**CBSE Sample Papers for Class 9 SA2 MathsÂ Solved 2016 SetÂ 2**

**SECTION-A**

**1.A hemispherical bowl has radius of its rim equal to 4.2 cm. Find the total capacity of 30 such bowls.**

**2.The class marks of a frequency distribution are as follows : 15, 20, 25, … . Find the class corresponding to the class mark 20.**

**4. There are 40 students in 9 class of a school. Total number of students passed the examination are 28. If a student is selected at random, then find the probability that student has failed in the examination.**

**SECTION-B**

**6. Find the mean of the following distribution :**

**7. Squares each of side 6 cm are cut off from the four comers of a sheet of tin measuring 42 cm by 30 cm. The remaining sheet of the tin is made into an open box by folding up the flaps. Find the volume of the open box so formed.**

**8. A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4 cm. Find its :**

**(i) inner curved surface area**

**(ii) outer curved surface area**

**(iii) total surface area**

**9. A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m ^{2}. How many revolutions did it make ?**

**10. The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them in kg are 51, 53, 57, 52, 55, 56. Find the weight of the seventh student.**

**SECTION-C**

**Question numbers 11 to 18 carry 3 marks each.**

**11. ABCD is a parallelogram in which side AB is produced to E so that BE = AB. Prove that ED bisects BC.**

**12. Prove that the angle bisectors of a parallelogram form a rectangle.**

**13. A recent survey found that the ages of workers in a factory is distributed as follows :**

**If a person is selected at random, find the probability that the person is :**

**(i) 40 years or moreÂ Â Â Â Â Â Â (ii) having age from 30 to 39 years**

**(iii) under 60 but over 39 years**

**14. Construct an isosceles triangle whose perimeter is 12 cm and altitude is 4 cm.**

**15. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm.**

**16. A semicircular sheet of metal of diameter 28 cm is bent to form an open conicalÂ**

**cup. Find the capacity of the cup.**

**18. Following are the ages (in years) of 360 patients getting medical treatment in a hospital :**

**Draw histogram and frequency polygon for the above data.**

**SECTION-D**

**Question numbers 19 to 28 carry 4 marks each.**

**19. Amrit prepare a poster in the form of aÂ parallelogram, as shown in figure.**

**(ii) Which mathematical concept is used in this problem ?**

**(iii) By writing a slogan on poster which value is depicted by Amrit ?**

**21. If each diagonal of a quadrilateral separates it into two triangles of equal area, then show that the quadrilateral is a parallelogram.**

**23. Of any two chords of a circle show that the one which is nearer to the centre is larger.**

**25. Manu Sweet Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboad is Rs. 4 for 1000 cm ^{2}, find the cost of cardboard required for supplying 250 boxes of each kind.**

**26. Find the cost of sinking a tubewell 280 m deep, having diameter 3 m at the rate of Rs. 20 per cubic metre. Find also the cost of cementing its inner curved surface at Rs.13 per square metre.**

**27. Over the past 200 working days, the number of defective parts produced by a machine is given in the following table :**

**Determine the probability that tomorrowâ€™s output will have :**

**(i) no defective part**

**(ii) at least one defective part**

**(iii) not more than 5 defective partsÂ more than 13 defective parts**

**28. The monthly profit (in Rs.) of 100 shops are distributed as follows :**

**Draw a cumulative frequency curve by less than type method.**

**SECTION – E**

**Question numbers 29 to 31 carry total of 10 marks based on Open Text Theme provided by CBSE from Linear Equations in Two Variables.**

**Answers**

**SECTION-A**

**Question numbers 1 to 4 carry 1 mark each.**

**1.A hemispherical bowl has radius of its rim equal to 4.2 cm. Find the total capacity of 30 such bowls.**

**2.The class marks of a frequency distribution are as follows : 15, 20, 25, … . Find the class corresponding to the class mark 20.**

**Sol.** Here, class size = 20 – 15 = 5

**4. There are 40 students in 9 class of a school. Total number of students passed the examination are 28. If a student is selected at random, then find the probability that student has failed in the examination.**

**Sol. Â Â Â Â Â Â Â Â Â Â Â Â **Â Total number of students = 40

Number of students passed the examination = 28

.*.Â Â Â Â Number of students failed = 40 – 28 = 12

P(student fail in examination) =12/40=3/10

**SECTION-B**

**Question numbers 5 to 10 carry 2 marks each.**

**6. Find the mean of the following distribution :**

**7. Squares each of side 6 cm are cut off from the four comers of a sheet of tin measuring 42 cm by 30 cm. The remaining sheet of the tin is made into an open box by folding up the flaps. Find the volume of the open box so formed.**

**Sol.** Here, measurements of tin sheet are 42 cm by 30 cm and side of each square is 6 cm.

Length of open box = 42-6-6

Breadth of open box = 30 cm

Height of open box = 6 cm

Therefore, Â Volume of open box = l x b x h

= 30 x 18 x 6

=3240Â cm^{3Â }

**8. A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4 cm. Find its :**

**(i) inner curved surface area**

**(ii) outer curved surface area**

**(iii) total surface area**

**Sol.** Here,Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â inner radius (r) = 2 cm

Outer radius (R) = 2.2 cm

Height (h) = 77 cm

**9. A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m ^{2}. How many revolutions did it make ?**

**Sol :**Length of a cylindrical roller = 2.5 m

Radius of the cylindrical roller = 1.75 m

Area covered in one revolution = 2 x 22/7 Â x 1.75 x 2.5 = 27.5 m

^{2}

Total area covered by the roller = 5500 m

^{2}

Number of revolutions = 5500/27.5=200

**10. The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them in kg are 51, 53, 57, 52, 55, 56. Find the weight of the seventh student.**

**Sol.**Mean weight of 7 students is 55 kg.

.â€™. Total weight of 7 students = 7 x 55 = 385 kg

Total weight of 6 students = 51 + 53 + 57 + 52 + 55 + 56 = 324 kg

.â€™. Weight of the seventh student = 385 – 324 = 61 kg

Hence, the weight of the seventh student is 61 kg.

**SECTION-C**

**Question numbers 11 to 18 carry 3 marks each.**

**11. ABCD is a parallelogram in which side AB is produced to E so that BE = AB. Prove that ED bisects BC.**

**Sol.**Since ABCD is a parallelogram.

**12. Prove that the angle bisectors of a parallelogram form a rectangle.**

**13. A recent survey found that the ages of workers in a factory is distributed as follows :**

**If a person is selected at random, find the probability that the person is :**

**(i) 40 years or moreÂ Â Â Â Â Â Â (ii) having age from 30 to 39 years**

**(iii) under 60 but over 39 years**

**Sol.**Total number of workers = 200

(i) P(40 years or more) =135/200 =0.675 [ therefore, 86+46+3]

(ii) P( from 30 to 39 years) = 27/200 = 0.135

(iii) P(under 60 but over 39 years) =132/200 = 0.66 [ Therefore, 86+46]

**14. Construct an isosceles triangle whose perimeter is 12 cm and altitude is 4 cm.**

**Sol. Steps of Construction :**

1.Draw a line segment XY = 12 cm.

2.Draw â€˜pâ€™ the perpendicular bisector of XY and let it intersect XY in M.

3. Cut off AM = 4 cm.

4. Join AX and AY.

5. Draw the perpendicular bisectors of XA and YA and let they intersect XY in B andC.

6.Join AB and AC.

Thus, AABC is the required triangle.

**15. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm.**

**Sol.**Given that radius of the base (r)= 6 m

Height of the cone (h)= 8 m

= 3.14 x 6x 10 == 188.4 m

^{2}

Width of the tarpaulin= 3 m

Stitching margins + Wastage required= 20 cm

Therefore, Â Total length of the tarpaulin required =188.4/3 +22/100

=(62.8 +0.2) m =63 m

Hence, the length of tarpaulin of width 3 m is 63 m.

**16. A semicircular sheet of metal of diameter 28 cm is bent to form an open conicalÂ**

**cup. Find the capacity of the cup.**

**18. Following are the ages (in years) of 360 patients getting medical treatment in a hospital :**

**Draw histogram and frequency polygon for the above data.**

**Sol.**Required histogram and frequency polygon are shown in the graph.

**SECTION-D**

**Question numbers 19 to 28 carry 4 marks each.**

**19. Amrit prepare a poster in the form of aÂ parallelogram, as shown in figure.**

**(ii) Which mathematical concept is used in this problem ?**

**(iii) By writing a slogan on poster which value is depicted by Amrit ?**

**Sol.**Since sum of adjacent angles of a parallelogram is supplementary.

(ii) Properties of parallelogram.

(iii) Value depicted by Amrit in his slogan is ” Save electricity, Save energy”.

**21. If each diagonal of a quadrilateral separates it into two triangles of equal area, then show that the quadrilateral is a parallelogram.**

**23. Of any two chords of a circle show that the one which is nearer to the centre is larger.**

**Sol.**

**Given :**Two chords AB and CD of a circle C(0, r), such that OL < OM, where OL and OM are perpendiculars from O on AB and CD respectively.

**To Prove**: AB > CD

**Construction**: Join OA and OC.

**Â Proof**: Since the perpendicular from the centre of a circle to a chord bisects the chord.

**Justification**: Clearly, A lies on the perpendicular bisector ofCD.

Therefore, AC = AD

Now, Â Â Â Â Â Â Â Â Â Â BD = 6.8 cm

=>Â Â Â Â Â Â Â Â Â Â Â Â Â BA + AD = 6.8 cm

=>Â Â Â Â Â Â Â Â Â Â Â Â Â AB + AC = 6.8 cm

Hence, AABC is the required triangle.

**25. Manu Sweet Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboad is Rs. 4 for 1000 cm**

^{2}, find the cost of cardboard required for supplying 250 boxes of each kind.**Â Sol.Â**= 2 (lb + bh + hi)= 2(15 x 12 + 12 x 5 + 5 x 15)

= 2(180 + 60 + 75) – 630 cm

^{2 }

= 2 (lb + bh + hi)

= 2 (25 x 20 + 20 x 5 + 5 x 25)

= 2(500 + 100 + 125)

= 1450 cm

^{2}

Total surface area of one smaller box and one bigger box = (630 + 1450) = 2080 cm

^{2 }Total surface area of 250 boxes of each kind = 250 x 2080 = 520000 cm

^{2}

Extra area of cardboard required for overlapping = x 520000 = 26000 cm

^{2}

Total area of cardboard required for 250 boxes of each kind including overlapping

= 520000 + 26000 = 546000 cm

^{2}

Now, total cost of cardboard required at the rate of Rs. 4 for 1000 cm

^{2}

= Rs. 4/1000 x 546000 = Rs. 2184

**26. Find the cost of sinking a tubewell 280 m deep, having diameter 3 m at the rate of Rs. 20 per cubic metre. Find also the cost of cementing its inner curved surface at Rs.13 per square metre.**

**Sol.**Â Â Radius of the tubewell (r) = 3/2 m

Depth of the tubewell (h) = 280 m

**27. Over the past 200 working days, the number of defective parts produced by a machine is given in the following table :**

**Determine the probability that tomorrowâ€™s output will have :**

**(i) no defective part**

**(ii) at least one defective part**

**(iii) not more than 5 defective partsÂ more than 13 defective parts**

**Sol.**Total number of working days = 200Â

(i) Number of no defective part = 50

Therefore, Required probability = 50/ 200 =1/4

(ii) Number of at least one defective part = 150

Therefore, Required probability = 150/ 200 = 3/4

(iii) Number of not more than 5 defective parts = 50+320+22+18+12+12=146

Therefore, Required probability = 146/200 =73/100

(iv) Number of more than 13 defective parts = 0

Therefore, Required probability = 0

**28. The monthly profit (in Rs.) of 100 shops are distributed as follows :**

**Draw a cumulative frequency curve by less than type method.**

**Sol.**Let us prepare the cumulative frequency table by less than type method as given below :

Mark the upper limit on X-axis and cumulative frequency along Y-axis. Plot the points (50, 12), (100, 30), (150, 57), (200, 77), (250, 94) and (300, 100).

Join them to get requiredÂ cumulative frequency curve as shown in the graph.

**SECTION – E**

**Question numbers 29 to 31 carry total of 10 marks based on Open Text Theme provided by CBSE from Linear Equations in Two Variables.**

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