Maths
Section -A
Q.1. If HCF (336,54) = 6, find LCM (336, 54),
Answer : Given, HCF (336, 54) = 6
We know HCF × LCM = one number × other number
6 × LCM = 336 × 54
LCM = 336×54/6 = 336 × 9 = 3024
Q.2. Find the nature of roots of the quadratic equation 2x2 – 4x + 3 = 0.
Answer: Given, 2x2 – 4x + 3 = 0
Comparing it with quadratic equation ax2 + bx + c = 0
Here, a = 2, b = -4 and c = 3
D = b2 – 4ac = (-4)2 – 4 × (2)(3) = 16 – 24 = -8 < 0
Hence, D < 0 this shows that roots will be imaginary.
Q.3. Find the common difference of the Arithmetic Progression (A.P.).
Answer :
Q.4. Evaluate: sin2 60° + 2 tan 45° – cos2 30°
OR
Q.4. If sin A = ¾, Calculate sec A.
Answer :
We know,
Q.5. Write the coordinates of a point P on the x-axis which is equidistant from point A(-2, 0) and B(6, 0).
Answer: Let coordinates of P on x-axis is (x, 0)
Given, A(-2, 0) and B(6, 0)
Here, PA = PB
On squaring both sides, we get
(x + 2)2 = (x – 6)2
⇒ x2 + 4 + 4x = x2 + 36 – 12x
⇒ 4 + 4x = 36 – 12x
⇒ 16x = 32
⇒ x = 2
Coordinates of P are (2, 0)
Q.6. Find the 21st term of the A.P. -4½, – 3, -1½ ………
Answer :
Section- B
Q.7. For what value of k, will the following pair of equations have infinitely many Answers:
2x + 3y = 7 and (k + 2)x – 3(1 – k)y = 5k + 1 [2]
Answer: Given, The system of equations is
2x + 3y = 7 and (k + 2) x – 3 (1 – k) y = 5k +1
These equations are of the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
where, a1 = 2, b1 = 3, c1 = -7
a2 = (k + 2), b2 = -3(1 – k), c2 = -(5k + 1)
Since, the given system of equations have infinitely many Answers.
Hence, the given system of equations has infinitely many Answers when k = 4.
Q.8. Find a relation between x and y if the points A(x, y), B(-4, 6) and C(-2, 3) are collinear.
OR
Q.8. Find the area of a triangle whose vertices are given as (1, -1) (-4, 6) and (-3, -5).
Answer: Given, A(x, y), B(-4, 6), C(-2, 3)
x1 = x, y1 = y, x2 = -4, y2 = 6, x3 = -2, y3 = 3
If these points are collinear, then area of triangle made by these points is 0.
Q.9. The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is ⅕. The probability of selecting a black marble at random from the same jar is ¼. If the jar contains 11 green marbles, find the total number of marbles in the jar.
Answer: Let the probability of selecting a blue marble, black marble and green marbles are P(x), P(y), P(z) respectively.
P(x) = ⅕ , P (y) = ¼ Given
We Know,
P(x) + P(y) + P(z) = 1
Q.10. Find the value(s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique Answer.
Answer: Given, x + 2y = 5, 3x + ky + 15 = 0
Comparing above equations with
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0,
We get,
a1 = 1, b1 = 2, c1 = -5
a2 = 3, b2 = k, c2 = 15
Condition for the pair of equations to have unique Answer is
k can have any value except 6.
Q.11. The larger of the two supplementary angles exceeds the smaller by 18°. Find the angles.
OR
Q.11. Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present?
Answer: Let two angles A and B are supplementary.
A + B = 180° …(i)
Given, A = B + 18°
On putting A = B + 18° in equation (i),
we get B + 18° + B = 180°
⇒ 2B + 18° = 180°
⇒ 2B = 162°
⇒ B = 81°
A = B + 18°
⇒ A = 99°
OR
Answer: Let age of Sumit be x years and age of his son be y years.Then, according to question we have, x = 3y …… (i)
Five years later, x + 5 = 2½ (y + 5)……….(ii)
On putting x = 3y in equation (ii)
Q.12. Find the mode of the following frequency distribution:
Answer :
Here, the maximum frequency is 50.
So, 35 – 40 will be the modal class.
l = 35, f0 = 34, f1 = 50, f2 = 42 and h = 5
Section – C
Q.13. Point A lies on the line segment XY joining X(6, -6) and Y (-4, -1) in such a way that XA/XY = ⅖.
If Point A also lies on the line 3x + k (y + 1) = 0, find the value of k.
Answer: Given,
Since, point A(2, -4) lies on line 3x + k (y + 1) = 0.
Therefore it will satisfy the equation.
On putting x = 2 and y = -4 in the equation, we get
3 × 2 + k(-4 + 1) = 0
⇒ 6 – 3k = 0
⇒ 3k = 6
⇒ k = 2
Q.14. Solve for x: x2 + 5x – (a2 + a – 6) = 0 [3]
Answer: Taking (a2 + a – 6)
= a2 + 3a – 2a – 6
= a(a + 3) – 2 (a + 3)
= (a + 3) (a – 2)
x2 + 5x – (a + 3) (a – 2) = 0
⇒ x2 + (a + 3)x – (a – 2)x – (a + 3)(a – 2) = 0
⇒ x[x + (a + 3)] – (a – 2) [x + (a + 3)] = 0
⇒ (x – a + 2)(x + a + 3) = 0
Hence, x – a + 2 = 0 and x + a + 3 = 0
x = a – 2 and x = -(a + 3)
Required values of x are (a – 2), -(a + 3).
Q.15. Find A and B if sin (A + 2B) = √3/2 and cos (A + 4B) = 0, where A and B are acute angles.
Answer: Given
Sin (A + 2B) = √3/2 and cos (A + 4B) = 0
⇒ sin (A + 2B) = 60° (∵ sin 60° = √3/2)
A + 2B =60 …(i)
cos (A + 4B) = cos 90° (∵ cos 90° = 0)
⇒ A + 4B = 90° …(ii)
On solving equation (i) and (ii), we get
B = 15° and A = 30°
Q.16. Find the ratio in which the line x – 3y = 0 divides the line segment joining the points (-2, -5) and (6, 3). Find the coordinates of the point of intersection.
Answer: Let the required ratio be k : 1
By section formula, we have
Q.17. Evaluate:
Answer :
Q.18. In Figure 4, a square OABC is inscribed in a quadrant OPBQ. If OA = 15 cm, find the area of the shaded region. (Use π = 3.14)
OR
Q.18. In Figure 5, ABCD is a square with side 2√2 cm and inscribed in a circle. Find the area of the shaded region. (Use π = 3.14)
Answer: Given, OABC is a square with OA = 15 cm
OB = radius = r
Let side of square be a then,
a2 + a2 = r2
⇒ 2a2 = r2
⇒ r = √2 a
⇒ r = 15√2 cm (∵ a = 15 cm)
Area of square = Side × Side = 15 × 15 = 225 cm2
Area of shaded region = Area of quadrant OPBQ – Area of square
= 353.25 – 225 = 128.25 cm
OR
Answer: Given, ABCD is a square with side 2√2 cm
BD = 2r
In ∆BDC,
BD2 = DC2 + BC2
⇒ 4r2 = 2(DC)2 (∵ DC = CB = Side = 2√2 )
⇒ 4r2 = 2 × 2√2 × 2√2
⇒ 4r2 = 8 × 2
⇒ 4r2 = 16
⇒ r2 = 4
⇒ r = 2 cm
Area of square BCDA = Side × Side = DC × BC = 2√2 × 2√2 = 8 cm2
Area of circle = πr2 = 3.14 × 2 × 2 = 12.56 cm2
Area of shaded region = Area of circle – Area of square. = 12.56 – 8 = 4.56 cm2
Q.19. A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume of the solid. (Use π = 22/7).
Answer: ABCD is a cylinder and BFC and AED are two hemispheres which has radius (r) = 7/2 cm
Total volume of solid = Volume of two hemisphere + Volume of cylinder
= 179.67 + 500.5 = 680.17 cm
Q.20. The marks obtained by 100 students in an examination are given below:
Find the mean marks of the students.
Answer :
Q.21. For what value of k, is the polynomial f(x) = 3x4 – 9x3 + x2 + 15x + k completely divisible by 3x2 – 5?
OR
Q.21. Find the zeroes of the quadratic polynomial 7y2 – 11/2y – ⅔ and verify the relationship between the zeroes and the coefficients.
Answer: Given, f(x) = 3x4 – 9x3 + x2 + 15x + k
It is completely divisible by 3x2 – 5
Q.22. Write all the values of p for which the quadratic equation x2 + px + 16 = 0 has equal roots. Find the roots of the equation so obtained.
Answer: Given, equation is x2 + px + 16 = 0
This is of the form ax2 + bx + c = 0
where, a = 1, b = p and c = 16
D = b2 – 4ac = p2 – 4 × 1 × 16 = p2 – 64
for equal roots, we have D = 0
p2 – 64 = 0
⇒ p2 = 64
⇒ p = ±8 Putting p = 8 in given equation we have,
x2 + 8x + 16 = 0
⇒ (x + 4)2 = 0
⇒ x + 4 = 0⇒ x = -4
Now, putting p = -8 in the given equation, we get
x2 > – 8x + 16 = 0
⇒ (x – 4)2 = 0
⇒ x = 4
Required roots are -4 and -4 or 4 and 4.
Section – D
Q.23. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares on their corresponding sides.
Answer: Given, ΔABC ~ ΔDEF
Q.24. Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point P between them on the road, the angle of elevation of the top of a pole is 60° and the angle of depression from the top of the other pole of point P is 30°. Find the heights of the poles and the distance of the point P from the poles.
Answer: Let AC is the road of 80 m width. P is the point on road AC and height of poles AB and CD is h m.
h———–(i)
Equating the values of h from equation (i) and (ii) we get
⇒ x√3 =
⇒ 3x = 80 – x
⇒ 4x = 80
⇒ x = 20m
On putting x = 20 in equation (i), we get
h = √3 × 20 = 20√3
h = 20√3 m
Thus, height of poles is 20√3 m and point P is at a distance of 20 m from left pole and (80 – 20) i.e., 60 m from right pole.
Q.25. The total cost of a certain length of a piece of cloth is ₹ 200. If the piece was 5 m longer and each metre of cloth costs ₹ 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per metre?
Answer: Let the original length of the piece of cloth is x m and rate of cloth is ₹ y per metre.
Then according to question, we have
x × y = 200 …(i)
and if length be 5 m longer and each meter of cloth be ₹ 2 less than
(x + 5) (y – 2) = 200
⇒ (x + 5) (y – 2) = 200
⇒ xy – 2x + 5y – 10 = 200 …(ii)
On equating equation (i) and (ii), we have
xy = xy – 2x + 5y – 10
⇒ 2x – 5y = -10 …… (iii)
⇒ y = 200/x from equation (i)
⇒ 2x – 5 × 200/x = -10
= 2x – 1000/x -10
⇒ 2x2 – 1000 = -10x
⇒ 2x2 + 10x – 1000 = 0
⇒ x2 + 5x – 500 = 0
⇒ x2 + 25x – 20x – 500 = 0
⇒ x(x + 25) – 20 (x + 25) = 0
⇒ (x + 25) (x – 20) = 0
⇒ x = 20 (x ≠ -25 length of cloth can never be negative)
∴ x × y = 200
20 × y = 200
y = 10
Thus, the length of the piece of cloth is 20 m and original price per metre is ₹ 10.
Q.26. Construct an equilateral ∆ABC with each side 5 cm. Then construct another triangle whose sides are ⅔ times the corresponding sides. Draw two concentric circles of radii 2 cm and 5 cm. Take a point P on the outer circle and construct a pair of tangents PA and PB to the smaller circle. Measure PA.
Answer: Steps for construction are as follows:
- Draw a line segment BC = 5 cm
- At B and C construct ∠CBX = 60° and ∠BCX = 60°
- With B as centre and radius 5 cm, draw an arc cutting ray BX at A. On graph paper, we take the scale.
- Join AC. Thus an equilateral ∆ABC is obtained.
- Below BC, make an acute angle ∠CBY
- Along BY, mark off 3 points B1, B2, B3 Such that BB1, B1B2, B2B3 are equal.
- Join B3C
- From B2 draw B2D || B3C, meeting BC at D
- From D, draw DE || CA, meeting AB at E.
Then ∆EBD is the required triangle, each of whose sides is ⅔ of the corresponding side of ∆ABC.
Q.27. Change the following data into ‘less than type’ distribution and draw its olive:
Answer :
Q.28. Prove that:
Answer :
Q.29. Which term of the Arithmetic Progression -7, -12, -17, -22,…..will be -82? Is -100 any term of the A.P.? Give a reason for your answer.
OR
Q.29. How many terms of the Arithmetic Progression 45, 39, 33, … must be taken so that their sum is 180? Explain the double answer.
Answer:-7, -12, -17, -22, …….
Here a = -7, d = -12 – (-7) = -12 + 7 = -5
Let Tn = -82Tn = a + (n – 1) d
⇒ -82 = -7 + (n – 1) (-5)
⇒ -82 = -7 – 5n + 5
⇒ -82 = -2 – 5n
⇒ -82 + 2 = -5n
⇒ -80 = -5n
⇒ n = 16
Therefore, 16th term will be -82.
Let Tn = -100
Again, Tn = a + (n -1) d
⇒ -100 = -7 + (n – 1) (-5)
⇒ -100 = -7 – 5n + 5
⇒ -100 = – 2 – 5n
⇒ -100 + 2 = -5n
⇒ -98 = -5n
⇒ n = 98/5
But the number of terms can not be in fraction.
So, -100 can not be the term of this A.P.
OR
Answer: 45, 39, 33, …..
Here a = 45, d = 39 – 45 = -6
Let Sn = 180
⇒ n/2 [ 2a + (n – 1) d] = 180
⇒ n/2 [2 × 45 + (n – 1) (-6)] = 180
⇒ n/2 [90 – 6n + 6] = 180
⇒ n/2 [96 – 6n] = 180
⇒ n(96 – 6n) = 360
⇒ 96n – 6n2 = 360
⇒ 6n2 – 96n + 360 = 0
On dividing the above equation by 6
⇒ n2 – 16n + 60 = 0
⇒ n2 – 10n – 6n + 60 = 0
⇒ n(n – 10) – 6 (n – 10) = 0
⇒ (n – 10) (n – 6) = 0
⇒ n = 10, 6
Sum of first 10 terms = Sum of first 6 terms = 180
This means that the sum of all terms from 7th to 10th is zero.
Q.30. In a class test, the sum of Aran’s marks in Hindi and English is 30. Had he got 2 marks more in Hindi and 3 marks less in English, the product of the marks would have been 210. Find his marks in the two subjects.
Answer: Let Aran marks in Hindi be x and marks in English be y.
Then, according to question, we have
x + y = 30 …(i)(x + 2)(y – 3) = 210 …(ii)
from equation (i) put x = 30 – y in equation (ii)
(30 – y + 2) (y – 3) = 210
⇒ (32 – y) (y – 3) = 210
⇒ 32y – 96 – y2 + 3y = 210
⇒ y2 – 35y + 306 = 0
⇒ y2 – 18y – 17y + 306 = 0
⇒ y(y – 18) – 17(y – 18) = 0
⇒ (y – 18) (y – 17) = o
⇒ y = 18, 17
Put y = 18 and 17 in equation (i), we get x = 12, 13
Hence his marks in hindi can be 12 and 13 and in english his marks can be 18 and 17.