Maths Class 7 RS Aggarwal Solutions

RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 14 Properties of Parallel Lines Ex 14.

Question 1.
Solution:
A transversal t intersects two parallel lines l and m.
∠ 1 = ∠ 5 (corresponding angles)
But ∠ 5 = 70° (given)
∠ 1 = 70°

But ∠ 3 = ∠ 5 (Alternate angles)
∠ 3 = 70°
∠4 + ∠5 = 180° (Sum of co-interior angles)
⇒ ∠4 + 70° = 180°
⇒ ∠4 = 180° – 70°
⇒ ∠4 = 110°
But ∠ 4 = ∠ 8 (corresponding angles)
∠ 8 = 110°
Hence ∠ 1 = 70°, ∠3 = 70°, ∠4 = 110° and ∠ 8 = 110°

Question 2.
Solution:
A transversal t intersects two parallel lines l and m
∠1 : ∠2 = 5 : 7
But ∠ 1 + ∠ 2 = 180° (Linear pair)


But ∠ 3 = ∠ 1 (vertically opposite angles)
∠ 3 = 75°
∠ 8 = ∠ 4 (corresponding angles)
and ∠ 4 = ∠ 2 (vertically opposite angles)
∠8 = ∠2 = 105°
Hence ∠ 1 = 75°, ∠2 = 105°, ∠3 = 75° and ∠ 8 = 105°

Question 3.
Solution:
A transversal t intersects two parallel lines l and m interior angles of the same side of t are (2x – 8)° and (3x – 7)°
(2x – 8)° + (3x – 7)° = 180° (sum of co-interior angles)
⇒ 2x – 8 + 3x – 7 = 180°
⇒ 5x – 15° = 180°
⇒ 5x = 180° + 15°
⇒ 5x = 195°

⇒ x = \(\frac { 195 }{ 5 }\) = 39°
First angle = 2x – 8° = 2 x 39° – 8° = 78° – 8° = 70°
Second angle = 3x – 7 = 3 x 39° – 7° = 117° – 7° = 110°

Question 4.
Solution:
l || m and two transversals intersect these lines but s is not parallel to t.

∠ 5 = ∠ 1 (vertically opposite angles)
∠ 5 = 50°
But l || m and s the transversal
∠ 5 + ∠ 2 = 180° (sum of co-interior angles)
⇒ 50° + x = 180°
⇒ x = 180° – 50° – 130°
x = 130°
∠ 4 = ∠ 6 (vertically opposite angles)
∠ 6 = y
But l || m and t is the transversal
∠ 6 + ∠ 3 = 180° (sum of co-interior angles)
⇒ y + 65° = 180°
⇒ y = 180° – 65° = 115°
y = 115°
Hence x = 130° and y = 115°

Question 5.
Solution:
In the figure, ABC is a triangle, DAE || BC
∠B = 65°, ∠C = 45°
∠ DAB = x° and ∠ EAC = y°
DAE || BC and AB is transversal
∠ DAB = ∠ B (Alternate angles)

⇒ x° = 65°
Similarly ∠ EAC = ∠ C (Alternate angles)
y° = 45°
Hence x = 65° and y = 45°

Question 6.
Solution:
In ∆ABC, AB || CE
∠BAC = 80°, ∠ECD = 35°
AB || CE and BCD is the transversal
∠ABC = ∠ECD (corresponding angles)

⇒ ∠ABC = 35° (∠ECD = 35°)
Again AB || CE and AC is the transversal
∠ BAC = ∠ ACE (alternate angles)
∠ACE = 80° (∠BAC = 80°)
In ∆ABC
∠A + ∠B + ∠ACB = 180° (Sum of angles of a triangle)
∠ 80° + ∠ 35° + ∠ACB = 180°
⇒ ∠ACB + ∠ 115° = 180°
⇒ ∠ACB = 180° – 115° = 65°
Hence ∠ ACE = 80°, ∠ ACB = 65° and ∠ ABC = 35°

Question 7.
Solution:
In the figure,
AO || CD, DB || CE and ∠AOB = 50°
AO || CD and CD is the transversal
∠ AOB = ∠ CDB (corresponding angles)

∠ CDB = 50° (∠ AOB = 50°)
Similarly CE || OB and CD in transversal
∠ECD + ∠CEB = 180° (sum of co-interior angles)
⇒ ∠ECD + 50° = 180°
⇒ ∠ECD = 180° – 50° = 130°
∠ECD = 130°

Question 8.
Solution:
In the fig, AB || CD
∠ABO = 50° and ∠CDO = 40°
From O, draw EOF || AB or CD
AB || EF and BO is the transversal

∠ABO = ∠ 1 (Alternate angles) …(i)
∠ CDO = ∠ 2 (Alternate angles) …(ii)
Similarly, EF || CD and OD is the transversal
Adding (i) and (ii),
∠ 1 + ∠ 2 = ∠ABO + ∠CDO
⇒ ∠BOD = 50° + 40° = 90°
Hence ∠ BOD = 90°

Question 9.
Solution:
Given : In the figure, AB || CD and EF is a transversal which intersects them at G and H respectively
GL and HM are the angle bisectors or ∠ AGH and ∠ GHD respectively.
To prove : GL || HM.
Proof : AB || CD and EF is a transversal
∠ AGH = ∠ CHD (Alternate angles)
GL is the bisector of ∠ AGH
∠ 1 = ∠2 = \(\frac { 1 }{ 2 }\) ∠ AGH

Similarly, HM is the bisectors of ∠ GHD
∠3 = ∠4 = \(\frac { 1 }{ 2 }\) ∠ GHD
∠ AGH = ∠ GHD (proved)
∠ 1 = ∠3
But, these are alternate angles
BL || HM
Hence proved.

Question 10.
Solution:
In the given figure,
AB || CD
∠ ABE = 120° and ∠ECD = 100° ∠ BEC = x°
From E, draw FG || AB or CD.
AB || EF

∠ABE + ∠1 = 180° (sum of co-interior angles)
⇒ 120° + ∠1 = 180°
⇒ ∠1 = 180°- 120° = 60°
Similarly CD || EG
∠ECD + ∠2 = 180°
⇒ 100° + ∠2 = 180°
⇒ ∠2 = 180° – 100°
∠ 2 = 80°
But ∠1 + ∠x + ∠2 = 180° (Angles on one side of a straight line)
⇒ 60° + x + 80° = 180°
⇒ x + 140° = 180°
⇒ x = 180° – 140° = 40°
x = 40°

Question 11.
Solution:
Given : In the figure, ABCD is a quadrilateral in which AB || DC and AD || BC
To prove : ∠ADC = ∠ABC
Proof : AB || DC and DA is the transversal
∠ADC + ∠ DAB = 180° (co-interior angles)

Similarly, AD || BC and AB is the transversal
∠DAB + ∠ABC = 180° …(ii)
from (i) and (ii),
∠ ADC + ∠ DAB = ∠DAB + ∠ABC
⇒ ∠ADC = ∠ABC
Hence ∠ ADC = ∠ ABC
Hence proved.

Question 12.
Solution:
In the figure,
l || m and p || q.
∠1 = 65°
∠ 2 = ∠ 1 (vertically opposite angles)

∠ 2 = 65°
⇒ a = 65°
p || q and l is the transversal
∠ 2 + ∠ 3 = 180° (co-interior angles)
⇒ a + b= 180°
⇒ 65° + b = 180°
⇒ b = 180° – 65° = 115°
Again l || m and p is the transversal
∠ 3 + ∠4 = 180°
⇒ b + c = 180°
⇒ 115° + c = 180°
⇒ c = 180° – 115° = 65°
l || m and q is the transversal
∠ 2 + ∠ 5 = 180°
⇒ a + d = 180°
⇒ 65° + d = 180°
⇒ d = 180° – 65° = 115°
Hence a = 65°, b = 115°, c = 65° and d = 115°

Question 13.
Solution:
In the given figure, AB || DC and AD || BC and AC is the diagonal of parallelogram ABCD.

∠BAC = 35°, ∠CAD = 40°, ∠ACB = x° and ∠ ACD = y°. .
AB || DC and CA is the transversal
∠ DCA = ∠ CAB (Alternate angles)
⇒ y = 35°
and similarly AD || BC and AC is the transversal
∠ CAD = ∠ ACB (Alternate angles)
⇒ 40° = x°
x = 40° and y = 35°

Question 14.
Solution:
In the figure, AB || CD and CD has been produced to E so that
∠ BAE = 125° ∠ BAC = x°, ∠ ABD = x°, ∠ BDC = y° and ∠ ACD = z°
DAE is a straight line and AB stands on it.

∠ BAD + ∠ BAE = 180° (Linear pair)
⇒ x + 125° = 180°
⇒ x = 180° – 125° = 55°
But ∠ABC = x = 55°
DC || AB and CB is the transversal
∠ABC + ∠ BCD = 180° (co-interior angles)
⇒ x + y = 180°
⇒ 55° + y = 180°
⇒ y = 180° – 55° = 125°
Again DC || AB and DAE is its transversal
∠ CDA = ∠ BAE (corresponding angles).
z = 125°
Hence x = 55°, y = 125° and z = 125°

Question 15.
Solution:
Given : In each figure,
l and m are two lines and t is the transversal
To prove : l || m or not
Proof:
(i) fig. (i)
A transversal t intersects two lines l and m

and ∠ 1 = 40°, ∠2 = 130°
But ∠ 1 + ∠3 = 180° (Linear pair)
⇒ 40° + ∠ 3 = 180°
⇒ ∠3 = 180° – 40° = 140°
l || m,
If ∠ 3 = ∠ 2
⇒ 140° = 130°
Which is not possible.
l is not parallel to m.
(ii) fig. (ii)
Transversal t, intersects l and m and ∠ 1 = 35°, ∠2 = 145°
But ∠ 1 = ∠ 3 (vertically opposite angles).
∠3 = 35°
l || m,
if ∠3 + ∠2 = 180°
if 35° + 145° = 180°

if 180°= 180°
which is true
l || m
(iii) Transversal t, intersects l and m.
∠ 1 = 125°, ∠ 2 = 60°

But ∠ 1 = ∠ 3 (vertically opposite angles)
∠ 3 = 125°
l || m
If ∠3 + ∠2 = 180° (co-interior angles)
If 125° + 60° = 180°
If 185° =180°
which is not possible.
Hence l is not parallel to m.

Hope given RS Aggarwal Solutions Class 7 Chapter 14 Properties of Parallel Lines Ex 14 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 7 Solutions Chapter 13 Lines and Angles Ex 13

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 13 Lines and Angles Ex 13.

Question 1.
Solution:
(i) The given angle = 35°
Let x be its complementary, then
x + 35° = 90°
⇒ x = 90° – 35° = 55°
Complement angle = 55°
(ii) The given angle = 47°
Let x be its complement, then
x + 47° = 90 ⇒ x = 90° – 47° = 43°
Complement angle = 43°
(iii) The given angles = 60°
Let x be its complement angle
x + 60° = 90° ⇒ x = 90° – 60° = 30°
Complement angle = 30°
(iv) The given angle = 73°
Let x be its complement angle
x + 73° = 90°
⇒ x = 90° – 73° = 17°
Complement angle = 17°

Question 2.
Solution:
(i) Given angle = 80°
Let x be its supplement angle, then
x + 80° = 180°
⇒ x = 180° – 80° = 100°
Supplement angle = 100°
(ii) Given angle = 54°
Let x be its supplement angle, then
x + 54° = 180°
⇒ x = 180° – 54° = 126°
Supplement angle = 126°
(iii) Given angle = 105°
let x be its supplement angle, then
x + 105° = 180°
⇒ x = 180° – 105° = 75°
Supplement angle = 75°
(iv) Given angle = 123°
Let x be its supplement angle, then
x + 123° = 180°
⇒ x = 180° – 123° = 57°
⇒ Supplement angle = 57°

Question 3.
Solution:
Let smaller angle =x
Then larger angle = x + 36°
But x + x + 36° = 180° (Angles are supplementary)
2x = 180° – 36°= 144°
x = 72°
Smaller angle = 72°
and larger angle = 72° + 36° = 108°

Question 4.
Solution:
Let angle be = x
Then other supplement angle = 180°- x
x = 180° – x
⇒ x + x = 180°
⇒ 2x = 180°
⇒ x = 90°
Hence angles are 90°, and 90°

Question 5.
Solution:
Sum of two supplementary angles is 180°
If one is acute, then second will be obtuse or both angles will be equal
Hence both angles can not be acute or obtuse
Both can be right angles only

Question 6.
Solution:
In the given figure,
AOB is a straight line and the ray OC stands on it.
∠AOC = 64° and ∠BOC = x°

∠AOC + ∠BOC = 180° (Linear pair)
⇒ 64° + x = 180°
⇒ x = 180° – 64° = 116°
Hence x = 116°

Question 7.
Solution:
AOB is a straight line and ray OC stands on it ∠AOC = (2x – 10)°, ∠BOC = (3x + 20)°

∠AOC + ∠BOC = 180° (Linear pair)
⇒ 2x – 10° + 3x + 20° = 180°
⇒ 5x + 10° = 180°
⇒ 5x = 170°
⇒ x = 34°
∠AOC = (2x – 10)° = 2 x 34° – 10 = 68° – 10° = 58°
∠BOC = (3x + 20)° = 3 x 34° + 20° – 102° + 20° = 122°

Question 8.
Solution:
AOB is a straight line and rays OC and OD stands on it ∠AOC = 65°, ∠BOD = 70° and ∠COD = x

But ∠AOC + ∠COD + ∠BOD = 180° (Angles on one side of the straight line)
⇒ 65° + x + 70° = 180°
⇒ 135° + x = 180°
⇒ x = 180° – 135°
⇒ x = 45°
Hence x = 45°

Question 9.
Solution:
Two straight lines AB and CD intersect each other at O.

∠AOC = 42°
AB and CD intersect each other at O.
∠AOC = ∠BOD (Vertically opposite angles)
and ∠AOD = ∠BOC
But ∠AOC = 42°
∠BOD = 42°
AOB is a straight line and OC stands on it
∠AOC + ∠BOC = 180°
⇒ 42° = ∠BOC = 180°
⇒ ∠BOC = 180° – 42° = 138°
But ∠AOD = ∠BOC (vertically opposite angles)
∠AOD = 138°
Hence ∠AOD = 138°, ∠BOD = 42° and ∠COB =138°

Question 10.
Solution:
Two straight lines PQ and RS intersect at O.
∠POS = 114°
Straight lines,

PQ and RS intersect each other at O
∠POS = ∠QOR (Vertically opposite angles)
But ∠POS = 114°
∠QOR = 114° or ∠ROQ = 114°
But ∠POS + ∠POR = 180° (Linear pair)
⇒ 114° + ∠POR = 180°
⇒ ∠POR = 180° – 114° = 66°
But ∠QOS = ∠POR (vertically opposite angles)
∠QOS = 66°
Hence ∠POR = 66°, ∠ROQ =114° and ∠QOS = 66°

Question 11.
Solution:
In the given figure, rays OA, OB, OC and OD meet at O and ZAOB – 56°,
∠BOC = 100°, ∠COD = x and ∠DOA = 74°

But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360° (Angles at a point)
56° + 100° + x° + 74° = 360°
⇒ 230° + x° = 360°
⇒ x° = 360° – 230° = 130°
⇒ x = 130°

Hope given RS Aggarwal Solutions Class 7 Chapter 13 Lines and Angles Ex 13 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 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

• RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest Ex 12A
• RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest Ex 12B
• RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:
Let the sum be ₹ x

Question 3.
Solution:
P = ₹ 3625, A = ₹ 4495, T = 2 years
S.I. = A – P = ₹ 4495 – ₹ 3625 = ₹ 870

Question 4.
Solution:
P = ₹ 3600, A = ₹ 4410, R = 9%
S.I. = A – P = ₹ 4410 – ₹ 3600 = ₹ 810

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

Question 6.
Solution:
Let the sum be ₹ 4

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

Question 8.
Solution:
(c) ₹ 3500
A = ₹ 3626, R = 6%,

Question 9.
Solution:
(a) 9 months
P = ₹ 6000, A = ₹ 6360
S.I. = A – P = 6300 – 6000 = ₹ 360

Question 10.
Solution:
(c) 12%
Let the sum be ₹ x

Question 11.
Solution:

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

Question 13.
Solution:

Question 14.
Solution:
(i) False

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

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

• RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest Ex 12A
• RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest Ex 12B
• RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest CCE Test Paper

Question 1.
Solution:
Principal (P) = Rs. 6250
Rate of (R) = 4% p.a.
Period (T) = 6 months = \(\frac { 1 }{ 2 }\) year

Question 2.
Solution:
Amount (A) = Rs. 3605
Rate (R) = 5% p.a.

Question 3.
Solution:
Let sum (P) = Rs. 100

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.

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

Question 6.
Solution:
S.I. = Rs. x
Rate (R) = x% p.a.
Time (T) = x year

Rs. \(\frac { 100 }{ x }\) (c)

Question 7.
Solution:
Let sum (P) = Rs. 100

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

Question 9.
Solution:
Amount (A) = Rs. 720
Principal (P) = Rs. 600
S.I. = A – P = Rs. 720 – 600 = Rs. 120
Period (T) = 4 years

Question 10.
Solution:

Question 11.
Solution:
Let sum (P) = Rs. 100
Their S.I. = 0.125 of Rs. 100

Question 12.
Solution:
S.I. = Rs. 210

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

• RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest Ex 12A
• RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest Ex 12B
• RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest CCE Test Paper

Find the simple interest and the amount when :
Question 1.
Solution:
Principal (P) = Rs. 6400
Rate (r) = 6% p.a.
Time (t) = 2 years

Question 2.
Solution:
Principal (P) = Rs. 2650
Rate (r) = 8% p.a.

Question 3.
Solution:
Principal (P) = Rs. 1500
Rate (r) = 12% p.a.

Question 4.
Solution:
Principal (P) = Rs. 9600

Question 5.
Solution:
Principal (P) = Rs. 5000
Rate (r) = 9% p.a.

Find the time when :
Question 6.
Solution:
Principal (P) = Rs. 6400
S.I. = Rs. 1152
Rate (r) = 6% p.a.

Question 7.
Solution:
Principal (P) = Rs. 9540
S.I. = Rs. 1908
Rate (r) = 8% p.a.

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.

Find the rate when :
Question 9.
Solution:
Principal (P) = Rs. 8250
S.I. = Rs. 1100
Time (t) = 2 years

Question 10.
Solution:
Principal (P) = Rs. 5200
S.I. = Rs. 975

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

Question 12.
Solution:
Principal (P) = Rs. 6000
Rate (r) = 12% p.a.

Question 13.
Solution:
Principal = Rs. 12600
Rate (A) = 15% p.a.
Time (t) = 3 years

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

Question 15.
Solution:
Amount (A) = Rs. 3920

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

Question 17.
Solution:
Amount = Rs. 12122
Let principal (P) = Rs. 100
Rate (r) = 8% p.a.
Time (t) = 2 years

Question 18.
Solution:
Amount (A) = Rs. 4734
Principal (P) = Rs. 3600
S.I. = A – P = Rs. 4734 – Rs. 3600 = Rs. 1134

Question 19.
Solution:
In first case,
Amount (A) = Rs. 768
Principal (P) = Rs. 640
S.I. = A – P = Rs. 768 – 640 = Rs. 128

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.

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

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

Question 23.
Solution:
Amount in 5 years = Rs. 5475
Amount in 3 years = Rs. 4745
Interest for 2 years = Rs. 5475 – 4745 = Rs. 730

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

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.

⇒ 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

Hope given RS Aggarwal Solutions Class 7 Chapter 12 Simple Interest Ex 12A 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.