Question Set D

CBSE 9th Standard Mathematics Linear Equations in Two Variables Important Questions And Answers Study Material - QB365

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Questions:

Multiple Choice Question

12 x 1 = 12

The area of the four walls of a room is 300 m². Its length and height are 15 m and 6 m respectively. Find its breadth.

(a)

10 m

(b)

5 m

(c)

20 m

(d)

15 m

The side of an equilateral triangle is 6 cm.The area of the triangle is

(a)

\(6\sqrt { 3 } \) cm²

(b)

\(9\sqrt { 3 } \) cm²

(c)

\(16\sqrt { 3 } \) cm²

(d)

\(3\sqrt { 3 } \) cm²

In the figure, AOB is a diameter of the semicircle. If \(\angle A=60°\) , then \(\angle B\) is equal

(a)

60°

(b)

30°

(c)

50°

(d)

40°

In the following figure, ABCD is a parallelogram, \(AE\bot DC\) and \(CF\bot AD\). If AB = 16 cm, AB = 8 cm and CF = 10 cm, then AD =

(a)

16 cm

(b)

12.8 cm

(c)

8 cm

(d)

10 cm

In figure AB || CD and t is a transversal the value of x is equal to:

(a)

\(50^{ 0 }\)

(b)

\(70^{ 0 }\)

(c)

\(35^{ 0 }\)

(d)

\(20^{ 0 }\)

In figure, the value of an angle q is

(a)

\(60^{ 0 }\)

(b)

\(90^{ 0 }\)

(c)

\(50^{ 0 }\)

(d)

\(40^{ 0 }\)

Where does the line -2x+y-7=0 intersect y-axis?

(a)

at (0,7)

(b)

at(7,0)

(c)

at(7,7)

(d)

at(-7,-7)

If (x+2,4)=(5,y-2), then the coordinates (x,y) are:

(a)

(7,12)

(b)

(6,3)

(c)

(3,6)

(d)

(2,1)

For what value of b, is the polynomial \(x^3-3x^2+bx-6\) divisible by x-3?

(a)

(b)

(c)

(d)

-3

10.

Degree of polynomial \((x^3+5)(4-x^5)\) is:

(a)

(b)

(c)

(d)

11.

The value of \(\sqrt [ 4 ]{ { \left( 64 \right) }^{ -2 } } \) is

(a)

1/8

(b)

1/2

(c)

(d)

1/64

12.

\(\sqrt { 12 } \times \sqrt { 18 } \) is equal to:

(a)

\(2\sqrt { 6 } \)

(b)

\(3\sqrt { 6 } \)

(c)

\(4\sqrt { 6 } \)

(d)

\(6\sqrt { 6 } \)

Generic 4

6 x 4 = 24

13.

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that:
(i) \(\triangle AMC\cong \triangle BMD\)
(ii) \(\triangle\)DBC is a right angle

14.

Find a and b, if \(\frac { 2\sqrt { 5 } +\sqrt { 3 } }{ 2\sqrt { 5 } -\sqrt { 3 } } +\frac { 2\sqrt { 5 } -\sqrt { 3 } }{ 2\sqrt { 5 } +\sqrt { 3 } } =a+\sqrt { 15 } b\)

15.

Find the value of p is mean of following distribution is 20.

x	f
15	2
17	3
19	4
20 + p	5p
23	6

16.

(i) Construct a triangle PQR with base PQ = 8.4 cm, LP = 45° and PR - QR = 2.8 cm.
(ii) Measure PR.
(iii) Measure QR.
(iv) Verify that PR - QR = 2.8 cm.
(v) Gaffar says that L PQR = 85°. Is he correct? Which value is depicted by his statement?

17.

Parallelograms on the same base and between same parallels are equal in area. Prove this.

18.

Let \(R_1\) and \(R_2\)are the remainders when the polynomials \({ x }^{ 3 }+2{ x }^{ 2 }-5ax-7\) and \({ x }^{ 3 }+2{ x }^{ 2 }-5ax-7\) are divided by (x+1) and (x-2) respectively. If 2\(R_1\)+\(R_2\)=6, Find the value of a.

Generic 2

6 x 2 = 12

19.

In the adjoining figure if ㄥDAB=60^o and ㄥACB=70^o, find the measure of ㄥDBA

20.

In the given figure, A, B, C and D are four points on a circle. AC and BD intersect at E such that \(\angle\) BEC = 130° and \(\angle\)ECD = 20°. Find \(\angle\)BAC

21.

In the given figure, if \(\angle A=60^o\) and \(\angle B=70^o\), then find \(\angle ACD\)

22.

In the figure AB || CD and CD || EF Also \(EA\bot AB\) if \(\angle BEF=55^{ 0 }\) find the values of x,y and z

23.

(i) In figure ,AO \(\bot \)OB Find \(\angle \)AOC and \(\angle \) BOC

(ii) In figure,\(\angle \) AOB ;\(\angle \)BOC=2:3
If \(\angle \)AOC=\(75^{ 0 }\) then find the measure of ,\(\angle \) AOB ;\(\angle \)BOC

24.

The polynomials \(kx^3+3x^2-8 \) and \(3x^3-5x+8\) are divided by x+2. If the remainder in each case is the same, find the value of k.

Generic 3

6 x 3 = 18

25.

A random survey of the number of children of various age groups playing in a park was found as follows:

Age (in years)	Number of children
1-2	5
2-3	3
3-5	6
5-7	12
7-10	9
10-15	10
15-17	4

Draw a histogram to represent the data above.

26.

The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface area.

27.

A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and side 6cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it?

28.

A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

29.

The Auto fare in a city is as follows:
For the first kilometer the fare is Rs.5 and for the successive distance it is Rs.1 per kilometer.Taking a distance covered as x kilometer and total fare as Rs.y, write a linear equation for the above said data and draw its graph

30.

Plot the points A(0,3), B(5,3), C(4,0) and D(-1,0) on the graph paper.Identify the figure ABCD and find whether the point (2,2) lies inside the figure or not?

1 Marks

7 x 1 = 7

31.

A die is thrown, what will be the probability of getting an even number?

32.

When a coin tossed, the probability of getting a head is?

33.

The radii of two right circular cylinders are in the ratio 2:3 and their heights are in the ratio 5:4, then the ratio of their volumes will be _______________

34.

For the given data: 11,15, 17, y+1, 19, y-2, 3; if the mean is 14, find the value of y.

35.

Compute the curved surface area of a hemishpere whose diameter is 14 cm.

36.

\(\triangle PQR\cong \triangle ABC\), if PQ = 5 cm, \(\angle\)Q = 40° and \(\angle\)P = 80°, calculate the value of \(\angle\)C.

37.

Factorize: 8y³-125x³.

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Answers Key:

Multiple Choice Question

12 x 1 = 12

Number of cubes = \(\frac { { \left( 20 \right) }^{ 3 } }{ { \left( 5 \right) }^{ 3 } } =64\)

\(\frac { \sqrt { 3 } }{ 4 } { (6) }^{ 2 }=9\sqrt { 3 }\) cm².

\(\angle ACB=90°\)

(b)
12.8 cm

\(2x+40^{ 0 }=x+90^{ 0 }\)

\(x+50^{ 0 }\)
\(y+90^{ 0 }+x=180^{ 0 }\quad 50^{ 0 }+90^{ 0 }+x=180^{ 0 }\)
\(\Rightarrow x=40^{ 0 }\)
\(q=x=40^{ 0 }\)

Put x=0 -2(0)+y-7=0 ⇒ y=7

\(f(x)=x^3+3x^2+bx-6\)
\(f(3)=0\)
\(\Rightarrow \ 3^3-3 \times3^2+b\times 3-6=0\)
\(\Rightarrow\ b=2\)

10.

Highest power of x=3+2=5

11.

(a)
1/8

12.

Generic 4

6 x 4 = 24

13.

(i) In \(\triangle\) s AMC and \(\triangle\)BMD, we have
AM = BM
(\(\because\) M is the mid-point of AB)
\(\angle\) AMC = \(\angle\)BMD
(Vertically opp. angles)
and CM = MD (Given)
\(\therefore\) By SAS criterion of congruence, we have
\(\therefore \ \triangle AMC\cong \triangle BMD\)
(ii) Now, \(\triangle AMC\cong \triangle BMD\)
\(\therefore\) BD = CA and \(\angle\)BDM = \(\angle\)ACM
(\(\therefore\)Corresponding parts of congruent triangles are equal)
Thus, transversal CD intersects CA and BD at C and D respectively such that the alternate angle \(\angle\)BDM and \(\angle\)ACM are equal. Therefore BD II CA.
\(\therefore\) \(\angle\)CBD + \(\angle\)BCA = 1800
(\(\therefore\) Sum of the interior angles on the same side of transversal = 180°)
\(\angle\)CBD + 90° = 180°
\(\Rightarrow\) DBC=90°.

14.

\(LHS=\frac { { \left( 2\sqrt { 5 } +\sqrt { 3 } \right) }^{ 2 }+{ \left( 2\sqrt { 5 } -\sqrt { 3 } \right) }^{ 2 } }{ \left( 2\sqrt { 5 } -\sqrt { 3 } \right) \left( 2\sqrt { 5 } +\sqrt { 3 } \right) } \)
\(=\frac { 4\times 5+3+2\times 2\sqrt { 5 } \times \sqrt { 3 } +4\times 5+3-2\times 2\sqrt { 5 } \times \sqrt { 3 } }{ { \left( 2\sqrt { 5 } \right) }^{ 2 }-{ \left( \sqrt { 3 } \right) }^{ 2 } } \)
\(=\frac { 20+3+4\sqrt { 15 } +20+3-4\sqrt { 15 } }{ 20-3 } \)
\(=\frac { 46 }{ 17 } =\frac { 46 }{ 17 } +\sqrt { 15 } \times \left( 0 \right) \)
\(\therefore \frac { 46 }{ 17 } +\sqrt { 15 } \times \left( 0 \right) =a\sqrt { 15 } b=RHS\)
Comparing both sides, we get
a=\(\frac{46}{17}\) , b = 0

15.

x	f	fx
15	2	30
17	3	51
19	4	76
20 + p	5p	5p(20 + p)
23	6	138
Total	5p + 15	295 + 100p+5p²

Mean \(= \frac {\sum fx}{\sum f}\)
\(\Rightarrow\) \(2= \frac {295 + 100p + 5p^2}{5p+15}\)
\(\Rightarrow\) 20(5p+15) = 295 +100p +5p²
\(\Rightarrow\) 100p + 300 = 295 + 100p + 5p²
\(\Rightarrow\) 5p² = 5
\(\Rightarrow\) p² = 1
\(\Rightarrow\) p = 1

16.

(i) Steps of Construction
1. Draw the base PQ = 8.4 cm.
2. At point P make an angle say XPQ=45°.
3. Cut the line segment PD = 2.8 ern from rayPX.
4. Join DQ and draw the perpendicular bisector of DQ.
5. Let it intersect PX at a point R. Join RQ. Then PQR is the required triangle.

(ii) By measurement, PR = 10cm
(iii) By measurement, QR = 7.2cm
(iv) PR - QR = 10-7.2 = 2.8 cm
(v) By measurement,
ㄥPQR = 54°
∴ Gaffar is correct.
∴ The value 'intelligence' is depicted by his statement.

17.

Given: Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels AF and DC.
To Prove: ar(ABCD) = ar(EFCD)

In \(\Delta ADE\) and \(\Delta BCF\)
\(\angle DAE=\angle CBF\quad \quad ...(1)\)
Corresponding angles
(AD || BC and a transversal AF intersects them)
\(\angle AED=\angle BFC\quad \quad ...(2)\)
| Corresponding angles
(ED || FC and a transversal AF intersects them)
\(\angle ADE=\angle BCF\quad \quad ...(2)\)
Angle sum property of a triangle
Also AD+BC
| Opposite sides of the parallelogram ABCD
\(\Delta ADE\cong \Delta BCF\)
| By ASA congruence rule, using (1),(3) and (4)
\(ar(\Delta ADE)=ar(\Delta BCF)\)
|Congruent figures have equal areas
\(\Rightarrow \ ar(\Delta ADE)+ar(EDCB)\)
\(\\ =ar(\Delta BCF)+ar(EDCB)\)
|Adding ar(EDCB) to both sides
\(\Rightarrow \ ar(ABCD)=ar(EDCB)\)
So, parallelograms ABCD and EFCD are eual in area.

18.

Let \(f(x)={ x }^{ 3 }+2{ x }^{ 2 }-5ax-7\)
and \(g(x)={ x }^{ 3 }-a{ x }^{ 2 }-12x+6\)
By remainder theorem,
\(f(-1)={ R }_{ 1 }\ \ \ \ ........(1)\)
\(x+1=0\Rightarrow x=-1\)
\(\Rightarrow { (-1) }^{ 3 }+2{ (-1) }^{ 2 }-5a(-1)-7={ R }_{ 1 }\)
\(\Rightarrow \ -1+2+5a-7={ R }_{ 1 }\)
\(\Rightarrow { R }_{ 1 }=a-6 \ \ ................(2)\)
and \(g(2)={ R }_{ 2 }\ \ ...........(3)\)
\(x-2=0\Rightarrow x=2\)
\(\Rightarrow { (2) }^{ 3 }+a{ (2) }^{ 2 }-12(2)+6={ R }_{ 2 }\)
\(\Rightarrow 8+4a-24+6={ R }_{ 2 }\)
\(\Rightarrow { R }_{ 2 }=4a-10\ \ \ .....(4)\)
According to the question,
\(2{ R }_{ 1 }+{ R }_{ 2 }=6\)
\(\Rightarrow 2(5a-6)+(4a-10)=6\)
\(\Rightarrow -12+4a-10=6\)
\(\Rightarrow 14a=28\)
\(\Rightarrow a=2\)

Generic 2

6 x 2 = 12

19.

ㄥACB= 70^o
ㄥADB=ㄥACB
⇒ ㄥADB=70^o(Angles in the same segment of a circle)
In AB,
ㄥDAB+ㄥADB+ㄥDBA=180^o(Angle sum property of triangles)
⇒ 60^o+70^o+ㄥDBA=180^o
⇒ㄥDBA=50^o

20.

In \(\Delta\)EDC, \(\angle\)EDC + \(\angle\)ECD = \(\angle\)BEC (exterior angle of a \(\Delta\) is equal to the sum of two opposite angles)
\(\Rightarrow\) \(\angle\)EDC + 20° = 130°
\(\Rightarrow\) \(\angle\)EDC = 110° or \(\angle\)BDC =110°
\(\angle\)BAC = \(\angle\)BDC = 110°. (Angle in the same segment)

21.

\(\angle ACD=\angle A+\angle B\) [Exterior angle is the sum of the two interior opposite angles]
=60+70
\(\angle ACD=130^o\)

22.

Given AB||CD, CD||EF, \(EA\bot AB\)
\(\angle BEF=55^o\)
y = 180^o- 55^o= 125^o(Co-interior angles)
x = y = 125^o(Corresponding angles)
\(EA\bot AB\)
\(EA\bot AB\)
Z + 55^o= 90^o
Z = 90^o- 55^o
= 35^o

23.

Let \(\angle AOB=2x\) and \(\angle BOC=3x\)
\(\angle AOB+\angle BOC=\angle AOC\)
2x+3x=75^o
\(\Rightarrow x=\frac{75^o}{5}=15^o\)
\(\angle AOB=2x=30^o\) and \(\angle BOC=3x=45^o\)

24.

\(\frac{5}{4}\)

Generic 3

6 x 3 = 18

25.

Modified Table
[Minimum class-size = 1]

Age (in years)	Number of children (frequency)	Width of the class	Length of the rectangle
1-2	5	1	\(\frac {5}{1}\times 1 = 5\)
2-3	3	1	\(\frac {3}{1}\times 1 = 3\)
3-5	6	2	\(\frac {6}{2}\times 1 = 3\)

Age (in years)	Number of children (frequency)	Width of the class	Length of the rectangle
5-7	12	2	\(\frac {12}{2}\times 1 = 6\)
7-10	9	3	\(\frac {9}{3}\times 1 = 3\)
10-15	10	5	\(\frac {10}{5}\times 1 = 2\)
15-17	4	2	\(\frac {4}{2}\times 1 = 2\)

26.

Let the diameter of the earth be 2r.
Then diameter of the moon \(=\frac { 1 }{ 4 } \left( 2r \right) =\frac { r }{ 2 } \)
\(\therefore\) Radius of the earth = \(\frac { 2r }{ 2 } =r\)
and, Radius of the moon \(=\frac { 1 }{ 2 } \left( \frac { r }{ 2 } \right) =\frac { r }{ 4 } \)
\(\therefore\) Surface area of the earth \(=4\pi { r }^{ 2 }\)
and, Surface area of the moon
\(=4\pi { \left( \frac { r }{ 4 } \right) }^{ 2 }=\frac { 1 }{ 4 } \pi { r }^{ 2 }\)
\(\therefore\) Ratio of their surface area
\(=\frac { Surface\ area\ of\ the\ moon }{ Surface\ area\ of\ the\ earth } \)
\(\\ =\frac { \frac { 1 }{ 4 } \pi { r }^{ 2 } }{ 4\pi { r }^{ 2 } } =\frac { 1 }{ 16 } =1:16.\)

27.

Area of paper of shade I \(2\times \left( \frac { 1 }{ 2 } \times 16\times 16 \right) =256\)cm²
Similarly, Area of paper of shade II = 256 cm²
For area of paper of shade III
a = 8 cm, b = 6 cm, c = 6cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 8+6+6 }{ 2 } =10\) cm
\(\therefore \) Area of paper of shade III = \(\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 10(10-8)(10-6)(10-6) } \)
\(=\sqrt { (10)(2)(4)(4) } =8\sqrt { 5 } \)
= 17.89 cm²

28.

Let BD = x m

Then in right triangle ODB,
OB²= OD²+ BD²
By Pythagoras Theorem
⇒ (20)²= OD²+ x²
⇒ OD²= 400 - x²⇒ OD =\(\sqrt { 400-x^{ 2 } } \)
Again, area of equilateral triangle ABC
= Area of \(\Delta OBC\) + Area of \(\Delta OCA\) + Area of \(\Delta OAB\)
= 3 Area of \(\Delta OBC\)= 3\(\frac { \left( BC \right) \left( OD \right) }{ 2 } \)
\(=3x\sqrt { 400-x^{ 2 } } \)   ....(2)
⇒ \(\sqrt { 3 } \sqrt { 400-x^{ 2 } } =x\)
Squaring both sides,
3(400 - x²) = x²
⇒ 1200 - 3x²= x²
⇒ 4x²= 1200   ⇒ x²= 300
⇒ \(x=10\sqrt { 3 } \)  ⇒ BD=\(10\sqrt { 3 } \)
⇒  \(2BD=20\sqrt { 3 } \)  ⇒ \(BC=20\sqrt { 3 } \)
Hence, the length of string of each phone is \(20\sqrt { 3 } \) m.

29.

y=5+2(x-1) ⇒ y=3+2x

30.

Parallelogram; yes

1 Marks

7 x 1 = 7

31.

()
Favourable number of outcomes = 3(2,4,6)
Total number of outcomes = 6
Required probability\(=\frac{3}{6}=\frac{1}{2}\)

32.

()
When a coin is tossed, total number of outcomes = 2(Head or Tail)
P(getting a head) = \(\frac{1}{2}\)

33.

()
Let radii of cylinders be 2x and 3x and heights be 5y and 3y respectively.
\(\therefore\) Ratio of volumes = \(\frac { \pi { (2x) }^{ 2 }\times 5y }{ \pi { (3x) }^{ 2 }\times 3y } \)
\(=\frac { { 4x }^{ 2 }\times 5 }{ { 9x }^{ 2 }\times 3 } \)
= 20:27.

34.

()
\(14=\frac{11+15+17+y+1+19+y-2+3}{7}\)
⇒ 98 = 64+2y
⇒ 2y = 34
⇒ y = 17

35.

()
Given diameter of hemisphere = 14 cm
\(\therefore\) radius = 7 cm
\(\therefore\) Curved surface area = 2\(\pi\)r²
\(=2\times \frac { 22 }{ 7 } \times 7\times 7\)
= 308 cm²