SEBA SCERT Solutions for Class 7
Maths Chapter 1 Integers Exercise 1.2
SEBA SCERT Solutions for Class 7 Maths Chapter 1 Exercise 1.2 Integers in English and Assamese Medium modified and updated for academic year 2024-25. All the question-answers and solutions are revised on the basis of new syllabus and latest SEBA SCERT textbooks issued for curriculum 2024-25.
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All the question answers with solutions are done according to latest SEBA SCERT Books for 2024-25.
SEBA SCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.2
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Class: 7- |
Mathematics |
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Chapter: 1 |
Exercise 1.2 |
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Session: |
SEBA 2024 - 25 |
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Medium: |
English, Assamese & Hindi |
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SEBA SCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.2
Exercise -1.2
1. Find the product.
(i) 5× (-2)
(ii) (-3) ×7
(iii) (-4) ×(-3)
(iv) (-129) × (-1)
(v) (-12) ×0×(-17)
(vi) (-22) ×(-11) ×10
(vii) 13×(-5) ×(-3)
(viii) (-27) ×(-31) ×(-2)
(ix) (-3) ×(-1) ×(-2) ×5
Solution:
(i) 5× (−2) =−10
(ii) (−3) ×7=−21
(iii) (−4) × (−3) =12
(iv) (−129) × (−1) =129
(v) (−12)×0×(−17)=0 (anything multiplied by 0 is 0)
(vi) (−22) ×(−11)×10
=22×11×10
=2420
(vii) 13×(−5)×(−3)
=13×15
=195
(viii) (−27)×(−31)×(−2)
=27×31×(−2)
=837×(−2)
=−1674
(ix) (−3)×(−1)×(−2)×5
=3×1×(−2)×5
=3×(−2)×5
=−6×5
=−30
2. Verify whether true or false.
(i) 27×{(−5) +10}=27×(−5) +27×10
(ii) (−25) ×{(−16)+( −24)}=( −25) ×(−16) ×( −24)
(iii) a– (–b) = a+b, where a = (–75), b = (–20)
Solution:
(i) 27×{(−5)+10}=27×(−5)+27×10
LHS = 27×{(−5)+10}
=27×5
=135
RHS = 27×(−5)+27×10
=−135+270
=135
Since LHS = RHS, the statement is true.
(ii) (−25)×{(−16)+(−24)}=(−25)×(−16)+(−25)×(−24)
LHS = (−25)×{(−16)+(−24)}
= (−25)×(−40)
=1000
RHS = ( −25) ×(−16) ×( −24)
= 400×( −24)
= −9600
The left-hand side is 1000, and the right-hand side is -9600
Since LHS ≠ RHS, the statement is false.
(iii) a−(−b)=a+b, where 𝑎 = (−75), and 𝑏=(−20)
Substitute 𝑎 and b into the equation,
(−75)−(−(−20))
=(−75)+20
= −55
Simplify the equation,
(−75)+20
=−55
So, the statement is true.
3. (i) Product of any two integers −33. If one of them is 11, what is the other number?
(ii) Product of any two integers is 51. If one of them is −1, what is the other number?
(iii) What is the value of (−1×a) for any integer ‘a’?
Solution:
(i) Let the other number = x
Given,
`11\times x\=\-33`
`\Rightarrow x=\frac{-33}{11}`
`\Rightarrow x=-3`
So, the other number is −3.
(ii) Let the other number = x
Given,
`-1\times x=51`
`\Rightarrow x=51/(-1)`
`\Rightarrow x=-51`
(iii) For any integer 𝑎,
−1×a=−a
So, the value of (−1×𝑎) is −a, which is the negation of 𝑎
4. Find the product applying appropriate property –
(i) 125×(−54) ×8
(ii) (−25) ×(−97) ×4
(iii) (−27) ×(−33)
(iv) 25×(−58)+( −58) ×(−35)
(v) 15×(−25) ×(−4) ×(−10) (vi) (−57) ×(−19) ×57
Solution:
(i) 125×(−54)×8
Using the associative property of multiplication
= (125×8)×(−54)
= 1000×(−54)
= −54000
So, the product is −54000
(ii) (−25)×(−97)×4
Using the associative property of multiplication
= (−25)×4×(−97)
= −100×(−97)
= 9700
So, the product is 9700.
(iii) (−27)×(−33)
= (−27)×(−33)
= 27×33
= 891
So, the product is 891.
(iv) 25×(−58)+(−58)×(−35)
Using the distributive property,
= 25×(−58)+(−58)×(−35)
= (−58)×{25+(−35)}
=(−58)×(−10)
= 580
So, the product is 580.
(v) 15×(−25)×(−4)×(−10)
Using the associative property of multiplication
= {15×(−25)}×(−4)×(−10)
= (−375)×(−4)×(−10)
= 1500×(−10)
= −15000
So, the product is −15000.
(vi) (−57)×(−19)×57
Using the commutative and associative properties
(−57)×57×(−19)
=(−57×57)×(−19)
=−3249×(−19)
=3249×19
=61731
So, the product is 61731
5. Evaluate with the help of distributive and associative property:
(i) 125×(54) ×8
(ii) (−25) ×75×8×(−4)
(iii) 225×67×3
Solution:
(i) 125×54×8
Using the associative property of multiplication,
(125×8)×54
=1000×54
=54000
So, the product is 54000.
(ii) (−25)×75×8×(−4)
Using the associative property of multiplication,
[(−25)×(−4)]×(75×8)
=100×600
=60000
So, the product is 60000.
(iii) 225×67×3
Using the associative property of multiplication,
225×(67×3)
=225×201
We can break this down using the distributive property,
225×201
=225×(200+1)
=225×200+225×1
=45000+225
=45225
So, the product is 45225.
6. Evaluate using distributive property:
(i) 172×25+172×35
(ii) 159×82+159×16+159×2
(iii) 67×78+67×(−43)+67×(−25)
(iv) 999×99+99
(v) 58×47+94
Solution:
(i) 172×25+172×35
Using the distributive property,
172×(25+35)
= 172×60
= 10320
(ii) 159×82+159×16+159×2
Using the distributive property,
159×(82+16+2)
= 159×100
= 15900
(iii) 67×78+67×(−43)+67×(−25)
Using the distributive property,
67×(78+(−43)+(−25))
= 67×(78−43−25)
= 67×10
= 670
(iv) 999×99+99
= 99×(999+1)
= 99×1000
= 99000
(v) 58×47+94
We cannot directly apply the distributive property here because the second term does not have a common factor with the first term. Instead, let's compute it directly,
58×47+94
= 2726+94
= 2820
7. Justify whether right or wrong:
(i) (−7) ×15×(−4)=( −7) ×15+(−7) ×(−4)
(ii) (−6)×23×(−2) = (−2)×(−6)×23
(iii) (−5)×{(−3)×2} = {(−5)×( −3)}×2
(iv) (−175) × (−1) = −175
(v) (−25) × ( − 4) × 0 = 100
Solution:
(i) (−7) ×15×(−4)=( −7) ×15+(−7) ×(−4)
LHS: = (−7) ×15×(−4)
= (−105)×(−4)
= 420
RHS: = (−7)×15+(−7)×(−4)
= −105+28
= −77
Since 420 ≠ −77, the statement is wrong.
(ii) (−6)×23×(−2) = (−2)×(−6)×23
Both sides are equivalent because of the commutative property of multiplication. Let's compute the product,
(−6) ×23×(−2)=(−6)×(−2)×23
⇒ 12×23 = 276
Both sides are indeed equal. So, the statement is right.
(iii) (−5)×{(−3)×2} = {(−5)×( −3)}×2
LHS: = (−5)×{(−3)×2}
= (−5) ×(−6)
= 30
RHS: = {(−5)×(−3)}×2
= 15×2
= 30
Both sides are equal, so the statement is right.
(iv) (−175) × (−1) = −175
Compute the product,
(−175)×(−1)=175
Since 175 ≠ −175, the statement is wrong.
(v) (−25) × ( − 4) × 0 = 100
Compute the product,
(−25)×(−4)×0=100
Since any number multiplied by zero is zero,
(−25)×(−4)×0=0
Since 0 ≠ 100, the statement is wrong.
