RD Sharma Solutions Class 7 Integers Exercise 1.4

RD Sharma Class 7 Solutions Chapter 1 Ex 1.4 PDF Free Download

RD Sharma Solutions Class 7 Chapter 1 Exercise 1.4

Exercise 1.4

Simplify each of the following:

(i) \(3-(5-6\div 3)\)

Solution:

\(3-(5-6\div 3)\)

= 3 – (5 – 2)

= 3 – 3

= 0

(ii) \(-25+14\div (5-3)\)

Solution:

\(-25+14\div (5-3)\) = \(-25+14\div (2)\)

= \(-25+\frac{14}{2}\)

= –25 + 7

= –18

(iii) \(25-\frac{1}{2}(5+4-(3+2-\overline{1+3}))\)

Solution:

\(25-\frac{1}{2}[5+4-(3+2-\overline{1+3})]\)

= \(25-\frac{1}{2}[5+4-(5-4)]\)

= \(25-\frac{1}{2}[5+4-1]\)

= \(25-\frac{1}{2}[8]\)

= 25 – 4

= 21

(iv) \(27-[38-(46-(15-\overline {13-2}))]\)

Solution:

\(27-[38-(46-(15-\overline {13-2}))]\)

= \(27-[38-(46-(15-11))]\)

= 27 – [38 – (46 – 4)]

= 27 – [38 – 42]

= 27 – [–4]

= 27 + 4

= 31

(v)  36−[18−(14−(15−4÷2×2))]

Solution:

36−[18−(14−(15−4÷2×2))]

= 36−[18−(14−(15 – 2×2))]

= 36−[18−(14−(15 – 4))]

= 36 – [18 – (14 – 11)]

= 36 – [18 – 3]

= 36 – 15

= 21

(vi) \(45-[38-(60\div 3-(6-9\div 3)\div 3)]\)

Solution:

\(45-[38-(60\div 3-(6-9\div 3)\div 3)]\)

= \(45-[38-(20-(6-3)\div 3)]\)

= \(45-[38-(20-3\div 3)]\)

= 45 – [38 – (20 – 1)]

= 45 – [38 – 19]

= 45 – [19]

= 26

(vii) \(23-[23-(23-(23-\overline{23-23}))]\)

Solution:

\(23-[23-(23-(23-\overline{23-23}))]\)

= 23 – [23 – (23 – (23 – 0))]

= 23 – [23 – (23 – 23)]

= 23 – [23 – 0]

= 23 – 23

= 0

(viii) \(2550-[510-(270-(90-\overline{80+70}))]\)

Solution:

\(2550-[510-(270-(90-\overline{80+70}))]\)

= 2550 – [510 – (270 – (90 – 150))]

= 2550 – [510 – (270 – (–60))]

= 2550 – [510 – 330]

= 2550 – [180]

= 2550 – 180

= 2370

(ix) \(4+\frac{1}{5}[(-10\times (25-\overline{13-3}))\div (-5)]\)

Solution:

\(4+\frac{1}{5}[(-10\times (25-\overline{13-3}))\div (-5)]\)

= \(4+\frac{1}{5}[(-10\times (25-10))\div (-5)]\)

= \(4+\frac{1}{5}[(-10\times (15))\div (-5)]\)

= \(4+\frac{1}{5}[(-150)\div (-5)]\)

= \(4+\frac{1}{5}[30]\)

= 4 + 6

= 10

(x) \(22-\frac{1}{4}(-5-(-48)\div (-16))\)

Solution:

\(22-\frac{1}{4}(-5-(-48)\div (-16))\)

= \(22-\frac{1}{4}(-5-(\frac{-48}{-16}))\)

= \(22-\frac{1}{4}(-5-\frac{48}{16})\)

= \(22-\frac{1}{4}(-5-3)\)

= \(22-\frac{1}{4}(-8)\)

= \(22+\frac{8}{4}\)

= 22 + 2

= 24

(xi) \(63-[(-3)(-2-\overline{8-3})]\div [3(5+(-2)(-1))]\)

Solution:

\(63-[(-3)(-2-\overline{8-3})]\div [3(5+(-2)(-1))]\)

= \(63-[(-3)(-2-5)]\div [3(5+2)]\)

= \(63-[(-3)(-7)]\div [3(7)]\)

= \(63-(21)\div (21)\)

= 63 – 1

= 62

(xii) \([29-(-2)(6-(7-3))]\div [3\times (5+(-3)\times (-2))]\)

Solution:

\([29-(-2)(6-(7-3))]\div [3\times (5+(-3)\times (-2))]\)

= \([29-(-2)(6-4)]\div [3\times (5+(3\times 2))]\)

= \([29-(-2)(2)]\div [3\times (5+6)]\)

= \([29+4]\div [3\times 11]\)

= \((33)\div (33)\)

= 1

(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.

Solution:

(i) 9 (2 + 5)

(ii) \(12\div (1+3)\)

(iii) \(20\div (7-2)\)

(iv) \((2\times 3)-8\)

(v) \(40\div (1+(9+10))\)

(vi) \(2\times [(19-6)-1]\)

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