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

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

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 10D.

Other Exercises

Question 1.
Solution:
Answer = (c)
C.P. of toy Rs. = 75
S.P. = Rs. 100
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 1.1

Question 2.
Solution:
Answer = (b)
C.P. of bat = Rs. 120
S.P. = Rs. 105
Loss = Rs. 120 – Rs. 105 = Rs. 15
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 2.1

Question 3.
Solution:
Answer = (b)
S.P. of book = Rs. 100
gain = Rs. 20
C.P. = 100 – 20 = Rs. 80
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 3.1

Question 4.
Solution:
SP of an article = Rs. 48
Loss = 20%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 4.1

Question 5.
Solution:
First time gain = 10%
Let SP – Rs. 100
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 5.1

Question 6.
Solution:
Let no. of bananas bought = 6
Now C.P. of bananas at the sale of 3
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 6.1

Question 7.
Solution:
SP of 10 pens = CP of 12 pens
= Rs. 100 (suppose)
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 7.1

Question 8.
Solution:
Gain on 100 pencils = SP of 20 pencils
SP of 100 pencils gains = CP of 100 pencils
=> SP of 100 pencils – SP of 20 pencils = CP of 100 pencils
=> SP of 80 pencils – CP of 100 pencils = Rs 100 (suppose)
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 8.1

Question 9.
Solution:
SP of 5 toffees = Re. 1
SP of 2 toffees = Re. 1
Now CP of 1 toffee = Rs. \(\\ \frac { 1 }{ 5 } \)
and SP of 1 toffee = Rs. \(\\ \frac { 1 }{ 2 } \)
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 9.1

Question 10.
Solution:
CP of 5 oranges = Rs. 10
SP of 6 oranges = Rs. 15
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 10.1

Question 11.
Solution:
SP of a radio = Rs. 950
Loss = 5%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 11.1

Question 12.
Solution:
Let CP of an article = Rs. 100
SP = \(\\ \frac { 6 }{ 5 } \) of CP = \(\\ \frac { 6 }{ 5 } \) x 100 = Rs. 120
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 12.1

Question 13.
Solution:
SP of a chair = Rs. 720
Loss = 25%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 13.1

Question 14.
Solution:
Ratio in CP and SP = 20 : 21
Let CP = Rs. 20
and SP = Rs. 21
Gain = SP – CP = Rs. 21 – 20 = Re. 1
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 14.1

Question 15.
Solution:
SP of first chair = Rs. 500
Gain = 20%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 15.1
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 15.2

Question 16.
Solution:
Gain % SP of Rs. 625 = Loss on SP of Rs. 435
CP of an article = x
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 16.1

Question 17.
Solution:
CP of an article = Rs. 150
Overhead expenses = 10% of CP
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 17.1

Question 18.
Solution:
In first case, gain = 5%
and in second case, loss = 5%
and difference = Rs. 5 more
But difference in % = 5 + 5 = 10%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 18.1

Question 19.
Solution:
Let CP of an article = Rs. 100
List price = Rs. 100 + 20% of Rs. 100
= Rs. 100 + 20 = Rs. 120
Discount = 10%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 19.1

Question 20.
Solution:
Let CP of an article = Rs. 100
Then Marked price
= Rs. 100 + 10% of 100
= Rs. 100 + 10 = Rs. 110
Discount = 10%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 20.1

Question 21.
Solution:
Price of watch including VAT = Rs. 825
VAT% = 10%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10D 21.1

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

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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

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RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.1

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RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.1

Other Exercises

Question 1.
Fill in the blanks: [NCERT]
(i) All points lying inside / outside a circle are called …….. points / ………. points.
(ii) Circles having the same centre and different radii are called …….. circles.
(iii) A point whose distance from the centre of a circle is greater than its radius lies in …….. of the circle.
(iv) A continuous piece of a circle is …….. of the circle.
(v) The longest chord of a circle is a ……… of the circle.
(vi) An arc is a …….. when its ends are the ends of a diameter.
(vii) Segment of a circle is the region between an are and ……..of the circle.
(viii)A circle divides the plane, on which it lies, in …….. parts.
Solution:
(i) All points lying inside / outside a circle are called interior points / exterior points.
(ii) Circles having the same centre and different radii are called concentric circles.
(iii) A point whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.
(iv) A continuous piece of a circle is arc of the circle.
(v) The longest chord of a circle is a diameter of the circle.
(vi) An arc is a semi-circle when its ends are the ends of a diameter.
(vii) Segment of a circle is the region between an arc and centre of the circle.
(viii) A circle divides the plane, on which it lies, in three parts.

Question 2.
Write the truth value (T/F) of the following with suitable reasons: [NCERT]
(i) A circle is a plane figure.
(ii) Line segment joining the centre to any point on the circle is a radius of the circle.
(iii) If a circle is divided into three equal arcs each is a major arc.
(iv) A circle has only finite number of equal chords.
(v) A chord of a cirlce, which is twice as long is its radius is a diameter of the circle.
(vi) Sector is the region between the chord and its corresponding arc.
(vii) The degree measure of an arc is the complement of the central angle containing the arc.
(viii)The degree measure of a semi-circle is 180°.
Solution:
(i) True.
(ii) True.
(iii) True.
(iv) False. As it has infinite number of equal chords.
(v) True.
(vi) False. It is a segment not sector.
(vii) False. As total degree measure of a circle is 360°.
(viii) True.

 

 

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NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion

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NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion

MULTIPLE CHOICE QUESTIONS

Question 1.
A particle is moving in a circular path of radius r. The displacement after half a circle would be :
(a) Zero       (b) πr            (c) 2r                  (d) 2πr.
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 1
Answer:
(c) Explanation : Particle is just opposite to the initial position on the circle.

More Resources

Question 2.
A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is,
(a) u/g             (b) u2/2g                (c) u2/g                (d) u/2g.
Answer:
(b) Explanation : u2 – v2 = – 2gh or 0 – u2 = – 2gh (∴ u = 0 at highest point).
∴ h = u2/2g.

Question 3.
The numerical ratio of displacement to distance for a moving object is
(a) always less than 1
(b) always equal to 1
(c) always more than 1
(d) equal or less than 1
Answer:
(d) Explanation : Displacement ≤ distance.

Question 4.
If the displacement of an object is proportional to square of time, then the object moves with
(a) uniform velocity
(b) uniform acceleration
(c) increasing acceleration
(d) decreasing acceleration.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 2

Question 5.
From the given u -t graph (Fig. 1), it can be inferred that the object is
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 3
(a) in uniform motion
(b) at rest
(c) in non-uniform motion
(d) moving with uniform acceleration.
Answer:
(a) Explanation : Velocity of object is constant.

Question 6.
Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 m s-1. It implies that the boy is
(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity.
Answer:
(c) Explanation : Velocity of the boy changes continuously due to the change in direction of motion in circular path.

Question 7.
Area under a v — t graph represents a physical quantity which has the unit
(a) m2          (b) m                (c) m3              (d) m s-1.
Answer:
(b) Explanation : Area = v x t = (m/s) x s = m.

Question 8.
Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in Fig. 2 Choose the correct statement Q
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 4
(a) Car A is faster than car D
(b) Car B is the slowest
(c) Car D is faster than car C
(d) Car C is the slowest.
Answer:
(b) Explanation : Slope of distance – time graph = speed of object.

Question 9.
Which of the following figures represents uniform motion of a moving object correctly ?
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 5
Answer:
(a) Explanation : For uniform motion, distance ∞ time.

Question 10.
Slope of a velocity – time graph gives
(a) the distance
(b) the displacement
(c) the acceleration
(d) the speed.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 6

Question 11.
In which of the following cases of motions, the distance moved and the magnitude of displacement are equal ?
(a) If the car is moving on straight road
(b) If the car is moving in circular path
(c) The pendulum is moving to and fro
(d) The earth is revolving around the Sun.
Answer:
(a)

SHORT ANSWER QUESTIONS

Question 12.
The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero ? Justify your answer.
Answer:
No. When object moves in a circular path of radius r, then displacement of object .after completing a circle is zero but distance travelled = 2πr.

Question 13.
How will the equations of motion for an object moving with a uniform velocity change ?
Answer:
Equations of motion of a uniformly accelerated motion of an object are

  1. V = u + at,
  2. S = ut + ½ at2,
  3. v2 – u2 = 2aS.

When object moves with a uniform velocity, its acceleration, a = 0. Hence equations of motion become

  1. V = u,
  2. S = ut,
  3. v2 = u2.

Question 14.
A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement-time graph is shown in Fig. 4. Plot a velocity-time graph for the same.
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 7
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 8

Question 15.
A car starts from rest and moves along the x-axis with constant acceleration 5 m s-2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 9
Velocity after 8 s,V = u + at = 0 + 5 x 8 = 40 ms-1.
Distance travelled in last 4 s moving with constant velocity, x2 = v t = 40 x 4 = 160 m
∴Total distance = x1 + x2 = 320 m.

Question 16.
A motorcyclist drives from A to B with a uniform speed of 30 km h-1 and returns back with a speed of 20 km h-1. Find its average speed.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 10
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 11

Question 17.
The velocity-time graph (Fig. 6) shows the motion of a cyclist. Find
(i) its acceleration
(ii) its velocity and
(iii) the distance covered by the cyclist in 15 seconds.
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 12
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 13

Question 18.
Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height
Answer:
When stone is thrown vertically upwards, it has some initial velocity (u). The velocity of the stone goes on decreasing as it goes upwards and becomes zero at the maximum height. There after, stone begins to fall and its velocity goes on increasing but in opposite direction and becomes equal to the initial velocity (u) when it reaches the point of projection. v – t graph of a stone is shown in figure 7 A. The speed-time graph of the stone is shown in figure 7 B.
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 14

Question 19.
An object is dropped from rest at the height of 150m and simultaneously another object is dropped from rest at the height of 100m.
What is the difference in their heights after 2 s if both the objects drop with same accelerations ? How does the difference in heights vary with time ?
Answer:
Distance moved by the object dropped from height of 150m in 2 seconds can be calculated using
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 15
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 16
Height of the object from ground after 2 seconds = 150 – 20 = 130 m Similarly, distance moved by the other object in 2 seconds = 20 m
Height of the second object from ground after 2 seconds = 100 – 20 = 80 m
∴ Difference in their heights after 2 seconds = 130 m – 80 m = 50 m
Difference in height of both objects remains the same with time.

Question 20.
An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s. What will be the velocity after 7 s from the start ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 17

Question 21.
Using following data, draw time-displacement graph for a moving object :

Time (s)     0     246810121416
Displacement (m) 024446420

Use this graph to find average velocity for first 4 s, for next 4 s and for last 6 s.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 18
Displacement-time graph is shown in figure 8.
Average velocity for first 4s = Slope of displacement-time graph
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 19

Question 22.
An electron moving with a velocity of 5 x 104 ms-1 enters into a uniform electric field and acquires a uniform acceleration of 104 ms-2 in the direction of its initial motion.
(i) Calculate the time in which the electron would acquire a velocity double of its initial velocity.
(ii) How much distance the electron would cover in this time ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 20

Question 23.
Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds.
Answer:
We know, distance travelled by a uniformly accelerated object in time t is given by
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 21

Question 24.
Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of u12 : u22 (Assume upward acceleration is -g and downward acceleration to be +g).
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion image - 22

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

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

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 10B.

Other Exercises

Question 1.
Solution:
Marked price of cooler = Rs 4650
Rate of discount = 18%
Selling price
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 1.1

Question 2.
Solution:
Marked price = Rs 960
Selling price = Rs 816
Total Discount = M.P. – S.P.
= Rs 960 – Rs 816
= Rs 144
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 2.1

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 3.
Solution:
S.P. of shirt = Rs 1092
Discount = Rs 208
M.P. of shirt = S.P. + discount
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 3.1

Question 4.
Solution:
S.P. of toy = Rs 216.20
Discount = 8%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 4.4

Question 5.
Solution:
S.P. of tea set = Rs 528
Rate of discount = 12%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 5.1

Question 6.
Solution:
Let C.P. of goods = Rs 100
Marked price = Rs 100 + 35
= Rs 135
Rate of discount = 20%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 6.1

Question 7.
Solution:
Let C.P. of phone = Rs 100
.’. Marked price = Rs 100 + 40
= Rs 140
Rate of discount = 30%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 7.1

Question 8.
Solution:
C.P. of fan = Rs. 1080
Gain = 25%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 8.1

Question 9.
Solution:
C.P of refrigerator = Rs. 11515
and gain % = 20%
S.P. of refrigerator
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 9.1

Question 10.
Solution:
C.P. of ring = Rs. 1190
Gain = 20%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 10.1

Question 11.
Solution:
Let Marked price = Rs. 100
Discount = 10%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 11.1

Question 12.
Solution:
Let C.P. = Rs. 100
Gain = 8%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 12.1

Question 13.
Solution:
Marked price of TV = Rs. 18500
Series of two successive discounts = 20% and 5%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 13.1

Question 14.
Solution:
Let M.P. = Rs. 100
First discount = 20%
and second discount = 5%
RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10B 14.1

 

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