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Page No 1.10:

Question 11:

Find the value of
(−15) + 4 ÷ (5 − 3)

Answer:

 On applying the BODMAS rule, we get:

   (− 15) + 4 ÷ (5 − 3)

= (− 15) + 4 ÷ 2  (On simplifying brackets)

= (− 15) + 2        (On performing division)

= − 13

Page No 1.10:

Question 12:

Find the value of
(−40) × (−1) + (−28) ÷ 7

Answer:

On applying the BODMAS rule, we get:

(− 40) × (− 1) + (− 28) ÷ 7

=  40 + (− 4)                           (On performing division and multiplication)

=  36 

Page No 1.10:

Question 13:

Find the value of
(−3) + (−8) ÷ (−4) −2 × (−2)

Answer:

On applying the BODMAS rule, we get:

(− 3) + (− 8) ÷ (− 4) − 2 × (− 2)

=  (− 3) + 2 + 4                (On performing division and multiplication)

= (− 3) + 6                       (On performing addition)

= 3                                  (On performing subtraction)

Page No 1.10:

Question 14:

Find the value of
(−3) × (−4) ÷ (−2) + (−1).

Answer:

On applying the BODMAS rule, we get:

(−3) ​× (−4) ÷ (−2) + (−1)

= (−3) × 2 + (−1)                    (On performing division)

=  −6 − 1                                (On performing multiplication)

= −7                                       (On performing addition)



Page No 1.12:

Question 1:

Simplify each of the following:
3 − (5 − 6 ÷ 3)

Answer:

On applying the BODMAS rule, we get:

3 − (5 − 6 ÷ 3)

= 3 − (5 − 2)     (On performing division)

= 3 − 3              (On performing subtraction)

= 0

Page No 1.12:

Question 2:

Simplify each of the following:
−25 + 14 ÷ (5 − 3)

Answer:

On applying the BODMAS rule, we get:

−25 + 14 ÷ (5 − 3)

= −25 + 14 ÷ 2       (On simplifying brackets)

= −25 + 7               (On performing division)

= −18

Page No 1.12:

Question 3:

Simplify each of the following:
25-125+4-(3+2-1+3)

Answer:

On applying the BODMAS rule, we get:

​25 - 1/2 {5+4-( 3 + 2 − 1+3 )}
 = 25- 12{9- (3+2-4)} [Removing vinculum]=  25- 12{9-(5-4)}    [Performing addition]=25- 12{8}    [Performing subtraction]= 25- 4= 21

Page No 1.12:

Question 4:

Simplify each of the following:
27-38-46-(15-13-2)

Answer:

On applying the BODMAS rule, we get:

27 - [38 - {46 - (15 -11)}]    (On simplifying vinculum)

= 27 - [38- {46 - 4}]              (On simplifying parentheses)

= 27 - [38 - 42]                       (On simplifying braces)

= 27 - (-4) = 31

Page No 1.12:

Question 5:

Simplify each of the following:
36-18-{14-(15-4 ÷2×2)}

Answer:

On applying the BODMAS rule, we get:

36 - [18 - { 14 - (15 - 4 ÷ 2 ​× 2)}]
= 36 − [18 − {14 − (15 − 2 × 2)}]                (On performing division)
= 36 - [18 - {14 - (15 - 4)}]                   (On performing multiplication)
= 36 - [18 - {14 - 11}]                            (On simplifying parentheses)
= 36 - [18 - 3]                                          (On simplifying braces)
= 36 - 15
= 21

Page No 1.12:

Question 6:

Simplify each of the following:
45-38-{60÷3-(6-9 ÷3)÷3}

Answer:

On applying the BODMAS rule, we get:

    45 - [38 - { 60 ÷  3 - (6 - 9 ÷ 3) ÷ 3}]
=  45 - [38 - {60 ÷ 3 - (6 - 3) ÷ 3}]           (On performing division)
=  45 - [38 - {60 ÷ 3 - 3 ÷ 3}]                    (On simplifying parentheses)
=  45 - [38 - {60 ÷ 3 - 1}]                          (On performing division)
=  45 - [38 - {20 - 1}]                                (On performing division)
=  45 - [38 - 19]                                          (On performing subtraction)
=  45 - 19   
= 26

Page No 1.12:

Question 7:

Simplify each of the following:
23-23-{23-(23-23-23)}

Answer:

On applying the BODMAS rule, we get:

   23 - [23 - {23 - (23 - 23-23)}]
= 23 - [23 - {23 - (23 - 0}]            (On simplifying vinculum)
= 23 - [23 - {23 - 23}]                    (On simplifying parentheses)
= 23 - [23 - 0]                                  (On simplifying braces)
= 23 - 23 = 0

Page No 1.12:

Question 8:

Simplify each of the following:
2550-510-{270-(90-80+70)}

Answer:

On applying the BODMAS rule, we get:

   2550 - [510 - {270 - (90 - 80+70)}]    
= 2550 - [510 - {270 - (90 - 150)}]        (On simplifying vinculum)
= 2550 - [510 - { 270 - (- 60)}]              (On simplifying parentheses)
= 2550 - [510 - 330]                                 (On simplifying braces)
= 2550 - 180
= 2370

Page No 1.12:

Question 9:

Simplify each of the following:
4+15-10×(25-13-3)÷(-5)

Answer:

On applying the BODMAS rule, we get:
 
     4 + 15[{ - 10 × ( 25 - 13-3)} ÷ (-5)]
=  4 + 15[{- 10 × (25 - 10)} ÷ (- 5)]        (On simplifying vinculum)
=  4 + 15[{- 10 × 15} ÷ (-5 )]                   (On simplifying parentheses)
= 4 + 15[30]                                               (On simplifying braces)
= 4 + 6
= 10

Page No 1.12:

Answer:

On applying the BODMAS rule, we get:
 
22-14{-5-(-48)÷(-16)}= 22- 14{-5-3}    [Performing division]= 22-14{-8}=22-(-2)  [Removing braces]= 22+2 = 24

Page No 1.12:

Question 11:

Simplify each of the following:

63-(-3)-2-8-3÷3{5+(-2)(-1)}

Answer:

On applying the BODMAS rule, we get:

    63 - (- 3) {- 2 - 8-3} ÷ 3 {5 + (- 2) (-1)}
=  63 - (- 3) {- 2 - 5} ÷ 3 {5 + 2}       (On simplifying vinculum)
= 63 - (- 3) (- 7 ) ÷ 3 × 7                     (On simplifying braces)
= 63 - (21÷21)
= 63 - 1
= 62

Page No 1.12:

Question 12:

Simplify each of the following:
29-(-2)6-(7-3)÷3×{5+(-3)×(-2)}

Answer:

On applying the BODMAS rule, we get:

   [29 - (- 2) {6 - (7 - 3)}] ÷ [3 × { - 3) × (- 2)}]
= [29 - (- 2) {6 - 4}] ÷ [3 × { 5 + 6}]                       (On simplifying parentheses)
= [29 - (- 2) (2)] ÷ [3 × 11]                                        (On performing subtraction and addition)
= [29 + 4] ÷ 33                                                           (On performing multiplication)
= 33 ÷ 33
= 1

Page No 1.12:

Question 13:

Using brackets, write a mathematical expression for each of the following:
(i) Nine multiplied by the sum of two and five.
(ii) Twelve divided by the sum of one and three.
(iii) Twenty divided by the difference of seven and two.
(iv) Eight subtracted from the product of two and three.
(v) Forty divided by one more than the sum of nine and ten.
(vi) Two multiplied by one less than the difference of nineteen and six.

Answer:

(i) 9 (2 + 5)
(ii) 12 ÷ (1 + 3)
(iii) 20 ÷ (7 - 2) 
(iv) (2 × 3 ) - 8 
(v) 40 ÷ {(9 + 10) + 1}
(vi) 2 × {(19 - 6) - 1}



Page No 1.4:

Question 1:

Determine each of the following products:
(i) 12 ☓ 7
(ii) (−15) ☓ 8
(iii) (−25) ☓ (−9)
(iv) 125 ☓ (−8)

Answer:

(i) 12 × 7 = 84

(ii) (15) × 8 =  120

(iii) (25) × (9) =  225

(iv) 125 × (8) =  1000

Page No 1.4:

Question 2:

Find each of the following products:
(i) 3 ☓ (−8) ☓ 5
(ii) 9 ☓ (−3) ☓ (−6)
(iii) (−2) ☓ 36 ☓ (−5)
(iv) (−2) ☓ (−4) ☓ (−6) ☓ (−8)

Answer:

(i) 3 × (−8) × 5 = -3×(8×5) = -120

(ii) 9 × (−3) × (−6) = 9× (3×6) = 162

(iii) (−2) × 36 × (−5) = 36 × (2×5) = 360

(iv) (−2) × (−4) × (−6) × (−8) =  (2×4×6×8) = 384

Page No 1.4:

Question 3:

Find the value of:
(i) 1487 × 327 + (−487) × 327
(ii) 28945 × 99 − (−28945)

Answer:

(i) 1487 × 327 + (−487) × 327 = 327 (1487-487) = 327 ×1000 =327000

(ii) 28945 × 99 − (−28945) = 28945 (99 -(-1)) = 28945 (99+1) = 2894500



Page No 1.5:

Question 4:

Complete the following multiplication table:

Is the multiplication table symmetrical about the diagonal joining the upper left corner to the lower right corner?

Answer:

 ×  −4  −3  −2  −1  0    1   2   3   4
 −4   16   12   8   4  0  −4  −8  −12  −16
 −3  12   9   6   3  0  −3  −6  −9  −12
 −2   8   6   4   2  0  −2  −4  −6  −8
 −1   4   3   2   1  0  −1  −2  −3  −4
0 0 0 0 0 0 0 0 0 0
   1  −4  −3  −2  −1 0   1   2   3   4
   2  −8  −6  −4  −2 0   2   4   6   8
   3  −12  −9  −6  −3 0   3    6   9  12
   4  −16  −12  −8  −4 0   4   8   12   16

Yes, the table is symmetrical along the diagonal joining the upper left corner to the lower right corner.

Page No 1.5:

Question 5:

Determine the integer whose product with '−1' is
(i) 58
(ii) 0
(iii) −225

Answer:

The integer, whose product with −1 is the given number, can be found by multiplying the given number by −1.

Thus, we have:

(i) 58 × (−1) =  −58

(ii) 0 ​× (−1) = - (0×1) = 0

(iii) (−225) ​× (−1) =  225

Page No 1.5:

Question 6:

What will be the sign of the product if we multiply together
(i) 8 negative integers and 1 positive integer?
(ii) 21 negative integers and 3 positive integers?
(iii) 199 negative integers and 10 positive integers?

Answer:

Negative numbers, when multiplied even number of times, give a positive number. However, when multiplied odd number of times, they give a negative number. Therefore, we have:

(i) (negative) 8 times × (positive)  1 time = positive × positive = positive integer

(ii) (negative) 21 times  × (positive) 3 times = negative ×positive  = negative integer

(iii) (negative) 199 times × (positive) 10 times = negative × positive = negative integer

Page No 1.5:

Question 7:

State which is greater:
(i) (8 + 9) × 10 and 8 + 9  × 10
(ii) (8 − 9) × 10 and 8 − 9 × 10
(iii) {(−2) − 5} × (−6) and (−2) −5 × (−6)

Answer:

(i) ( 8 + 9) × 10 = 170   >   8 + 90 = 98

(ii) (8 − 9) ​× 10 = −10  >  8 − 90 = − 82

(iii) {(−2) − 5 } ​× (−6) = −7 ​× (−6) = 42  >   (−2) − 5 × (−6)  = ( −2 ) −  (−30)  = −2 + 30 = 28

Page No 1.5:

Question 8:

(i) If a × (−1) = −30, is the integer a positive or negative?
(ii) If a × (−1) = 30, is the integer a positive or negative?

Answer:

(i) × (−1) = −30  

When multiplied by a negative integer, a gives a negative integer. Hence, a should be a positive integer.

a = 30

(ii) × (−1) = 30   

​When multiplied by a negative integer, a gives a positive integer. Hence, a should be a negative integer​.

a = −30

Page No 1.5:

Question 9:

Verify the following:
(i) 19 × {7 + (−3)} = 19 × 7 + 19 × (−3)
(ii) (−23) {(−5) + (+19)} = (−23) × (−5) + (−23) × (+19)

Answer:

(i)
LHS = 19 ​× { 7 + (−3) } = 19 × {4} =  76
   
RHS =  19 × 7 + 19 × (−3) = 133 + (−57) = 76

Because LHS is equal to RHS, the equation is verified.

(ii)
LHS = (−23) {(−5) + 19} = (−23) { 14} = −322

RHS = (−23) × (−5) + (−23) × 19 = 115 + (−437) = −322

Because LHS is equal to RHS, the equation is verified.

Page No 1.5:

Question 10:

Which of the following statements are true?
(i) The product of a positive and a negative integer is negative.
(ii) The product of three negative integers is a negative integer.
(iii) Of the two integers, if one is negative, then their product must be positive.
(iv) For all non-zero integers a and b, a × b is always greater than either a or b.
(v) The product of a negative and a positive integer may be zero.
(vi) There does not exist an integer b such that for a> 1, a × b = b × a = b.

Answer:

(i) True. Product of two integers with opposite signs give a negative integer.

(ii) True. Negative integers, when multiplied odd number of times, give a negative integer.

(iii) False. Product of two integers, one of them being a negative integer, is not necessarily positive. For example, (−1) × 2 = −2

(iv) False. For two non-zero integers a and b, their product is not necessarily greater than either a or b. For example, if a = 2 and  b = −2, then, × b = −4, which is less than both 2 and −2.

(v) False. Product of a negative integer and a positive integer can never be zero.

(vi) True. If a > 1, then, a×b  b×a b



Page No 1.8:

Question 1:

Divide:
(i) 102 by 17
(ii) −85 by 5
(iii) −161 by −23
(iv) 76 by −19
(v) 17654 by −17654
(vi) (−729) by (−27)
(vii) 21590 by −10
(viii) 0 by −135

Answer:

(i) 102  ÷ 17 = 10217 = 10217 = 6
(ii) −85  ÷  5 = --855 = -855 = − 17
(iii) −161  ÷ ( − 23)  = -161-23 =16123=  7
(iv) 76   ÷  − 19 =  -76-19  = -7619 = − 4
(v) 17654   ÷  (− 17654) =  - 17654-17654= -1765417654= − 1
(vi) (−729)  ÷  (− 27) = -729-27 = 72927=  27
(vii) 21590   ÷  − 10 = -21590-10 = -2159010= − 2159
(viii) 0 ​ ÷ ​(−135) = -0-135= -0135= 0

Page No 1.8:

Question 2:

Fill in the blanks:
(i) 296 ÷ ... = −148
(ii) −88 ÷ ... = 11
(iii) 84 ÷ ... = 12
(iv) ....... ÷ −5 = 25
(v) ....... ÷ 156 = −2
(vi) ....... ÷ 567 = −1

Answer:

(i) 296  ÷ -148 = -296-148= -296148=-296148= -2
∴ 296 ÷ (-2) = -148

(ii) − 88  ÷ 11 = --8811=-8811=-8811= -8
-88 ÷ -8 = 11

(iii) 84  ÷ 12 = 8412 = 8412=7
84÷7 = 12

(iv) 25×(-5) = -125

-125÷-5 = 25

(v) 156×(-2) = -312

∴ -312÷156 = -2

(vi) 567×(-1) = -567

-567 ÷ 567 = -1

Page No 1.8:

Question 3:

Which of the following statements are true?
(i) 0 ÷ 4 = 0
(ii) 0 ÷ (−7) = 0
(iii) −15 ÷ 0 = 0
(iv) 0 ÷ 0 = 0
(v) (−8) ÷ (−1) = −8
(vi) −8 ÷ (−2) = 4

Answer:

(i)
LHS =04=04=0         = RHS

Because LHS is equal to RHS, the equation is true.

(ii)
LHS = -0-7=-07=-0=0         = RHS

Because LHS is equal to RHS, the equation is true.

(iii)
LHS = --150=-150= Not defined       RHS 
Because LHS is not equal to RHS, the equation is false.

(iv)
LHS = 00= 00= Not Defined         RHS

Because LHS is not equal to RHS, the equation is false.

(v)
LHS =-8-1= 81=8        RHS
Because LHS and RHS are not equal, the equation is false.

(vi)
LHS = -8-2 = 82=4        = RHS

Because LHS is equal to RHS, the equation is true.



Page No 1.9:

Question 1:

Find the value of
36 ÷ 6 + 3

Answer:

On applying the BODMAS rule, we get:

36  ÷  6  + 3

= 6 + 3  (On performing division)

 = 9

Page No 1.9:

Question 2:

Find the value of
24 + 15 ÷ 3

Answer:

On applying the BODMAS rule, we get:

24 + 15 ÷ 3

= 24 + 5  (On performing division)

= 29

Page No 1.9:

Question 3:

Find the value of
120 − 20 ÷ 4

Answer:

On applying the DMAS rule, we get:

120 − 20 ÷ 4

= 120 − 5 (On performing division)

= 115

Page No 1.9:

Question 4:

Find the value of
32 − (3 × 5) + 4

Answer:

On applying the DMAS rule, we get:

32 − ( 3 ​× 5 ) + 4

= 32 − 15 + 4    (On performing multiplication)

=  36 − 15         (On performing addition)

= 21                  (On performing ​subtraction)

Page No 1.9:

Question 5:

Find the value of
3 − (5 − 6 ÷ 3)

Answer:

On applying the DMAS rule, we get:

3 - ( 5 - 6 ÷ 3)

= 3 - ( 5 − 2)       (On performing division)

= 3 - 3                (On performing subtraction)

= 0

Page No 1.9:

Question 6:

Find the value of
21 − 12 ÷ 3 × 2

Answer:

On applying the DMAS rule, we get:

21 − 12 ÷ 3 × 2

= 21 − 4 × 2         (On performing division)

= 21 − 8               (On performing multiplication)

= 13                      (On performing subtraction)

Page No 1.9:

Question 7:

Find the value of
16 + 8 ÷ 4 − 2 × 3

Answer:

On applying the DMAS rule, we get:

16 + 8 ÷ 4 − 2 ​× 3

= 16 + 2 − 6            (On performing division and multiplication)

= 18 − 6

= 12

Page No 1.9:

Question 8:

Find the value of
28 − 5 × 6 + 2

Answer:

On applying the DMAS rule, we get:

28 - 5 × 6 + 2

= 28 - 30 + 2 (On performing multiplication)

= 30 - 30       (On performing addition)

= 0                 (On performing subtraction)

Page No 1.9:

Question 9:

Find the value of
(−20) × (−1) + (−28) ÷ 7

Answer:

On applying the DMAS rule, we get:

(− 20) ​× (− 1) + (−28) ÷ 7

=  20 + (− 4)   (On performing division and multiplication)

= 20 − 4 

= 16

Page No 1.9:

Question 10:

Find the value of
(−2) + (−8) ÷ (−4)

Answer:

On applying the DMAS rule, we get:

(− 2) + (− 8) ÷ (− 4)

= (− 2) + 2  (On performing division)

= 0             (On performing addition)



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