Integers

integers class 7 selina

Table of Contents

Exercise: 1-A

Q1: Evaluate

i. 427 × 8 + 2 × 427
Step 1: Identify the common factor
427 × 8 + 2 × 427
= (427 × 8) + (427 × 2)     [Rewriting terms]
Step 2: Apply distributive property:
= 427 × (8 + 2)     [a × b + a × c = a × (b + c)]
= 427 × 10
Step 3: Multiply:
= 4270

ii. 394 × 12 + 394 × (-2)
Step 1: Identify the common factor
= (394 × 12) + (394 × -2)
Step 2: Apply distributive property:
= 394 × (12 + (-2))
= 394 × 10
Step 3: Multiply:
= 3940

iii. 558 × 27 + 3 × 558
Step 1: Rewriting:
= (558 × 27) + (558 × 3)
Step 2: Apply distributive property:
= 558 × (27 + 3)
= 558 × 30
Step 3: Multiply:
= 16740


Q2: Evaluate

i. 673 × 9 + 673
= 673 × 9 + 673 × 1
= 673 × (9 + 1)
= 673 × 10
= 6730

ii. 1925 × 101 – 1925
= 1925 × 101 – 1925 × 1
= 1925 × (101 – 1)
= 1925 × 100
= 192500


Q3: Verify

i. 37 × {8 + (-3)} = 37 × 8 + 37 × (-3)
LHS = 37 × (8 – 3) = 37 × 5 = 185
RHS = 296 + (-111) = 185
Verified

ii. (-82) × {(-4) + 19} = (-82) × (-4) + (-82) × 19
LHS = (-82) × 15 = -1230
RHS = 328 + (-1558) = -1230
Verified

iii. {7 – (-7)} × 7 = 7 × 7 – (-7) × 7
LHS = (7 + 7) × 7 = 14 × 7 = 98
RHS = 49 – (-49) = 49 + 49 = 98
Verified

iv. {(-15) – 8} × (-6) = (-15) × (-6) – 8 × (-6)
LHS = (-23) × (-6) = 138
RHS = 90 + 48 = 138
Verified


Q4: Evaluate

i. 15 × 8 = 120

ii. 15 × (-8) = -120

iii. (-15) × 8 = -120

iv. (-15) × (-8) = 120


Q5: Evaluate

i. 4 × 6 × 8 = 192

ii. 4 × 6 × (-8) = -192

iii. 4 × (-6) × 8 = -192

iv. (-4) × 6 × 8 = -192

v. 4 × (-6) × (-8) = 192

vi. (-4) × (-6) × 8 = 192

vii. (-4) × 6 × (-8) = 192

viii. (-4) × (-6) × (-8) = -192


Q6: Evaluate

i. 2 × 4 × 6 × 8 = 384

ii. 2 × (-4) × 6 × 8 = -384

iii. (-2) × 4 × (-6) × 8 = 384

iv. (-2) × (-4) × 6 × (-8) = -384

v. (-2) × (-4) × (-6) × (-8) = 384


Q7: Determine the integer whose product with ‘-1” is:

i. -4747

ii. 63-63

iii. -11

iv. 00


Q8: Eighteen integers are multiplied together. What will be the sign of their product, if :

i. 15 of them are negative and 3 are positive?
→ Odd negatives → Negative


ii. 12 of them are negative and 6 are positive?
→ Even negatives → Positive


iii. 9 of them are positive and the remaining are negative?
→ Odd negatives → Negative


iv. all are negative?
→ Even negatives → Positive



Q9: Find which is greater?

i. (8+10)×15 = 18×15 = 270
8 + 10×15 = 8 + 150 = 158
→ (8+10)×15 is greater

ii. 12×(6-8) = 12×(-2) = -24
12×6 – 8 = 72 – 8 = 64
→ 12×6 – 8 is greater

iii. {(-3)-4}×(-5) = (-7)×(-5) = 35
(-3) – 4×(-5) = -3 – (-20) = -3 + 20 = 17
→ {(-3)-4}×(-5) is greater


Q10: State true or false

i. Product of two different integers can be zero → True

ii. Product of 120 negative and 121 positive integers is negative → True

iii. a × (b + c) = a × b + c → False

iv. (b – c) × a = b – c × a → False

previous
next
Share the Post:

Related Posts

Fractions

Fractions

Step by Step solutions of RS Aggarwal ICSE Class-6 Maths chapter 4- Fractions by Goyal Brothers Prakashan is provided.

Read More

Leave a Comment

Your email address will not be published. Required fields are marked *

Join Our Newsletter

Scroll to Top