ar(ΔPQS)
⇒   ar(ΔPQR) – ar(ΔPOQ) = ar(ΔPQS) – ar(ΔPOQ)
⇒                 ar(ΔQOR) = ar(ΔPOS)

Q28. Construct a triangle ABC in which BC = 3.4 cm , AB­­–AC = 1.5 cm and ∠B = 45°.

Ans:

Given:- BC=3.4 cm, AB-AC=1.5 cm and ∠B=45°
To Construct:- A ΔABC.

Steps of Construction
Step 1. Draw a line segment BC of length 3.4cm
Step 2. Draw an angle of 45 degree from point B
Step 3. From Ray AX cut off the line segment BD = 1.5cm
Step 4. Join B to C
Step 5. Draw side bisector of DC
Step 6 Extend side bisector of DC it intersect the Ray BX at point A
Step 7 Join A to C, ABC is the required triangle.

Q29. Find the value of a and b in the figure given below.

Ans:

Since ,we have∠POT=a=2∠PQT=2×60°=120°∵The angle subtended by an arc at the centre is doublethe angle subtended by it any point on the remaining part of the circle.∠PRS=b=12×120°+30°=75°∵The angle subtended by an arc at the centre is doublethe angle subtended by it any point on the remaining part of the circle.

Q30. A,B,C and D are four consecutive points on a circle such that AB = CD. Prove that AC = BD.

Ans:

Since, we have⇒AB=CD⇒ minor are AB = minor arc CD ∵If two arcs of a circle are congruent,then corresponding chords are equal.⇒ minor arc AB+minor arc BC = minor arc CD + minor arc BC⇒ArcABC≅ArcBCD⇒ChordAC=ChordBD∵ If two arcs of a circle are congruent,then corresponding chords are equal.⇒ Ac = BD

Q31. The median of the below data arranged in ascending order is 22. Find x.

8, 11, 13, 15, x+1, x+3, 30, 35, 40, 43

Ans:

Number of observations n = 10 which is evenMedian = value of n2th observation + value of n2+1th observation2Median = 5th observation + 6th observation2Median = x+1+x+32⇒ 22 = 2x+42⇒ 2x+4 = 44⇒ 2x = 40⇒ x = 20

Q32. Mean of 18 numbers is 57. If 9 is added to each number, find the new mean.

Ans:

Total numbers = 18 Mean = 57Mean = Sum of numbersTotal numbers57 = Sum of numbers18Sum of numbers = 57×18Now on adding 9 to each numberNew sum = 57×18+9×18New Mean = New sumTotal Numbers = 66×1818 = 66

Q33. In the figure, side QR of ΔPQR has been produced S, if P : Q : R = 3 : 2 : 1 and RT ⊥ PR, then ∠TRS will be

Ans:

Given, ΔPQR, side QR is us extended upto S and TR ⊥ PR,∠P:∠Q:∠R=3:2:1orP3=Q2=R1=KP=3K,Q=2K,R=K∠P+∠Q+∠R=180°6K=180°K=30°∠P=3K=90°∠Q=2K=60°∠R=K=30°⇒∠P=90°,∠Q=60°and​ ∠R=30°Hence, we have to find the value of ∠TRS.From the figure, we have∠PRT=90∘∴∠TRS=180°−∠PRQ+∠PRT=180°−30°+90°=180°−120°=60°

Q34. If x + y + z = 6 and xy + yz + zx = 11, then find the value of x² + y² + z²_.

Ans:”

Given:x+y+z=6 and xy+yz+zx=11Squaringbothsides,wegetx+y+z2=62⇒x2+y2+z2+2xy+2yz+2zx=36⇒x2+y2+z2+2xy+yz+zx=36⇒x2+y2+z2+211=36Since xy+yz+zx=​ 11⇒x2+y2+z2+22=36⇒x2+y2+z2=36−22=14

Q35. ABC and DBC are two isosceles triangles on the same base BC. Show that ∠ABD = ∠ACD.

Ans:

Join the line AD.
In ΔABD and ΔACD,
AB = AC        (Given)                                     ..(1)
AD = AD        (Common Side)                          ..(2)
Also, BD = CD                                               ..(3)
So, ΔABD ≅ ΔACD            (Using (1), (2), (3) and SSS rule).
This gives ∠ABD = ∠ACD             (By CPCT)

Q36. Draw the graph of x+y = 7.

Ans:

To draw the graph, we need at least two solutions.x+y = 7 ⇒ y = 7 −x x = 0 ⇒ y = 7−0⇒ y = 7 x = 7 ⇒ y = 7−7 ⇒ y = 00,7 and 7,0 are solutions of x+y = 7We​ can use the following table to draw the graph:x07y70

Q37. Solve the equation 2x + 1 = x – 3, and represent the solution(s) on
(i) the number line,
(ii) the Cartesian plane.

Ans:

We solve 2x+1 = x–3 and get x = –4
(i) The representation of the solution on the number line is shown in Fig, where x = – 4 is treated as an equation in one variable.

(ii)
We know that  x=-4 can be written as x+0.y=-4
To draw the graph, we need at least two solutions. Hence, two solutions of the given equation are
x = –4, y = 0 and x = –4, y = –2.

Q38. An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below:

Monthly income Vehicles per family (in Rs)	Vehicles per family 0 1 2 Above 2
Less than 7000 7000 – 10000 10000 – 13000 13000 – 16000 16000 or more	10 0 1 2 1 160 305 535 469 579 25 27 29 59 82 0 2 1 25 88

Suppose a family is chosen. Find the probability that the family chosen is
(i) earning Rs 10000 – 13000 per month and owning exactly 2 vehicles.
(ii) earning Rs 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than Rs 7000 per month and does not own any vehicle.
(iv) earning Rs 13000 – 16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.

Ans:

Total number of families=2400(i)​ Number of families earning Rs 10000– 13000 per month and owning exactly 2 vehicles=29P(exactly two vehicles)=292400(ii) Number of families earning Rs 16000 or more per month and owning exactly 1 vehicle=579P(exactly two vehicles)=5792400(iii) Number of families earning less than Rs 7000 per month and does not own any vehicle=10P(does not own any vehicle)=102400=1240(iv) Number of families earning Rs 13000–16000 per month and more than 2 vehicles=25P(more than 2 vehicles)=252400=196(v) Number of families owning not more than 1 vehicle=10+1+2+1+160+305+535+469+579=2062P(owning not more than 1 vehicle) =20622400=10311200

Q39. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
(i) ABCD is a square.
(ii) diagonal BD bisects ∠B as well as ∠D.

Ans:

Given:ABCD is a rectangle. Diagonal AC bisects ∠A as well as ∠C.To prove:(i) ABCD is a square.(ii)Diagonal BD bisects ∠B as well as ∠D. Proof:(i)Since, ∠1=∠2 and ∠3=∠4∵AB∥CD∴∠2=∠3[Alternate interior angles]⇒∠1=∠3⇒CD=AD [Opposite sides of equal angles are equal.] Since,adjecent sides of a rectangle are equal, so ABCD is a square.(ii)As ABCD is a square. So, AB=AD⇒∠ADB=∠ABD ...(i) [Opposite angles of equal sides are equal.]Since,AB∥CDSo,∠ABD=∠CDB ...(ii) [Alternate interior angles]⇒∠ADB=∠CDB MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@2BFD@ ⇒BD bisects ∠D.Similarly, we can prove that BD bisects ∠B.Therefore, BD bisects ∠B as well as ∠D. Hence proved.

Q40. Case Study – 2

Sam buys a plot that is in the shape of a parallelogram. He divides the plot into three parts, which are also in the shape of three different parallelograms, to plant wheat, rice, and sugar cane. He places the scarecrow at the corner of the field to scare the birds away. After visiting the land, a few questions came to his mind. Give answers to his questions by looking at the figure representing the division of plot given below.

i. What is the measure of ∠BAH and ∠AGD?

​ii. If BE = 70 ft and HG = GF = 15 ft, what is the length of the side AH of the wheat plot?

Ans:

i.  Angles on the same side of the transversal add up to 180^o.

So, ∠ABC + ∠HAB = 180^o

∠HAB = 180^o – 60^o

= 120^o

Further, since ABCH and HCDG are parallelograms, ABDG is a parallelogram. In a parallelogram, opposite angles are equal.

So, ∠ABD = ∠AGD = 60^o

ii. Given: ABEF is a parallelogram.

Opposite sides of a parallelogram are equal.

So, AF = BE = 70 ft

Given: HG = GF = 15 ft

AF = AH + HG + GF

70 = AH + 15 + 15

AH = 70 – 30

AH = 40 ft