GSEB Class 7 Maths Solutions Chapter 4 સાદા સમીકરણ Exercise 4.1

 

Question 1. આપેલ કોષ્ટકનું છેલ્લું ખાનું પૂર્ણ કરોઃ
Answer:

In simple words: To check if an equation is right, put the given number into the left side. If that side then matches the right side, the equation is solved. If they don't match, it's not solved.

Exam Tip: Always replace the variable with the given value on the left side of the equation. If the result matches the right side, the equation is satisfied; otherwise, it is not.

 

Question 2. કૌસમાં આપેલી કિંમતો આપેલાં સમીકરણનો ઉકેલ છે કે નહીં તે તપાસોઃ
(a) \( n + 5 = 19 \) (n = 1)
Answer: Here, for the equation \( n + 5 = 19 \) and \( n = 1 \), the Left Hand Side (LHS) is \( n + 5 \). When we substitute \( n = 1 \), we get \( 1 + 5 = 6 \). This value \( 6 \) is not equal to the Right Hand Side (RHS), which is \( 19 \). Therefore, \( n = 1 \) is not a solution for the given equation.
In simple words: Put \( n = 1 \) into the left part of \( n + 5 = 19 \). You get \( 1 + 5 = 6 \). Since \( 6 \) is not \( 19 \), \( n = 1 \) doesn't solve the equation.

Exam Tip: Always check if the value satisfies the equation by substituting it into the LHS and comparing with the RHS. If they are equal, the value is a solution.

 

Question 2. (b) \( 7n + 5 = 19 \) (n = -2)
Answer: For the equation \( 7n + 5 = 19 \) with \( n = -2 \), the Left Hand Side (LHS) is \( 7n + 5 \). Upon substituting \( n = -2 \), we calculate \( 7(-2) + 5 = -14 + 5 = -9 \). Since this result \( -9 \) is not equal to the Right Hand Side (RHS) of \( 19 \), it means \( n = -2 \) is not a solution for the given equation.
In simple words: When \( n = -2 \) is put into \( 7n + 5 \), it becomes \( 7 \times (-2) + 5 \), which is \( -14 + 5 = -9 \). As \( -9 \) doesn't match \( 19 \), \( n = -2 \) isn't the correct answer.

Exam Tip: Remember to apply the rules of integer multiplication (positive times negative equals negative) when substituting negative values for variables.

 

Question 2. (c) \( 7n + 5 = 19 \) (n = 2)
Answer: For the equation \( 7n + 5 = 19 \) with \( n = 2 \), the Left Hand Side (LHS) is \( 7n + 5 \). When we substitute \( n = 2 \), we compute \( 7(2) + 5 = 14 + 5 = 19 \). This result \( 19 \) is equal to the Right Hand Side (RHS), which is also \( 19 \). Therefore, \( n = 2 \) is a solution for the given equation.
In simple words: If we put \( n = 2 \) into \( 7n + 5 \), it becomes \( 7 \times 2 + 5 \), which is \( 14 + 5 = 19 \). Since \( 19 \) matches the right side, \( n = 2 \) is indeed the answer.

Exam Tip: Showing all steps of substitution and calculation clearly helps avoid errors and demonstrates understanding of the process.

 

Question 2. (d) \( 4p - 3 = 13 \) (p = 1)
Answer: For the equation \( 4p - 3 = 13 \) with \( p = 1 \), the Left Hand Side (LHS) is \( 4p - 3 \). Substituting \( p = 1 \), we get \( 4(1) - 3 = 4 - 3 = 1 \). This value \( 1 \) is not equal to the Right Hand Side (RHS), which is \( 13 \). Therefore, \( p = 1 \) is not a solution for the given equation.
In simple words: When \( p = 1 \) is placed in \( 4p - 3 \), it turns into \( 4 \times 1 - 3 \), which equals \( 1 \). Because \( 1 \) is not \( 13 \), \( p = 1 \) isn't the correct solution.

Exam Tip: Be careful with subtraction after multiplication. Ensure the order of operations is followed correctly (PEMDAS/BODMAS).

 

Question 2. (e) \( 4p - 3 = 13 \) (p = -4)
Answer: For the equation \( 4p - 3 = 13 \) with \( p = -4 \), the Left Hand Side (LHS) is \( 4p - 3 \). Upon substituting \( p = -4 \), we calculate \( 4(-4) - 3 = -16 - 3 = -19 \). This result \( -19 \) is not equal to the Right Hand Side (RHS), which is \( 13 \). Therefore, \( p = -4 \) is not a solution for the given equation.
In simple words: If we put \( p = -4 \) into \( 4p - 3 \), it becomes \( 4 \times (-4) - 3 \). This works out to \( -16 - 3 = -19 \). Since \( -19 \) is not \( 13 \), \( p = -4 \) doesn't solve it.

Exam Tip: Pay close attention to signs when multiplying and subtracting negative numbers. A common mistake is mismanaging double negatives.

 

Question 2. (f) \( 4p - 3 = 13 \) (p = 0)
Answer: For the equation \( 4p - 3 = 13 \) with \( p = 0 \), the Left Hand Side (LHS) is \( 4p - 3 \). When we substitute \( p = 0 \), we get \( 4(0) - 3 = 0 - 3 = -3 \). This value \( -3 \) is not equal to the Right Hand Side (RHS), which is \( 13 \). Therefore, \( p = 0 \) is not a solution for the given equation.
In simple words: Putting \( p = 0 \) into \( 4p - 3 \) makes it \( 4 \times 0 - 3 \), which is \( 0 - 3 = -3 \). Because \( -3 \) doesn't match \( 13 \), \( p = 0 \) is not the correct solution.

Exam Tip: Remember that any number multiplied by zero is zero. This simplifies calculations when a variable is zero.

 

Question 3. નીચેનાં સમીકરણો પ્રયત્ન અને ભૂલની રીતથી ઉકેલો:
(i) \( 5p + 2 = 17 \)
Answer: To solve \( 5p + 2 = 17 \) using the trial and error method, we substitute different integer values for \( p \):

• For \( p = 0 \), LHS = \( 5(0) + 2 = 0 + 2 = 2 \), which is not equal to 17.
• For \( p = 1 \), LHS = \( 5(1) + 2 = 5 + 2 = 7 \), which is not equal to 17.
• For \( p = 2 \), LHS = \( 5(2) + 2 = 10 + 2 = 12 \), which is not equal to 17.
• For \( p = 3 \), LHS = \( 5(3) + 2 = 15 + 2 = 17 \), which is equal to 17 (RHS).

Thus, the equation is satisfied when \( p = 3 \). Therefore, \( p = 3 \) is the solution for \( 5p + 2 = 17 \).
In simple words: We try different numbers for \( p \). When \( p = 3 \), \( 5 \times 3 + 2 \) gives \( 15 + 2 \), which is \( 17 \). This matches the other side of the equation, so \( p = 3 \) is the answer.

Exam Tip: When using trial and error, start with small, simple integer values and gradually increase or decrease them based on whether your LHS is too small or too large.

 

Question 3. (ii) \( 3m - 14 = 4 \)
Answer: To solve \( 3m - 14 = 4 \) by the trial and error method, we test different integer values for \( m \):

• For \( m = 1 \), LHS = \( 3(1) - 14 = 3 - 14 = -11 \), which is not equal to 4.
• For \( m = 2 \), LHS = \( 3(2) - 14 = 6 - 14 = -8 \), which is not equal to 4.
• For \( m = 3 \), LHS = \( 3(3) - 14 = 9 - 14 = -5 \), which is not equal to 4.
• For \( m = 4 \), LHS = \( 3(4) - 14 = 12 - 14 = -2 \), which is not equal to 4.
• For \( m = 5 \), LHS = \( 3(5) - 14 = 15 - 14 = 1 \), which is not equal to 4.
• For \( m = 6 \), LHS = \( 3(6) - 14 = 18 - 14 = 4 \), which is equal to 4 (RHS).

Thus, the equation is satisfied when \( m = 6 \). Therefore, \( m = 6 \) is the solution for \( 3m - 14 = 4 \).
In simple words: We try numbers for \( m \). When \( m = 6 \), \( 3 \times 6 - 14 \) becomes \( 18 - 14 \), which is \( 4 \). This matches the equation's right side, so \( m = 6 \) is the answer.

Exam Tip: Organize your trials clearly, showing each substitution and the resulting LHS value. This helps track your progress and pinpoint the correct solution.

 

Question 4. નીચે આપેલાં વિધાનોને સમીકરણ સ્વરૂપે લખોઃ
(i) x અને 4નો સરવાળો 9 છે.
(ii) yમાંથી 2 બાદ કરતાં 8 મળે.
(iii) aના 10 ગણા 70 છે.
(iv) એક સંખ્યા bને 5 વડે ભાગતાં 6 મળે.
(v) tનો \( \frac{3}{4} \) ભાગ 15 છે.
(vi) mના 7 ગણામાં 7 ઉમેરતાં 77 મળે.
(vii) કોઈ સંખ્યા ના એક ચતુર્થાંશમાંથી 4 બાદ કરતાં 4 મળે.
(viii) yના 6 ગણામાંથી 6 બાદ કરતાં 60 મળે છે.
(ix) જો તમે zના ત્રીજા ભાગમાં ૩ ઉમેરો, તો તમને 30 મળે છે.
Answer:
(i) \( x + 4 = 9 \)
(ii) \( y - 2 = 8 \)
(iii) \( 10a = 70 \)
(iv) \( \frac{b}{5} = 6 \)
(v) \( \frac{3t}{4} = 15 \)
(vi) \( 7m + 7 = 77 \)
(vii) \( \frac{x}{4} - 4 = 4 \)
(viii) \( 6y - 6 = 60 \)
(ix) \( \frac{z}{3} + 3 = 30 \)
In simple words: Just write down what each sentence says using numbers, letters, and math signs.

Exam Tip: Clearly identify the unknown quantity (variable) and translate keywords like "sum," "difference," "times," "divided by," "more than," or "less than" into their corresponding mathematical operations.

 

Question 5. નીચે આપેલાં સમીકરણોને વિધાનના સ્વરૂપે લખો :
(i) \( p + 4 = 15 \)
(ii) \( m - 7 = 3 \)
(iii) \( 2m = 7 \)
(iv) \( \frac{m}{5} = 3 \)
(v) \( \frac{3m}{5} = 6 \)
(vi) \( 3p + 4 = 25 \)
(vii) \( 4p - 2 = 18 \)
(viii) \( \frac{p}{2} + 2 = 8 \)
Answer:
(i) pમાં 4 ઉમેરતાં 15 મળે છે. (Adding 4 to p gives 15.)
(ii) mમાંથી 7 બાદ કરતાં 3 મળે છે. (Subtracting 7 from m gives 3.)
(iii) mના 2 ગણા 7 થાય છે. (2 times m is 7.)
(iv) mનો પાંચમો ભાગ 3 છે. (One-fifth of m is 3.)
(v) mના ત્રણ ગણાને 5 વડે ભાગતાં 6 મળે છે. (Three times m divided by 5 gives 6.)
(vi) pના ત્રણ ગણામાં 4 ઉમેરતાં 25 મળે છે. (Adding 4 to three times p gives 25.)
(vii) pના ચાર ગણામાંથી 2 બાદ કરતાં 18 મળે છે. (Subtracting 2 from four times p gives 18.)
(viii) pના અડધા ભાગમાં 2 ઉમેરતાં 8 મળે છે. (Adding 2 to half of p gives 8.)
In simple words: Turn the math equations back into regular sentences that say what happens to the number.

Exam Tip: Practice translating common mathematical operations into clear, concise verbal statements. For example, "multiplied by" can often be simplified to "times" or "[number] of."

 

Question 6. (i) ઇરફાને કહ્યું કે તેની પાસે પરમિત પાસેની લખોટીના 5 ગણા કરતાં 7 લખોટી વધુ છે. ઇરફાન પાસે 37 લખોટી છે. (પરમિત પાસેની લખોટીની સંખ્યા માટે m ધારો.)
Answer: Let \( m \) be the number of marbles Parmit has. According to the problem, Irfan states that he possesses 7 more marbles than 5 times the number of marbles Parmit has. This means Irfan's marbles can be expressed as \( 5m + 7 \). We are also told that Irfan has a total of 37 marbles. Therefore, we can set up the equation: \( 5m + 7 = 37 \).
In simple words: Let \( m \) be how many marbles Parmit has. Irfan says he has 5 times Parmit's marbles, plus 7 more, which is \( 5m + 7 \). Since Irfan actually has 37 marbles, the equation is \( 5m + 7 = 37 \).

Exam Tip: Clearly define your variable first. Then break down the word problem into smaller phrases that can be translated into mathematical expressions, building up to the full equation.

 

Question 6. (ii) લક્ષ્મીના પિતા 49 વર્ષના છે. તે લક્ષ્મીની ઉંમરના ત્રણ ગણાથી 4 વર્ષ મોટા છે. (લક્ષ્મીની ઉંમર માટે y ધારો.)
Answer: Let \( y \) represent Lakshmi's age in years. The problem states that Lakshmi's father is 4 years older than three times Lakshmi's age, which can be expressed as \( 3y + 4 \). We also know that Lakshmi's father's actual age is 49 years. Therefore, the equation representing this situation is: \( 3y + 4 = 49 \).
In simple words: If Lakshmi's age is \( y \), her father's age is 3 times \( y \) plus 4, so \( 3y + 4 \). Since her father is 49 years old, the equation is \( 3y + 4 = 49 \).

Exam Tip: Be careful with phrases like "older than" or "more than"; they imply addition. "Times" or "gull" means multiplication.

 

Question 6. (iii) શિક્ષકે વર્ગમાં કહ્યું કે સૌથી વધારે ગુણ મેળવનાર વિદ્યાર્થીના ગુણ વર્ગના સૌથી ઓછા ગુણ મેળવનાર વિદ્યાર્થીના ગુણના બે ગણાથી 7 વધારે છે. સૌથી વધારે ગુણ 87 છે. (સૌથી ઓછા ગુણ માટે l ધારો.)
Answer: Let \( l \) denote the lowest score achieved in the class. The teacher explained that the highest score is 7 more than twice the lowest score, which can be written as \( 2l + 7 \). We are also provided that the highest score is 87. Therefore, the equation representing this situation is: \( 2l + 7 = 87 \).
In simple words: Let \( l \) be the lowest score. The highest score is two times \( l \) plus 7, which is \( 2l + 7 \). Since the highest score is 87, the equation is \( 2l + 7 = 87 \).

Exam Tip: Differentiate between "more than" (addition) and "times" (multiplication) to correctly form expressions. The word "is" often signifies equality.

 

Question 6. (iv) એક સમદ્વિબાજુ ત્રિકોણમાં શિરોબિંદુકોણ એ આધારકોણ કરતાં બે ગણો છે. (આધારકોણનું માપ b ધારો. યાદ રાખો કે ત્રિકોણના ખૂણાઓનાં. માપનો સરવાળો 180 અંશ છે.)
Answer: Let \( b \) represent the measure of each base angle of the isosceles triangle. Since it is an isosceles triangle, both base angles are equal. The problem states that the vertex angle is twice the base angle, so the vertex angle will be \( 2b \). We know that the sum of the angles in any triangle is \( 180^\circ \). Therefore, adding all three angles, we get the equation: \( b + b + 2b = 180^\circ \), which simplifies to \( 4b = 180^\circ \).
In simple words: If each base angle of the triangle is \( b \), then the top angle is \( 2b \). Because all angles in a triangle add up to \( 180^\circ \), we add \( b + b + 2b \), which equals \( 4b \). So, \( 4b = 180^\circ \).

Exam Tip: For geometry problems, recall fundamental properties like the sum of angles in a triangle and properties of specific shapes (e.g., isosceles triangles having two equal base angles).