RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B

RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15B.

Other Exercises

Question 1.
Solution:
In ∆ABC,
∠A = 75°, ∠B = 45°
side BC is produced to D
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 1
forming exterior ∠ ACD
Exterior ∠ACD = ∠A + ∠B (Exterior angle is equal to sum of its interior opposite angles)
= 75° + 45° = 120°

Question 2.
Solution:
In ∆ABC, BC is produced to D forming an exterior angle ACD
∠ B = 68°, ∠ A = x°, ∠ ACB = y° and ∠ACD = 130°
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 2
In triangle,
Exterior angles is equal to sum of its interior opposite angles
∠ACD = ∠A + ∠B
⇒ 130° = x + 68°
⇒ x = 130° – 68° = 62°
But ∠ACB + ∠ACD = 180° (Linear pair)
⇒ y + 130° = 180°
⇒ y = 180° – 130° = 50°
Hence x = 62° and y = 50°

Question 3.
Solution:
In ∆ABC, side BC is produced to D forming exterior angle ACD.
∠ACD = 65°, ∠A = 32°
∠B = x, ∠ACB = y
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 3
In a triangle, the exterior angles is equal to the sum of its interior opposite angles
∠ACD = ∠A + ∠B
⇒ 65° = 32° + x
⇒ x = 65° – 32° = 33°
But ∠ ACD + ∠ ACB = 180° (Linear pair)
⇒ 65° + y = 180°
⇒ y = 180°- 65° = 115°
x = 33° and y = 115°

Question 4.
Solution:
In ∆ABC, side BC is produced to D forming exterior ∠ ACD
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 4
∠ACD = 110°, and ∠A : ∠B = 2 : 3
In a triangle, exterior angles is equal to the sum of its interior opposite angles
⇒ ∠ACD = ∠A + ∠B
⇒ ∠A + ∠B = 110°
But ∠A : ∠B = 2 : 3
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 5
But ∠A + ∠B + ∠C = 180° (sum of angles of a triangle)
⇒ 44° + 66° + ∠C = 180°
⇒ 110° + ∠C = 180°
⇒ ∠C = 180° – 110° = 70°
Hence ∠ A = 44°, ∠ B = 66° and ∠ C = 70°

Question 5.
Solution:
In ∆ABC, side BC is produced to forming exterior angle ACD.
∠ACD = 100° and ∠A = ∠B
Exterior angle of a triangle is equal to the sum of its interior opposite angles.
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 6
∠ACD = ∠A + ∠B But ∠A = ∠B
∠A + ∠A = ∠ACD = 100°
⇒ 2 ∠A = 100°
⇒ ∠A = 50°
∠B = ∠A = 50°
But ∠A + ∠B + ∠ ACB = 180° (sum of angles of a triangle)
⇒ 50° + 50° + ∠ ACB = 180°
⇒ 100° + ∠ ACB = 180°
⇒ ∠ ACB = 180° – 100° = 80°
Hence ∠ A = 50°, ∠ B = 50° and ∠ C = 80°

Question 6.
Solution:
In ∆ABC, side BC is produced to D From D, draw a line meeting AC at E so that ∠D = 40°
∠A = 25°, ∠B = 45°
In ∆ABC,
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 7
Exterior ∠ACD = ∠A + ∠B = 25° + 45° = 70°
Again, in ∆CDE,
Exterior ∠ AED = ∠ ECD + ∠ D = ∠ACD + ∠D = 70° + 40° = 110°
Hence ∠ACD = 70° and ∠AED = 110°

Question 7.
Solution:
In ∆ABC, sides BC is produced to D and BA to E
∠CAD = 50°, ∠B = 40° and ∠ACB = 100°
∠ ACB + ∠ ACD = 180° (Linear pair)
⇒ 100° + ∠ ACD = 180°
⇒ ∠ ACD = 180° – 100° = 80°
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 8
In ∆ACD,
∠ CAD + ∠ ACD + ∠ ADC = 180° (sum of angles of a triangle)
⇒ 50° + 80° + ∠ ADC = 180°
⇒ 130° + ∠ ADC = 180°
⇒ ∠ ADC = 180° – 130° = 50°
Now, in ∆ABD, BA is produced to E
Exterior ∠DAE = ∠ACD + ∠ADC = 80° + 50° = 130°
Hence ∠ ACD = 80°, ∠ ADC = 50° and ∠DAE = 130°

Question 8.
Solution:
In ∆ABC, BC is produced to D forming exterior ∠ ACD
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 9
∠ACD = 130°, ∠A = y°, ∠B = x° and ∠ACB = z°.
x : y = 2 : 3
Now, in ∆ABC,
Exterior ∠ACD = ∠A + ∠B
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 10
But ∠A + ∠B + ∠ACB = 180° (sum of angles of a triangle)
⇒ 78° + 52° + ∠ACB = 180°
⇒ 130° + ∠ACB = 180°
⇒ ∠ACB = 180° – 130°
⇒ ∠ACB = 50°
⇒ ∠ = 50°
Hence x = 52°, y = 78° and z = 50°

Hope given RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15B are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15A

RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15A.

Other Exercises

Question 1.
Solution:
In ∆ABC,
∠A = 72°, ∠B = 63°
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ 72° + 63° + ∠C = 180°
⇒ 135° + ∠C = 180°
⇒ ∠C= 180°- 135° = 45°

Question 2.
Solution:
In. ∆PQR,
∠E = 105°, and ∠F = 40°
But ∠D + ∠E+ ∠F= 180° (sum of angles of a triangle)
⇒ ∠D + 105°+ 40°= 180°
⇒ ∠ D + 145° = 180°
⇒ ∠D = 180°- 145°
⇒ ∠D = 35°

Question 3.
Solution:
In ∆XYZ,
∠ X = 90°, ∠ Z = 48°
But ∠X + ∠Y + ∠Z = 180° (Sum of angles of a triangle)
⇒ 90° + ∠ Y + 48° = 180°
⇒ 138°+ ∠ Y = 180°
⇒ ∠Y = 180° – 138° = 42°
⇒ ∠Y = 42°

Question 4.
Solution:
Sum of angles of a triangle = 180°
and ratio in the three angles = 4 : 3 : 2
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15A 1

Question 5.
Solution:
In a right triangle
Sum of the two acute angles = 90°
One angle = 30°
Second angle = 90° – 36° = 54°

Question 6.
Solution:
In a right triangle
Sum of two acute angles = 90°
and ratio of these two angles = 2 : 1
Let first angle = 2x
Then second angle = x
2x + x = 90°
⇒ 3x = 90°
⇒ x = \(\frac { 90 }{ 3 }\) = 30°
First angle = 2x = 2 x 30° = 60°
and second angle = x = 1 x 30° = 30°

Question 7.
Solution:
In a triangle,
Measure of one angle = 100°
Sum of other two angles = 180° – 100° = 80°
(Sum of angles of a triangles)
But, these two angles are equal.
Measure of each angle = \(\frac { 80 }{ 2 }\) = 40°

Question 8.
Solution:
Sum of angles of a triangle = 180°
Let third angle = x
then, each equal angles = 2x
x + 2x + 2x = 180°
⇒ 5x = 180°
⇒ x = \(\frac { 180 }{ 5 }\) = 36°
Each equal angle = 2x = 2 x 36° = 72°
and third angle = 36°

Question 9.
Solution:
In a triangle ABC,
Let ∠A = ∠B + ∠C
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ ∠A + ∠A = 180° (∠B + ∠C = ∠A)
⇒ 2A = 180°
⇒ ∠ A = \(\frac { 180 }{ 2 }\) = 90°
∠ A = 90°
Hence, ∆ABC is a right triangle.

Question 10.
Solution:
In a ∆ABC,
2 ∠A = 3 ∠B = 6 ∠C = 1 (suppose)
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15A 2

Question 11.
Solution:
In an equilateral triangle,
All sides are equal.
All angles are also equal.
Each angle = \(\frac { 180 }{ 3 }\) = 60°
(Sum of angles of a triangle = 180°)

Question 12.
Solution:
In the given figure,
ABC is a triangle in which DE || BC,
∠A = 65° and ∠B = 55°
DE || BC and ADB is the transversal
⇒ ∠ ADE = ∠ B (corresponding angles) = 55° (∠B = 55°)
In ∆ADE,
∠A + ∠ADE + ∠AED = 180° (sum of angles of a triangle)
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15A 3
⇒ 65° + 55° + ∠AED = 180°
⇒ ∠ 120° + ∠AED = 180°
⇒ ∠AED = 180°- 120° = 60°
∠AED = 60°
D || BC and AEC is the transversal
∠ C = ∠ AED (A corresponding angles)
∠C = 60°
Hence ∠ADE = 55°, ∠AED = 60° and ∠ C = 60°

Question 13.
Solution:
(i) No. In a triangle, only one right angle is possible as if there are two right angles, then The third angle will be ∠ero which is not possible.
(ii) No. In a triangle only one obtuse angle is possible as if there are two obtuse angles, then the sum of these two angles will be greater than 180° which is not possible.
(iii) Yes. two acute can arc possible.
(iv) No. The sum of these three angles will be greater than 180° which is not possible in a triangle.
(v) No. The sum of these angles will be less than 180° which is not possible.
(vi) Yes. The sum of there three angle will be in 180° which is possible.

Question 14.
Solution:
(i) Yes, it can be a right triangle also
(ii) Yes, if right triangle has its sides different then it is possible.
(iii) No, a right triangle cannot be an equilateral triangle as an equilateral triangle has each side 60°.
(iv) Yes, it is possible, if its sides opposite to acute angles are equal.

Question 15.
Solution:
(i) A right triangle cannot have an obtuse angle.
(ii) The acute angles of a right triangle are complementary.
(iii) Each acute angle of an isosceles right triangle measures 45°.
(iv) Each angle of an equilateral triangle measures 60°.
(v) The side opposite the right angle of the right triangle is called the hypotenuse.
(vi) The sum of the lengths of the sides of a triangle is called its perimeter.

 

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RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17C

RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17C.

Other Exercises

Mark (✓) against the correct answer in each of the following:
Question 1.
Solution:
(c) Supplement of 45° is 135°
135°+ 45° = 180°

Question 2.
Solution:
(b) Complement of 80° is 10°
10° + 80° = 90°

Question 3.
Solution:
(b) The angle is its own complement.
The measure of the angles will be 45° (45° + 45° = 90°)

Question 4.
Solution:
(a) The angle is one-fifth of its supplement
Let angle be x, then
x + 5x = 180°
⇒ 6x = 180°
⇒ x = 30°
Angle is 30°

Question 5.
Solution:
(b) Let angle is x
Then its complement angle=x-24° But x + x- 24° = 90°
⇒ 2x = 90° + 24° = 114°
⇒ x = 57°
The required angle is 57°

Question 6.
Solution:
(b) Let required angle = x
Then its supplement angle = x + 32
But x + x + 32° = 180°
⇒ 2x = 180° – 32 = 148°
⇒ x = 74°
Required angle = 74°

Question 7.
Solution:
(c) Two supplementary angle are in the ratio = 3 : 2
Let first angle = 3x
Second angle = 2x
But 3x + 2x = 180°
⇒ 5x = 180°
⇒ x = 36°
Smaller angle = 2x = 2 x 36° = 72°

Question 8.
Solution:
(b) In the figure ∠BOC = 132°
But ∠AOC + ∠BOC =180° (Linear pair)
⇒ ∠AOC + 132° = 180°
⇒ ∠AOC = 180° – 132° = 48°

Question 9.
Solution:
(c) In the figure, ∠AOC = 68°
But ∠AOC + ∠BOC = 180° (Linear pair)
⇒ 68° + x = 180°
⇒ x = 180° – 68° = 112°

Question 10.
Solution:
(b) In the figure,
AOB is a straight line
∠AOC + ∠BOC = 180° (Linear pair)
⇒ 2x – 10° + 3x + 15° = 180°
⇒ 5x = 180° + 10° – 15° = 175°
⇒ x = 35°
x = 35

Question 11.
Solution:
(d) In the figure,
AOB is a straight line
∠AOC + ∠COD + ∠DOB = 180°
⇒ 55° + x + 45° = 180°
⇒ x + 100° = 180°
⇒ x = 180° – 100° = 80°

Question 12.
Solution:
(a) AOB is a straight line
x + y = 180°
But 4x = 5y
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17C 1

Question 13.
Solution:
(b) AB and CD intersect each other at O and ∠AOC = 50°
∠BOD = ∠AOC = 50° (Vertically opposite angles)

Question 14.
Solution:
(a) AOB is a straight line
∠AOC + ∠COD + ∠DOB = 180°
⇒ 3x – 8° + 50° + x + 10° = 180°
⇒ 4x = 180° + 8° – 50° – 10°
⇒ 4x = 128°
⇒ x = 32°

Question 15.
Solution:
(b) In ∆ABC, side BC is produced to D
∠ACD = 132° and ∠A = 54°
Ext. ∠ACD = ∠A + ∠B
⇒ 132° = 54° + ∠B
⇒ ∠B = 132° – 54° = 78°

Question 16.
Solution:
(c) In ∆ABC,
Side BC is produced to D
∠A = 45°, ∠B = 55°
Ext. ∠ACD = ∠A + ∠B = 45° + 55° = 100°

Question 17.
Solution:
(b) In ∆ABC, side BC is produced to D
∠ABC = 70° and ∠ACD = 120°
Ext. ∠ACD = ∠BAC + ∠ABC
⇒ 120° = ∠BAC + 70°
⇒ ∠BAC = 120° – 70° = 50°

Question 18.
Solution:
(c) In the figure,
∠AOB = 50°, ∠BOC = 90°
∠COD = 70°, ∠AOD = x.
But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360° (Angles at a point)
⇒ 50° + 90° + 70° + x = 360°
⇒ 210 + x = 360°
⇒ x = 360° – 210°
⇒ x = 150°

Question 19.
Solution:
(c) In the figure,
Side BC of ∆ABC is produced to D
CE || BA is drawn
∠A = 50° and ∠ECD = 60°
AB || CE
∠ABC = ∠ECD (corresponding angle) = 60°
But in ∆ABC,
∠A + ∠B + ∠ACB = 180° (Angles of a triangles)
⇒ 50° + 60° + ∠ACB = 180°
⇒ ∠ACB = 180° – 50° – 60° = 70°

Question 20.
Solution:
(b) In ∆ABC,
∠A = 65°, ∠C = 85°
But ∠A + ∠B + ∠C = 180° (Angles of a triangle)
⇒ 65° +∠B+ 85° = 180°
⇒ 150° + ∠B = 180°
⇒ ∠B = 180° – 150° = 30°

Question 21.
Solution:
(d) Sum of angles of a triangle = 180°

Question 22.
Solution:
(c) Sum of angles of a quadrilateral = 360°

Question 23.
Solution:
(b) In the figure, AB || CD
∠OAB = 150°, ∠OCD = 120°
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17C 2
From O, draw OE || AB or CD
AB || DE
∠OAB + ∠AOE =180°
⇒ 150° + ∠AOE = 180°
⇒ ∠AOE = 180° – 150° = 30°
Similarly DE || CD
∠EOC + ∠OCD = 180°
⇒ ∠EOC + 120° = 180°
⇒ ∠EOC = 180° – 120° = 60°
Now ∠AOC = ∠AOE + ∠EOC = 30° + 60° = 90°

Question 24.
Solution:
(a) In the given figure,
PQ || RS,
∠PAB = 60° and ∠ACS = 100°
PQ || RS
∠ABC = ∠PAB (alternate angles) = 60°
But Ext. ∠ACS = ∠BAC + ∠ABC
⇒ 100° = ∠BAC + 60°
⇒ ∠BAC = 100° – 60° = 40°

Question 25.
Solution:
(c) In the figure, AB || CD || EF
∠ABG =110° and ∠GCD = 100°
∠BGC = x°
AB || EF
∠ABG + ∠BGE = 180°
⇒ 110° + ∠BGE = 180°
⇒ ∠BGE = 180° – 110° = 70°
Similarly CD || EF
∠GCD + ∠CGF = 180°
⇒ 100° + ∠CGF = 180°
⇒ ∠CGF = 180° – 100° = 80°
But ∠BGE + ∠BGC + ∠CGF = 180°
⇒ 70° + x + 80° = 180°
⇒ 150° + x = 180°
⇒ x = 180° – 150° = 30°

Question 26.
Solution:
(d) Sum of any two sides of a triangle is always greater than the third side

Question 27.
Solution:
(d) The diagonals of a rhombus always bisect each other at right angles.

Question 28.
Solution:
(c) In ∆ABC, ∠B = 90°
AB = 5 cm and AC = 13 cm
But AC² = AB² + BC² (By Pythagoras Theorem)
⇒ (13)² = (5)² + BC²
⇒ 169 = 25 + BC2
⇒ BC² = 169 – 25 = 144 = (12)²
BC = 12 cm

Question 29.
Solution:
(c) In ∆ABC, ∠B = 37°, ∠G = 29°
But ∠A + ∠B + ∠C = 180° (angles of a triangle)
⇒ ∠A + 37° + 29° = 180°
⇒ ∠A + 66° = 180°
⇒ ∠A = 180° – 66° = 114°

Question 30.
Solution:
(c) The ratio of angles of a triangle is 2 : 3 : 7
But sum of angles of a triangle = 180°
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17C 3

Question 31.
Solution:
In ∆ABC,
Let 2∠A = 3∠B = 6∠C = x
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17C 4

Question 32.
Solution:
(a) In ∆ABC,
∠A + ∠B = 65°, ∠B + ∠C = 140°
∠A = 65°
∠C = 140° – ∠B
But ∠A + ∠B + ∠C = 180° (Angles of a triangle)
⇒ 65° – ∠B + 140° – ∠B + ∠B = 180°
⇒ 205° – ∠B = 180°
⇒ ∠B = 205° – 180° = 25°

Question 33.
Solution:
(b) In ∆ABC, ∠A – ∠B = 33°
and ∠B – ∠C = 18°
∠A = 33° + ∠B and ∠C = ∠B – 18°
But ∠A + ∠B + ∠C = 180°
⇒ 33° + ∠B + ∠B + ∠B – 18° = 180°
⇒ 3∠B = 180° – 33° + 18° = 165°
⇒ ∠B = 55°

Question 34.
Solution:
(c) In ∆ABC
∠A + ∠B + ∠C= 180° (Sum of angles of a triangle)
But angles are (3x)°, (2x – 7)° and (4x – 11)°
3x + (2x – 7) + (4x – 11)° = 180°
⇒ 3x + 2x – 7 + 4x – 11° = 180°
⇒ 9x – 18° = 180°
⇒ 9x = 180° + 18° = 198°
⇒ x = 22°

Question 35.
Solution:
(c) ∆ABC is a right angled, ∠A = 90°
AB = 24 cm, AC = 7 cm
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17C 5
but BC² = AB² + AC²
⇒ BC² = (24)² + (7)² = 576 + 49 = 625 = (25)²
BC = 25 cm

Question 36.
Solution:
(b) Let AB is a ladder and A is the window
BC = 15 m, AC = 20 m
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17C 6
Now in right ∆ABC
AB² = BC² + AC² = (15)² + (20)² = 225 + 400 = 625 = (25)²
AB = 25 m
Length of ladder = 25 m

Question 37.
Solution:
(a) Let AB and CD are two poles such that
AB = 6 m, CD = 11 m
and distance between two poles BD = 12m
From A, draw AE || BD
AE = BD = 12m
CE = CD – ED = 11 – 6 = 5 m
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17C 7
Now in right ∆AEC
AC² = AE² + CE² = (12)² + (5)² = 144 + 25 = 169 = (13)²
AC = 13 m
Distance between tops of poles = 13 m

Question 38.
Solution:
(d) ∆ABC is an isosceles triangle
∠C = 90°,
AC = 5 cm
BC = AC = 5 cm
RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17C 8
In right ∆ABC
AB² = AC² + BC² = (5)² + (5)² = 25 + 25 = 50 = 2 x 25
AB = √(2 x 25) = 5√2 cm

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RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper

RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest CCE Test Paper.

Other Exercises

Question 1.
Solution:
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 1

Question 2.
Solution:
Let the sum be ₹ x
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 2

Question 3.
Solution:
P = ₹ 3625, A = ₹ 4495, T = 2 years
S.I. = A – P = ₹ 4495 – ₹ 3625 = ₹ 870
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 3

Question 4.
Solution:
P = ₹ 3600, A = ₹ 4410, R = 9%
S.I. = A – P = ₹ 4410 – ₹ 3600 = ₹ 810
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 4

Question 5.
Solution:
Let the sum be ₹ x
Amount = ₹ 2x
S.I. = (2x – x) = ₹ x
Time = 12 years
P = x, S.I. = x, T = 12 years
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 5

Question 6.
Solution:
Let the sum be ₹ 4
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 6

Mark (✓) against the correct answer in each of the following :
Question 7.
Solution:
(c) 9%
Let the sum be ₹ x
A = \(\frac { 49x }{ 40 }\)
We know:
A = P + S.I.
S.I. = A – P
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 7

Question 8.
Solution:
(c) ₹ 3500
A = ₹ 3626, R = 6%,
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 8
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 9

Question 9.
Solution:
(a) 9 months
P = ₹ 6000, A = ₹ 6360
S.I. = A – P = 6300 – 6000 = ₹ 360
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 10

Question 10.
Solution:
(c) 12%
Let the sum be ₹ x
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 11

Question 11.
Solution:
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 12
P = ₹ \(\frac { 100 }{ x }\)

Question 12.
Solution:
(b) 10%
Let the sum be ₹ x
Amount ₹ 2x
Time =10 years
S.I. = A – P = 2x – x = ₹ x
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 13

Question 13.
Solution:
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 14
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 15
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 16

Question 14.
Solution:
(i) False
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 17
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest CCE Test Paper 18

Hope given RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest CCE Test Paper are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12B

RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest Ex 12B.

Other Exercises

Question 1.
Solution:
Principal (P) = Rs. 6250
Rate of (R) = 4% p.a.
Period (T) = 6 months = \(\frac { 1 }{ 2 }\) year
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12B 1

Question 2.
Solution:
Amount (A) = Rs. 3605
Rate (R) = 5% p.a.
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12B 2
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12B 3

Question 3.
Solution:
Let sum (P) = Rs. 100
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12B 4

Question 4.
Solution:
Principal (P) = Rs. 8000
Amount (A) – Rs. 8360
S.I. = A – P = Rs. 8360 – Rs. 8000 = Rs. 360
Rate (R) = 6% p.a.
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12B 5

Question 5.
Solution:
Let sum (P) = Rs. 100
Then amount (A) = Rs. 100 x 2 = Rs. 200
S.I. = A – P = Rs. 200 – Rs. 100 = Rs. 100
Period (T) = 10 years
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12B 6

Question 6.
Solution:
S.I. = Rs. x
Rate (R) = x% p.a.
Time (T) = x year
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12B 7
Rs. \(\frac { 100 }{ x }\) (c)

Question 7.
Solution:
Let sum (P) = Rs. 100
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12B 8

Question 8.
Solution:
A’s principal (P) = Rs. 8000
Rate (R) = 12% p.a.
B’s principal = Rs. 9100
Rate = 10%
Let after x years, then amount will be equal
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12B 9
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12B 10

Question 9.
Solution:
Amount (A) = Rs. 720
Principal (P) = Rs. 600
S.I. = A – P = Rs. 720 – 600 = Rs. 120
Period (T) = 4 years
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12B 11

Question 10.
Solution:
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12B 12

Question 11.
Solution:
Let sum (P) = Rs. 100
Their S.I. = 0.125 of Rs. 100
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12B 13

Question 12.
Solution:
S.I. = Rs. 210
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12B 14

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RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A

RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest Ex 12A.

Other Exercises

Find the simple interest and the amount when :
Question 1.
Solution:
Principal (P) = Rs. 6400
Rate (r) = 6% p.a.
Time (t) = 2 years
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 1

Question 2.
Solution:
Principal (P) = Rs. 2650
Rate (r) = 8% p.a.
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 2

Question 3.
Solution:
Principal (P) = Rs. 1500
Rate (r) = 12% p.a.
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 3

Question 4.
Solution:
Principal (P) = Rs. 9600
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 4

Question 5.
Solution:
Principal (P) = Rs. 5000
Rate (r) = 9% p.a.
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 5

Find the time when :
Question 6.
Solution:
Principal (P) = Rs. 6400
S.I. = Rs. 1152
Rate (r) = 6% p.a.
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 6

Question 7.
Solution:
Principal (P) = Rs. 9540
S.I. = Rs. 1908
Rate (r) = 8% p.a.
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 7

Question 8.
Solution:
Amount (A) = Rs. 6450
Principal (P) = Rs. 5000
S.I. = A – P = Rs. (6450 – 5000) = Rs. 1450
Rate (r) = 12% p.a.
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 8

Find the rate when :
Question 9.
Solution:
Principal (P) = Rs. 8250
S.I. = Rs. 1100
Time (t) = 2 years
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 9

Question 10.
Solution:
Principal (P) = Rs. 5200
S.I. = Rs. 975
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 10

Question 11.
Solution:
Principal (P) = Rs. 3560
Amount (A) = Rs. 4521.20
S.I. = A – P = Rs. 4521.20 – 3560 = Rs. 961.20
Time (t) = 3 years
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 11

Question 12.
Solution:
Principal (P) = Rs. 6000
Rate (r) = 12% p.a.
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 12

Question 13.
Solution:
Principal = Rs. 12600
Rate (A) = 15% p.a.
Time (t) = 3 years
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 13
Amount = P + S.I. = Rs. 12600 + Rs. 5670 = Rs. 18270
Amount paid in cash = Rs. 7070
Balance = Rs. 18270 – 7070 = Rs. 11200
Price of goat = Rs. 11200

Question 14.
Solution:
S.I. = Rs. 829.50
Rate (r) = 10% p.a.
Time (t) = 3 years
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 14

Question 15.
Solution:
Amount (A) = Rs. 3920
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 15

Question 16.
Solution:
Amount = Rs. 4491
Let principal (P) = Rs. 100
Rate (r) =11% p.a.
Time (t) = 2 years 3 months = 2\(\frac { 1 }{ 4 }\) = \(\frac { 9 }{ 4 }\) years
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 16

Question 17.
Solution:
Amount = Rs. 12122
Let principal (P) = Rs. 100
Rate (r) = 8% p.a.
Time (t) = 2 years
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 17
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 18

Question 18.
Solution:
Amount (A) = Rs. 4734
Principal (P) = Rs. 3600
S.I. = A – P = Rs. 4734 – Rs. 3600 = Rs. 1134
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 19

Question 19.
Solution:
In first case,
Amount (A) = Rs. 768
Principal (P) = Rs. 640
S.I. = A – P = Rs. 768 – 640 = Rs. 128
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 20

Question 20.
Solution:
Principal (P) = Rs. 5600
Amount (A) = Rs. 6720
S.I. = A – P = Rs. 6720 – 5600 = Rs. 1120
Rate (r) = 8% p.a.
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 21

Question 21.
Solution:
Let principal (P) = Rs. 100
then amount (A) = Rs. 100 x \(\frac { 8 }{ 5 }\) = Rs. 160
S.I. = A – P = Rs. 160 – 100 = Rs. 60
Time (t) = 5 years
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 22

Question 22.
Solution:
Amount in 3 years = Rs. 837
Amount in 2 years = Rs. 783
Difference = Rs. 837 – Rs. 783 = Rs. 54
Rs. 54 is interest for 1 year
Interest for 2 years = 2 x 54 = Rs. 108
Principal = Rs. 783 – 108 = Rs. 675
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 23

Question 23.
Solution:
Amount in 5 years = Rs. 5475
Amount in 3 years = Rs. 4745
Interest for 2 years = Rs. 5475 – 4745 = Rs. 730
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 24

Question 24.
Solution:
Total sum = Rs. 3000
Let first part = Rs. x
Then second part = Rs. (3000 – x)
Now, interest on first part at the rate of
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 25

Question 25.
Solution:
Total sum =Rs. 3600
Let first part = Rs. x
Then second part = Rs. (3600 – x)
Interest on first part for 1 year at 9% p.a.
RS Aggarwal Class 7 Solutions Chapter 12 Simple Interest Ex 12A 26
⇒ 9x + 10 (3600 – x) = 33300
⇒ 9x + 36000 – 10x = 33300
⇒ -x = 33300 – 36000
⇒ -x = – 2700
⇒ x = 2700
First part = Rs. 2700
and second part = Rs. 3600 – 2700 = Rs. 900

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RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss CCE Test Paper

RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss CCE Test Paper.

Other Exercises

Question 1.
Solution:
SP of the chair = ₹ 1375
Gain% = 10
CP of the chair
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss CCE Test Paper 1

Question 2.
Solution:
Let the cost of each pen be ₹ 1
CP of 10 pens = ₹ 10
SP of 10 pens = CP of 14 pens = ₹ 14
Gain = SP – CP = 14 – 10 = ₹ 4
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss CCE Test Paper 2

Question 3.
Solution:
Let the cost price of the fan be ₹ x
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss CCE Test Paper 3

Question 4.
Solution:
Cost price of six lemons = ₹ 10
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss CCE Test Paper 4

Question 5.
Solution:
SP of the bat = ₹ 486
Loss = 10%
CP of the bat
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss CCE Test Paper 5

Mark (✓) against the correct answer in each of the following :
Question 6.
Solution:
(b) ₹ 85
SP of a football = ₹ 100
Gain = ₹ 15
Gain = SP – CP
⇒ 15 = 100 – CP
⇒ CP = ₹ (100 – 15)
⇒ CP = ₹ 85

Question 7.
Solution:
(c) 15.2%
Cost price of 12 bananas = ₹ 25
Cost price of one banana = ₹ \(\frac { 25 }{ 12 }\)
Cost price of five bananas = 5 x \(\frac { 25 }{ 12 }\)
= \(\frac { 125 }{ 12 }\) = 10.42
He shells five bananas at the cost (SP) of ₹ 12.
Gain = SP – CP = ₹ (12 – 10.42) = ₹ 1.58
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss CCE Test Paper 6

Question 8.
Solution:
(c) ₹ 196
Let the cost price of the jug be ₹ x
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss CCE Test Paper 7
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss CCE Test Paper 8

Question 9.
Solution:
(c) 66\(\frac { 2 }{ 3 }\) %
Let the cost price of each banana be ₹ 1
Cost price of three bananas = ₹ 3
SP of three bananas = CP of five bananas = ₹ 5
Gain = SP – CP = ₹ (5 – 3) = ₹ 2
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss CCE Test Paper 9

Question 10.
Solution:
(i) Loss = (CP) – (SP).
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss CCE Test Paper 10

Question 11.
Solution:
(i) False
Gain or loss is always reckoned on the cost price.
(ii) True
(iii) True
(iv) True

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RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11B

RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss Ex 11B.

Other Exercises

Objective Questions :
Mark (✓) against the correct answer in each of the following :
Question 1.
Solution:
(b)
C.P. = Rs. 80
S.P. = Rs. 100
Gain = S.P. – C.P. = Rs. 100 – 80 = Rs. 20
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11B 1

Question 2.
Solution:
(a)
C.P. = Rs. 120
S.P. = Rs. 105
Loss = C.P. – S.P. = Rs. 120 – 105 = Rs. 15
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11B 2

Question 3.
Solution:
(b)
S.P. = Rs. 100
Gain = Rs. 20
C.P. = S.P. – gain = Rs. 100 – 20 = Rs. 80
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11B 3

Question 4.
Solution:
(a)
S.P. = Rs. 198
Gain = 10%
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11B 4

Question 5.
Solution:
(a)
First S.P. of jug = Rs. 144
Loss = \(\frac { 1 }{ 7 }\) of C.P.
Let C.P. = x
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11B 5

Question 6.
Solution:
(d)
First S.P. of a pen = Rs. 48
Loss = 20%
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11B 6

Question 7.
Solution:
(a)
C.P. of 12 pencils = S.P. of 15 pencils = Re 1 (suppose)
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11B 7

Question 8.
Solution:
(d)
Let CP of 4 toffees = SP of 3 toffee = Rs. 12
(LCM of 4, 3 = 12)
CP of 1 toffee = Rs. \(\frac { 12 }{ 4 }\) = Rs. 3
SP of 1 toffee = Rs. \(\frac { 12 }{ 3 }\) = 4
Gain = Rs. 4 – 3 = Re 1
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11B 8

Question 9.
Solution:
(c) SP of an article = Rs. 144
Loss% = 10%
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11B 9

Question 10.
Solution:
(a)
CP of 6 lemons = Re 1
CP of 1 lemon = Re \(\frac { 1 }{ 6 }\)
SP of 4 lemons = Re 1
SP of 1 lemon = Re \(\frac { 1 }{ 4 }\)
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11B 10

Question 11.
Solution:
(d)
SP of chair = Rs. 720
Gain% = 20%
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11B 11

Question 12.
Solution:
(c)
SP of stool = Rs. 630
Loss% = 10%
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11B 12

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RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A

RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 11 Profit and Loss Ex 11A.

Other Exercises

Question 1.
Solution:
(i) C.P. = Rs. 950
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 1
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 2
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 3

Question 2.
Solution:
(i) CP = Rs. 2400
SP = Rs. 2592
Gain = SP – CP = Rs. 2592 – Rs. 2400 = Rs. 192
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 4
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 5

Question 3.
Solution:
(i) S.P. = Rs. 924
Gain% = 10%
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 6
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 7
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 8

Question 4.
Solution:
C.P. of an almirah = Rs. 13600
Amount spent on transportation = Rs. 400
Total cost price (CP) = Rs. 13600 + Rs. 400 = Rs. 14000
SP = Rs. 16800
Total gain = SP – CP
= Rs. 16800 – Rs. 14000 = Rs. 2800
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 9

Question 5.
Solution:
CP of an old house = Rs. 765000
Amount spent on repairs = Rs. 115000
Total cost price (CP) = Rs. 765000 + Rs. 115000 = Rs. 880000
Gain = 5%
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 10

Question 6.
Solution:
CP of 1 dozen or 12 lemons = Rs. 25
SP of 5 lemons = Rs. 12
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 11

Question 7.
Solution:
SP of 12 pens = CP of 15 pens
Let CP of 1 pen = Re. 1
The CP of 15 pens = Rs. 15
and SP of 12 pens = Rs. 12
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 12

Question 8.
Solution:
SP of 16 spoons = CP of 15 spoons
Let CP of 1 spoon = Re. 1
Then CP of 15 spoons = Rs. 15
and SP of 16 spoons = Rs. 15
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 13

Question 9.
Solution:
For Manoj
CP of video = Rs. 12000
Gain = 10%
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 14
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 15

Question 10.
Solution:
SP of sofa set = Rs. 21600
Gain = 8%
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 16

Question 11.
Solution:
SP of a watch = Rs. 11400
Loss = 5%
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 17

Question 12.
Solution:
SP of calculator = Rs. 1325
Gain = 6%
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 18
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 19

Question 13.
Solution:
SP of computer = Rs. 24480
Loss = 4%
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 20

Question 14.
Solution:
In first case gain =15%
and in second case, gain = 20%
Difference = 20 – 15 = 5%
Now 5% of the CP = Rs. 108
CP = \(\frac { 108 x 100 }{ 5 }\) = Rs. 2160

Question 15.
Solution:
In first case, loss = 8%
and in second case gain = 6%
Difference = 8 + 6 = 14%
Now 14% of CP = Rs. 3360
CP = Rs. \(\frac { 3360 x 100 }{ 14 }\) = Rs. 24000

Question 16.
Solution:
SP of first cycle = Rs. 2376
Gain = 10%
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 21
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 22

Question 17.
Solution:
SP of an exhaust fan = Rs. 7350
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 23

Question 18.
Solution:
Let CP of watch for Mohit = Rs. 100
Gain =10%
SP for Mohit = Rs. 100 + 10 = Rs. 110
CP for Karim = Rs. 110
Gain = 4%
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 24

Question 19.
Solution:
Retail price or S.P. of the machine for retailer = Rs. 37950
Gain = 25%
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 25
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 26

Question 20.
Solution:
CP of video = Rs. 20000
and CP of television = Rs. 30000
Loss on video = 5%
and gain on TV = 8%
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 27
Total CP of video and TV = Rs. 20000 + 30000 = Rs. 50000
and SP of video and TV = Rs. 190000 + 32400 = RS. 51400
Gain = SP – CP = Rs. 51400 – 50000 = Rs. 1400
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 28
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 29

Question 21.
Solution:
SP of 36 oranges = CP of 36 oranges – loss = CP of 36 oranges – SP of 4 oranges
⇒ SP of 36 oranges + SP of 4 oranges = CP of 3 6 oranges
⇒ SP of 40 oranges = CP of 36 oranges
Let CP of each orange = Re. 1
CP of 36 oranges = Rs. 36
and SP of 40 of 40 oranges = Rs. 36
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 30

Question 22.
Solution:
We know that C.P. = S.P. – gain
C.P. of 8 dozen pencils = S.P. of 8 dozen pencils – S.P. of 1 dozen pencils
= S.P. of 7 dozen pencils = Rs. 100 (suppose)
C.P. of 1 dozen = Rs.\(\frac { 100 }{ 8 }\)
RS Aggarwal Class 7 Solutions Chapter 11 Profit and Loss Ex 11A 31

 

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RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper

RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 10 Percentage CCE Test Paper.

Other Exercises

Question 1.
Solution:
We have :
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 1

Question 2.
Solution:
(i) Let x% of 1 kg be 125 g
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 2
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 3

Question 3.
Solution:
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 4

Question 4.
Solution:
(i) Let x be the required number.
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 5

Question 5.
Solution:
Let x be the number
The number is increased by 10%
Increased number = 110% of x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 6
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 7

Question 6.
Solution:
The present value of the machine = ₹ 10000
The decrease in its value after the 1 st year = 10% of ₹ 10000
\(\frac { 10 }{ 100 }\) x 10000 = ₹ 1000
The depreciated value of the machine after the 1 st year = ₹ (10000 – 1000) = ₹ 9000
The decrease in its value after th 2nd year = 10% of ₹ 9000
\(\frac { 10 }{ 100 }\) x 9000 = 900
The depreciated value of the machine after the 2nd year = ₹ (9000 – 900) = ₹ 8100
Hence, the value of the machine after two years will be ₹ 8100.

Question 7.
Solution:
The present population of the town = 16000
Increase in population after 1 year = 5% of 16000
= (\(\frac { 5 }{ 100 }\) x 16000) = 800
Thus, population after one year = 16000 + 800 = 16800
Increase in population after 2 years = 5% of 16800
= \(\frac { 5 }{ 100 }\) x 16800 = 840
Population after two years = 16800 + 840 = 17640
Hence, the population of the town after two years will be 17,640.

Question 8.
Solution:
Let us assume that the original price of the tea set is Increase in price = 5%
So, value increased on the tea set = 5% of ₹ x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 8
Hence, the original price of the tea set is ₹ 420.

Mark (✓) against the correct answer in each of the following :
Question 9.
Solution:
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 9

Question 10.
Solution:
(c) 12
Given that x% of 75 = 12
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 10

Question 11.
Solution:
(c) 25
Let the number be x. Then, we have:
120% of x = increased number
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 11

Question 12.
Solution:
(d) 180
Let the required number be x.
Then, we have :
5% of x = 9
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 12

Question 13.
Solution:
(a) 60
Let the number be x According to question, we get:
(35 % of x) + 39 = x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 13
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 14

Question 14.
Solution:
(c) 500
Let x be the maximum marks Pass marks = (160 + 20) = 180
36 % of x = 180
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 15

Question 15.
Solution:
(i) 3 : 4 = (75) %
Explanation:
3 : 4 = \(\frac { 3 }{ 4 }\)
= (\(\frac { 3 }{ 4 }\) x 100) %
= (3 x 25)% = 75%
(ii) 0.75 = (75)%
Explanation : (0.75 x 100)% = 75%
(iii) 6% = 0.06 (express in decimals)
Explanation :
6% = \(\frac { 6 }{ 100 }\) = 0.06
(iv) If x decreased by 40% gives 135, then x = 225
Explanation :
Let the number be x.
According to question, we have :
x – 40% of x = 135
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 16

Question 16.
Solution:
(i) True (T)
RS Aggarwal Class 7 Solutions Chapter 10 Percentage CCE Test Paper 17

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RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C

RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 10 Percentage Ex 10C.

Other Exercises

Objective Questions :
Mark (✓) against the correct answer in each of the following :
Question 1.
Solution:
(b) \(\frac { 3 }{ 4 }\) = \(\frac { 3 }{ 4 }\) x 100 = 75 %

Question 2.
Solution:
(c)
2 : 5 = \(\frac { 2 }{ 5 }\) = \(\frac { 2 }{ 5 }\) x 100 = 40%

Question 3.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 1

Question 4.
Solution:
(c) x% of 75 = 9
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 2

Question 5.
Solution:
(d)
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 3

Question 6.
Solution:
(b)
Let x% of 1 day = 36 minutes
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 4

Question 7.
Solution:
(a)
Let x be the required number
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 5

Question 8.
Solution:
(b)
Let x be the required number, then
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 6

Question 9.
Solution:
(d)
Let ore = x, then
5% of x = 400g
\(\frac { x x 5 }{ 100 }\) 400
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 7

Question 10.
Solution:
(b)
Let gross value of T.V = x
Commission = 10%
After deducting commission, the value
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 8

Question 11.
Solution:
(b)
Increase in salary = 25%
Let original salary = x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 9

Question 12.
Solution:
(c)
Let x be the number of total examinees
No. of examinees passed
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 10

Question 13.
Solution:
(c)
Let total number of apples = x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 11

Question 14.
Solution:
(b)
Present value of machine = Rs. 25000
Rate of depreciation = 10% p.a.
Value of machine after one year
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 12
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 13

Question 15.
Solution:
(c) Let x be numbers
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 14

Question 16.
Solution:
(c) 60% of 450
= \(\frac { 60 }{ 100 }\) x 450 = 270

Question 17.
Solution:
(d) Rate of reduction = 6%
Price after reduction = Rs. 658
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 15

Question 18.
Solution:
(b) Boys = 70% of students
No. of girls = 240
Girls percentage = 100 – 70 = 30%
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 16

Question 19.
Solution:
(c) Let number = x
11% of x – 7% of x = 18
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 17

Question 20.
Solution:
(a) Let number = x
35% of x + 39 = x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 18

Question 21.
Solution:
(c) Pass marks = 36%
A students get =145 marks
But failed by 3 5 marks
Then pass marks = 145 + 35 = 180
Maximum marks = \(\frac { 180 x 100 }{ 36 }\) = 500

Question 22.
Solution:
(d) Let number be = x
Then decreasing by 40%
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 19

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