Arithmetic ExpressionsClass 7 Mathematics NCERT Solutions

To Find: The missing numbers in the blanks.

(a) $13+4=$ ______ + 6 Solution: First, evaluate the Left Hand Side (LHS): LHS = $13 + 4 = 17$ Now, the Right Hand Side (RHS) must also be equal to 17. RHS = ______ + 6 So, 17 = ext{______} + 6 To find the missing number, subtract 6 from 17: $17 - 6 = 11$ Final Answer: $13+4 = \underline{11} + 6$

(b) $22+$ ______ $=6 \times 5$ Solution: First, evaluate the Right Hand Side (RHS): RHS = $6 \times 5 = 30$ Now, the Left Hand Side (LHS) must also be equal to 30. LHS = $22 +$ ______ So, 22 + ext{______} = 30 To find the missing number, subtract 22 from 30: $30 - 22 = 8$ Final Answer: $22 + \underline{8} = 6 \times 5$

(c) $8 \times$ ______ $=64 \div 2$ Solution: First, evaluate the Right Hand Side (RHS): RHS = $64 \div 2 = 32$ Now, the Left Hand Side (LHS) must also be equal to 32. LHS = $8 \times$ ______ So, 8 \times ext{______} = 32 To find the missing number, divide 32 by 8: $32 \div 8 = 4$ Final Answer: $8 \times \underline{4} = 64 \div 2$

(d) $34-$ ______ $=25$ Solution: We have the equation 34 - ext{______} = 25. To find the missing number, subtract 25 from 34: $34 - 25 = 9$ Final Answer: $34 - \underline{9} = 25$

To Find: The ascending order of the given expressions based on their values.

Solution: First, we need to find the value of each expression.

(a) $67-19 = 48$

(b) $67-20 = 47$

(c) $35+25 = 60$

(d) $5 \times 11 = 55$

(e) $120 \div 3 = 40$

The values are 48, 47, 60, 55, and 40.

Now, we arrange these values in ascending (increasing) order: $40 < 47 < 48 < 55 < 60$

Finally, we write the corresponding expressions in this order. $120 \div 3 < 67-20 < 67-19 < 5 \times 11 < 35+25$

Final Answer: The expressions in ascending order of their values are: (e) $120 \div 3$, (b) $67-20$, (a) $67-19$, (d) $5 \times 11$, (c) $35+25$

To Compare: The pairs of expressions using reasoning.

(a) $245+289$ [ ] $246+285$ Reasoning: We are comparing $245+289$ with $246+285$. The second expression can be rewritten by relating it to the first expression. $246 = 245 + 1$ $285 = 289 - 4$ So, $246+285 = (245+1) + (289-4) = 245+289+1-4 = 245+289-3$. Since we are subtracting 3 from the first expression to get the second, the first expression is greater. Answer: $245+289 > 246+285$

(b) $273-145$ [ ] $272-144$ Reasoning: We are comparing $273-145$ with $272-144$. In the second expression, both the number being subtracted from (minuend) and the number being subtracted (subtrahend) are decreased by 1. $272-144 = (273-1) - (145-1) = 273-1-145+1 = 273-145$. Since both expressions simplify to the same value, they are equal. Answer: $273-145 = 272-144$

(c) $364 + 587$ [ ] $363 + 589$ Reasoning: We are comparing $364 + 587$ with $363 + 589$. The second expression can be rewritten by relating it to the first expression. $363 = 364 - 1$ $589 = 587 + 2$ So, $363 + 589 = (364-1) + (587+2) = 364+587-1+2 = 364+587+1$. Since we are adding 1 to the first expression to get the second, the second expression is greater. Answer: $364 + 587 < 363 + 589$

(d) $124+245$ [ ] $129+245$ Reasoning: The number 245 is being added to both 124 and 129. Since $124 < 129$, adding the same number to both will maintain the inequality. Therefore, $124+245$ is less than $129+245$. Answer: $124+245 < 129+245$

(e) $213-77$ [ ] $214-76$ Reasoning: We are comparing $213-77$ with $214-76$. The second expression can be rewritten by relating it to the first expression. $214 = 213 + 1$ $76 = 77 - 1$ So, $214-76 = (213+1) - (77-1) = 213+1-77+1 = 213-77+2$. Since we are adding 2 to the first expression to get the second, the second expression is greater. Answer: $213-77 < 214-76$

To Do: Complete the given table by identifying the terms of each expression.

Solution: Terms are parts of an expression separated by '+' signs. Subtraction is treated as the addition of a negative number.

To Find: The value of each expression by first identifying its terms.

(a) $28-7+8$ Solution: Terms: 28, -7, 8 Value = $28 - 7 + 8 = 21 + 8 = 29$ Final Answer: 29

(b) $39-2 \times 6+11$ Solution: Terms: 39, $-2 \times 6$, 11 First, evaluate each term: $39, -12, 11$ Value = $39 - 12 + 11 = 27 + 11 = 38$ Final Answer: 38

(c) $40-10+10+10$ Solution: Terms: 40, -10, 10, 10 Value = $40 - 10 + 10 + 10 = 30 + 10 + 10 = 50$ Final Answer: 50

(d) $48-10 \times 2+16 \div 2$ Solution: Terms: 48, $-10 \times 2$, $16 \div 2$ First, evaluate each term: $48, -20, 8$ Value = $48 - 20 + 8 = 28 + 8 = 36$ Final Answer: 36

(e) $6 \times 3-4 \times 8 \times 5$ Solution: Terms: $6 \times 3$, $-4 \times 8 \times 5$ First, evaluate each term: $18, -160$ Value = $18 - 160 = -142$ Final Answer: -142

To Do: Create a story for each expression and find its value.

(a) $89+21-10$ Story: A fruit seller had 89 mangoes in a basket. He received a new delivery of 21 mangoes. Later in the day, he sold 10 mangoes. How many mangoes does he have now? Value: $89 + 21 - 10 = 110 - 10 = 100$ Final Answer: The fruit seller has 100 mangoes.

(b) $5 \times 12-6$ Story: Sunita bought 5 notebooks, and each notebook has 12 pages. She used 6 pages for her homework. How many unused pages are left? Value: $5 \times 12 - 6 = 60 - 6 = 54$ Final Answer: There are 54 unused pages left.

(c) $4 \times 9+2 \times 6$ Story: For a school event, students are arranged in rows. There are 4 rows with 9 students each and 2 rows with 6 students each. What is the total number of students at the event? Value: $4 \times 9 + 2 \times 6 = 36 + 12 = 48$ Final Answer: There are 48 students in total.

To Do: Write an expression, identify terms, and find the value for each situation.

(a) Gold Coins Given: Elsa started with 100 coins and doubled them. Anna started with 100 coins and has half left. Solution: Number of coins Elsa has = $100 \times 2$ Number of coins Anna has = $100 \div 2$ Total coins they have together = $(100 \times 2) + (100 \div 2)$ Expression: $100 \times 2 + 100 \div 2$ Terms: $100 \times 2$ and $100 \div 2$ Value: $200 + 50 = 250$ Final Answer: Together they have 250 gold coins.

(b) Metro Tickets Given: Adult ticket = ₹40, Child ticket = ₹20. (i) for four adults and three children? Solution: Cost for 4 adults = $4 \times 40$ Cost for 3 children = $3 \times 20$ Total cost = $(4 \times 40) + (3 \times 20)$ Expression: $4 \times 40 + 3 \times 20$ Terms: $4 \times 40$ and $3 \times 20$ Value: $160 + 60 = 220$ Final Answer: The total cost is ₹220.

(ii) for two groups having three adults each? Solution: Total number of adults = $2 \times 3 = 6$ Total cost = $6 \times 40$ Expression: $6 \times 40$ or $(2 \times 3) \times 40$ Terms: The expression has a single term, $6 \times 40$. Value: $240$ Final Answer: The total cost is ₹240.

(c) Height of the window Solution: The question refers to a picture which is not provided in the source text. Therefore, it is not possible to write an expression or find the total height of the window.

To Do: Fill in the blanks and boxes with numbers and operation signs.

(a) $24+(6-4)=24+6$ ☐ 4 Solution: When brackets preceded by a '+' sign are removed, the signs of the terms inside do not change. So, $24+(6-4) = 24+6-4$. Answer: $24+(6-4)=24+6 \boxminus 4$

(b) $38+($ ______ ☐ ______ $) = 38+9-4$ Solution: The right side is $38+9-4$. To create the expression on the left, we can put $9-4$ inside brackets preceded by a '+' sign. Answer: $38+(\underline{9} \boxminus \underline{4}) = 38+9-4$

(c) $24-(6+4)=24$ ☐ 6-4 Solution: When brackets preceded by a '-' sign are removed, the signs of the terms inside change. The term $+6$ becomes $-6$ and the term $+4$ becomes $-4$. So, $24-(6+4)=24-6-4$. Answer: $24-(6+4)=24 \boxminus 6-4$

(d) $24-6-4=24-(6$ ☐ ______) Solution: To put the terms $-6-4$ inside brackets preceded by a '-' sign, we must change their signs inside the bracket. $-6$ becomes $+6$ and $-4$ becomes $+4$. So, $24-6-4 = 24-(6+4)$. Answer: $24-6-4=24-(6 \boxplus \underline{4})$

(e) $27-(8+3)=27$ ☐ 8 ☐ 3 Solution: Removing the brackets preceded by a '-' sign changes the signs of both $+8$ and $+3$ to $-8$ and $-3$. Answer: $27-(8+3)=27 \boxminus 8 \boxminus 3$

(f) $27-($ ______ ☐ ______ $) = 27-8+3$ Solution: The right side is $27-8+3$. To create the expression on the left, we need to put $-8+3$ inside brackets preceded by a '-' sign. We must change their signs. $-8$ becomes $+8$ and $+3$ becomes $-3$. Answer: $27-(\underline{8} \boxminus \underline{3}) = 27-8+3$

To Do: Rewrite each expression without brackets.

(a) $14+(12+10)$ Solution: Brackets are preceded by '+'. Signs inside do not change. Answer: $14+12+10$

(b) $14-(12+10)$ Solution: Brackets are preceded by '-'. Signs of terms inside change. $+12$ becomes $-12$, $+10$ becomes $-10$. Answer: $14-12-10$

(c) $14+(12-10)$ Solution: Brackets are preceded by '+'. Signs inside do not change. Answer: $14+12-10$

(d) $14-(12-10)$ Solution: Brackets are preceded by '-'. Signs of terms inside change. $+12$ becomes $-12$, $-10$ becomes $+10$. Answer: $14-12+10$

(e) $-14+12-10$ Solution: There are no brackets in this expression. Answer: $-14+12-10$

(f) $14-(-12-10)$ Solution: Brackets are preceded by '-'. Signs of terms inside change. $-12$ becomes $+12$, $-10$ becomes $+10$. Answer: $14+12+10$

To Do: Evaluate and compare each pair of expressions.

(a) $(6+10)-2$ and $6+(10-2)$ Guess: These might be equal because they involve the same numbers and operations. This looks like the associative property. Evaluation: LHS: $(6+10)-2 = 16-2 = 14$ RHS: $6+(10-2) = 6+8 = 14$ Conclusion: The expressions have the same value. They are equal. This is an example of the associative property of addition, as it can be written as $(6+10)+(-2) = 6+(10+(-2))$.

(b) $16-(8-3)$ and $(16-8) - 3$ Guess: These might not be equal because subtraction is not associative. Evaluation: LHS: $16-(8-3) = 16-5 = 11$ RHS: $(16-8)-3 = 8-3 = 5$ Conclusion: The expressions do not have the same value. $11 \neq 5$. They are not equal.

(c) $27-(18+4)$ and $27+(-18-4)$ Guess: These might be equal. The first expression involves subtracting a sum. The second involves adding negative numbers. Evaluation: LHS: $27-(18+4) = 27-22 = 5$ RHS: $27+(-18-4) = 27+(-22) = 27-22 = 5$ Conclusion: The expressions have the same value. They are equal. This is because removing the bracket in the LHS gives $27-18-4$, and the RHS is also equivalent to $27-18-4$.

To Do: Identify expressions with the same value using properties of operations.

(a) $319+537, 319-537, -537+319, 537-319$ Solution: We analyze the terms of each expression.

Comparing the sets of terms, expressions 2 and 3 have the same terms {319, -537}. Final Answer: $319-537$ and $-537+319$ have the same value.

(b) $87+46-109, 87+46-109, 87+46-109, 87-46+109, 87 -(46+109), (87-46)+109$ Solution: Let's analyze the terms of each expression. The first three expressions are identical.

To Do: Add brackets to make the equations true.

(a) $34-9+12=13$ Solution: If we calculate from left to right: $34-9+12 = 25+12=37$. This is not 13. Let's try grouping the addition first: $34-(9+12)$. $34-(9+12) = 34-21 = 13$. This is correct. Final Answer: $34-(9+12)=13$

(b) $56-14-8=34$ Solution: If we group the first two numbers: $(56-14)-8$. $(56-14)-8 = 42-8=34$. This is correct. If we group the last two numbers: $56-(14-8) = 56-6=50$. This is incorrect. Final Answer: $(56-14)-8=34$

(c) $-22-12+10+22=-22$ Solution: Let's try grouping the last three terms with a minus sign in front: $-22-(12-10-22)$. $-22-(12-10-22) = -22-(2-22) = -22-(-20) = -22+20 = -2$. This is incorrect. Let's try grouping differently: $-22-(12+10-22)$. $-22-(12+10-22) = -22-(22-22) = -22-0 = -22$. This is correct. Final Answer: $-22-(12+10-22)=-22$

To Do: Fill in the blanks using reasoning.

(a) $423+$ __ $=419+$ __ Reasoning: The number 423 on the LHS is 4 more than the number 419 on the RHS ($423 = 419+4$). To keep the equation balanced, the number added to 419 on the RHS must be 4 more than the number added to 423 on the LHS. Let the first blank be $x$ and the second be $y$. Then $423+x = 419+y$. Substituting $423=419+4$, we get $419+4+x = 419+y$, which means $y=x+4$. We can choose any number for the first blank. Let's choose 10. Then the second blank must be $10+4=14$. Final Answer: An example solution is $423+\underline{10} = 419+\underline{14}$.

(b) $207-68=210-$ __ Reasoning: On the LHS, we have $207-68$. On the RHS, the first number is 210, which is 3 more than 207 ($210=207+3$). To keep the difference the same, we must also increase the number being subtracted by 3. So, the number in the blank should be $68+3=71$. Let's check: $207-68 = 139$. And $210-71 = 139$. The values are equal. Final Answer: $207-68=210-\underline{71}$

To Do: Create expressions using 2, 3, 5, +, -, and brackets.

Solution: Here are some possible expressions and their values:

Final Answer: Possible values obtained are 10, 6, 4, 0, -4, -6.

(a) Do you think she always gets the correct answer? Why? Answer: Yes, she always gets the correct answer. Reasoning: Jasoda's method is to replace $-9$ with $-10+1$. Since $-10+1 = -9$, the value of the expression does not change. For any number $x$, the expression $x-9$ is mathematically equivalent to $x-10+1$. This is a valid mental math strategy.

(b) Can you think of other similar strategies? Give some examples. Answer: Yes, there are many similar strategies. Examples:

To Do: Match the given expressions with equivalent expressions from the collection.

Expression a) $73-14+1$ Analysis: The terms are 73, -14, and 1. Let's evaluate the collection: (a) $73-(14+1) = 73-14-1$. (Not equal) b) $73-(14-1) = 73-14+1$. (Equal) (c) $73+(-14+1) = 73-14+1$. (Equal) d) $73+(-14-1) = 73-14-1$. (Not equal) Final Answer for a): The expressions equal to $73-14+1$ are b) $73-(14-1)$ and c) $73+(-14+1)$.

Expression b) $73-14-1$ Analysis: The terms are 73, -14, and -1. Let's evaluate the collection: (a) $73-(14+1) = 73-14-1$. (Equal) b) $73-(14-1) = 73-14+1$. (Not equal) (c) $73+(-14+1) = 73-14+1$. (Not equal) d) $73+(-14-1) = 73-14-1$. (Equal) Final Answer for b): The expressions equal to $73-14-1$ are a) $73-(14+1)$ and d) $73+(-14-1)$.

To Do: Fill in the blanks based on the distributive property.

(c) $3 \times(5+8)=3 \times 5 \boxplus 3 \times \underline{8}$ (d) $(9+2) \times 4=9 \times 4 \boxplus 2 \times \underline{4}$ (e) $3 \times(\underline{5}+4)=3 \times \underline{5} +3 \times 4$ (f) $(\underline{13}+6) \times 4=13 \times 4+ \underline{6} \times 4$ (g) $3 \times(\underline{5}+\underline{2}) =3 \times 5+3 \times 2$ (h) $(\underline{2}+\underline{3}) \times \underline{4} =2 \times 4+3 \times 4$ (i) $5 \times(9-2)=5 \times 9-5 \times \underline{2}$ (j) $(5-2) \times 7=5 \times 7-2 \times \underline{7}$ (k) $5 \times(8-3)=5 \times 8 \boxminus 5 \times \underline{3}$ (l) $(8-3) \times 7=8 \times 7 \boxminus 3 \times \underline{7}$ (m) $5 \times(12-\underline{1}) = 5 \times \underline{12} \boxminus 5 \times \underline{1}$ (This is one possible answer, others exist.) (n) $(15-\underline{6}) \times 7= \underline{15 \times 7} \boxminus 6 \times 7$ (o) $5 \times(\underline{9}-\underline{4}) =5 \times 9-5 \times 4$ (p) $(\underline{17}-\underline{9}) \times \underline{7} =17 \times 7-9 \times 7$

To Do: Compare the expressions using reasoning.

(a) $(8-3) \times 29$ ☐ $(3-8) \times 29$ Reasoning: LHS: $(8-3) \times 29 = 5 \times 29$. This is a positive number. RHS: $(3-8) \times 29 = -5 \times 29$. This is a negative number. A positive number is always greater than a negative number. Answer: $(8-3) \times 29 > (3-8) \times 29$

(b) $15+9 \times 18$ ☐ $(15+9) \times 18$ Reasoning: LHS: $15+9 \times 18$. The terms are 15 and $9 \times 18$. This means 15 is added to 9 groups of 18. RHS: $(15+9) \times 18 = 24 \times 18$. This means there are 24 groups of 18. Clearly, 24 groups of 18 is much larger than 9 groups of 18 plus 15. Answer: $15+9 \times 18 < (15+9) \times 18$

(c) $23 \times(17-9)$ ☐ $23 \times 17+23 \times 9$ Reasoning: Using the distributive property on the RHS: $23 \times 17+23 \times 9 = 23 \times (17+9)$. So we are comparing $23 \times (17-9)$ with $23 \times (17+9)$. Since $(17-9)$ is less than $(17+9)$, the LHS is smaller than the RHS. Answer: $23 \times(17-9) < 23 \times 17+23 \times 9$

(d) $(34-28) \times 42$ ☐ $34 \times 42-28 \times 42$ Reasoning: This is a direct application of the distributive property, which states that $(a-b) \times c = a \times c - b \times c$. Here, $a=34, b=28, c=42$. The expressions on both sides are equivalent. Answer: $(34-28) \times 42 = 34 \times 42-28 \times 42$

To Do: Find different ways to write 14 in the given format.

Solution: We need to find a number and a sum of two numbers whose product is 14. The factors of 14 are 1, 2, 7, 14.

(a) Using the factor 2: We need the sum in the bracket to be 7. $2 \times (3 + 4) = 14$

(b) Using the factor 7: We need the sum in the bracket to be 2. $7 \times (1 + 1) = 14$

(c) Using the factor 1: We need the sum in the bracket to be 14. $1 \times (10 + 4) = 14$

(d) We can also use fractions. For example, if the first number is 4, we need the sum to be $14/4 = 3.5$. $4 \times (2 + 1.5) = 14$ Another integer example: $2 \times (5+2) = 14$

To Do: Find the total number of items in the described picture in two different ways using expressions.

Description of the Picture: The picture shows a parade formation. There is a group of boy scouts arranged in 4 rows with 5 scouts in each row. Next to them is a group of girl guides arranged in 3 rows with 5 guides in each row.

Method 1: Calculate each group separately and then add. First, find the number of boy scouts. Since there are 4 rows of 5 scouts, the total is $4 \times 5$. Next, find the number of girl guides. Since there are 3 rows of 5 guides, the total is $3 \times 5$. The total number of participants is the sum of these two amounts. Expression: $4 \times 5 + 3 \times 5$ Calculation: $20 + 15 = 35$

Method 2: Combine the groups first. Since both groups have 5 participants in each row, we can think of this as a single large group. The total number of rows is the sum of the scout rows and guide rows, which is $4 + 3$. Each of these combined rows has 5 participants. So, the total number of participants is the total number of rows multiplied by the number of participants per row. Expression: $(4 + 3) \times 5$ Calculation: $7 \times 5 = 35$

Final Answer: Both methods give a total of 35 participants. The expressions are $4 \times 5 + 3 \times 5$ and $(4 + 3) \times 5$.

To Do: Write expressions and find values for each situation.

(a) Mango Supply Solution: Method 1: Find the total daily supply and multiply by the number of days in a week. Total mangoes supplied per day = Rahim's supply + Shyam's supply = $9 + 11$ kg. Total supply in a week (7 days) = $7 \times (9 + 11)$. Expression: $7 \times (9 + 11)$ Value: $7 \times 20 = 140$ kg.

Method 2: Find the total weekly supply for each person and then add them. Rahim's supply in a week = $7 \times 9$ kg. Shyam's supply in a week = $7 \times 11$ kg. Total supply = $(7 \times 9) + (7 \times 11)$. Expression: $7 \times 9 + 7 \times 11$ Value: $63 + 77 = 140$ kg. Final Answer: The total amount of mangoes supplied in a week is 140 kg.

(b) Binu's Savings Solution: First, find Binu's total monthly expenses. Total expenses = Rent + Food + Other expenses = $5000 + 5000 + 2000$. Next, find her monthly savings. Monthly savings = Monthly earnings - Total monthly expenses = $20000 - (5000 + 5000 + 2000)$. Finally, find her annual (12 months) savings. Annual savings = $12 \times (20000 - (5000 + 5000 + 2000))$. Expression: $12 \times (20000 - (5000 + 5000 + 2000))$ Value: $12 \times (20000 - 12000) = 12 \times 8000 = 96000$. Final Answer: Binu will save ₹96,000 by the end of a year.

(c) The Snail's Climb Solution: Net distance covered by the snail each day = (distance climbed) - (distance slipped) = $3 - 2 = 1$ cm. So, at the end of each day, the snail is 1 cm higher than the previous day. Let's track its progress: End of Day 1: 1 cm End of Day 2: 2 cm ... End of Day 7: 7 cm On the morning of Day 8, the snail starts at a height of 7 cm. It then climbs 3 cm. Height reached on Day 8 = $7 + 3 = 10$ cm. Since it reaches the top (10 cm) on Day 8, it gets the treat. Expression: We are looking for the smallest integer 'd' (days) such that the height at the end of day (d-1) plus the day's climb is at least 10 cm. Height at end of day (d-1) is $(d-1) \times (3-2) = d-1$. The condition is $(d-1) + 3 \ge 10$. This simplifies to $d+2 \ge 10$, or $d \ge 8$. The minimum number of days is 8. Final Answer: The snail will get the treat in 8 days.

To Do: Determine the number of stories read and identify the correct expression(s).

Now let's analyze the given expressions to see which one describes the number of stories (which is the total number of reading days).

Final Answer: The correct expressions are (b) $(7-2) \times 8$ and (g) $7 \times 8-2 \times 8$.