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,
Answer:
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. In Figure 1, ABC is an isosceles triangle right angled at C with AC = 4 cm. Find the length of AB.
OR
Q.6. In Figure 2, DE || BC. Find the length of side AD, given that AE = 1.8 cm, BD = 7.2 cm and CE = 5.4 cm.
Answer:
Given, ∠C = 90° and AC = 4 cm, AB = ?
∆ABC is an isosceles triangle so, BC = AC = 4 cm
On applying Pythagoras theorem, we have
AB2 = AC2 + BC2
⇒ AB2 = AC2 + AC2 (∵ BC = AC)
⇒ AB2 = 42 + 42 = 16 + 16 = 32
⇒ AB = √32 = 4√2 cm
OR
Answer: Given, DE || BCOn applying, Thales theorem, we have
Section – B
Q.7. Write the smallest number which is divisible by both 306 and 657.
Answer: Smallest number which is divisible by 306 and 657 is,
LCM (657, 306)
657 = 3 × 3 × 73
306 = 3 × 3 × 2 × 17
LCM =3 × 3 × 73 × 2 × 17 = 22338
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. Prove that 2 + 5√3 is an irrational number, given that √3 is an irrational number.
OR
Q.13. Using Euclid’s Algorithm, find the HCF of 2048 and 960.
Answer: Let 2 + 5√3 = r, where, r is rational.
⇒ (2 + 5√3)2 = r2
⇒ 4 + 75 + 20√3 = r2
⇒ 79 + 20√3 = r2
⇒ 20√3 = r2 – 79
⇒ √3 = r2 – 79/20
Now, r2 – 79/20 is a rational number. So, √3 must also be a rational number.
But √3 is an irrational number (Given).
So, our assumption is wrong.
2 + 5√3 is an irrational number.
Hence Proved.
OR
Answer: Step I: Here 2048 > 960 so, On applying Euclid’s algorithm, we get 2048 = 960 × 2 + 128
Step II: Because remainder 128 ≠ 0, so, On applying Euclid’s algorithm between 960 and 128, we get
960 = 128 × 7 + 64
Step III: Again remainder 64 ≠ 0, so
128 = 64 × 2 + 0
Here the remainder is 0. So, the process ends here. And the dividend is 64 so, required HCF is 64.
Q.14. Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side of BC. If AC and BD intersect at P, prove that AP × PC = BP × DP.
OR
Q.14. Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3RS. Find the ratio of the areas of triangles POQ and ROS.
Answer: Given, ∆ABC, ∆DBC are right-angle triangles, right-angled at A and D, on the same side of BC.
AC & BD intersect at P.
In ∆APB and ∆PDC,
∠A = ∠D = 90°
∠APB = ∠DPC (Vertically opposite)
∆APB ~ ∆PDC (By AA Similarity)
AP/BP = PD/PC (by c.s.s.t.)
⇒ AP × PC = BP × PD.
Hence Proved.
OR
Answer: Given, PQRS is a trapezium where PQ || RS and diagonals intersect at O and PQ = 3RS
In ∆POQ and ∆ROS, we have
∠ROS = ∠POQ (vertically opposite angles)
∠OQP = ∠OSR (alternate angles)
Hence, ∆POQ ~ ∆ROS by AA similarity then,
If two triangles are similar, then the ratio of areas is equal to the ratio of square of its corresponding sides. Then,
Q.15. In Figure 3, PQ and RS are two parallel tangents to a circle with centre O and another tangent AB with the point of contact C intersecting PQ at A and RS at B. Prove that ∠AOB = 90°.
Answer: Given, PQ || RS
To prove: ∠AOB = 90°
Construction: Join O and C, D and E
In ∆ODA and ∆OCA
OD = OC (radii of circle)
OA = OA (common)
AD = AC (tangent drawn from the same point)
By SSS congruence
∆ODA = ∆OCA
Then, ∠DOA = ∠AOC …(i)
Similarly, in ∆EOB and ∆BOC, we have
∆EOB = ∆BOC∠EOB = ∠BOC …(ii)
EOD is a diameter of the circle, therefore it is a straight line.
Hence, ∠DOA + ∠AOC + ∠EOB + ∠BOC = 180°
⇒ 2(∠AOC) + 2(∠BOC) = 180°
⇒ ∠AOC + ∠BOC = 90°
⇒ ∠AOB = 90°.Hence Proved.
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. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
Answer: Given, A ∆ABC in which DE || BC and DE intersect AB and AC at D and E respectively.
To prove: AD/DB = AE/EC
Construction: Join BE and CD
Draw EL ⊥ AB and DM ⊥ AC
Proof: we have
area (∆ADE) = ½ × AD × EL
and area (∆DBE) = ½ × DB × EL (∵ ∆ = ½ × b × h)
Now, ∆DBE and ∆ECD, being on same base DE and between the same parallels DE and BC, We have
area (∆DBE) = area (∆ECD) …..(iii)
from equations (i), (ii) and (iii), we have
AD/DC = AE/EC
Hence Proved.
Q.24. Amit, standing on a horizontal plane, finds a bird flying at a distance of 200 m from him at an elevation of 30°. Deepak standing on the roof of a 50 m high building, finds the angle of elevation of the same bird to be 45°. Amit and Deepak are on opposite sides of the bird. Find the distance of the bird from Deepak.
Answer: Let Amit be at C point and the bird is at A point. Such that ∠ACB = 30°. AB is the height of bird from point B on ground and Deepak is at D point, DE is the building of height 50 m.
Hence, the distance of bird from Deepak is 50√2 m.
Q.25. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 gm mass. (Use π = 3.14)
Answer: Let AB be the iron pole of height 220 cm with base radius 12 cm and there is the other cylinder CD of height 60 cm whose base radius is 8 cm.
Volume of AB pole = πr1h1 = 3.14 × 12 × 12 × 220 = 99475.2 cm3
Volume of CD pole = πr2h2 = 3.14 × 8 × 8 × 60 = 12057.6 cm3
Total volume of the poles = 99475.2 + 12057.6 = 111532.8 cm3
It is given that,
Mass of 1 cm3 of iron = 8 gm
Then mass of 111532.8 cm3 of iron = 111532.8 × 8 gm
Then the total mass of the pole is = 111532.8 × 8 gm = 892262.4 gm = 892.2624 kg
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.