time allowed : 3 hours maximum marks : 90 €¦ · in a certain test, nikita scored marks x and y...

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MATHEMATICS / ªÁáÊÃ Class – IX / ∑§ˇÊÊ - IX

ÁŸœÊ¸Á⁄UÃ ‚◊ÿ — 3 ÉÊá≈U •Áœ∑§Ã◊ •¥∑§ — 90

Time allowed : 3 hours Maximum Marks : 90

‚Ê◊Êãÿ ÁŸŒ̧‡Ê —

(i) ‚÷Ë ¬˝‡Ÿ •ÁŸflÊÿ¸ „Ò¥–

(ii) ß‚ ¬˝‡Ÿ ¬òÊ ◊¥ 31 ¬˝‡Ÿ „Ò¥, Á¡ã„¥ ¬Ê¥ø πá«UÊ¥ •, ’, ‚, Œ ÃÕÊ ÿ ◊¥ ’Ê¥≈UÊ ªÿÊ „Ò– πá«U-• ◊¥ 4 ¬˝‡Ÿ „Ò¥ Á¡Ÿ◊¥ ¬˝àÿ∑§ 1 •¥∑§ ∑§Ê „Ò, πá«U-’ ◊¥ 6 ¬˝‡Ÿ „Ò¥ Á¡Ÿ◊¥ ¬˝àÿ∑§ ∑§ 2 •¥∑§ „Ò¥, πá«U-‚ ◊¥ 8 ¬˝‡Ÿ „Ò¥ Á¡Ÿ◊¥ ¬˝àÿ∑§ ∑§ 3 •¥∑§ „Ò¥, πá«U-Œ ◊¥ 10 ¬˝‡Ÿ „Ò¥ Á¡Ÿ◊¥ ¬à̋ÿ∑§ ∑§ 4 •¥∑§ „Ò¥ ÃÕÊ πá«U-ÿ ◊¥ ◊ÈQ ¬Ê∆ ¬⁄U •ÊœÊÁ⁄UÃ 3-3 •¥∑§Ù¥ ∑§ 2 ¬˝‡Ÿ ÃÕÊ 4 •¥∑§Ê¥ ∑§Ê 1 ¬˝‡Ÿ „Ò–

(iii) ß‚ ¬˝‡Ÿ ¬òÊ ◊¥ ∑§Êß ̧÷Ë ‚flÊ¬̧Á⁄U Áfl∑§À¬ Ÿ„Ë¥ „Ò– (iv) ∑Ò§‹∑È§‹≈U⁄U ∑§Ê ¬˝ÿÊª flÁ¡¸Ã „Ò– General Instructions :

(i) All questions are compulsory. (ii) The question paper consists of 31 questions divided into five sections A, B, C ,D and E.

Section-A comprises of 4 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 and Section-D comprises of 10 questions of 4 marks each. Section E comprises of two questions of 3 marks each and 1 question of 4 marks from Open Text theme.

(iii) There is no overall choice. (iv) Use of calculator is not permitted.

πá«U-• / SECTION-A

¬˝‡Ÿ ‚¥ïÿÊ 1 ‚ 4 ◊¥ ¬˝àÿ∑§ ∑§Ê 1 •¥∑§ „Ò– Question numbers 1 to 4 carry one mark each.

1 ;fn 3 9,x ky+ = ∑§Ê •Ê‹π fcUnq (1, –2) ls xqtjrÊ gS] rks k dk eku Kkr dhft,A 1

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If the graph of 3x + ky = 9, passes through the point (1, – 2), find k.

2 fcUnq (0] 0) ls nks pj okys fdrus jSf[kd lehdj.kksa ds vkys[k xqtj ldrs gSa \

How many graphs of linear equations in two variables can pass through point (0, 0) ?

1

3 ‚◊Ê¥Ã⁄U øÃÈ÷¸È¡ ∑§ ©‚ Áfl‡Ê· ¬˝∑§Ê⁄U ∑§Ê ŸÊ◊ ’ÃÊß∞, Á¡‚∑§ Áfl∑§áÊ¸ ’⁄UÊ’⁄U „ÊÃ „Ò¥ ÃÕÊ ¬⁄US¬⁄U ‚◊Ám÷ÊÁ¡Ã ∑§⁄UÃ

„Ò¥– ‚ÊÕ „Ë, •ÊSÊãŸ ÷È¡Ê•Ê¥ ∑§Ê ∞∑§ ÿÈÇ◊ •‚◊ÊŸ „Ò– ©‚∑§ ∞∑§ •ÊÒ⁄U ªÈáÊ ∑§Ê ∑§ÕŸ ŒËÁ¡∞–

Name the special type of parallelogram whose diagonals are equal and bisect each other. Also,

a pair of adjacent sides is unequal. State its one more property.

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4 •ÊœÊ⁄U ÁòÊíÿÊ r •ÊÒ⁄U ™°§øÊß¸ h flÊ‹ ’‹Ÿ ∑§Ê ¡Ê◊ÈŸË ¬¥≈U ‚ ÷⁄UË „Èß¸ ’ÊÀ≈UË ◊¥ ™§äflÊœ̧⁄U •ÊœÊ «ÈU’ÊÿÊ ªÿÊ– ©‚

¬Îc∆UËÿ ˇÊòÊ»§‹ ∑§Ê ôÊÊÃ ∑§ËÁ¡∞ ¡Ê Á∑§ ¬¥≈U „Ê ªÿÊ–

A cylinder of base radius r and height h is dipped vertically to half the height in a bucket full

of purple paint. Find the area of the surface which gets painted.

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πá«U-’ / SECTION-B

¬˝‡Ÿ ‚¥ïÿÊ 5 ‚ 10 ◊¥ ¬˝àÿ∑§ ∑§ 2 •¥∑§ „¥Ò–

Question numbers 5 to 10 carry two marks each.

5 ŒË „Èß¸ •Ê∑Î§ÁÃ ◊¥, ∆ ABC ∑§Ê ˇÊòÊ»§‹ 18 cm2 „Ò– ÿÁŒ D •ÊÒ⁄U E ∑§̋◊‡Ê— ÷È¡Ê BC •ÊÒ⁄U ◊ÊÁäÿ∑§Ê AD ∑§ ◊äÿ-

Á’¥ŒÈ „Ò¥, ÃÊ ∆BED ∑§Ê ˇÊòÊ»§‹ ôÊÊÃ ∑§ËÁ¡∞–

In the given figure, area of ∆ABC=18 cm2. If D and E are the mid - points of the side BC and

median AD respectively, find the area of the ∆BED.

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6 M§‹⁄U •ÊÒ⁄U ¬⁄U∑§Ê⁄U ∑§ ¬˝ÿÊª ‚ 4∠XYZ ∑§Ë ⁄UøŸÊ ∑§ËÁ¡∞, ¡’Á∑§ ∠XYZ=20� „Ò–

Using ruler and compass, construct 4∠XYZ if ∠XYZ=20�.

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7 PQR ∞∑§ ÁòÊ÷È¡ „Ò– ÿÁŒ P, Q •ÊÒ⁄U R ‚ „Ê∑§⁄U ∑˝§◊‡Ê— ÷È¡Ê•Ê¥ QR, PR •ÊÒ⁄U PQ ∑§ ‚◊Ê¥Ã⁄U ⁄UπÊ∞° πË¥øË ¡ÊÃË

„Ò¥, Á¡‚‚ ∆ABC ¬Ê̋åÃ „ÊÃÊ „Ò, ¡Ò‚Ê Á∑§ •Ê∑Î§ÁÃ ◊¥ Œ‡ÊÊ¸ÿÊ ªÿÊ „Ò, ÃÊ Œ‡ÊÊß̧∞ Á∑§ 1

PQ AB2

= „Ò–

PQR is a triangle. If lines drawn through P, Q and R are parallel respectively to the sides QR,

PR and PQ and form ∆ABC as shown in the figure, show that 1

PQ AB.2

=

2

8 ∞∑§ ∆UÊ‚ œÊÃÈ ∑§ ’‹Ÿ ∑§Ë ÁòÊíÿÊ 10.5 cm •ÊÒ⁄U ™°§øÊß ̧60 cm „Ò– ’‹Ÿ ∑§Ê Œ˝√ÿ◊ÊŸ ôÊÊÃ ∑§ËÁ¡∞, ¡’Á∑§ œÊÃÈ ∑§Ê

ÉÊŸàfl 5 ª̋Ê◊ ¬˝ÁÃ cm3 „Ò–

The radius of a solid metal cylinder is 10.5 cm and height is 60 cm. Calculate the mass of

cylinder, if the density of metal of cylinder is 5 gm per cm3.

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9 ∞∑§ ¬Ê¥‚Ê 250 ’Ê⁄U »¥§∑§Ê ªÿÊ ÃÕÊ ¬Á⁄UáÊÊ◊ ß‚ ¬˝∑§Ê⁄U ¬˝ÊåÃ „È•Ê —

¬˝ÊåÃ ‚¥ïÿÊ 1 2 3 4 5 6

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’Ê⁄Uï’⁄UÃÊ 40 45 35 38 52 40

¬˝ÊÁÿ∑§ÃÊ ôÊÊÃ ∑§ËÁ¡∞–

(i) ‚◊ ‚¥ïÿÊ ∑§Ë

(ii) 3 ∑§ ªÈáÊ¡ ∑§Ë A dice is rolled 250 times and its outcomes are recorded as below :

Outcome 1 2 3 4 5 6

Frequency 40 45 35 38 52 40

Find the probability of getting :

(i) An even number

(ii) A multiple of 3

10 ∞∑§ Á‚Ä∑§ ∑§Ê ÁŸïŸ ’Ê⁄¥U’Ê⁄UÃÊ•Ê¥ ∑§ ‚ÊÕ 1000 ’Ê⁄U ©¿UÊ‹Ê ªÿÊ—

ÁøÃ — 455, ¬≈U — 545

¬˝àÿ∑§ ÉÊ≈UŸÊ ∑§Ë ¬Ê̋Áÿ∑§ÃÊ •Á÷∑§Á‹Ã ∑§ËÁ¡∞–

A coin is tossed 1000 times with the following frequencies:

Head : 455, Tail : 545

Compute the probability for each event.

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πá«U-‚ / SECTION-C

¬˝‡Ÿ ‚¥ïÿÊ 11 ‚ 18 ◊¥ ¬˝àÿ∑§ ∑§ 3 •¥∑§ „Ò¥–

Question numbers 11 to 18 carry three marks each.

11 2x−y=0 ∑§ •Ê‹π ¬⁄U ÁSÕÃ ©Ÿ Á’¥ŒÈ•Ê¥ ∑§Ê ¬„øÊÁŸ∞ Á¡Ÿ∑§ ÁŸŒ¸‡ÊÊ¥∑§ ÁŸïŸ ÃÊÁ‹∑§Ê ◊¥ ÁŒ∞ ª∞ „Ò–

x −1 0 1 4 3

y −2 0 −2 2 6

‚◊Ë∑§⁄UáÊ ∑§Ê •Ê‹π πË¥Áø∞–

Identify the points whose co-ordinates are given in the following table which lie on the graph of 2x−y=0

x −1 0 1 4 3

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y −2 0 −2 2 6

Draw the graph of the equation.

12 ‚◊Ë∑§⁄UáÊ 4x=6 (1−y)+3x ∑§Ê ax+by=c ∑§ M§¬ ◊¥ Á‹Áπ∞ •ÊÒ⁄U ©Ÿ Á’¥ŒÈ•Ê¥ ∑§ ÁŸŒ¸‡ÊÊ¥∑§ ÷Ë ôÊÊÃ ∑§ËÁ¡∞ ¡„Ê°

©‚∑§Ê •Ê‹π ŒÊŸÊ¥ •ˇÊÊ¥ ∑§Ê ∑§Ê≈UÃÊ „Ò–

Write the equation 4x=6 (1−y)+3x in the form ax+by=c and also find the coordinates of the

points where its graph cuts the two axes ?

3

13 ŒË „Èß¸ •Ê∑Î§ÁÃ ◊¥, PQRS ∞∑§ ‚◊Ê¥Ã⁄U øÃÈ÷È¡̧ „Ò, Á¡‚∑§ Áfl∑§áÊ¸ PR •ÊÒ⁄U SQ ¬⁄US¬⁄U O ¬⁄ ¬Á̋Ãë¿UŒ ∑§⁄UÃ „Ò¥–

Œ‡ÊÊ¸ß∞ Á∑§ ar (∆SOR) = ar (∆POQ) „Ò–

In the given figure, PQRS is a parallelogram in which diagonals PR and SQ intersect each

other at O. Show that ar (∆SOR) = ar (∆POQ).

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14 ŒË „Èß¸ •Ê∑Î§ÁÃ ◊¥, Á’¥ŒÈ A, D, P, C •ÊÒ⁄U B ∑§ãŒ˝ O flÊ‹ ∞∑§ flÎûÊ ¬⁄U ÁSÕÃ „Ò¥– ÿÁŒ ∠BOD=150� „Ò, ÃÊ

∠BPD, ∠BCD •ÊÒ⁄U ∠BAD ∑§ ◊Ê¬ ôÊÊÃ ∑§ËÁ¡∞–

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In the given figure, points A, D, P, C and B lie on a circle with centre O. If ∠BOD=150�, find

the measures of ∠BPD, ∠BCD and ∠BAD.

15 7 cm ◊Ê¬ ∑§Ê ∞∑§ ⁄UπÊπ¥«U PQ πË¥Áø∞– ß‚∑§ ‹¥’ ‚◊Ám÷Ê¡∑§ ∑§Ë ⁄UøŸÊ ∑§ËÁ¡∞ ÃÕÊ flÊSÃÁfl∑§ ◊Ê¬Ÿ mÊ⁄UÊ ‚àÿÊ¬Ÿ ∑§ËÁ¡∞– Draw a line segment PQ of measure 7 cm. Construct its perpendicular bisector and verify it by actual measurement.

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16 ∞∑§ ‚◊Ê¥Ã⁄U øÃÈ÷È¸¡ PQRS ◊¥, ∠P=( 2x+45)� •ÊÒ⁄U ∠Q=(3x−15)� „Ò– x ∑§Ê ◊ÊŸ ôÊÊÃ ∑§ËÁ¡∞– ß‚ ‚◊Ê¥Ã⁄U

øÃÈ÷È¸¡ PQRS ∑§ ‚÷Ë ∑§ÊáÊ ÷Ë ôÊÊÃ ∑§ËÁ¡∞–

In a parallelogram PQRS, ∠P=( 2x+45)� and ∠Q=(3x−15)�⋅ Find the value of x. Also, find

all the angles of the parallelogram PQRS.

3

17 L§‹⁄U ÃÕÊ ¬⁄U∑§Ê⁄U ∑§Ë ‚„ÊÿÃÊ ‚ 52

1

2

� ∑§Ê ∑§ÊáÊ ’ŸÊß∞–

Construct angle of 521

2

�, using compass and ruler.

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18 ŒÊ ªÊ‹Ê¥ ∑§ •ÊÿÃŸÊ¥ ◊¥ 27 : 8 ∑§Ê •ŸÈ¬ÊÃ „Ò– ©Ÿ∑§ ¬Îc∆UËÿ ˇÊòÊ»§‹Ê¥ ∑§Ê •ŸÈ¬ÊÃ ôÊÊÃ ∑§ËÁ¡∞– 3



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The ratio of the volumes of two spheres is 27 : 8. Find the ratio of their surface areas.

πá«U-Œ / SECTION-D

¬˝‡Ÿ ‚¥ïÿÊ 19 ‚ 28 ◊¥ ¬˝àÿ∑§ ∑§ 4 •¥∑§ „Ò¥–

Question numbers 19 to 28 carry four marks each.

19 Á∑§‚Ë ≈US≈U ◊¥, ÁŸÁ∑§ÃÊ Ÿ ªÁáÊÃ •ÊÒ⁄U •¥ª˝$¡Ë ◊¥ ∑˝§◊‡Ê— x •ÊÒ⁄U y •¥∑§ ¬Ê̋åÃ Á∑§∞– ÿÁŒ ©‚∑§ ∑È§‹ ¬Ê̋åÃÊ¥∑§ 70 „Ò¥, ÃÊ

ß‚∑§ Á‹∞ ŒÊ ø⁄UÊ¥ flÊ‹Ë ∞∑§ ⁄UÒÁπ∑§ ‚◊Ë∑§⁄UáÊ ’ŸÊß∞– ß‚∑§Ê •Ê‹π ÷Ë πË¥Áø∞– •Ê‹πËÿ M§¬ ‚ ©‚∑§ •ãÿ

Áfl·ÿ ◊¥ ¬˝ÊåÃ •¥∑§ ôÊÊÃ ∑§ËÁ¡∞, ÿÁŒ ©‚∑§ —

(a) ªÁáÊÃ ◊¥ •¥∑§ 30 „Ò¥– (b) •¥ª˝$¡Ë ◊¥ •¥∑§ 20 „Ò¥–

In a certain test, Nikita scored marks x and y respectively in Mathematics and English. If her

total score is 70, then form a linear equation in two variables for this. Also, draw its graph.

Find graphically, her score in other subject if :

(a) score in Mathematics is 30. (b) score in English is 20.

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20 ŸËø ÁŒ∞ •Ê‹π ◊¥ ŒË ªß ̧⁄UπÊ•Ê¥ p •ÊÒ⁄U q ∑§ ‚◊Ë∑§⁄UáÊ Á‹Áπ∞ —

∞∑§ ÁfllÊÕË¸ Ÿ ⁄UπÊ q ∑§Ë ‚◊Ë∑§⁄UáÊ ∑§Ê x+y=1 ’ÃÊÿÊ– ÄÿÊ ©‚Ÿ ‚„Ë ©ûÊ⁄U ÁŒÿÊ? ⁄UπÊ•Ê¥ p, q •ÊÒ⁄U r ∑§

’Ëø ÁÉÊ⁄UÊ ˇÊòÊ»§‹ ÷Ë ôÊÊÃ ∑§ËÁ¡∞–

Write the equations of the lines p and r in following graph :

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A student answered equation of line ‘q’ as x+y=1. Did he answer correctly ? Also, find the

area enclosed between lines p, q and r.

21 ŒË „Èß¸ •Ê∑Î§ÁÃ ◊¥, Á’¥ŒÈ D •ÊÒ⁄U E ÁòÊ÷È¡ ABC ∑§ •ÊœÊ⁄U BCU ∑§Ê ‚◊ÁòÊ÷ÊÁ¡Ã ∑§⁄UÃ „Ò¥– ‚ÊÕ „Ë, AF=FB •ÊÒ⁄U

AM ⊥ EF „Ò– Á‚h ∑§ËÁ¡∞ Á∑§ ar (∆BEF)=ar (∆ADE) =3 ar (∆AOF) „Ò–

In the given figure, points D and E trisect the base BC of ∆ABC. Also, AF=FB and AM ⊥ EF.

Prove that ar (∆BEF)=ar (∆ADE) =3 ar (∆AOF).

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22 ÿÁŒ Á∑§‚Ë ø∑˝§Ëÿ øÃÈ÷¸È¡ PQRS ∑§ ‚ï◊Èπ ∑§ÊáÊÊ¥ P •ÊÒ⁄U R ∑§ ‚◊Ám÷Ê¡∑§ ‚¥ªÃ flÎûÊ ∑§Ê A •ÊÒ⁄U B ¬⁄U ¬˝ÁÃë¿UŒ 4



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∑§⁄UÃ „Ò¥, ÃÊ Á‚h ∑§ËÁ¡∞ Á∑§ AB flÎûÊ ∑§Ê ∞∑§ √ÿÊ‚ „Ò–

In the bisectors of the opposite angle P and R of a cyclic quadrilateral PQRS intersect the

corresponding circle at A and B respectively, then prove that AB is a diameter of the circle.

23 ∞∑§ ‚◊’Ê„È ÁòÊ÷È¡ ∑§Ë ⁄UøŸÊ ∑§ËÁ¡∞ Á¡‚∑§Ê ‡ÊË·¸‹¥’ 5.5 cm „Ò–

Construct an equilateral triangle of altitude 5.5 cm.

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24 ABCD ∞∑§ øÃÈ÷È¸¡ „Ò, Á¡‚∑§ Áfl∑§áÊ¸ ’⁄UÊ’⁄U „Ò¥ ÃÕÊ ¬⁄US¬⁄U ‚◊∑§ÊáÊ ¬⁄U ‚◊Ám÷ÊÁ¡Ã ∑§⁄UÃ „Ò¥– Á‚h ∑§ËÁ¡∞ Á∑§

ABCD ∞∑§ flª¸ „Ò–

ABCD is a quadrilateral whose diagonals are equal and bisect each other at right angles. Prove

that ABCD is a square.

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25 ∞∑§ ∑§Ê‹ÊŸË ÁŸflÊÁ‚ÿÊ¥ Ÿ ∑§Ê‹ÊŸË ∑§ ¬Ê∑§̧ ◊¥ ÉÊÊ‚ ∑§Ê≈UŸ ÃÕÊ ß‚∑§ øÊ⁄UÊ¥ •Ê⁄U ÃÊ⁄U ‹ªÊŸ ∑§Ê ∑§Êÿ¸ •¬Ÿ •Ê¬

Á∑§ÿÊ– ¬Ê∑¸§ •ÊÿÃÊ∑§Ê⁄U „Ò ‹Á∑§Ÿ ß‚∑§Ë ŒÊŸÊ¥ øÊÒ«∏Êß¸ÿÊ° •œ¸ ªÊ‹Ê∑§Ê⁄U „Ò–

(a) ß‚ ¬Ê∑§̧ ∑§Ë ÉÊÊ‚ ∑§Ê ` 100 ¬Á̋Ã 50 flª¸ ◊Ë≈U⁄U ∑§Ë Œ⁄U ‚ ∑§≈UflÊŸ ∑§Ê πø ̧ôÊÊÃ ∑§ËÁ¡∞–

(b) ß‚ ¬Ê∑§̧ ∑§ øÊ⁄UÊ¥ •Ê⁄U 3 ’Ê⁄U ÃÊ⁄U ‹ªflÊŸ ¬⁄U ÃÊ⁄U ∑§Ë ‹ï’Êß ̧ôÊÊÃ ∑§ËÁ¡∞–

(c) ∑§Ê‹ÊŸË ÁŸflÊÁ‚ÿÊ¥ Ÿ§ Á∑§Ÿ ◊ÍÀÿÊ¥ ∑§Ê Œ‡ÊÊ¸ÿÊ?

A colony decides to take care of their common park by cutting their grasses and fencing the whole

area. The park is in the shape of rectangle adjoint with the semicircle on the both width sides.

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(a) Find the cost of cutting the grass at rate of ` 100 per 50 m2.

(b) How much length of wire is used to fence the whole area if it is rounded 3 times ?

(c) Which value is depicted by colony ?

26 ∞∑§ 12 m ‹¥’ ∑§◊⁄U ◊¥ ŒËflÊ⁄UÊ¥ ¬⁄U 1.35 L§. ¬˝ÁÃ m2 ∑§Ë Œ⁄U ‚ ¬¬⁄U ‹ªflÊŸ ∑§Ê √ÿÿ

340.20 L§. „Ò ÃÕÊ $»§‡Ê¸ ¬⁄U 85 ¬Ò‚ ¬Á̋Ã m2 ∑§Ë Œ⁄U ‚ ◊Ò≈U Á’¿UflÊŸ ∑§Ê √ÿÿ 91.80 L§. „Ò– ∑§◊⁄U ∑§Ë ™°§øÊß¸ ôÊÊÃ

∑§ËÁ¡∞–

The cost of papering the walls of the room 12 m long at the rate of Rs. 1.35 per m2 is Rs. 340.20

and the cost of matting the floor at the rate of 85 paisa per m2 is

Rs. 91.80. Find the height of the room.

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27 ∞∑§ •œ¸flÎûÊÊ∑§Ê⁄U ‡ÊË≈U ∑§Ë ÁòÊíÿÊ 21 cm „Ò– ß‚∑§Ê ◊Ê«∏∑§⁄U ∞∑§ πÈ‹Ê ‡Ê¥ÄflÊ∑§Ê⁄U ∑§¬ ’ŸÊÿÊ ªÿÊ „Ò– ß‚ ∑§¬ ∑§Ê

•ÊÿÃŸ •ÊÒ⁄U ª„⁄UÊß¸ ôÊÊÃ ∑§ËÁ¡∞–

A semi- circular sheet of paper has radius 21 cm. It is bent to form an open conical cup. Find

the volume and depth of the cup.

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28 ÃËŸ Á‚P§Ê¥ ∑§Ê ∞∑§ ‚ÊÕ 300 ’Ê⁄U ©¿UÊ‹Ê ªÿÊ– ¬Ê̋# ¬Á⁄UáÊÊ◊Ê¥ ∑§Ê ’¥≈UŸ ß‚ ¬˝∑§Ê⁄U „Ò — (i) ÃËŸ ¬≈U —40, (ii) ŒÊ ¬≈UU ; 90 (iii) ∞∑§ ¬≈U ; 90, (iv) ∑§Êß¸ ¬≈U Ÿ„Ë¥ — 80, ¬˝àÿ∑§ ¬Á⁄UáÊÊ◊ ∑§Ë ¬˝ÊÁÿ∑§ÃÊ ôÊÊÃ Á∑§Á¡∞ •ÊÒ⁄U ¡Ê°ø Á∑§Á¡∞ Á∑§ ‚÷Ë ¬Ê̋Áÿ∑§ÃÊ•Ê¥ ∑§Ê ÿÊª 1 „Ò– Three coins are tossed simultaneously 300 times. The distribution of various outcomes is listed below. (i) Three tails : 40 (ii) Both tails :90, (iii) one tail :90, (iv) No tails : 80 Find the respective probability of each event and check that the sum of all probabilities is 1.

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πá«U-य/SECTION-E

(◊ÈQ ¬Ê∆ /Open Text)

(·Ñi;k lqfuf'pr dj ysa fd miÿ¸ÈDr fo"k; dh eqDr ikB~;lkexzh bl iz'ui=k ∑ lkFk layXu gS)

(* Please ensure that open text of the given theme is supplied with this question paper.)

fo"k; % vfrfFk nsoks Hko% Theme : Atithidevo Bhavah



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29 ÁøòÊ 2 ‚ ÁŸïŸ ¬˝‡ŸÊ¥ ∑§ ©ûÊ⁄U ŒËÁ¡∞ —

(i) ÷ÍÁ◊ mÊ⁄UÊ ÁflŒ‡ÊË ¬ÿ¸≈U∑§Ê¥ ∑§Ê •Êª◊Ÿ ∑§◊ „Ò– ©‚∑§ ÄÿÊ ∑§Ê⁄UáÊ „¢Ò ?

(ii) Á∑§Ÿ Œ‡ÊÊ¥ ‚ ÁflŒ‡ÊË ¬ÿ¸≈U∑§ ÷ÍÁ◊ ÿÊÃÊÿÊÃ ‚ ‚’‚ •Áœ∑§ •ÊÃ „Ò¢?

Refer to Table-2 and answer the following questions :

(i) The percentage of people travelling by land is less. What do you think are the factors responsible for it ?

(ii) FTA by land route is mostly from which countries ?

3

30 fl·¸ 2006-2011 ◊¥ •Á¡¸Ã ÁflŒ‡ÊË ◊ÈŒÊ̋ (∑§⁄UÊ«∏ ∑§Ê Œ‚ „$¡Ê⁄U ∑§ ªÈáÊÊ¥∑§ ◊) ∑§Ê Œá«U ÁøòÊ mÊ⁄UÊ Œ‡ÊÊ¸ß∞–

Draw a bar graph to represent Foreign exchange earnings (in Rs. Core to the nearest ten thousand) during the years 2006-2011.

3

31 (a) vfrfFk nsoks Hko% ds ikB ls vkidks fdu fdu ckrksa dk Kku gqvk gS\ xf.kr blds fy, fdruk

lgk;d gS \

(b) lkj.kh 2 ls 2011 rFkk 2012 esa vyx vyx {ks=kksa ls fons'kh i;ZVd vkxeu dk ekè; Kkr dhft,

A 2012 ds ekè; esa 2011 ds ekè; ls fdrus izfr'kr dh of̀¼ gqbZ gS \ (a) What insight does this text gives to the reader? How is mathematics relevant to it ? (b) Refer to table-2 : Calculate mean of foreign tourists arrivals from different regions in 2011 and 2012. Find the percentage increase of mean in 2012 as compared to 2011 ?

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Marking Scheme Mathematics (Class – IX)

General Instructions:

1. The Marking Scheme provides general guidelines to reduce subjectivity and maintain uniformity.

The answers given in the marking scheme are the best suggested answers.

2. Marking be done as per the instructions provided in the marking scheme. (It should not be done

according to one’s own interpretation or any other consideration).

3. Alternative methods be accepted. Proportional marks be awarded.

4. If a question is attempted twice and the candidate has not crossed any answer, only first attempt be

evaluated and ‘EXTRA’ be written with the second attempt.

5. In case where no answers are given or answers are found wrong in this Marking Scheme,

correct answers may be found and used for valuation purpose.

/ SECTION-A

1 4 1

Question numbers 1 to 4 carry one mark each.

1 3(1)2k 9

2k 6

k 3.

1

2 Infinite.

1

3 Rectangle

Property : Each angle is a right angle.

1



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4 Radius of base of cylinderr

Height of cylinderh

Half the heighth

2

Area which gets paintedr (hr)

1

/ SECTION-B

5 10 2

Question numbers 5 to 10 carry two marks each.

5 ar (ABD)

1

2 ar(ABC) ___________ (1) AD is median

ar (BED) 1

2 ar (ABD) ___________ (2) BE is the median

using (1) and (2) we get

ar (BED) 1

4 ar (ABC)

2

6 Construction of XYZ200 (½)

Constructing 4 XYZ (1½)

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7

PQAR is a parallelogram (opposite sides are parallel)

PQAR ------(1)

PQRB is a parallelogram (opposite sides are parallel)

PQRB ------(2)

Adding (1) and (2)

PQPQARRB

i.e, 2PQAB

i.e, 1

PQ AB2

2

8 Volume of cylinderr2h

22

7(10.5)2

60

20790 cm3

Mass of the cylinderVolumedensity

207905

103950 gm

103.95 kg

2

9 (i) An even number

P(E) 123

250

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(ii) A multiple of 3

P(E) 75 3

250 10

10 455P(Head) 0.455

1000545

P(Tail) 0.5451000

2

/ SECTION-C

11 18 3

Question numbers 11 to 18 carry three marks each.

11 2xy0

y2x

Now (1, 2), (0, 0) and (3, 6) lie on the above line.

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12 4x6(1y)3x

x6y6

The line cuts x-axis at the point (6, 0)

and y-axis at the point (0, 1)

3

13

ar (SPR) ar (SPQ) __________ (1)

( ‘s on same base and between the same parallels)

Subtracting SPO from both sides of (1)

ar (SPR) ar (SPO)ar (SPQ) ar (SPO)

ar (SOR) ar (POQ)

3

14 For major arc DB

reflex DOB2BPD (angle subtended by an arc at the centre is twice the angle at

remaining circumference)

BPD1

2 (360150)

1

2210

105

BCDBPD105 (angles in the same segment)

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BADBCD180 (opp. angles of cyclic quadrilateral)

BAD105180

BAD18010575

15 Draw tine segment of given length Bisecting tine segment Measuring each part

3

16

P Q 180 ( Co-interior ‘s )

i.e. 2x 45 3x15 180

i.e. 5x 30 180

i.e 5x18030

x 150

5

x30

P (2x 45 ) 2 30 45 105

Q (3x 15 ) 3 3015 75

R P105 Opposite angles of parallelogram are equal

And S Q 75

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17 Construction (2)

Steps of construction (1)

3

18 Volume of sphere I 27

Volume of sphere II 8

33

1

3

2

4 r

33 4 2 r3

1

2

r 3

r 2

22

1I 12

II 22

4 rS r 9

S 44 r r

Thus ratio of the surface areas is 9 :4

3

/ SECTION-D

19 28 4

Question numbers 19 to 28 carry four marks each.

19 xy70 4



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x 30 40 50

y 40 30 20

(a) Score in English40

(b) Score in Mathematics50

20 p : x1

r : y2

Yes, equation of q is xy1

Area1

244

8 sq. units

4

21 ar (BEF)

1

2 (ar ABE) __________ (1)

( QMedian divides a into two ‘s of equal area)

Similarly ar (ADE) 1

2 ar (ABE) _________ (2)

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RHS (1) RHS (2)

ar (BEF) ar (ADE) __________ (3)

also FO : OE 1 : 2

(Medians divide each other in the ratio 2 : 1)

FO1

3 EF

ar (FOA) 1

2 FOAM

1 1

2 3EFAM

ar (FOA) 1

3 ar (AEF)

ar (AEF)3 ar (AOF)

ar (BEF)3 ar (AOF) _________ (4)

From (3) and (4) we get

ar (BEF) ar (ADE)ar (BEF)3 ar (AOF)

Hence proved

22 Given : A cyclic quadrilateral PQRS in which bisectors of P and R meet the corresponding

circle at A and B respectively.

To prove : AB is a diameter of the circle

Construction : Join AB, PB and BS

Proof : In cyclic quadrilateral PQRS

SPQSRQ180 (opposite angles of cyclic quadrilateral are

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supplementary)

1

2 SPQ

1

2 SRQ

1

2180

APSBRS90 --------- (1)

But BRS and BPS are angles in same segment

BPSBRS ---------- (2)

from (1) and (2) we get

APSBPS90

APB90

AB is the diameter (angle in semicircle is 90)

23 Construction (3)

Steps of construction (1)

4

24

In AOB and COD,

AOCO (Given)

AOBCOD90 (Given)

OBOD (Given)

AOB COD (By SAS Congruence Rule)

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ABCD By cpct

and 12

But these are alternate interior angles,

ABCD

In quadrilateral ABCD,

ABCD and ABCD

ABCD is a parallelogram -----(1)

In DAB and CBA,

DACB (Opposite side of gm)

ABBA (Common)

DBCA (Given)

DAB CBA (By SSS Congruence Rule)

DABCBAx (By cpct)

But, DABCBA180 (co-interior angles)

ie, xx180

ie, 2x180

x180

2

x90

DABCBA90 -----(2)

In AOB and COB,

AOCO (Given)

AOBCOB90 (Given)

OBOB (Common)

AOB COB (By SAS Congruence Rule)



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ABBC (By cpct) -----(3)

ABCD is a gm with each angle as 90 and ABBC (From (1), (2) and (3)).

ABCD is a square.

25 (a) r0.7 km 700 m l3000 m b1400 m

Area of grassr2l b

5740000

Cost100

5740000 50

` 11480000 (b) Perimeter(2r2 l)3 31200 m 31.2 km (c) Social value Co-operation Caring by people of their surrounding

4

26

340.20Areas of walls 2 b h 252 Sq m -----------(i)

1.3591.80

Areas of floor b 108 Sq m.85

12 b 108

b 9 m -----------(ii)

(i) and (ii)

2(12 9) h 252

252h 6m

2 21

l

l

4

27 Radius of circular sheet 21cm 4



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Slant height of cone, l21cm

Let ‘r’ be the radius of the base of cone and ‘h’

be the height of the cone

Circumference of base circumference of semi circular sheet

2r 21

21

r cm2

Now, l2h2r2

2

2 2 2121 h

2

2 441441 h

4

2441441 h

4

2441 4 441 h

4

2 3h 441

4

21 3 21 1 732

h 18 19cm2 2

..

Volume of conical cup 21 r h3

1 22

21 21 18 193 7

. cm3

8403.78 cm3

28 (i) P (three tails)

40 2 300 15

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(ii) P (2 tails) 90 3

300 10

(iii) P (one tail) 90 3

300 10

(iv) P (no tail) 80 4

300 15

Sum2 3 3 4 12 18

115 10 10 15 30 30

/SECTION-E

( /Open Text)

(* Please ensure that open text of the given theme is supplied with this question paper.)

Theme : Empower to learn

29 (i) ………………………………………………………………………….. [1.5]

(ii) …………………………………………………………………………. [1.5]

3

30 Converting ………………………………………………………… [ ½ ]

Drawing of bar graph ……………………………………………. [2]

Labelling, scale and heading……………………… …………….. [ ½ ]

3

31 (a) Text gives in sight into the number of foreign tourist arrivals in India during various seasons. It also shows that tourist are all of age groups. Data is supported by statistics (which is part of mathematics) showing bar graphs, frequency distribution table etc.

(b) Mean of FTA’s for year 2011

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Mean of FTA’s for year 2012

% increase =

= 11.9 %

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time allowed : 3 hours maximum marks : 90 €¦ · in a certain test, nikita scored marks x and y...

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