Selina Concise Class 7 Math Chapter 1 Integer Exercise 1A Solution
EXERCISE 1A
(1) Evaluate:
(i) 427 × 8 + 2 × 427
= 427 (8 + 2)
= 427 × 10
= 4270
(ii) 394 × 12 + 394 × (- 2)
= 394 (12 – 2)
= 394 × 10
= 3940
(iii) 558 × 27 + 3 × 558
= 558 (27 + 3)
= 558 × 30
= 16740
(2) Evaluate:
(i) 673 × 9 + 673
= 673 (9 + 1)
= 673 × 10
= 6730
(ii) 1925 × 101 – 1925
= 1925 (101 – 1)
= 1925 × 100
= 192500
(3) Verify:
(i) 37 × {8 + (- 3)} = 37 × 8 + 37 × (- 3)
⇒ 37 × (8 – 3) = 37 × (8 – 3)
⇒ 37 × 5 = 37 × 5
(ii) (- 82) × {(- 4) + 19} = (- 82) × (- 4) + (- 82) × 19
⇒ (- 82) × (– 4 + 19) = (- 82) × (- 4 + 19)
⇒ – 82 × 15 = – 82 × 15
(iii) {7 – (- 7)} × 7 = 7 × 7 – (- 7) × 7
⇒ (7 + 7) × 7 = 7 × (7 + 7) × 7
⇒ 14 × 7 ≠ 7 × 14 × 7
(iv) {(- 15) – 8} × (- 6) = (- 15) × (- 6) – 8 × (- 6)
⇒ – 23 × (- 6) = (- 15) × (- 14) × (- 6)
⇒ 138 ≠ – 1260
(4) Evaluate:
(i) 15 × 8 = 120
(ii) 15 × (- 8) = – 120
(iii) (- 15) × 8 = – 120
(iv) (- 15) × (- 8) = 120
(5) Evaluate:
(i) 4 × 6 × 8 = 24 × 8 = 192
(ii) 4 × 6 × (- 8) = 24 × (- 8) = – 192
(iii) 4 × (- 6) × 8 = – 24 × 8 = – 192
(iv) (- 4) × 6 × 8 = – 24 × 8 = – 192
(v) 4 × (- 6) × (- 8) = – 24 × (- 8) = 192
(vi) (- 4) × (- 6) × 8 = 24 × 8 = 192
(vii) (- 4) × 6 × (- 8) = – 24 × (- 8) = 192
(viii) (- 4) × (- 6) × (- 8) = – 24 × (- 8) = 192
(6) Evaluate:
(i) 2 × 4 × 6 × 8
= 8 × 6 × 8
= 48 × 8
= 384
(ii) 2 × (- 4) × 6 × 8
= – 8 × 6 × 8
= – 48 × 8
= – 384
(iii) (- 2) × 4 × (- 6) × 8
= – 8 × (- 6) × 8
= 48 × 8
= 384
(iv) (- 2) × (- 4) × 6 × (- 8)
= 8 × 6 × (- 8)
= 48 × (- 8)
= – 384
(v) (- 2) × (- 4) × (- 6) × (- 8)
= 8 × (- 6) × (- 8)
= – 48 × (- 8)
= 384
(7) Determine the integer whose product with ‘- 1’ is:
(i) – 47 = (- 1) × 47 = – 47
Hence, the integer is 47.
(ii) 63 = (- 10 × (- 63) = 63
Hence, the integer is 63.
(iii) – 1 = (- 1) × 1 = – 1
Hence, the integer is 1.
(iv) 0 = (- 1) × 0 = 0
Hence, the integer is 0.
(8) (i) 15 of them are negative, which is odd number. Hence the sign of the product will be negative.
(ii) 12 of them are negative, which is even number. Hence the sign of the product will be positive.
(iii) 9 of them are negative, which is odd number. Hence the sign of the product will be negative.
(iv) All are negative, which is ever number. Hence the sign of the product will be positive.
(9) Find which is greater?
(i) (8 + 10) × 15 or 8 + 10 × 15
= 18 × 15 or 8 + 150
= 270 > 158
(ii) 12 × (6 – 8) or 12 × 6 – 8
= 12 × 2 or 72 – 8
= 24 < 64
(iii) {(- 3) – 4} × (- 5) or (- 3) – 4 × (- 5)
= (- 3 – 4) × (- 5) or (- 3) + 20
= -7 × (- 5) or 17
= 35 > 17
(10) State, true or false:
(i) True (5 × 0 = 0, 0 × (- 8) = 0)
(ii) False
(iii) False
(iv) False.
Wrong Solution given
In (5)Evaluate -part (viii)
Either of one number should have +ve sign
(-4) x (-6) x (+8) = 192
No dear its correct ….actually in integers the rule says that if the integers are in even sequence then only the answer is positive….for ex(-2)×(-4)….if in odd sequence I.e.(-2)×(-5)(-8)…the answer will come negative…hope uh understand it…
Yes, the correct answer is :
24×(-8)= -192
This is very helpful
No as there are 3 negative numbers….and according to the rule if integers multiplyed in even sequence then only answer is positive…for ex(-2)×(-4)…and if in odd sequence then answer is negative…
It makes me eazy to learn this problems
Thanks Riddhi Please share to your friends