During revision, students quickly go through Class 7 Maths Extra Questions Chapter 4 Expressions using Letter Numbers Class 7 Extra Questions with Answers for clarity.
Class 7 Expressions using Letter Numbers Extra Questions
Class 7 Maths Chapter 4 Expressions using Letter Numbers Extra Questions
Class 7 Maths Chapter 4 Extra Questions – Expressions using Letter Numbers Extra Questions Class 7
Question 1.
Identify the following shape sequence. Can you identify the number pattern involved in it? Write it. And, extend the pattern by drawing next two shapes.
Solution:
Starting from 1 each shape of the given pattern is obtained by adding 1 vertex to the previous shape.
So, the number pattern involved is 1, 2, 3, 4, 5, 6, 7,…
And, next two shapes are:
Question 2.
Study the given patterns and complete the table that follows:
Solution:
It can be observed from the given figures that 2 matchsticks are fixed and 3 matchsticks are added at each step. The general term for the number of matchsticks in a figure will be 3n + 2.
The complete table is
Solution:
By observing the matchsticks pattern, the general term for the number of matchsticks will be 7n + 1.
Solution:
By observing the matchsticks pattern, the genera term for the number of matchsticks will be 3n + 2.
Question 3.
Make the following patterns using matchsticks and find the general rule. Use a letter-number to write the rule.
Solution:
In the given figure there are 4 matchsticks. To get the two similar figures we need 8 matchsticks, for three similar figures 12 matchsticks, and so on.
So, the number of matchsticks in «th figure is An.
Solution:
Similar as part (a).
General term for n number of figures is 5n.
Solution:
To get the two similar figures, we just need to add 4 more matchsticks and so on.
The general rule is 4n + 2.
Solution:
To get the two similar figures, we just need to add 5 more matchsticks in the given figure, and so on.
The general rule is 5n + 2.
Solution:
To get the two similar figures, we just need to add 4 more matchsticks in the given figure, and so on.
The general rule is 4n + 1.
Question 4.
Look at the following and answer the questions:
What is the length of the knife?
Solution:
The length of the knife = x cm + 4 cm = (x + 4) cm
What is length of the handle of the screwdriver?
Solution:
The length of the handle of the screwdriver
= x cm – 3 cm
= (x – 3) cm
What is perimeter of the notebook?
Solution:
Perimeter of the notebook = 2(length + breadth)
= 2(4x + 2y) units
Area of the notebook = (length × breadth) sq units
= 4x × 2y sq units
= 8xy sq units
If there are x apples and y bananas in the baskets. What is the total cost of the fruits altogether?
Solution:
Total cost of the fruits = total cost of apples + total cost of bananas
= ₹ 5 × x + ₹ 2 × y
= ₹ 5x + ₹ 2y
= ₹ (5x + 2y)
Question 5.
Write expression for the following:
Take Ruhi’s present age to be x years.
(a) What will be her age after 5 years from now?
(b) What was her age 3 years back?
(c) Ruhi’s grandfather is 6 times her age. What is the age of her grandfather?
(d) Grandmother is 2 years younger than grandfather. What is grandmother’s age?
(e) Ruhi’s father’s age is 5 years more than 3 times Ruhi’s age. What is her father’s age?
Solution:
(a) Ruhi’s age after 5 years from now = Her present age + 5 years
= x years + 5 years
= (x + 5) years
(b) Ruhi’s age 3 years back = Her present age – 3 years
= x years – 3 years
= (x – 3) years
(c) Grandfather’s age = 6 × Ruhi’s age
= 6 × x years
= 6x years
(d) Grandmother’s age = Grandfather’s age – 2 years = 6x years – 2 years
= (6x – 2) years
(e) Ruhi’s father’s age = 3 × Ruhi’s age + 5 years
= 3 × x years + 5 years
= 3x years + 5 years
= (3x + 5) years
Question 6.
Find the values of the following expressions for x = 2.
(a) 2x – 5
(b) 4 + 3x
Solution:
(a) Putting x = 2, 2x – 5 = 2 × 2 – 5 = 4 – 5 = -1
(b) Putting x = 2, 4 +3x = 4 + 3 × 2 = 4 + 6 = 10
Question 7.
Add the following:
(a) ab + bc and bc + ca
(b) 3x – y + z, 2y – 5z and 3z – 4x
(c) 3p – q + 5r, q – 2r and 3r – 2p + 7q
Solution:
(a) (ab + be) + (bc + ca)
= ab + (bc + be) + ca
= ab + (1 + 1 )bc + ca
= ab + 2 be + ca
(b) 3x – y + z,2y – 5z and 3z – 4x (3x – y + z) + (2y – 5z) + (3z – 4x)
= (3x – 4x) + (-y + 2y) + (z – 5z + 3z)
= (3 – 4)x + (-1 + 2)y + (1 -5 + 3)z
= -x+y-z
(c) 3p – q + 5r, q – 2r and 3r – 2p + 7q
= (3p – q + 5r) + (q – 2r) + (3r – 2p + 7q)
= (3p – 2p) + (—q + q + 7g) + (5r – 2r + 3r)
= -p + 7q + 6r
Question 8.
Subtract the following:
(a) 7a + 8b – c from 5a – b + 3c
(b) 2x – 3y from z – x
(c) 3pq – 2p + q from 4pq + 7p – 2p
(d) 11 xy – 5y + 2x from 15xy + y – x
Solution:
(a) (5a – b + 3c) – (la + 8b – c)
= 5a – b + 3c – la-8b + c
= 5a – la – b – 8b + 3c + c
(Rearranging like terms) = (5 – 7)a + (-1 -8)b + (3 + 1)c
= -2a -9b + 4c.
(b) (z – x) – (2x – 3y) = z – x – 2x + 3y
= z – 3x + 3y
(c) (4pqr + 7p – 2p) – (3pq – 2p + q)
= 4pq + 7p – 2p – 3pq + 2p – q
= 4pq – 3pq + 7p – 2p + 2p – q
= pq + 7p – q
(d) (15xy + y-x) – (11xy — 5y + 2x)
= 15 xy + y – x- 11 xy + 5y – 2x
= 15 xy – 11xy – x – 2x + y + 5y
= 4xy – 3x + 6y
Question 9.
In the given question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option as:
(a) Both assertion and reason are true and the reason is the correct explanation of assertion.
(b) Both assertion and reason are true but the reason is not the correct explanation of the assertion.
(c) Assertion is true and the reason is false.
(d) Assertion is false and the reason is true.
(i) Assertion (A): Every algebraic expression has at least one letter-numbers.
Reason (R): A letter-number is a symbol used to represent an unknown value in an expression.
(ii) Assertion (A): If Aditi is x years of age now. 5 years ago her age was 10 years, then her present age is 15 years.
Reason (R): We can express the given situation in form of expression as, x – 5 = 15.
Solution:
(i) (a) Explanation: Every algebraic expression has at least one letter-numbers which is used to represent an unknown value in an expression. Hence, both assertion and reason are true.
(ii) (c) Explanation: Since, Aditi present age is of x. 5 years ago her age was 10 years, then her present age is 15 years. We can express this situation by the expression, x – 5 = 10, so the value of x is 15 years.
Hence, the assertion is true, but the reason is false.
Question 10.
Case-Based Question: Rita is organising a school fair. She is in charge of budgeting for various stalls. She notes the following expenses:
- The cost of each food stall is ₹ x.
- The cost of each game stall is ₹ y.
- She plans to have 4 food stalls and 3 game stalls.
There is a fixed decoration cost of ₹ 500.
Based on the above information, answer the following questions.
(a) Write an expression for the total cost to organise the school fair.
(b) What does the term 4x represent?
(c) If the cost of each food stall is ₹200 and each game stall is ₹ 150, what will be the total cost?
Solution:
(a) Since the cost of each food stall = ₹ x
So, the cost of 4 food stall = 4x
Since, the cost of each game stall = ₹ y
So, the cost of 3 game stall = 3y
And, the fixed decoration cost = ₹ 500
So, expression for the total cost to organise the school fair = 4x + 3y + 500.
(b) The term 4x represents cost of 4 food stalls.
(c) Since the expression for the total cost to organise the school fair is 4x + 3y + 500
So, the total cost when the cost of each food stall is ₹ 200 and each game stall is ₹ 150 is
4 × 200 + 3 × 150 + 500
= 800 + 450 + 500 = ₹ 1750.
Very Short Answer Type Questions
Question 1.
Write algebraic expressions using letters of your choice.
(i) 13 less than 12 times a number
(ii) 3 more than a number
Answer:
Let the letter-number be x, to get the algebraic expressions.
(i) 13 less than 12 times a number = 12x – 13
(ii) 3 more than a number = x + 3
Question 2.
Simplify:
(i) 20 + 20 × 20 – 20 + 20 × 20
(ii) 38 – 83 + 8 × 3 – 3 × 3 + 8 – 3
Answer:
(i) We have, 20 + 20 × 20-20 + 20 × 20
= 20 + 400-20 + 400
= 800
(ii) We have, 38 – 83 + 8 × 3 – 3 × 3 + 8 – 3
= 38 – 83 + 24 – 9 + 5
= 38 + 24 + 5 – 9 – 83
= 67 – 92
= – 25
Question 3.
If t = – 3 and b = 2, then find the value of 2 t + b + 24.
Answer:
If
t = -3 and b = 2, then 2t + b + 24 = 2 × (-3) + 2 + 2
= -6 + 2 + 24
= -6 + 26
= 20
Question 4.
Find the value of the expressions for p = 1, q = 2 and r = 3, where the expressions given as
(i) p + q + r
(ii) 7 p – 2 q
Answer:
Given, p = 1, q = 2 and r = 3.
(i) The value of p + q + r is given as
p + q + r = 1 + 2 + 3 = 6
(ii) The value of 7 p – 2 q is given as
7p – 2q = 7 × 1 – 2 × 2
= 7 – 4
= 3
Question 5.
Simplify the expression 4x + 5 – 3 x + 1 + 2x + 3 and find the value when x = 2.
Answer:
We have, 4x + 5 – 3x + 1 + 2x + 3
= 4x – 3x + 2x + 5 + 1 + 3
= 3x + 9
On putting x = 2 in 3x + 9, we get
3x + 9 = 3(2) + 9 = 6 + 9 = 15
Hence, the simplified expression is 3 x + 9 and its required value is 15.
Question 6.
Simplify :
(i) p – p + 3p – 5 q + 3q – 14p – q
(ii) x + y(3 – 34) – (y + 14) – (3x – 4y)
Answer:
(i) We have,
p – p + 3p – 5q + 3q – 14p – q
= p – p + 3p – 14p + 3q – 5q – q
= – 11p – 3q
(ii) We have,
x + y(3 – 34) – (y + 14)-(3x – 4y)
= x + 3y – 34y – y – 14 – 3x + 4y
= x – 3x + 3y – 34y – y + 4y – 14
= -2x – 28y -14
Question 7.
Riya is 5 years elder to her brother. If her brother’s age is x years. Then, what is Riya’s age?
Answer:
Let Riya’s brother’s age be x years.
Since, Riya is 5 years elder to her brother.
Then, age of Riya =(x + 5) years.
Short Answer Type Questions
Question 1.
If Aditya’s father’s age is 4 times of Aditya’s age, then write an algebraic expression representing sum of their ages.
Answer:
Let Aditya’s age be x years.
Since, the age of Aditya’s father is 4 times the age of Aditya.
∴ The age of Aditya’s father = 4x years.
Then, the algebraic expression representing the sum of their ages is x + 4x = 5x.
Question 2.
Aadil runs a shop of grocery items. He sells rice and sugar mainly. The price of rice is ₹ 50 per kilogram and the price of sugar is ₹ 40 per kilogram.
Then, find an algebraic expression for the total price of the grocery : (x + 5) kg rice and (y + 6) kg sugar.
Answer:
Given, the price of 1 kg rice = ₹ 50
and the price of 1 kg sugar = ₹ 40
∴ The price of (x + 5) kg rice = ₹ 50 × (x + 5)
= ₹(50x + 250)
and the price of (y + 6) kg sugar = ₹ 40 × (y + 6)
= ₹(40y + 240)
Therefore, the total price = ₹(50x + 250 + 40y + 240)
= ₹(50x + 40y + 490)
∴ The required algebraic expression is 50x + 40y + 490.
Question 3.
Observe the given sequence
5,8,11,14,17, ….. .
(i) Find an algebraic expression to get the nth term of this sequence.
(ii) What is 46th term of this sequence?
Answer:
Given sequence is
5,8,11,14,17, …..
We observe that
(i) To get the nth term of this sequence, we formulate it as
3 × 1 + 2, 3 × 2 + 2,3 × 3 + 2, …… .
∴ The nth term = 3 n + 2 ; n = 1,2,3, …..
(ii) The 46th term of this sequence is
3 × 46 + 2 = 138 + 2 = 140
Question 4.
Find the perimeter of a quadrilateral ABCD having sides AB = a units, BC = a + b units, CD = 2 a units and DA = a + b – 1 units.
Answer:
We know that the perimeter of A B C D will be given by
AB + BC + CD + DA
∴ AB + BC + CD + DA
= a + (a + b) + 2a + (a + b – 1)
= (a + a + 2a + a) + (b + b) – 1
= 5a + 2b – 1
∴ The perimeter is 5 a + 2b – 1 units.
Question 5.
What is the sum of the numbers in the picture (unknown values are denoted by letter – numbers)? Try it adding in two different ways. Is the sum same?
Answer:
Method I On adding row wise, we get
(2 r + s + (-s) + r) + (-2 + 4) + (4 + (-2))
+ (2r + s + (-s) + r)
= (3r + s – s) + 2 + 2 + (3r + s – s)
= 3r + 4 + 3r
= 6r + 4
Method II Now, on adding upper half and then doubling, we get
2 × (2r + s + (-s) + r + (-2) + 4)
= 2 × (3r + s – s + 2)
= 6r + 4
Clearly, we observe that the sum is same in both the ways.
Long Answer Type Questions
Question 1.
Anshu is 6 years elder to his sister, but his father’s age is 6 times of his age. Then,
(i) What are their ages in terms of letter – number x ?
(ii) What is the sum of their ages after simplifying?
(iii) If age of his sister is 6 years, then what is Anshu’s father’s age?
Answer:
(i) Let the age of Anshu’s sister be x years.
Since, Anshu is 6 years elder to his sister.
∴ The age of Anshu = (x + 6) years.
Also, Anshu’s father’s age is 6 times of Anshu’s age.
∴ The age of Anshu’s father = 6 × (x + 6) years
= (6x + 36) years.
(ii) Now, the sum of their ages = x + (x + 6) + (6 x + 36)
= (x + x + 6x) + (6 + 36)
= (8 x + 42)
(iii) Since, the age of Anshu’s sister is 6 years i.e. if x = 6, then, the age of Anshu’s father
= 6 x + 36 = 6 × 6 + 36
= 36 + 36
= 72 years old.
Question 2.
Aastha has notes of ₹ 100, ₹ 20 and ₹ 10.
She has x notes of ₹ 100, (x – 5) notes of ₹ 20 and (x + 1) notes of ₹ 10.
(i) What will be an algebraic expression describing the total amount that she has?
(ii) Simplify that expression.
(iii) If she has 5 notes of ₹ 100 , then how much total amount has she have?
Answer:
(i) Given, Aastha has x notes of ₹ 100,(x – 5) notes of ₹ 20 and (x + 1) notes of ₹ 10.
∴ The total amount she has
= 100 × x + 20 ×(x – 5) + 10 ×(x + 1)
= 100x + 20(x – 5) + 10(x + 1)
(ii) We have, 100 x + 20(x-5) + 10(x + 1)
= 100x + 20x – 100 + 10x + 10
= (100x + 20x + 10x) + (10-100)
= 130x – 90
(iii) Since, she has 5 notes of ₹ 100 i.e. x = 5
then the total amount = ₹(130x – 99)
= ₹(130 × 5 – 90)
= ₹(650 – 90)
= ₹ 560
Question 3.
Fill in the blanks by replacing the letter – numbers by numbers.
Answer:
Question 4.
Consider a set of numbers from the calendar (having endless rows) forming under the following shape.
(i) Find the sum of all the numbers.
(ii) Compare the sum with the number at the centre?
(iii) Will this always happen? If yes, try to show this.
Answer:
(i) The sum of all the numbers = 22 + 28 + 29 + 30 + 36 = 145
(ii) The number at the centre is 29 . Clearly, the sum 145 is 5 times 29.
(iii) Yes, this will happen always that the sum is 5 times the number at centre.
Let the letter-number to be a at the centre.
Then, the grid of 3 × 3 will be looked like this,
Now, the sum of all the entries
= (a – 7) + (a – 1) + a + (a + 1) + (a + 7)
= a + a + a + a + a + 7 – 7 + 1 – 1
= 5 a
∴ It is cleared that the sum of all the five entries will be 5 times the number at the centre.
Question 5.
Look at the picture below. It is a pattern using bold lines. Can you identify the pattern?
(i) Can you write an algebraic expression representing this pattern?
(ii) How many lines will be used in step 10?
Answer:
It the given figure,
In step 1 the number of bold lines used = 4
In step 2, the number of bold lines used = 6
In step 3, the number of bold lines used = 8 and so on.
Yes, we observe a pattern of number of lines as,
(i) Yes, we can write an algebraic expression representing this pattern.
The required algebraic expression for the number of lines used in k th step is 2 k + 2, k = 1,2,3, ……
(ii) In step 10th, the number of lines used will be
2 × 10 + 2 = 20 + 2 = 22
Question 6.
Simplify the following expressions.
(i) 5a + 3b – 2a + 7 – 4 b – 9
(ii) 4(x – 2) + 3x + 10
(iii) 9m – (4m – 3) + 2
(iv) 6(2p – 1) – 3(p + 4)
(v) 2y + 5(3y – 4) – 6
(vi) 10 – 3(2x – 5) + x
Answer:
(i) We have, 5a + 3b – 2a + 7 – 4b – 9
= (5a – 2a) + (3b – 4b) + 7 – 9
= 3a – b – 2
(ii) We have, 4(x – 2) + 3x + 10
= 4x – 8 + 3x + 10
= (4x + 3x) + (-8 + 10)
= 7x + 2
(iii) We have, 9m – (4m – 3) + 2
= 9m – 4 m + 3 + 2
= 5m + 5
(iv) We have, 6(2p – 1) – 3(p + 4)
= 12p – 6 – 3p – 12
= (12p – 3p)-(6 + 12)
= 9p – 18
(v) We have, 2y + 5(3y – 4) -6
= 2y + 15y – 20 – 6
= 17y – 26
(vi) We have, 10 – 3(2x – 5) + x
= 10 – 6x + 15 + x
= (-6x + x) + 15 + 10
= -5x + 25
Skill Based Questions
Question 1.
At age of 2 yr , a cat or a dog is considered 24 ‘human’ years old. Each year, after age 2 is equivalent to 4 ‘human’ years. Fill in the expression [24 + ….(a – 2)], so that it represents the age of a cat or dog in human years. Also, you need to determine for what ‘a’ stands for. Copy the chart and use your expression to complete it.
Answer:
1. The expression is [24 + 4(a – 2)].
Here, ‘a’ represents the present age of dog or cat.
Question 2.
Observe the following nutritional chart carefully
Write an algebraic expression for the amount of carbohydrates (in grams) for
(i) y units of potatoes and 2 units of rajma.
(ii) 2x units tomatoes and y units apples.
Answer:
(i) 22y + 120
(ii) 8x + 14y
Question 3.
Answer:
(i)
\(\frac{87}{2}\)
(ii)
1
Class 7 Maths Chapter 4 Extra Questions in Hindi अक्षर संख्याओं के उपयोगी व्यंजक
Expressions using Letter Numbers Class 7 Extra Questions in Hindi
A. दक्षता आधारित प्रश्न (MCQs)
प्रश्न 1.
x इकाई भुजा की लंबाइयों वाले एक समषटभुज का परिमाप है:
(a) (x + 6 ) इकाई
(b) x × 6 इकाई
(c) (x ÷ 6 ) इकाई
(d) (x – 6 ) इकाई
उत्तर :
(b) x × 6 इकाई
प्रश्न 2.
3 में से 7 बार घटाने का परिणाम है:
(a) 7x – y
(b) x – 7y
(c) 7y – x
(d) y – 7x
उत्तर :
(d) y – 7x
प्रश्न 3.
यदि p = – 3 है, तो 5p – 10 का मान है :
(a) -25
(c) 5
(b) 25
(d) –5
उत्तर :
(a) -25
प्रश्न 4.
5 (x + y) – 5 (x – y) बराबर हैं :
(a) 0
(b) 10x
(c) 10y
(d) 10x – 10y
उत्तर :
(c) 10y
प्रश्न 5.
x + y + y + y + xxy बराबर है:
(a) x + 2y
(b) x – 2y
(c) x + y
(d) x y
उत्तर :
(a) x + 2y
B. स्थिति अध्ययन
प्रश्न 1.
वर्षों में आयु
पाँच वर्ष पूर्व, अहमद की आयु वर्ष थी। उसकी बहन उससे 3 वर्ष छोटी है तथा उसका भाई उससे 2 वर्ष बड़ा है।
रिक्त स्थानों को भरिए:
(a) अहमद की वर्तमान आयु _________ वर्ष है।
उत्तर :
x + 5
(b) उसके भाई की वर्तमान आयु _________ वर्ष है।
उत्तर :
x + y + 5
(c) 7 वर्ष पूर्व उसके भाई की आयु _________ वर्ष थी।
उत्तर :
x + y – 2
(d) अहमद के भाई और बहन की आयु का अंतर _________ वर्ष है।
उत्तर :
y + 3
(e) तीनों की वर्तमान आयु का योग है :
(i) (3x + y + 12) वर्ष
(ii) (3x + y 12) वर्ष
(iii) (3x + y + 7) वर्ष
(iv) (3x + y – 7 ) वर्ष
उत्तर :
(i)
प्रश्न 2.
मित्रों का भार तीन मित्रों A, B और C का कुल भार 15z कि. ग्रा. है। A का भार कुल भार का \(\frac{1}{3}\) + 1 कि.ग्रा. है तथा B का भार A के भार से 2 कि.ग्रा. अधिक है।
रिक्त स्थानों को भरिए:
(a) A का भार _________ कि.ग्रा. है।
उत्तर :
5z + 1
(b) B का भार _________ कि.ग्रा. है।
उत्तर :
5z + 3
(c) B का भार कुल भार का \(\frac{1}{3}\) + _________ .कि.ग्रा. है।
उत्तर :
3
(d) C का भार (कि.ग्रा. में) है:
(i) 5z + 3
(ii) 5z – 3
(iii) 5z – 4
(iv) 52 + 4
उत्तर :
(iii) 5z – 4
(e) A और C के भारों का अंतर 4 कि.ग्रा है। (सत्य / असत्य)
उत्तर :
असत्य।
C. मिलान कीजिए
प्रश्न 1.
स्तंभ I का स्तंभ II से मिलान कीजिए :
| स्तंभ I | स्तंभ II |
| (i) व्यंजक 2 y + 7 में चर | (a) x + 24 |
| (ii) (x + 3 ) + 7 × 3 | (b) 3x + 30 |
| (iii) (x + 3 + 7 ) × 3 | (c) 4x – 3y |
| (iv) 3x – 2y + (3 – 2) × x – y | (d) 9x + 2y |
| (v) 15x + 2x – 8x + 2y | (e) y |
उत्तर :
| स्तंभ I | स्तंभ II |
| (i) व्यंजक 2 y + 7 में चर | (e) y |
| (ii) (x + 3 ) + 7 × 3 | (a) x + 24 |
| (iii) (x + 3 + 7 ) × 3 | (b) 3x + 30 |
| (iv) 3x – 2y + (3 – 2) × x – y | (c) 4x – 3y |
| (v) 15x + 2x – 8x + 2y | (d) 9x + 2y |
D. अभिकथन – कारण प्रश्न
निम्नलिखित में से प्रत्येक प्रश्न में दो कथन दिए गए हैं। एक को अभिकथन (A) और दूसरे को कारण (R) कहा गया है। आपको दिए गए चार विकल्पों में से सही विकल्प का चुनाव करना है।
(a) दोनों कथन (A) और (R) सही हैं, तथा (A) का (R) सही स्पष्टीकरण है।
(b) दोनों कथन (A) और (R) सही हैं, परंतु (A) का (R) सही स्पष्टीकरण नहीं है।
(c) (A) सही है, परंतु (R) सही नहीं है।
(d) (R) सही है, परंतु (A) सही नहीं है।
प्रश्न 1.
(A) : 4 से.मी. भुजाओं की लंबाई वाले समपंचभुज का परिमाप 20 से.मी. है।
(R) : m इकाई भुजाओं वाले n भुजाओं के समबहुभुज का परिमाप nm इकाई होता है।
उत्तर :
(a) दोनों कथन (A) और (R) सही हैं, तथा (A) का (R) सही स्पष्टीकरण है।
प्रश्न 2.
(A) : व्यंजक 6 (x – 3 ) सरल करने पर 6x – 3 है।
(R) : वितरण गुण के अनुसार, a ( b + c) = ab + ac
उत्तर :
(d) (R) सही है, परंतु (A) सही नहीं है।
प्रश्न 3.
(A) : यदि एक व्यक्ति की वर्तमान आयु x वर्ष है, तो 6 वर्ष बाद उसकी आयु (x + 6) वर्ष होगी।
(R) : किसी व्यक्ति की 2 वर्ष के बाद आयु ज्ञात करने के लिए, हम उसकी वर्तमान आयु में से 2 वर्ष घटा देते हैं।
उत्तर :
(c) (A) सही है, परंतु (R) सही नहीं है।
प्रश्न 4.
(A) : व्यंजक 5x – 7y + 22 में पदों की संख्या 3 है।
(R) : इस व्यंजक में दो चर x और y हैं।
उत्तर :
(b) दोनों कथन (A) और (R) सही हैं, परंतु (A) का (R) सही स्पष्टीकरण नहीं है।
प्रश्न 5.
(A) : व्यंजक 9x – 8y + 32 + 4u + 9 में चरों की संख्या 4 है।
(R) : यहाँ x, y, z और u अक्षर संख्याएँ हैं ।
उत्तर :
(a) दोनों कथन (A) और (R) सही हैं, तथा (A) का (R) सही स्पष्टीकरण है।
E. सत्य / असत्य
प्रश्न 1.
भुजा लंबाई x इकाई वाले वर्ग का क्षेत्रफल 4x वर्ग इकाई है।
उत्तर :
असत्य,
प्रश्न 2.
यदि हम x को 3 से गुणा करें तथा इस प्राप्त गुणनफल को 5 में से घटाएँगे, तो परिणामी व्यंजक 5 – 3x होगा।
उत्तर :
सत्य,
प्रश्न 3.
x = 1 पर 5x और 5x – 5 के मान क्रमश: 5 और 0 हैं।
उत्तर :
सत्य,
प्रश्न 4.
y और 3 का योग 3y है ।
उत्तर:
असत्य ।
F. रिक्त स्थानों को भरिए.
1.40 कि.मी. प्रति घंटा की अचर चाल से 7 घंटे में तय की दूरी _________ है।
उत्तर :
40h
2. यदि p कि.ग्रा. आलू ₹70 में खरीदे जाएँ, तो 1 कि.ग्रा. आलू का मूल्य ₹ _________ है।
उत्तर :
\(\frac{70}{p}\)
3. w सप्ताहों में दिनों की संख्या _________ है।
उत्तर :
7w
4. 7x से 11 अधिक को व्यंजक _________ के रूप में व्यक्त किया जा सकता है।
उत्तर :
7x + 11
5. 4x + 3y – 9 + 2x – y = _________ है।
उत्तर :
6ox + 2y – 9
G. अति संक्षिप्त उत्तरीय प्रश्न
प्रश्न 1.
भुजा 2 से.मी. वाले वर्ग का क्षेत्रफल ज्ञात कीजिए।
हल:
क्षेत्रफल = m × m वर्ग से.मी. है।
प्रश्न 2.
p विद्यार्थियों वाली एक कक्षा ने पिकनिक जाने की योजना बनाई। प्रत्येक विद्यार्थी से ₹50 की धनराशि ली गई। इस राशि में से ₹1800 यात्रा शुल्क के रूप में अग्रिम दिए गए। अन्य वस्तुओं पर व्यय करने के लिए अब उनके पास क्या धनराशि शेष रही?
हल :
5p – 1800 (₹ में)
प्रश्न 3.
किसी व्यक्ति का भार 55 कि.ग्रा. था। उसका भार एक वर्ष में p कि.ग्रा. बढ़ जाता है। उसके वर्तमान भार के लिए (कि.ग्रा. में), एक व्यंजक लिखिए।
हल:
भार = 55 + p है।
प्रश्न 4.
x = 3 और y = 2 पर 2x + y + 24 का मान ज्ञात कीजिए।
हल:
2x + y + 24 = 2 × ( 3 ) + 2 + 24
= -6 + 2 + 24
= 20
H. संक्षिप्त उत्तरीय प्रश्न
प्रश्न 1.
चावल का मूल्य ₹ 70 प्रति कि.ग्रा. है तथा चीनी का मूल्य ₹40 प्रति कि.ग्रा. है । (x + 5) कि.ग्रा. चावल और (y + 6) कि.ग्रा. चीनी के मूल्य के लिए (₹ में) एक व्यंजक लिखिए तथा उसे सरल कीजिए ।
हल:
चावल का मूल्य (₹ में) = 70 × (x + 5)
चीनी का मूल्य (₹ में) = 40 × (y + 6)
इस प्रकार, वाँछित व्यंजक
= 70x (x + 5) + 40x (y +6)
= 70x + 350 + 40y + 240
= 70x + 40y + 350 + 240
= 70x + 40y + 590
प्रश्न 2.
इस व्यंजक को सरल कीजिए:
x + y (3 – 27) – (y + 12 ) – ( 4x – 3y)
हल:
दिया हुआ व्यंजक
= x + y (3 – 27) – (y + 12 ) – (4x – 3y )
= x + y (– 24 ) – y – 12 – 4x + 3y
= x – 24y – y – 12 – 4x + 3y
= x – 4x – 24y – y + 3y – 12
= 5x – 22y – 12 है।
प्रश्न 3.
रमेश के पिता की आयु (वर्षो में) रमेश की आयु के 4 गुने से 2 वर्ष अधिक है। उनकी आयु का योग (वर्षों में) ज्ञात कीजिए ।
हल :
मान लीजिए कि रमेश की आयु = x वर्ष है।
अतः, उनके पिता की आयु = (4x + 2) वर्ष
उनकी आयु का योग (वर्षों में) = x + (4x + 2)
= x + 4x + 2
= 5x + 2 है।
प्रश्न 4.
6 (2x – 1) – 3 (x + 4) को सरल कीजिए ।
हल :
6 (2x – 1 ) – 3 (x + 4)
= 12x – 6 – 3x – 12
= 12x – 3x – 6 – 12
= 9x – 18 है।
प्रश्न 5.
नीचे माचिस की तीलियों की विभिन्न संख्याओं से रचित किया गया पैटर्न दिया गया है:
(i) प्रत्येक चित्र में उपयोग की गई तीलियों की संख्या को इंगित करने वाले पैटर्न की पहचान कीजिए।
हल :
पैटर्न 2, 4, 6, 8, है।
अत:, nवें चित्र में, उपयोग की जाने वाली तीलियों की संख्या 2 × n = 2n होगी।
(ii) छठे चित्र में कितनी तीलियाँ उपयोग की जाएँगी?
हल :
छठे चित्र में उपयोग की गई तीलियों की संख्या 2 × 6 = 12 होगी।
I. दीर्घ उत्तरीय प्रश्न
प्रश्न 1.
3x – y + z, 2y – 52 और 3z – 2x का योग ज्ञात कीजिए ।
हल:
हमें प्राप्त है : 3x – y + 2
प्रश्न 2.
7x + 8y – 2 को 5x y + 32 में से घटाइए ।
हल:
हमें प्राप्त है :
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