Selina Concise Maths Solutions Class 7 Chapter 1 Integers PDF

Downloaded from https://www.ncertbooksolutions.com/ Integers EXERCISE 1 (A) Question 1. Evaluate: 1. 427 x 8 + 2 x 427 2. 394 x 12 + 394 x (-2) 3. 558 x 27 + 3 x 558...

Downloaded from https://www.ncertbooksolutions.com/ Integers EXERCISE 1 (A) Question 1. Evaluate: 1. 427 x 8 + 2 x 427 2. 394 x 12 + 394 x (-2) 3. 558 x 27 + 3 x 558 om Solution:.c 1. 427 x 8 + 2 x 427 = 427 x (8 + 2) (Distributive property) ns = 427 x 10 io = 4270 t 2. 394 x 12 + 394 x (-2) = 394 x (12-2) (Distributive property) = 394 x 10 lu so = 3940 k 3. 558 x 27 + 3 x 558 = 558 x (27 + 3) (Distributive property) oo = 558 x 30 rtb = 16740 ce Question 2..n Evaluate: w w 1. 673 x 9 + 673 //w 2. 1925 x 101 – 1925 s: tp Solution: ht 1. 673 x 9 + 673 = 673 x (9 + 1) (Distributive property) = 673 x 10 = 6730 2. 1925 x 101 – 1925 = 1925 x (101 – 1) (Distributive property) = 1925 x 100 = 192500 Question 3. Verify: 1. 37 x {8 +(-3)} = 37 x 8 + 37 x – (3) 2. (-82) x {(-4) + 19} = (-82) x (-4) + (-82) x 19 3. {7 – (-7)} x 7 = 7 x 7 – (-7) x 7 4. {(-15) – 8} x -6 = (-15) x (-6) – 8 x (-6) Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ Solution: 1. 37 x {8 + (-3)} = 37 x 8 + 37 x – (3) L.H.S. = 37 x {8 + (-3)} = 37 x {8-3} = 37 x {5} = 37 x 5 = 185 R.H.S. = 37 x 8 + 37 – 3 = 37 x (8 – 3) = 37 x 5 = 185 Hence, L.H.S. = R.H.S. om 2. (-82) x {(-4) + 19} = (-82) x (-4) + (-82) x 19 L.H.S. = (-82) x {(_4) + 19}.c = (-82) x {-4 + 19} ns = (-82)x {15} io = -82 x 15 t =-1230 R.H.S. = (-82) x (-4) + (-82) x 19 lu so = -82 x (-4 + 19) k = -82 x 15 oo =-1230 rtb Hence, L.H.S. = R.H.S. 3. {7 – (-7)}. x 7 = 7 x 7 – (-1) x 7 ce L.H.S. = {7 – (-7)} x 7.n = {7 + 7} x 7 w = {14} x 7 w = 14 x 7 //w = 98 R.H.S. = 7 x 7 – (-7) x 7 s: =7 x 7+7 x 7 = tp 7 x (7 + 7) ht = 7 x (14) = 98 Hence, L.H.S. = R.H.S. 4. {(-15) – 8} x -6 = (-15) x (-6) – 8 x (-6) L.H.S. = {(-15)-8} x-6 = {-15-8} x-6 = {-23} x-6 = -23 x- 6 = 138 R.H.S. = (-15) x (-6) – 8 x (-6) = -6 x (-15-8) = -6 x -23 Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ = 138 Hence, L.H.S. = R.H.S. Question 4. Evaluate: 1. 15 x 8 2. 15 x (-8) 3. (-15) x 8 4. (-15) x -8 Solution: om 1. 15 x 8= 120.c 2. 15 x (-8) = -120 ns 3. (-15) x 8 = -120 4. (-15) x -8 = 120 io (Since the number of negative integers in the product is even) t lu so Question 5. k Evaluate: oo rtb 1. 4x6x8 2. 4 x 6 x (-8) ce 3. 4 x (-6) x 8.n 4. (-4) x 6 x 8 w 5. 4 x (-6) x (-8) w 6. (-4) x (-6) x 8 //w 7. (-4) x 6 x (- 8) 8. (-4) x (-6) x (-8) s: tp Solution: ht 1. 4 x 6 x 8 = 192 2. 4 x 6 x (-8) = -192 (It have one negative factor) 3. 4 x (-6) x 8 = -192 (It have one negative factor) 4. (-4 )x 6 x 8 = -192 (It have one negative factor) 5. 4 x (-6) x (-8) = 192 (It have two negative factors) 6. (-4) x (-6) x 8 = 192 (It have two negative factors) Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ 7. (-4) x 6 x (-8) = 192 (It have two negative factors) 8. (-4) x (-6) x (-8) = -192 (It have three negative factors) Question 6. Evaluate: 1. 2x4x6x8 2. 2 x (-4) x 6 x 8 3. (-2) x 4 x (-6) x 8 4. (-2) x (-4) X 6 x (-8) 5. (-2) x (-4) x (-6) x (-8) om Solution:.c ns 1. 2 x 4 x 6 x 8 = 384 io 2. 2 x (-4) x 6 x 8 = -384 t lu (Number of negative integer in the product is odd) so 3. (-2) x 4 x (-6) x 8 = 384 (Number of negative integer in the product is even) k oo 4. (-2) x (-4) x 6 x (-8) = -384 (Number of negative integer in the product is odd) rtb 5. (-2) x (-4) x (-6) x (-8) = 384 ce (Number of negative integer in the product is even).n w Question 7. Determine the integer whose product with ‘-1’ is: w //w 1. -47 s: 2. 63 tp 3. -1 ht 4. 0 Solution: 1. -1 x 47 = -47 Hence, integer is 47 2. -1 x -63 = 63 Hence, integer is -63 3. -1 x 1 = -1 Hence, integer is 1 4. -1 x 0 = 0 Hence, integer is 0 Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ Question 8. Eighteen integers are multiplied together. What will be the sign of their product, if: 1. 15 of them are negative and 3 are positive? 2. 12 of them are negative and 6 are positive? 3. 9 of them are positive and the remaining are negative? 4. all are negative? Solution: 1. Since out of eighteen integers, 15 of them are negative, which is odd number. Hence, sign of product will be negative (-). om 2. Since out of eighteen integers 12 of them are negative, which is even number. Hence sign of product will be positive (+)..c 3. Since out of eighteen integers 9 of them are negative, which is odd number. ns Hence, sign of product will be negative (-). io 4. Since all are negative, which is even number. Hence sign of product will be t lu positive (+). so k Question 9. oo Find which is greater? rtb 1. (8 + 10) x 15 or 8 + 10 x 15 ce 2. 12 x (6 – 8) or 12 x 6 – 8.n 3. {(-3) – 4} x (-5) or (-3) – 4 x (-5) w w Solution: //w s: 1. (8 + 10) x 15 or 8 + 10 x 15 tp (8 + 10) x 15 = 18 x 15 = 270 8 + 10 x 15 = 8 + 150 = 158 ht ∴(8 + 10) x 15 > 8 + 10 x 15 2. 12 x (6 – 8) or 12 x 6 – 8 12 x (6 – 8) = 12 (-2) = -24 12 x 6 – 8 = 72 – 8 = 64 ∴12 x 6 – 8 > 12 x (6-8) 3. {(-3) – 4} x (-5) or (-3) – 4 x (-5) {(-3) – 4} x (-5) = {-3 – 4} x (-5) = -7 x -5 = 35 (-3) – 4 x (-5) = -7 x (-5) = 35 ∴{(-3) – 4} x (-5) = (-3) – 4 x (-5) Question 10. State, true or false : Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ 1. product of two integers can be zero. 2. product of 120 negative integers and 121 positive integers is negative. 3. a x (b + c) = a x b + c 4. (b – c) x a=b – c x a Solution: om.c 1. False. ns 2. False. Correct : Since 120 integers are even numbers, hence product will be positive and io for 121 integers are positive in numbers, hence product will be positive. t 3. False. lu so Correct :a x (b + c) ≠ a x b + c ab + ac ≠ ab + c k oo 4. False. Correct: (b – c) x a ≠ b – c x a rtb ab – ac ≠ b – ca ce.n EXERCISE 1 (B) w w Question 1. //w Divide: (i) 117 by 9 s: (ii) (-117) by 9 tp (iii) 117 by (-9) ht (iv) (-117) by (-9) (v) 225 by (-15) (vi) (-552) ÷ 24 (vii) (-798) by (-21) (viii) (-910) ÷ – 26 Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ Solution : om.c ns tio lu k so oo rtb ce.n w w //w s: tp ht Question 2. Evaluate: (i) (-234) ÷ 13 (ii) 234 ÷ (-13) (iii) (-234) ÷ (-13) (iv) 374 ÷ (-17) (v) (-374) ÷ 17 (vi) (-374) ÷ (-17) (vii) (-728) ÷ 14 (viii) 272 ÷ (-17) Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ Solution: om.c ns io t lu k so oo rtb ce.n w w //w Question 3. s: Find the quotient in each of the following divisions: tp (i) 299 ÷ 23 ht (ii) 299 ÷ (-23) (iii) (-384) ÷ 16 (iv) (-572) ÷ (-22) (v) 408 ÷ (-17) Solution: Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ om Question 4. Divide:.c (i) 204 by 17 ns (ii) 152 by-19 io (iii) 0 by 35 t (iv) 0 by (-82) (v) 5490 by 10 lu so (vi) 762800 by 100 k oo Solution: rtb ce.n w w //w s: tp ht Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ Question 5. State, true or false : 1. 0 ÷ 32 = 0 2. 0 ÷ (-9) = 0 3. (-37) ÷ 0 = 0 om 4. 0÷0=0.c Solution: ns io 1. True. t lu 2. True. so 3. False. Correct: It is not meaningful (defined) k oo 4. False. Correct: It is not defined. rtb ce Question 6..n Evaluate: (i) 42 ÷ 7 + 4 w w (ii) 12+18 ÷ 3 (iii) 19 – 20 ÷ 4 //w (iv) 16 – 5 x 3+4 s: (v) 6 – 8 – (-6) ÷ 2 tp (vi) 13 -12 ÷ 4 x 2 ht (vii) 16 + 8 ÷ 4- 2 x 3 (viii) 16 ÷ 8 + 4 – 2 x 3 (ix) 16 – 8 + 4 ÷ 2 x 3 (x) (-4) + (-12) ÷ (-6) (xi) (-18) + 6 ÷ 3 + 5 (xii) (-20) x (-1) + 14 – 7 Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ Solution: om.c ns tio lu k so oo rtb ce.n w w //w s: tp ht Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ om.c ns t io lu k so oo rtb ce.n w w //w s: tp EXERCISE 1 (C) ht Question 1. Evaluate: 18-(20- 15 ÷ 3) Solution: 18-(20- 15 ÷ 3) = 18 – = 18 – (20 – 5) = 18 – 20 + 5 = 18 + 5 – 20 = 23 – 20 =3 Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ Question 2. -15+ 24÷ (15-13) Solution: -15+ 24÷ (15- 13) = -15 + 24 ÷ 2 = -15 + 12 = -3 Question 3. 35 – [15 + {14-(13 + )}] Solution: 35- [15 + {14-(13 + )}] = 35-[15+ 14-(13+4)] = 35 — [15 + 14 – (13 + 4}] om = 35-{15 + 14-17].c = 35-15-14+ 17 ns = 35 + 17-15-14 = 52 – 29 io = 23 t lu so Question 4. k 27- [13 + {4-(8 + 4 – oo )}] Solution: rtb 27- [13 + {4-(8 + 4 – )}] = 27-[13 +{4-(8+ 4-4)}] ce = 27-[13 + {4-8}].n = 27 – [13 + (-4)] w = 21 – w = 27-9 //w = 18 s: tp Question 5. ht 32 – [43-{51 -(20 – )}] Solution: 32 – [43 – {51 – (20 – )}] = 32-[43 – {51 -(20- 11)}] = 32-[43-{51 -9}] = 32-[43 -42] = 32-1 =31 Question 6. 46-[26-{14-(15-4÷ 2 x 2)}] Solution: 46 – [26 – {14 – (15 – 4 ÷ 2 x 2)}] Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ = 46-[26- {14-(15-2 x 2)}] = 46-[26- {14-(15 -4)}] = 46-[26- {14- 11}] = 46 – [26 – 3] = 46 – 23 = 23 Question 7. 45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}] Solution: 45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}] = 45-[38- {60 ÷ 3-(6-3)÷ 3}] = 45-[38 -{20-3 ÷ 3}] = 45-[38- {20-1}] om = 45-[38- 19].c = 45-19 ns = 26 tio lu Question 8. 17- [17 — {17 — (17 – so )}] Solution: k 17- [17-{17-(17 – )}] oo = 17-[17-{17-(17-0)}] rtb = 17 – [17 – {17 — 17}] = 17 — [17 — 0] ce = 17-17.n =0 w w //w Question 9. 2550 – [510 – {270 – (90 – )}] s: Solution: tp 2550- [510-{270-(90- )}] ht = 2550 – [510 – {270 – (90 – 87)}] = 2550 -[510- {270 -3}] = 2550-[510-267] = 2550 – 243 = 2307 Question 10. 30+ [{-2 x (25- )}] Solution: 30+ [{-2 x (25- )}] = 30 + [{-2 x (25 – 10)}] = 30 + [{-2 x 15}] = 30 + [-30] Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ = 30-30 =0 Question 11. 88-{5-(-48)+ (-16)} Solution: 88- {5-(-48)+ (-16)} =88 – = 88 – {5-3} = 88 – 2 = 86 om Question 12. 9 x (8- ) – 2 (2 + ).c Solution: ns 9 x (8- ) -2(2 + ) io = 9 x (8 – 5) – 2(2 + 6) t =9x3–2x8 = 27- 16 lu so = 11 k oo Question 13. rtb 2 – [3 – {6 – (5 – )}] ce Solution: 2 – [3 – {6 – (5 – )}].n ⇒ 2 – [3 – {6 – (5 – 1)}] w ⇒ 2 – [3 – {6 – 4}] w ⇒2 – (3 – 2) //w ⇒2-1 = 1 s: tp EXERCISE 1 (D) ht Question 1. The sum of two integers is -15. If one of them is 9, find the other. Solution: Sum of two integers = -15 One integer = 9 ∴ Second integer = -15 – 9 = -(15 + 9) = -24 Question 2. The difference between an integer and -6 is -5. Find the values of x. Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ Solution: The difference between an integer = x-(-6) = -5 ∴ Value of ⇒ x – (-6) = -5 ⇒ x + 6 = -5 x = -5 – 6 x = -11 Question 3. The sum of two integers is 28. If one integer is -45, find the other. Solution: Sum of two integers = 28 om One integer = -45 ∴ Second integer = 28 – (-45).c = 28 + 45 ns = 73 io t Question 4. lu so The sum of two integers is -56. If one integer is -42, find the other. k Solution: oo Sum of two integers = -56 rtb One integer = -42 ∴Second integer = -56 – (-42) ce = -56+ 42.n =-14 w w Question 5. //w The difference between an integer x and (-9) is 6. Find all possible values ofx. s: Solution: The difference between an integer x – (-9) = 6 or -9 – x = 6 tp ∴ Value of x ht ⇒ x – (-9) = 6 or ⇒ -9 – x = 6 ⇒ x + 9 = 6 or Answer-x = 6 + 9 ⇒ x = 6 – 9 or ⇒ -x = 15 ⇒x = -3 or ⇒ x = -15 Hence, possible values ofx are -3 and -15. Question 6. Evaluate: 1. (-1) x (-1) x (-1) x ….60 times. 2. (-1) x (-1) x (-1) x (-1) x …. 75 times. Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ Solution: 1. 1 (because (-1) is multiplied even times.) 2. -1 (because (-1) is multiplied odd times.) Question 7. Evaluate: 1. (-2) x (-3) x (-4) x (-5) X (-6) 2. (-3) x (-6) x (-9) x (-12) 3. (-11) x (-15) + (-11) x (-25) 4. 10 x (-12) + 5 x (-12) om Solution:.c ns 1. (-2) x (-3) x (-4) x (-5) x (-6) ⇒ 6 x 20 x (-6) = 120 x (-6) io = -720 t 2. (-3) x (-6) x (-9) x (-12) lu so ⇒ 18 x 108 k = 1944 oo 3. (-11) x (-15) + (-11) x (-25) rtb ⇒ 165 + 275 = 440 ce 4. 10 x (-12) + 5 x (-12).n ⇒ -120-60 w = -180 w //w Question 8. s: tp 1. If x x (-1) = -36, is x positive or negative? 2. If x x (-1) = 36, is x positive or negative? ht Solution: 1. x x (-1) = -36 -lx = -36 x= x = 36 ∵ x = 36 ∴ It is a positive integer. 2. x x (-1) = 36 -1x = 36 Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ x= x = -36 ∵x = -36 ∴It is a negative integer. Question 9. Write all the integers between -15 and 15, which are divisible by 2 and 3. Solution: The integers between -15 and 15 are : -12, -6, 0, 6 and 12 That are divisible by 2 and 3. Question 10. om Write all the integers between -5 and 5, which are divisible by 2 or 3. Solution:.c The integers between -5 and 5 are : ns -4, -3, -2, 0, 0, 2, 3 and 4 That are divisible by 2 or 3. tio Question 11. lu so Evaluate: k oo 1. (-20) + (-8) ÷ (-2) x 3 rtb 2. (-5) – (-48) ÷ (-16) + (-2) x 6 ce 3. 16 + 8 ÷ 4- 2 x 3 4. 16 ÷ 8 x 4 – 2 x 3.n 5. 27 – [5 + {28 – (29 – 7)}] w 6. 48 – [18 – {16 – (5 – )}] w 7. -8 – {-6 (9 – 11) + 18 = -3} //w 8. (24 ÷ – 12) – (3 x 8 ÷ 4 + 1) s: tp Solution: ht We know that, if these type of expressions that has more than one fundamental operations, we use the rule of DMAS i.e., First of all we perform D (division), then M (multiplication), then A (addition) and in the last S (subtraction). 1. (-20) + (-8) ÷ (-2) x 3 ⇒ -20 + 4 x 3 ⇒ -20+ 12 =-8 2. (-5) – (-48) ÷ (-16) + (-2) x 6 ⇒ (-5) – 3 + (-2) x 6 ⇒ -5 – 3 – 12 Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ ⇒ -8- 12 = -20 3. 16 + 8 ÷ 4 – 2 x 3 ⇒ 16 + 2 – 2 x 3 ⇒16 + 2 – 6 ⇒ 18-6 = 12 4. 16 ÷ 8 x 4 – 2 x 3 ⇒2x4–2x3 ⇒8–6 =2 5. 27 – [5 + {28 – (29 – 7)}] ⇒ 27 – [5 + {28 – 22}] om ⇒ 27 – [5 + 6] ⇒ 27 — 11.c = 16 ns 6. 48-[18-{16-(5 – )}] io ⇒ 48-[18-{16-(5-5)}] t lu ⇒ 48-[18- {16-0)}] so ⇒ 48-[18- 16] ⇒ 48 – 2 k oo = 46 7. -8 – {-6 (9 – 11) + 18 ÷ -3} rtb ⇒ -8 – {-6 (-2) – 6} ce ⇒ -8- {12-6} ⇒ -8 – {6}.n ⇒ -8-6 w = -14 w 8. //w (24 ÷ – 12) – (3 x 8 = 4 + 1) ⇒ (24 ÷ 3-12)-(3 x 2 + 1) s: ⇒ (8- 12)-(6+ 1) tp ⇒ —4 — 7 ht = —11 Question 12. Find the result of subtracting the sum of all integers between 20 and 30 from the sum of all integers from 20 to 30. Solution: Required number = (Sum of all integers between 20 and 30 – Integers between 20 and 30) (20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30) – (21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 ) ⇒ 20 + 30 = 50 ∴ Required number = 50 Downloaded from https://www.ncertbooksolutions.com/  Downloaded from https://www.ncertbooksolutions.com/ Question 13. Add the product of (-13) and (-17) to the quotient of (-187) and 11. Solution: (-13) x (-17)+ (-187- 11) ⇒ (-13) x (-17) + (-17) ⇒ 221 – 17 = 204 Question 14. om The product of two integers is-180. If one of them is 12, find the other..c Solution: The product of two integers = -180 One integer = 12 ns ∴ Second integer = -180 – 12 = -15 io t Question 15. lu k so 1. A number changes from -20 to 30. What is the increase or decrease in the oo number? rtb 2. A number changes from 40 to -30. What is the increase or decrease in the ce number?.n Solution: w w //w 1. ∵A number changes from = -20 to 30 ⇒ -20 – 30 = -50 s: ∴-50, it will be increases. tp 2. ∵A number changes from = 40 to -30 ht ⇒ 40 – (-30) 40 + 30 = 70 ∴70, it will be decreases Downloaded from https://www.ncertbooksolutions.com/