{"id":21692,"date":"2022-01-12T16:59:17","date_gmt":"2022-01-12T11:29:17","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=21692"},"modified":"2022-01-18T10:06:48","modified_gmt":"2022-01-18T04:36:48","slug":"ap-board-7th-class-maths-solutions-chapter-3-intext-questions-new-syllabus","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-7th-class-maths-solutions-chapter-3-intext-questions-new-syllabus\/","title":{"rendered":"AP Board 7th Class Maths Solutions Chapter 3 Simple Equations InText Questions"},"content":{"rendered":"

SCERT\u00a0AP 7th Class Maths Solutions Pdf<\/a> Chapter 3 Simple Equations InText Questions and Answers.<\/p>\n

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations InText Questions<\/h2>\n

Check your Progress (Page No: 43)<\/span><\/p>\n

Question 1.
\nWrite simple equations for the following verbal statements.
\n(i) The sum of five times of x and 3 is 28.
\nAnswer:
\nGiven number = x
\nFive times of a number = 5 \u2219 x
\nBy adding 3 the result = 5x + 3
\n\u2234 5x + 3 = 28<\/p>\n

\"AP<\/p>\n

(ii) Taking away 7 from p gives 21.
\nAnswer:
\nGiven number = p
\nBy subtracting 7 from the number = p – 7
\n\u2234 p – 7 = 21<\/p>\n

(iii) If you add one third to m, then you get 25.
\nAnswer:
\nGiven number = m
\nBy adding \\(\\frac{1}{3}\\) to m = m + \\(\\frac{1}{3}\\)
\nthen, m + \\(\\frac{1}{3}\\) = 25.<\/p>\n

(iv) Sum of angles x, (x + 20) is a straight angle.
\nAnswer:
\nGiven angles are x, x + 20
\nSum of angles = x + x + 20
\nSum of angles is straight angle (180\u00b0).
\n\u21d2 x + x + 20 = 180\u00b0
\n\u2234 2x + 20 = 180\u00b0<\/p>\n

(v) Perimeter of a rectangle whose length is 2 cm more than its width is 16 cm.
\nAnswer:
\nLet width of rectangle = x
\nLength of rectangle
\n= 2 more than width = x + 2
\nPerimeter = 2 (length + width)
\n= 2(x + 2 + x) = 2(2x + 2)
\nGiven Perimeter = 16 cm
\n\u2234 4x + 4 = 16<\/p>\n

Question 2.
\nWrite the following equations in state-ment form.
\n(i) x + 4 = 9
\nAnswer:
\n4 is added to a number is 9.<\/p>\n

(ii) 2y = 15
\nAnswer:
\nTwo times (twice) a number is 15.<\/p>\n

(iii) 3m – 13 = 25
\nAnswer:
\n13 is taking away from 3 times of m is 25.<\/p>\n

(iv) \\(\\frac{n}{4}\\) = 5
\nAnswer:
\nOne fourth of n is 5.<\/p>\n

Let’s Do Activity (Page No: 45)<\/span><\/p>\n

Describe the balances given below in two ways.
\n(a) Using words (b) Using mathematical symbols.
\nAnswer:
\n\"AP<\/p>\n

\"AP<\/p>\n

Lets Explore (Page No: 46)<\/span><\/p>\n

Develop equations from the pictures given below and solve pictorially.<\/p>\n

(i)
\n\"AP
\nAnswer:
\nx + x + x = 5 + 5 + 1 + 1
\n3x = 12<\/p>\n

(ii)
\n\"AP
\nAnswer:
\ny + y + 1 = 5 + 5 + 5
\n2y + 1 = 15<\/p>\n

Let’s Think (Page No: 47)<\/span><\/p>\n

An equation is multiplied or divided by two different numbers on either side What will happen to the equality?
\nAnswer:
\nConsider an equation 7x = 14
\nIf we multiply by two different numbers 5 and 8 on either side.
\n7x \u00d7 5 = 14 \u00d7 8 \u21d2 35x = 112
\nSo, the equation 7x = 14 is not equal to 35x = 112<\/p>\n

(ii) Consider an equation 3x = 15
\nIf we divide by two different numbers 3 and 5 on either side,
\n3x \u00f7 3 = 15 \u00f7 5
\n\\(\\frac{3 x}{3}\\) = \\(\\frac{15}{5}\\) \u21d2 x = 3
\nSo, the equation 3x = 15 is not equal to x = 3 either side, the equality will be changes.<\/p>\n

Puzzle Time (Page No: 48)<\/span><\/p>\n

Question 1.
\nSolve the puzzle using simple equations.
\n(i)
\n\"AP
\nAnswer:
\n\"AP<\/p>\n

\"AP<\/p>\n

(ii)
\n\"AP
\nAnswer:
\n\"AP<\/p>\n

(iii)
\n\"AP
\nAnswer:
\n\"AP<\/p>\n

(iv)
\n\"AP
\nAnswer:
\n\"AP<\/p>\n

\"AP<\/p>\n

Check Your Progress (Page No: 51)<\/span><\/p>\n

Find three equivalent equations having the same solutions x = – 1.
\nAnswer:
\n\"AP
\nHere x + 5 = 4, 8 – x = 9 and 4x + 5 = 1 are equivalent equations because they have the same solution x = – 1<\/p>\n

Puzzle Time (Page No. 52)<\/span><\/p>\n

Question 1.
\nHere is a puzzle to reveal the name of great mathematician. Solve the equations and fill the below box with the letter opposite to the solution of the equation and finally it reveals the name of mathematician.
\n\"AP
\nAnswer:
\n\"AP
\n\"AP<\/p>\n

\"AP<\/p>\n

Check Your Progress (Page No. 53)<\/span><\/p>\n

Find the value of variable x from the below diagrams.<\/p>\n

(i)
\n\"AP
\nAngle at ‘O’ is straight angle
\nAnswer:
\nStraight angle = 180\u00b0
\n\u2220 AOB = 180\u00b0
\n\u2220 AOB + \u2220 BOD = 180\u00b0
\nx + 30\u00b0 + x = 180\u00b0
\n\u21d2 2x + 30\u00b0 = 180\u00b0
\n\u21d2 2x + 30\u00b0 – 30\u00b0 – 180\u00b0- 30\u00b0 (Subtract 30\u00b0 on both sides)
\n\u21d2 2x = 150\u00b0
\n\u21d2 \\(\\frac{2 x}{2}\\) = \\(\\frac{150^{\\circ}}{2}\\) (Divide by 2 on both sides)
\n\u2234 x = 75\u00b0<\/p>\n

(ii)
\n\"AP
\nPerimeter = 36 cm
\nAnswer:
\nPerimeter of rectangle = 36 cm
\n2(length + breadth) = 36
\n\u21d2 2(x + 4 + x) = 36
\n\u21d2 2(2x + 4) = 36
\n\u21d2 \\(\\frac{2(2 x+4)}{2}\\) = \\(\\frac{36}{2}\\) (Divide by 2 on both sides)
\n\u21d2 2x + 4 = 18
\n\u21d2 2x + 4 – 4 = 18 – 4 (Subtract 4 on both sides)
\n\u21d2 2x = 14 .
\n\u21d2 \\(\\frac{2x}{2}\\) = \\(\\frac{14}{2}\\) (Divide by 2 on both sides)
\n\u2234 x = 7<\/p>\n

\"AP<\/p>\n

Puzzle Time (Page No: 59)<\/span><\/p>\n

Collect the minimum and maximum temperature (in degrees) of any 5 cities of India of any month and convert the temperature into Fahrenheit and Kelvin Scale.<\/p>\n

\"AP<\/p>\n

\"AP
\n\"AP<\/p>\n

\"AP<\/p>\n

Examples<\/span><\/p>\n

Question 1.
\nConvert the mathematical statements into simple equations.
\n(i) 5 added to a number is 9.
\nAnswer:
\nLet the number = x
\nBy adding 5 to the number = 5 + x
\n\u2234 5 + x = 9<\/p>\n

(ii) 4 times a number decreased by 3 is 5.
\nAnswer:
\nLet the number = m
\n4 times the number = 4m
\nBy decreasing 3 the result = 4m – 3
\n\u2234 4m – 3 = 5<\/p>\n

(iii) The sum of 3 times of n and 7 is 13.
\nAnswer:
\nLet the number = n
\n3 times the number = 3n
\nBy adding 7 the result = 3n + 7
\n\u2234 3n + 7 = 13<\/p>\n

(iv) Length of rectangle is 3 m more than its breadth and its perimeter is 24 m.
\nAnswer:
\nBreadth of rectangle = x
\nLength of rectangle = x + 3
\nPerimeter = 2(x + 3 + x) = 4x + 6
\nGiven Perimeter = 24 m
\n\u2234 4x + 6 = 24<\/p>\n

Question 2.
\nConvert simple equations into statements.
\n(i) y – 7 = 11
\nAnswer:
\nTaking away 7 from \u2018y\u2019 is 11.<\/p>\n

(ii) 8m = 24
\nAnswer:
\n8 times of number \u2018m\u2019 is 24.<\/p>\n

(iii) 2x + 13 = 25
\nAnswer:
\nIf you add 13 to 2 times of number \u2018x\u2019 is 25.<\/p>\n

(iv) \\(\\frac{y}{4}\\) – 7 = 1
\nAnswer:
\n7 is subtracted from one fourth of y is 1.<\/p>\n

\"AP<\/p>\n

Question 3.
\nCheck whether the value given in the brackets is a solution to the given equation or not.
\n2k – 11 = 5 (k = 7)
\nAnswer:
\nWhen k = 7
\nLHS: 2k – 11 = 2(7) – 11 = 14 – 11 = 3
\nRHS: 5
\nHere LHS \u2260 RHS,
\nSo k = 7 is not a solution.<\/p>\n

Question 4.
\nSolve 6n – 1 = 29 trial and error method.
\nAnswer:
\n\"AP
\nFor n = 5, LHS = RHS. So, n = 5 is the solution of given equation.<\/p>\n

Question 5.
\nSolve x + 5 = 8 by using common balance idea.
\nAnswer:
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 6.
\nSolve 2x – 5 = 9 pictorially.
\nAnswer:
\n\"AP<\/p>\n

Question 7.
\nSolve: 3k + 4 = 28
\nAnswer:
\n3k + 4 = 28
\n\u21d2 3k + 4 – 4 = 28 – 4 (Add both sides \u2018- 4\u2019)
\n\u21d2 3k = 24
\n\u21d2 \\(\\frac{3 \\mathrm{k}}{3}\\) = \\(\\frac{24}{3}\\) (Divide both sides by \u20183\u2019)
\n\u21d2 k = 8<\/p>\n

Check: Substitute k = 8 in the given equation.
\nLHS = 3k + 4
\n= 3(8) + 4
\n= 24 + 4 = 28 = RHS
\nHence verified.<\/p>\n

Question 8.
\nSolve: – 4(x – 1) = 16
\nAnswer:
\n– 4(x – 1) = 16
\n\u21d2 4x + 4 = 16 (Distributive property)
\n\u21d2 – 4x + 4 – 4 = 16 – 4 (Subtract both sides \u20184\u2019)
\n\u21d2 – 4x = 12
\n\u21d2 (- 4x) \u00d7 (- 1) = 12 \u00d7 (- 1)
\n(Multiply with \u2018- 1 \u2018on both sides)
\n\u21d2 4x = – 12
\n\u21d2 \\(\\frac{4 x}{4}\\) = \\(\\frac{-12}{4}\\) (Divide both sides by \u20184\u2019)
\n\u21d2 x = – 3<\/p>\n

Check: Substitute x = – 3 in the given equation.
\nLHS = – 4(x – 1)
\n= – 4(- 3 – 1)
\n= – 4 (- 4)
\n= 16
\n= RHS
\nHence verified.<\/p>\n

\"AP<\/p>\n

Question 9.
\nSolve: 2(b + 3) + 13 = 27
\nAnswer:
\n2(b + 3) + 13 = 27
\n\u21d2 2b + 6 + 13 = 27 (distributive law)
\n\u21d2 2b + 19 = 27
\n\u21d2 2b = 27 – 19 (\u2235+ 19 transposed and becomes – 19)
\n\u21d2 2b = 8
\n\u21d2 b = \\(\\frac{8}{2}\\) (\u2235 \u00d7 2 transposed and becomes \u00f7 2)
\n\u21d2 b = 4<\/p>\n

Check:
\nSubstitute b = 4
\nLHS = 2(b + 3) + 13
\n= 2(4 + 3) + 13
\n= 2(7) + 13
\n= 14 + 13
\n= 27 = RHS
\nHence verified.<\/p>\n

Question 10.
\nSolve : 5(x + 1) – 2(x – 7) = 13
\nAnswer:
\n5(x + 1) – 2(x – 7) = 13
\n\u21d2 5x + 5 – 2x + 14 = 13 (distributive law)
\n\u21d2 (5x – 2x) + (5 + 14) = 13 (regrouping like terms)
\n\u21d2 3x +19 = 13
\n\u21d2 3x = 13 – 19 (\u2235 +19 transposed arid becomes
\n\u21d2 3x = – 6
\n\u21d2 x = \\(\\frac{-6}{3}\\) (\u2235 \u00d7 3 transposed and becomes \u00f7 3)
\n\u21d2 x = – 2<\/p>\n

Question 11.
\nSolve: 12 = 13 + 7 (y-6)
\nAnswer:
\n12 = 13 + 7 (y – 6)
\nAn equation remains the same,
\nwhen LHS, RHS are interchanged.
\n\u21d2 13 + 7(y – 6) = 12 .
\n\u21d2 3 + 7y – 42 = 12 (distributive law)
\n\u21d2 7y – 29 = 12
\n\u21d2 7y = 12 + 29 (\u2235 – 29 transposed and becomes + 29)
\n\u21d2 7y = 41
\n\u21d2 y = \\(\\frac{41}{7}\\) (\u2235 \u00d7 7 transposed and becomes \u00f7 7)<\/p>\n

Check:
\nSubstitute y = \\(\\frac{41}{7}\\)
\nRHS = 13 + 7(y – 6)
\n= 13 + 7(\\(\\frac{41}{7}\\) – 6)
\n= 13 + 7\\(\\left(\\frac{41-42}{7}\\right)\\)
\n= 13 + 7\\(\\left(\\frac{-1}{7}\\right)\\)
\n= 13 – 1
\n= 12 = LHS
\nHence verified.<\/p>\n

\"AP<\/p>\n

Question 12.
\nSolve \\(\\frac{m}{2}-\\frac{2 m}{7}-\\frac{m}{10}=\\frac{8}{5}\\)
\nAnswer:
\n\"AP<\/p>\n

Check:
\nSubstitute m = 14
\nLHS = \\(\\frac{\\mathrm{m}}{2}-\\frac{2 \\mathrm{~m}}{7}-\\frac{\\mathrm{m}}{10}\\)
\n= 7 – 4 – \\(\\frac{7}{5}\\)
\n= 3 – \\(\\frac{7}{5}\\)
\n= \\(\\frac{15-7}{5}\\)
\n= \\(\\frac{8}{5}\\)
\n= RHS
\nHence verified.<\/p>\n

Question 13.
\nFind the value of variable \u2018x\u2019 from the following.
\n(i) x + 20% of x = 60
\nAnswer:
\nx + 20% of x = 60
\n\u21d2 x + \\(\\frac{20 x}{100}\\) = 60
\n\u21d2 x + \\(\\frac{x}{5}\\) = 60
\n\u21d2 \\(\\frac{6x}{5}\\) = 60
\n\u21d2 6x = 60 \u00d7 5
\n\u21d2 x = \\(\\frac{(60 \\times 5)}{6}\\)
\n\u21d2 x = 50<\/p>\n

(ii) If two numbers are in the ratio 2 : 3 and their difference is 5, then find largest number.
\nAnswer:
\nLet largest number = 3x and smallest number = 2x
\nDifference = 3x – 2x = x
\nBut as per the problem difference = 5
\n\u21d2 x = 5
\n\u2234 Largest number = 3x = 3 \u00d7 5 = 15<\/p>\n

(iii) Find the value of x in the following picture.
\n\"AP
\nAnswer:
\nSum of two angles is 90
\n2x + 25 + x – 10 = 90
\n\u21d2 3x + 15 = 90
\n\u21d2 3x = 90 – 15
\n\u21d2 3x = 75
\n\u21d2 x = \\(\\frac{75}{3}\\)
\n\u21d2 x = 25<\/p>\n

\"AP<\/p>\n

(iv) If the perimeter of the following tri-angle is 15 cm, then find the value of x.
\n\"AP
\nAnswer:
\nPerimeter of a triangle = 15 cm
\nPerimeter = x + x + 2 + 2x – 3 = 4x – 1
\n\u2234 4x – 1 = 15
\n\u21d2 4x = 15 + 1
\n\u21d2 4x = 16
\n\u21d2 x = 4<\/p>\n

Question 14.
\nFind two consecutive natural numbers whose sum is 125.
\nAnswer:
\nLet the number = x
\nConsecutive number of x = x + 1
\nGiven that, x + (x + 1) = 125
\n\u21d2 2x + 1 = 125
\n\u21d2 2x = 125 – 1
\n\u21d2 2x = 124
\n\u21d2 x = \\(\\frac{124}{2}\\)
\nx = 62 and x + 1 = 62 + 1 = 63
\n\u2234 Required numbers are 62 and 63.<\/p>\n

Check:
\nNumbers: 62, 63
\nSum = 62 + 63 = 125
\nHence verified.<\/p>\n

Question 15.
\nThe sum of two numbers is 35. One of the numbers exceeds the other by 7. Find the numbers.
\nAnswer:
\nLet the first number = x
\nThen the second number = x + 7 (exceeds first number by 7)
\nSum of two numbers = 35
\nAccording to problem, x + x + 7 = 35
\n\u21d2 2x + 7 = 35
\n\u21d2 2x = 35 – 7
\n\u21d2 2x = 28
\n\u21d2 x = \\(\\frac{28}{2}\\)
\n\u21d2 x = 14
\nx + 7 = 14 + 7 = 21
\n\u2234 The two numbers are 14 and 21<\/p>\n

Check:
\nNumbers: 14, 21
\nSum = 14 + 21
\n= 35
\nHence verified.<\/p>\n

\"AP<\/p>\n

Question 16.
\nA person has \u20b9 1400 in denominations of \u20b920, \u20b910 and \u20b95 notes. The number of notes in each denomination is equal, then find the number of notes of each denomination ?
\n\"AP
\nAnswer:
\nLet the number of notes in each denomination be x.
\n\u20b920 denomination amount = 20x
\n\u20b910 denomination amount = 10x
\n\u20b95 denomination amount = 5x
\nTotal amount = \u20b9 1400
\n\u21d2 20x + 10x + 5x = 1400
\n\u21d2 35x = 1400
\n\u21d2 x = \\(\\frac{1400}{35}\\)
\n\u21d2 x = 40
\n\u2234 Number of notes in each denomination is 40.<\/p>\n

Check:
\nNumber of notes = 40
\nAmount = (20 \u00d7 40 + 10 \u00d7 40 + 5 \u00d7 40)
\n= 800 + 400 + 200
\n= \u20b9 1400
\nHence Verified.<\/p>\n

Question 17.
\nThe length of a rectangle is five meters more than twice of its breadth. If the perimeter is 148 meter, then find the length and breadth of the rectangle.
\n\"AP
\nAnswer:
\nLet the breadth of the rectangle = x,
\nlength of a rectangle is five more than twice of its breadth.
\nThen, length of the rectangle = 2x + 5
\nPerimeter of the rectangle = 148m
\n2 (Length + Breadth) = 148
\n\u21d2 2(2x + 5 + x) = 148
\n\u21d2 2(3x + 5) = 148
\n\u21d2 6x + 10 = 148
\n\u21d2 6x = 148 – 10
\n\u21d2 6x = 138
\n\u21d2 x = \\(\\frac{138}{6}\\)
\n\u21d2 x = 23
\n\u2234 Breadth of rectangle = 23 m
\nLength of the rectangle = 2x + 5 = (2 \u00d7 23) + 5
\n= 46 + 5 = 51 m<\/p>\n

Check:
\nLength = 51 m, Breadth = 23 m
\nPerimeter = 2(l + b)
\n= 2(51 + 23)
\n= 2(74)
\n= 148 m
\nHence Verified.<\/p>\n

\"AP<\/p>\n

Question 18.
\nOn his birthday Yakshith\u2019s grand father has given \u20b9 2000. He used some amount for purchasing books for needy children and thrice of that amount for purchasing food items for orphanage children and the remaining \u20b9 200 used for purchasing chocolates . for his friends. Find the amount spent for purchasing books and food items for orphanage children.
\nAnswer:
\nLet the amount spent for purchasing books for needy children = \u20b9 x
\nAmount spent for purchasing food items for orphanage children = \u20b9 3x
\nAmount spent for purchasing chocolates for friends = \u20b9 200
\nTotal amount spent = \u20b9 2000<\/p>\n

\u21d2 x + 3x + 200 = 2000
\n\u21d2 4x + 200 = 2000
\n\u21d2 4x = 2000 – 200
\n\u21d2 4x = 1800
\n\u21d2 x = \\(\\frac{1800}{4}\\)
\n\u21d2 x = 450
\n\u2234 Amount spent for purchasing books for needy children = \u20b9 450
\nAmount spent for purchasing food items for orphanage children = \u20b9 3x
\n= 3 \u00d7 450 = \u20b9 1350<\/p>\n

Check:
\nAmount = \u20b9 450 + \u20b9 1350 + \u20b9 200
\n= \u20b9 2000
\nHence Verified.<\/p>\n

Question 19.
\nA school bus starts with full strength of 40 students. It drops some students at the first bus stop. At the second bus stop, twice the number of students get down from the bus. 8 students get down at the third bus stop and the number of students remaining in the bus is only 5. How many students got down at the first stop and second stop?
\n\"AP
\nAnswer:
\nLet us take the number of students get down at first stop = x
\nNo.of students get down at the second bus stop = 2x
\nNo. of students get down at the third bus stop = 8
\nRemaining students in the bus = 5
\n\u21d2 x + 2x +8 + 5 = 40
\n\u21d2 3x + 13 = 40
\n\u21d2 3x = 40 – 13
\n\u21d2 3x = 27
\n\u21d2 x = \\(\\frac{27}{3}\\)
\n\u21d2 x = 9
\n\u2234 The number of students got down in the first bus stop = 9.
\nNo. of students got down in second stop = 2x = 2 \u00d7 9 = 18.<\/p>\n

\"AP<\/p>\n

Practice Questions (Page No: 62)<\/span><\/p>\n

In the questions given below, equations have become wrong due to wrong order of signs (+, -, \u00d7, \u00f7, =). Choose the correct order of signs from the alternatives given under each question so that the equation becomes right.<\/p>\n

Question 1.
\n7 + 2 = 2 \u00d7 3
\n(a) = \u00d7 +
\n(b) = + \u00d7
\n(c) = + \u00f7
\n(d) + \u00d7 =
\nAnswer:
\n(a) = \u00d7 +<\/p>\n

Explaination
\nGiven 7 + 2 = 2 \u00d7 3 ,
\n\u21d2 7 = 2 \u00d7 2 + 3
\n\u21d2 7 = 4 + 3
\n\u2234 7 = 7<\/p>\n

Question 2.
\n7 + 2 \u00d7 6 = 20
\n(a) = \u00d7 +
\n(b) \u00d7 – =
\n(c) \u00d7 + =
\n(d) \u00f7 + =
\nAnswer:
\n(c) \u00d7 + =<\/p>\n

Explaination:
\nGiven 7 + 2 \u00d7 6 = 20
\n\u21d2 7 \u00d7 2 + 6 = 20
\n\u21d2 14 + 6 = 20
\n\u2234 20 = 20<\/p>\n

\"AP<\/p>\n

Question 3.
\n15 \u00f7 5 = 2 \u00d7 1
\n(a) \u00f7 \u00d7 =
\n(b) \u00f7 = \u00d7
\n(c) \u00d7 = +
\n(d) \u00f7 = +
\nAnswer:
\n(d) \u00f7 = +<\/p>\n

Explaination:
\nGiven 15 \u00f7 5 = 2 \u00d7 1
\n\u21d2 15 \u00f7 5 = 2 + 1
\n\u2234 3 = 3<\/p>\n

Question 4.
\n6 = 3 – 6 \u00f7 12
\n(a) = \u00d7 +
\n(b) \u00f7 = \u00d7
\n(c) + = –
\n(d) \u00f7 \u00d7 =
\nSol.
\n(d) \u00f7 \u00d7 =<\/p>\n

Explaination:
\nGiven 6 = 3 – 6 \u00f7 12
\n\u21d2 6 \u00f7 3 \u00d7 6 =12
\n\u21d2 2 \u00d7 6 = 12
\n\u2234 12 = 12<\/p>\n

Question 5.
\n3 + 1 \u00f7 4 = 16
\n(a) – = \u00d7
\n(b) \u00d7 + =
\n(c) + \u00d7 =
\n(d) = \u00d7 +
\nAnswer:
\n(c) + \u00d7 =<\/p>\n

Explaination:
\nGiven 3 + 1 \u00f7 4 = 16
\n\u21d23 + 1 \u00d7 4 = 16
\n\u21d2 4 \u00d7 4 = 16
\n\u2234 16 = 16<\/p>\n

\"AP<\/p>\n

Question 6.
\n8 \u00f7 4 = 2 + 1
\n(a) \u00f7 = +
\n(b) \u00f7 = \u00d7
\n(c) \u00f7 \u00d7 =
\n(d) = \u00f7 \u00d7
\nAnswer:
\n(b) \u00f7 = \u00d7<\/p>\n

Explaination:
\nGiven 8 \u00f7 4 = 2 + 1
\n\u21d2 8 \u00f7 4 = 2 \u00d7 1
\n\u2234 2 = 2<\/p>\n

Question 7.
\n2 \u00d7 2 + 2 = 2
\n(a) \u00d7 \u00f7 =
\n(b) \u00d7 = \u00f7
\n(c) + \u00d7 =
\n(d) \u00d7 + =
\nAnswer:
\n(a) \u00d7 \u00f7 =<\/p>\n

Explaination:
\nGiven 2 \u00d7 2 + 2 = 2
\n\u21d2 2 \u00d7 2 \u00f7 2 = 2
\n\u21d2 2 \u00d7 1 = 2
\n\u2234 2 = 2<\/p>\n

Question 8.
\n5 – 6 + 8 = 3
\n(a) + – =
\n(b) + = –
\n(c) – = \u00d7
\n(d) \u00f7 \u00d7 =
\nAnswer:
\n(a) + – =<\/p>\n

Explaination:
\nGiven 5 – 6 + 8 = 3
\n\u21d2 5 + 6 – 8 = 3
\n\u21d2 11 – 8 = 3
\n\u2234 3 = 3<\/p>\n

\"AP<\/p>\n

Question 9.
\n8 \u00f7 2 = 2 \u00d7 8
\n(a) + – =
\n(b) + \u00d7
\n(c) \u00f7 = \u00d7
\n(d) \u00d7 = \u00f7
\nAnswer:
\n(b) + \u00d7<\/p>\n

Explaination:
\nGiven 8 \u00f7 2 = 2 \u00d7 8
\n\u21d2 8 \u00f7 2 \u00d7 2 = 8
\n\u21d2 4 \u00d7 2 – 8
\n\u22348 = 8<\/p>\n

Question 10.
\n3 = 3 – 7 + 0
\n(a) – + =
\n(b) + \u00d7 =
\n(c) – \u00d7 =
\n(d) = \u00d7 –
\nAnswer:
\n(c) – \u00d7 =<\/p>\n

Explaination:
\nGiven 3 = 3 – 7 + 0
\n\u21d2 3 – 3 \u00d7 7 = 0
\n\u21d2 0 \u00d7 7 = 0
\n\u2234 0 = 0<\/p>\n","protected":false},"excerpt":{"rendered":"

SCERT\u00a0AP 7th Class Maths Solutions Pdf Chapter 3 Simple Equations InText Questions and Answers. AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations InText Questions Check your Progress (Page No: 43) Question 1. Write simple equations for the following verbal statements. (i) The sum of five times of x and 3 is 28. …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[9],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/21692"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=21692"}],"version-history":[{"count":6,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/21692\/revisions"}],"predecessor-version":[{"id":22592,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/21692\/revisions\/22592"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=21692"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=21692"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=21692"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}