Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Basic with Solutions Set 8 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Basic Set 8 with Solutions
Time: 3 Hours
Maximum Marks: 80
General Instructions:
Read the following instructions very carefully and strictly follow them:
- This question paper contains two parts A and B.
- Both Part A and Part B have internal choices.
Part A
- Consists of two Sections, I and II.
- Section I has 16 questions of 1 mark each. Internal choices are provided in 5 questions.
- Section II has 4 questions on case study (Q.No. 17 – 20). Each question has 5 sub-parts. An examinee is to attempt any 4 out of 5 sub-parts. Each is of 1 mark.
Part B
- Consists of three sections III, IV and V.
- Section III has 6 questions No. 21 to 26 of Very-short Answer Type of 2 marks each.
- Section IV has 7 questions No. 27 to 33 of Short Answer Type of 3 marks each.
- Section V has 3 questions No. 34 to 36 of Long Answer Type of 5 marks each.
- Internal choice is provided in 2 questions in Section III, 2 questions in Section IV and 1 question in Section V.
PART – A
Section – I (16 marks)
Question 1.
Find the distance between the points 1
A(-\(\frac{7}{3}\), 5) and B (\(\frac{2}{3}\), 5).
Answer:
Given points are A(-\(\frac{7}{3}\), 5) and B (\(\frac{2}{3}\), 5).
Distance between A and B using distance formula,
d = \(\sqrt{\left(\frac{2}{3}-\left(-\frac{7}{3}\right)\right)^2+(5-5)^2}\)
= \(\sqrt{\left(\frac{2}{3}+\frac{7}{3}\right)^2+0}=\sqrt{(3)^2}\) = 3 units
Question 2.
Express 288 as product of its prime factors.1
Answer:
Prime factorisation of 288
288 = 2 × 2 × 2 × 2 × 2 × 3 × 3
Question 3.
Write the common difference of the A.P.:
\(\frac{1}{5}\), \(\frac{4}{5}\), \(\frac{7}{5}\), \(\frac{10}{5}\), ….
OR
Find the 8th term of the A.P. whose first term is -2 and common difference is 3. 1
Answer:
Given A.P. is \(\frac{1}{5}\), \(\frac{4}{5}\), \(\frac{7}{5}\), \(\frac{10}{5}\), ….
d = \(\frac{4}{5}-\frac{1}{5}=\frac{3}{5} ; \frac{7}{5}-\frac{4}{5}=\frac{3}{5} ; \frac{10}{5}-\frac{7}{5}=\frac{3}{5}\)
OR
nth term of an with first term a and common difference d is given as,
an = a + (n – 1) d
Hence, a = – 2 and d = 3
So, 8th term, a8 = – 2 + (8 – 1) × 3
= -2 + 7 × 3 = -2 + 21
= 19
Question 4.
Find the sum and product of zeroes of the polynomial, p(x) = x2 + 5x + 6. 1
Answer:
Given polynomial is p(x) = x2 + 5x + 6
Sum of zeroes = –\(\frac{5}{1}\) = -5
And product of zeroes = \(\frac{6}{1}\) = 6
Question 5.
If 2 cos θ = √3, then find the value of θ. 1
Answer:
Given: 2cosθ = √3
⇒ cosθ = \(\frac{\sqrt{3}}{2}\) = cos 30°
⇒ θ = 30°
Question 6.
In the given figure ΔABC ~ ΔPQR. Write similarity criterion by which ΔABC and ΔPQR are similar.
Answer:
In ∠ABC and ∠PQR
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{2}{4}=\frac{1}{2}\) and \(\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{3.5}{7}=\frac{1}{2}\)
and ∠ABC = ∠PQR = 50°
Here, two sides are in proportion and angle included between two sides are equal, so
AABC ~ APQR by SAS similarity criterion.
Question 7.
The graph of y = p(x) is shown in figure for some polynomial p(x). Find the number of zeroes of p(x). 1
Answer:
Number of zeroes of p(x) is 0 because y ≠ 0 at any value of x.
Question 8.
Find the discriminant of the quadratic equation 2x2 – 5x – 6 = 0. 1
Answer:
Given quadratic equation: 2x2 – 5x – 6 = 0 Discriminant,
D = b2 – 4ac
= (-5)2 – 4 × 2 × (-6)
= 25 + 48 = 73
Question 9.
A card is drawn at random from a well- shuffled pack of 52 playing cards. Find the probability of getting a red face card. 1
Answer:
Total number of possible outcomes = 52
Red face cards are 2 kings, 2 queens, 2 jacks Number of red face cards = 6
Therefore, P(getting a red face card) = \(\frac{6}{52}\) = \(\frac{3}{26}\)
Question 10.
Show that the tangents drawn at the ends of a diameter of a circle are parallel. 1
Answer:
In the figure, PQ is diameter of the given circle and O is its centre.
Let tangents AB and CD be drawn at the end points of the diameter PQ.
Since the tangent at a point to a circle is perpendicular to the radius through the point.
PQ ⊥ AB
⇒ ∠APQ = 90°
And PQ ⊥ CD
⇒ ∠PQD = 90°
⇒ ∠APQ = ∠PQD
They form a pair of alternate angles.
AB || CD
Question 11.
If PL and PM are two tangents to a circle with centre O from an external point P and PL = 4 cm, find the length of OP, where radius of the circle is 3 cm.
OR
Find the distance between two parallel tangents of a circle of radius 2.5 cm. 1
Answer:
Since, the tangent to a circle is perpendicular to the radius through the point of contact.
∴ ∠OLP = 90°
Now, in the right ΔOLP, we have:
OP2 = OL2 + LP2
⇒ OP2 = (3)2 + (4)2
⇒ OP2 = 9 + 16
⇒ OP2 = 25
⇒ OP = 5 cm
Thus, the length of OP is 5 cm.
OR
Distance between two parallel tangents is equal to diameter of the circle.
So, distance = 2 × 2.5 cm = 5 cm
Question 12.
Two different coins are tossed simultaneously. Write all the possible outcomes.
OR
A die is thrown once. Write the probability of getting a number less than 7. 1
Answer:
When two different coins are tossed, all possible outcomes are (HH, HT, TH, TT).
OR
When a die is thrown, possible outcomes are 1, 2, 3, 4, 5, 6.
Number less than 7 = 1, 2, 3, 4, 5, 6
So, probability of getting number less than 7
= \(\frac{6}{6}\) = 1
Question 13.
Write the expression for the volume of a cone of radius Y and height three times the radius V.
OR
Write the expression for the total surface area of a solid hemisphere of radius ‘r’. 1
Answer:
Since, volume of cone,
V = \(\frac{1}{3}\) πr2h
Given, height is 3 times the radius,
h = 3r
then V = \(\frac{1}{3}\)πr2 × 3r = πr3
OR
Total surface area of solid hemisphere,
S = curved surface area + area of circle
S = 2πr2 + πr2 = 3πr2
Question 14.
A vertical pole is 100 metres high. Find the angle subtended by the pole at a point on the ground 100√3 meters from the base of the pole. 1
Answer:
Let AB be the pole, AC be the distance of a point C from the base of the pole.
Then, ∠ACB = θ, ∠CAB = 90°,
AB = 100 m and AC = 100√3 m.
tanθ = \(\frac{100}{100 \sqrt{3}}=\frac{1}{\sqrt{3}}\) = tan30°
Thus, the angle subtended by the pole at a point on the ground 100√3 meters from the base of the pole is 30°.
Question 15.
In ΔABC, right-angled at A, if AB = 7 cm and AC = 24 cm, then find sin B and tan C. 1
Answer:
In ΔABC, AB = 7cm, AC = 24 cm
Using Pythagoras theorem,
BC = AC2 + AB2
= (2 4)2 + (7)2
= 576 + 49
= 625
BC = 25 cm
Now sin B = \(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{24}{25}\)
and tan C = \(\frac{A B}{A C}=\frac{7}{24}\)
Section – II
Case study based questions (Q. No. 17 – 20) are compulsory. Attempt any 4 sub-parts from each question. Each sub-part carries 1 mark.
Question 16.
During the lockdown period, many families got bored of watching TV all the time. Out of these families, one family of 6 members decided to play a card game. 17 cards numbered 1, 2, 3, 4,…, 17 are put in a box and mixed thoroughly.
One card is drawn by one member at random and other family members bet for the chances of drawing the number either prime, odd or even etc.
Based on the above information, answer the following questions:
(A) The first member of the family draws a card at random and another member bets that it is an even prime number. What is the probability of his winning the bet?
(a) \(\frac{2}{17}\)
(b) \(\frac{3}{17}\)
(c) \(\frac{1}{17}\)
(d) \(\frac{4}{17}\)
Answer:
(a) \(\frac{2}{17}\)
Explanation: Total possible outcomes = 1, 2, 3 ………, 17, which is 17 in numbers
Even prime number = 2
P(getting even prime number) = \(\frac{2}{17}\)
(B) The second member of the family draws a card at random and some other member bets that it is an even number. What is the probability of his winning the bet?
(a) \(\frac{7}{17}\)
(b) \(\frac{8}{17}\)
(c) \(\frac{9}{17}\)
(d) \(\frac{5}{17}\)
Answer:
(b) \(\frac{8}{17}\)
Explanation: Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, which is 8 in number.
P(getting even number) = \(\frac{8}{17}\)
(C) What is the probability that the number on the card drawn at random is divisible by 5?
(a) \(\frac{5}{17}\)
(b) \(\frac{4}{17}\)
(c) \(\frac{3}{17}\)
(d) \(\frac{2}{17}\)
Answer:
(c) \(\frac{3}{17}\)
Explanation: Numbers divisible by 5 are 5, 10, 15, which is 3 in number.
P(getting number divisible by 5) = \(\frac{3}{17}\)
(D) What is the probability that the number on the card drawn at random is a multiple of 3?
(a) \(\frac{5}{17}\)
(b) \(\frac{6}{17}\)
(c) \(\frac{7}{17}\)
(d) \(\frac{8}{17}\)
Answer:
(a) \(\frac{5}{17}\)
Explanation: Numbers multiple of 3 are 3, 6, 9, 12, 15,
which is 5 in number.
P(number multiple of 3) = \(\frac{5}{17}\)
(E) What is the probability that the number on the card is a factor of 9?
(a) \(\frac{9}{17}\)
(b) \(\frac{3}{17}\)
(c) \(\frac{8}{17}\)
(d) \(\frac{1}{17}\)
Answer:
(b) \(\frac{3}{17}\)
Explanation: Factors of 9 are 1, 3, 9, which is 3 in number.
P(getting a factor of number) = \(\frac{3}{17}\)
Question 17.
Roshni being a plant lover decides to start a nursery. She bought few plants with pots. She placed the pots in such a way that the number of pots in the first row is 2, in the second is 5, in the third row is 8 and so on.
Based on the above, answer the following questions:
(A) How many pots were placed in the 7th row? 1
(a) 20
(b) 23
(c) 77
(d) 29
Answer:
(a) 20
Explanation: Pots are arranged in an AP, in which first term, a = 2 and common difference,
d = 5 – 2 = 8 – 5 = 3
Number of pots in 7th row,
a7 = a + (7 – 1) × d
= 2 + 6 × 3 = 2 + 18 = 20.
(B) If Roshni wants to place 100 pots in total, then total number of rows formed in the arrangement will be: 1
(a) 8
(b) 9
(c) 10
(d) 12
Answer:
(a) 8
Explanation: Total number of pots = 100
Let total number of rows = n
We have,
(C) How many pots are placed in the last row? 1
(a) 20
(b) 23
(c) 26
(d) 29
Answer:
(b) 23
Explanation: Number of pots in last row means number of pots in 8th row
a8 = a + (8 – 1 )d
= 2 + 7 × 3 = 23
(D) If Roshni has sufficient space for 12 rows, then how many total number of pots are placed by her with the same arrangement? 1
(a) 222
(b) 155
(c) 187
(d) 313
Answer:
(a) 222
Explanation: Number of pots till 12th row,
S12 = \(\frac{12}{2}\)[2 × 2+ (12 – 1) × 3]
= 6[4 + 33] = 6 × 37 = 222
(E) The difference in number of pots placed in the 4th row and the 2nd row, is: 1
(a) 3
(b) 4
(c) 6
(d) 8
Answer:
(c) 6
Explanation: Difference in number of pots in 4th row and 2nd row
= a4 – a2
= (a + (4 – 1) d) – (a + (2 – 1) d)
= a + 3d – a – d = 2d
= 2 × 3 = 6
Question 18.
To explain how trigonometry can be used to measure the height of an inaccessible object, a teacher gave the following example to students:
A TV tower stands vertically on the bank of a canal. From a point on the other bank directly opposite the tower, the angle of the elevation of the top of the tower is 60°. From another point 20 m away from this point on the Line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (as shown in figure).
Based on the above information, answer the following questions:
(A) The width of the canal is: 1
(a) 10√3 m
(b) 20√3 m
(c) 10 m
(d) 20 m
Answer:
(A) (c) 10 m
Explanation: Given, DC = 20 m
Let BC = x m and A =h m.
Now, in right ΔABD,
\(\frac{\mathrm{AB}}{\mathrm{BD}}\) = tan30°
⇒ \(\frac{\mathrm{AB}}{20+x}=\frac{1}{\sqrt{3}}\)
⇒ h = \(\frac{20+x}{\sqrt{3}}\) …… (i)
Also, in right ΔABC,
\(\frac{\mathrm{AB}}{\mathrm{BC}}\) = tan 60° BC
⇒ \(\frac{A B}{x}\) = √3
⇒ h = √3x ….. (ii)
From (i) and (ii), we get
\(\frac{20+x}{\sqrt{3}}\) = √3x
⇒ 20 + x = 3x
⇒ 2x = 20
⇒ x = 10m
(B) Height of the tower is: 1
(a) 10√3 m
(b) 10 m
(c) 20√3 m
(d) 20 m
Answer:
(a) 10√3m
Explanation: From equation (i) in part (A), we get
h = √3x = 10√3m
(C) Distance of the foot of the tower from the point D is: 1
(a) 20 m
(b) 30 m
(c) 10 m
(d) 20√3 m
Answer:
(b) 30 m
Explanation: Distance of the foot of the tower from the point
D = DC + BC
= 20 m + 10 m = 30 m
(D) The angle formed by the line of sight with the horizontal when it is above the horizontal line is known as: 1
(a) angle of depression
(b) line of sight
(c) angle of elevation
(d) obtuse angle
Answer:
(c) angle of elevation.
(E) In figure, measure of angle XAC is: 1
(a) 30°
(b) 60°
(c) 90°
(d) 45°
Answer:
(b) 60°
Explanation: ∠XAC = ∠ACB = 60° [Alternate angle]
Question 19.
A children’s park is in the triangular shape as shown in Figure below. In the middle of the park, there is a circular region for younger children to play. It is fenced with three layers of wire. The radius of the circular region is 3 m.
Based on the above information, answer the following questions:
(A) The perimeter (or circumference) of the circular region is: 1
(a) 3π m
(b) 18π m
(c) 6π m
(d) 9π m
Answer:
(c) 6π m
Explanation: Given that radius of circular region = 3 m.
Circumference of circular region
= 2πr = 2 × π × 3m = 6π m
(B) The total length of wire used is: 1
(a) 9π m
(b) 18π m
(c) 54π m
(d) 27π m
Answer:
(b) 18π m2
Explanation: Circular region is fenced with 3 layers of wire.
So, total length of wire
= 3 × circumference of circular region
= 3 × 6π m = 18π m.
(C) The area of the circular region is: 1
(a) 54π m2
(b) 3π m2
(c) 18π m2
(d) 9π m2
Answer:
(d) 9π m2
Explanation: Area of circle = πr2, where r is radius of circle.
So, area of circular region
= π × (3)2 = 9π m2
(D) If BD = 6 m, DC = 9 m and ar (ABC) = 54 m2, then the Length of sides AB and AC, respectiveLy, are: 1
(a) 9 m, 12 m
(b) 12m, 9m
(c) 10m, 12 m
(d) 12m, 10m
Answer:
(a) 9 m, 12 m
Explanation:
Let x be the length of AE.
Tangents from an external point to a circle are equal in length,
So BD = BE = 6m;
AE = AF = x and DC = CF = 9 m.
Perimeter of triangle,
P = AB + BC + AC
= (6 + x)m + (15m) + (9 + x) m
= (30 + 2x) m
Semi-perimeter
s = \(\frac{P}{2}\) = (15 + x)m
According to Heron’s formula, area of triangle, with sides a, b, c and semi¬perimeter s
= \(\sqrt{s(s-a)(s-b)(s-c)}\)
Or, ar(∆ABC)
= \(\sqrt{\begin{array}{r}
(15+x)\{(15+x)-15\}\{(15+x)-(x+6)\} \\
\{(15+x)-(x+9)\}
\end{array}}\)
= \(\sqrt{(15+x) \times x \times 9 \times 6}\) = \(\sqrt{(15+x) \times x \times 54}\)
Given area of triangle = 54 m2
⇒ \(\sqrt{(15+x) \times x \times 54}\) =54
Squaring both sides,
(15 + x) × x × 54 = 54 × 54
⇒ x(15 + x) = \(\frac{54 \times 54}{54}\) = 54
⇒ x2 + 15x – 54 = 0
⇒ x2 + 18 – 3x – 54 = 0
⇒ x(x + 18) – 3(x + 18) = 0
⇒ (x + 18) (x – 3) =0
⇒ x = -18 or 3
Side cannot be negative, so x = 3
Thus, AB = (x + 6)m = (3 + 6)m
= 9 m;
and AC = (x + 9) m
= (3 + 9) m
= 12 m
(E) The perimeter of ABC is: 1
(a) 28m
(b) 37m
(c) 36m
(d) 38m
Answer:
(c) 36m
Explanation: Perimeter of triangle ABC
= AB + BC + AC
= 9m + 15m + 12m
= 36 m
PART – B
Section – III (12 marks)
All questions are compulsory. In case of internal choices, attempt any one.
Question 20.
Find the LCM and HCF of two numbers 26 and 91 by the method of prime factorisation. 2
Answer:
Prime factorisation of 26 and 91
26 = 2 × 13;
91 = 7 × 13
LCM = 2 × 7 × 13 = 182 and HCF = 13
Question 21.
If sin (A + B) = \(\frac{\sqrt{3}}{2}\), sin (A – B) = \(\frac{1}{2}\), where 0° < A + B < 90°; A > B, then find the values of A and B.
OR
Simplify:
\(\frac{\sin 30^{\circ}+\tan 45^{\circ}-{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\) 2
Answer:
Given,
sin (A + B) = \(\frac{\sqrt{3}}{2}\)
⇒ sin (A + B) = sin 60°
⇒ A + B = 60° ……… (i)
and sin (A – B) = \(\frac{1}{2}\)
⇒ sin (A – B) = sin 30°
⇒ A – B = 30° …….. (ii)
Adding (i) and (ii), we get
2A = 90°
⇒ A = 45°
Putting value of A in (i), we get
45° + B = 60°
⇒ B = 15°
Therefore, A = 45° and B = 15°.
OR
Question 22.
A quadrilateral ABCD is drawn to circumscribe a circle (see Figure). Prove that AB + CD = AD + BC. 2
Answer:
Since the sides of quadrilateral ABCD, i.e., AB, BC, CD and DA touch the circle at P, Q, R and S respectively, and the lengths of two tangents to a circle from an external point are equal.
∴ AP = AS
BP = BQ
DR = DS
CR = CQ
Adding them, we get
(AP + BP) + (CR + RD) = (BQ + QC) + (DS + SA)
⇒ AB + CD = BC + DA
or, AB + CD = AD + BC
Hence, proved.
Question 23.
The greater of two supplementary angles exceeds the smaller by 18°. Find the two angles. 2
Answer:
Let one angle be x and its supplementary angle be y.
Let x > y
1st condition: x + y = 180° ………… (i)
2nd conditions: x – y = 18°
⇒ x = 18° + y ……….. (ii)
Putting the value of x in equation (i), we get
18° + y + y= 180°
⇒ 18° + 2y = 180°
2y = 162°
⇒ y = 81°
From (ii), x = 18° + 81° = 99°
∴ One angle 81° and another angle 99°.
Question 24.
Find the coordinates of the point which divides the line segment joining the points A(7, -1) and B(-3, -4) in the ratio 2 : 3. 2
Answer:
Let C (x, y) divide the line segment joining the points A (7, -1) and B (-3, -4) in the ratio 2 : 3.
Thus, the point C is
[\(\frac{2 \times(-3)+3 \times 7}{2+3}\), \(\frac{2 \times(-4)+3 \times(-1)}{2+3}\)]
C[latex]\frac{21-6}{5}, \frac{-8-3}{5}[/latex] = C(3, \(\frac{-11}{5}\))
So, coordinate of the point is (3, \(\frac{-11}{5}\))
Question 26.
Find whether the following pair of linear equations are consistent or inconsistent. 2
5x – 3y= 11, -10x + 6y = 22
OR
Solve for x and y:
x + y = 6, 2x – 3y = 4
Answer:
Given equations are
5x – 3y = 11
and -10x + 6y = 22
Here a1 = 5, b1 = -3, c1 = -11
a2 = -10, b2 = 6, c2 = -22
Now \(\frac{a_1}{a_2}=\frac{5}{-10}=-\frac{1}{2}\)
\(\frac{b_1}{b_2}=\frac{-3}{6}=-\frac{1}{2}\)
\(\frac{c_1}{c_2}=\frac{-11}{-22}=\frac{1}{2}\)
Since, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), so equations has no solutions. So, equations are inconsistent.
OR
Given equations are:
x + y = 6 …….. (i)
and 2x – 3y = 4 ……… (ii)
Multiplying (i) by 2 and subtracting it from (ii),
we get
2x – 3y – (2x + 2y) = 4 – 12
-5y = -8
y = \(\frac{8}{5}\)
Putting value of y in (i), we get
x + \(\frac{8}{5}\) = 6
x = 6 – \(\frac{8}{5}\)
x = \(\frac{22}{5}\)
So, x = \(\frac{22}{5}\) and y = \(\frac{8}{5}\)
Section – IV (21 Marks)
Question 25.
Prove that 7√2 is an irrational number, given that √2 is an irrational number. 3
Answer:
Let 7√2 be rational.
Then, 7√2 = \(\frac{p}{q}\) where p and q are non-zero integers, having no common factor other than 1.
Now, 7√2 = \(\frac{p}{q}\)
⇒ √2 = \(\frac{p}{7 q}\) …….. (i)
But, p and lq both are rational and 7q ≠ 0.
∴ \(\frac{p}{7 q}\) is rational.
Thus, from (i), it follows that √2 is rational.
This contradicts the fact that √2 is irrational.
The contradiction arises by assuming that √2 is rational.
Hence, 7√2 is irrational.
Question 26.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre. 3
Answer:
Since, ∠OAP = 90° and ∠OBP = 90°
[Angle between the tangent and the radius of the circle is 90°]
In quadrilateral OAPB,
∠OAP + ∠APB + ∠OBP + ∠AOB = 360° [Angles sum property]
⇒ 90° + ∠APB + 90° + ∠AOB = 360°
⇒ ∠APB + ∠AOB = 360° – 180° = 180°
Hence, proved.
Question 27.
Prove that
sec θ(1 -sin θ) (sec θ + tan θ) = 1
Prove that
\(\frac{1+\sec A}{\sec A}=\frac{\sin ^2 A}{1-\cos A}\) 3
Answer:
LHS = secθ(1 – sinθ)(secθ + tanθ)
OR
Question 28.
Show that the points A(1, 7), B(4, 2), C(-1, 1) and D(-4, 4) are the vertices of a square ABCD. 3
Answer:
The given four points are A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4).
So,
AB = BC = CD = AD and the diagonal AC = BD
Therefore, ABCD is a square.
Question 29.
If a, b are zeroes of the quadratic polynomial x2 + 9x + 20, form a quadratic polynomial whose zeroes are (α + 1) and (β + 1). 3
Answer:
Given polynomial is: x2 + 9x + 20
x2 + 9x + 20 = x2 + 5x + 4x + 20
= x(x + 5) + 4 (x +5)
= (x + 5) (x + 4)
To find zeroes of polynomials On comparing with 0,
x + 5 = 0 or x + 4 = 0
∴ x = -5, -4
So, α = -5, β = -4
Now, α + 1 = -5 + 1 = -4,
β + 1 = -4 + 1 = -3
Polynomial whose zeroes are (α + 1) and (β + 1):
x2 -(- 4)x – 3 = x2 + 4x – 3
Section – V (15 Marks)
Question 30.
The table shows the daily expenditure on food of 25 households in a locality:
Daily Expenditure (₹) | Number of Households |
100 – 150 | 4 |
150 – 200 | 5 |
200 – 250 | 12 |
250 – 300 | 2 |
300 – 350 | 2 |
Find the mean daily expenditure on food. Also, find the modal expenditure. 5
Answer:
We know that, mean \(\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{5275}{25}\) = 211
Thus, the mean daily expenditure of food = ₹ 211.
For Mode
Class | Frequency |
100 – 150 | 4 |
150 – 200 | 5 |
200 – 250 | 12 |
250 – 300 | 2 |
300 – 350 | 2 |
Highest frequency = 12
Modal class = 200 – 250
l = 200, F0 = 5, F1 = 12, F2 = 2, h = 50
Model = l + (\(\frac{F_1-F_0}{2 F_1-F_0-F_2}\)) × h
= 200 + (\(\frac{12-5}{2 \times 12-5-2}\)) × 50
= 200 + (\(\frac{7}{24-7}\)) × 50
= 200 + (\(\frac{7}{17}\)) × 50 = 200 + (\(\frac{350}{17}\)) × 50
= 200 + 20.6 = 220.6 (Approx)
Thus, modal expenditure = ₹ 220.6.
Question 31.
The diagonal of a rectangular field is 60 metres longer than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
OR
The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages (in years) was 124. Determine their present ages. 5
Answer:
Let x be the longer side and y be the shorter side of the rectangle.
Also, diagonal,
d = \(\sqrt{x^2+y^2}\)
According to the given condition:
d = 60 + y
⇒ \(\sqrt{x^2+y^2}\) = 60 + y ………. (i)
Also, x = y + 30
Putting value of x in (i),
\(\sqrt{(y+30)^2+y^2}\) = 60 + y
⇒ \(\sqrt{y^2+900+60 y+y^2}\) = 60 + y
= \(\sqrt{2 y^2+900+60 y}\) = 60 + y
Squaring both sides,
2y2 + 900 + 60y = (60 + y)2
⇒ 2y2 + 900 + 60y = y2 + 120y + 3600
⇒ y2 – 60y – 2700 = 0
⇒ y2 – 90y + 30y – 2700 = 0
⇒ y(y – 90) + 30(y – 90) = 0
⇒ (y + 30) (y – 90) = 0
⇒ y = -30,90
Length of side cannot be negative, so shorter side = 90 m
Longer side = 90 m + 30 m = 120 m
Thus, longer side = 120 m and shorter side = 90 m
OR
Let father’s present age = x years and Son’s present age = y years.
According to question: x + y = 45 ……… (i)
Five years ago: Father’s age = (x – 5) years and Son’s age = (y – 5) years
According to question:
⇒ (x – 5) × (y – 5) = 124
⇒ xy – 5x – 5y + 25 = 124
⇒ xy – 5(x + y)= 124 – 25
⇒ xy – 5 × 45 = 99
⇒ xy = 99 + 225 = 324 ……. (ii)
From (i), y = 45-x
Putting value of y in (ii), we get
⇒ x × (45 -x) = 324
⇒ 45x – x2 = 324
⇒ x2 – 45x + 324 = 0
⇒ x2 – 36x-9x + 324 = 0
⇒ x(x – 36) – 9(x – 36) = 0
⇒ (x – 9)(x – 36) = 0
⇒ x = 9, 36
∵ x cannot be 9 as x is father’s age.
∴ x = 36, then y = 45 – 36 = 9
Thus, father’s age = 36 years son’s present age = 9 years.