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CHAPTER – 13 Exponents and Powers | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Get Chapter Wise MCQ Questions for Class 7 Maths with Answers prepared here according to the latest CBSE syllabus and NCERT curriculum. Students can practice CBSE Class 7 Maths MCQs Multiple Choice Questions with Answers to score good marks in the examination. Students can also visit the most accurate and elaborate NCERT Solutions for Class 7 Maths. Every question of the textbook has been answered here.

Chapter - 13 Exponents and Powers

MCQs

Question 1.
The exponential form of 10000 is
(a) 10³
(b) 10⁴
(c) 10⁵
(d) none of these

Answer

Answer: (b) 10⁴
Hint:
10000 = 10 × 10 × 10 × 10 = 10⁴

Question 2.
The exponential form of 100000 is
(a) 10³
(b) 10⁴
(c) 10⁵
(d) none of these

Answer

Answer: (c) 10⁵
Hint:
100000 = 10 × 10 × 10 × 10 × 10 = 10⁵

Question 3.
The exponential form of 81 is
(a) 3⁴
(b) 3³
(c) 3²
(d) none of these

Answer

Answer: (a) 3⁴
Hint:
81 = 3 × 3 × 3 × 3 = 3⁴

Question 4.
The exponential form of 125 is
(a) 5⁴
(b) 5³
(c) 5²
(d) none of these

Answer

Answer: (b) 5³
Hint:
125 = 5 × 5 × 5 = 5³

Question 5.
The exponential form of 32 is
(a) 2³
(b) 2⁴
(c) 2⁵
(d) none of these

Answer

Answer: (c) 2⁵
Hint:
32 = 2 × 2 × 2 × 2 × 2 = 2⁵

Question 6.
The exponential form of 243 is
(a) 3⁵
(b) 3⁴
(c) 3³
(d) 3²

Answer

Answer: (a) 3⁵
Hint:
243 = 3 × 3 × 3 × 3 × 3 = 3⁵

Question 7.
The exponential form of 64 is
(a) 2⁵
(b) 2⁶
(c) 2⁷
(d) 2⁸

Answer

Answer: (b) 2⁶
Hint:
64 = 2 × 2 × 2 × 2 × 2 × 2 = 2⁶

Question 8.
The exponential form of 625 is
(a) 5²
(b) 5³
(c) 5⁴
(d) 5⁵

Answer

Answer: (c) 5⁴
Hint:
625 = 5 × 5 × 5 × 5 = 5⁴

Question 9.
The exponential form of 1000 is
(a) 10¹
(b) 10²
(c) 10³
(d) 10⁴

Answer

Answer: (c) 10³
Hint:
1000 = 10 × 10 × 10 = 10³

Question 10.
The value of (- 2)³ is
(a) 8
(b) -8
(c) 16
(d) -16

Answer

Answer: (b) -8
Hint:
(-2)³ = (-2) × (-2) × (-2) = -8

Question 11.
The value of (- 2)⁴ is
(a) 8
(b) -8
(c) 16
(d) -16

Answer

Answer: (c) 16
Hint:
(-2)⁴ = (-2) × (-2) × (-2) × (-2) = 16

Question 12.
What is the base in 8²?
(a) 8
(b) 2
(c) 6
(d) 10

Answer

Answer: (a) 8

Question 13.
What is the exponent in 8²?
(a) 8
(b) 2
(c) 16
(d) 6

Answer

Answer: (b) 2

Question 14.
(-1)^{even number} =
(a) -1
(b) 1
(c) 0
(d) none of these

Answer

Answer: (b) 1

Question 15.
(-1)^{odd number} =
(a) -1
(b) 1
(c) 0
(d) none of these

Answer

Answer: (a) -1

Question 16.
0 × 10⁴ =
(a) 0
(b) 10⁴
(c) 1
(d) none of these

Answer

Answer: (a) 0

Question 17.
If 2³ × 2⁴ = 2^?, then ? =
(a) 3
(b) 4
(c) 1
(d) 7

Answer

Answer: (d) 7
Hint:
2³ × 2⁴ = 2³⁺⁴ = 2⁷

Important Questions

Question 1.
Express 343 as a power of 7.
Solution:
We have 343 = 7 × 7 × 7 = 7³
Thus, 343 = 7³

Question 2.
Which is greater 3² or 2³?
Solution:
We have 3² = 3 × 3 = 9
2³ = 2 × 2 × 2 = 8
Since 9 > 8
Thus, 3² > 2³

Question 3.
Express the following number as a powers of prime factors:
(i) 144
(ii) 225
Solution:
(i) We have
144 = 2 × 2 × 2 × 2 × 3 × 3 = 2⁴ × 3²
Thus, 144 = 2⁴ × 3²

(ii) We have
225 = 3 × 3 × 5 × 5 = 3² × 5²
Thus, 225 = 3² × 5²

Question 4.
Find the value of:
(i) (-1)¹⁰⁰⁰
(ii) (1)²⁵⁰
(iii) (-1)¹²¹
(iv) (10000)⁰
Solution:
(i) (-1)¹⁰⁰⁰ = 1 [∵ (-1)^{even number} = 1]
(ii) (1)²⁵⁰ = 1 [∵ (1)^{even number} = 1]
(iii) (-1)¹²¹ = -1 [∵ (-1)^{odd number} = -1]
(iv) (10000)⁰ = 1 [∵ a⁰ = 1]

Question 5.
Express the following in exponential form:
(i) 5 × 5 × 5 × 5 × 5
(ii) 4 × 4 × 4 × 5 × 5 × 5
(iii) (-1) × (-1) × (-1) × (-1) × (-1)
(iv) a × a × a × b × c × c × c × d × d
Solution:
(i) 5 × 5 × 5 × 5 × 5 = (5)⁵
(ii) 4 × 4 × 4 × 5 × 5 × 5 = 4³ × 5³
(iii) (-1) × (-1) × (-1) × (-1) × (-1) = (-1)⁵
(iv) a × a × a × b × c × c × c × d × d = a³b¹c³d²

Question 6.
Express each of the following as product of powers of their prime factors:
(i) 405
(ii) 504
(iii) 500
Solution:
(i) We have
405 = 3 × 3 × 3 × 3 × 5 = 3⁴ × 5¹
Thus, 405 = 3⁴ × 5¹

(ii) We have
504 = 2 × 2 × 2 × 3 × 3 × 7 = 2³ × 3² × 7¹
Thus, 504 = 2³ × 3² × 7¹

(iii) We have
500 = 2 × 2 × 5 × 5 × 5 = 2² × 5³
Thus, 500 = 2² × 5³

Question 7.
Simplify the following and write in exponential form:
(i) (5²)³
(ii) (2³)³
(iii) (a^b)^c
(iv) [(5)²]²
Solution:
(i) (5²)³ = 5^2×3 = 5⁶
(ii) (2³)³ = 2^3×3 = 2⁹
(iii) (a^b)^c = a^b×c = a^bc
(iv) [(5)²]² = 5^2×2 = 5⁴

Question 8.
Verify the following:

Solution:

Question 9.
Simplify:

Solution:

Question 10.
Simplify and write in exponential form:

Solution:

Exponents and Powers Class 7 Extra Questions Short Answer Type

Question 11.
Express each of the following as a product of prime factors is the exponential form:
(i) 729 × 125
(ii) 384 × 147
Solution:
(i) 729 × 125 = 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5 = 3⁶ × 5³
Thus, 729 × 125 = 3⁶ × 5³

(ii) 384 × 147 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7 = 2⁷ × 3² × 7²
Thus, 384 × 147 = 2⁷ × 3² × 7²

Question 12.
Simplify the following:
(i) 10³ × 9⁰ + 3³ × 2 + 7⁰
(ii) 6³ × 7⁰ + (-3)⁴ – 9⁰
Solution:
(i) 10³ × 9⁰ + 3³ × 2 + 7⁰
= 1000 + 54 + 1
= 1055
(ii) 6³ × 7⁰ + (-3)⁴ – 9⁰
= 216 × 1 + 81 – 1
= 216 + 80
= 296

Question 13.
Write the following in expanded form:
(i) 70,824
(ii) 1,69,835
Solution:
(i) 70,824
= 7 × 10000 + 0 × 1000 + 8 × 100 + 2 × 10 + 4 × 10⁰
= 7 × 10⁴ + 8 × 10² + 2 × 10¹ + 4 × 10⁰
(ii) 1,69,835
= 1 × 100000 + 6 × 10000 + 9 × 1000 + 8 × 100 + 3 × 10 + 5 × 10⁰
= 1 × 10⁵ + 6 × 10⁴ + 9 × 10³ + 8 × 10² + 3 × 10¹ + 5 × 10⁰

Question 14.
Find the number from each of the expanded form:
(i) 7 × 10⁸ + 3 × 10⁵ + 7 × 10² + 6 × 10¹ + 9
(ii) 4 × 10⁷ + 6 × 10³ + 5
Solution:
(i) 7 × 10⁸ + 3 × 10⁵ + 7 × 10² + 6 × 10¹ + 9
= 7 × 100000000 + 3 × 100000 + 7 × 100 + 6 × 10 + 9
= 700000000 + 300000 + 700 + 60 + 9
= 700300769
(ii) 4 × 10⁷ + 6 × 10³ + 5
= 4 × 10000000 + 6 × 1000 + 5
= 40000000 + 6000 + 5
= 40006005

Question 15.
Find the value of k in each of the following:

Solution:

Question 16.
Find the value of
(a) 3⁰ ÷ 4⁰
(b) (8⁰ – 2⁰) ÷ (8⁰ + 2⁰)
(c) (2⁰ + 3⁰ + 4⁰) – (4⁰ – 3⁰ – 2⁰)
Solution:
(a) We have 3⁰ ÷ 4⁰ = 1 ÷ 1 = 1 [∵ a⁰ = 1]
(6) (8⁰ – 2⁰) ÷ (8⁰ + 2⁰) = (1 – 1) ÷ (1 + 1) = 0 ÷ 2 = 0
(c) (2⁰ + 3⁰ + 4⁰) – (4⁰ – 3⁰ – 2⁰)
= (1 + 1 + 1) – (1 – 1 – 1) [∵ a⁰ = 1]
= 3 – 1
= 2

Question 17.
Express the following in standard form:
(i) 8,19,00,000
(ii) 5,94,00,00,00,000
(iii) 6892.25
Solution:
(i) 8,19,00,000 = 8.19 × 10⁷
(ii) 5,94,00,00,00,000 = 5.94 × 10¹¹
(iii) 6892.25 = 6.89225 × 10³

Question 18.
Evaluate:

Solution:

Exponents and Powers Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 19.
Find the value of x, if

Solution:

Question 20.

Solution:

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CHAPTER – 12 Algebraic Expressions | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 12 Algebraic Expressions

MCQs

Question 1.
How many terms are there in the expression 2x²y?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (a) 1

Question 2.
How many terms are there in the expression 2y + 5?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (b) 2

Question 3.
How many terms are there in the expression 1.2ab – 2.4b + 3.6a?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (c) 3

Question 4.
How many terms are there in the expression – 2p³ – 3p² + 4p + 7?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (d) 4

Question 5.
What is the coefficient of x in the expression 4x + 3y?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (d) 4

Question 6.
What is the coefficient of x in the expression 2 – x + y?
(a) 2
(b) 1
(c) -1
(d) None of these

Answer

Answer: (c) -1

Question 7.
What is the coefficient of x in the expression y²x + y?
(a) y²
(b) y
(c) 1
(d) 0

Answer

Answer: (a) y²

Question 8.
What is the coefficient of x in the expression 2z – 3xz?
(a) 3
(b) z
(c) 3z
(d) -3z

Answer

Answer: (d) -3z

Question 9.
What is the coefficient of x in the expression 1 + x + xz?
(a) z
(b) 1 + z
(c) 1
(d) 1 + x

Answer

Answer: (b) 1 + z

Question 10.
What is the coefficient of x in the expression 2x + xy²?
(a) 2 + y²
(b) 2
(c) y²
(d) None of these

Answer

Answer: (a) 2 + y²

Question 11.
What is the coefficient of y² in the expression 4 – xy²?
(a) 4
(b) x
(c) -x
(d) None of these

Answer

Answer: (c) -x

Question 12.
What is the coefficient of y² in the expression 3y² + 4x?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (c) 3

Question 13.
What is the coefficient of y² in the expression 2x²y – 10xy² + 5y²?
(a) 5 – 10x
(b) 5
(c) -10 x
(d) None of these

Answer

Answer: (a) 5 – 10x

Question 14.
What is the coefficient of x in the expression ax³ + bx² + d?
(a) a
(b) b
(c) d
(d) 0

Answer

Answer: (d) 0

Question 15.
What is the coefficient of x² in the expression ax + b?
(a) a
(b) b
(c) a + b
(d) 0

Answer

Answer: (d) 0

Imporatant Questions

Question 1.
Identify in the given expressions, terms which are not constants. Give their numerical coefficients.
(i) 5x – 3
(ii) 11 – 2y²
(iii) 2x – 1
(iv) 4x²y + 3xy² – 5
Solution:

Question 2.
Group the like terms together from the following expressions:
-8x²y, 3x, 4y, −32x , 2x²y, -y
Solution:
Group of like terms are:
(i) -8x²y, 2x²y
(ii) 3x, −32x
(iii) 4y, -y

Question 3.
Identify the pairs of like and unlike terms:
(i) −32x, y
(ii) -x, 3x
(iii) −12y2x, 32xy²
(iv) 1000, -2
Solution:
(i) −32x, y → Unlike Terms
(ii) -x, 3x → Like Terms
(iii) −12y2x, 32xy² → Like Terms
(iv) 1000, -2 → Like Terms

Question 4.
Classify the following into monomials, binomial and trinomials.
(i) -6
(ii) -5 + x
(iii) 32x – y
(iv) 6x² + 5x – 3
(v) z² + 2
Solution:
(i) -6 is monomial
(ii) -5 + x is binomial
(iii) 32x – y is binomial
(iv) 6x² + 5x – 3 is trinomial
(v) z² + z is binomial

Question 5.
Draw the tree diagram for the given expressions:
(i) -3xy + 10
(ii) x² + y²
Solution:

Question 6.
Identify the constant terms in the following expressions:
(i) -3 + 32x
(ii) 32 – 5y + y²
(iii) 3x² + 2y – 1
Solution:
(i) Constant term = -3
(ii) Constant term = 32
(iii) Constant term = -1

Question 7.
Add:
(i) 3x²y, -5x²y, -x²y
(ii) a + b – 3, b + 2a – 1
Solution:
(i) 3x²y, -5x²y, -x²y
= 3x²y + (-5x²y) + (-x²y)
= 3x²y – 5x²y – x²y
= (3 – 5 – 1 )x²y
= -3x²y
(ii) a + b – 3, b + 2a – 1
= (a + b – 3) + (b + 2a – 1)
= a + b – 3 + b + 2a – 1
= a + 2a + b + b – 3 – 1
= 3a + 2b – 4

Question 8.
Subtract 3x² – x from 5x – x².
Solution:
(5x – x²) – (3x² – x)
= 5x – x² – 3x² + x
= 5x + x – x² – 3x²
= 6x – 4x²

Question 9.
Simplify combining the like terms:
(i) a – (a – b) – b – (b – a)
(ii) x² – 3x + y² – x – 2y²
Solution:
(i) a – (a – b) – b – (b – a)
= a – a + b – b – b + a
= (a – a + a) + (b – b – b)
= a – b
(ii) x² – 3x + y² – x – 2y²
= x² + y² – 2y² – 3x – x
= x² – y² – 4x

Algebraic Expressions Class 7 Extra Questions Short Answer Type

Question 10.
Subtract 24xy – 10y – 18x from 30xy + 12y – 14x.
Solution:
(30xy + 12y – 14x) – (24xy – 10y – 18x)
= 30xy + 12y – 14x – 24xy + 10y + 18x
= 30xy – 24xy + 12y + 10y – 14x + 18x
= 6xy + 22y + 4x

Question 11.
From the sum of 2x² + 3xy – 5 and 7 + 2xy – x² subtract 3xy + x² – 2.
Solution:
Sum of the given term is (2x² + 3xy – 5) + (7 + 2xy – x²)
= 2x² + 3xy – 5 + 7 + 2xy – x²
= 2x² – x² + 3xy + 2xy – 5 + 7
= x² + 5xy + 2
Now (x² + 5xy + 2) – (3xy + x² – 2)
= x² + 5xy + 2 – 3xy – x² + 2
= x² – x² + 5xy – 3xy + 2 + 2
= 0 + 2xy + 4
= 2xy + 4

Question 12.
Subtract 3x² – 5y – 2 from 5y – 3x² + xy and find the value of the result if x = 2, y = -1.
Solution:
(5y – 3x² + xy) – (3x² – 5y – 2)
= 5y – 3x² + xy – 3x² + 5y + 2
= -3x² – 3x² + 5y + 5y + xy + 2
= -6x² + 10y + xy + 2
Putting x = 2 and y = -1, we get
-6(2)² + 10(-1) + (2)(-1) + 2
= -6 × 4 – 10 – 2 + 2
= -24 – 10 – 2 + 2
= -34

Question 13.
Simplify the following expressions and then find the numerical values for x = -2.
(i) 3(2x – 4) + x² + 5
(ii) -2(-3x + 5) – 2(x + 4)
Solution:
(i) 3(2x – 4) + x² + 5
= 6x – 12 + x² + 5
= x² + 6x – 7
Putting x = -2, we get
= (-2)² + 6(-2) – 7
= 4 – 12 – 7
= 4 – 19
= -15
(ii) -2(-3x + 5) – 2(x + 4)
= 6x – 10 – 2x – 8
= 6x – 2x – 10 – 8
= 4x – 18
Putting x = -2, we get
= 4(-2) – 18
= -8 – 18
= -26

Question 14.
Find the value of t if the value of 3x² + 5x – 2t equals to 8, when x = -1.
Solution:
3x² + 5x – 2t = 8 at x = -1
⇒ 3(-1)² + 5(-1) – 2t = 8
⇒ 3(1) – 5 – 2t = 8
⇒ 3 – 5 – 2t = 8
⇒ -2 – 2t = 8
⇒ 2t = 8 + 2
⇒ -2t = 10
⇒ t = -5
Hence, the required value of t = -5.

Question 15.
Subtract the sum of -3x³y² + 2x²y³ and -3x²y³ – 5y⁴ from x⁴ + x³y² + x²y³ + y⁴.
Solution:
Sum of the given terms:

Required expression

Question 16.
What should be subtracted from 2x³ – 3x²y + 2xy² + 3y² to get x³ – 2x²y + 3xy² + 4y²? [NCERT Exemplar]
Solution:
We have

Required expression

Question 17.
To what expression must 99x³ – 33x² – 13x – 41 be added to make the sum zero? [NCERT Exemplar]
Solution:
Given expression:
99x³ – 33x² – 13x – 41
Negative of the above expression is
-99x³ + 33x² + 13x + 41
(99x³ – 33x² – 13x – 41) + (-99x³ + 33x² + 13x + 41)
= 99x³ – 33x² – 13x – 41 – 99x³ + 33x² + 13x + 41
= 0
Hence, the required expression is -99x³ + 33x² + 13x + 41

Algebraic Expressions Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 18.
If P = 2x² – 5x + 2, Q = 5x² + 6x – 3 and R = 3x² – x – 1. Find the value of 2P – Q + 3R.
Solution:
2P – Q + 3R = 2(2x² – 5x + 2) – (5x² + 6x – 3) + 3(3x² – x – 1)
= 4x² – 10x + 4 – 5x² – 6x + 3 + 9x² – 3x – 3
= 4x² – 5x² + 9x² – 10x – 6x – 3x + 4 + 3 – 3
= 8x² – 19x + 4
Required expression.

Question 19.
If A = -(2x + 3), B = -3(x – 2) and C = -2x + 7. Find the value of k if (A + B + C) = kx.
Solution:
A + B + C = -(2x + 13) – 3(x – 2) + (-2x + 7)
= -2x – 13 – 3x + 6 – 2x + 7
= -2x – 3x – 2x – 13 + 6 + 7
= -7x
Since A + B + C = kx
-7x = kx
Thus, k = -7

Question 20.
Find the perimeter of the given figure ABCDEF.

Solution:
Required perimeter of the figure
ABCDEF = AB + BC + CD + DE + EF + FA
= (3x – 2y) + (x + 2y) + (x + 2y) + (3x – 2y) + (x + 2y) + (x + 2y)
= 2(3x – 2y) + 4(x + 2y)
= 6x – 4y + 4x + 8y
= 6x + 4x-4y + 8y
= 10x + 4y
Required expression.

Question 21.
Rohan’s mother gave him ₹ 3xy² and his father gave him ₹ 5(xy² + 2). Out of this total money he spent ₹ (10 – 3xy²) on his birthday party. How much money is left with him? [NCERT Exemplar]
Solution:
Money give by Rohan’s mother = ₹ 3xy²
Money given by his father = ₹ 5(xy² + 2)
Total money given to him = ₹ 3xy² + ₹ 5 (xy² + 2)
= ₹ [3xy² + 5(xy² + 2)]
= ₹ (3xy² + 5xy² + 10)
= ₹ (8xy² + 10).
Money spent by him = ₹ (10 – 3xy)²
Money left with him = ₹ (8xy² + 10) – ₹ (10 – 3xy²)
= ₹ (8xy² + 10 – 10 + 3x²y)
= ₹ (11xy²)
Hence, the required money = ₹ 11xy²

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CHAPTER – 11 Perimeter and Area | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 11 Perimeter and Area

MCQs

Question 1.
Perimeter of a square =
(a) side × side
(b) 3 × side
(c) 4 × side
(d) 2 × side

Answer

Answer: (c) 4 × side
Hint:
Formula

Question 2.
Perimeter of a rectangle of length Z and breadth 6 is
(a) l + b
(b) 2 × (l + b)
(c) 3 × (l + b)
(d) l × b

Answer

Answer: (b) 2 × (l + b)
Hint:
Formula

Question 3.
Area of a square =
(a) side × side
(b) 2 × side
(c) 3 × side
(d) 4 × side

Answer

Answer: (a) side × side
Hint:
Formula

Question 4.
Area of a rectangle of length l and breadth b is
(a) l × b
(b) l + b
(c) 2 × (l + b)
(d) 6 × (l + b)

Answer

Answer: (a) l × b
Hint:
Formula

Question 5.
Area of a parallelogram =
(а) base × height
(b) 12 × base × height
(c) 13 × base × height
(d) 14 × base × height

Answer

Answer: (а) base × height
Hint:
Formula

Question 6.
Area of a triangle =
(а) base × height
(b) 12 × base × height
(c) 13 × base × height
(d) 14 × base × height

Answer

Answer: (b) 12 × base × height
Hint:
Formula

Question 7.
The circumference of a circle of radius r is
(a) πr
(b) 2πr
(c) πr²
(d) 14 πr²

Answer

Answer: (b) 2πr
Hint:
Formula

Question 8.
The circumference of a circle of diameter d is
(a) πd
(b) 2πd
(c) 12 πd
(d) πd²

Answer

Answer: (a) πd
Hint:
Formula

Question 9.
If r and d are the radius and diameter of a circle respectively, then
(a) d = 2 r
(b) d = r
(C) d = 12 r
(d) d = r²

Answer

Answer: (a) d = 2 r
Hint:
Formula

Question 10.
The area of a circle of radius r is
(a) πr²
(b) 2πr²
(c) 2πr
(d) 4πr²

Answer

Answer: (a) πr²
Hint:
Formula

Question 11.
The area of a circle of diameter d is
(a) πd²
(b) 2πd²
(c) 14 πd²
(d) 2πd

Answer

Answer: (c) 14 πd²
Hint:
Formula

Question 12.
1 cm² =
(a) 10 mm²
(b) 100 mm²
(c) 1000 mm²
(d) 10000 mm²

Answer

Answer: (b) 100 mm²
Hint:
Formula

Question 13.
1 m² =
(a) 10 cm²
(b) 100 cm²
(c) 1000 cm²
(d) 10000 cm²

Answer

Answer: (d) 10000 cm²
Hint:
Formula

Question 14.
1 hectare =
(a) 10 m²
(b) 100 m²
(c) 1000 m²
(d) 10000 m²

Answer

Answer: (d) 10000 m²
Hint:
Formula

Important Questions

Question 1.
The side of a square is 2.5 cm. Find its perimeter and area.
Solution:
Side of the square = 2.5 cm
Perimeter = 4 × Side = 4 × 2.5 = 10 cm
Area = (side)² = (4)² = 16 cm²

Question 2.
If the perimeter of a square is 24 cm. Find its area.
Solution:
Perimeter of the square = 24 cm
Side of the square = 244 cm = 6 cm
Area of the square = (Side)² = (6)² cm² = 36 cm²

Question 3.
If the length and breadth of a rectangle are 36 cm and 24 cm respectively. Find
(i) Perimeter
(ii) Area of the rectangle.
Solution:
Length = 36 cm, Breadth = 24 cm
(i) Perimeter = 2(l + b) = 2(36 + 24) = 2 × 60 = 120 cm
(ii) Area of the rectangle = l × b = 36 cm × 24 cm = 864 cm²

Question 4.
The perimeter of a rectangular field is 240 m. If its length is 90 m, find:
(i) it’s breadth
(ii) it’s area.
Solution:
(i) Perimeter of the rectangular field = 240 m
2(l + b) = 240 m
l + b = 120 m
90 m + b = 120 m
b = 120 m – 90 m = 30 m
So, the breadth = 30 m.
(ii) Area of the rectangular field = l × b = 90 m × 30 m = 2700 m²
So, the required area = 2700 m²

Question 5.
The length and breadth of a rectangular field are equal to 600 m and 400 m respectively. Find the cost of the grass to be planted in it at the rate of ₹ 2.50 per m².
Solution:
Length = 600 m, Breadth = 400 m
Area of the field = l × b = 600 m × 400 m = 240000 m²
Cost of planting the grass = ₹ 2.50 × 240000 = ₹ 6,00,000
Hence, the required cost = ₹ 6,00,000.

Question 6.
The perimeter of a circle is 176 cm, find its radius.
Solution:
The perimeter of the circle = 176 cm

Question 7.
The radius of a circle is 3.5 cm, find its circumference and area.
Solution:
Radius = 3.5 cm
Circumference = 2πr

Question 8.
Area of a circle is 154 cm², find its circumference.
Solution:
Area of the circle = 154 cm²

Question 9.
Find the perimeter of the figure given below.

Solution:
Perimeter of the given figure = Circumference of the semicircle + diameter
= πr + 2r
= 227 × 7 + 2 × 7
= 22 + 14
= 36 cm
Hence, the required perimeter = 36 cm.

Question 10.
The length of the diagonal of a square is 50 cm, find the perimeter of the square.

Solution:
Let each side of the square be x cm.
x² + x² = (50)² [Using Pythagoras Theorem]
2x² = 2500
x² = 1250

x = √1250 = 2×5×5×5×5−−−−−−−−−−−−−√
x = 5 × 5 × √2 = 25√2
The side of the square = 25√2 cm
Perimeter of the square = 4 × side = 4 × 25√2 = 100√2 cm

Perimeter and Area Class 7 Extra Questions Short Answer Type

Question 11.
A wire of length 176 cm is first bent into a square and then into a circle. Which one will have more area?
Solution:
Length of the wire = 176 cm
Side of the square = 176 ÷ 4 cm = 44 cm
Area of the square = (Side)² = (44)² cm² = 1936 cm²
Circumference of the circle = 176 cm

Since 2464 cm² > 1936 cm²
Hence, the circle will have more area.

Question 12.
In the given figure, find the area of the shaded portion.

Solution:
Area of the square = (Side)² = 10 cm × 10 cm = 100 cm²
Area of the circle = πr²
= 227 × 3.5 × 3.5
= 772 cm²
= 38.5 cm²
Area of the shaded portion = 100 cm² – 38.5 cm² = 61.5 cm²

Question 13.
Find the area of the shaded portion in the figure given below.

Solution:
Area of the rectangle = l × b = 14 cm × 14 cm = 196 cm²
Radius of the semicircle = 142 = 7 cm
Area of two equal semicircle = 2 × 12 πr²
= πr²
= 227 × 7 × 7
= 154 cm²
Area of the shaded portion = 196 cm² – 154 cm² = 42 cm²

Question 14.
A rectangle park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.

Solution:
Length of the rectangular park = 45 m
Breadth of the park = 30 m
Area of the park = l × 6 = 45m × 30m = 1350 m²
Length of the park including the path = 45 m + 2 × 2.5 m = 50 m
Breadth of the park including the path = 30 m + 2 × 2.5 m = 30m + 5m = 35m
Area of the park including the path = 50 m × 35 m = 1750 m²
Area of the path = 1750 m² – 1350 m² = 400 m²
Hence, the required area = 400 m².

Question 15.
In the given figure, calculate:
(а) the area of the whole figure.
(b) the total length of the boundary of the field.

Solution:
Area of the rectangular portions = l × b = 80 cm × 42 cm = 3360 cm²
Area of two semicircles = 2 × 12 πr² = πr²
= 227 × 21 × 21
= 22 × 3 × 21
= 1386 cm²
Total area = 3360 cm² + 1386 cm² = 4746 cm²
Total length of the boundary of field = (2 × 80 + πr + πr) cm
= (160 + 227 × 21 + 227 × 21)
= (160 + 132) cm
= 292 cm
Hence, the required (i) area = 4746 cm² and (ii) length of boundary = 292 cm.

Question 16.
How many times a wheel of radius 28 cm must rotate to cover a distance of 352 m?
(Take π = 227)
Solution:
Radius of the wheel = 28 cm
Circumference = 2πr = 2 × 227 × 28 = 176 cm
Distance to be covered = 352 m or 352 × 100 = 35200 m
Number of rotation made by the wheel to cover the given distance = 35200176 = 200
Hence, the required number of rotations = 200.

Perimeter and Area Class 7 Extra Questions Long Answer Type

Question 17.
A nursery school playground is 160 m long and 80 m wide. In it 80 m × 80 m is kept for swings and in the remaining portion, there are 1.5 m wide path parallel to its width and parallel to its remaining length as shown in Figure. The remaining area is covered by grass. Find the area covered by grass. (NCERT Exemplar)

Solution:
Area of the playground = l × b = 160 m × 80 m = 12800 m²
Area left for swings = l × b = 80m × 80m = 6400 m²
Area of the remaining portion = 12800 m² – 6400 m² = 6400 m²
Area of the vertical road = 80 m × 1.5 m = 120 m²
Area of the horizontal road = 80 m × 1.5 m = 120 m²
Area of the common portion = 1.5 × 1.5 = 2.25 m²
Area of the two roads = 120 m² + 120 m² – 2.25 m² = (240 – 2.25) m² = 237.75 m²
Area of the portion to be planted by grass = 6400 m² – 237.75 m² = 6162.25 m²
Hence, the required area = 6162.25 m².

Question 18.
Rectangle ABCD is formed in a circle as shown in Figure. If AE = 8 cm and AD = 5 cm, find the perimeter of the rectangle. (NCERT Exemplar)

Solution:
DE (Radius) = AE + AD = 8 cm + 5 cm = 13 cm
DB = AC = 13 cm (Diagonal of a rectangle are equal)
In right ∆ADC,
AD² + DC² = AC² (By Pythagoras Theorem)
⇒ (5)² + DC² = (13)²
⇒ 25 + DC² = 169
⇒ DC² = 169 – 25 = 144
⇒ DC = √144 = 12 cm
Perimeter of rectangle ABCD = 2(AD + DC)
= 2(5 cm + 12 cm)
= 2 × 17 cm
= 34 cm

Question 19.
Find the area of a parallelogram-shaped shaded region. Also, find the area of each triangle. What is the ratio of the area of shaded portion to the remaining area of the rectangle?

Solution:
Here, AB = 10 cm
AF = 4 cm
FB = 10 cm – 4 cm = 6 cm
Area of the parallelogram = Base × Height = FB × AD = 6 cm × 6 cm = 36 cm²
Hence, the required area of shaded region = 36 cm².
Area ∆DEF = 12 × b × h
= 12 × AF × AD
= 12 × 4 × 6
= 12 cm²
Area ∆BEC = 12 × b × h
= 12 × GC × BC
= 12 × 4 × 6
= 12 cm²
Area of Rectangle ABCD = l × b = 10 cm × 6 cm = 60 cm²
Remaining area of Rectangle = 60 cm² – 36 cm² = 24 cm²
Required Ratio = 36 : 24 = 3 : 2

Question 20.
A rectangular piece of dimension 3 cm × 2 cm was cut from a rectangular sheet of paper of dimensions 6 cm × 5 cm. Find the ratio of the areas of the two rectangles. (NCER T Exemplar)

Solution:
Length of the rectangular piece = 6 cm
Breadth = 5 cm
Area of the sheet = l × b = 6 cm × 5 cm = 30 cm²
Area of the smaller rectangular piece = 3 cm × 2 cm = 6 cm²
Ratio of areas of two rectangles = 30 cm² : 6 cm² = 5 : 1

Perimeter and Area Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 21.
In the given figure, ABCD is a square of side 14 cm. Find the area of the shaded region.
(Take π = 227)

Solution:
PQ = 12 AB = 12 × 14 = 7 cm
PQRS is a square with each side 7 cm
Radius of each circle = 72 cm
Area of the quadrants of each circle = 14 × πr²
Area of the four quadrants of all circles

Area of the square PQRS = Side × Side = 7 cm × 7 cm = 49 cm²
Area of the shaded portion = 49 cm² – 38.5 cm² = 10.5 cm²
Hence, the required area = 10.5 cm².

Question 22.
Find the area of the following polygon if AB = 12 cm, AC = 2.4 cm, CE = 6 cm, AD = 4.8 cm, CF = GE = 3.6 cm, DH = 2.4 cm.

Solution:
BE = AB – AE
= 12 cm – (AC + CE)
= 12 cm – (2.4 cm + 6 cm)
= 12 cm – 8.4 cm
= 3.6 cm

Area of the polygon AFGBH = Area of ∆ACF + Area of rectangle FCEG + Area of ∆GEB + Area of ∆ABH
= 3.6 cm² + 4.32 cm² + 21.6 cm² + 6.48 cm² + 14.4 cm²
= 50.40 cm²
Hence, the required area = 50.40 cm².

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CHAPTER – 10 Practical Geometry | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 10 Practical Geometry

MCQs

Question 1.
In ΔRST, R = 5 cm, and ∠SRT = 45° and ∠RST = 45°. Which criterion can be used to construct ΔRST?
(a) A.S.A. criterion
(b) S.A.S. criterion
(c) S.S.S. criterion
(d) R.H.S. criterion

Answer

Answer: (a) A.S.A. criterion
Hint:

Clearly, from the figure two angles and the included side are given. So, A.S.A. criterion can be used to construct ARST.

Question 2.
Identify the criterion of construction of the equilateral triangle LMN given LM = 6 cm.
(a) S.A.S. criterion
(b) R.H.S. criterion
(c) A.S.A. criterion
(d) S.S.S. criterion

Answer

Answer: (d) S.S.S. criterion
Hint:
Since ALMN is equilateral the measurement of one side is used for the other two sides of the triangle. Hence ALMN can be constructed by S.S.S. criterion.

Question 3.
The idea of equal alternate angles is used to construct which of the following?
(a) A line parallel to a given line
(b) A triangle
(c) A square
(d) Two triangles

Answer

Answer: (a) A line parallel to a given line.

Question 4.
A Given AB = 3 cm, AC = 5 cm,and ∠B = 30°, ΔABC cannot be uniquely constructed, with AC as base, why?
(a) Two sides and included angle are given.
(b) The other two angles are not given.
(c) The vertex B cannot be uniquely located.
(d) The vertex A coincides with the vertex C.

Answer

Answer: (c) The vertex B cannot be uniquely located.

Question 5.
A line panda point X not on it are given. Which of the following is used to draw a line parallel to p through X?
(a) Equal corresponding angles.
(b) Congruent triangles.
(c) Angle sum property of triangles.
(d) Pythagoras’ theorem.

Answer

Answer: (a) Equal corresponding angles.
Hint:
Corresponding angles of parallel lines are equal.

Question 6.
Δ PQR is such that ∠P = ∠Q = ∠R = 60° which of the following is true?
(a) Δ PQR is equilateral.
(b) Δ PQR is acute angled.
(c) Both [a] and [b]
(d) Neither [a] nor [b]

Answer

Answer: (c) Both [a] and [b]
Hint:
In ΔPQR since all the angles are acute, it is acute angled. Also since all the angles are equal, it is equilateral.

Question 7.
Which vertex of ΔABC is right angled if AB¯¯¯¯¯¯¯¯ = 8 cm, AC¯¯¯¯¯¯¯¯ = 6 cm,and BC¯¯¯¯¯¯¯¯ = 10 cm,?
(a) ∠C
(b) ∠A
(c) ∠B
(d) A or C

Answer

Answer: (b) ∠A
Hint:
From the given measurements, BC¯¯¯¯¯¯¯¯ is the hypotenuse. The angle opposite to BC¯¯¯¯¯¯¯¯ is ∠A which is a right angle.

Question 8.
An isosceles triangle is constructed as shown in the figure.

Which of the given statements is incorrect?
(a) PR¯¯¯¯¯¯¯¯ is the hypotenuse of ΔPQR.
(b) ΔPQR is an equilateral triangle.
(c) ΔPQR is a right angled triangle.
(d) If right angled ΔPQR has its equal angles measuring 45° each.

Answer

Answer: (b) ΔPQR is an equilateral triangle.

Question 9.
ΔPQR is constructed with all its angles measuring 60° each. Which of the following is correct?
(a) ΔPQR is an equilateral triangle.
(b) ΔPQR is isosceles triangle.
(c) ΔPQR is a scalene triangle.
(d) ΔPQR is a right angled triangle.

Answer

Answer: (a) ΔPQR is an equilateral triangle.

Question 10.
How many perpendicular lines can be drawn to a line from a point not on it?
(a) 1
(b) 2
(c) 0
(d) Infinite

Answer

Answer: (a) 1
Hint:

As can be seen from the given figure, one and only one perpendicular line can be drawn to a given line from a point not on it.

Question 11.
Identify the false statement.
(a) A triangle with three equal sides is called an equilateral triangle.
(b) A triangle with a right angle is called a right angled triangle.
(c) A triangle with two equal sides is called a scalene triangle.
(d) A right angled triangle has two acute angles and a right angle.

Answer

Answer: (c) A triangle with two equal sides is called a scalene triangle.

Question 12.
ΔPQR is constructed such that PQ = 5 cm, PR = 5 cm and ∠RPQ = 50° Identify the type of triangle constructed.
(a) An isosceles triangle
(b) An acute angled triangle
(c) An obtuse angled triangle
(d) Both [a] and [b]

Answer

Answer: (d) Both [a] and [b]

Question 13.
Which of the following is NOT constructed using a ruler and a set square?
(a) A perpendicular to a line from a point not on it.
(b) A perpendicular bisector of a line segment.
(c) A perpendicular to a line at a point on the line.
(d) A line parallel to a given line through a given point.

Answer

Answer: (b) A perpendicular bisector of a line segment.

Important Questions

Question 1.
State whether the triangle is possible to construct if
(a) In ΔABC, m∠A = 80°, m∠B = 60°, AB = 5.5 cm
(b) In ΔPQR, PQ = 5 cm, QR = 3 cm, PR = 8.8 cm
Solution:
(a) m∠A = 80°, m∠B = 60°
m∠A + m∠B = 80° + 60° = 140° < 180°
So, ΔABC can be possible to construct.
(b) PQ = 5 cm, QR = 3 cm, PR = 8.8 cm
PQ + QR = 5 cm + 3 cm = 8 cm < 8.8 cm
or PQ + QR < PR
So, the ΔPQR can not be constructed.

Question 2.
Draw an equilateral triangle whose each side is 4.5 cm.

Solution:
Steps of construction:
(i) Draw AB = 4.5 cm.
(ii) Draw two arcs with centres A and B and same radius of 4.5 cm to meet each other at C.
(iii) Join CA and CB.
(iv) ΔCAB is the required triangle.

Question 3.
Draw a ΔPQR, in which QR = 3.5 cm, m∠Q = 40°, m∠R = 60°.

Solution:
Steps of construction:
(i) Draw QR = 3.5 cm.
(ii) Draw ∠Q = 40°, ∠R = 60° which meet each other at P.
(iii) ΔPQR is the required triangle.

Question 4.
There are four options, out of which one is correct. Choose the correct one:
(i) A triangle can be constructed with the given measurement.
(a) 1.5 cm, 3.5 cm, 4.5 cm
(b) 6.5 cm, 7.5 cm, 15 cm
(c) 3.2 cm, 2.3 cm, 5.5 cm
(d) 2 cm, 3 cm, 6 cm
(ii) (a) m∠P = 40°, m∠Q = 60°, AQ = 4 cm
(b) m∠B = 90°, m∠C = 120° , AC = 6.5 cm
(c) m∠L = 150°, m∠N = 70°, MN = 3.5 cm
(d) m∠P = 105°, m∠Q = 80°, PQ = 3 cm
Solution:
(i) Option (a) is possible to construct.
1.5 cm + 3.5 cm > 4.5 cm
(ii) Option (a) is correct.
m∠P + m∠Q = 40° + 60° = 100° < 180°

Question 5.
What will be the other angles of a right-angled isosceles triangle?

Solution:
In right angled isosceles triangle ABC, ∠B = 90°
∠A + ∠C = 180° – 90° = 90°
But ∠A = ∠B
∠A = ∠C = 902 = 45°
Hence the required angles are ∠A = ∠C = 45°

Question 6.
What is the measure of an exterior angle of an equilateral triangle?

Solution:
We know that the measure of each interior angle = 60°
Exterior angle = 180° – 60° = 120°

Question 7.
In ΔABC, ∠A = ∠B = 50°. Name the pair of sides which are equal.

Solution:
∠A = ∠B = 50°
AC = BC [∵ Sides opposite to equal angles are equal]
Hence, the required sides are AC and BC.

Question 8.
If one of the other angles of a right-angled triangle is obtuse, whether the triangle is possible to construct.
Solution:
We know that the angles other than right angle of a right-angled triangle are acute angles.
So, such a triangle is not possible to construct.

Question 9.
State whether the given pair of triangles are congruent.

Solution:
Here, AB = PQ = 3.5 cm
AC = PR = 5.2 cm
∠BAC = ∠QPR = 70°
ΔABC = ΔPQR [By SAS rule]

Practical Geometry Class 7 Extra Questions Short Answer Type

Question 10.
Draw a ΔABC in which BC = 5 cm, AB = 4 cm and m∠B = 50°.

Solution:
Steps of construction:
(i) Draw BC = 5 cm.
(ii) Draw ∠B = 50° and cut AB = 4 cm.
(iii) Join AC.
(iv) ΔABC is the required triangle.

Question 11.
Draw ΔPQR in which QR = 5.4 cm, ∠Q = 40° and PR = 6.2 cm.

Solution:
Steps of construction:
(i) Draw QR = 5.4 cm.
(ii) Draw ∠Q = 40°.
(iii) Take R as the centre and with radius 6.2 cm, draw an arc to meet the former angle line at P.
(iv) Join PR.
(v) ΔPQR is the required triangle.

Question 12.
Construct a ΔPQR in which m∠P = 60° and m∠Q = 30°, QR = 4.8 cm.

Solution:
m∠Q = 30°, m∠P = 60°
m∠Q + m∠P + m∠R = 180° (Angle sum property of triangle)
30° + 60° + m∠R = 180°
90° + m∠R = 180°
m∠R = 180° – 90°
m∠R = 90°
Steps of construction:
(i) Draw QR = 4.8 cm.
(ii) Draw ∠Q = 30°.
(iii) Draw ∠R = 90° which meets the former angle line at P.
(iv) ∠P = 180° – (30° + 90°) = 60°
(v) ΔPQR is the required triangle.

Practical Geometry Class 7 Extra Questions Higher Order Thinking Skills [HOTS] Type

Question 13.
Draw an isosceles right-angled triangle whose hypotenuse is 5.8 cm.
Solution:
Right angled triangle is an isosceles triangle
Each of its acute angles = 902 = 45°

Steps of construction:
(i) Draw AB = 5.8 cm.
(ii) Construct ∠A = 45° and ∠B = 45° to meet each other at C.
(iii) ∠C = 180° – (45° + 45°) = 90°
(iv) ΔACB is the required isosceles right angle triangle.

Question 14.
Construct a ΔABC such that AB = 6.5 cm, AC = 5 cm and the altitude AP to BC is 4 cm.

Solution:
Steps of construction:
(i) Draw a line l and take any point P on it.
(ii) Construct a perpendicular to l at P.
(iii) Cut AP = 4 cm.
(iv) Draw two arcs with centre A and radii 6.5 cm and 5 cm to cut the line l at B and C respectively.
(v) Join AB and AC.
(vi) ΔABC is the required triangle.

Question 15.
Construct an equilateral triangle whose altitude is 4.5 cm.

Solution:
Steps of construction:
(i) Draw any line l and take a point D on it.
(ii) Construct a perpendicular to l at D and cut AD = 4.5 cm.
(iii) Draw the angle of 30° at on either side of AD to meet the line l at B and C.
(iv) ΔABC is the required equilateral triangle.

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CHAPTER -9 Rational Numbers | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 9 Rational Numbers

MCQs

Question 1.
The numerator of the rational number 35 is
(a) 3
(b) 5
(c) 2
(d) 8

Answer

Answer: (a) 3

Question 2.
The numerator of the rational number 1100 is
(a) 100
(b) 1
(c) 10
(d) 99

Answer

Answer: (b) 1

Question 3.
The denominator of the rational number 47 is
(a) 7
(b) 4
(c) 3
(d) 11

Answer

Answer: (a) 7

Question 4.
The denominator of the rational number 713 is
(a) 13
(b) 7
(c) 6
(d) 91

Answer

Answer: (a) 13

Question 5.
The numerator of the rational number −34 is
(a) -3
(b) 3
(c) 4
(d) -4

Answer

Answer: (a) -3

Question 6.
The numerator of the rational number −29 is
(a) -2
(b) 2
(c) -9
(d) 9

Answer

Answer: (a) -2

Question 7.
The denominator of the rational number 5−3 is
(a) 5
(b) -3
(c) 3
(d) 8

Answer

Answer: (b) -3

Question 8.
The denominator of the rational number 3−7 is
(a) 7
(b) -7
(c) 3
(d) -3

Answer

Answer: (b) -7

Question 9.
The numerator of the rational number −2−5 is
(a) 2
(b) -2
(c) 5
(d) -5

Answer

Answer: (b) -2

Question 10.
the numerator of the rational number −5−3 is
(a) -5
(b) 5
(c) -3
(d) 3

Answer

Answer: (a) -5

Question 11.
The denominator of the rational number −2−9 is
(a) -2
(b) 2
(c) 9
(d) -9

Answer

Answer: (d) -9

Question 12.
The denominator of the rational number −6−5 is
(a) 6
(b) -6
(c) 5
(d) -5

Answer

Answer: (d) -5

Question 13.
The denominator of the rational number −13−11 is
(a) -13
(b) 13
(c) 11
(d) -11

Answer

Answer: (d) -11

Question 14.
The numerator of the rational number 0 is
(a) 0
(b) 1
(c) 2
(d) 3

Answer

Answer: (a) 0

Question 15.
The denominator of the rational number 0 is
(a) 0
(b) 1
(c) -1
(d) any non-zero integer

Answer

Answer: (d) -1

Question 16.
The numerator of a rational number 8 is
(a) 2
(b) 4
(c) 6
(d) 8

Answer

Answer: (d) 8

Question 17.
The denominator of the rational number 8 is
(a) 8
(b) 2
(c) 4
(d) 1

Answer

Answer: (d) 1

Question 1.
Find three rational numbers equivalent to each of the following rational numbers.
(i) −25
(ii) 37
Solution:

Question 2.
Reduce the following rational numbers in standard form.
(i) 35−15
(ii) −36−216
Solution:

Question 3.
Represent 32 and −34 on number lines.
Solution:

Question 4.
Which of the following rational numbers is greater?
(i) 34, 12
(ii) −32, −34
Solution:

Question 5.
Find the sum of

Solution:

Question 6.
Subtract:

Solution:

Question 7.
Find the product:

Solution:

Rational Numbers Class 7 Extra Questions Short Answer Type

Question 8.
If the product of two rational numbers is −916 and one of them is −415, find the other number.
Solution:
Let the required rational number be x.

Question 9.
Arrange the following rational numbers in ascending order.

Solution:

Question 10.
Insert five rational numbers between:

Solution:

Rational Numbers Class 7 Extra Questions Long Answer Type

Question 11.
Evaluate the following:

Solution:

Question 12.
Subtract the sum of −56 and -135 from the sum 223 and -625.
Solution:

Question 13.

Solution:
We have

Question 14.
Divide the sum of -21517 and 3534 by their difference.
Solution:

Question 15.
During a festival sale, the cost of an object is ₹ 870 on which 20% is off. The same object is available at other shops for ₹ 975 with a discount of 623 %. Which is a better deal and by how much?
Solution:
The cost of the object = ₹ 870
Discount = 20% of ₹ 870 = 20100 × 870 = ₹ 174
Selling price = ₹ 870 – ₹ 174 = ₹ 696
The same object is available at other shop = ₹ 975

Selling price = ₹ 975 – ₹ 65 = ₹ 910
Since ₹ 910 > ₹ 696
Hence, deal at first shop is better and by ₹ 910 – ₹ 696 = ₹ 214

Rational Numbers Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 16.
Simplify:
21.5 ÷ 5 – 15 of (20.5 – 5.5) + 0.5 × 8.5
Solution:
Using BODMAS rule, we have
21.5 ÷ 5 – 15 of (20.5 – 5.5) + 0.5 × 8.5
= 21.5 ÷ 5 – 15 of 15 + 0.5 × 8.5
= 21.5 × 15 – 15 × 15 + 0.5 × 8.5
= 4.3 – 3 + 4.25
= 4.3 + 4.25 – 3
= 8.55 – 3
= 5.55

Question 17.
Simplify:

Solution:
Using BODMAS rule, we have
2.3 – [1.89 – {3.6 – (2.7 – 0.77)}]
= 2.3 – [1.89 – {3.6 – 1.93}]
= 2.3 – [1.89 – 1.67]
= 2.3 – 0.22
= 2.08

Question 18.

Solution:
Using BODMAS rule, we have

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CHAPTER -8 Comparing Quantities | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 8 Comparing Quantities

MCQs

Question 1.
The ratio of Rs 10 to 50 paise is
(a) 20 : 1
(b) 10 : 1
(c) 5 : 1
(d) 1 : 20

Answer

Answer: (a) 20 : 1
Hint:
Rs 10 : 50 paise = 10 × 100 paise: 50 paise
= 1000 : 50 = 20 : 1

Question 2.
The ratio of 6 kg to 400 g is
(a) 10 : 1
(b) 15 : 1
(c) 12 : 1
(d) 6 : 1

Answer

Answer: (b) 15 : 1
Hint:
6 kg : 400 g = 6 × 1000 g : 400 g
= 6000 : 400 = 15 : 1.

Question 3.
The ratio of 3 m to 60 cm is
(a) 4 : 1
(b) 3 : 1
(c) 5 : 1
(d) 2 : 1

Answer

Answer: (c) 5 : 1
Hint:
3 m : 60 cm = 3 × 100 cm : 60 cm
= 300 : 60 = 5 : 1.

Question 4.
The ratio of 15 days to 72 hours is
(a) 2 : 1
(b) 3 : 1
(c) 4 : 1
(d) 5 : 1

Answer

Answer: (d) 5 : 1
Hint:
15 days : 72 hours
= 15 × 24 hours : 72 hours
= 15 × 24 : 72 = 5 : 1.

Question 5.
The ages of father and son are 45 years and 10 years. The ratio of their ages is
(a) 3 : 2
(b) 5 : 2
(c) 9 : 2
(d) 15 : 2

Answer

Answer: (c) 9 : 2
Hint:
45 years : 10 years ⇒ 9 : 2

Question 6.
The cost of 3 envelopes is ₹ 15. Find the cost of 5 envelopes.
(a) ₹ 20
(b) ₹ 25
(c) ₹ 30
(d) ₹ 40

Answer

Answer: (b) ₹ 25
Hint:
3 : 5 = 15 : x
⇒ 35 = 15x ⇒ x = 25

Question 7.
The cost of 7 kg of potatoes is ₹ 42. How many kg of potatoes can be purchased for ₹ 96?
(a) 10 kg
(b) 12 kg
(c) 15 kg
(d) 16 kg

Answer

Answer: (d) 16 kg
Hint:
7. 7 : x = 42 : 96
⇒ 7x = 4296 ⇒ x = 16

Question 8.
The cost of 4 m of cloth is Rs 40. Find the cost of 9 m of cloth.
(a) ₹ 90
(b) ₹ 60
(c) ₹ 50
(d) ₹ 40

Answer

Answer: (a) ₹ 90
Hint:
8. 4 : 9 = 40 : x
⇒ 49 = 40x ⇒ x = 90

Question 9.
The cost of 8 pencils is ₹ 10. Find the cost of 20 pencils.
(a) ₹ 20
(b) ₹ 25
(c) ₹ 24
(d) ₹ 30

Answer

Answer: (b) ₹ 25
Hint:
8 : 20 = 10 : x
⇒ 820 = 10x ⇒ x = 25

Question 10.
A motorcycle goes 120 km in 3 l of petrol. How much petrol will be required to go 600 km?
(a) 10 l
(b) 12 l
(c) 15 l
(d) 20 l

Answer

Answer: (c) 15 l
Hint:
120 : 600 = 3 : x
⇒ 120600 = 3x ⇒ x = 15

Question 11.
14 as per cent is
(a) 20%
(b) 25%
(c) 30%
(d) 12 12%

Answer

Answer: (b) 25%
Hint:
14 = 14 × 100% = 25%

Question 12.
18 as per cent is
(a) 25%
(b) 121%
(c) 20%
(d) 16%

Answer

Answer: (b) 121%
Hint:
18 = 18 × 100% = 1212%

Question 13.
52 as per cent is
(a) 125%
(b) 150%
(c) 200%
(d) 250%

Answer

Answer: (d) 250%
Hint:
52 = 52 × 100% = 250%

Question 14.
0.1 as per cent is
(a) 1%
(b) 10%
(c) 100%
(d) 0.1%

Answer

Answer: (b) 10%
Hint:
0.1 = 110 = 110 × 100% = 10%

Question 15.
0.04 as per cent is
(a) 10%
(b) 20%
(c) 25%
(d) 4%

Answer

Answer: (d) 4%
Hint:
0.04 = 4100 = 4100 × 100% = 4%

Question 16.
22 as per cent is
(a) 19%
(b) 10%
(c) 100%
(d) none of these

Answer

Answer: (c) 100%
Hint:
22 = 22 × 100% = 100%

Question 17.
Out of 40 children in a class, 10 are boys. What is the percentage of boys?
(a) 10%
(b) 40%
(c) 4%
(d) 25%

Answer

Answer: (d) 25%
Hint:
Percentage of boys = 1040 × 100% = 25%

Important Questions

Question 1.
Find the ratio of:
(a) 5 km to 400 m
(b) 2 hours to 160 minutes
Solution:
(a) 5 km = 5 × 1000 = 5000 m
Ratio of 5 km to 400 m
= 5000 m : 400 m
= 25 : 2
Required ratio = 25 : 2
(b) 2 hours = 2 × 60 = 120 minutes
Ratio of 2 hours to 160 minutes
= 120 : 160
= 3 : 4
Required ratio = 3 : 4

Question 2.
State whether the following ratios are equivalent or not?
(a) 2 : 3 and 4 : 5
(b) 1 : 3 and 2 : 6
Solution:
(a) Given ratios = 2 : 3 and 4 : 5

Hence 2 : 3 and 4 : 5 are not equivalent ratios.
(b) Given ratios = 1 : 3 and 2 : 6
LCM of 3 and 6 = 6

Hence, 1 : 3 and 2 : 6 are equivalent ratios.

Question 3.
Express the following ratios in simplest form:
(a) 615 : 213
(b) 42 : 56
Solution:

Question 4.
Compare the following ratios:
3 : 4, 5 : 6 and 3 : 8
Solution:
Given: 3 : 4, 5 : 6 and 3 : 8
or 34 , 56 and 38
LCM of 4, 6 and 8 = 24

Hence, 3 : 8 < 3 : 4 < 5 : 6

Question 5.
State whether the following ratios are proportional or not:
(i) 20 : 45 and 4 : 9
(ii) 9 : 27 and 33 : 11
Solution:
(i) 20 : 45 and 4 : 9
Product of extremes = 20 × 9 = 180
Product of means = 45 × 4 = 180
Here, the product of extremes = Product of means
Hence, the given ratios are in proportion.
(ii) 9 : 27 and 33 : 11
Product of extremes = 9 × 11 = 99
Product of means = 27 × 33 = 891
Here, the product of extremes ≠ Product of means
Hence, the given ratios are not in proportion.

Question 6.
24, 36, x are in continued proportion, find the value of x.
Solution:
Since, 24, 36, x are in continued proportion.
24 : 36 :: 36 : x
⇒ 24 × x = 36 × 36
⇒ x = 54
Hence, the value of x = 54.

Question 7.
Find the mean proportional between 9 and 16.
Solution:
Let x be the mean proportional between 9 and 16.
9 : x :: x : 16
⇒ x × x = 9 × 16
⇒ x² = 144
⇒ x = √144 = 12
Hence, the required mean proportional = 12.

Question 8.
Find:
(i) 36% of 400
(ii) 1623% of 32
Solution:

Question 9.
Find a number whose 614% is 12.
Solution:
Let the required number be x.

Hence, the required number = 192.

Question 10.
What per cent of 40 kg is 440 g?
Solution:
Let x% of 40 kg = 440 g

Hence, the required Percentage = 1.1%

Comparing Quantities Class 7 Extra Questions Short Answer Type

Question 11.
Convert each of the following into the decimal form:
(а) 25.2%
(b) 0.15%
(c) 25%
Solution:

Question 12.
What per cent of
(a) 64 is 148.48?
(b) 75 is 1225?
Solution:

Question 13.
A machine costs ₹ 7500. Its value decreases by 5% every year due to usage. What will be its price after one year?
Solution:
The cost price of the machine = ₹ 7500
Decrease in price = 5%
Decreased price after one year

= 75 × 95
= ₹ 7125
Hence, the required price = ₹ 7125.

Question 14.
What sum of money lent out at 12 per cent p.a. simple interest would produce ₹ 9000 as interest in 2 years?
Solution:
Here, Interest = ₹ 9000
Rate = 12% p.a.
Time = 2 years
Principal = ?

Hence, the required principal amount = ₹ 37500.

Question 15.
Rashmi obtains 480 marks out of 600. Rajan obtains 560 marks out of 700. Whose performance is better?
Solution:
Rashmi obtains 480 marks out of 600
Marks Percentage = 480600 × 100 = 80%
Rajan obtains 560 marks out of 700
Marks Percentage = 560700 × 100 = 80%
Since, both of them obtained the same per cent of marks i.e. 80%.
So, their performance cannot be compared.

Question 16.
₹ 9000 becomes ₹ 18000 at simple interest in 8 years. Find the rate per cent per annum.
Solution:
Here, Principal = ₹ 9000
Amount = ₹ 18000
Interest = Amount – Principal = ₹ 18000 – ₹ 9000 = ₹ 9000

Hence, the required rate of interest = 1212%.

Question 17.
The cost of an object is increased by 12%. If the current cost is ₹ 896, what was its original cost?
Solution:
Here, rate of increase in cost = 12%
Increased Cost = ₹ 896
Original Cost = ?
Let the Original Cost be ₹ x

Hence, the required cost = ₹ 800.

Comparing Quantities Class 7 Extra Questions Long Answer Type

Question 18.
Radhika borrowed ₹ 12000 from her friends. Out of which ₹ 4000 were borrowed at 18% and the remaining at 15% rate of interest per annum. What is the total interest after 3 years? (NCERT Exemplar)
Solution:
Total amount borrowed by Radhika = ₹ 12,000
The amount borrowed by her at 18% p.a. = ₹ 4000

Total interest = ₹ 2160 + ₹ 3600 = ₹ 5760
Hence, the total interest = ₹ 5760.

Question 19.
Bhavya earns ₹ 50,000 per month and spends 80% of it. Due to pay revision, her monthly income increases by 20% but due to price rise, she has to spend 20% more. Find her new savings. (NCERT Exemplar)
Solution:
Monthly income of Bhavya = ₹ 50,000
Money spent by her = 80% of ₹ 50,000
= 80100 × 50,000 = ₹ 40,000
Due to pay revision, income is increased by 20%

So, the new savings = ₹ 60,000 – ₹ 48,000 = ₹ 12,000

Question 20.
The simple interest on a certain sum at 5% per annum for 3 years and 4 years differ by ₹ 82. Find the sum.
Solution:
Let the required sum be ₹ P.
Simple interest for 3 years

Alternate Method
Simple Interest gained from 3rd to 4th year = ₹ 82
Time (4th year – 3rd year) = 1 year

Required sum = ₹ 1640

Comparing Quantities Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 21.
Rajan’s monthly income is 20% more than the monthly income of Sarita. What per cent of Sarita’s income is less than Rajan’s monthly income?
Solution:
Let the monthly income of Sarita be ₹ 100.
Rajan’s monthly income

Now, Sarita’s monthly income is less than the monthly income of Raj an by = ₹ 120 – ₹ 100 = ₹ 20
Per cent of less in Rajan’s monthly income
= 20×100120 = 503% = 1623%
Hence, the required per cent = 1623%

Question 22.
If 10 apples are bought for ₹ 11 and sold at the rate of 11 apples for ₹ 10. Find the overall gain or loss per cent in these transactions.
Solution:
CP of 10 apples = ₹ 11
CP of 1 apple = ₹ 1110
SP of 11 apples = ₹ 10
SP of 1 apple = ₹ 1011

Question 23.
If 25 men can do a work in 36 hours, find the number of men required to do the same work in 108 hours.
Solution:
Let the number of men required to be x.
Men : Hours :: Men : Hours
25 : 36 :: x : 108
Product of extremes = 25 × 108
Product of means = 36 × x
Product of means = Product of extremes
36 × x = 25 × 108
⇒ x = 25 × 3 = 75
Hence, the required number of men = 75.

Question 24.
A machine is sold by A to B at a profit of 10% and then B sold it to C at a profit of 20%. If C paid ₹ 1200 for the machine, what amount was paid by A to purchase the machine?
Solution:
Cost price of machine for C = Selling price of the machine for B = ₹ 1200

Hence, the required cost price = ₹ 9091011 or ₹ 909.09 (approx)

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CHAPTER -7 Congruence of Triangles | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 7 Congruence of Triangles

MCQs

1. ‘Under a given correspondence, two triangles are congruent if the three sides of the one are equal to the three corresponding sides of the other.’
The above is known as
(a) SSS congruence of two triangles
(b) SAS congruence of two triangles
(c) ASA congruence of two triangles
(d) RHS congruence of two right-angled triangles

Answer

Answer: (a)

2. ‘Under a given correspondence, two triangles are congruent if two sides and the angle included between them in one of the triangles are equal to the corresponding sides and the angle included between them of the other triangle.’
The above is known as
(a) SSS congruence of two triangles
(b) SAS congruence of two triangles
(c) ASA congruence of two triangles
(d) RHS congruence of two right-angled triangles

Answer

Answer: (b)

3. ‘Under a given correspondence, two triangles are congruent if two angles and the side included between them in one of the triangles are equal to the corresponding angles and the side included between them of the other triangle.’
The above is known as
(а) SSS congruence of two triangles
(b) SAS congruence of two triangles
(c) ASA congruence of two triangles
(d) RHS congruence of two right-angled triangles

Answer

Answer: (c)

4. ‘Under a given correspondence, two right-angled triangles are congruent if the hypotenuse and a leg of one of the triangles are equal to the hypotenuse and the corresponding leg of the other triangle.’
The above is known as
(а) SSS congruence of two triangles
(b) SAS congruence of two triangles
(c) ASA congruence of two triangles
(d) RHS congruence of two right-angled triangles

Answer

Answer: (d)

5. For two given triangles ABC and PQR, how many matchings are possible?
(a) 2
(b) 4
(c) 6
(d) 3

Answer/Explanation

Answer: (c)
Explanation : ABC ↔ PQR, ABC ↔ PRQ,
ABC ↔ QRP, ABC ↔ QPR,
ABC ↔ RPQ, ABC ↔ RQP.

6. The symbol for congruence is
(a) ≡
(b) ≅
(c) ↔
(d) =

Answer

Answer: (b)

7. The symbol for correspondence is
(a) =
(b) ↔
(c) ≡
(d) ≅

Answer

Answer: (b)

8. If ∆ ABC = ∆ PQR, then AB¯¯¯¯¯¯¯¯ corresponds to
(a) PQ¯¯¯¯¯¯¯¯
(b) QR¯¯¯¯¯¯¯¯
(c) RP¯¯¯¯¯¯¯¯
(d) none of these

Answer

Answer: (a)

9. If ∆ ABC = ∆ PQR, then BC¯¯¯¯¯¯¯¯ corresponds to
(a) PQ¯¯¯¯¯¯¯¯
(b) QR¯¯¯¯¯¯¯¯
(c) RP¯¯¯¯¯¯¯¯
(d) none of these

Answer

Answer: (b)

10. If ∆ ABC = ∆ PQR, then CA¯¯¯¯¯¯¯¯ corresponds to
(a) PQ¯¯¯¯¯¯¯¯
(b) QR¯¯¯¯¯¯¯¯
(c) RP¯¯¯¯¯¯¯¯
(d) none of these

Answer

Answer: (c)

11. If ∆ ABC = ∆ PQR, then ∠A corresponds to
(a) ∠P
(b) ∠Q
(c) ∠R
(d) none of these

Answer

Answer: (a)

12. If ∆ ABC = ∆ PQR, then ∠B corresponds to
(a) ∠ P
(b) ∠ Q
(c) ∠ R
(d) none of these

Answer

Answer: (b)

13. If ∆ ABC= ∆ PQR, then ∠C corresponds to
(a) ∠ P
(b) ∠ Q
(c) ∠ R
(d) none of these

Answer

Answer: (c)

14. We want to show that ∆ ART = ∆ PEN and we have to use SSS criterion. We have AR = PE and RT = EN. What more we need to show?
(a) AT = PN
(b) AT = PE
(c) AT = EN
(d) none of these

Answer

Answer: (a)

15. We want to show that ∆ ART = ∆ PEN. We have to use SAS criterion. We have ∠ T = ∠ N, RT = EN. What more we need to show?
(a) PN = AT
(b) PN = AR
(c) PN = RT
(d) None of these

Answer

Answer: (a)

16. We want to show that ∆ ART = ∆ PEN. We have to use ASA criterion. We have AT = PN, ∠ A = ∠ P. What more we need to show?
(a) ∠T = ∠N
(b) ∠T = ∠E
(c) ∠T = ∠P
(d) None of these

Answer

Answer: (a)

17. Which congruence criterion do you use in the following?

Given AC = DF
AB = DE
BC = EF
So, ∆ ABC ≅ ∆ DEF
(a) SSS
(b) SAS
(c) ASA
(d) RHS

Answer

Answer: (a)

18. Which congruence criterion do you use in the following?

Given : ZX = RP
RQ = ZY
∠ PRQ = ∠ XZY
So, ∆ PRQ = ∆ XYZ
(a) SSS
(b) SAS
(c) ASA
(d) RHS

Answer

Answer: (b)

Important Questions

Question 1.
In the given figure, name
(a) the side opposite to vertex A
(b) the vertex opposite A to side AB
(c) the angle opposite to side AC
(d) the angle made by the sides CB and CA.

Solution:
(a) The side opposite to vertex A is BC.
(b) The vertex opposite to side AB is C.
(c) The angle opposite to side AB is ∠ACB.
(d) The angle made by the sides CB and CA is ∠ACB.

Question 2.
Examine whether the given triangles are congruent or not.
Solution:
Here,
AB = DE = 3 cm
BC = DF = 3.5 cm
AC = EF = 4.5 cm
ΔABC = ΔEDF (By SSS rule)

So, ΔABC and ΔEDF are congruent.

Question 3.
In the given congruent triangles under ASA, find the value of x and y, ΔPQR = ΔSTU.

Solution:
Given: ΔPQR = ΔSTU (By ASA rule)
∠Q = ∠T = 60° (given)
QR¯ = TU¯ = 4 cm (given)
∠x = 30° (for ASA rule)
Now in ΔSTU,
∠S + ∠T + ∠U = 180° (Angle sum property)
∠y + 60° + ∠x = 180°
∠y + 60° + 30° = 180°
∠y + 90° = 180°
∠y = 180° – 90° = 90°
Hence, x = 30° and y = 90°.

Question 4.
In the following figure, show that ΔPSQ = ΔPSR.

Solution:
In ΔPSQ and ΔPSR
PQ¯ = PR¯ = 6.5 cm (Given)
PS¯ = PS¯ (Common)
∠PSQ = ∠PSR = 90° (Given)
ΔPSQ = ΔPSR (By RHS rule)

Question 5.
Can two equilateral triangles always be congruent? Give reasons.
Solution:
No, any two equilateral triangles are not always congruent.
Reason: Each angle of an equilateral triangle is 60° but their corresponding sides cannot always be the same.

Question 6.
In the given figure, AP = BQ, PR = QS. Show that ΔAPS = ΔBQR

Solution:
In ΔAPS and ΔBQR
AP = BQ (Given)
PR = QS (Given)
PR + RS = QS + RS (Adding RS to both sides)
PS = QR
∠APS = ∠BQR = 90° (Given)
ΔAPS = ΔBQR (by SAS rule)

Question 7.
Without drawing the figures of the triangles, write all six pairs of equal measures in each of the following pairs of congruent triangles.
(i) ΔABC = ADEF
(ii) ΔXYZ = ΔMLN
Solution:
(i) Given: ΔABC = ΔDEF
Here AB = DE
BC = EF
AC = DF
∠A = ∠D, ∠B = ∠E and ∠C = ∠F
(ii) Given ΔXYZ = ΔMLN
Here XY = ML
YZ = LN
XZ = MN
∠X = ∠M, ∠Y = ∠L and ∠Z = ∠N

Question 8.
Lengths of two sides of an isosceles triangle are 5 cm and 8 cm, find the perimeter of the triangle.
Solution:
Since the lengths of any two sides of an isosceles triangle are equal, then
Case I: The three sides of the triangle are 5 cm, 5 cm and 8 cm.
Perimeter of the triangle = 5 cm + 5 cm + 8 cm = 18 cm
Case II: The three sides of the triangle are 5 cm, 8 cm and 8 cm.
Perimeter of the triangle = 5 cm + 8 cm + 8 cm = 21 cm
Hence, the required perimeter is 18 cm or 21 cm.

Question 9.
Write the rule of congruence in the following pairs of congruent triangles.

Solution:
(i) Here, AB = ST = 3 cm
BC = TU = 4.5 cm
∠ABC = ∠STU = 110°
ΔABC = ΔSTU (By SAS rule)
(ii) Here ∠PQR = ∠MNL = 90°
hypt. PR = hypt. ML
QR = NL = 3 cm
ΔPQR = ΔMNL (By RHS rule)

Question 10.
In the given figure, state the rule of congruence followed by congruent triangles LMN and ONM.

Solution:
In ΔLMN and ΔONM
LM = ON
LN = OM
MN = NM
ΔLMN = ΔONM

Congruence of Triangles Class 7 Extra Questions Short Answer Type

Question 11.
In the given figure, PQR is a triangle in which PQ = PR. QM and RN are the medians of the triangle. Prove that
(i) ΔNQR = ΔMRQ
(ii) QM = RN
(iii) ΔPMQ = ΔPNR

Solution:
ΔPQR is an isosceles triangle. [∵ PQ = PR]
⇒ 12 PQ = 12 PR
⇒ NQ = MR and PN = PM
(i) In ΔNQR and ΔMRQ
NQ = MR (Half of equal sides)
∠NQR = ∠MRQ (Angles opposite to equal sides)
QR = RQ (Common)
ΔNQR = ΔMRQ (By SAS rule)
(ii) QM = RN (Congruent parts of congruent triangles)
(iii) In ΔPMQ and ΔPNR
PN = PM (Half of equal sides)
PR = PQ (Given)
∠P = ∠P (Common)
ΔPMQ = ΔPNR (By SAS rule)

Question 12.
In the given figure, PQ = CB, PA = CR, ∠P = ∠C. Is ΔQPR = ΔBCA? If yes, state the criterion of congruence.

Solution:
Given:
PQ = CB, PA = CR
and ∠P = ∠C
In ΔQPR and ΔBCA,
PQ = CB (Given)
∠QPR = ∠BCA (Given)
PA = CR (Given)
PA + AR = CR + AR (Adding AR to both sides)
or PR = CA
ΔQPR = ΔBCA (By SAS rule)

Question 13.
In the given figure, state whether ΔABC = ΔEOD or not. If yes, state the criterion of congruence.

Solution:
In ΔABC and ΔEOD
AB = OE
∠ABC = ∠EOD = 90°
AC = ED
ΔABC = ΔEOD
Hence, ΔABC = ΔEOD
RHS is the criterion of congruence.

Question 14.
In the given figure, PQ || RS and PQ = RS. Prove that ΔPUQ = ΔSUR.

Solution:
In ΔPUQ and ΔSUR
PQ = SR = 4 cm
∠UPQ = ∠USR (Alternate interior angles)
∠PQU = ∠SRU (Alternate interior angles)
ΔPUQ = ΔSUR (By ASA rule)

Congruence of Triangles Class 7 Extra Questions Long Answer Type

Question 15.
In the given figure ΔBAC = ΔQRP by SAS criterion of congruence. Find the value of x and y.

Solution:
Given: ΔBAC = ΔQRP (By SAS rule)
So, BA = QR
⇒ 3x + 10 = 5y + 15 ……(i)
∠BAC = ∠QRP
⇒ 2x + 15° = 5x – 60° ……(ii)
From eq. (ii), we have
2x + 15 = 5x – 60
⇒ 2x – 5x = -15 – 60
⇒ -3x = -7 5
⇒ x = 25
From eq. (i), we have
3x + 10 = 5y + 15
⇒ 3 × 25 + 10 = 5y + 15
⇒ 75 + 10 = 5y + 15
⇒ 85 = 5y + 15
⇒ 85 – 15 = 5y
⇒ 70 = 5y
⇒ y = 14
Hence, the required values of x andy are 25 and 14 respectively.

Question 16.
Observe the figure and state the three pairs of equal parts in triangles ABC and DCB.
(i) Is ΔABC = ΔDCB? Why?
(ii) Is AB = DC? Why?
(iii) Is AC = DB? Why? (NCERT Exemplar)

Solution:
(i) In ΔABC and ΔDCB
∠ABC = ∠DCB = 70° (40° + 30° = 70°) (Given)
∠ACB = ∠DCB = 30° (Given)
BC = CB (Common)
ΔABC = ΔDCB (By ASA rule)
(ii) Yes,
AB = DC (Congruent parts of congruent triangles)
(iii) Yes,
AC = DB (Congruent parts of congruent triangles)

Question 17.
In the given figure, ΔQPS = ΔSRQ. Find each value.
(a) x
(b) ∠PQS
(c) ∠PSR

Solution:
(a) ΔQPS = ΔSRQ
∠QPS = ∠SRQ (Congruent part of congruent triangles)
106 = 2x + 12
⇒ 106 – 12 = 2x
⇒ 94 = 2x
⇒ x = 47
∠QRS = 2 × 47 + 12 = 94 + 12 = 106°
So, PQRS is a parallelogram.
∠QSR = 180° – (42° + 106°) = 180° – 148° = 32°
(b) ∠PQS = 32° (alternate interior angles)
(c) ∠PSQ = 180° – (∠QPS + ∠PQS) = 180° – (106° + 32°) = 180° – 138° = 42°
∠PSR = 32° + 42° = 74°

Congruence of Triangles Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 18.
In ΔABC, medians BD and CE are equal and intersect each other at O. Prove that ΔABC is an isosceles triangle.

Solution:
We know that the medians of a triangle intersect each other in the ratio 2 : 1.
BD = CE (Given)
23 BD = 23 CE
⇒ OB = OC
13 BD = 13 CE
⇒ OE = OD
In ΔBOE and ΔCOD,
OB = OC
OE = OD
∠BOE = ∠COD (Vertically opposite angles)
ΔBOE = ΔCOD (By SAS rule)
BE = CD (Congruent parts of congruent triangles)
2BE = 2CD
⇒ AB = AC
Hence ΔABC is an isosceles triangle.

Question 19.
Prove that the lengths of altitudes drawn to equal sides of an isosceles triangle are also equal.
(i) ∠TRQ = ∠SQR?
(ii) If ∠TRQ = 30°, find the base angles of the ΔPQR.
(iii) Is ΔPQR an equilateral triangle?

Solution:
In ΔQTR and ΔRSQ
∠QTR = ∠RSQ = 90° (Given)
∠TQR = ∠SRQ (Base angle of an isosceles triangle)
∠QRT = ∠RQS (Remaining third angles)
QR = QR (Common)
ΔQTR = ΔRSQ (By ASA rule)
QS = RT (Congruent parts of congruent triangles)
Hence proved.
(i) ∠TRQ = ∠SQR (Congruent parts of congruent triangles)
(ii) In ΔQTR,
∠TRQ = 30° (Given)
∠QTR + ∠TQR + ∠QRT = 180° (Angle sum property)
⇒ 90° + ∠TQR + 30° = 180°
⇒ 120° + ∠TQR = 180°
⇒ ∠TQR = 180° – 120° = 60°
⇒ ∠TQR = ∠SRQ = 60°
Each base angle = 60°
(iii) In ΔPQR,
∠P + ∠Q + ∠R = 180° (Angle sum property)
⇒ ∠P + 60° + 60° = 180° (From ii)
⇒ ∠P + 120° = 180°
⇒ ∠P = 180° – 120° = 60°
Hence, ΔPQR is an equilateral triangle.

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CHAPTER -6 The Triangle and its Properties | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 6 The Triangle and its Properties

MCQs

1. How many elements are there in a triangle?
(a) 3
(b) 6
(c) 4
(d) None of these.

Answer/Explanation

Answer: (b)
Explanation : See a triangle.

2. How many vertices does a triangle have?
(a) 1
(b) 2
(c) 3
(d) 4

Answer/Explanation

Answer: (c)
Explanation : See a triangle.

3. How many sides are there in a triangle?
(a) 1
(b) 2
(c) 3
(d) 4

Answer/Explanation

Answer: (c)
Explanation : See a triangle.

4. How many angles are there in a triangle?
(a) 1
(b) 2
(c) 3
(d) 4

Answer/Explanation

Answer: (c)
Explanation : See a triangle.

5. If two sides of a triangle are not equal, the triangle is called
(a) scalene
(b) isosceles
(c) equilateral
(d) right-angled

Answer/Explanation

Answer: (a)
Explanation : Definition of a scalene triangle.

6. If two sides of a triangle are equal, the triangle is called
(a) isosceles
(b) equilateral
(c) scalene
(d) right-angled

Answer/Explanation

Answer: (a)
Explanation : Definition of an isosceles triangle.

7. If all the three sides of a triangle are equal, the triangle is called
(a) equilateral
(b) right-angled
(c) isosceles
(d) scalene

Answer/Explanation

Answer: (a)
Explanation : Definition of an equilateral triangle.

8. If all the angles of a triangle are acute, the triangle is called
(a) obtuse-angled
(b) acute-angled
(c) right-angled
(d) none of these

Answer/Explanation

Answer: (b)
Explanation : Definition of an acute-angled triangle.

9. If one angle of a triangle measures 90°, the triangle is called
(a) acute-angled
(b) obtuse-angled
(c) right-angled
(d) none of these

Answer/Explanation

Answer: (c)
Explanation : Definition of a right triangle.

10. If one angle of a triangle is obtuse, the triangle is called
(a) acute-angled
(b) obtuse-angled
(c) right-angled
(d) none of these

Answer/Explanation

Answer: (b)
Explanation : Definition of an obtuse angled triangle.

11, How many medians can a triangle have?
(a) 1
(b) 2
(c) 3
(d) 4

Answer/Explanation

Answer: (c)
Explanation : Draw medians and count.

12. How many altitudes can a triangle have?
(a) 1
(b) 2
(c) 3
(d) 4

Answer/Explanation

Answer: (c)
Explanation : Draw altitudes and count.

13. The total measure of the three angles of a triangle is
(a) 360°
(b) 90°
(c) 180°
(d) none of these

Answer/Explanation

Answer: (c)
Explanation : Angle Sum Property of a triangle.

14. The measure of each angle of an equilateral triangle is
(a) 30°
(b) 45°
(c) 90°
(d) 60°

Answer/Explanation

Answer: (d)
Explanation : x° + x° + x° = 180° ⇒ x° = 60°.

15. Which of the following statements is true?
(a) A triangle can have two right angles
(b) A triangle can have two obtuse angles
(c) A triangle can have two acute angles
(d) A triangle can have all the three angles less than 60°

Answer/Explanation

Answer: (c)
Explanation :

Important Questions

Question 1.
In ∆ABC, write the following:
(a) Angle opposite to side BC.
(b) The side opposite to ∠ABC.
(c) Vertex opposite to side AC.

Solution:
(a) In ∆ABC, Angle opposite to BC is ∠BAC
(b) Side opposite to ∠ABC is AC
(c) Vertex opposite to side AC is B

Question 2.
Classify the following triangle on the bases of sides

Solution:
(i) PQ = 5 cm, PR = 6 cm and QR = 7 cm
PQ ≠ PR ≠ QR
Thus, ∆PQR is a scalene triangle.
(ii) AB = 4 cm, AC = 4 cm
AB = AC
Thus, ∆ABC is an isosceles triangle.
(iii) MN = 3 cm, ML = 3 cm and NL = 3 cm
MN = ML = NL
Thus, ∆MNL is an equilateral triangle.

Question 3.
In the given figure, name the median and the altitude. Here E is the midpoint of BC.
Solution
In ∆ABC, we have

AD is the altitude.
AE is the median.

Question 4.
In the given diagrams, find the value of x in each case.

Solution:
(i) x + 45° + 30° = 180° (Angle sum property of a triangle)
⇒ x + 75° – 180°
⇒ x = 180° – 75°
x = 105°
(ii) Here, the given triangle is right angled triangle.
x + 30° = 90°
⇒ x = 90° – 30° = 60°
(iii) x = 60° + 65° (Exterior angle of a triangle is equal to the sum of interior opposite angles)
⇒ x = 125°

Question 5.
Which of the following cannot be the sides of a triangle?
(i) 4.5 cm, 3.5 cm, 6.4 cm
(ii) 2.5 cm, 3.5 cm, 6.0 cm
(iii) 2.5 cm, 4.2 cm, 8 cm
Solution:
(i) Given sides are, 4.5 cm, 3.5 cm, 6.4 cm
Sum of any two sides = 4.5 cm + 3.5 cm = 8 cm
Since 8 cm > 6.4 cm (Triangle inequality)
The given sides form a triangle.

(ii) Given sides are 2.5 cm, 3.5 cm, 6.0 cm
Sum of any two sides = 2.5 cm + 3.5 cm = 6.0 cm
Since 6.0 cm = 6.0 cm
The given sides do not form a triangle.

(iii) 2.5 cm, 4.2 cm, 8 cm
Sum of any two sides = 2.5 cm + 4.2 cm = 6.7 cm
Since 6.7 cm < 8 cm
The given sides do not form a triangle.

Question 6.
In the given figure, find x.

Solution:
In ∆ABC, we have
5x – 60° + 2x + 40° + 3x – 80° = 180° (Angle sum property of a triangle)
⇒ 5x + 2x + 3x – 60° + 40° – 80° = 180°
⇒ 10x – 100° = 180°
⇒ 10x = 180° + 100°
⇒ 10x = 280°
⇒ x = 28°
Thus, x = 28°

Question 7.
One of the equal angles of an isosceles triangle is 50°. Find all the angles of this triangle.
Solution:
Let the third angle be x°.
x + 50° + 50° = 180°
⇒ x° + 100° = 180°
⇒ x° = 180° – 100° = 80°
Thus ∠x = 80°

Question 8.
In ΔABC, AC = BC and ∠C = 110°. Find ∠A and ∠B.

Solution:
In given ΔABC, ∠C = 110°
Let ∠A = ∠B = x° (Angle opposite to equal sides of a triangle are equal)
x + x + 110° = 180°
⇒ 2x + 110° = 180°
⇒ 2x = 180° – 110°
⇒ 2x = 70°
⇒ x = 35°
Thus, ∠A = ∠B = 35°

The Triangles and its Properties Class 7 Extra Questions Short Answer Type

Question 9.
Two sides of a triangle are 4 cm and 7 cm. What can be the length of its third side to make the triangle possible?
Solution:
Let the length of the third side be x cm.
Condition I: Sum of two sides > the third side
i.e. 4 + 7 > x ⇒ 11 > x ⇒ x < 11
Condition II: The difference of two sides less than the third side.
i.e. 7 – 4 < x ⇒ 3 < x ⇒ x > 3
Hence the possible value of x are 3 < x < 11
i.e. x < 3 < 11

Question 10.
Find whether the following triplets are Pythagorean or not?
(a) (5, 8, 17)
(b) (8, 15, 17)
Solution:
(a) Given triplet: (5, 8, 17)
17² = 289
8² = 64
5² = 25
8² + 5² = 64 + 25 = 89
Since 89 ≠ 289
5² + 8² ≠ 17²
Hence (5, 8, 17) is not Pythagorean triplet.

(b) Given triplet: (8, 15, 17)
17² = 289
15² = 225
8² = 64
15² + 8² = 225 + 64 = 289
17² = 15² + 8²
Hence (8, 15, 17) is a Pythagorean triplet.

Question 11.
In the given right-angled triangle ABC, ∠B = 90°. Find the value of x.

Solution:
In ΔABC, ∠B = 90°
AB² + BC² = AC² (By Pythagoras property)
(5)² + (x – 3)² = (x + 2)²
⇒ 25 + x² + 9 – 6x = x² + 4 + 4x
⇒ -6x – 4x = 4 – 9 – 25
⇒ -10x = -30
⇒ x = 3
Hence, the required value of x = 3

Question 12.
AD is the median of a ΔABC, prove that AB + BC + CA > 2AD (HOTS)

Solution:
In ΔABD,
AB + BD > AD …(i)
(Sum of two sides of a triangle is greater than the third side)
Similarly, In ΔADC, we have
AC + DC > AD …(ii)
Adding (i) and (ii), we have
AB + BD + AC + DC > 2AD
⇒ AB + (BD + DC) + AC > 2AD
⇒ AB + BC + AC > 2AD
Hence, proved.

Question 13.
The length of the diagonals of a rhombus is 42 cm and 40 cm. Find the perimeter of the rhombus.

Solution:
AC and BD are the diagonals of a rhombus ABCD.
Since the diagonals of a rhombus bisect at the right angle.
AC = 40 cm
AO = 402 = 20 cm
BD = 42 cm
OB = 422 = 21 cm
In right angled triangle AOB, we have
AO² + OB² = AB²
⇒ 20² + 21² = AB²
⇒ 400 + 441 = AB²
⇒ 841 = AB²
⇒ AB = √841 = 29 cm.
Perimeter of the rhombus = 4 × side = 4 × 29 = 116 cm
Hence, the required perimeter = 116 cm

Question 14.
The sides of a triangle are in the ratio 3 : 4 : 5. State whether the triangle is right-angled or not.
Solution:
Let the sides of the given triangle are 3x, 4x and 5x units.
For right angled triangle, we have
Square of the longer side = Sum of the square of the other two sides
(5x)² = (3x)² + (4x)²
⇒ 25x² = 9x² + 16x²
⇒ 25x² = 25x²
Hence, the given triangle is a right-angled.

Question 15.
A plane flies 320 km due west and then 240 km due north. Find the shortest distance covered by the plane to reach its original position.
Solution:
Here, OA = 320 km
AB = 240 km
OB = ?

Clearly, ∆OBA is right angled triangle
OB² = OA² + AB² (By Pythagoras property)
⇒ OB² = 320² + 240²
⇒ OB² = 102400 + 57600
⇒ OB² = 160000
⇒ OB = √160000 = 400 km.
Hence the required shortest distance = 400 km.

The Triangles and its Properties Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 16.
In the following figure, find the unknown angles a and b, if l || m.

Solution:
Here, l || m
∠c = 110° (Corresponding angles)
∠c + ∠a = 180° (Linear pair)
⇒ 110° + ∠a = 180°
⇒ ∠a = 180° – 110° = 70°
Now ∠b = 40° + ∠a (Exterior angle of a triangle)
⇒ ∠b = 40° + 70° = 110°
Hence, the values of unknown angles are a = 70° and b = 110°

Question 17.
In figure (i) and (ii), Find the values of a, b and c.

Solution:
(i) In ∆ADC, we have
∠c + 60° + 70° = 180° (Angle sum property)
⇒ ∠c + 130° = 180°
⇒ ∠c = 180° – 130° = 50°
∠c + ∠b = 180° (Linear pair)
⇒ 50° + ∠b = 180°
⇒ ∠ b = 180° – 50° = 130°
In ∆ABD, we have
∠a + ∠b + 30° = 180° (Angle sum property)
⇒ ∠a + ∠130° + 30° = 180°
⇒ ∠a + 160° = 180°
⇒ ∠a = 180° – 160° = 20°
Hence, the required values are a = 20°, b = 130° and c = 50°

(ii) In ∆PQS, we have
∠a + 60° + 55° = 180°(Angle sum property)
⇒ ∠a + 115° = 180°
⇒ ∠a = 180° – 115°
⇒ ∠a = 65°
∠a + ∠b = 180° (Linear pair)
⇒ 65° + ∠b = 180°
⇒ ∠b = 180° – 65° = 115°
In ∆PSR, we have
∠b + ∠c + 40° = 180° (Angle sum property)
⇒ 115° + ∠c + 40° = 180°
⇒ ∠c + 155° = 180°
⇒ ∠c = 180° – 155° = 25°
Hence, the required angles are a = 65°, b = 115° and c = 25°

Question 18.
I have three sides. One of my angle measure 15°. Another has a measure of 60°. What kind of a polygon am I? If I am a triangle, then what kind of triangle am I? [NCERT Exemplar]
Solution:
Since I have three sides.
It is a triangle i.e. three-sided polygon.
Two angles are 15° and 60°.
Third angle = 180° – (15° + 60°)
= 180° – 75° (Angle sum property)
= 105°
which is greater than 90°.
Hence, it is an obtuse triangle.

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CHAPTER -5 Lines and Angles | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 5 Lines and Angles

MCQs

1. When the sum of the measures of two angles is 90°, the angles are called
(а) supplementary angles
(b) complementary angles
(c) adjacent angles
(d) vertically opposite angles

Answer/Explanation

Answer: (b)
Explanation : Definition of complementary angles

2. The sum of the measures of two complementary angles is
(a) 180°
(b) 60°
(c) 45°
(d) 90°

Answer/Explanation

Answer: (d)
Explanation : Definition of complementary angles

3. The measure of the complement of the angle 30° is
(а) 30°
(b) 16°
(c) 60°
(d) 160°

Answer/Explanation

Answer: (c)
Explanation : 90° – 30° = 60°.

4. Which of the following statements is true?
(a) Two acute angles can be complementary to each other
(b) Two obtuse angles can be complementary to each other
(c) Two right angles can be complementary to each other
(d) One obtuse angle and one acute angle can be complementary to each other

Answer

Answer: (a)

5. The measure of the complement of the angle 46° is
(a) 90°
(b) 46°
(c) 16°
(d) 136°

Answer/Explanation

Answer: (b)
Explanation : 90°- 45° = 45°.

6. What is the measure of the complement of the angle 80°?
(a) 10°
(b) 100°
(c) 36°
(d) 20°

Answer/Explanation

Answer: (a)
Explanation : 90° – 80° = 10°.

7. Which pair of the following angles are complementary?

Answer/Explanation

Answer: (a)
Explanation : 60°+ 30° = 90°.

8. The measure of the angle which is equal to its complement is
(a) 30°
(b)60°
(c) 46°
(d) 90°

Answer/Explanation

Answer: (c)
Explanation : x° + x° = 90° ⇒ x° = 45°.

9. Which of the following pairs of angles is not a pair of complementary angles?
(a) 60°, 30°
(b) 66°, 34°
(c) 0°, 90°
(d) 160°, 30°

Answer/Explanation

Answer: (d)
Explanation : 150° + 30° = 180° ≠ 90°.

10. What is the measure of the complement of the angle 90°?
(а) 90°
(b) 0°
(c) 180°
(d) 46°

Answer/Explanation

Answer: (b)
Explanation : 90° – 90° = 0°.

11. When the sum of the measures of two angles is 180°, the angles are called
(a) adjacent angles
(b) complementary angles
(c) vertically opposite angles
(d) supplementary angles

Answer/Explanation

Answer: (d)
Explanation : Definition of supplementary angles.

12. The sum of the measures of two supplementary angles is
(a) 90°
(b) 180°
(c) 360°
(d) none of these

Answer/Explanation

Answer: (b)
Explanation : Definition of supplementary angles.

13. The measure of the supplement of the angle 120° is
(a) 30°
(b)45°
(c) 60°
(d) 90°

Answer/Explanation

Answer: (c)
Explanation : 180° – 120° = 60°.

14. Which of the following statements is true?
(a) Two acute angles can be supplementary.
(b) Two right angles can be supplementary.
(c) Two obtuse angles can be supplementary.
(d) One obtuse angle and one acute angle cannot be supplementary

Answer

Answer: (b)

15. The measure of the supplement of the angle 90° is
(a) 45°
(b) 60°
(c) 30°
(d) 90°

Answer/Explanation

Answer: (d)
Explanation : 180° – 90° = 90°.

16. The measure of the angle which is equal to its supplement is
(a) 30°
(b) 45°
(c) 90°
(d) 60°

Answer/Explanation

Answer: (c)
Explanation : x° + x° = 180° ⇒ x° = 90°.

Important Questions

Question 1.
Find the angles which is 15 of its complement.
Solution:
Let the required angle be x°
its complement = (90 – x)°
As per condition, we get

Question 2.
Find the angles which is 23 of its supplement.
Solution:
Let the required angle be x°.
its supplement = (180 – x)°
As per the condition, we get
23 of (180 – x)° = x°

Question 3.
Find the value of x in the given figure.

Solution:
∠POR + ∠QOR = 180° (Angles of linear pair)
⇒ (2x + 60°) + (3x – 40)° = 180°
⇒ 2x + 60 + 3x – 40 = 180°
⇒ 5x + 20 = 180°
⇒ 5x = 180 – 20 = 160
⇒ x = 32
Thus, the value of x = 32.

Question 4.
In the given figure, find the value of y.

Solution:
Let the angle opposite to 90° be z.
z = 90° (Vertically opposite angle)
3y + z + 30° = 180° (Sum of adjacent angles on a straight line)
⇒ 3y + 90° + 30° = 180°
⇒ 3y + 120° = 180°
⇒ 3y = 180° – 120° = 60°
⇒ y = 20°
Thus the value of y = 20°.

Question 5.
Find the supplements of each of the following:
(i) 30°
(ii) 79°
(iii) 179°
(iv) x°
(v) 25 of right angle
Solution:
(i) Supplement of 30° = 180° – 30° = 150°
(ii) Supplement of 79° = 180° – 79° = 101°
(iii) Supplement of 179° = 180° – 179° = 1°
(iv) Supplement of x° = (180 – x)°
(v) Supplement of 25 of right angle
= 180° – 25 × 90° = 180° – 36° = 144°

Question 6.
If the angles (4x + 4)° and (6x – 4)° are the supplementary angles, find the value of x.
Solution:
(4x + 4)° + (6x – 4)° = 180° (∵ Sum of the supplementary angle is 180°)
⇒ 4x + 4 + 6x – 4 = 180°
⇒ 10x = 180°
⇒ x = 18°
Thus, x = 18°

Question 7.
Find the value of x.

Solution:
(6x – 40)° + (5x + 9)° + (3x + 15) ° = 180° (∵ Sum of adjacent angles on straight line)
⇒ 6x – 40 + 5x + 9 + 3x + 15 = 180°
⇒ 14x – 16 = 180°
⇒ 14x = 180 + 16 = 196
⇒ x = 14
Thus, x = 14

Question 8.
Find the value of y.

Solution:
l || m, and t is a transversal.
y + 135° = 180° (Sum of interior angles on the same side of transversal is 180°)
⇒ y = 180° – 135° = 45°
Thus, y = 45°

Lines and Angles Class 7 Extra Questions Short Answer Type

Question 9.
Find the value ofy in the following figures:

Solution:
(i) y + 15° = 360° (Sum of complete angles round at a point)
⇒ y = 360° – 15° = 345°
Thus, y = 345°
(ii) (2y + 10)° + 50° + 40° + 130° = 360° (Sum of angles round at a point)
⇒ 2y + 10 + 220 = 360
⇒ 2y + 230 = 360
⇒ 2y = 360 – 230
⇒ 2y = 130
⇒ y = 65
Thus, y = 65°
(iii) y + 90° = 180° (Angles of linear pair)
⇒ y = 180° – 90° = 90°
[40° + 140° = 180°, which shows that l is a straight line]

Question 10.
In the following figures, find the lettered angles.

Solution:
(i) Let a be represented by ∠1 and ∠2
∠a = ∠1 + ∠2
∠1 = 35° (Alternate interior angles)
∠2 = 55° (Alternate interior angles)
∠1 + ∠2 = 35° + 55°
∠a = 90°
Thus, ∠a = 90°

Question 11.
In the given figure, prove that AB || CD.

Solution:
∠CEF = 30° + 50° = 80°
∠DCE = 80° (Given)
∠CEF = ∠DCE
But these are alternate interior angle.
CD || EF ……(i)
Now ∠EAB = 130° (Given)
∠AEF = 50° (Given)
∠EAB + ∠AEF = 130° + 50° = 180°
But these are co-interior angles.
AB || EF …(ii)
From eq. (i) and (ii), we get
AB || CD || EF
Hence, AB || CD
Co-interior angles/Allied angles: Sum of interior angles on the same side of transversal is 180°.

Question 12.
In the given figure l || m. Find the values of a, b and c.

Solution:
(i) We have l || m
∠b = 40° (Alternate interior angles)
∠c = 120° (Alternate interior angles)
∠a + ∠b + ∠c = 180° (Sum of adjacent angles on straight angle)
⇒ ∠a + 40° + 120° = 180°
⇒ ∠a + 160° = 180°
⇒ ∠a = 180° – 160° = 20°
Thus, ∠a = 20°, ∠b = 40° and ∠c = 120°.
(ii) We have l || m
∠a = 45° (Alternate interior angles)
∠c = 55° (Alternate interior angles)
∠a + ∠b + ∠c = 180° (Sum of adjacent angles on straight line)
⇒ 45 + ∠b + 55 = 180°
⇒ ∠b + 100 = 180°
⇒ ∠b = 180° – 100°
⇒ ∠b = 80°

Question 13.
In the adjoining figure if x : y : z = 2 : 3 : 4, then find the value of z.

Solution:
Let x = 2s°
y = 3s°
and z = 4s°
∠x + ∠y + ∠z = 180° (Sum of adjacent angles on straight line)
2s° + 3s° + 4s° = 180°
⇒ 9s° = 180°
⇒ s° = 20°
Thus x = 2 × 20° = 40°, y = 3 × 20° = 60° and z = 4 × 20° = 80°

Question 14.
In the following figure, find the value of ∠BOC, if points A, O and B are collinear. (NCERT Exemplar)

Solution:
We have A, O and B are collinear.
∠AOD + ∠DOC + ∠COB = 180° (Sum of adjacent angles on straight line)
(x – 10)° + (4x – 25)° + (x + 5)° = 180°
⇒ x – 10 + 4x – 25 + x + 5 = 180°
⇒ 6x – 10 – 25 + 5 = 180°
⇒ 6x – 30 = 180°
⇒ 6x = 180 + 30 = 210
⇒ x = 35
So, ∠BOC = (x + 5)° = (35 + 5)° = 40°

Question 15.
In given figure, PQ, RS and UT are parallel lines.
(i) If c = 57° and a = c3, find the value of d.
(ii) If c = 75° and a = 25c , find b.

Solution:
(i) We have ∠c = 57° and ∠a = ∠c3
∠a = 573 = 19°
PQ || UT (given)
∠a + ∠b = ∠c (Alternate interior angles)
19° + ∠b = 57°
∠b = 57° – 19° = 38°
PQ || RS (given)
∠b + ∠d = 180° (Co-interior angles)
38° + ∠d = 180°
∠d = 180° – 38° = 142°
Thus, ∠d = 142°
(ii) We have ∠c = 75° and ∠a = 25 ∠c
∠a = 25 × 75° = 30°
PQ || UT (given)
∠a + ∠b = ∠c
30° + ∠b = 75°
∠b = 75° – 30° = 45°
Thus, ∠b = 45°

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CHAPTER -4 Simple Equations | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 4 Simple Equations

MCQs

1. Write the following statement in the form of an equation:
The sum of three times x and 10 is 13.
(a) 3x + 10 = 13
(b) 3x – 10 = 13
(c) 3x + 13 = 10
(d) none of these

Answer

Answer: (a)

2. Write the following statement in the form of an equation:
If you subtract 3 from 6 times a number, you get 9
(a) 3x – 6 = 9
(b) 6x – 3 = 9
(c) 6x + 3 = 9
(d) 3x + 6 = 9

Answer

Answer: (b)

3. Write the following statement in the form of an equation:
One fourth of n is 3 more than 2
(a) n4 – 2 = 3
(b) n4 + 2 = 3
(c) n2 – 4 = 3
(d) n2 + 4 = 3

Answer

Answer: (a)

4. Write the following statement in the form of an equation:
One third of a number plus 2 is 3
(a) m3 – 2 = 3
(b) m3 + 2 = 3
(c) m2 – 3 = 3
(d) m2 + 3 = 3

Answer

Answer: (b)

5. Write the following statement in the form of an equation:
Taking away 5 from x gives 10
(a) x – 5 = 10
(b) x + 5 = 10
(c) x – 10 – 5
(d) none of these

Answer

Answer: (a)

6. Write the following statement in the form of an equation:
Four times a number p is 8.
(a) 4P = 8
(b) P + 4 = 8
(c) p – 4 = 8
(d) p ÷ 4 = 8

Answer

Answer: (a)

7. Write the following statement in the form of an equation:
Add 1 to three times n to get 7
(a) 3n + 1 = 7
(b) 3n – 1 = 7
(c) 3n + 7 = 1
(d) none of these

Answer

Answer: (a)

8. Write the following statement in the form of an equation:
The number b divided by 6 gives 5.
(a) b6 = 5
(b) b – 5 = 6
(c) 5b = 6
(d) b + 5 = 6

Answer

Answer: (a)

9. The solution of the equation x + 3 = 0 is
(a) 3
(b) – 3
(c) 0
(d) 1

Answer

Answer: (b)

10. The solution of the equation x – 6 = 1 is
(a) 1
(b) 6
(c) – 7
(b) 7

Answer

Answer: (d)

11. The solution of the equation 5x = 10 is
(a) 1
(b) 2
(c) 5
(d) 10

Answer

Answer: (b)

12. The solution of the equation m2 = 3 is
(a) 2
(b) 3
(c) 12
(d) 6

Answer

Answer: (d)

13. The solution of the equation 7n + 5 = 12 is
(a) 0
(b) – 1
(c) 1
(d) 5

Answer

Answer: (c)

14. The solution of the equation 4p – 3 = 9 is
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (c)

15. The solution of the equation 5p + 2 = 7 is
(a) 0
(b) 1
(c) – 1
(d) 2

Answer

Answer: (b)

16. The solution of the equation 3p – 2 = 4 is
(a) 0
(b) 1
(c) 2
(d) 3

Answer

Answer: (c)

Important Questions

Question 1.
Write the following statements in the form of equations.
(a) The sum of four times a number and 5 gives a number five times of it.
(b) One-fourth of a number is 2 more than 5.
Solution:
(a) Let the number be x.
Sum of 4x and 5 = 4x + 5
The sum is 5x.
The equation is 4x + 5 = 5x as required.
(b) Let the number be x.
14x = 5 + 2
⇒ 14x = 7 as required.

Question 2.
Convert the following equations in statement form:
(a) 5x = 20
(b) 3y + 7 = 1
Solution:
(a) Five times a number x gives 20.
(b) Add 7 to three times a number y gives 1.

Question 3.
If k + 7 = 10, find the value of 9k – 50.
Solution:
k + 7 = 10
⇒ k = 10 – 7 = 3
Put k = 3 in 9k – 50, we get
9 × 3 – 50 = 27 – 50 = -23
Thus the value of k = -23

Question 4.
Solve the following equations and check the answers.

Solution:

Question 5.
Solve the following equations:
3(y – 2) = 2(y – 1) – 3
Solution:
3(y – 2) = 2(y – 1) – 3
⇒ 3y – 6 = 2y – 2 – 3 (Removing the brackets)
⇒ 3y – 6 = 2y – 5
⇒ 3y – 2y = 6 – 5 (Transposing 6 to RHS and 2y to LHS)
⇒ y = 1
Thus y = 1

Question 6.
If 5 is added to twice a number, the result is 29. Find the number.
Solution:
Let the required number be x.
Step I: 2x + 5
Step II: 2x + 5 = 29
Solving the equation, we get
2x + 5 = 29
⇒ 2x = 29 – 5 (Transposing 5 to RHS)
⇒ 2x = 24
⇒ x = 12 (Dividing both sides by 2)
⇒ x = 12
Thus the required number is 12.

Question 7.
If one-third of a number exceeds its one-fourth by 1, find the number.
Solution:
Let the required number be x.

Simple Equations Class 7 Extra Questions Short Answer Type

Question 8.
The length of a rectangle is twice its breadth. If its perimeter is 60 cm, find the length and the breadth of the rectangle.
Solution:
Let the breadth of the rectangle be x cm.
its length = 2x
Perimeter = 2 (length + breadth) = 2(2x + x) = 2 × 3x = 6x
As per the condition of the question, we have
6x = 60 ⇒ x = 10
Thus the required breadth = 10 cm
and the length = 10 × 2 = 20 cm.

Question 9.
Seven times a number is 12 less than thirteen times the same number. Find the number.
Solution:
Let the required number be x.
7x = 13x – 12
⇒ 7x – 13x = -12 (Transposing 13x to LHS)
⇒ -6x = -12
⇒ x = 2
Thus, the required number is 2.

Question 10.
The present age of a son is half the present age of his father. Ten years ago, the father was thrice as old as his son. What are their present age?
Solution:
Let the present age of a father be x years.
Son’s age = 12x years
10 years ago, father’s age was (x – 10) years
10 years ago, son’s age was (x2 – 10) years
As per the question, we have

Question 11.
The sum of three consecutive multiples of 2 is 18. Find the numbers.
Solution:
Let the three consecutive multiples of 2 be 2x, 2x + 2 and 2x + 4.
As per the conditions of the question, we have
2x + (2x + 2) + (2x + 4) = 18
⇒ 2x + 2x + 2 + 2x + 4 = 18
⇒ 6x + 6 = 18
⇒ 6x = 18 – 6 (Transposing 6 to RHS)
⇒ 6x = 12
⇒ x = 2
Thus, the required multiples are
2 × 2 = 4, 4 + 2 = 6, 6 + 2 = 8 i.e., 4, 6 and 8.

Question 12.
Each of the 2 equal sides of an isosceles triangle is twice as large as the third side. If the perimeter of the triangle is 30 cm, find the length of each side of the triangle. [NCERT Exemplar]
Solution:
Let the length of the third side be x cm.
Each equal side = 2x cm.
As per the condition of the question, we have
Perimeter = x + 2x + 2x = 30
⇒ 5x = 30
⇒ x = 6
Thus, the third side of the triangle = 6 cm
and other two equal sides are 2 × 6 = 12 cm each

Question 13.
A man travelled two-fifth of his journey by train, one-third by bus, one-fourth by car and the remaining 3 km on foot. What is the length of his total journey? [NCERT Exemplar]
Solution:
Let the total length of total journey be x km.
Distance travelled by train = 25x km
Distance travelled by bus = 13x km
Distance travelled by car = 14x km
Remaining distance = 3 km
As per the question, we have

Thus, the required journey = 180 km.

Author Archives: Rohit

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MCQs

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Chapter - 6 The Triangle and its Properties

MCQs

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MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 5 Lines and Angles

MCQs

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Chapter - 4 Simple Equations

MCQs

Simple Equations Class 7 Extra Questions Short Answer Type

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