NCERT Solutions for Class 9 Science Chapter 8 Motion

NCERT Solutions for Class 9 Science Chapter 8 Motion

These Solutions are part of NCERT Solutions for Class 9 Science. Here we have given NCERT Solutions for Class 9 Science Chapter 8 Motion. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 9 Science (Biology) Chapter 8 – Motion solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 8 – Motion Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

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NCERT TEXT BOOK QUESTIONS

IN TEXT QUESTIONS

Question 1.
An object has moved through a distance. Can it have zero displacement ? If yes, support your answer with an example. (CBSE 2015)
Answer:
Yes, The displacement of the object can be zero. Let a boy completes one round of a circular track in 5 minutes. The distance travelled by the boy = circumference of the circular track. However, displacement of the boy is zero because his initial and final positions are same.

Question 2.
A farmer moves along the boundary of a square field of side 10 m in 40 s.
What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds ? (CBSE 2010, 2015)
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 1
Answer:
ABCD is a square field of side 10 m.
The farmer moves along the boundary of the field from the corner A via the corners B, C and D.
After every 40 s, the farmer is again at the corner A, so his displacement after every 40 s is zero.
At the end of 2 minutes 20 seconds = (2 x 60 + 20) = 140 s, the farmer will be at the corner C.
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 2

Question 3.
Distinguish between speed and velocity. (CBSE 2010, 2012, 2013, 2015)
Answer:

Speed

Velocity
1. Distance travelled by an object per unit time is known as its speed. The distance travelled by an object in a particular direction (i.e. displacement) per unit time is known as its velocity.
2. Average speed of a moving object cannot be zero. Average velocity of a moving object can be zero.
3. tells how fast an object moves. Velocity tells how fast an object moves and in which direction it moves.
4. Speed is a scalar quantity. Velocity is a vector quantity.
5. Speed of an object is always positive. Velocity of an object can be positive or negative.

Question 4.
Under what conditions) is the magnitude of the average velocity of an object equal to its average speed ? (CBSE 2012, 2013)
Answer:
When an object moves in one direction along a straight line.

Question 5.
What does the odometer of an automobile measure ? (CBSE 2010, 2011, 2014, 2015)
Answer:
Odometer of an automobile measures distance travelled by the automobile.

Question 6.
What does the path of an object look like when it is in uniform motion ? (CBSE 2012)
Answer:
Straight path.

Question 7.
During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station ? The signal travels at the speed of light, that is, 3 x 108 m s-1
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 3

Question 8.
When will you say a body is in

  1. uniform acceleration ?
  2. non-uniform acceleration (CBSE 2012, 2013)

Answer:

  1. A body has uniform acceleration if its velocity changes by an equal amount in equal intervals of time. .
  2. A body has non-uniform acceleration if its velocity changes by unequal amount in equal intervals of time.

Question 9.
A bus decreases its speed from 80 kmh-1 to 60 kmh-1 in 5 s. Find the acceleration of the bus.
(CBSE 2010, 2011, 2012, 2015)
Answer:
Here,
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 4
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 5

Question 10.
A train starting from a railway station and moving with a uniform acceleration attains a speed of 40 km h-1 n 10 minutes. Find its acceleration.
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 6

Question 11.
What is the nature of the distance-time graphs for uniform and non-uniform motion of an object ?
(CBSE 2012)
Answer:

  1. For uniform motion, distance-time graph is a straight line having constant gradient or slope as shown in figure.
  2. For non-uniform motion, distance-time graph is a curve having increasing gradient or slope or decreasing gradient as shown in figures.
    NCERT Solutions for Class 9 Science Chapter 8 Motion image - 7

Question 12.
What can you say about the motion of an object whose distance-time graph is a straight line parallel to
the time axis ? (CBSE 2010, 2012)
Answer:
When distance-time graph is a straight line parallel to the time axis, it means, the distance of the object is not changing with time. Thus, the object is stationary.

Question 13.
What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time-axis ?
Answer:
In this case, speed of the object is constant. That is, the object travels equal distances in equal intervals of time along a straight line. Hence, motion of the object is uniform motion.

Question 14.
What is the quantity which is measured by the area occupied below the velocity-time graph ?
(CBSE 2010, 2012)
Answer:
Magnitude of the displacement of a body is measured by the area under the velocify-time graph.

Question 15.
A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find
(a) the speed acquired,
(b) the distance travelled. (CBSE 2012)
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 8

Question 16.
A train is travelling at a speed of 90 km h-1. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s-2 . Find how far the train will go before it is brought to rest ?
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 9

Question 17.
A trolley, while going down an inclined plane has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start ?
Answer:
Here, u = 0,v = ?, a = 2 cm s-2, t = 3 s
Using, v = u + at, we get
v = 0 + 2 x 3 = 6 cm s-2.

Question 18.
A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start ?
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 10

Question 19.
A stone is thrown in a vertically upward direction with a velocity of 5 m s-1 If the acceleration of the stone during its motion is 10 m s-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there ?
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 11

NCERT CHAPTER END EXERCISE

Question 1.
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes and 20 s ?
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 12
(ii) After every 40 s, athlete reaches his starting point, so after 40 s, his displacement is zero. It means, the athlete completes the circular track 3 times in 120 s and in the next 20 s, he is just opposite to his starting point. Therefore, the magnitude of the displacement of the athlete at the end of the 140 s (or 2 minutes 20 s) = Diameter of the circular track = 200 m.

Question 2.
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging
(a) from A to B and
(b) from A to C ? (CBSE 2010, Term I)
Answer:
Distance from A to B – 300 m
Displacement from A to B = 300 m
Time taken to go from A to B =2 minutes 50 s = 170 s
T Solutions for Class 9 Science Chapter 8 Motion image - 13

Question 3.
Abdul, while driving to school, computes the average speed for his trip to be 20 km h-1. On his return trip along the same route, there is less traffic and the average speed is 40 km h-1. What is the average speed for Abdul’s trip ? (CBSE 2011)
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 14

Question 4.
A motor boat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s-2 for 8.0 s. How far does the boat travel during this time ? (CBSE 2011, 2013)
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 15

Question 5.
A driver of a car travelling at 52 km h-1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h-1 in another car applies his breaks slowly and stops in 10 s. On the same graph paper plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied ?
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 16

Question 6.
Figure shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions :
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 17
(a) Which of the three is travelling the fastest ?
(b) Are all three ever at the same point on the road ?
(c) How far has C travelled when B passes A ?
(d) How far has B travelled by the time it passes C ?
Answer:
(a) Speed = Slope of distance-time graph.
Since slope of distance-time graph for object B is the greatest, so object B is travelling the fastest.
(b) All the three objects will be at the same point on the road if all the three distance-time graphs intersect each other at a time. Since, all the three distance-time graphs do not intersect each other at a time, so they are never at the same point on the road.
(c) When B passes A, distance travelled by C = 9.6 – 2 = 7.6 km.
(d) Distance travelled by B by the time it passes C = 6 km.

Question 7.
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground ? After what time will it strike the ground ?
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 18

Question 8.
The speed time graph for a car as shown in figure
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 19
(a) Find how far does the car travel in the firdt four seconds.
Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
Answer:
(a)
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 20
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 21
(b) The straight part of the curve parallel to time axis represents the uniform motion of the car.

Question 9.
State which of the following situations are possible and give an example for each of these :
(a) an object with a constant acceleration but zero velocity.
(b) an object moving with an acceleration but with uniform speed.
(c) an object moving in a certain direction with an acceleration in the perpendicular direction.
(CBSE 2012)
Answer:
(a) When an object is thrown vertically upward, its velocity at the highest point is zero but it has constant acceleration = 9.8 m s-2
(b) Uniform circular motion.
(c) An object moving in a circular path with uniform speed, its acceleration is always perpendicular to the direction of the motion of the object.

Question 10.
An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 . hours to revolve around the earth. (CBSE 2011)
Answer:
NCERT Solutions for Class 9 Science Chapter 8 Motion image - 22

NCERT Solutions for Class 9 Science Chapter 8 Motion

Hope given NCERT Solutions for Class 9 Science Chapter 8 are helpful to complete your science homework.

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HOTS Questions for Class 9 Science Chapter 14 Natural Resources

HOTS Questions for Class 9 Science Chapter 14 Natural Resources

These Solutions are part of HOTS Questions for Class 9 Science. Here we have given HOTS Questions for Class 9 Science Chapter 14 Natural Resources

Question 1.
(a) Fill in the blanks : A, B and C.
HOTS Questions for Class 9 Science Chapter 14 Natural Resources image - 1
(b) Identify the biogeochemical cycle.
Answer:
(a) Fill in the blanks :
A- Organic molecules (C6H12O6).
B- Photosynthesis.
C- Respiration.
(b) Identification: Oxygen cycle.

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Question 2.
(a) Fill in the blanks: A to F.
HOTS Questions for Class 9 Science Chapter 14 Natural Resources image - 2
(b) What will happen if the step A does not take place ?
(c) What will happen if the step F does not take place ?
Answer:
(a) Fill in the blanks :
A- Ammonification.
B- Ammonia.
C- Nitrites.
D- Nitrates.
E- Nitrification.
F- Denitrification.
(b) Absence of step A: No breakdown of proteins to release ammonia.
(c) Absence of step F: There will be no replenishment of atmospheric nitrogen.

Hope given HOTS Questions for Class 9 Science Chapter 14 Natural Resources are helpful to complete your science homework.

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RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS

RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS

Other Exercises

Mark the correct alternative in each of the following:
Question 1.
Two parallelograms are on the same base and between the same parallels. The ratio of their areas is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 1
(d) 3 : 1
Solution:
Two parallelograms which are on the same base and between the same parallels are equal in area
∴ Ratio in their areas =1 : 1 (c)

Question 2.
A triangle and a parallelogram are on the same base and between the same parallels. The ratio of the areas of triangle and parallelogram is
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Solution:
A triangle and a parallelogram which are on the same base and between the same parallels, then area of triangle is half the area of the parallelogram
∴ Their ratio =1:2 (c)

Question 3.
Let ABC be a triangle of area 24 sq. units and PQR be the triangle formed by the mid-points of sides of ∆ABC. Then the area of ∆PQR is
(a) 12 sq. units
(b) 6 sq. units
(c) 4 sq. units
(d) 3 sq. units
Solution:
Area of ∆ABC = 24 sq. units
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q4.1
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q4.2

Question 4.
The median of a triangle divides it into two
(a) congruent triangle
(b) isosceles triangles
(c) right triangles
(d) triangles of equal areas
Solution:
The median of a triangle divides it into two triangles equal in area (d)

Question 5.
In a ∆ABC, D, E, F are the mid-points of sides BC, CA and AB respectively. If
ar(∆ABC) = 16 cm2, then ar(trapezium FBCE) =
(a) 4 cm²
(b) 8 cm²
(c) 12 cm²
(d) 10 cm²
Solution:
In ∆ABC, D, E and F are the mid points of sides BC, CA and AB respectively
ar(∆ABC) = 16 cm²
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q5.1

Question 6.
ABCD is a parallelogram. P is any point on CD. If ar(∆DPA) = 15 cm² and ar(∆APC) = 20 cm², then ar(∆APB) =
(a) 15 cm²
(b) 20 cm²
(c) 35 cm²
(d) 30 cm²
Solution:
In ||gm ABCD, P is any point on CD
AP, AC and PB are joined
ar(∆DPA) =15 cm²
ar(∆APC) = 20 cm²
Adding, ar(∆ADC) = 15 + 20 = 35 cm²
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q6.1
∵ AC divides it into two triangles equal in area
∴ ar(∆ACB) = ar(∆ADC) = 35 cm²
∵ ∆APB and ∆ACB are on the same base
AB and between the same parallels
∴ ar(∆APB) = ar(∆ACB) = 35 cm²(c)

Question 7.
The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 16 cm and 12 cm is
(a) 28 cm²
(b) 48 cm²
(c) 96 cm²
(d) 24 cm²
Solution:
In rhombus ABCD,
P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively and are joined in order to get a quad. PQRS
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q7.1

Question 8.
A, B, C, D are mid points of sides of parallelogram PQRS. If ar(PQRS) = 36 cm²,then ar(ABCD) =
(a) 24 cm²
(b) 18 cm²
(c) 30 cm²
(d) 36 cm²
Solution:
A, B, C and D are the mid points of a ||gm PQRS
Area of PQRS = 36 cm²
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q8.1
The area of ||gm formed by joining AB, BC, CD and DA
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q8.2

Question 9.
The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is
(a) a rhombus of area 24 cm²
(b) a rectangle of area 24 cm²
(c) a square of area 26 cm²
(d) a trapezium of area 14 cm²
Solution:
Let P, Q, R, S be the mid points of sides of a rectangle ABCD. Whose sides 8 cm and 6 cm
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q9.1
Their PQRS is a rhombus
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q9.2

Question 10.
If AD is median of ∆ABC and P is a point on AC such that ar(∆ADP) : ar(∆ABD) = 2:3, then ar(∆PDC) : ar(∆ABC) is
(a) 1 : 5
(b) 1 : 5
(c) 1 : 6
(d) 3 : 5
Solution:
AD is the median of ∆ABC,
P is a point on AC such that
ar(∆ADP) : ar(∆ABD) = 2:3
Let area of ∆ADP = 2×2
Then area of ∆ABD = 3×2
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q10.1
But area of AABD = \(\frac { 1 }{ 2 }\) area AABC
∴ Area ∆ABC = 2 x area of ∆ABD
= 2 x 3x² = 6x²
and area of ∆PDC = area ∆ADC – (ar∆ADP) = area ∆ABD – ar ∆ADP
= 3x² – 2x² = x²
∴ Ratio = x² : 6x²
= 1 : 6 (c)

Question 11.
Medians of AABC, intersect at G. If ar(∆ABC) = 27 cm2, then ar(∆BGC) =
(a) 6 cm2
(b) 9 cm2
(c) 12 cm2
(d) 18 cm2
Solution:
In ∆ABC, AD, BE and CF are the medians which intersect each other at G
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q11.1

Question 12.
In a ∆ABC if D and E are mid-points of BC and AD respectively such that ar(∆AEC) = 4 cm², then ar(∆BEC) =
(a) 4 cm²
(b) 6 cm²
(c) 8 cm²
(d) 12 cm²
Solution:
In ∆ABC, D and E are the mid points of BC and AD
Join BE and CE ar(∆AEC) = 4 cm²
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q12.1
In ∆ABC,
∵ AD is the median of BC
∴ ar(∆ABD) = ar(∆ACD)
Similarly in ∆EBC,
ED is the median
∴ ar(∆EBD) = ar(∆ECD)
and in ∆ADC, CE is the median
∴ ar(∆FDC) = ar(∆AEC)
= 4 cm
∴ar∆EBC = 2 x ar(∆EDC)
= 2 x 4 = 8 cm (c)

Question 13.
In the figure, ABCD is a parallelogram. If AB = 12 cm, AE = 7.5 cm, CF = 15 cm, then AD =
(a) 3 cm
(b) 6 cm
(c) 8 cm
(d) 10.5 cm
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q13.1
Solution:
In ||gm ABCD, AB = 12 cm AE = 7.5 cm
∴ Area of ||gm ABCD = base x height = AB x AE = 12 x 7.5 cm² = 90 cm²
Now area ||gm ABCD = 90 cm²
and altitude CF = 15 cm
∴ Base AD = \(\frac { Area }{ Altitude }\) = \(\frac { 90 }{ 15 }\) = 6 cm (b)

Question 14.
In the figure, PQRS is a parallelogram. If X and Y are mid-points of PQ and SR respectively and diagonal SQ is joined. The ratio ar(||gm XQRY) : ar(∆QSR) =
(a) 1 : 4
(b) 2 : 1
(c) 1 : 2
(d) 1 : 1
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q14.1
Solution:
In ||gm PQRS, X and Y are the mid points of PQ and SR respectively XY and SQ are joined.
∵ XY bisects PQ and SR
∴ PXYS and XQRY are also ||gms and ar(∆PXYS) = nr(∆XQRY)
∵ ||gm PQRS and AQSR are on the same base and between the same parallel lines
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q14.2

Question 15.
Diagonal AC and BD of trapezium ABCD, in which AB || DC, intersect each other at O. The triangle which is equal in area of ∆AOD is
(a) ∆AOB
(b) ∆BOC
(c) ∆DOC
(d) ∆ADC
Solution:
In trapezium ABCD, diagonals AC and BD intersect each other at O. AB || DC
∆ABC and ∆ABD are on the same base and between the same parallels
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q15.1
∴ ar(∆ABC) = or(∆ABD)
Subtracting ar(∆AOB)
ar(∆ABC) – ar(∆AOB) = ar(∆ADB) – ar(∆AOB)
⇒ ar(∆BOC) = ar(∆AOD)
ar(∆AOD) = ar(∆BOC) (c)

Question 16.
ABCD is a trapezium in which AB || DC. If ar(∆ABD) = 24 cm² and AB = 8 cm, then height of ∆ABC is
(a) 3 cm
(b) 4 cm
(c) 6 cm
(d) 8 cm
Solution:
In trapezium ABCD, AB || DC
AC and BD are joined
ar(∆ABD) = 24 cm2
AB = 8 cm,
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q16.1

Question 17.
ABCD is a trapezium with parallel sides AB = a and DC = b. If E and F are mid-points of non-parallel sides AD and BC respectively, then the ratio of areas of quadrilaterals ABFE and EFCD is
(a) a : b
(b) (a + 3b) : (3a + b)
(c) (3a + b) : (a + 3b)
(d) (2a + b) : (3a + b)
Solution:
In quadrilateral ABCD, E and F are the mid points of AD and BC
AB = a, CD = b
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q17.1
Let h be the height of trapezium ABCD then height of each quadrilateral
ABFE = altitude of quadrilateral EFCD = \(\frac { h }{ 2 }\)
Now area of trap. ABFE = \(\frac { 1 }{ 2 }\) (sum of parallel sides) x altitude
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q17.2

Question 18.
ABCD is a rectangle with O as any point in its interior. If or(∆AOD) = 3 cm2 and ar(∆BOC) = 6 cm2, then area of rectangle ABCD is
(a) 9 cm2
(b) 12 cm2
(c) 15 cm2
(d) 18 cm2
Solution:
In rectangle ABCD, O is any point
ar(∆AOD) = 3 cm2
and ar(∆BOC) = 6 cm2
Join OA, OB, OC and OD
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q18.1
We know that if O is any point in ABCD Then ar(AOB) + ar(COD) = ar(AOB) + ar(BOC)
= 3 + 6 = 9 cm
∴ ar(rect. ABCD) = 2 x 9 = 18 cm (d)

Question 19.
The mid-points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q19.1
Solution:
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q19.2
P,Q and R the mid points of the sides of a ∆ABC then area of any parallelogram formed by the mid points and one vertex of the given triangle has area = \(\frac { 1 }{ 2 }\) area ∆ABC (b)

Question 20.
In the figure, ABCD and FECG are parallelograms equal in area. If ar(∆AQE) = 12 cm2, then ar(||gm FGBQ) =
(a) 12 cm2
(b) 20 cm2
(c) 24 cm2
(d) 36 cm2
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q20.1
Solution:
In the figure, ABCD and EFCG are parallelograms equal in area and ar(∆AQE) = 12 cm2
In ||gm AQED, AE is the diagonal
∴ ar(∆AQE) = \(\frac { 1 }{ 2 }\) ar(||gm AQED)
⇒ 12 cm2 = \(\frac { 1 }{ 2 }\) ar(||gm AQED)
∴ ar(||gm AQED) = 24 cm2
∵ ar ||gm ABCD = ar ||gm FECG
⇒ ar(||gm ∆QED) + ar(|| gm QBCE)
= ar(||gm QBCE) + ar(||gm FGBQ)
⇒ ar(||gm ∆QED) = ar(||gm FGBQ)
= 24 cm2 (c)

Hope given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8C

RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8C.

Other Exercises

Objective Questions :
Tick the correct answer in each of the following:

Question 1.
Solution:
2x – 3 = x + 2
=> 2x – x
= 2 + 3
= 5 (c)

Question 2.
Solution:
5x + \(\\ \frac { 7 }{ 2 } \) = \(\\ \frac { 3 }{ 2 } \) x – 14
=> 5x – \(\\ \frac { 3 }{ 2 } \) x = – 14 – \(\\ \frac { 7 }{ 2 } \)
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8C 2.1

Question 3.
Solution:
z = \(\\ \frac { 4 }{ 5 } \)(z + 10)
=> 5z = 4z + 40
=> 5z – 4z = 40
=> z = 40 (a)

Question 4.
Solution:
3m = 5m – \(\\ \frac { 8 }{ 5 } \)
=> 3m – 5m = \(\\ \frac { -8 }{ 5 } \)
=> – 2m = \(\\ \frac { -8 }{ 5 } \)
=> m = \(\\ \frac { -8 }{ -5×2 } \) = \(\\ \frac { 4 }{ 5 } \) (c)

Question 5.
Solution:
5t – 3 = 3t, – 5
=> 5t – 3t = – 5 + 3
=> 2t = – 2
=> t = \(\\ \frac { -2 }{ 2 } \) = – 1 (b)

Question 6.
Solution:
2y + \(\\ \frac { 5 }{ 3 } \) = \(\\ \frac { 26 }{ 3 } \) – y
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8C 6.1

Question 7.
Solution:
\(\\ \frac { 6x+1 }{ 3 } \) +1 = \(\\ \frac { x-3 }{ 6 } \)
\(\\ \frac { 12x+2+6=x-3 }{ 6 } \)
12x – x = – 3 – 2 – 6
11x – 11
=> x = \(\\ \frac { -11 }{ 11 } \) = – 1 (b)

Question 8.
Solution:
\(\\ \frac { n }{ 2 } \) – \(\\ \frac { 3n }{ 4 } \) + \(\\ \frac { 5n }{ 6 } \) = 21
\(\\ \frac { 6n-9n+10n= 252 }{ 12 } \)
LCM of 2, 4, 6 = 12
16n – 9n = 252
=> 7n = 252
=> n = \(\\ \frac { 252 }{ 7 } \) = 36 (c)

Question 9.
Solution:
\(\\ \frac { x+1 }{ 2x+3 } \) = \(\\ \frac { 3 }{ 8 } \)
=> 8 (x + 1) = 3 (2x + 3)
(By cross multiplication)
x + 8x + 9 = 8x – 6x = 9 – 8
=> 2x = 1
=> x = \(\\ \frac { 1 }{ 2 } \)
x = \(\\ \frac { 1 }{ 2 } \) (d)

Question 10.
Solution:
\(\\ \frac { 4x+8 }{ 5x+8 } \) = \(\\ \frac { 5 }{ 6 } \)
6(4x + 8) = 5(5x + 8)
(By cross multiplication)
24x + 48 = 25x + 40
=> 24x – 25x = 40 – 48
=> – x = – 8
=> x = 8 (c)

Question 11.
Solution:
\(\\ \frac { n }{ n+15 } \) = \(\\ \frac { 4 }{ 9 } \)
9n = 4n + 60
(By cross multiplication)
9n – 4n = 60
=> 5n = 60
=> n = \(\\ \frac { 60 }{ 5 } \) = 12
n = 12 (d)

Question 12.
Solution:
3(t – 3) = 5 (2t + 1)
3t – 9 = 10t + 5
=> 3t – 10t = 5 + 9
=> – 7t = 14
=> t = \(\\ \frac { 14 }{ -7 } \) = – 2
t = – 2 (a)

Question 13.
Solution:
Let number = x
Then \(\\ \frac { 4 }{ 5 } \)x = \(\\ \frac { 3 }{ 4 } \)x + 4
=> \(\\ \frac { 16x=15x+80 }{ 20 } \)
16x – 15x = 80
=> x = 80
:. Number = 80 (c)

Question 14.
Solution:
Ages of A : B = 5 : 7
Let A’s age = 5x
Then B’s age = 7x
After 4 years
A’s age = 5x + 4
and B’s age = 7x + 4
\(\\ \frac { 5x+4 }{ 7x+5 } \) = \(\\ \frac { 3 }{ 4 } \)
=> 3(7x + 4) = 4(5x + 4)
21x + 12 = 20x + 16
=>21x – 20x = 16 – 12
x = 4
B’s age = 7x
= 7 x 4
= 28 years (b)

Question 15.
Solution:
Perimeter of an isosceles triangle = 16 cm
and base = 6 cm
Let each equal side = x cm
x + x + 6 = 16
=> 2x = 16 – 6 = 10
=> x = \(\\ \frac { 10 }{ 2 } \) = 5
Each equal side = 5 cm (b)

Question 16.
Solution:
Let first number = x
Then second number = x + 1
and third number = x+ 2
x + x + 1 + x + 2 = 51
=> 3x + 3 = 51
=> 3x = 51 – 3 = 48
=> x = \(\\ \frac { 48 }{ 3 } \) = 16
Middle number = x + 1 = 16 + 1 = 17 (b)

Question 17.
Solution:
Let first number = x
Then second number = x + 15
x + x + 15 = 95
=> 2x = 95 – 15 = 80
=> x= \(\\ \frac { 80 }{ 2 } \) = 40
=> Smaller number = 40

Question 18.
Solution:
Ratio in boys and girls in a class = 7:5
Let no. of boys = 7x
Then no. of girls = 5x
7x – 5x = 8
=> 2x = 8
x = \(\\ \frac { 8 }{ 2 } \) = 4
Total strength = 7x + 5x = 12x
= 12 x 4
= 48 (c)

Hope given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7E

RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7E

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7E.

Other Exercises

OBJECTIVE QUESTIONS.
Tick the correct answer in each of the following :

Question 1.
Solution:
7a2 – 63b2 = 7 (a2 – 9b2)
= 7 {(a)2 – (3b)2}
= 7 (a – 3b) (a + 36) (d)

Question 2.
Solution:
2x – 32x3 = 2x (1 – 16x2)
= 2x {(1)2 – (4x)2}
= 2x (1 – 4x) (1 + 4x) (d)

Question 3.
Solution:
x3 – 144x
= x (x2 – 144)
= x {(x)2 – (12)2}
= x (x – 12) (x + 12) (c)

Question 4.
Solution:
2 – 50x2 = 2 (1 – 25x2)
= 2 {(1)2 – (5x)2}
= 2 (1 – 5x) (1 + 5x) (d)

Question 5.
Solution:
a2 + bc + ab + ac
= a2 + ab + ac + bc
= a (a + b) + c (a + b)
= (a + b) (a + c)   (a)

Question 6.
Solution:
pq2 + q (p – 1) – 1
= pq2 + pq – q – 1
= pq (q + 1) – 1 (q + 1)
= (q + 1) (pq – 1)
= (pq – 1) (q + 1) (d)

Question 7.
Solution:
ab – mn + an – bm
= ab + an – bm – mn
= a (b + n) – m (b + n)
= (b + n) (a – m) (b)

Question 8.
Solution:
ab – a – b + 1
= a (b – 1) – 1 (b – 1)
= b – 1) (a – 1) (a)

Question 9.
Solution:
x2 – xz + xy – yz
= x (x – z) + y (x – z)
= (x – z) (x + y) (c)

Question 10.
Solution:
12m2 – 21
= 3 (4m2 – 9)
= 3 {(2m)2 – (3)2}
= 3 (2m – 3) (2m + 3) (c)

Question 11.
Solution:
x3 – x
= x (x2 – 1)
= x (x – 1) (x + 1) (d)

Question 12.
Solution:
1 – 2ab – (a2 + b2)
= 1 – 2ab – a2 – b2
= 1 – (a2 + b2 + 2ab)
= 1 – (a + b)2
= (1 + a + b) (1 – a – b) (c)

Question 13.
Solution:
x2 + 6x + 8
= x2 + 4x + 2x + 8
{8 = 4 x 2, 6 = 4 + 2}
= x (x + 4) + 2 (x + 4)
= (x + 4) (x + 2) (c)

Question 14.
Solution:
x2 + 4x – 21
= x2 + 7x – 3x – 21
{ – 21 = + 7 x ( – 3), 4 = 7 – 3}
= x (x + 7) – 3 (x + 7)
= (x + 7) (x – 3) (b)

Question 15.
Solution:
y2 + 2y – 3
{ – 3 = 3 x ( – 1), 2 = 3 – 1}
= y2 + 3y – y – 3
= y (y + 3) – 1 (y + 3)
= (y + 3) (y – 1) (a)

Question 16.
Solution:
40 + 3x – x2
= 40 + 8x – 5x -x2
{40 = 8 x ( – 5), 3 = 8 – 5}
= 8(5 + x) – x(5 + x)
= (5 + x)(8 – x) (c)

Question 17.
Solution:
2x2 + 5x + 3
= 2x2 + 2x + 3x + 3
{2 x 3 = 6, 6 = 2 x 3, 5 = 2 + 3}
= 2x(x + 1) + 3(x + 1)
= (x + 1)(2x + 3) (b)

Question 18.
Solution:
6a2 – 13a + 6
= 6a2 – 9a – 4a + 6
{6 x 6 = 36, 36 = ( – 9)x( – 4), – 13 = – 9 – 4}
= 3a (2a – 3) – 2 (2a – 3)
= (2a – 3) (3a – 2) (c)

Question 19.
Solution:
4z2 – 8z + 3
= 4z2 – 6z – 2z + 3
{4 x 3 = 12,12 = ( – 6)x( – 2), – 8 = – 6 – 2}
= 2z (2z – 3) – 1 (2z – 3)
= (2z – 3) (2z – 1) (a)

Question 20.
Solution:
3 + 23y – 8y2
= 3 + 24y – y – 8y2
{3 x ( – 8) = – 24, – 24 = 24 x ( – 1), 23 = 24 – 1}
= 3(1 + 8y) – y(1 + 8y)
= (1 + 8y) (3 – y) (b)

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7E are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C

RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10C.

Other Exercises

Question 1.
Solution:
List price of refrigerator = Rs. 14650
Sales tax = 6%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 1.1

Question 2.
Solution:
(i) Lost of tie = Rs. 250
ST = 6%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 2.1
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 2.2

Question 3.
Solution:
Price of watch including VAT = Rs. 1980
Rate of VAT = 10%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 3.1

Question 4.
Solution:
Price of shirt including VAT = Rs. 133750
Rate of VAT = 7%
∴ Original price of the shirt
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 4.1

Question 5.
Solution:
Sale price of 10 g gold including VAT = Rs. 15756
Rate of VAT = 1%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 5.1

Question 6.
Solution:
Sale price of computer including VAT = Rs. 37960
Rate of VAT = 4%
∴ Original price of computer
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 6.1

Question 7.
Solution:
Sale price of car parts including VAT = Rs. 20776
Rate of VAT = 12%
∴ Original price of car parts
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 7.1

Question 8.
Solution:
Sale price of TV set including VAT = Rs. 27000
Rate of VAT = 8%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 8.1

Question 9.
Solution:
Sale price of shoes including VAT = Rs. 882
Original price = Rs 840
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 9.1

Question 10.
Solution:
Sale price of VCR including VAT = Rs. 19980
Original price = Rs. 18500
∴Amount of VAT
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 10.1

Question 11.
Solution:
Sale price of car including VAT = Rs. 382500
Basic price of the car = Rs. 340000
Amount of VAT = Rs. 382500 – 340000
= Rs. 42500
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10C 11.1

Hope given RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10C are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A

RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 9 Percentage Ex 9A.

Other Exercises

Question 1.
Solution:
(i) 48% = \(\\ \frac { 48 }{ 100 } \) = \(\\ \frac { 12 }{ 25 } \)
(ii) 220% = \(\\ \frac { 220 }{ 100 } \) = \(\\ \frac { 11 }{ 5 } \)
(iii) 2.5% = \(\\ \frac { 2.5 }{ 100 } \) = \(\\ \frac { 25 }{ 10X100 } \) = \(\\ \frac { 1 }{ 40 } \)

Question 2.
Solution:
(i) 6% = \(\\ \frac { 6 }{ 100 } \) = 0.06
(ii) 72% = \(\\ \frac { 72 }{ 100 } \) = 0.72
(iii) 125% = \(\\ \frac { 125 }{ 100 } \) = 1.25

Question 3.
Solution:
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 3.1

You can also Download NCERT Solutions for Class 8 Maths to help you to revise complete Syllabus and score more marks in your examinations.

Question 4.
Solution:
Ratio is 4:5 or \(\\ \frac { 4 }{ 5 } \)
\(\\ \frac { 4 }{ 5 } \)
= \(\\ \frac { 4X20 }{ 5X20 } \)
= \(\\ \frac { 80 }{ 100 } \)
= 80%

Question 5.
Solution:
125% = \(\\ \frac { 125 }{ 100 } \)
= \(\\ \frac { 5 }{ 4 } \) (dividing by 25)
Ratio = 5:4

Question 6.
Solution:
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 6.1
We see that 15% or \(\\ \frac { 3 }{ 20 } \) is the largest

Question 7.
Solution:
(i) Let x% of 150
= 96
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 7.1
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 7.2

Question 8.
Solution:
\(6\frac { 2 }{ 3 } %\) of Rs 3600
= Rs \(\\ \frac { 9 }{ 2 } \) x \(\\ \frac { 3600 }{ 100 } \)
= Rs 162

Question 9.
Solution:
16% of a number = 72
Number = \(\\ \frac { 72 }{ 16 }\)%
= \(\\ \frac { 72X100 }{ 16 } \)
= 450

Question 10.
Solution:
Let monthly income = Rs x
Savings = 18%
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 10.1

Question 11.
Solution:
Let total games = x
Then game which a team wins = 73
and it is 35% of total games
35% of x = 7
=> x = \(\\ \frac { 7X100 }{ 35 } \) = 20
Number of total games = 20

Question 12.
Solution:
Let salary = Rs x
Increment = 20%
Total salary = x + 20% of x
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 12.1

Question 13.
Solution:
No. of days, Sonal attended = 204 days
and her attendance = 85% of total days
85% of total days = 204
Total number of days = \(\\ \frac { 204X100 }{ 85 } \)
= 240 days

Question 14.
Solution:
Let B’s income = Rs. 100
Then A’s income = 20% less than B’s
= 100 – 20 = Rs. 80
Difference = 100 – 80 = 20
B’s more income that A’s = \(\\ \frac { 20 }{ 80 } \) x 100
= 25%

Question 15.
Solution:
Increase in price of petrol = 10%
Let first price = Rs. 100 p.l.
Increased price = 100 + 10 = Rs. 110
Now reduction in consumption = 110 – 100= 10
Percentage reduced consumption
= \(\\ \frac { 10X100 }{ 110 } \)
= \(\\ \frac { 100 }{ 11 } \)
= \(9\frac { 1 }{ 11 } %\)

Question 16.
Solution:
Present population of a town = 54000
Rate of increase = 8% annually
Population a year ago = \(\\ \frac { 54000X100 }{ 100+8 } \)
= \(\\ \frac { 54000X100 }{ 108 } \)
= 50000

Question 17.
Solution:
Depreciation in the value of machine = 20%
Present value = Rs. 160000
Then value of machine one year ago
= \(\\ \frac { 160000X100 }{ 100-20 } \)
= \(\\ \frac { 160000X100 }{ 80 } \)
= Rs 200000

Question 18.
Solution:
In an alloy,
Copper = 40%
Nickel = 32%
Zinc = 100 – (40% + 32%)
= 100 – 72 = 28%
Then mass of zinc in one kg of alloy
= 1 kg X 28%
= \(\\ \frac { 1000X28 }{ 100 } \) g
= 280 gm

Question 19.
Solution:
In balance diet,
Protein = 12%
Fats = 25%
Carbohydrates = 63%
Total number of dories = 2600
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 19.1

Question 20.
Solution:
In gunpowder,
Nitre = 75%
Sulphur = 10%
(i) Amount of gunpowder if nitre is 9 kg
= \(\\ \frac { 100 }{ 75 } \) x 9
= 12 kg
(ii) Amount of gunpowder if sulphur is 2.5kg 100
= \(\\ \frac { 100 }{ 10 } \) x 2.5
= 25 kg

Question 21.
Solution:
Let C get = Rs. x
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 21.1
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 21.2

Question 22.
Solution:
24-carat gold is 100% pure
22 parts out of 24 part in 22-carat gold
RS Aggarwal Class 8 Solutions Chapter 9 Percentage Ex 9A 22.1

Question 23.
Solution:
Let present salary = Rs. 100
Increase = 25%
Increased salary = Rs. 100 + 25
= Rs. 125
To receive the original salary, amount to be decreased = Rs. 125 – 100 = Rs. 25
∴ % decrease = \(\\ \frac { 25X100 }{ 125 } \) = 20%

 

Hope given RS Aggarwal Solutions Class 8 Chapter 9 Percentage Ex 9A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

NCERT Solutions for Class 9 Science Chapter 14 Natural Resources

NCERT Solutions for Class 9 Science Chapter 14 Natural Resources

These Solutions are part of NCERT Solutions for Class 9 Science. Here we have given NCERT Solutions for Class 9 Science Chapter 14 Natural Resources. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 9 Science (Biology) Chapter 14 – Natural Resources solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 14 – Natural Resources Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

More Resources

NCERT TEXT BOOK QUESTIONS

IN TEXT QUESTIONS

Question 1.
How is our atmosphere different from the atmosphere of Venus and Mars ?
Answer:

Atmosphere over Earth Atmosphere over Venus and Mars
1. Nitrogen and Oxygen. It contains nitrogen and oxygen. The two are absent
2. Carbon Dioxide. Carbon dioxide content is little (0-03%). Carbon dioxide is the major component of atmosphere forming 95-97%.
3. Water Vapours. The atmosphere contains water vapours which form a component of water cycle. The atmosphere does not contain water vapours. Living beings, being absent, have no role in
4. living Beings. Composition of atmosphere is maintained by living beings. determining composition of atmosphere.

Question 2.
How does atmosphere act as a blanket ?
Answer:
Atmosphere or air is a bad conductor of heat. It, therefore, functions as a blanket.

  1. It does not allow sudden increase in temperature during the daylight hours when sun shines overhead.
  2. There is no sudden cooling during night. Atmosphere slows down the escape of heat to the outer space from the area of earth under darkness.
  3. Atmosphere maintains the average temperature of the earth fairly steady not only during the day but throughout the year.

Question 3.
What causes winds ?
Answer:
Winds are basically caused by heating of air in certain parts. The hot air rises upwards. This creates an area of low pressure. Cooler air from adjacent higher pressure areas passes into this area. This creates wind. The factors which control movement of winds in different directions in different parts of the earth are

  1. Uneven heating of land in different parts of earth
  2. Differences in heating and cooling of land and water.
  3. Barrier of mountains
  4. Rotation of earth.

Question 4.
How are clouds formed ? (CCE 2011)
Answer:
Clouds are wet air masses that float in the direction of prevailing wind. They develop when water vapours are formed in large number. There is evaporation from the surface of water bodies and wet areas due to their heating during the daytime. Plants also give out water vapours in transpiration while animals do so in exhaled air and perspiration. Air also heats up during daytime. The hot air along with water vapour rises up. At a height, air expands and becomes cool. Cooling causes the water vapours to condense. Suspended particles of dust and other materials function as nuclei around which water vapours condense. When a large wet air mass collects, cloud is formed.

Question 5.
Answer:
List any three human activities that you think would lead to air pollution.

  1. Burning of fossil fuels in industries, vehicles and thermal plants.
  2. Processing industries like textiles, asbestos, flour mills.
  3. Stone crushing.

Question 6.
Why do organisms need water ? (CCE 2011)
Answer:
Organisms need water due to following reasons :

  1. Component of Living Matter: Water is a major component (60-90%) of living matter.
  2. Solvent: Water is a general solvent for chemicals found in the living world.
  3. Reaction Medium: All biochemical reactions occur in the medium of water.
  4. Transport: Substances are transported in the body of a living organisms only in the dissolved state.
  5. Turgidity: Cells, cell organelles, tissues and other structures maintain their shape only when they contain sufficient water to make them turgid.
  6. Temperature Buffer: Water protects the body from sudden changes of temperature.
  7. Wastes: It helps in separation and elimination of metabolic wastes.

Question 7.
What is the major source of fresh water in the city/town/village where you live ?
Answer:
Ground water which is pumped out by tube wells (In some it is local reservoir, canal or river.

Question 8.
Do you know of any activity which may be polluting this water source ?
Answer:
Dumping of industrial wastes from where pollutants seep into soil to reach ground water (sewage and industrial effluents in case of canal or river water).

Question 9.
How is soil formed ? What is the function of humps in soil ? (CCE 2011, 2012)
Answer:
Soil is formed through two processes of weathering and humification.
Weathering
It is pulverisation of rocks or breaking of rocks into fine particles. There are three types of weathering — physical (atmospheric changes and mechanical forces), chemical and biological. Sun, water, wind and living organisms perform them.

  1. Sun: It causes expansion of rocks by heating. Cooling causes their contraction. Different parts of rocks expand and contract at different rates. Uneven expansion and contraction produces cracks leading to fragmentation of rocks.
  2. Water:
    1. Wetting and Drying: Certain rock components can pick up and lose moisture. They undergo swelling and contraction resulting in fragmentation of rocks.
    2. Frost Action: Water seeping in cracks would swell up and exert a great pressure if it freezes due to low temperature. The rock would undergo fragmentation.
    3. Abrasion: Running water carrying rock fragments would break and grind rocks occurring in the pathway. Rain and hail also cause rock breaking.
  3. Wind: Dust and fine sand carried by wind cause abrasion of the rock surface when wind strikes the same.
  4. Living Organisms: Lichens secrete chemicals to dissolve minerals from the rock surface. This produces crevices where dust collects. Mosses grow there. They cause deepening of crevices and development of small cracks. Roots of short lived plants widen these cracks. Roots of larger plants cause fragmentation of rocks by entering the cracks and growing in size.

Humification
Partially decomposed organic matter or humus mixes with weathered rock particles to form soil. Humus helps in formation of soil crumbs which are essential for maintaining proper hydration and aeration of soil.

Question 10.
What are the methods of preventing or reducing soil erosion ?
Answer:
Soil Erosion: It is removal of top soil by agency of wind or water. Wind and water are also the agencies which cause weathering of rocks and carrying the fine particles to other places for the formation of soil. Removal of top soil by water or wind leaves the underneath subsoil and rocky base exposed. Very little plant growth occurs there.
Factors Promoting Soil Erosion

  1. It destroys herbs, grasses and seedlings. The soil is exposed. Trampling by animals causes compaction of soil which reduces its porosity and percolation.
  2. Removal of litter or scraping of forest floor leaves the ground bare for action of agencies causing erosion.
  3. There is decreased absorption of water as the latter does not stay for long on the slope. Run off water passing along the slope gathers speed and develops high cutting and carrying capacity.
  4. Felling of Trees: Felling of trees in excess of regeneration capacity of a forest causes deforestation. It also leaves large area bare for action of wind and water. Deforestation or clearing of forest not only destroys biodiversity but also leads to large scale soil erosion.
  5. Clean Tilling: Clean tilling of crop fields exposes the soil to erosion.
  6. Heavy Rain and Strong Winds. Uncovered soil is eroded quickly by heavy rain and strong wind.

Question 11.
What are two different states in which water is found during the water cycle ?
Answer:
Liquid and vapour, occasionally solid (snow) as well.

Question 12.
Name two biologically important compounds that contain both oxygen and nitrogen. . (CCE 2014)
Answer:
Proteins and nucleic acids.

Question 13.
List any three human activities which would lead to an increase in the carbon dioxide content of air.
Answer:

  1. Increasing combustion of fossil fuels (coal, petroleum, natural gas) in homes, industries, transportation and power projects.
  2. Increasing use of wood for cooking and heating.
  3. Deforestation leading to reduced utilisation of carbon dioxide in photosynthesis.

Question 14.
What is green house effect ? (CCE 2011, 2012, 2013)
Answer:
Green house effect is keeping an area warm by allowing the solar radiations to pass in but preventing long waves to escape due to presence of radiatively active gases and glass panes.

Question 15.
What are the two forms of oxygen found in the atmosphere ? (CCE 2011, 2014)
Answer:

  1. Diatomic oxygen, O2
  2. Triatomic oxygen or ozone, O3.

NCERT CHAPTER END EXERCISES

Question 1.
Why is the atmosphere essential for life ? (CCE 2012)
Answer:

  1. Oxygen: Atmosphere contains oxygen which is essential for combustion and respiration of most organisms.
  2. Carbon Dioxide: Atmosphere provides carbon dioxide for photosynthesis of plants.
  3. Protection: Atmosphere filters out lethal cosmic rays and high energy ultraviolet rays.
  4. Temperature Buffer: Atmosphere does not allow daytime temperature to rise abnormally nor does it allow night time temperature to fall down drastically. This provides favourable temperature for the living organisms.
  5. Other Functions: Air currents help in dispersal of spores and other dissemules. Water cycle operates through atmosphere and produces rain to replenish fresh water over land.

Question 2.
Why is water essential for life ?
Answer:

Organisms need water due to following reasons :

  1. Component of Living Matter: Water is a major component (60-90%) of living matter.
  2. Solvent: Water is a general solvent for chemicals found in the living world.
  3. Reaction Medium: All biochemical reactions occur in the medium of water.
  4. Transport: Substances are transported in the body of a living organisms only in the dissolved state.
  5. Turgidity: Cells, cell organelles, tissues and other structures maintain their shape only when they contain sufficient water to make them turgid.
  6. Temperature Buffer: Water protects the body from sudden changes of temperature.
  7. Wastes: It helps in separation and elimination of metabolic wastes.

Question 3.
How are living organisms dependent on soil ? Are organisms that live in water totally independent of soil as a resource ? ‘
Answer:
All terrestrial organisms depend upon plants for their food and its contained energy. Plants are dependent on soil for anchorage, absorption of water and nutrients. Without them plants cannot manufacture food. So, all living terrestrial organisms depend upon soil.
Aquatic organisms are apparently not connected with soil. However, aquatic autotrophs require inorganic nutrients for manufacture of food. Nutrients reach water bodies only when rain water passes over and inside the soil. Therefore, aquatic organisms are not totally independent of soil as a resource.

Question 4.
You have seen weather reports on television and in newspapers. How do you think we are able to predict the weather ?
Answer:
Weather reports depict areas of low and high pressure, prevailing direction of wind, dryness or wetness of air masses, clouds and their progress, presence and progress of any cyclone, etc. Predictions are then made whether a particular area will or will not receive rain, have calm weather or high speed wind and dust storm.

Question 5.
We know that many human activities lead to increasing levels of pollution of air, water bodies and soil. Do you think that isolating these activities to specific and limited areas would help in reducing pollution ?
Answer:
Restricting pollution related activities to specific and limited areas will not reduce pollution in those areas. The same may rather increase. There are two benefits of such a practice

  1. Joint pollution treatment plants can be installed
  2. The residential and commercial areas away from such pollution generating regions will be comparatively free from pollution.

Question 6.
Write a note on how forests influence the quality of our air, soil and water resources. (CCE 2012)
Answer:
Air Resources:

  1. Oxygen-Carbon Dioxide Balance. Forests maintain the optimum level of oxygen and carbon dioxide. They function as sink for excess carbon dioxide being produced due to excessive combustion. Forests also release a lot of more oxygen to compensate for excess being consumed elsewhere in respiration and combustion.
  2. Control of Air Pollution. Both suspended particles and gaseous pollutants are picked up by forest plants.

Soil Resources: Roots of the forest plants hold the soil firmly. Forest cover protects the soil from direct pounding by rain drops. Forest soil is also sufficiently porous to reduce run off and increase infiltration of rain water. All the three factors prevent soil erosion.
Water Resources:

  1. Rainfall. Forests help in increasing the amount and periodicity of rain fall.
  2. Forest trees retain a lot of water at their bases. Percolation of water into interior of earth produces springs which form rivulets with perennial flow of water.

SELECTION TYPE QUESTIONS

Alternate Response Type Questions :
(True/False, Right/Wrong, Yes/No)

Question 1.
As 75% of earth’s surface is covered with water, the outer crust of earth is called hydrosphere.
Question 2.
Combustion consumes oxygen and liberates carbon dioxide.
Question 3.
Winds develop due to uneven heating of earth.
Question 4.
Carbon monoxide and carbon dioxide produce acid rain.
Question 5.
The amount of rainfall directly influences the abundance and diversity of like forms.
Question 6.
Soil has no role in supplying nutrients to aquatic biota
Question 7.
Pesticides and fertilizers are harmful to soil as they kill the microorganisms involved in recycling of nutrients.
Question 8.
Green house gases are the ones which allow the heat emitted by earth to pass out.

Matching Type Questions :

Question 9.
Match the articles of the column I and column II (Single Matching)

Column I Column II
(a) Chlorofluorocarbons (i) Bacteria
(b) Carbon dioxide (ii) Fossil fuels
(c) Nitrogen fixation (iii) Ozone depletion
(d) Nitrogen and sulphur oxides (iv) Green house gas

Question 10.
Match the contents of columns I, II and III (Double matching)

Column I Column II Column III
(a) Food (i) Sun, water and wind (p) Soil
(b) Abiotic (ii) Resources (q) Living organisms
(c) Carbon dioxide (iii) Air and Water (r) Energy
(d) Paedogenesis (iv) Photosynthesis (s) Shells

Question 11.
Match the pollutants with the type of pollution—air (A), water (W) and soil (S) (Key or check List matching)

Pollutants Pollution

(a) Hydrogen sulphide

(b) SPM

(c) Algal bloom

(d) Raw manure

Question 12.
Match Stimulus with Appropriate Response.

Conservation Practice Soil-A Water-B Air-C

(i) Pollution under control certificate

(ii) Vegetation cover

(iii) Terracing

(iv) Sewage treatment

Fill In the Blanks

Question 13. The earth ……………… and energy from sun are necessary to meet basic requirements of life forms.
Question 14. Surface temperature of moon varies from …………….. to 110°C.
Question 15. …………….. makes soil porous and allows water and air to penetrate deep underground.
Question 16. Carbon occurs in elemental form in ………………. and graphite.
Question 17. CFCs are carbon compounds having both ……………. and chlorine.

Answers:

SOME TYPICAL QUESTIONS

Question 1.
Name the different physical divisions of biosphere.
Answer:
Three – atmosphere (air), hydrosphere (water) and lithosphere (land).

Question 2.
Differentiate between renewable and non-renewable resources.
Answer:

Renewable Resources Non-renewable Resources
1.     Replacement. The resources are replenished within reasonable time.

2.     Use. The resources can be used forever provided the use is limited.

3.     Life. They are both abiotic and biotic.

4.     Availability. It can be increased only by enhancing replenishment.

 Replenishment is absent.

The resources will ultimately dwindle, and get exhausted.

They are abiotic.

This is not possible. Increased exploitation will result in quick exhaustion.

Question 3.
Name the parts of India falling under wet zone.
Answer:
Wet zone (rainfall over 200 cm/yr) comprises Western Ghats, Andaman-Nicobar and North East India.

Question 4.
Name two diseases caused by
(a) Infectious agents in polluted water,
(b) Toxic chemicals in polluted water.
Answer:
(a) Infectious Agents. Jaundice, dysentery.
(b) Toxic Agents. Minamata (mercury), itai-itai (cadmium).

Question 5.
What is ozone hole ? Where is it found ? What is its effect ? (CCE 2011)
Answer:
Ozone hole is thinning of ozone in the stratosphere where it is normally present in high concentration as ozone layer. Ozone hole is formed during spring time over antarctica and to a small extent over north pole. Thinning of ozone layer or ozone hole increases the passage of harmful ultravoilet rays to earth. This has increased incidence of skin cancers, defective eye sight, reduced immunity, increased number of mutations and reduced crop yield in southern countries of southern hemisphere.

NCERT Solutions for Class 9 Science Chapter 14 Natural Resources

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RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B

RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 8 Linear Equations Ex 8B.

Other Exercises

Question 1.
Solution:
The numbers be 8x and 3x (∵ Ratio is 8 : 3)
Sum = 143
According to the condition
8x + 3x = 143
=> 11x = 143
=> x = \(\\ \frac { 143 }{ 11 } \) = 13
First number = 8x = 8 x 13 = 104
and second number = 3x = 3 x 13 = 39 Ans.

Question 2.
Solution:
Let the number = x
According to the condition,
x – \(\frac { 2 }{ 3 } x\) = 20
=> \(\\ \frac { 3x-2x }{ 3 } \) = 20
=> \(\\ \frac { x }{ 3 } \) = 20
=> x = 20 x 3 = 60
Hence original number = 60 Ans.

You can also Download NCERT Solutions for Class 8 Maths to help you to revise complete Syllabus and score more marks in your examinations.

Question 3.
Solution:
Let the number = x
According to the condition
\(\frac { 4 }{ 5 } x-\frac { 2 }{ 3 } x\) = 10
=> \(\\ \frac { 12x-10x }{ 15 } \) = 10
=> \(\\ \frac { 2x }{ 15 } \) = 10
=> 2x = 10 x 15
x = \(\\ \frac { 10 x 15 }{ 2 } \) = 75
Hence number = 75 Ans.

Question 4.
Solution:
Let first part = x
then second part = 24 – x (∵ Sum = 24)
According to the condition,
7x + 5 (24 – x) = 146
=> 7x + 120 – 5x = 146
=> 2x = 146 – 120 = 26
=> x = \(\\ \frac { 26 }{ 2 } \) = 13
First part = 13
and second part = 24 – 13 = 11 Ans.

Question 5.
Solution:
Let number = x
According to the condition
\(\frac { x }{ 5 } +5=\frac { x }{ 4 } -5\)
=> \(\\ \frac { x }{ 5 } \) – \(\\ \frac { x }{ 4 } \)= – 5 – 5
=> \(\\ \frac { 4x-5x }{ 20 } \) = – 10
=> \(\\ \frac { -x }{ 20 } \) = – 10 = \(\\ \frac { x }{ 20 } \) = 10
=> x = 10 × 20 = 200
Hence number = 200 Ans.

Question 6.
Solution:
Ratio between three numbers = 4:5:6
Let the largest number = 6x
smallest number = 4x
and third number = 5x
According to the condition,
6x + 4x = 5x + 55
=> 10x = 5x + 55
=> 10x – 5x = 55
=> 5x = 55
=> x = \(\\ \frac { 55 }{ 5 } \) = 11
Numbers will be 4x = 4 × 11 = 44,
5x = 5 × 31 = 55 and 6x = 6 × 11 = 66
Hence numbers are 44, 55, 66 Ans.

Question 7.
Solution:
Let the number = x
According to the condition,
4x + 10 = 5x – 5
=> 4x – 5x = – 5 – 10
=> – x = – 15
x = 15
Hence number = 15 Ans.

Question 8.
Solution:
Ratio between two numbers 3: 5
Let first number = 3x
and second number = 5x
Now according to the condition.
3x + 10 : 5x + 10 = 5 : 7
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 8.1

Question 9.
Solution:
Let first odd number = 2x + 1
second number 2x + 3
and third number = 2 + 5
According to the condition.
2x + 1 + 2x + 3 + 2x + 5 = 147
=> 6x + 9 = 147
=> 6x = 147 – 9 = 138
=> x = \(\\ \frac { 138 }{ 6 } \) = 23
Hence first odd number 2x + 1
= 23 x 2 + 1 = 46 + 1 = 47
Second number 47 + 2 = 49
and third number = 49 + 2 = 51 Ans.

Question 10.
Solution:
Let first even number = 2x
second number = 2x + 2
and third number = 2x + 4
According to the condition,
2x + 2x + 2 + 2x + 4 = 234
=> 6x + 6 = 234
=> 6x = 234 – 6
=> 6x = 228
=> x = \(\\ \frac { 228 }{ 6 } \) = 38
First even number = 2x = 38 x 2 = 76
second number = 76 + 2 = 78
and third number 78 + 2 = 80 Ans.

Question 11.
Solution:
The sum of two digits = 12
Let the ones digit of the number = x
then tens digit = 12 – x
and number = x + 10 (12 – x)
= x + 120 – 10x = 120 – 9x
Reversing the digits,
ones digit of new number = 12 – x
and tens digit = x
the number = 12 – x + 10x = 12 + 9x
According to the condition,
12 + 9x = 120 – 9x + 54
=> 9x + 9x
=> 120 + 54 – 12
=> 174 – 12
=> 18x = 162
=> x = \(\\ \frac { 162 }{ 18 } \) = 9
Original number = 120 – 9x
= 120 – 9 x 9
= 120 – 81
= 39
Hence number 39 Ans.
Check :Original number= 39
Sum of digits = 3 + 9 = 12
Now reversing its digit the new number
will be = 93
and 93 – 39 = 54 which is given.

Question 12.
Solution:
Let units digit of the number = x
then tens digit = 3x
and number = x + 10
3x = x + 30x = 31x
on reversing the digits.
units digit = 3x
and tens digit = x
then number 3x + 10x = 13x
According to the condition,
31x – 36 = 13x
=> 31x – 13x = 36
=> 18 x 36
=> x = \(\\ \frac { 36 }{ 18 } \) = 2
The original number = 31x = 31 x 2 = 62
Hence number = 62 Ans.
Check : Number = 62
tens digit = 2 x 3 = 6
On reversing the digit, the new number will be = 26
62 – 26 = 36 which is given.

Question 13.
Solution:
Let numerator of a rational number = x
Then its denominator = x + 7
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 13.1

Question 14.
Solution:
Let numerator of a fraction = x
The denominator = 2x – 2
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 14.1

Question 15.
Solution:
Let breadth of the rectangle = x cm
then length = x + 7
Area = l × b = (x + 7) × x
In second case,
Length of the new rectangle = x + 7 – 4
= x + 3 cm
and breadth = x + 3
Area = (x + 3)(x + 3)
According to the condition,
(x + 3)(x + 3) = x(x + 7)
x2 + 3x + 3x + 9 = x2 + 7x
=> x2 + 6x – 7x – x2 = – 9
=> x = – 9
=> x = 9
Length of the original rectangle
= > x + 7 = 9 + 7 = 16 cm
and breadth = x = 9 cm. Ans.

Question 16.
Solution:
Let length of rectangle = x m
then width = \(\frac { 2 }{ 3 } x\)m
Perimeter = 2 (l + b) m
=> \(2\left( x+\frac { 2 }{ 3 } x \right) =180\)
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 16.1

Question 17.
Solution:
Let the length of the base of the triangle = x cm
then altitude = \(\frac { 5 }{ 3 } x\)cm
Area = \(\\ \frac { 1 }{ 2 } \) base x altitude
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 17.1
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 17.2

Question 18.
Solution:
Let ∠A, ∠B and ∠C are the three angles of a triangle and
Let ∠A + ∠B = ∠C
But ∠A + ∠B + ∠C = 180°
∠C + ∠C = 180°
=> 2∠C = 180°
∠C = 90°
and ∠A + ∠B = 90°
Let ∠A = 4x and ∠B = 5x
4x + 5x = 90°
=> 9x = 90°
x = \(\frac { { 90 }^{ o } }{ 9 } \) = 10°
∠A = 4x = 4 x 10° = 40°
and ∠B = 5x = 5 x 10° = 50°
and ∠C = 90° Ans.

Question 19.
Solution:
Time taken downstream = 9 hour
and time taken upstream = 10 hour
Speed of stream = 1 km/h
Let the speed of a steamer in still water
= x km/h.
Distance downstream = 9(x + 1) km.
and upstream = 10 (x – 1) km.
According to the condition,
10(x – 1) = 9(x + 1)
l0x – 10 = 9x + 9
=> 10x – 9x = 9 + 10
=> x = 19
Hence speed of steamer in still water = 19 km/h
and distance = 9(x + 1)
= 9(19 + 1)
= 9 x 20km
= 180km. Ans.

Question 20.
Solution:
The distance between two stations = 300 km
Let the speed of first motorcyclists = x km/h
and speed of second motorcyclists = (x + 7)km/h
Distance covered by the first = 2x km
and distance covered by the second = 2 (x + 7) km
= 2x + 14 km
Distance uncovered by them = 300 – (2x + 2x + 14)kms.
According to the condition,
300 – (4x + 4) = 34
=> 300 – 4x – 14 = 34
=> 300 – 14 – 34 = 4x
=> 4x = 300 – 48
=> 4x = 252
=> x = \(\\ \frac { 252 }{ 4 } \) = 63
Speed of the first motorcyclists = 63km/h
and speed of second = 63 + 7 = 70 km/h
Check. Distance covered by both of them
= 2 x 63 + 2 x 70 = 126 + 140 = 266
Total distance = 300 km.
Distance between them = 300 – 266 = 34 km.
which is given.

Question 21.
Solution:
Sum of three numbers = 150
Let first number = x
then second number = \(\frac { 5 }{ 6 } x\)
and third number = \(\\ \frac { 4 }{ 5 } \) of second
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 21.1

Question 22.
Solution:
Sum of two pans = 4500
Let first part = x
then second part = 4500 – x
According to the condition,
5% of x = 10% of (4500 – x)
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 22.1

Question 23.
Solution:
Let age of Rakhi = x years
then her mother’s age = 4x
After 5 years,
Rakhi’s age = x + 5
and her mother’s age = 4x + 5
According to the conditions,
4x + 5 = 3 (x + 5)
=> 4x + 5 = 3x + 15
=> 4x – 3x = 15 – 5
=> x = 10
Rakhi ‘s present age = 10 years
and her mother’s age = 4 x 10
= 40 years Ans.

Question 24.
Solution:
Let age of Monu = x year
His father’s age = x + 29
and his grandfather’s age = x + 29 + 26
= x + 55
and sum of their ages = 135 years
Now,
x + x + 29 + x + 55 = 135
=> 3x + 84 = 135
=> 3x = 135 – 84 = 51
=> x = \(\\ \frac { 51 }{ 3 } \) = 17 years
Monu’s age = 17 years
His father age = 17 + 29 = 46 years
and his grandfather’s age = 46 + 26 = 72 years

Question 25.
Solution:
Let age of grandson = x year
Then his age = 10x
But 10x = x + 54
=> 10x – x = 54
=> 9x = 54
=> x = \(\\ \frac { 54 }{ 9 } \) = 6
Grand’s son age = 6 years
and his age = 6 x 10 = 60 years

Question 26.
Solution:
Let age of elder cousin = x years
and age of younger = (x – 10) years.
15 years ago,
age of older, cousin = x – 15 years
and age of younger = x – 10 – 15
= (x – 25) years.
According to the condition,
x – 15 = 2(x – 25)
=> x – 15 = 2x – 50
=> 2x – x = 50 – 15
=> x = 35
Age of elder cousin = 35 years
and age of younger = 35 – 10
= 25 years Ans.

Question 27.
Solution:
Let number of deer = x
RS Aggarwal Class 8 Solutions Chapter 8 Linear Equations Ex 8B 27.1

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RS Aggarwal Class 8 Solutions Chapter 5 Playing with Numbers Ex 5C

RS Aggarwal Class 8 Solutions Chapter 5 Playing with Numbers Ex 5C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5C.

Other Exercises

Replace A, B, C by suitable numerals.

Question 1.
Solution:
Here A can be as 6 + 7 = 13
Now 1 + 5 + 8 = 14
∴C = 1, B = 4, A = 6

Question 2.
Solution:
Here A can be 7, as 6+7 = 13
1 + B + 9 = 10 + B
∴B can be 7
∴10 + 7 = 17
1 + C + 6 = 7 + C
∴C can be 4
∴1 + 4 + 6 = 11
and 1 + 4 + 3 = 8
∴A = 7, B = 7, C = 4

Question 3.
Solution:
Here A + A + A = A
∴A can = 5
∴5 + 5 + 5 = 15
∴B = 1
Hence A = 5, B = 1

Question 4.
Solution:
6 – A = 3
1 + 5 – A = 3
5 – A = 3
∴A = 5 – 3 = 2
Now 2 – B = 7
=>12 – B = 7
∴B = 5
Hence A = 2, B = 5

Question 5.
Solution:
– 5 – A = 9 =>A = 5 – 9 or 15 – 9
= 6
∴A = 6
Now B – 1 – 8 = 5 =>B – 9 = 5
=>B = 5 + 9 = 14
∴B = 4
Now C – 1 – 2 = 2 =>C – 3 = 2
C = 2 + 3 = 5
∴A = 6, B = 4, C = 5

Question 6.
Solution:
B x 3 = B
∴B can be 5 or 0
∴5 x 3 = 15 => B = 5 or 3 x 0 =0
If B = 0, then A can be 5
∴3 x 5 = 15
∴A = 5 and C = 1
Hence A = 5, B = 0, C = 1

Question 7.
Solution:
RS Aggarwal Class 8 Solutions Chapter 5 Playing with Numbers Ex 5C 7.1
∴AB = B
=>A = 1
and A² + B² – 1 + B² + C
∴B² +1 = C
∴B² in one digit
If B = 3
∴3² + 1 = 9 + 1 = 10 = C
∴C = 0
B x 1 + 1 = B + 1 = 3 + 1
Hence A = 1, B = 3, C = 0

Question 8.
Solution:
Here we see that 6 x 9 = 54
∴A – 4 = 3 => A = 3 + 4 = 7
and 6 x 6 = 36
3B = 36 => B = 6
and C = 6
Hence A = 7, B = 6, C = 6

Question 9.
Solution:
Product of two numbers = 1 -digit number
and sum = 2-digit numbers
Let first number = x
and second number = y
∴x X y = 1-digit number
x + y = 2-digit number
By hit and hail, we sec that
1 x 9 = 9 which is I-digit number
and 1 + 9 = 10 which is 2-digit number

Question 10.
Solution:
By hit and trail method, we see that
1 + 2 + 3 = 6 and 1 x 2 x 3 = 6
1, 2 and 3 are the required whole numbers
whose sum and product is same

Question 11.
Solution:
In the given square, we have to interest the numbers from 1 to 9, such that the sum in each raw, column on diagonal to be 15
So, we complete it as given here

RS Aggarwal Class 8 Solutions Chapter 5 Playing with Numbers Ex 5C 11.1

Question 12.
Solution:
We shall complete the triangle by intersecting the numbers from 1 to 6 without repetition so that the sum in each side be 12
RS Aggarwal Class 8 Solutions Chapter 5 Playing with Numbers Ex 5C 12.1

Question 13.
Solution:
The given numbers are
a, b (a + b), (a + 2b), (2a + 3b), (3a + 5b), (5a + 8b), (8a + 13b), (13a + 21b), and (21a + 34b)
Sum of there numbers = 11 (5a + 8b)
= 11 x 7th number
Now taking a = 8, b = 13, then the 10 number be 8, 13, 21, 34, 55, 89, 144, 233, 377, 610
Whose 7th number = 144
By adding these 10 numbers, we get the
sum
= 8+ 13 + 21 + 34 + 55 + 89 + 144 + 233 + 377 + 610 = 1584 and 11 x 7th number =11 x 144 , = 1584
Which is same in each case

Question 14.
Solution:
We see that in the magic box sum of 0 + 11 + 7 + 12 = 30
Now we shall complete this magic square, to get 30 as the sum in each row and column and also diagonal wise
RS Aggarwal Class 8 Solutions Chapter 5 Playing with Numbers Ex 5C 14.1

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RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2

RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2

Other Exercises

Question 1.
In the figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD. [NCERT]
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2 Q1.1
Solution:
In ||gm ABCD,
Base AB = 16 cm
and altitude AE = 8 cm
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2 Q1.2
∴ Area = Base x Altitude
= AB x AE
= 16 x 8 = 128 cm2
Now area of ||gm ABCD = 128 cm2
Altitude CF = 10 cm
∴ Base AD = \(\frac { Area }{ Altitude }\) = \(\frac { 128 }{ 10 }\) = 12.8cm

Question 2.
In Q. No. 1, if AD = 6 cm, CF = 10 cm, AE = 8 cm, find AB.
Solution:
Area of ||gm ABCD,
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2 Q2.1
= Base x Altitude
= AD x CF
= 6 x 10 = 60 cm2
Again area of ||gm ABCD = 60 cm2
Altitude AE = 8 cm
∴ Base AB =\(\frac { Area }{ Altitude }\) = \(\frac { 60 }{ 8 }\) = \(\frac { 15 }{ 2 }\) cm = 7.5 cm

Question 3.
Let ABCD be a parallelogram of area 124 cm2. If E and F are the mid-points of sides AB and CD respectively, then find the area of parallelogram AEFD.
Solution:
Area of ||gm ABCD = 124 cm2
E and F are the mid points of sides AB and CD respectively. E, F are joined.
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2 Q3.1
Draw DL ⊥ AB
Now area of ||gm ABCD = Base x Altitude
= AB x DL = 124 cm2
∵ E and F are mid points of sides AB and CD
∴ AEFD is a ||gm
Now area of ||gm AEFD = AE x DL
= \(\frac { 1 }{ 2 }\)AB x DL [∵ E is mid point of AB]
= \(\frac { 1 }{ 2 }\) x area of ||gm ABCD
= \(\frac { 1 }{ 2 }\) x 124 = 62 cm2

Question 4.
If ABCD is a parallelogram, then prove that ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = \(\frac { 1 }{ 2 }\)ar( ||gm ABCD).
Solution:
Given : In ||gm ABCD, BD and AC are joined
RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2 Q4.1
To prove : ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = \(\frac { 1 }{ 2 }\)ar(||gm ABCD)
Proof: ∵ Diagonals of a parallelogram bisect it into two triangles equal in area When BD is the diagonal, then
∴ ar(∆ABD) = ar(∆BCD) = \(\frac { 1 }{ 2 }\)ar(||gm ABCD) …(i)
Similarly, when AC is the diagonal, then
ar(∆ABC) = ar(∆ADC) = \(\frac { 1 }{ 2 }\)ar(||gm ABCD) …(ii)
From (i) and (ii),
ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = \(\frac { 1 }{ 2 }\) ar(||gm ABCD)

Hope given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2 are helpful to complete your math homework.

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