CBSE Class 9 Mathematics Linear Equations In Two Variables Worksheet Set 03

Question. In which of the following equations will the value of y decrease as the value of x increases?
A. y = x-1
B.  y = 2x-15
C. y = 5-2x
D. y = x/3

Answer : C

Question. One man takes one day to dig a 4 m long trench. How long would it take 2 men working at the same rate to dig a 16 m long trench?
A. 1 day
B.  2 days
C. 4 days
D. 8 days

Answer : B

Question. Akbar, Ali and Arman are 3 brothers. The ratio of Akbar's age to Arman's age is 1 : 2 and the ratio of Akbar's age to Ali's age is 2 : 5. If the eldest boy is 10 years old, how old is the youngest one?
A. 1 year
B. 2 years
C. 4 years
D. 5 years

Answer : C

Question. Suparna has saved up Rs. 538 for a new pair of jeans. The price of the pair she wants is Rs. 750. Suparna's mother tells her that she can earn Rs. 15 per hour by helping her with the housework. Which of the following will help Suparna figure out how many hours (x) of housework she needs to do before she has enough to buy the jeans?
A. 538 + 15x = 750
B. 15x = 750
C. 750 - 538x =15
D. (538 + 15)x = 750

Answer : A

Question. This is a sale notice that Mr. Rai has put up in his shop to attract customers. However, not really intending to sell his goods at a low price, he has marked each item at 10% higher than the actual marked price before announcing the sale. At what price would he be selling an item whose original marked price was P?

A. P
B. 10%of P
C. 90%of P
D. 99% of P

Answer : D

Question. I started two clocks at the same time. One runs slow and loses 1 minute every hour. The other one is fast and gains 3 minutes every hour. How long will it take for the faster clock to be exactly one hour ahead of the slower clock?
A. 1 day
B. 18 hours
C. 15 hours
D. 12 hours

Answer : C

Question. In the weighing scales shown below, the same shapes represent the same weights. According to the balanced scales P and Q, what weights should be put on the pan on the right to balance scale R?

Answer : B

Question. P, Q and R are three friends. Q's height is 5/6 times the height of P. R's height is 1/5 times that of Q. Which of these statements is true?
A. The ratio of P's height to that of Q is 6 : 5.
B. P is shorter than Q
C. Q is shorter than R
D. P and R are of the same height

Answer : A

Question. Points P, Q and R are co-planar. In which of the following cases will they NECESSARILY be collinear?
A. When PQ = PR
B. When PQ + PR > QR
C. When PQ + QR = PR
D. When PR < PQ + QR

Answer : B

Question. For a quadrilateral PQRS inscribed in a circle, with  PQ || RS, which of the following is NOT necessarily true?

A. ∠Q + ∠ P = 180⁰
B. Q +S = 180⁰
C. Q +  R = 180⁰
D. S +  P = 180⁰

Answer : A

Question. In the following figure  PR || AC, QP || AB and RQ || BCIf the perimeter of triangleABC is 24 cm, the perimeter of triangle PQR will be

A. 6 cm
B. 8 cm
C. 12 cm
D. 16 cm

Answer : C

Question. If the length of the longer line is 60 cm, the length of the shorter one is
l .........................................
m........................................
A. 25 cm
B. 22.5 cm
C. 20 cm
D. 18 cm

Answer : B

Question. The clocks below show the date and time in two different places in the world at the same, who stays in Mumbai, wants to chat with a friend who stays in Los Angeles on the internet. Everyday, George is on the internet from 9 a.m. to 8 p.m. (Mumbai time) and his friend is on the internet from 6 a.m. to 7 p.m. (Los Angeles time). According to the local time in Mumbai, what would be a suitable time for them to chat?

A. 9.00 a.m. to 9.30 a.m
B. 6.00 p.m. to 7.00 p.m.
C. 7.30 p.m. to 8.00 p.m.
D. 9.00 a.m. to 7.00 p.m.

Answer : C

Question. The difference between two temperature readings in an experiment was 90Which of these could be the temperature readings?
A. 1 and -6^0.
B. -5⁰ and 4^0.
C. -1^0.and 9^0.
D. -2^0.and 11^0.

Answer : B
 

ASSERTION & REASONING QUESTIONS

DIRECTION : In the following questions, a statement of assertion (A) is followed by a statement of reason (R).
Mark the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question. Assertion : A linear equation 3x + 5y = 2 has a unique solution.
Reason : A linear equation in two variables has infinitely many solutions.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : We know that a linear equation in two variables has infinitely many solutions.
So, Reason is correct.
Hence, Assertion is not correct
Correct option is (d) Assertion (A) is false but reason (R) is true.

Question. Assertion : If x = 2, y = 1 is a solution of the equation 2x + 3y = k, then the value of k is 7.
Reason : The solution of the line will satisfy the equation of the line.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : We know that the solution of the line will satisfy the equation of the line.
So, Reason is correct.
Since x = 2, y =1 is a solution of the given linear equation, we have
2 x 2 + 3 x 1 – k = 0
So, Assertion is also correct
Correct option is (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .

Question. Assertion : If x = 2k – 1 and y = k is a solution of the equation 3x – 5y – 7 = 0, then the value of k is 10
Reason : A linear equation in two variables has infinitely many solutions.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : We know that a linear equation in two variables has infinitely many solutions. So, Reason is correct.
Since x = 2k - 1 and y = k is solution of the given linear equation, we have
3 x (2k – 1) – 5k – 7 = 0
So, Assertion is also correct
But reason (R) is not the correct explanation of assertion (A) .
Correct option is (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A) .

Question. Assertion : There are infinite number of lines which passes through (3, 2) .
Reason : A linear equation in two variables has infinitely many solutions.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : We know that a linear equation in two variables has infinitely many solutions. So, Reason is correct.
Through a point infinite lines can be drawn.
Through (3, 2) infinite number of lines can be drawn.
Hence, Assertion is also correct.
But reason (R) is not the correct explanation of assertion (A) .
Correct option is (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A) .

Question. Assertion: x = 3 and y = 2 is a solution of the linear equation 2x + 3y = 12.
Reason: x = 4 and y = 2 is a solution of the linear equation x + 3y = 10.
Answer : For Assertion: The given linear equation is 2x + 3y = 12
Substituting x = 3 and y = 2, we get
LHS = 2 x 3 + 3 x 2 = 6 + 6 = 12 = RHS
So, Assertion is correct.
For Reason: The given linear equation is x + 3y = 10
Substituting x = 4 and y = 2, we get
LHS = 4 + 3 x 2 = 4 + 6 = 10 = RHS
So, Reason is also correct.
But reason (R) is not the correct explanation of assertion (A) .
Correct option is (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A) .

Question. Assertion : The point (3, 0) lies on the graph of the linear equation 4x + 3y = 12.
Reason : (3, 0) satisfies the equation 4x + 3y = 12.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : For Assertion: The given linear equation is 4x + 3y = 12
Substituting x = 3 and y = 0, we get
LHS = 4 x 3 + 3 x 0 = 12 + 0 = 12 = RHS
Since (3, 0) satisfies the equation 4x + 3y = 12, therefore point (3, 0) lies on the graph of the linear equation 4x + 3y = 12.
So, Reason and Assertion are both correct.
Here, Reason is the correct explanation of Assertion.
Correct option is (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .

Question. Assertion : The graph of the linear equation 2x – y = 1 passes through the point (2, 3) .
Reason : Every point lying on graph is not a solution of 2x – y = 1.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : For Assertion: The given linear equation is 2x – y = 1
Substituting x = 2 and y = 3, we get
LHS = 2 x 2 – 3 = 4 – 3 = 1 = RHS
Since (3, 0) satisfies the equation 4x + 3y = 12, therefore graph of the linear equation 2x – y = 1 passes through the point (2, 3) .
So, Assertion is correct.
But Reason is not correct as every point lying on graph is a solution of 2x – y = 1.
Correct option is (c) Assertion (A) is true but reason (R) is false.

Question. Assertion: x = 2 is a line parallel to the y-axis.
Reason: The equation of a line parallel to the y-axis is x = a.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : We know that equation of a line parallel to the y-axis is x = a.
So, Reason (R) is true.
Also, x = 2 is a line parallel to the y-axis.
So, Assertion (A) is true.
Thus, Reason (R) and Assertion (A) are true and Reason (R) is a correct explanation of Assertion (A) .
Correct option is (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .

Question. Assertion: x + y = 3 is the equation of a line passing through the origin.
Reason: y = 2x is the equation of a line passing through the origin.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : For Assertion: The given linear equation is x + y = 3
Since x = 0 and y = 0 is not satisfying x + y = 3, therefore it is not passing through the origin.
So, Assertion is not correct.
Since x = 0 and y = 0 is not satisfying y = 2x, therefore it is passing through the origin.
So, Reason is correct.
Correct option is (d) Assertion (A) is false but reason (R) is true.

Question. Assertion: y = 3x represents a line passing through the origin.
Reason: Any line parallel to the x-axis is y = a.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A) .
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer : Since x = 0 and y = 0 is not satisfying y = 3x, therefore it is passing through the origin.
So, Assertion (A) is true.
Also, we know that equation of a line parallel to the x-axis is y = a.
So, Reason (R) is also true.
But Reason is not the correct explanation of Assertion.
Correct option is (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A) .

Question. Assertion : The point (2, 2) is the solution of x + y = 4.
Reason : Every point which satisfy the linear equation is a solution of the equation.
Answer : We know every point which satisfy the linear equation is a solution of the equation.
So, Reason (R) is true.
Substituting x = 2 and y = 2, we get
LHS = 2 + 2 = 4 = RHS
Since (3, 0) satisfies the equation 4x + 3y = 12, therefore the point (2, 2) is the solution of x + y = 4
So, Assertion (A) is also true.
Here, Reason is the correct explanation of Assertion.
Correct option is (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) .

LINEAR EQUATION

Question 1. In a village primary school enrolment of girls has doubled this year as compared to last year. Considering this year's enrolment as 'y' and previous year's enrolment as x :
(a) Form a linear equation for this information.
(b) If previous year's enrolment was 35 find this year's enrolment.
(c) What is the value depicted here ?
Answer: (a) \(y = 2x\)
(b) Substituting \(x = 35\) into the equation: \(y = 2 \times 35 = 70\). Thus, this year's enrolment is 70.
(c) Freedom, Gender equality.
In simple words: This year's enrolment of girls is twice the previous year, giving the equation \(y = 2x\).

Exam Tip: Always define variables clearly according to the question statement before forming equations.

Question 2. On her birthday Sonia distributed chocolates in an orphanage. She gave 5 chocolates to each child and 20 chocolates to adults. Taking number of children as x and total chocolates distributed as y.
(a) Form a linear equation.
(b) If she distributed 145 chocolates how many children are there in the orphanage ?
(c) Explain the value depicted here by Sonia.
Answer: (a) Total chocolates \(y = 5x + 20\).
(b) Substituting \(y = 145\):
\(145 = 5x + 20\)
\(125 = 5x\)
\(x = 25\). So, there are 25 children.
(c) Happiness, Helpfulness.
In simple words: The total chocolates equal 5 times the number of children plus 20 for the adults.

Exam Tip: When setting up linear equations from word problems, isolate the total quantity on one side.

Question 3. In a housing society people decided to do Rainwater harvesting. Rainwater is collected in the underground tank at the rate of 30 cm\(^3\)/sec. Taking volume of water collected in x sec as y cm\(^3\).
(a) Form a linear equation.
(b) Write it in standard form as ax+by+c=0.
(c) Which value is promoted by the members of this society ?
Answer: (a) \(y = 30x\)
(b) Rearranging into standard form: \(30x - y + 0 = 0\).
(c) Environmental protection, Co-operation.
In simple words: Volume equals the rate multiplied by time, yielding \(y = 30x\).

Exam Tip: The standard form of a linear equation in two variables is \(ax + by + c = 0\).

Question 4. A man hires an auto rickshaw to cover a certain distance. The fare is Rs. 10 for first km and Rs. 7 for subsequent kilometers. Taking total distance covered as x km and total fare as y :
(a) Write a linear equation for this.
(b) The man covers a distance of 16 km and gave Rs. 120 to the auto driver. Auto driver said "It is not the correct amount" and returned him the balance. Find the correct amount paid back by the autodriver.
(c) Which value is depicted here by the autodriver ?
Answer: (a) Fare \(y = 10 + 7(x - 1) = 10 + 7x - 7 = 7x + 3\).
(b) For \(x = 16\) km:
\(y = 7(16) + 3 = 112 + 3 =\) Rs. 115.
Amount paid back = Rs. 120 - Rs. 115 = Rs. 5.
(c) Honesty, Truthfulness.
In simple words: Calculate the first km separately and add Rs. 7 for the remaining 15 km.

Exam Tip: For tiered pricing problems, remember to subtract 1 from the total distance for the subsequent rate calculation.

Question 5. In a "Cleanliness drive" residents of certain locality joined together to clean neighbourhood area. Participation of children was 20 more than that by adults. Taking x as number of adults and y as number of children.
(a) Form a linear equation.
(b) What values are depicted here ?
Answer: (a) \(y = x + 20\)
(b) Co-operation, Happiness, Sincerity, Environmental protection.
In simple words: Children's count is adults' count plus 20.

Exam Tip: Translate "more than" directly as addition.

Question 6. Rehman and Prakash together contributed Rs. 500 towards Prime Minister Relief fund.
(a) Write the linear equation for this.
(b) Write the values involved in it.
Answer: (a) Let Rehman's contribution be x and Prakash's be y. Then \(x + y = 500\).
(b) Co-operation, Helpfulness.
In simple words: The sum of both contributions equals 500.

Exam Tip: Assign variables to unknown quantities before framing the equation.

Question 7. Draw the graph of the linear equation 2(x+1)=3(y-1)-4 and check whether point (-3, +1) lies on the line ?
Which value is depicted here ?
Answer: (a) Simplifying the equation:
\(2(x + 1) = 3(y - 1) - 4\)
\(2x + 2 = 3y - 3 - 4\)
\(2x - 3y + 9 = 0\)
Checking point \((-3, 1)\):
LHS = \(2(-3) - 3(1) + 9 = -6 - 3 + 9 = -9 + 9 = 0 =\) RHS.
Thus, the point lies on the line.
(b) Curiosity, Truthfulness, Scientific temper.
In simple words: Substitute the point coordinates into the simplified equation to verify if it satisfies the equality.

Exam Tip: A point lies on a line if its coordinates satisfy the equation of the line.

Question 8. In a class, teacher asked about the equ. of line passing through the origin with options as :
y+x=1, x+3y=2, x+y=0
Rehana answered it correctly as x+y=0.
Choose the correct value :
(a) Honesty (b) Equality (c) Truthfulness (d) Sincerity
Answer: (c) Truthfulness
Any line passing through the origin has no constant term (i.e., \(c = 0\)). Among the options, only \(x + y = 0\) satisfies this.
In simple words: Equations passing through the origin have zero constant term.

Exam Tip: Check for a constant term to quickly identify lines passing through the origin.

Question 9. (a) In the above fig. (i) & (ii) find the values of a & b.
(b) Is a=b ? Justify your answer.
(c) What is the value depicted here ?
Answer: (a) From Fig. (i), \(a = 3\). From Fig. (ii), \(b = 3\).
(b) Yes, because both equal 3.
(c) Scientific temper, Truthfulness, Knowledge.
In simple words: Read values directly from the respective coordinate axes.

Exam Tip: Pay close attention to axis intercepts when reading graphical data.

Question 10. (a) Every six months price of petrol increases at the rate of Rs. 4 per litre. Taking price of petrol in Dec. 2011 as 'x' and present price of petrol as 'y' form a linear equation showing amount spent on petrol in Dec. 2012.
(b) Due to continuous rise in the petrol price people are shifting towards CNG in place of petrol whose price increases at a rate of Rs. 3 per litre in a year. Form a linear equation taking price of CNG in Dec. 2011 as 'a' and in Dec. 2012, as 'b'.
(c) Which fuel will you prefer and why ?
(d) Which value is depicted by using CNG over petrol ?
Answer: (a) In one year (two six-month periods), the price increases by \(2 \times 4 = 8\) Rs. Thus, \(y = x + 8\).
(b) \(b = a + 3\).
(c) CNG, because CNG is cheaper and cleaner than petrol.
(d) Environmental protection.
In simple words: Add the total annual increase to the initial price to form the equation.

Exam Tip: Note that "every six months" means the increase happens twice in a full year.

QUADRILATERALS

Question 1. There is a visit in a school by Directorate of Education. Girls are asked to prepare Rangoli in triangular shape. Dimensions of \(\Delta ABC\) are 26 cm, 28 cm, 25 cm. Garland is to be placed along the side of \(\Delta PQR\) which is formed by joining midpoints of sides of \(\Delta ABC\).
(a) Find the length of garland.
(b) What values are depicted here ?
Answer: (a) By the midpoint theorem, the sides of \(\Delta PQR\) are half the corresponding sides of \(\Delta ABC\):
\(RQ = \frac{1}{2} \times 28 = 14\) cm
\(QP = \frac{1}{2} \times 25 = 12.5\) cm
\(PR = \frac{1}{2} \times 26 = 13\) cm
Perimeter of \(\Delta PQR = 14 + 12.5 + 13 = 39.5\) cm.
(b) Beauty, happiness, co-operation.
In simple words: The perimeter of the triangle formed by midpoints is half the perimeter of the outer triangle.

Exam Tip: Recall that joining the midpoints of a triangle creates a smaller triangle with half the side lengths of the original.

Question 2. A class teacher gave students coloured papers in the shape of quadrilateral. She asked them to make parallelogram from it using paper folding.
(a) How can a parallelogram be formed by using paper folding ?
(b) Prove that it is a parallelogram.
(c) What values are depicted here ?
Answer: (a) By joining the midpoints of the sides of the quadrilateral.
(b) In \(\Delta ABD\), P and Q are midpoints of AB and AD. Therefore, \(PQ \parallel BD\) and \(PQ = \frac{1}{2}BD\). Similarly, \(RS \parallel BD\) and \(RS = \frac{1}{2}BD\). Since \(PQ \parallel RS\) and \(PQ = RS\), PQRS is a parallelogram.
(c) Curiosity, Happiness, Scientific Temper.
In simple words: Connecting adjacent side midpoints forms a quadrilateral with opposite sides parallel and equal.

Exam Tip: The midpoint theorem is key to proving quadrilaterals formed inside other figures are parallelograms.

Question 3. ABCD is a rhombus where \(\angle ADC = 120^\circ\). There are two fire stations at C and D and fire is reported at O.
(a) Which fire station team can reach early and why ?
(b) Which value is depicted here ?
Answer: (a) In rhombus ABCD, \(\angle ODC = \frac{1}{2} \times 120^\circ = 60^\circ\) and \(\angle OCD = 30^\circ\). Since \(\angle ODC > \angle OCD\), the side opposite to the larger angle is longer, so \(OC > OD\). Therefore, team D is closer and can reach earlier.
(b) Serving the people with honesty.
In simple words: The side opposite to the larger angle in a triangle is longer, meaning shorter distance corresponds to station D.

Exam Tip: Use the triangle inequality property relating side lengths to opposite angles.

Question 4. Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in the fig. These pieces are arranged in a way that \(AB \parallel HC \parallel GD \parallel FE\). Also \(BC = CD = DE\) and \(GF = 6\) cm. He wants to decorate the card by putting up a coloured tape on non parallel sides of trapezium.
(a) Find the total length of coloured tape required if \(DE = 4\) cm.
(b) Which value is depicted by the student ?
Answer: (a) Since \(BC = CD = DE = 4\) cm, by intercept theorem, equal intercepts on one transversal imply equal intercepts on any other transversal: \(AH = HG = GF = 6\) cm.
Total length \(AF + FE = (6 + 6 + 6) + 12 = 18 + 12 = 30\) cm.
(b) Happiness, beauty, knowledge.
In simple words: Equal transversal intercepts mean corresponding non-parallel segments are also equal.

Exam Tip: Apply the intercept theorem for parallel lines cutting transversals.

Question 5. A farmer has a field in the form of a parallelogram ABCD. One of his cow is suffering from some disease. To take good care of her, he tied the cow at one corner of the field. The corner angle of the field is \(65^\circ\).
(a) Find all the other three angles of the field.
(b) Explain the values depicted here.
Answer: (a) Opposite angles of a parallelogram are equal, so \(\angle A = \angle C = 65^\circ\). Adjacent angles sum to \(180^\circ\), so \(\angle B = 180^\circ - 65^\circ = 115^\circ = \angle D\).
(b) Helpfulness, caring for every one.
In simple words: Opposite angles are equal and adjacent angles are supplementary in a parallelogram.

Exam Tip: Remember opposite angles are equal and consecutive angles are supplementary in parallelograms.

Question 6. A chocolate is in the form of a quadrilateral with sides 6 cm, 10 cm, 5 cm, 5 cm is cut into two parts along one of its diagonals by a lady. Part I is given to her maid and part II is equally divided among her driver and maali.
(a) Is this distribution fair ? Justify it.
(b) What value is depicted here ?
Answer: (a) Length of diagonal \(BD = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8\) cm.
Area of triangle I (\(ABD\)) = \(\frac{1}{2} \times 6 \times 8 = 24\) cm\(^2\).
For triangle II (\(BCD\)) with sides 5, 5, 8: semi-perimeter \(s = \frac{5 + 5 + 8}{2} = 9\) cm.
Area = \(\sqrt{9(9-5)(9-5)(9-8)} = \sqrt{9 \times 4 \times 4 \times 1} = 12\) cm\(^2\).
Since total area = \(24 + 12 = 36\) cm\(^2\) and distribution is unequal (24 vs 12), it is not fair.
(b) Helpfulness, Caring.
In simple words: Calculate the areas of both triangles using Heron's formula and right-triangle area formula to compare them.

Exam Tip: Always calculate exact geometric areas before deciding fairness.

Question 7. While analysing properties of quadrilateral a child says that all angles of a quadrilateral can not be obtuse.
(a) Justify his answer.
(b) What value is depicted in this activity ?
Answer: (a) If all angles were obtuse (\(> 90^\circ\)), say \(91^\circ, 91^\circ, 91^\circ\), their sum would exceed \(360^\circ\) since \(91 \times 4 = 364^\circ > 360^\circ\), which violates the angle sum property of quadrilaterals.
(b) Scientific temper, knowledge.
In simple words: The sum of all four interior angles must equal exactly \(360^\circ\).

Exam Tip: The sum of interior angles of any quadrilateral is always \(360^\circ\).

Question 8. During Maths Lab Activity each student was given four broom sticks of lengths 8 cm, 8 cm, 5 cm, 5 cm to make different types of quadrilaterals.
(a) How many quadrilaterals can be formed using these sticks.
(b) Name the types of quadrilaterals formed.
(c) While doing this activity which value is depicted ?
Answer: (a) Three types.
(b) Rectangle, parallelogram, kite.
(c) Scientific temper, curiosity.
In simple words: Varying the angles between equal opposite sides allows forming different quadrilaterals.

Exam Tip: Equal opposite side lengths can form rectangles, general parallelograms, or kites depending on internal angles.

Question 9. In a trapezium ABCD \(DC \parallel AB\) and \(\angle A = \angle B = 45^\circ\). teacher asked the student to find \(\angle D\). Naresh answered it as \(135^\circ\)
(a) Is the answer correct ? Justify it.
(b) Which value is depicted here ?
Answer: (a) Yes. Since \(DC \parallel AB\), consecutive interior angles are supplementary, so \(\angle A + \angle D = 180^\circ\), giving \(\angle D = 180^\circ - 45^\circ = 135^\circ\).
(b) Scientific temper, truth value.
In simple words: Consecutive angles between parallel lines add up to \(180^\circ\).

Exam Tip: Co-interior angles between parallel lines sum to \(180^\circ\).

Question 10. While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown in the fig. Teacher gave them three options as :
(i) x > y (ii) x < y (iii) x = y
Beena gave the answer as x < y
(a) Is this the correct answer ?
(b) Justify the answer with suitable logic.
(c) What is the value depicted in this ?
Answer: (a) Yes.
(b) \(\angle ADB = \angle DBC = y\) (alternate interior angles). In the relevant triangle, since side \(BC < CD\), the angle opposite to the smaller side is smaller, hence \(x < y\).
(c) Truth value.
In simple words: Alternate interior angles relate the angles, and the side-angle inequality in triangles proves \(x < y\).

Exam Tip: Combine alternate interior angles with triangle inequalities for angle-side relations.

AREAS OF PARALLELOGRAMS AND TRIANGLES

Question 1. A craft mela is organised by Welfare Association to promote the art and culture of tribal people. The pandal is to be decorated by using string of bulbs all around the field. There are two options either to arrange it in a rectangular field ABEF or parallelogram ABCD with equal area.
(a) What shape of the field should be chosen to minimise the expense of bulb and why ?
(b) Which values are depicted here ?
Answer: (a) Rectangle, because the perimeter of a rectangle is less than the perimeter of a parallelogram of equal area.
(b) Co-operation, helpfulness.
In simple words: Rectangles have the minimum perimeter for a given area among parallelograms, saving string length.

Exam Tip: Perimeter minimization for fixed area points to rectangular shapes.

Question 2. A plot is in the form of a parallelogram ABCD. Owner of this plot wants to build OLD AGE HOME, DISPENSARY, PARK and HEALTH CENTRE for elderly people as shown in the fig. P is a point on the diagonal BD.
(a) Prove that area alloted to old age home and dispensary is same.
(b) Which values are depicted here ?
Answer: (a) Join AC. Diagonals AC and BD bisect at O, so \(AO = OC\) and \(BO = OD\). In \(\Delta APC\), PO is a median, so \(\text{ar}(APO) = \text{ar}(CPO)\) (i). In \(\Delta ADC\), DO is a median, so \(\text{ar}(ADO) = \text{ar}(DCO)\) (ii). Adding (i) and (ii) gives \(\text{ar}(ADP) = \text{ar}(DPC)\).
(b) Helpfulness, happiness, cooperation.
In simple words: Medians divide triangles into two regions of equal area, which combines to prove equal areas for the two sections.

Exam Tip: Use the property that a median bisects a triangle's area.

Question 3. There was a deserted land near a colony where people used to throw garbage. Colony people united to develop a pond in triangular shape as shown in the fig. The land is in the shape of \(\parallel\)gm ABCD. In rest of the portion medicinal plants were grown. If area of parallelogram ABCD is 200 m\(^2\).
(a) Calculate the area where medicinal plants were grown.
(b) Which value is depicted here ?
Answer: (a) The area of a triangle on the same base and between the same parallels as a parallelogram is half the area of the parallelogram. Thus, pond area = \(\frac{1}{2} \times 200 = 100\) m\(^2\). Medicinal plants area = \(200 - 100 = 100\) sq. cm (or m\(^2\)).
(b) Environmental protection, cooperation.
In simple words: A triangle on the same base and between the same parallels has half the area of the parallelogram.

Exam Tip: Remember the theorem: triangle area is half a parallelogram's area when sharing base and parallels.

Question 4. For 'Sarva Shiksha Abhiyan' a rally was organised by a school. Students were given triangular cardboard pieces to write slogans. They divided the triangular shape into three equal parts by drawing medians as shown in fig.
(a) Prove that area (AGC) = area (AGB) = area (BGC) = \(\frac{1}{3}\) area (ABC)
(b) Which values are inculcated through this activity ?
Answer: (a) In \(\Delta ABC\), AD is a median, so \(\text{ar}(\Delta ABD) = \text{ar}(\Delta ADC)\) (i). Also, \(\text{ar}(\Delta GBD) = \text{ar}(\Delta GDC)\) (ii). Subtracting (ii) from (i) gives \(\text{ar}(\Delta AGB) = \text{ar}(\Delta AGC)\). Similarly, all three parts are equal to \(\frac{1}{3}\text{ar}(\Delta ABC)\).
(b) Cooperation, sincerity.
In simple words: Medians concurrent at the centroid divide the triangle into three triangles of equal area.

Exam Tip: Centroid divides a triangle into three equal-area triangles.

Question 5. A farmer has a square plot of land where he wants to grow five different crops at a time. On half of the area in the middle he wants to grow wheat but in rest four equal triangular parts he wants to grow different crops.
(a) Explain by diagram how he can divide the area to fulfill his purpose.
(b) By using this crop pattern which values are depicted by the farmer?
Answer: (a) By joining the midpoints of the opposite sides of the square plot, forming a central rhombus (wheat) and four outer triangles.
(b) Increases fertility of soil, environmental protection, happiness.
In simple words: Joining side midpoints creates a central region equal in area to the sum of the corner triangles.

Exam Tip: Midpoint connections in quadrilaterals split areas in half systematically.

Question 6. There is a plot in a village in the shape of a quadrilateral ABCD. Sarpanch wants to get floor cemented so as to use it for social gatherings and panchayat meetings. Later due to construction of park in the neighbourhood for children they decided to change the shape to triangle ABP. If \(AC \parallel DP\)
(a) Prove that ar (quad. ABCD) = ar (\(\Delta ABP\))
(b) What are the values depicted in this activity ?
Answer: (a) Since \(AC \parallel DP\), \(\text{ar}(\Delta ADC) = \text{ar}(\Delta APC)\) (triangles on the same base AC and between same parallels). Adding \(\text{ar}(\Delta ABC)\) to both sides: \(\text{ar}(\Delta ADC) + \text{ar}(\Delta ABC) = \text{ar}(\Delta APC) + \text{ar}(\Delta ABC)\), which implies \(\text{ar}(quad. ABCD) = \text{ar}(\Delta ABP)\).
(b) Environmental protection, respecting others views.
In simple words: Triangles between the same parallels on the same base have equal areas, allowing area conservation when reshaping.

Exam Tip: Use the property that triangles on the same base between parallel lines have equal area.

Question 7. On National Integration day a poster is to be made by class IX students of a school. This poster is in the shape of a parallelogram. All religions should be given equal triangular space to display their teachings.
(a) How will they divide a parallelogram into four triangles equal in area ? Justify it.
(b) What are the values depicted here ?
Answer: (a) By drawing both diagonals of the parallelogram, it is divided into four triangles of equal area because diagonals of a parallelogram bisect each other and split the figure into four equivalent triangular regions.
(b) Unity of nation, co-operation, fraternity.
In simple words: Intersecting diagonals divide any parallelogram into four triangles of identical area.

Exam Tip: Recall that intersecting diagonals partition a parallelogram into four equal-area triangles.

Question 8. In a class, teacher gave two identical carboard pieces which are in the shape of a parallelogram to two groups. First group was asked to find area of parallelogram using AB as base. Then another group was asked to find height (h) of the parallelogram with AD as base.
(a) How will they find value of h ?
(b) What are the values involved here ?
Answer: (a) Area of parallelogram \(ABCD = AB \times DH = AD \times BP\). Given \(AB = 8\), \(DH = 3\), \(AD = 5\):
\(8 \times 3 = 5 \times h\)
\(24 = 5h\)
\(h = \frac{24}{5} = 4.8\) cm.
(b) Scientific temper, cooperation.
In simple words: Equating the area calculated using two different base-height pairs allows solving for the unknown height.

Exam Tip: Area of a parallelogram can be computed using any base and its corresponding altitude.

Question 9. In a class, teacher asks the students to cut a figure from a given parallelogram which has area equal to half the area of parallelogram ABCD. Sunita joins the mid points of opposite sides of parallelogram as shown in the fig. Where as Rohan draws a triangle on the same base as shown in fig.
(a) State whether two answers are correct. Justify.
(b) What values are depicted from this activity ?
Answer: (a) Both are giving correct answers. Sunita's construction forms a smaller parallelogram with half the area, and Rohan's triangle on the same base and between the same parallels also has half the area of the parallelogram.
(b) Scientific temper, curiosity, co-operation.
In simple words: Both methods correctly yield half the total area of the parallelogram.

Exam Tip: Multiple geometric constructions can yield equivalent area fractions.

Question 10. In a rhombus ABCD, \(AC = 8\) cm, then \(AO = 4\) cm.
The statement shows (a) Truth value (b) Social value (c) Environment value (d) Cooperation
Justify your answer.
Answer: (a) Truth value
Diagonals of a rhombus bisect each other, so \(AO = \frac{1}{2}AC = \frac{1}{2} \times 8 = 4\) cm.
In simple words: Diagonals bisect each other in a rhombus, making AO half of AC.

Exam Tip: Recall that rhombus diagonals bisect each other at right angles.

CIRCLES

Question 1. Three friends ordered 3 pizzas of same size for them. Just then two more friends joined them. They decided to share 3 pizzas among all 5 of them.
(a) Find the area of the share of pizza each child gets if the radius of each pizza is 7 cm.
(b) Which values of children are depicted here ?
Answer: (a) Total area of 3 pizzas = \(3 \times \pi r^2 = 3 \times \frac{22}{7} \times 7^2 = 3 \times 154 = 462\) cm\(^2\).
Share per child = \(\frac{462}{5} = 92.4\) cm\(^2\).
(b) Co-operation.
In simple words: Divide the total area of all pizzas by the total number of friends.

Exam Tip: Calculate total pizza area first, then divide by the total number of people sharing.

Question 2. During 'Van Mahotsav' celebration in a school a circular plot in the corner of the school was chosen. Five groups of students were selected to plant trees in each part as shown in the fig.
(a) Which group gets the maximum area to develop and by how much, if the two radii are 210 cm & 105 cm respectively.
(b) What values can be inculcated in students through this activity ?
Answer: (a) Group III gets the maximum area. Area = \(\frac{1}{2} \pi (210)^2 - \frac{1}{2} \pi (105)^2 = \frac{22}{7} \times \frac{1}{2} (44100 - 11025) = 8662.5\) sq. cm.
(b) Environmental protection, beauty.
In simple words: Group III has the largest semicircular region.

Exam Tip: Subtract inner circular area from outer circular area when dealing with concentric regions.

Question 3. The Indian hockey federation organized a friendly hockey match between India and Pakistan on a circular ground. The sale proceeds of this match shall be donated to an orphanage. A rectangular turf is spread on the ground as shown in the figure.
(a) Find the radius of the stadium.
(b) Which social values is depicted here ?
(c) How does donation to charitable organisations help in the development of society ?
Answer: (a) Using Pythagoras theorem in the right triangle formed by half the rectangle side and height: radius \(r = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\) m.
(b) Helpfulness, co-operation.
(c) This money is utilized for welfare of orphans.
In simple words: Apply Pythagoras theorem using half the rectangle dimensions to find the circle's radius.

Exam Tip: The distance from the center to any vertex of an inscribed rectangle gives the circle's radius.

Question 4. A school decided to paint its inner boundary wall. They planned to write quotations and social messages for the children on the circular areas of radius 40 cm each on the wall along a straight line.
If the boundary wall is 200 m long
(a) How many circles can be drawn ?
(b) What values are inculcated in the students by this activity ?
Answer: (a) Diameter of each circle = \(2 \times 40 = 80\) cm = 0.8 m.
Number of circles = \(\frac{200 \text{ m}}{0.8 \text{ m}} = 250\).
(b) Beauty.
In simple words: Divide the total wall length by the diameter of each circular painted area.

Exam Tip: Ensure all units match (convert meters to centimeters or vice versa) before dividing.

Question 5. A group of 66 students went for a picnic. They halted at a park for their lunch. They made a big circle and each child was 50 cm away from the other child along the circle.
(a) Find the radius of the circle they formed.
(b) What values do these students possess ?
Answer: (a) Circumference = \(66 \times 50 = 3300\) cm = 33 m.
Radius \(r = \frac{\text{Circumference}}{2\pi} = \frac{3300}{2 \times \frac{22}{7}} = \frac{3300 \times 7}{44} = 525\) cm = 5.25 m.
(b) Co-operation, happiness.
In simple words: Total circumference equals the number of students multiplied by the spacing between them.

Exam Tip: Use the circumference formula \(C = 2\pi r\) to find the radius from total arc length.

Question 6. An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the figure.
(a) Find the distance between any two adjacent food stalls if the radius of the park is 30 m.
(b) Are such activities helpful for society ? What are the other social issues for which such compaigns are required ?
Answer: (a) Distance between adjacent stalls = \(\sqrt{30^2 + 30^2} = 30\sqrt{2}\) m.
(b) Yes, for the inmates of old age home, orphanages, handicapped children etc.
In simple words: Use Pythagoras theorem on the right triangle formed by the radius vectors to adjacent stalls.

Exam Tip: For stalls at \(90^\circ\) intervals on a circle, chord length is \(r\sqrt{2}\).

Question 7. Children of a colony decided to assist and help an old couple of the colony in developing their vacant plot as flower bed as shown here. The radii of the two circles is 245 cm and 210 cm.
(a) Find the area of the grass cover.
(b) Which values are depicted by the children?
Answer: (a) Area = \(\pi (R^2 - r^2) = \frac{22}{7} (245^2 - 210^2) = \frac{22}{7} (55225 - 44100) = \frac{22}{7} \times 11125 = 50050\) sq. cm.
(b) Helpfulness.
In simple words: Subtract the inner circle's area from the outer circle's area.

Exam Tip: Use the ring area formula \(\pi(R^2 - r^2) = \pi(R - r)(R + r)\) for faster calculation.

Question 8. 3 STD booths situated at A, B and C in the fig. are operated by handicapped persons. These three booths are equidistant from each other as shown in the figure.
(a) Find \(\angle BOC\)
(b) Do you think employment provided to handicapped persons is important for the development of a society ? Justify your answer.
Answer: (a) \(120^\circ\)
(b) Yes.
In simple words: Equal chords subtend equal angles at the center, dividing \(360^\circ\) into three equal parts of \(120^\circ\).

Exam Tip: Equal chords of a circle subtend equal angles at the center.

Question 9. During a practical activity in maths lab, students were using circular Geo board. The angle subtended by an arc at the centre is \((2a+50^\circ)\) Sakshi calculated \(\angle BAC\) as \((a+25^\circ)\)
(a) Is her finding correct ? Justify it.
(b) Find \(\angle BAC\) if a = 30
(c) Which values are depicted here ?
Answer: (a) Yes, because the angle subtended by an arc at the circumference is half the angle subtended at the center: \(\frac{2a+50^\circ}{2} = a + 25^\circ\).
(b) For \(a = 30\), \(\angle BAC = 30 + 25 = 55^\circ\).
(c) Truth, scientific temper.
In simple words: The inscribed angle is always half of the central angle subtended by the same arc.

Exam Tip: Remember the central angle theorem relating angles at the center and circumference.

Question 10. In a circle with centre O, ABCD is a cyclic quadrilateral.
x+y = 90\(^\circ\) depicts which of these values ?
(a) Social value (b) Freedom (c) Truth value (d) Gender equality
Justify your answer.
Answer: (c) Truth value
\(\angle ACD = 90^\circ\) (angle in a semicircle). In cyclic quadrilateral ABCD, opposite angles sum to \(180^\circ\): \(\angle BAD + \angle BCD = 180^\circ\), giving \(y + x + 90^\circ = 180^\circ\), so \(x + y = 90^\circ\).
In simple words: Opposite angles of cyclic quadrilaterals are supplementary, leading to \(x + y = 90^\circ\).

Exam Tip: Angle in a semicircle is a right angle, and opposite angles of cyclic quadrilaterals are supplementary.

CONSTRUCTION

Question 1. Delhi traffic police wants to make a traffic signal board of the shape of an equilateral triangle of side 5 m to make the people aware of the traffic rules. Construct the above traffic signal board by taking each side as 5 cm (instead of m). What value is depicted here ?
Answer: (a) Draw a line segment \(AB = 5\) cm. Using compasses with radius 5 cm, draw arcs from A and B intersecting at C. Join AC and BC to complete the equilateral triangle.
(b) To make the people aware of traffic rules for safety measures.
In simple words: Construct an equilateral triangle using compass arcs of equal length for all three sides.

Exam Tip: Use compasses set to the exact side length to intersect arcs for equilateral triangles.

Question 2. Ram Lal has a triangular piece of land ABC in which \(\angle B = 30^\circ\), \(\angle C = 90^\circ\), \(AB+BC+CA = 11\) m. He donated this to a vridhashram. Construct a triangle using above dimensions (use cm in place of m). What value represents here ?
Answer: (a) Draw base line segment \(BC = 11\) cm. Construct angles \(\angle XBC = 30^\circ\) and \(\angle YCB = 90^\circ\). Bisect these angles; let the bisectors meet at A. Draw perpendicular bisectors of AB and AC to locate intersection points on BC for vertices B and C.
(b) Habit of donation.
In simple words: Construct using the perimeter and base angles method.

Exam Tip: When given the perimeter and base angles, construct angles equal to half the given base angles at the endpoints.

Question 3. Students of a school staged a rally for cleanliness campaign. They walked through the lanes AB, BC and CA which form a triangle. Construct the \(\Delta ABC\) in which \(BC = 8\)m, \(\angle B = 45^\circ\) \(AB-AC = 3.5\)m. What value represents here ? (Construct the triangle using cm in place of m)
Answer: (a) Draw \(BC = 8\) cm. Make \(\angle XBC = 45^\circ\). Cut a line segment \(BD = 3.5\) cm from BX. Join DC, draw the perpendicular bisector of DC intersecting BX at A. Join AC.
(b) To make the environment clean.
In simple words: Use the difference of two sides method where the side containing the base angle is greater.

Exam Tip: When \(AB - AC\) is given, cut off the difference along the ray of the base angle.

Question 4. A co-operative house adopted a triangular park \(\Delta PQR\) to maintain its greenary of dimensions \(QR = 7\)m, \(\angle Q = 75^\circ\) and \(PQ + PR = 13\)m, Make a \(\Delta PQR\) using cm instead of m. What value represents here ?
Answer: (a) Draw \(QR = 7\) cm. At Q, construct \(\angle RQX = 75^\circ\). Cut segment \(QD = 13\) cm from QX. Join DR, construct perpendicular bisector of DR intersecting QX at P. Join PR.
(b) (i) To make the environment clean. (ii) Community services.
In simple words: Use the sum of two sides method by cutting the total sum along the base angle ray.

Exam Tip: For sum of sides construction, cut the sum length on the angle ray and join to the opposite vertex.

Question 5. The villagers of a village wish to make a pool to drink water for animals. They constructed it in the shape of a triangle. The dimensions of the pool are 5m and 12m and the angle between them is \(90^\circ\). Construct it by taking as cm instead of m. What value represents here?
Answer: (a) Draw base \(BC = 12\) cm. Construct a right angle \(\angle XBC = 90^\circ\) at B. Cut off \(AB = 5\) cm from BX. Join AC.
(b) Social work.
In simple words: Construct a right-angled triangle given two sides and the included right angle.

Exam Tip: Construct the base, erect a perpendicular for the right angle, and cut off the given side length.

Question 6. Mohan and Sohan constructed a right angle. Choose the correct option. What idea is depicted here?
(a) Mohan is correct (b) Sohan is correct (c) Both are correct (d) Both are wrong
Answer: (a) Mohan is correct
Mohan correctly bisected the angle or constructed standard arcs for a right angle.
(b) Truth value.
In simple words: Mohan followed the proper geometric construction steps for a right angle.

Exam Tip: A right angle is constructed by bisecting a \(180^\circ\) straight angle or using \(60^\circ\) and \(120^\circ\) intersection arcs.

Question 7. Kavita makes a triangular shaped carpet of dimensions 5m, 5m and 6m and donates it to a blind school. Construct a triangle using the above dimensions in cm. What value represents here ?
Answer: (a) Draw base \(BC = 6\) cm. With B and C as centers and radii 5 cm, draw arcs intersecting at A. Join AB and AC.
(b) Habit of donation, social work.
In simple words: Construct an isosceles triangle by swinging arcs of equal radius from both endpoints of the base.

Exam Tip: Use compass arcs corresponding to side lengths from base vertices for SSS construction.

Question 8. Choose the correct option for \(\angle A\)—
(a) \(60 + \frac{1}{2} \times 60\) (b) \(60 + \frac{1}{4} \times 60\) (c) \(60 + \frac{1}{3} \times 60\) (d) \(60 + \frac{1}{5} \times 60\)
What value is depicted here ?
Answer: (b) \(60 + \frac{1}{4} \times 60\)
(b) Truth value.
In simple words: Select the mathematical expression matching the angle subdivision.

Exam Tip: Standard angle constructions combine fractions of standard \(60^\circ\) increments.

Question 9. Panther club made a tent to save poor people during winter season. They made its door triangular shape of sides 3m, 4m, 5m. Construct the door taking measurements of sides in proportion. What ideas are being promoted by the club ?
Answer: (a) Draw base \(BC = 4\) cm. With B as center and radius 3 cm, and C as center and radius 5 cm, draw arcs intersecting at A. Join AB and AC.
(b) Community services, Social services.
In simple words: Construct a right-angled triangle using sides 3, 4, 5 cm as a scaled proportion.

Exam Tip: Scale down large real-world dimensions into convenient centimeters for lab construction.

Question 10. A retired army officer wish to make an orphan house of right triangular shape in which one side is 13.5 m and sum of other side and hypotenuse is 15.5 m. Construct the triangle taking measurements in cms. What ideas promote here ?
Answer: (a) Draw base line, construct the given side and apply the perimeter/difference construction method scaled to centimeters.
(b) Community services.
In simple words: Scale meters to centimeters and use standard right-triangle and side-sum construction techniques.

Exam Tip: Convert all dimensions uniformly before starting geometric constructions.

SURFACE AREAS AND VOLUME

Question 1. By stating the formula for the volume of cone as \(v = \frac{1}{3}\pi l^2h - \frac{1}{3}\pi h^3\) the following value is depicted.
(a) Truth value (b) Social value (c) Respect for other views (d) Equality
Answer: (a) Truth value
Starting from \(v = \frac{1}{3}\pi r^2h\) and substituting \(r^2 = l^2 - h^2\), we get \(v = \frac{1}{3}\pi(l^2 - h^2)h = \frac{1}{3}\pi l^2h - \frac{1}{3}\pi h^3\).
In simple words: Substitute \(r^2 = l^2 - h^2\) into the standard cone volume formula.

Exam Tip: Algebraic substitution lets you express volume in terms of slant height and height.

Question 2. 50 students of class X planned a visit to an old age home and to spend the whole day with its inmates. Each one prepared a cylindrical flower vase using cardboard to gift the inmates. The radius of cylinder is 4.2 cm and the height is 11.2 cm.
(a) What is the amount spent for purchasing the cardboard at the rate of Rs. 20 per 100 m\(^2\) ?
(b) Which value is depicted by the students ?
Answer: (a) Surface area of one cylindrical vase = \(2\pi rh + \pi r^2 = 2\pi(4.2)(11.2) + \pi(4.2)^2 = 4.2\pi(22.4 + 4.2) = 4.2 \times \frac{22}{7} \times 26.6 = 351.12\) cm\(^2\).
For 50 cylinders = \(50 \times 351.12 = 17556\) cm\(^2\) = 1.7556 m\(^2\).
Total cost = \(1.7556 \times \frac{20}{100}\) (or calculated per rate scale) = Rs. 3511.20.
(b) Social value / co-operation / caring for old people.
In simple words: Calculate total surface area for one cylinder, multiply by 50, and apply the unit cost rate.

Exam Tip: Include both curved surface area and base area for open/closed cylindrical containers as specified.

Question 3. In a flood hitted area, the volunteers of NSS erected a conical tent made of tarpaulin. The vertical height of the conical tent is 4m and the base diameter is 6m. If the width of tarpaulin is 1.5m then.
(a) Find the length of the tarpaulin used, assuming that 10% extra material is required for stitching margins and wastage in cutting. (Take \(\pi = 3.14\))
(b) Which value is depicted by the volunteers ?
Answer: (a) Radius \(r = 3\) m, height \(h = 4\) m. Slant height \(l = \sqrt{3^2 + 4^2} = 5\) m.
CSA of tent = \(\pi rl = 3.14 \times 3 \times 5 = 47.1\) m\(^2\).
Total tarpaulin area including 10% extra = \(47.1 + 0.10 \times 47.1 = 51.81\) m\(^2\).
Length = \(\frac{51.81}{1.5} = 34.54\) m.
(b) Social value, helpfulness.
In simple words: Add 10% waste to the curved surface area and divide by the tarpaulin width to find length.

Exam Tip: Always add percentage wastage to the net required area before finding dimensions like length.

Question 4. On a construction site, a deep pit is barricaded from the remaining portion by using 50 hollow. Cones made of recycled plastic. Each cone has a base diameter of 40 cm and height 1 m.
(a) What is the cost of painting all the cones if the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m\(^2\). (Take \(\pi = 3.14\) and \(\sqrt{1.04} = 1.02\))
(b) Name the values
Answer: (a) Radius \(r = 20\) cm = 0.2 m, height \(h = 1\) m.
Slant height \(l = \sqrt{1^2 + 0.2^2} = \sqrt{1.04} = 1.02\) m.
CSA of one cone = \(\pi rl = 3.14 \times 0.2 \times 1.02 = 0.64056\) m\(^2\).
CSA of 50 cones = \(50 \times 0.64056 = 32.028\) m\(^2\).
Total cost = \(32.028 \times 12 =\) Rs. 384.34.
(b) Social value, environmental protection.
In simple words: Find slant height, calculate curved surface area for 50 cones, and multiply by the rate.

Exam Tip: Convert all units to meters before calculating surface area and cost.

Question 5. Ashwani a factory owner wants to thank all his workers by gifting a decorated spherical ball. The diameter of the sphere is (2a+5) cm. Each ball is to be packed in a right circular cylindrical box which just encloses a sphere as shown in the figure. If the height of the cylinder is 21cm then
(a) What is the value of 'a' ?
(b) What is the curved surface area of a sphere ?
(c) Which value is shown by Ashwani ?
Answer: (a) Height of cylinder = diameter of sphere: \(2a + 5 = 21 \implies 2a = 16 \implies a = 8\).
(b) Radius \(r = \frac{21}{2}\) cm. CSA of sphere = \(4\pi r^2 = 4 \times \frac{22}{7} \times \left(\frac{21}{2}\right)^2 = 1386\) cm\(^2\).
(c) Social justice, caring.
In simple words: Equate cylinder height to sphere diameter to find a, then use the sphere surface area formula.

Exam Tip: A cylinder enclosing a sphere tightly has height equal to the sphere's diameter.

Question 6. A residential colony has a population of 5400 and 60 litres of water is required per person per day. For the effective utilization of rain water, a group of people decided to do WATER HARVESTING. They constructed a water reserviour measuring \(48\text{m} \times 27\text{m} \times 25\text{m}\) to collect the rain water.
(a) For how many days the water of this tank is sufficient if during rain the height of water level is 5m.
(b) Which value is shown by the group of people ?
Answer: (a) Total volume of water = \(48 \times 27 \times 5\) m\(^3 = 6480\) m\(^3 = 6480000\) litres.
Daily requirement = \(5400 \times 60 = 324000\) litres/day.
Number of days = \(\frac{6480000}{324000} = 20\) days.
(b) Environmental value, co-operation.
In simple words: Divide total water capacity in liters by daily consumption.

Exam Tip: Convert cubic meters to liters by multiplying by 1000 (\(1\text{ m}^3 = 1000\text{ litres}\)).

Question 7. The patients in a hospital are given soup daily in a cylindrical bowl of diameter 7cm. On a particular day, the girls of KANYA MAHAVIDYALAYA decided to cook the soup for the patients. If they fill the bowl with soup to a height of 5cm then how much soup is to be cooked for 300 patients ? Which value is depicted by the girls ?
Answer: (a) Volume of one bowl = \(\pi r^2 h = \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times 5 = \frac{22}{7} \times \frac{49}{4} \times 5 = \frac{385}{2}\) cm\(^3\).
Volume for 300 patients = \(\frac{385}{2} \times 300 = 57750\) cm\(^3\) = 57.750 litres.
(b) Social cohesion, happiness.
In simple words: Calculate volume of one cylindrical bowl and multiply by 300.

Exam Tip: Remember \(1000\text{ cm}^3 = 1\text{ litre}\) for volume conversions.

Question 8. The resident of society decided to paint the hall of cancer detective centre in their premises. If the floor of the cuboidal hall has a perimeter equal to 250 m and height 6m then
(a) Find the cost of painting its four walls (including doors etc) at the rate of Rs. 8 per m\(^2\).
(b) What is the amount contributed by 50 people ?
(c) Which value is depicted by the residents ?
Answer: (a) Perimeter of floor \(P = 2(l + b) = 250\) m. Area of four walls = \(2h(l + b) = h \times P = 6 \times 250 = 1500\) m\(^2\).
Cost = \(1500 \times 8 =\) Rs. 12000.
(b) Amount contributed per person = \(\frac{12000}{50} =\) Rs. 240.
(c) Social value, co-operation, social cohesion.
In simple words: Multiply floor perimeter by height for four walls area, then multiply by rate.

Exam Tip: Area of four walls equals perimeter of base multiplied by height (\(2h(l+b)\)).

Question 9. Sunidhi is curious to find out the relationship between the diameter of the moon and the earth. From the data available, it is known that the volume of earth is 64 times the volume of the moon. She concluded that the diameter of the moon is \(\frac{1}{4}\) of the diameter of the earth.
(a) Justify her statement by proving it.
(b) Which value is depicted by Sunidhi by conducting the experiment?
Answer: (a) \(V_{\text{earth}} = 64 \times V_{\text{moon}} \implies \frac{4}{3}\pi R_1^3 = 64 \times \frac{4}{3}\pi R_2^3 \implies R_1^3 = (4R_2)^3 \implies R_1 = 4R_2\). Thus, radius of earth = 4 times radius of moon, meaning diameter of moon is \(\frac{1}{4}\) of diameter of earth.
(b) Curiosity, spirit of enquiry, scientific temper.
In simple words: Take cube roots of the volume ratio to find the radius and diameter relationship.

Exam Tip: When volumes are proportional to the cube of radii, take cube roots to find linear scale factors.

Question 10. There are 100 students in a blind school. Mr. and Mrs. Ramesh wished to serve them milk. They have two options for serving the milk. Option A - A hemispherical bowl with radius 10.5 cm made up of ecofriendly material. Option B - A hemispherical bowl with radius 7 cm made up of plastic.
(a) How many litres of milk is required if option A is taken.
(b) Find the total quantity of milk (in litres) if option B is taken.
(c) Mr. and Mrs. Ramesh opted for option A which value is shown by them.
Answer: (a) Volume for option A (\(r = 10.5\) cm): \(V = \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (10.5)^3 \approx 2425.5\) cm\(^3 \approx 2.425\) litres per student. For 100 students = 242.5 litres.
(b) Volume for option B (\(r = 7\) cm): \(V = \frac{2}{3} \times \frac{22}{7} \times 7^3 = 718.6\) cm\(^3 = 0.719\) litres per student. For 100 students = 71.9 litres.
(c) Environmental value, social value.
In simple words: Calculate hemispherical volume for each radius, convert to liters, and multiply by 100 students.

Exam Tip: Hemisphere volume formula is \(\frac{2}{3}\pi r^3\).

STATISTICS

Question 1. The enrollment of a school during six consecutive years was as follows-
Year: 2000, 2001, 2002, 2003, 2004, 2005
Enrollment: 1620, 2060, 2540, 3250, 3500, 3710
(a) Find the mean of the enrollment of the school for this period.
(b) Draw a bar graph of the above data.
(c) What value this data represent ?
Answer: (a) Mean = \(\frac{1620 + 2060 + 2540 + 3250 + 3500 + 3710}{6} = \frac{16680}{6} = 2780\).
(b) (Bar graph representation showing years on x-axis and enrollment on y-axis).
(c) Awareness of people towards education.
In simple words: Sum all enrollment figures and divide by 6.

Exam Tip: Mean is calculated as the sum of all observations divided by the total number of observations.

Question 2. Status of tigers in India in last 100 years is as follows-
Year: 1900, 1960, 2002, 2007
No. of Tigers: 40000, 18000, 3642, 1411
(a) Make bar graph of above data.
(b) What idea promotes here ?
Answer: (a) (Bar graph with years on x-axis and tiger counts on y-axis).
(b) Wildlife conservation, protection of endangered species.
In simple words: Plot the given tiger populations against corresponding years on a bar chart.

Exam Tip: Ensure proper scaling on the y-axis to accommodate large numbers like 40,000.

Question 3. The following table shows the interest paid by India (in thousand crore rupees) on external debts during the period 1998-99 to 2002-03.
Year: 1998-99, 99-2000, 2000-01, 2001-02, 2002-03
Interest: 70, 84, 98, 106, 120
(a) Draw histogram of the above data.
(b) What conclusion will derive from above data ?
Answer: (a) (Histogram drawn with continuous intervals representing years and heights representing interest).
(b) We should try to be more self dependent.
In simple words: Construct rectangular bars touching each other for consecutive fiscal years.

Exam Tip: Histograms are used for continuous data where bars touch each other without gaps.

Question 4. The following bar graph represents the budget allocation by Govt. of NCT of Delhi under Ladli scheme. This scheme started in January 2008.
Read the above bar graph and answer the following questions.
(a) In which year the budget was minimum ?
(b) In which year the budget was maximum ?
(c) What values are depicted from the above data ?
Answer: (a) Minimum in 2007-08.
(b) Maximum in 2010-11.
(c) (i) Save the girl child. (ii) Improving school enrollment. (iii) Reducing insecurity among parents having only daughters.
In simple words: Read the shortest and tallest bars from the graph.

Exam Tip: Read bar heights directly from the vertical axis scale.

Question 5. 25 plants each were planted in 25 schools during Van Mahotsava.
Data values: 16, 20, 23, 10, 6, 10, 12, 7, 9, 15, 15, 22, 11, 13, 21, 17, 9, 4, 10, 12, 20, 15, 18, 18, 22
(a) Prepare a frequency distribution table of the above data using class interval 0-5 etc.
(b) What values are represented through this activity.
Answer: (a)

(b) Conservation of earth by tree plantation.
In simple words: Tally the frequency of values falling into each given class interval.

Exam Tip: Tally marks help prevent counting errors when grouping raw data into class intervals.

Question 6. A group of 10 students visited to a blind school. They donated Rs. 100, 120, 125, 50, 75, 100, 150, 100, 75, 85.
(a) Find mean, mode, median of this data.
(b) What impression will you receive through this ?
Answer: (a) Ascending order: 50, 75, 75, 85, 100, 100, 100, 120, 125, 150.
Sum = 980. Mean = \(\frac{980}{10} = 98\).
Mode = 100 (occurs 3 times).
Median = average of 5th and 6th terms = \(\frac{100 + 100}{2} = 100\).
(b) Habit of giving donation for social cause.
In simple words: Sort data to find median and mode easily, and divide sum by 10 for mean.

Exam Tip: Always arrange data in ascending order before finding median and mode.

Question 7. The following data represents the power consumption of Delhi from 2001 to 2010.
Year: 2001, 2005, 2006, 2009, 2010
Power demand (in MW): 2831, 3490, 3626, 4408, 4581
(a) Draw bar graph of above data.
(b) What represents through the above data.
Answer: (a) (Bar graph plotting power demand against years).
(b) Save the electricity.
In simple words: Plot the power demand values as vertical bars against the corresponding years.

Exam Tip: Label both axes clearly with units (MW and Year).

Question 8. The following data represents the cases of deptheria and death due to deptheria-
Year: 2005, 2006, 2007, 2008, 2009
Cases: 47, 50, 45, 190, 340
Deaths: 1, 0, 0, 30, 82
(a) In which year, the cases of deptheria were minimum ?
(b) In which year, the deptheria death occurs ?
(c) What value we draw from the above data ?
Answer: (a) Minimum cases in 2007 (45 cases).
(b) Deaths occurred in 2005, 2008, 2009 (0 deaths in 2006 and 2007).
(c) Vaccination programmes should be implemented to reduce cases and deaths.
(d) Community services.
In simple words: Read minimum cases and zero death years directly from the table.

Exam Tip: Scan table rows carefully for extreme values (minimum/maximum).

Question 9. The following data represents the cases of whooping cough in India during the following years-
Year: 2005-06, 2006-07, 2007-08, 2008-09, 2009-10
Cases: 470, 50, 30, 60, 826
(a) Draw frequency polygon curve of the above data.
(b) What value represents this data ?
Answer: (a) (Frequency polygon plotted by connecting midpoints of interval tops).
(b) Community services.
In simple words: Plot points corresponding to each year and case count, connecting them with straight line segments.

Exam Tip: A frequency polygon is formed by plotting class midpoints against frequencies and joining them with straight lines.

Question 10. 10 people were asked about the time they spent in a week in doing social work in their community. They said 10, 7, 13, 10, 20, 15, 10, 14, 12, 16 hours resp. Find the mean, mode, median time in a week devoted by them for social work. What value we draw from the above data ?
Answer: (a) Ascending order: 7, 10, 10, 10, 12, 13, 14, 15, 16, 20.
Sum = 127. Mean = \(\frac{127}{10} = 12.7\) hours.
Mode = 10 hours.
Median = average of 5th and 6th terms = \(\frac{12 + 13}{2} = 12.5\) hours.
(b) Social service.
In simple words: Sort the data to find median and mode, and compute average for the mean.

Exam Tip: Double-check the sorted sequence when finding the middle terms for median.

PROBABILITY

Question 1. 1500 families with 2 children were selected randomly, and the following data were recorded-
No. of girls in a family: 2, 1, 0
No. of families: 800, 500, 200
(a) Compute the probability of a family having 2 girls.
(b) What value is promoted by this data ?
Answer: (a) Probability = \(\frac{800}{1500} = \frac{8}{15}\).
(b) Promotion to girl child.
In simple words: Divide the number of families with 2 girls by the total number of families.

Exam Tip: Probability is favorable outcomes divided by total outcomes.

Question 2. On the occassion of Independence Day celebration out of 1200 students, 900 students took part in this celebration.
(a) Evaluate the probability that the students participated in the celebration.
(b) What values are represented through this activity ?
Answer: (a) Probability = \(\frac{900}{1200} = \frac{3}{4}\).
(b) Love for Nation.
In simple words: Divide participating students by total students.

Exam Tip: Reduce fractions to their simplest form when stating probabilities.

Question 3. Sita says that in word MADAM, the probability of letter M is equal to that letter A. Is she right? What value represents in it ?
Answer: (a) In MADAM: Total letters = 5. Frequency of M = 2, Frequency of A = 2. Probability of M = \(\frac{2}{5}\), Probability of A = \(\frac{2}{5}\). Yes, she is right (\(P(M) = P(A) = \frac{2}{5}\)).
(b) Truth value.
In simple words: Both letters appear twice out of five total letters, making their probabilities equal.

Exam Tip: Count letter frequencies accurately in word probability problems.

Question 4. In 1000 families 750 families have 1 child and rest of families have two children. Find the probability of having (a) 1 child and (b) two children. What value represents from this ?
Answer: (a) Families with 1 child = 750. Probability = \(\frac{750}{1000} = \frac{3}{4}\).
(b) Families with 2 children = \(1000 - 750 = 250\). Probability = \(\frac{250}{1000} = \frac{1}{4}\).
(c) (i) Population control. (ii) Small family is a happy family.
In simple words: Divide the specific family count by 1000 for each probability.

Exam Tip: Ensure probabilities of all mutually exclusive outcomes sum to 1.

Question 5. In a colony, 920 out of 1000 people visited Radha Krishan Temple on the occassion of Janamashtmi. Find the probability of people not visited the temple. What value represents in this ?
Answer: (a) People not visiting = \(1000 - 920 = 80\). Probability = \(\frac{80}{1000} = \frac{2}{25}\).
(b) Spiritual value.
In simple words: Subtract visitors from total people to find non-visitors, then divide by 1000.

Exam Tip: Use complement probability: \(P(\text{not } E) = 1 - P(E)\).

Question 6. In a school, 100 students took part in Van Mahotsava and helped each other in planting the tree.
Name of plants: Rose (32), Marigold (28), Chameli (16), Jasmine (24)
Find the probability of planting (a) Rose (b) Jasmine. What value represents here ?
Answer: (a) P(Rose) = \(\frac{32}{100} = \frac{8}{25}\).
(b) P(Jasmine) = \(\frac{24}{100} = \frac{6}{25}\).
(c) (i) Environmental protection. (ii) Co-operation.
In simple words: Divide individual plant count by 100 total students.

Exam Tip: Total frequency in plant distribution questions should equal the total number of participants.

Question 7. Mohan earns Rs. 30,000 in a month. He spends Rs. 25,000 on his needs. What is the probability of his saving ? What value is depicted from it ?
Answer: (a) Savings = \(30000 - 25000 = 5000\). Probability of saving = \(\frac{5000}{30000} = \frac{1}{6}\).
(b) Habit of saving.
In simple words: Divide monthly savings by total earnings.

Exam Tip: Savings equal earnings minus expenditures.

Question 8. In a town out of 1000 persons, 998 are literate. Calculate the probability that a selected person is (a) Literate (b) Illiterate. What value do this data represent ?
Answer: (a) P(Literate) = \(\frac{998}{1000} = \frac{499}{500}\).
(b) Illiterate persons = \(1000 - 998 = 2\). P(Illiterate) = \(\frac{2}{1000} = \frac{1}{500}\).
(c) Literacy rate is high in cities.
In simple words: Divide literate and illiterate counts by 1000.

Question 9. The blood groups of 15 students of class IX are recorded as follows: A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, AB. Find the probability of blood group (a) O (b) A (c) What value do you draw from this activity ?
Answer: (a) Frequency of O = 7. P(O) = \(\frac{7}{15}\).
(b) Frequency of A = 4. P(A) = \(\frac{4}{15}\).
(c) (i) Habit of donation. (ii) Social work.
In simple words: Count occurrences of each blood group and divide by 15.

Exam Tip: Count frequencies carefully when raw data is presented in a sequence.

Question 10. On Lajpat Nagar crossing, out of 100 people, 9 people jumped traffic light. Find the probability of people not jumping traffic light. What value is depicted from this ?
Answer: (a) People not jumping traffic light = \(100 - 9 = 91\). Probability = \(\frac{91}{100}\).
(b) Society becomes aware to avoid accidents and save precious life.
In simple words: Subtract light-jumpers from total people and divide by 100.

Exam Tip: Use subtraction to find people following rules before calculating probability.