Odisha HSC Paper 2014 Maths Part II

No. of Questions : 5
PART – II
No. of Printed Pages : 16
Booklet Sl. No. : ......................... SUBJECTIVE
Roll No. of the Candidate
2014
Time : 1 Hour 30 Minutes
Full Marks : 50
(Verified and found correct) AH
Script

Full signature of the Invigilator SET : A
Date of exam : ............................
REGULAR MTH
QUESTION-CUM ANSWER BOOKLET
AR – 15 MATHEMATICS
FÜÿç ¨÷ÉÓ
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: ¨Àÿêäæ$öêZÿ œÿçþ{;ÿ Óí`ÿœÿæ :
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FOR USE AT THE EVALUATION CENTRE
Q. No. Marks Full Signature of
Awarded Examiner
Full Signature of the Scrutiniser
01 Regd. No.
Regd. No.
02 Regd. No.
Full Signature of the Deputy Chief Examiner
03 Regd. No.
Regd. No.
04 Regd. No.
Full Signature of the Chief Examiner
05 Regd. No. Regd. No.

Total Mark in words (.....................................................................................)

Full Signature of the Examiner Who Entered The Total Marks
Regd. No. ................................ Date of Evaluation ......................
P.T.O.

Àÿüÿú ¨æBô ×æœÿ
SPACE FOR ROUGH

AR-15 / P - II / MTH 2 Contd.

xÿæÜÿæ~ ¨æQ{Àÿ ’ÿçAæ¾æB$ç¯ÿæ ÓóQ¿æSëÝLç ÿ þíàÿ¿æZÿ Óí`ÿæDdç >
The figures in the right-hand margin indicate marks.
ÓþÖ ¨÷ÉÀ§ ÿ DˆÿÀÿ œÿç{”öÉæœÿëÓæ{Àÿ {àÿQ >
Answer all questions as directed.
Time : 1 Hour 30 Minutes Full Marks : 50
22 22
π Àÿ þíàÿ¿ œÿçA (Take π = )
7 7
1. (i) Óþæ™æœÿ LÿÀÿ : 5
Solve :
6x + 5y = 7x + 3y + 1 = 2 (x + 6y – 1)
ç {SæsçF ÓóQ¿æÀÿ AZÿþæœÿZÿÀÿ ÓþÎç 10, Lÿç;ÿë AZÿSëÝLç ÿë ×æœÿ ¯ÿ’ÿÁÿæB
’ÿëB AZÿ ¯ÿçÉÎ
{àÿQ#{àÿ DŒŸ ÓóQ¿æsç þíÁÿ ÓóQ¿æÀÿ ’ÿëBSë~Àÿë 1 Lÿþú ÜÿëF > ÓóQ¿æsç œÿç‚ÿöß LÿÀÿ >
The sum of the digits of a two-digit number is 10. But if the number
is written interchanging the digits the number so formed is 1 less
than twice the original number. Find the number.

AR-15 / P - II / MTH 3 P.T.O.

(ii) ¨í‚ÿö ¯ÿSö{Àÿ ¨Àÿç~†ÿ LÿÀÿç Óþæ™æœÿ LÿÀÿ : 5
Solve by completing the squares :
3x2 – 13x + 12 = 0
x2 – px + q = 0 ÓþêLÿÀÿ~Àÿ {SæsçF þíÁÿ Aœÿ¿sçÀÿ 2 Së~ {Üÿ{àÿ, ¨÷þæ~ LÿÀÿ {¾
2p2 = 9q >
If one of the roots of the equation x 2 – px + q = 0 is double of the
other, then prove that 2p2 = 9q.

AR-15 / P - II / MTH 4 Contd.

2. (i) †ÿç{œÿæsç ÀÿæÉç Óþæ;ÿÀÿ ¨÷S†ÿç{Àÿ A¯ÿ×ç†ÿ > {ÓþæœÿZÿÀÿ {¾æSüÿÁÿ 18 F¯ÿó Së~üÿÁÿ
192 > ÓóQ¿æ †ÿç{œÿæsç œÿç‚ÿöß LÿÀÿ > 4
Three numbers are in arithmetic progression. Their sum is 18 and
product is 192. Find the numbers.
1 × 2 + 2 × 3 + 3 ×4 + ... + n (n + 1) Àÿ þæœÿ œÿç‚ÿöß LÿÀÿ >
Find the value of 1 × 2 + 2 × 3 + 3 ×4 + ... + n (n + 1).

AR-15 / P - II / MTH 5 P.T.O.

(ii) {SæsçF àÿëxÿë{SæsçLÿë ${Àÿ SÝæ Sàÿæ > "üÿÁÿ A¾ëS½ ¯ÿæ üÿÁÿ ≥ 3'Ws~æsçÀÿ Ó»æ¯ÿ¿†ÿæ
œÿç‚ÿöß LÿÀÿ > 4
A ludo die was thrown once. Find the probability of the event ‘‘result
odd or result ≥ 3’’.
¯ÿç¢ÿë†ÿ÷ß A, B, C Àÿ ×æœÿæZÿ ¾$æLÿ÷{þ (2, 3), (3, k) H (5, 9) > k Àÿ þæœÿ {Lÿ{†ÿ
{Üÿ{àÿ ¯ÿç¢ÿë†ÿ÷ß {SæsçF ÓÀÿÁÿ{ÀÿQæ{Àÿ ÀÿÜÿç{¯ÿ œÿç‚ÿöß LÿÀÿ >
The co-ordinates of the three points A, B, C are (2, 3), (3, k) and
(5, 9). Find the value of k so that the three points are collinear.

AR-15 / P - II / MTH 6 Contd.

(iii) 80 f~ ¨çàÿæ S~ç†ÿ{Àÿ ÀÿQ#$¯# ÿæ œÿºÀÿ œÿçþ§ ÓæÀÿ~ê{Àÿ ’ÿçAæ¾æBdç > ¨çàÿæþæ{œÿ ÀÿQ#$¯ÿæ
œÿºÀÿÀÿ þæšþæœÿ œÿç‚ÿöß LÿÀÿ > 4
The marks obtained by 80 students is given in the following table.
Find the mean marks obtained by them.
œÿºÀ ÿ 10Àÿë Lÿþú 20Àÿë Lÿþú 30Àÿë Lÿþú 40Àÿë Lÿþú 50Àÿë Lÿþú 60Àÿë Lÿþú
Score below 10 below 20 below 30 below 40 below 50 below 60
dæ†ÿ÷ ÓóQ¿æ 3 12 27 57 75 80
No. of pupils

D¨{Àÿ ’ÿçAæ¾æB$#¯ÿæ ÓæÀÿ~ê A;ÿµÿëNö ÿ †ÿ$¿æ¯ÿÁÿêÀÿ þšþæ œÿç‚ÿöß LÿÀÿ >
Find the median of the data given in the above table.

AR-15 / P - II / MTH 7 P.T.O.

3. (i) ¨÷þæ~ LÿÀÿ {¾, {SæsçF ¯ÿõˆÿÀÿ Óþæœÿ {’ÿðWö¿ ¯ÿçÉçÎ f¿æþæ{œÿ {Lÿ¢ÿ÷vÿæÀÿë
Óþ’ÿíÀÿ¯ÿˆÿöê > 5
Prove that chords of equal length in a circle are equidistant from the
centre.
¨÷þæ~ LÿÀÿ {¾, {Lÿò~Óç ¯ÿõˆÿÀÿ ¯ÿÜÿç × FLÿ ¯ÿç¢ÿëÀÿë DNÿ ¯ÿõˆÿ ¨÷†ÿç AZÿç†ÿ ØÉöLÿ Qƒ’ÿ´ßÀÿ
{’ÿðW¿ö Óþæœÿ >
Prove that the lengths of two tangent-segments drawn to a circle
from an external point are equal.

AR-15 / P - II / MTH 8 Contd.

(ii) AB = 8 {Ó.þç. ¯ÿçÉÎ
ç {ÀÿQæQƒ AZÿœÿ LÿÀÿ > A ¯ÿç¢ÿëLÿë {Lÿ¢ÿ÷ LÿÀÿç 3 {Ó.þç. ¯ÿ¿æÓæ•ö
ç FLÿ ¯ÿõˆÿ AZÿœÿ LÿÀÿ > B ¯ÿç¢ÿëÀÿë DNÿ ¯ÿõˆÿ ¨÷†ÿç ’ÿëBsç ØÉöLÿ AZÿœÿ LÿÀÿ >
¯ÿçÉÎ 5
Construct a line-segment AB = 8 cm. Taking A as cente construct a
circle of 3 cm radius. From B construct two tangents to the circle.
∆ ABC {Àÿ BC = 6 {Ó.þç., m ∠ A = 60° H þšþæ AD = 4.5 {Ó.þç. > †ÿ÷µç ÿëfsç
AZÿœÿ LÿÀÿ >
Construct ∆ ABC in which BC = 6 cm, m ∠ A = 60° and median
AD = 4.5 cm.

AR-15 / P - II / MTH 9 P.T.O.

4. (i) ç þú{Àÿÿ AB || CD > AC H BD Àÿ {d’ÿ¯ÿç¢ÿë O > ¨÷þæ~ LÿÀÿ {¾
ABCD s÷æ¨çfß
OC : AC = OD : BD > 5

In the trapezium ABCD, AB || CD . O is the point of intersection of
AC and BD . Prove that OC : AC = OD : BD.

¨÷þæ~ LÿÀÿ {¾ {SæsçF ¯ÿõˆÿæ;ÿàÿ}Q†# ÿ s÷æ¨çfß
ç þúÀÿ Lÿ‚ÿö’ÿ´ß ¨ÀÿØÀÿ Óþæœÿ >
Prove that the diagonals of a cyclic trapezium are equal.

AR-15 / P - II / MTH 10 Contd.

(ii) A + B + C = 90° {Üÿ{àÿ, ¨÷þæ~ LÿÀÿ {¾, 5
cot A + cot B + cot C = cotA . cotB . cotC
If A + B + C = 90°, then prove that
cot A + cot B + cot C = cotA . cotB . cotC
300 þçsÀÿ Daÿ FLÿ ¨æÜÿæÝ D¨Àÿë FLÿ Óþ†ÿÁÿ{Àÿ A¯ÿ×ç†ÿ {SæsçF Ö»Àÿ ÉêÌö H
¨æ’ÿ{’ÿÉÀÿ {Lÿò~çLÿ A¯ÿœÿ†ÿçÀÿ ¨Àÿçþæ~ ¾$æLÿ÷{þ 30° H 60° {Üÿ{àÿ Ö»Àÿ Daÿ†ÿæ
œÿç‚ÿöß LÿÀÿ >
From the top of a hill 300 metres high the angle of depression of the
top and bottom of a pillar, standing on the same plane, measure 30°
and 60° respectively. Find the height of the pillar.

AR-15 / P - II / MTH 11 P.T.O.

5. (i) {SæsçF ¯ÿõˆÿÀÿ {ä†ÿ÷üÿÁÿ 22176 ¯ÿSö {Ó.þç. > FÜÿæÀÿ 110 {Ó.þç. ’ÿêWö `ÿæ¨ ’ÿ´æÀÿæ
{Lÿ¢ÿ÷{Àÿ DŒŸ {ÜÿD$#¯ÿæ {Lÿæ~Àÿ ¨Àÿçþæ~ œÿç‚ÿöß LÿÀÿ > 4

Area of a circle is 22176 sq. cm. Find the measure of the angle
subtended at its centre by an arc of length 110 cm.

{SæsçF ¯ÿõˆÿÀÿ ¯ÿ¿æÓæ•ö 7 2 {Ó.þç. > FÜÿæÀÿ FLÿ ¯ÿõˆÿQƒ {Lÿ¢ÿ÷{Àÿ 90° {Lÿæ~ DŒŸ
LÿÀÿëdç > ¯ÿõˆÿQƒÀÿ {ä†ÿ÷üÿÁÿ œÿç‚ÿöß LÿÀÿ >
The radius of a circle is 7 2 cm. A segment of it subtends 90°
angle at the centre. Find the area of the segment.

AR-15 / P - II / MTH 12 Contd.

(ii) {SæsçF {LÿæœÿúÀÿ Aæ߆ÿœÿ 9240 Wœÿ {Ó.þç. > FÜÿæÀÿ µÿíþÀç ÿ ¯ÿ¿æÓæ•ö 21 {Ó.þç.
{Üÿ{àÿ, {LÿæœÿúÀÿ ¯ÿLÿ÷†ÿÁÿÀÿ {ä†ÿ÷üÿÁÿ œÿç‚ÿöß LÿÀÿ > 4

The volume of a cone is 9240 cubic cm. If the radius of the base is
21 cm, find the area of the curved surface of the cone.

{SæsçF {SæàÿLÿÀÿ ¨õφÿÁÿÀÿ {ä†ÿ÷üÿÁÿ 616 ¯ÿSö {Ó.þç. > FÜÿæÀÿ Aæ߆ÿœÿ œÿç‚ÿöß LÿÀÿ >
The surface area of a sphere is 616 sq. cm. Find its volume.

àÿ²æZÿ 5 6 7 8 9
Scores

AR-15 / P - II / MTH 13 P.T.O.

A†ÿçÀÿçNÿ ¨õÏæ
ADDITIONAL PAGE

AR-15 / P - II / MTH 14 Contd.

A†ÿçÀÿçNÿ ¨õÏæ
ADDITIONAL PAGE

AR-15 / P - II / MTH 15 P.T.O.

Àÿüÿú ¨æBô ×æœÿ
SPACE FOR ROUGH

AR-15 / P - II / MTH 16 Contd.