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These NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.3

Question 1.
Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent ¡n each case,
(a) Gardening shears bought for 250 and sold for ₹ 325 .
(b) A refrigerator bought for ₹ 12,000 and sold at 13,500.
(c) A cupboard bought for ₹ 2,500 and sold at 3,000.
(d) A skirt bought for ₹ 250 and sold at ₹ 150.
Answer:
(a) Cost Price (CP) = ₹ 250
Selling Price = ₹ 325
Profit = S.P – C.P
= ₹ 325 – 250
= ₹ 75
Profit% = \(\frac{75}{250}\) x 100%
= 3 x 100%
= 30%
Profit = ₹ 75; Profit % = 30%.

(b) Cost Price (CP) = ₹ 12,000
Selling Price (SP) = ₹ 13,500
Profit = SP – CP
= ₹ 13500 – ₹ 12000
= ₹ 1500
Profit% = \(\frac{\text { Profit }}{\text { C.P. }}\) x 100
= \(\frac{1500}{12000} \times\) x 100%
= \(\frac{150}{12} \%\)
= 12.5%
Profit = ₹ 1500 and Profit% = 12.5%.

(c) Cost Price (CP) = ₹ 2500
Selling Price (SP) = ₹ 3000
Profit = SP – CP
= ₹ 3000 – 2500
= ₹ 500
Profit% = \(\frac{\text { Profit }}{\text { C.P. }}\) x 100
= \(\frac{500}{2500}\)
= \(\frac{5}{25}\) x 100
= 5 x 4 = 20%
Profit = ₹ 500
and Profit% = 20%.

(d) Cost Price (CP) = ₹ 250
Selling Price (SP) = ₹ 150
Loss = CP – S.P
= ₹ 250 – 150
= ₹ 1oo
Loss% = \(\begin{aligned}
&\text { Loss } \\
&\hline \text { C.P. }
\end{aligned}\) x 100
= ₹ \(\frac{100}{250}\) x 100%
= \(\frac{10}{25}\) x 100%
= 10 x 4 = 40%
Loss = ₹ 100; loss% = 40%.

Question 2.
Convert each part of the ratio to percentage:
(a) 3 : 1
(b) 2 : 3 :5
(c) 1 : 4
(d) 1 : 2 : 5
Answer:
(a) 3:1
Total parts of the ratio = 3 + 1 = 4
Percentage of the first part
= \(\frac { 3 }{ 4 }\) x 100% = 3 x 25% = 75%
Percentage of the second part
= \(\frac { 1 }{ 4 }\) x 100 = 25%

(b) 2:3:5
Total parts of the ratio
= 2 + 3 + 5 = 10
Percentage of the first part
= \(\frac { 2 }{ 10 }\) x 100%
= 2 x 10% = 20%
Percentage of the second part
= \(\frac { 3 }{ 10 }\) x 100% = 3 x 10% = 30%
Percentage of the third part
= \(\frac { 5 }{ 10 }\) x 100% = 5 x 10% = 50%
1st part = 20%, 2nd part = 30%; 3rd part = 50%

(c) 1:4
Total parts of the ratio = 1+4 = 5
Percentage of the first part
= \(\frac { 1 }{ 5 }\) x 100% = 20%
Percentage of the second part
= \(\frac { 4 }{ 5 }\) x 100% = 80%
Percentage of first part = 20%
Percentage of second part = 80%

(d) 1:2:5
Total parts of the ratio = 1 + 2 + 5 = 8
Percentage of the first part
= \(\frac { 1 }{ 8 }\) x 100% = \(\frac { 25 }{ 2 }\)% = 12\(\frac { 1 }{ 2 }\) % or 12.5%
Percentage of the second part
= \(\frac { 2 }{ 8 }\) x 100% = 25%
Percentage of the third part
= \(\frac { 5 }{ 8 }\) x 100% = \(\frac { 125 }{ 2 }\) %
= 62\(\frac { 1 }{ 2 }\)% or 62.5%
Percentage of the first part = 12\(\frac { 1 }{ 2 }\) %
Percentage of the second part = 25%
Percentage of the third part = 62\(\frac { 1 }{ 2 }\) %
= \(\frac { 2 }{ 35 }\) x 100%
= \(\frac{2 \times 20}{7} \%=\frac{40}{7} \%=5 \frac{5}{7} \%\)
Increase percentage = 5\(\frac{5}{7}\)

Question 3.
The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Answer:
Initial population = 25000.
Decreased population = 24500
Decrease in population = 25000 – 24500 = 500
Percentage of decrease = \(\frac{\text { Decrease in population }}{\text { Initial population }}\) x 100
= \(\frac{500}{25000}\) x 100% = \(\frac{50 \times 1}{25}\) = 2%

Question 4.
Arun bought a car for ₹ 3,50,000. The next year, the price went upto ₹ 3,70,000. What was the percentage of price increase?
Answer:
Initial cost = ₹ 3,50,000
Increased cost = ₹ 3,70,000
Increase in price = ₹ 3,70,000 – 3,50,000
= ₹ 20,000
Increase% = \(\frac{\text { Increase in amount }}{\text { Initial cost }}\) x 100
= \(\frac{20,000}{3,50,000}\) x 100%
= \(\frac{2}{35}\) x 100%
= \(\frac{2 \times 20}{7} \%=\frac{40}{7} \%=5 \frac{5}{7} \%\)
Increase percentage = 5\(\frac{5}{7} \%\)

Question 5.
I buy a T.V. for ₹ 10,000 and sell it at a profit of 20%. How much money do I get for it?
Answer:
Cost price of the T.V = ₹ 10,000
Profit = 20% of C.P = \(\frac{20}{100}\) x 10,000
= \(\frac { 1 }{ 5 }\) x 10000 = ₹ 2000 5
Selling price of the T.V = ₹ 10,000 + ₹ 2000 = ₹ 12,000
Hence, I get ₹ 12,000

Question 6.
Juhi sells a washing machine for ₹ 13,500. She loses 20% in the bargain. What was the price at which she bought it?
Answer:
Let the cost price of the washing machine be Y
Selling price of the washing machine = ₹ 13,500
Loss = 20% of C.P 20
= \(\frac{20}{100}\) x x
= \(\frac{1}{5}\) x x = \(\frac{x}{5}\)
Selling price = Cost price – Loss
13500 = x – \(\frac{x}{5}\)
= \(\frac{5 x-x}{5}=\frac{4 x}{5}\)
\(\frac{4 \mathrm{x}}{5}\) = 13500
x = \(\frac{13500 \times 5}{4}\)
= ₹ 3,375 x 5 = ₹ 16,875
She bought it for ₹ 16,875.

Question 7.
(i) Chalk contains calcium, carbon and oxygen in the ratio 10 : 3 : 12. Find the percentage of carbon in chalk.
(ii) If in a stick of chalk, carbon is 3 g, what is the weight of the chalk stick?
Answer:
(i) The ratio of calcium : carbon :
oxygen = 10 : 3 : 12
Total ratios = 10 + 3 + 12 = 25
Percentage of carbon in chalk
= \(\frac { 3 }{ 25 }\) x 100% = 3 x 4% = 12%

(ii) Let the weight of the stick of chalk be ‘x’ g
12% of the chalk mixture is carbon 12% of x = 3
\(\frac{12}{100}\) x x = 3
x = \(\frac{3 \times 100}{12}=\frac{100}{4}\) = 25g
Weight of the chalk stick = 25 g

Question 8.
Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?
Answer:
Cost price of the book = ₹ 275
Loss = 15% of C.P
S.P of the book = ₹ 275 – ₹ 41.25
= \(\frac{15}{100}\) x 275 = ₹ 233.75
= ₹ \(\frac{15}{4}\) x 11
S.P of the book = ₹ 233.75
= ₹ 41.25

Question 9.
Find the amount to be paid at the end of 3 years in each case:
(a) Principal = ₹ 1200 at 12% p.a.
(b) Principal = ₹ 7500 at 5% p.a.
Answer:
(a) Principal (P) = ₹ 1200
Rate (R) = 12%
Time (T) = 3 years
T ,T. P x R x T
Interest (I) = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹ \(\frac{1200 \times 12 \times 3}{100}\)
= ₹ 12 x 12 x 3
= ₹ 432
Amount = Principal + Interest
= ₹ 1200 + 432
= ₹ 1632
Amount to be paid at the end of 3 years = ₹ 1632

(b) Principal (P) = ₹ 7500
Rate (R) = 5% p.a.
Time (T) = 3 years
Interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹ \(\frac{7500 \times 5 \times 3}{100}\)
= ₹ 75 x 5 x 3
= ₹ 1125
Amount= Principal + Interest
= ₹ 7500+ ₹ 1125 = ₹ 8625
Amount to be paid = ₹ 8625

Question 10.
What rate gives ₹ 280 as interest on a sum of ₹ 56,000 is 2 years?
Answer:
Principal (P) = ₹ 56000
280 = 560 x R x 2
Rate (R) = R
R = \(\frac{280}{560 \times 2}\) %
Time (T) = 2 years
Interest (I) = ₹ 280
R = \(\frac{1}{2 \times 2} \%=\frac{1}{4} \%\)
or 0.25%
Interest (I) = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
Thus, rate of interest = \(\frac{1}{4}\) % or 0.25%
280 = \(\frac{56000 \times \mathrm{R} \times 2}{100}\)

Question 11.
If Meena gives an interest of ₹ 45 for one year at 9% rate p.a. What is the sum she has borrowed?
Answer:
Let Principal be ‘P’
Rate (R) = 9% p.a.
45 = \(\frac{\mathrm{P} \times 9 \times 1}{100}\)
Interest (I) = ₹ 45
P x 9 = 45 x 100
Time (T) = 1 year
P = \(\frac{45 \times 100}{9}\)
Simple interest = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= ₹ 500
The sum borrowed is ₹ 500

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.3

Question 1.
Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q + r
(ii) ab, a – b
(iii) a + b, 7a²b²
(iv) a² – 9, 4a
(v) pq + qr + rp, 0
Solution:
(i) (4p) × (q + r)
= (4p × q) + (4p × r)
= 4pq + 4pr

(ii) (ab) × (a – b)
= (ab × a) – (ab × b)
= a²b – ab²

(iii) (a + b) (7a²b²)
= (a × 7a²b²) + (b × 7a²b²)
= 7a³b² + 7a²b³

(iv) (a² – 9) × 4a
= (a² × 4a) – (9 × 4a)
= 4a³ – 36a

(v) (pq + qr + rp) × 0
= (pq × 0) + (qr × 0) + (rp × 0)
= 0 + 0 + 0
= 0

Question 2.
Complete the table.

Solution:
(i) a × (b + c + d)
= (a × b) + (a × c) + (a × d)
= ab + ac + ad

(ii) (x + y – 5) (5xy)
= (x × 5xy) + (y × 5xy) – (5 × 5xy)
= 5x²y + 5xy² – 25xy

(iii) p × (6p² – 7p + 5)
= (p × 6p²) + p × (-7p) + p × 5
= 6p³ + (-7p²) + 5p
= 6p³ – 7p² + 5p

(iv) 4p²q² × (p² – q²)
= (4p²q² × p²) + (4p²q² × -q²)
= 4p⁴q² + (-4p²q⁴)
= 4p⁴q² – 4p²q⁴

(v) (a + b + c) × abc
= (a × abc) + (b × abc) + (c × abc)
= a²bc + ab²c + abc²

Question 3.
Find the product.
(i) (a²) × (2a²²) × (4a²⁶)
(ii) \(\left(\frac{2}{3} x y\right) \times\left(\frac{-9}{10} x^{2} y^{2}\right)\)
(iii) \(\left(-\frac{10}{3} \mathrm{pq}^{3}\right) \times\left(\frac{6}{5} \mathrm{p}^{3} \mathrm{q}\right)\)
(iv) x × x² × x³ × x⁴
Solution:
(i) (a²) × (2a²²) × (4a²⁶)
= (1 × 2 × 4) × (a² × a²² × a²⁶)
= 8 × (a^2+22+26)
= 8a⁵⁰

(iv) x × x² × x³ × x⁴
= x^1+2+3+4
= x¹⁰

Question 4.
(a) Simplify 3x(4x – 5) + 3 and find its values for
(i) x = 3
(ii) x = \(\frac{1}{2}\)
(b) Simplify a (a² + a + 1) + 5 and find its value for
(i) a = 0
(ii) a = 1
(iii) a = -1.
Solution:
(a) 3x (4x – 5) + 3
= (3x × 4x) + (3x × -5) + 3
= 12x² – 15x + 3
(i) when x = 3
12x² – 15x + 3
= 12 × (3)² – 15(3) + 3
= 12 × 9 – 45 + 3
= 108 – 45 + 3
= 111 – 45
= 66

(ii) When x = \(\frac{1}{2}\)
12x² – 15x + 3

(b) a(a² + a + 1) + 5
= (a × a²) + (a × a) + (a × 1) + 5
= a³ + a² + a + 5

(i) When a = 0
a³ + a² + a + 5
= (0)³ + (0)² + 0 + 5
= 0 + 0 + 0 + 5
= 5

(ii) When a = 1
a³ + a² + a + 5
= (1³) + (1²) + 1 + 5
= 1 + 1 + 1 + 5
= 8

(iii) When a = -1
a³ + a² + a + 5
= (-1)³ + (-1)² + (-1) + 5
= (-1) + 1 – 1 + 5
= -1 + 1 – 1 + 5
= 4

Question 5.
(a) Add: p(p – q), q(q – r) and r (r – p)
(b) Add: 2x(z – x – y) and 2y(z – y – x)
(c) Subtract: 3l(l – 4m + 5n) from 4l(10n – 3m + 2l)
(d) Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c (-a + b + c)
Solution:
(a) p(p – q) + q(q – r) + r(r – p)
= p × p – p × q + q × q + q(-r) + r × r + r × (-p)
= p² – pq + q² – qr + r² – rp
= p² + q² + r² – (pq + qr + rp)

(b) 2x (z – x – y) + 2y(z – y – x)
= 2x × z + 2x (-x) + 2x (-y) + 2yz + 2y (-y) + 2y(-x)
= 2xz – 2x² – 2xy + 2yz – 2y² – 2xy
= -2x² – 2y² – 4xy + 2yz + 2xz

(c) 4l(10n – 3m + 2l) – 3l(l – 4m + 5n)
= (4l × 10n) + (4l × -3m) + (4l × 2l) + (-3l × 1) + (-3l × -4m) + (-3l × 5n)
= 40ln + (-12lm) + 8l² + (-3l²) + 12lm + (-15ln)
= 40ln – 12lm + 8l² – 3l² + 12lm – 15ln
= 40ln – 15ln – 12lm + 12lm + 8l² – 3l²
= 25ln + 5l²
= 5l² + 25ln

(d) Simplify the I part we get 3a (a + b + c) – 2b(a – b + c)
= (3a × a) + (3a × b) + (3a × c) – [(2b × a) + 2b(-b) + (2b) × c]
= 3a² + 3ab + 3ac – (2ab – 2b² + 2bc)
= 3a² + 3ab + 3ac – 2ab + 2b² – 2bc
= 3a² + 2b² + 3ab – 2ab + 3ac – 2bc
= 3a² + 2b² + ab + 3ac – 2bc
Simplify the 2nd part 4c × (-a + b + c) = (4c × – a) + (4c × b) + (4c × c)
According to the given question = -4ac + 4bc + 4c²
2nd part – 1st part
= -4ac + 4bc + 4c² – (3a² + 2b² + ab + 3ac – 2bc)
= -4ac + 4bc + 4c² – 3a² – 2b² – ab – 3ac + 2bc
= -4ac – 3ac + 4bc + 2bc + 4c² – 3a² – 2b² – ab
= -7ac + 6bc + 4c² – 3a² – 2b² – ab
= -3a² – 2b² + 4c² – ab + 6bc – 7ac

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.2

Question 1.
Find the product of the following pairs of monomials
(i) 4, 7p
(ii) -4p, 7p
(iii) -4p, 7pq
(iv) 4p³, – 3p
(v) 4p, 0
Solution:
(i) 4 × 7p
= (4 × 7)p
= 28p

(ii) -4p × 7p
= {(-4) × 7} × (p × p)
= (-28) × p²
= -28p²

(iii) -4p × 7pq
= {(-4) × 7} × {p × (pq)}
= -28 × (p × p × q)
= -28p²q

(iv) 4p³ × -3p
= {4 × (-3)} × (p³ × p)
= -12 × (p⁴)
= -12p⁴

(v) 4p × 0
= (4 × 0) × p
= 0 × p
= 0

Question 2.
Find the areas of rectangles, with the following pairs of monomials as their lengths and breadths respectively.
(p, q); (10m, 5n); (20x², 5y²); (4x, 3x²); (3mn, 4np)
Solution:
(i) (p, q)
Area of the rectangle = length × breadth
= p x q
= pq

(ii) (10m, 5n)
Area of the rectangle = length × breadth
= 10m × 5n
= (10 × 5) × (m × n)
= 50 × mn
= 50mn

(iii) 20x², 5y²
Area of the rectangle = length × breadth
= 20x² × 5y²
= (20 × 5) × (x² × y²)
= 100 × x²y²
= 100x²y²

(iv) (4x, 3x²)
Area of the rectangle = length × breadth
= 4x × 3x²
= (4 × 3) × (x × x²)
= 12 × x³
= 12x³

(v) (3mn, 4np)
Area of the rectangle = length × breadth
= 3mn × 4np
= (3 × 4) × (mn × np)
= 12 × m × (n × n) × p
= 12 × m × n² × p
= 12mn²p

Question 3.
Complete the table products.

Solution:

Question 4.
Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) 5a, 3a², 7a4
(ii) 2p, 4q, 8r
(iii) xy, 2x²y, 2xy²
(iv) a, 2b, 3c
Solution:
(i) 5a, 3a², 7a⁴
Volume of the rectangular box = length × breadth × height
= (5a) × (3a²) × (7a⁴)
= (5 × 3 × 7) × (a × a² × a⁴)
= 105a⁷

(ii) 2p, 4q, 8r
Volume of the rectangular box = Length × Breadth × Height
= 2p × 4q × 8r
= (2 × 4 × 8) × (p × q × r)
= 64pqr

(iii) xy; 2x²y; 2xy²
Volume of the rectangular box = length × breadth × height
= xy × 2x²y × 2xy²
= (1 × 2 × 2) × (x × x² × x) × (y × y × y²)
= 4 × x⁴ × y⁴
= 4x⁴y⁴

(iv) a, 2b, 3c
Volume of the rectangular box = length × breadth × height
= a × 2b × 3c
= (1 × 2 × 3) × (a × b × c)
= 6abc

Question 5.
Obtain the product of
(i) xy, yz, zx
(ii) a, -a², a³
(iii) 2, 4y, 8y², 16y³
(iv) a, 2b, 3c, 6abc
(v) m, -mn, mnp
Solution:
(i) xy, yz, zx
(xy) × (yz) × (zx)
= (x × x) × (y × y) × (z × z)
= x² × y² × z²
= x²y²z²

(ii) a; -a²; a³
(a) × (-a²) × (a³)
= -(a × a² × a³)
= -a⁶

(iii) 2, 4y, 8y², 16y³
(2) × (4y) × (8y²) × (16y³)
= (2 × 4 × 8 × 16) × (y × y² × y³)
= 1024y⁶

(iv) a, 2b, 3c, 6abc
(a) × (2b) × (3c) × (6abc)
= (2 × 3 × 6) × (a × a) × (b × b) × (c × c)
= 36 × a² × b² × c²
= 36a²b²c²

(v) m, -mn, mnp
(m) × (-mn) × (mnp)
= -1 × (m × m × m) × (n × n) × p
= -1 × m³ × n² × p
= -m³n²p

These NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Exercise 7.2

Question 1.
Which congruence criterion do you use in the following?
(a) Given : AC = DF
AB = DE
BC = EF
So, ΔABC ≅ ΔDEF

(b) Given : ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ΔPQR ≅ ΔXYZ

(c) Given:
∠MLN = ∠FGH
∠NML = ∠GFH
ML = FG
So, ΔLMN ≅ ΔGFH

(d) Given : EB = DB
∠A = ∠C = 90°
AE = BC
So, ΔABE ≅ ΔCDB

Answer:
(a) Given: AC = DF
AB = DE
BC = EF
So, ΔABC ≅ ΔDEF
By SSS Congruence Criterion

(b) Given: ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ΔPQR ≅ ΔXYZ
By SAS Congruence Criterion

(c) Given:
∠MLN = ∠FGH
∠NML = ∠GFH
ML = FG
ΔLMN ≅ ΔGFH
By ASA Congruence Criterion

(d) Given: EB = DB
AE = BC
∠A = ∠C = 90°
So, ΔABE ≅ ΔCDB
By RHS Congruence Criterion.

Question 2.
You want to show that ΔART ≅ ΔPEN,
(a) If you have to use SSS criterion, then you need to show

(i) AR =
(ii) RT =
(iii) AT =

(b) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have
(i) RT = and
(ii) PN =

(c) If it is given that AT = PN and you are to use ASA criterion, you need to have
(i) ∠RAT =
(ii) ∠ATR =
Answer:
ΔART ≅ ΔPEN
(a) (i) AR = PE
(ii) RT = EN
(iii) AT = PN

(b) Given: ∠T = ∠N
(i) RT = EN
(ii) PN = AT

(c) (i) ∠RAT = ∠EPN
(ii) ∠ATR = ∠PNE

Question 3.
You have to show that ΔAMP = AMQ. In the following proof, provide the missing reasons.

Answer:

Question 4.
In ΔABC, ∠A =30°, ∠B = 40° and ∠C= 110°
In APQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°
A Student says that ΔABC = ΔPQR by AAA congruence criterion. Is he justified? Why or why not?
Answer:
No, he is not justified because AAA is not a congruence criterion.

Question 5.
In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ΔRAT ≅ ?

Answer:
We have N ↔T
O ↔ A
W ↔ R
∴ ΔRAT ≅ ΔWON

Question 6.
Complete the congruence statement:
ΔBCA ≅ ?

ΔQRS ≅ ?

(i) We have
A ↔ A
B ↔ B
T ↔ C
∴ ΔBCA ≅ ΔBTA

(ii) We have
R ↔ P
Q ↔ T
S ↔ Q
∴ ΔQRS ≅ ΔTPQ

Question 7.
In a squared sheet, draw two triangles of equal areas such that
(i) the triangles are congruent.
(ii) the triangles are not congruent.
What can you say about their perimeters?
Answer:

(i) Area of ΔABC = \(\frac { 1 }{ 2 }\) × 4 × 3 sq cm = 6 sq cm
Area of ΔCDE = = 6 sq cm
Perimeter of ΔABC = (3 + 4 + 5) cm = 12 cm
Perimeter of ΔCDE = (3 + 4 + 5) cm = 12 cm
The two triangles are congruent.
(∵ Perimeter of ΔABC = Perimeter of ΔCDE)
(ii)

Perimeter of ΔPQR = (3 + 4 + 5) cm = 12cm
Perimeter of ΔPRS = (4 + 3.5 + 4) cm = 11.5 cm
∴ The two triangles are not congruent.
(∵ Perimeter of ΔPRS ≠ Perimeter of ΔPQR)

Question 8.
Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.
Answer:
In ΔABC and ΔDEF

AB = 2 cm DF = 2cm
∴ AB = DF
BC = 4 cm, ED = 4 cm
∴ BC = ED
AC = 3 cm, EF = 3 cm
∴ AC = EF
∠BAC = ∠EDF
∠ABC = ∠DEF
But ΔABC is not congruent to ΔDEF.

Question 9.
If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Answer:
Given ΔABC = ΔPQR
∴ A ↔ P; B ↔ Q and C ↔ R
Two angles ∠B and ∠C of ΔABC are respectively equal to two angles ∠Q and
∠R of ΔPQR
If BC = QR then ΔABC ≅ ΔPQR (using ASA congruence criterion)
We use ASA congruence criterion.

Question 10.
Explain why AABC = AFED

Answer:
ZB = ZE (each 90°)
ZA = ZF (Given)
ZC = ZD
(3^rd angle are equal) BC = ED (Given)
Two angles ZB and ZC and included side BC of AABC are respectively equal to the angle ZE and ZD and the included side ED of ADEF.
∴ ΔABC ≅ ΔFED (ASA)

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.1

Question 1.
Identify the terms, their coefficients for each of the following expressions.
(i) 5xyz² – 3zy
(ii) 1 + x + x²
(iii) 4x²y² – 4x²y²z² + z²
(iv) 3 – pq + qr – rp
(v) \(\frac{x}{2}+\frac{y}{2}\) – xy
(vi) 0.3a – 0.6ab + 0.5b
Solution:
(i) 5xyz² – 3zy
Terms are 5xyz² and -3zy
Coefficients of 5xyz² is 5
Coefficients of-3zy is -3

(ii) 1 + x + x²
Terms are x², x, and 1
Coefficient of x² is 1
coefficient of x is 1
and constant term is 1

(iii) 4x²y² – 4x²y²z² + z²
Terms are 4x²y², -4x²y²z² and z²
Coefficient of x²y² is 4
Coefficient of -4x²y²z² is -4
Coefficient of z² is 1

(iv) 3 – pq + qr – rp
Terms are 3, -pq, qr and -rp
Coefficient of 3 is 3.
Coefficient of -pq is -1.
Coefficient of qr is 1 and coefficient of -rp is -1

(v) \(\frac{x}{2}+\frac{y}{2}\) – xy
Terms are \(\frac{x}{2}, \frac{y}{2}\) and -xy
Coefficient of \(\frac{\mathrm{x}}{2}\) is \(\frac{1}{2}\)
Coefficient of \(\frac{\mathrm{x}}{2}\) is \(\frac{1}{2}\)
and coefficient of -xy is -1

(vi) 0.3a – 0.6ab + 0.5b
Terms are 0.3a, -0.6ab and 0.5b
Coefficient of 0.3a is 0.3
Coefficient of -0.6ab is -0.6
Coefficient of 0.5b is 0.5

Question 2.
Classify the following polynomial as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
x + y, 1000, x + x² + x³ + x⁴, 7 + y + 5x, 2y – 3y², 2y – 3y² + 4y³, 5x – 4y + 3xy, 4z – 15z², ab + bc + cd + da, pqr, p²q + pq², 2p + 2q
Solution:
Monomials: 100pqr
Binomials: x + y; 2y – 3y²; 4z – 15z²; p²q + pq²; 2p + 2q
Trinomials: 7 + y + 5x; 2y – 3y² + 4y³; 5x – 4y + 3xy
Polynomials that do not fit in these categories: x + x² + x³ + x⁴ and ab + bc + cd + da
(Since the above polynomials has four terms)

Question 3.
Add the following.
(i) ab – bc, bc – ca, ca – ab
(ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p²q² – 3pq + 4, 5 + 7pq – 3p²q²
(iv) l² + m², m² + n², n² + l², 2lm + 2mn + 2nl
Solution:
(i) ab – bc; bc – ca; ca – ab

(ii) a – b + ab; b – c + bc; c – a + ac

(iii) 2p²q² – 3pq + 4; 5 + 7pq – 3p²q²

(iv) l² + m²; m² + n²; n² + l²; 2lm + 2mn + 2nl

= 2(l² + m² + n² + lm + mn + nl)

Question 4.
(a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p²q – 3pq + 5pq² – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq² + 5p²q
Solution:

These NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities Exercise 8.2

Question 1.
Convert the given fractional numbers to per cents
(i) \(\frac{1}{9}\)
(b) \(\frac{5}{4}\)
(c) \(\frac{3}{40}\)
(d) \(\frac{2}{7}\)
Answer:

Question 2.
Convert the given decimal fractions to per cents.
(a) 0.65
(b) 2.1
(c) 0.02
(d) 12.35
Answer:
(a) 0.65 : \(\frac{64}{100}=\frac{65}{100}\) × 100% = 65%
(b) 2.1 = \(\frac{21}{10}=\frac{21}{10}\) × 100% = 210%
(c) 0.02 = \(\frac{2}{100}=\frac{2}{100}\) × 100% =2 %
(d) 12.35 = \(\frac{1235}{100}=\frac{1235}{100}\) × 100% = 1235

Question 3.
Estimate what part of the figures is shaded and hence find the per cent which is shaded.

Answer:
(i) \(\frac{1}{4}\) part is shaded.
\(\frac{1}{4}=\frac{1}{4}\) × 100% = % = 25 %
This coloured part is 25%.

(ii) 3 parts out of 5 parts are shaded so,
\(\frac{3}{5}\) part is shaded
\(\frac{3}{5}=\frac{3}{5}\) × 100% = 3 × 20% = 60%
This coloured part is 60%.

(iii) Here 3 parts out of 8 parts are shaded 3
so, \(\frac{3}{8}\) part is shaded.
\(\frac{3}{8}=\frac{3}{8}\) × 100% = \(\frac{3}{2}\) x 25%
= 37\(\frac { 1 }{ 2 }\)% or 37.5%
This 37\(\frac { 1 }{ 2 }\) part is shaded.

Question 4.
Find:
(a) 15% of 250
(b) 1% of 1 hour
(c) 20% of ₹ 2500
(d) 75% of 1 kg
Answer:
(a) 15% of 250 = \(\frac{15}{100}\) of 250
= \(\frac{15}{100} \times 250=\frac{15 \times 250}{100}=\frac{75}{2}\)
= 37\(\frac{1}{2}\) or 37.5
∴ 15% of 250 = 37 \(\frac{1}{2}\) or 37.5 2

(b) 1% of 1 hour = \(\frac{1}{100}\) × 60 minutes
= \(\frac{1 \times 60}{100}=\frac{3}{5}\) minutes
= \(\frac { 3 }{ 5 }\) × 60 seconds = 36 seconds
∴ 1% of 1 hour = 36 seconds

(c) 20% of ₹ 2500 = \(\frac { 20 }{ 100 }\) of
2500 = \(\frac { 20 }{ 100 }\) x 2500
= ₹ \(\frac{20 \times 2500}{100}\) = ₹ 500
∴ 20% of ₹ 2500 = ₹ 500

(d) 75% of 1 kg = 75% of 1000 g
= \(\frac{75}{100} \times 100 \mathrm{~g}=\frac{75 \times 1000}{100}\) = 750 g
∴ 75% of 1 kg = 750g

Question 5.
Find the whole quantity if
(a) 5% of it is 600.
(b) 12% of it is ₹ 1080.
(c) 40% of it is 500 km.
(d) 70% of it is 14 minutes.
(e) 8% of it is 40 litres.
Answer:
(a) 5% of a quantity is 600.
Let the quantity be x
5% of x = 600
\(\frac { 5 }{ 100 }\) × x = 600
x = \(\frac{600 \times 100}{5}\)
= 120 × 100
= 12000
Thus the required quantity is 12000.

(b) 12% of it is ₹ 1080
Let the required amount be x
12% of x = 1080
\(\frac { 12 }{ 100 }\) × x = 1080
x = \(\frac{1080 \times 100}{12}\)
= \(\frac{1080 \times 25}{3}\)
The required amount = ₹ 9000
= 360 × 25 = ₹ 9000

(c) 40% of it is 500 km
Let the total quantity be x
40% of x = 500
\(\frac { 4 }{ 100 }\) × x = 500
x = \(\frac{500 \times 100}{40}\) km
= 50 × 25 km
= 1250 km
The required quantity is 1250 km.

(d) 70% of it is 14 minutes
Let the required time be x
70% of x = 14 minutes
\(\frac{70}{100}\) × x = 14
x = \(\frac{14 \times 100}{70}\)
= 20 minutes
Thus, the required quantity is 20 minutes.

(e) 8% of a quantity is 40 litres
Let the quantity be x
8% of x =40 litres
\(\frac { 8 }{ 100 }\) × x = 40
x = \(\frac{40 \times 100}{8}\) = 500litres
The required quantity is 500 litres.

Question 6.
Convert given per cents to decimal fractions and also to fractions in simplest forms:
(a) 25%
(b) 150%
(c) 20%
(d) 5%
Answer:
(a) 25% = \(\frac{25}{100}=\frac{1}{4}\)
Thus, 20% = \(\frac{1}{4}\) = 0.25

(b) 150% = \(\frac{150}{100}=\frac{3}{2}\)
Thus 150% = \(\frac{3}{2}\) (or) 1\(\frac{1}{2}\) = 1.5

(c) 20% = \(\frac{20}{100}=\frac{1}{5}\) = 0.2
Thus 20% = \(\frac{1}{5}\) = 0.2

(d) 5% = \(\frac{5}{100}=\frac{1}{20}\)
Thus 5% = \(\frac{1}{20}\) = 0.05

Question 7.
In a city, 30% are females, 40% are males and remaining are children. What per cent are children?
Answer:
Females are 30% and males are 40%
Number of children = 100% – (30% + 40%)
= 100% – 70% = 30%
Thus, children are 30% of the population.

Question 8.
Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?
Answer:
Total number of voters = 15,000
Part of voters who voted = 60%
Part of voters who did not vote = 100% – 60% = 40%
40% of 15000 = \(\frac{40}{100}\) × 15000
= 40 × 150 = 6000
Thus, 6000 voters did not vote.

Question 9.
Meeta saves ₹ 4000 from her salary. If this is 10% of her salary. What is her salary?
Answer:
Meeta saving = ₹ 4000
Let the salary be ₹ x
Saving = 10% of x
10% of x =4000
\(\frac{10}{100}\) × x = 4000
x = \(\frac{4000 \times 100}{10}\)
= ₹ 40,000
Meeta salary = ₹ 40,000/-

Question 10.
A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?
Answer:
Total number of matches played = 20
Part of matches won = 25%
25% of 20 = \(\frac{25}{100}\) × 20 = \(\frac{25 \times 20}{100}\) = 5
The team won 5 matches.