Home » CBSE » Arithmetic Expressions  – NCERT Solutions Class 7 Maths (Ganita Prakash)

Arithmetic Expressions  – NCERT Solutions Class 7 Maths (Ganita Prakash)

by

Arithmetic Expressions  – NCERT Solutions Class 7 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 7 Maths (Ganita Prakash).

NCERT Solutions Class 7

English Poorvi Hindi Malhar Maths Ganita Prakash Science Curiosity Social Exploring Society

Arithmetic Expressions – NCERT Solutions

Q.1: Mallika spends ₹25 every day for lunch at school. Write the expression for the total amount she spends on lunch in a week from Monday to Friday.

Solution:

Money spent every day = ₹25
Expression for the total amount spent from Monday to Friday = 5 ×× 25.

Q.2: Which is greater? 1023 + 125 or 1022 + 128?

Solution:

Imagining a situation could help us answer this without finding the values. Raja had 1023 marbles and got 125 more today. Now he has 1023 + 125 marbles. Joy had 1022 marbles and got 128 more today. Now he has 1022 + 128 marbles. Who has more? This situation can be represented as shown in the picture on the right. To begin with, Raja had 1 more marble than Joy. But Joy got 3 more marbles than Raja today. We can see that Joy has (two) more marbles than Raja now.
That is,
1023 + 125 < 1022 + 128.

Q.3: Which is greater? 113 – 25 or 112 – 24?

Solution:

Imagine a situation, Raja had 113 marbles and lost 25 of them. He has 113 – 25 marbles. Joy had 112 marbles and lost 24 today. He has 112 – 24 marbles. Who has more marbles left with them? Raja had 1 marble more than Joy. But he also lost 1 marble more than Joy did. Therefore, they have an equal number of marbles now.
That is,
113 – 25 = 112 – 24.

Q.4: Mallesh brought 30 marbles to the playground. Arun brought 5 bags of marbles with 4 marbles in each bag. How many marbles did Mallesh and Arun bring to the playground?

Solution:

Mallesh summarized this by writing the mathematical expression-
30 + 5 ×× 4.
Arithmetic Expressions 27 Without knowing the context behind this expression, Purna found the value of this expression to be 140. He added 30 and 5 first, to get 35, and then multiplied 35 by 4 to get 140.
Mallesh found the value of this expression to be 50. He multiplied 5 and 4 first to get 20 and added 20 to 30 to get 50.
In this case, Mallesh is right. But why did Purna get it wrong?
Just looking at the expression 30 + 5 ×× 4, it is not clear whether we should do the addition first or multiplication.
Just as punctuation marks are used to resolve confusions in language, brackets and the notion of terms are used in mathematics to resolve confusions in evaluating expressions.

Q.5: Irfan bought a pack of biscuits for ₹15 and a packet of toor dal for ₹56. He gave the shopkeeper ₹100. Write an expression that can help us calculate the change Irfan will get back from the shopkeeper.

Solution:

Irfan spent ₹15 on a biscuit packet and ₹56 on toor dal. So, the total cost in rupees is 15 + 56. He gave ₹100 to the shopkeeper. So, he should get back 100 minus the total cost. Can we write that expression as-
100 – 15 + 56?
Can we first subtract 15 from 100 and then add 56 to the result? We will get 141. It is absurd that he gets more money than he paid the shopkeeper!
We can use brackets in this case:
100 – (15 + 56).
Evaluating the expression within the brackets first, we get 100 minus 71, which is 29. So, Irfan will get back ₹29.

Q.6: Madhu is flying a drone from a terrace. The drone goes 6 m up and then 4 m down. Write an expression to show how high the final position of the drone is from the terrace.

Solution:

The drone is 6 – 4 = 2 m above the terrace. Writing it as sum of terms:

Will the sum change if we swap the terms?

It doesn’t in this case. We already know that swapping the terms does not change the sum when both the terms are positive numbers.

Q.7: Amu, Charan, Madhu, and John went to a hotel and ordered four dosas. Each dosa cost ₹23, and they wish to thank the waiter by tipping ₹5. Write an expression describing the total cost.

Solution:

Cost of 4 dosas = 4 × 23
Can the total amount with tip be written as 4 ×× 23 + 5? Evaluating it, we get

Thus, 4 × 23 + 5 is a correct way of writing the expression.

Q.8: Children in a class are playing “Fire in the mountain, run, run, run!”. Whenever the teacher calls out a number, students are supposed to arrange themselves in groups of that number. Whoever is not part of the announced group size, is out.

Solution:

Ruby wrote 6 × 5 + 3 because the expression (6 × 5) represents the total number of students who grouped themselves in teams of 5, and 3 represents the students who were left over, as they couldn’t form a complete group of 5.

Q.9: Raghu bought 100 kg of rice from the wholesale market and packed them into 2 kg packets. He already had four 2 kg packets. Write an expression for the number of 2 kg packets of rice he has now and identify the terms.

Solution:

He had 4 packets. The number of new 2 kg packets of rice is 100÷2100÷2, which we also write as 10021002.
The number of 2 kg packets he has now is 4+10024+1002. The terms are 

Q.10: Kannan has to pay ₹432 to a shopkeeper using coins of ₹1 and ₹5, and notes of ₹10, ₹20, ₹50 and ₹100. How can he do it?

Solution:

There is more than one possibility. For example,
432=4×100+1×20+1×10+2×1432=4×100+1×20+1×10+2×1
Meaning: 4 notes of ₹100, 1 note of ₹20, 1 note of ₹10 and 2 notes of ₹1
432=8×50+1×10+4×5+2×1432=8×50+1×10+4×5+2×1
Meaning: 8 notes of ₹50, 1 note of ₹10, 4 notes of ₹5 and 2 notes of ₹1

Q.11: Here are two pictures. Which of these two arrangements matches with the expression 5 ×× 2 + 3?

Solution:

Let us write this expression as a sum of terms.

This expression 5 × 2 + 3 can be understood as 3 more than 5 × 2, which describes the arrangement on the left.

Q.12: We also saw this earlier in the case of Irfan purchasing a biscuit packet (₹15) and a toor dal packet (₹56). When he paid ₹100, the change he gets in rupees is: 100 – (15 + 56) = 29.

Solution:

The change could also have been calculated as follows:

  1. First subtract the cost of the biscuit packet (15) from 100: 100 – 15 = 85. This is the amount the shopkeeper owes Irfan if he had purchased only the biscuits. As he has purchased toor dal also, its cost is taken from this remaining amount of 85.
  2. So, to find the change, we need to subtract the cost of toor dal from 85. 85 – 56 = 29. What we have done here is 100 – 15 – 56. So, 100 – (15 + 56) = 100 – 15 – 56. Notice how upon removing the brackets preceded by a negative sign, the signs of the terms inside the brackets change. Observe 3 cm 2 cm 5 cm Border Grill Gap Total Height Ganita Prakash | Grade 7 36 the signs of 40 and 3 in the first example, and that of 15 and 56 in the second.

Q.13: Consider the expression 500 – (250 – 100). Is it possible to write this expression without the brackets?

Solution:

To evaluate this expression, we need to subtract 250−100=150250−100=150 from 500:
500−(250−100)=500−150=350.500−(250−100)=500−150=350.
If we were to directly subtract 250 from 500, then we would have subtracted 100 more than what we needed to. So, we should add back that 100 to 500-250 to make the expression take the same value as 500 −(250−100)−(250−100). This sequence of operations is 500−250+100500−250+100. Thus,
500−(250−100)=500−250+100.500−(250−100)=500−250+100.
Check that 500−(250−100)500−(250−100) is not equal to 500−250−100500−250−100.
Notice again that when the brackets preceded by a negative sign are removed, the signs of the terms inside the brackets change. In this case, the signs of 250 and -100 change to -250 and 100.

Q.14: Hira has a rare coin collection. She has 28 coins in one bag and 35 coins in another. She gifts her friend 10 coins from the second bag. Write an expression for the number of coins left with Hira.

Solution:

This can be expressed by 28 + (35 – 10).
We know that this is the same as 28 + (35 + (-10)). Since the terms can be added in any order, this expression can simply be written as 28 + 35 + (-10), or 28 + 35 – 10. Thus,
28 + (35 – 10) = 28 + 35 – 10 = 53.
When the brackets are NOT preceded by a negative sign, the terms within them do not change their signs upon removing the brackets. Notice the sign of the terms 35 and -10 in the above expression.

Q.15: Lhamo and Norbu went to a hotel. Each of them ordered a vegetable cutlet and a rasgulla. A vegetable cutlet costs ₹43 and a rasgulla costs ₹24. Write an expression for the amount they will have to pay.

Solution:

As each of them had one vegetable cutlet and one rasagulla, each of their shares can be represented by 43 + 24.

Q.16: In the Republic Day parade, there are boy scouts and girl guides marching together. The scouts march in 4 rows with 5 scouts in each row. The guides march in 3 rows with 5 guides in each row (see the figure below). How many scouts and guides are marching in this parade?

Solution:

The number of boy scouts marching is 4 ×× 5. The number of girl guides marching is 3 ×× 5.
The total number of scouts and guides will be 4 ×× 5 + 3 ×× 5.
This can also be found by first finding the total number of rows, i.e., 4 + 3, and then multiplying their sum by the number of children in each row. Thus, the number of boys and girls can be found by (4 + 3) ×× 5.
Therefore, 4 × 5 + 3 ×× 5 = (4 + 3) ×× 5.
Computing these expressions, we get

(4 + 3) ×× 5 = 7 ×× 5 = 35

Q.17: Given 53×18=95453×18=954. Find out 63×1863×18.

Solution:

As 63×1863×18 means 63 times 18,
63×18=(53+10)×1863×18=(53+10)×18
=53×18+10×18=53×18+10×18
=954+180=954+180
=1134=1134

Q.18: Find an effective way of evaluating 97×2597×25.

Solution:

97×2597×25 means 97 times 25.
We can write it as (100−3)×25(100−3)×25
We know that this is the same as the difference of 100 times 25 and 3 times 25:
97×25=100×25−3×2597×25=100×25−3×25

Q.19: Choose your favourite number and write as many expressions as you can having that value.

Solution:

Let the number be 7.
Some expression with 7:
12 – 5 = 7
20 – 13 = 7
42 ×× 6 = 7
7 ×× 1 = 7
3 + (9 – 5) = 7
(26 – 13) – (10 – 4) = 7 etc.

Q.20: Use ‘>’ or ‘<’ or ‘=’ in each of the following expressions to compare them. Can you do it without complicated calculations? Explain your thinking in each case.

  1. 245 + 289 ________ 246 + 285
  2. 273 – 145 ________ 272 – 144
  3. 364 + 587 ________ 363 + 589
  4. 124 + 245 ________ 129 + 245
  5. 213 – 77 ________ 214 – 76

Solution:

  1. 245 + 289 ________ 246 + 285
    245 + 1 + 285 + 4 ________ 245 + 1 + 285
    245 + 285 + 5 ________ 1 + 245 + 285
    Removing the common terms,
    5(>)15(>)1
    ∴245+289(>)246+285.∴245+289(>)246+285.
  2. 273 – 145 ________ 272 – 144
    272 + 1 – 144 + 1 ________ 272 – 144
    272 – 144 + 2 ________ 272 – 144
    Removing the common terms,
    2(>)02(>)0
    ∴273−145(>)272−144.∴273−145(>)272−144.
  3. 364 + 587 ________ 363 + 589
    363 + 1 + 587 ________ 363 + 587 + 2
    363 + 587 + 1 ________ 2 + 363 + 587
    Removing the common terms,
    1(<)21(<)2
    ∴364+587(<)363+589.∴364+587(<)363+589.
  4. 124 + 245 ________ 129 + 245
    124 + 245 ________ 124 + 5 + 245
    124 + 245 ________ 5 + 124 + 245
    Removing the common terms,
    0(<)50(<)5
    ∴124+245(<)129+245.∴124+245(<)129+245.
  5. 213 – 77 ________ 214 – 76
    213 – 76 + 1 ________ 213 + 1 – 76
    213 – 76 + 1 ________ 213 – 76 + 1
    Removing the common terms,
    1(=)11(=)1
    ∴213−77(=)214−76.∴213−77(=)214−76.

Q.21: Check if replacing subtraction by addition in this way does not change the value of the expression, by taking different examples.

Solution:

17 – 9 = 17 + (-9)
8 – 5 = 8 + (-5)
12 – 4 + 6 ×× 3 = 12 + (-4) + 6 ×× 3.

Q.22: Can you explain why subtracting a number is the same as adding its inverse, using the Token Model of integers that we saw in the Class 6 textbook of mathematics?

Solution:

Expression: 8 – 3 = 8 + (-3)   or   (+8) – (+3) = (+8) + (-3)

Q.23: Complete the table.

ExpressionExpression as the sum of its termsTerms
13−2+613−2+613, -2, 6
5+6×35+6×3 
4+15−94+15−9 
23−2×4+1623−2×4+16 
28+19−828+19−8 

Solution:

ExpressionExpression as the sum of its termsTerms
13−2+613−2+613+(−2)+613+(−2)+613, -2, 6
5+6×35+6×35+(6×3)5+(6×3)5,(6×3)5,(6×3)
4+15−94+15−94+15+(−9)4+15+(−9)4, 15, -9
23−2×4+1623−2×4+1623+[−2×4]+1623+[−2×4]+1623, ( −2×4−2×4 ), 16
28 + 19-828+19+(−8)28+19+(−8)28, 19, -8

Q.24: Does changing the order in which the terms are added give different values?

Solution:

No, changing the order in which the terms are added doesn’t give different values.

Q.25: Will this also hold when there are terms having negative numbers as well? Take some more expressions and check.

Solution:

No, the sum will not change even on swapping the terms with negative numbers.
Example 1: (-3) + (-9) = -12. Also, (-9) + (-3) = -12.
Example 2: (-5) + 7 = 2. Also, 7 + (-5) = 2.

Q.26: Can you explain why this is happening using the Token Model of integers that we saw in the Class 6 textbook of mathematics?

Solution:

Using token model to explain, (-3) + (-6) = (-6) + (-3) = (-9)

  1. (-3) + (-6) = (-9)
  2. (-6) + (-3) = (-9)

This proves that the result remains the same even on swapping the positions of the terms with negative numbers.

Q.27: Will this also hold when there are terms having negative numbers as well? Take some more expressions and check it.

Solution:

Yes, even while adding negative integers, grouping them in any order gives the same sum.
Example:
{(-2) + (-5)} + (-7) = (-2) + {(-5) + (-7)}
L.H.S: {(-2) + (-5)} + (-7) = (-7) + (-7) = (-14)
R.H.S: (-2) + {(-5) + (-7)} = (-2) + (-12) = (-14)
Thus, we are getting the same value in both cases.

Q.28: Can you explain why this is happening using Token Model of integers that we saw in the Class 6 textbook of mathematics?

Solution:

{(-2) + (-3)} + (-5) = (-2) + {(-3) + (-5)} = (-10)
L.H.S: {(-2) + (-3)} + (-5)

R.H.S: (-2) + {(-3) + (-5)}

Both L.H.S. and R.H.S. are equal.

Q.29: Does adding the terms of an expression in any order give the same value? Take some more expressions and check. Consider expressions with more than 3 terms also.

Solution:

Yes, adding the terms of an expression in any order gives the same value.
Example:
{(-7) + 12} + (-13) + (-5) = (-7) + 12 + {(-13) + (-5)}
L.H.S: {(-7) + 12} + (-13) + (-5)
= 5 + (-13) + (-5)
= -8 + (-5)
= -13
R.H.S: (-7) + 12 + {(-13) + (-5)}
= (-7) + 12 + (-18)
= 5 + (-18)
= -13

Q.30: Can you explain why this is happening using Token Model of Integers that we saw in the Class 6 textbook of mathematics?

Solution:

Thus, the addition of terms in any order gives the same value.
Therefore, in an expression having only additions, it does not matter in what order the terms are added: they all give the same value.
Now let us consider expressions having multiplication and division also, without the order of operations specified by the brackets. The values of such expressions are found by first evaluating the terms. Once all the terms are evaluated, they are added.
For example, the expression 30 + 5 × 4 is evaluated as follows:

The expression 5 × (3 + 2) + 78 + 3 is evaluated as follows:

Where (3+2) is first evaluated and this sum is multiplied by 5 (= 25). The expression 7 × 8 is evaluated (= 56). This simplifies to 25 + 56 + 3 = 84

Q.31: Manasa is adding a long list of numbers. It took her five minutes to add them all and she got the answer 11749. Then she realised that she had forgotten to include the fourth number 9055. Does she have to start all over again? (Numbers list: 1342, 774, 8611, 9055, 1022)

Solution:

No, she doesn’t have to start over again. She can simply add the forgotten number (9055) to the previous sum (11749) to get the final correct answer.
Calculation: 11,749 + 9,055 = 20,804.

Q.32: Manasa is going outside to play. Her mother says, “Wear your hat and shoes!” Which one should she wear first?

Solution:

It doesn’t matter what she wears first; both pieces contribute equally to her final look. This situation is commutative.

Q.33: Amu, Charan, Madhu, and John went to a hotel and ordered four dosas. Each dosa cost ₹23, and they wish to thank the waiter by tipping ₹5.

Solution:

Cost of 1 dosa == Rs 23
Number of dosas bought = 4
Amount of tip = Rs 25
Now,
 Total cost = Cost of 1 dosa ×4+ Tip  Total cost = Cost of 1 dosa ×4+ Tip 
=23×4+5=23×4+5
=92+6=92+6
= Rs 97= Rs 97

Q.34: For each of the cases below, write the expression and identify its terms:

  1. If the teacher had called out ‘4’, Ruby would write ________
  2. If the teacher had called out ‘7’, Ruby would write ________
  3. Write expressions like the above for your class size.

Solution:

  1. Since 33 ÷ 4 gives a quotient of 8 and a remainder of 1. Therefore, Ruby could write ( 4 ×× 8 + 1 ).
  2. Since 33 ÷ 7 gives a quotient of 4 and a remainder of 5. Therefore, Ruby could write ( 7 ×× 4 + 5 ).
  3. Number of students in my class = 43
    If the teacher had called out 4, the expression would be ( 4 ×× 10 + 3 ).
    If the teacher had called out 7, the expression would be ( 7 ×× 6 + 1 ).

Q.35: Identify the terms in the two expressions above.

Solution:

  1. 432=4×100+1×20+1×10+2×1432=4×100+1×20+1×10+2×1
    Terms: (4×100),(1×20),(1×10)(4×100),(1×20),(1×10), and (2×1)(2×1).
  2. 432=8×50+1×10+4×5+2×1432=8×50+1×10+4×5+2×1
    Terms: (8×50),(1×10),(4×5)(8×50),(1×10),(4×5), and (2×1)(2×1).

Q.36: Can you think of some more ways of giving ₹432 to someone?

Solution:

Some ways are:

  1. 432 = 8 × 50 + 3 × 10 + 2 × 1.
  2. 432 = 20 × 20 + 2 × 10 + 2 × 5 + 2 × 1.

Q.37: What is the expression for the arrangement in the right making use of the number of yellow and blue squares?

Solution:

Do you recall the use of brackets? We need to use brackets for this.
2×(5+3)2×(5+3)
Notice that this arrangement can also be described using-
5+3+5+35+3+5+3
 OR  OR 
5×2+3×25×2+3×2

Q.38: What happens to the value of an expression if we increase or decrease the value of one of its terms? Some expressions are given in following three columns. In each column, one or more terms are changed from the first expression. Go through the example (in the first column) and fill the blanks, doing as little computation as possible.

Solution:

53+(−16)=3753+(−16)=3753+(−16)=3753+(−16)=37−87+(−16)=−103−87+(−16)=−103
54+(−16)=3854+(−16)=38
54 is one more than 53, so the value will be 1 more than 37.		52+(−16)=3652+(−16)=36
52 is one less than 53 , so the value will be 1 less than 37.		−88+(−15)=−103−88+(−15)=−103
(-88) is one less than (-87) and (−15)(−15) is one more than (-16). Since (−1)+1=0(−1)+1=0, the value stays the same: -103.
−86+(−18)=−104−86+(−18)=−104
(-86 ) is one more than (-87) and (-18) is two less than (-16). Since (+1)+(−2)=−1(+1)+(−2)=−1, the value becomes 1 less than -103, i.e., 104.
53 + (-15) = 38
(−15)(−15) is one more than (-16), so the value will be 1 more than 37.		53+(−17)=3653+(−17)=36
( -17 ) one less than -16, so the value will be 1 less than 37		−97+(−26)=−123−97+(−26)=−123
(-97) is 10 less than (-87) and (-26) is 10 less than (-16). Since (−10)+(−10)=−20(−10)+(−10)=−20, the value becomes 20 less than -103, i.e., 123.

Q.39: What about the total amount they have to pay? Can it be described by the expression: 2 ×× 43 + 24?

Solution:

Writing it as sum of terms gives:

This expression means 24 more than 2 × 43. But, we want an expression which means twice or double of 43 + 24. We can make use of brackets to write such an expression:
2 × (43 + 24).
So, we can say that together they have to pay 2 × (43 + 24). This is also the same as paying for two vegetable cutlets and two rasgullas:
2 × 43 + 2 × 24.
Therefore,
2 × (43 + 24) = 2 × 43 + 2 × 24.

Q.40: If another friend, Sangmu, joins them and orders the same items, what will be the expression for the total amount to be paid? [Expression for Lhamo and Norbu who each ordered a vegetable cutlet costing ₹43 and a rasgulla costing ₹24 is 2 ×× (43 + 24)]

Solution:

If another friend, Sangmu joins them and orders the same items.
Then, the total amount to be paid is 3 ×× (43 + 24).

Q.41: 5×4+3≠5×(4+3)5×4+3≠5×(4+3). Can you explain why?

Solution:

 L.H.S: 5×4+3=20+3=23 L.H.S: 5×4+3=20+3=23
 R.H.S: 5×(4+3)=5×4+5×3=20+15=35 R.H.S: 5×(4+3)=5×4+5×3=20+15=35
Since, 23≠3523≠35
∴5×4+3≠5×(4+3).∴5×4+3≠5×(4+3).

Q.42: Which other products might be quicker to find like the ones above?

Solution:

  1. 36×102=36×(100+2)36×102=36×(100+2) =36×100+36×2=3600+72=3672.=36×100+36×2=3600+72=3672.
  2. 75×98=75×(100−2)75×98=75×(100−2) =75×100−75×2=7500−150=7350.=75×100−75×2=7500−150=7350.
  3. 42×52=42×(50+2)42×52=42×(50+2) =42×50+42×2=2100+84=2184.=42×50+42×2=2100+84=2184.
  4. 995×67=(1000−5)×67995×67=(1000−5)×67 =67×1000−67×5=6700−335=6365.=67×1000−67×5=6700−335=6365.

Q.43: 13 + 4 = ________ + 6

Q.44: 22 + ________ = 6 ×× 5

Q.45: 8×8× ________ =64÷2=64÷2

Q.46: 34 – ________ = 25

Q.47: Arrange the following expressions in ascending (increasing) order of their values.

  1. 67−1967−19
  2. 67−2067−20
  3. 35+2535+25
  4. 5×115×11
  5. 120÷3120÷3

Solution:

  1. 67−19=4867−19=48
  2. 67−20=4767−20=47
  3. 35+25=6035+25=60
  4. 5×11=555×11=55
  5. 120÷3=40120÷3=40

Since, 40<47<48<55<6040<47<48<55<60
Therefore, 120÷3<67−20<67120÷3<67−20<67−19<5×11<35+25−19<5×11<35+25

Q.48: Find the values of the following expressions by writing the terms in each case.

  1. 28 – 7 + 8
  2. 39 – 2 ×× 6 + 11
  3. 40 – 10 + 10 + 10
  4. 48 – 10 ×× 2 + 16 ÷÷ 2
  5. 6 ×× 3 – 4 ×× 8 ×× 5

Solution:

  1. 28−7+8=28+(−7)+828−7+8=28+(−7)+8
    =21+8=29.=21+8=29.
    Terms: 28, (-7), 8.
  2. 39−2×6+11=39+{(−2)×6}+1139−2×6+11=39+{(−2)×6}+11
    =39+(−12)+11=39+(−12)+11
    =27+11=38=27+11=38
    Terms: 39,(−2×6),1139,(−2×6),11.
  3. 40−10+10+10=40+(−10)+10+1040−10+10+10=40+(−10)+10+10
    =30+10+10=30+10+10
    =30+20=50.=30+20=50.
    Terms: 40, (-10), 10, 10.
  4. 48−10×2+16÷248−10×2+16÷2 =48+(−10×2)+(16×2)=48+(−10×2)+(16×2)
    =48+(−20)+8=48+(−20)+8
    =28+8=36=28+8=36
    Terms: 48,(−10×2),(16×2)48,(−10×2),(16×2).
  5. 6×3−4×8×5=(6×3)+(−4×8×5)6×3−4×8×5=(6×3)+(−4×8×5)
    =18+(−160)=18+(−160)
    =−142=−142
    Terms: (6×3),(−4×8×5)(6×3),(−4×8×5)

Q.49: Write a story/situation for each of the following expressions and find their values.

  1. 89 + 21 – 10
  2. 5 ×× 12 – 6
  3. 4 ×× 9 + 2 ×× 6

Solution:

  1. Story: Ramesh had 89 marbles. He got 21 more marbles from his friend and later gave 10 marbles to another friend.
    Expression: 89+21−10=110−10=10089+21−10=110−10=100.
  2. Story: There were 5 classrooms, each having 12 desks. One day, 6 desks were found broken and were removed.
    Expression: 5×12−6=60−6=545×12−6=60−6=54.
  3. Story: A bakery baked 4 trays of cupcakes, with 9 cupcakes on each tray, and 2 trays of cookies, with 6 cookies on each tray.
    Expression: 4×9+2×6=36+12=484×9+2×6=36+12=48.

Q.50: For each of the following situations, write the expression describing the situation, identify its terms and find the value of the expression.

  1. Queen Alia gave 100 gold coins to Princess Elsa and 100 gold coins to Princess Anna last year. Princess Elsa used the coins to start a business and doubled her coins. Princess Anna bought jewellery and has only half of the coins left. Write an expression describing how many gold coins Princess Elsa and Princess Anna together have.
  2. A metro train ticket between two stations is ₹40 for an adult and ₹20 for a child. What is the total cost of tickets:
    1. for four adults and three children?
    2. for two groups having three adults each?
  3. Find the total height of the window by writing an expression describing the relationship among the measurements shown in the picture.

Solution:

  1. Princess Elsa and Princess Anna both had 100 gold coins each.
    Elsa doubled her coins, while Anna was left with half of hers.
    Expression: (2×100)+(100÷2)(2×100)+(100÷2)
    Terms: (2×1002×100), (100×2100×2).
    Value: (2×100)+(100÷2)=200+50=250(2×100)+(100÷2)=200+50=250.
    Therefore, together they have 250 gold coins.
  2. Ticket for adults =₹40=₹40
    Ticket for a child = ₹20
    1. For four adults and three children:
      Expression: (4×40)+(3×20)(4×40)+(3×20)
      Terms: (4×40),(3×20)(4×40),(3×20)
      Value: (4×40)+(3×20)=160+60=220(4×40)+(3×20)=160+60=220.
      Therefore, the total cost is ₹220.
    2. For two groups having three adults each:
      Expression: (3×40)+(3×40)(3×40)+(3×40)
      Terms: (3×40),(3×40)(3×40),(3×40)
      Value: (3×40)+(3×40)=120+120=240(3×40)+(3×40)=120+120=240.
      Therefore, the total cost is ₹240.
  3. Border =3 cm=3 cm
    Grill =2 cm=2 cm
    Gap =5 cm=5 cm
    Expression: (6×2)+(2×3)+(7×5)(6×2)+(2×3)+(7×5)
    Terms: (6×2),(2×3),(7×5)(6×2),(2×3),(7×5).
    Value: (6×2)+(2×3)+(7×5)(6×2)+(2×3)+(7×5) =12+6+35=53=12+6+35=53

Q.51: Fill in the blanks with numbers, and boxes with operation signs such that the expressions on both sides are equal.

  1. 24 + (6 – 4) = 24 + 6 ◻◻ ________
  2. 38 + (________ ◻◻ ________) = 38 + 9 – 4
  3. 24 – (6 +4) = 24 ◻◻ 6 – 4

Solution:

  1. 24+(6−4)=24+6−⎯⎯⎯4⎯⎯24+(6−4)=24+6−_4_
  2. 38+(9⎯⎯−⎯⎯⎯4⎯⎯)=38+9−438+(9_−_4_)=38+9−4
  3. 24−(6+4)=24−⎯⎯⎯6−424−(6+4)=24−_6−4

Q.52: Fill in the blanks with numbers, and boxes with operation signs such that the expressions on both sides are equal.

  1. 24 – 6 – 4 = 24 – 6 ◻◻ ________
  2. 27 – (8 + 3) = 27 ◻◻ 8 ◻◻ 3
  3. 27 – (________ ◻◻ ________) = 27 – 8 + 3

Solution:

  1. 24−6−4=24−6+(−⎯⎯⎯4⎯⎯)24−6−4=24−6+(−_4_)
  2. 27−(8+3)=27−⎯⎯⎯8−⎯⎯⎯327−(8+3)=27−_8−_3
  3. 27−(8−⎯⎯⎯3⎯⎯)=27−8+327−(8−_3_)=27−8+3

Q.53: Remove the brackets and write the expression having the same value.

  1. 14 + (12 + 10)
  2. 14 – (12 + 10)
  3. 14 + (12 – 10)

Solution:

  1. 14 + (12 + 10) = 14 + 12 + 10
  2. 14 – (12 + 10) = 14 – 12 – 10
  3. 14 + (12 – 10) = 14 + 12 – 10

Q.54: Remove the brackets and write the expression having the same value.

  1. 14 – (12 – 10)
  2. -14 + 12 – 10
  3. 14 – (-12 – 10)

Solution:

  1. 14−(12−10)=14−12+1014−(12−10)=14−12+10
  2. −14+12−10=−14+12−10−14+12−10=−14+12−10
  3. 14−(−12−10)=14+12+1014−(−12−10)=14+12+10

Q.55: Find the values of the following expressions. For each pair, first try to guess whether they have the same value. When are the two expressions equal?

  1. (6+10)−2 and 6+(10−2)(6+10)−2 and 6+(10−2)
  2. 16−(8−3) and (16−8)−316−(8−3) and (16−8)−3
  3. 27−(18+4) and 27+(−18−4)27−(18+4) and 27+(−18−4)

Solution:

  1. (6+10)−2 and 6+(10−2)(6+10)−2 and 6+(10−2)
    6+10−26+10−26+10−26+10−2
    16−216−216−216−2
    14(=)1414(=)14
    Here, the two expressions are equal.
  2. 16−(8−3) and (16−8)−316−(8−3) and (16−8)−3
    16−8+3_16−8−316−8+3_16−8−3
    8+3_8−38+3_8−3
    11(≠)411(≠)4
    Here, the two expressions are not equal.
  3. 27−(18+4) and 27+(−18−4)27−(18+4) and 27+(−18−4)
    27−18−4_27−18−427−18−4_27−18−4
    9−49−49−49−4
    5(=)55(=)5
    Here, the two expressions are equal.

Q.56: In each of the sets of expressions below, identify those that have the same value. Do not evaluate them, but rather use your understanding of terms.

  1. 319+537,319−537,−537+319,537−319319+537,319−537,−537+319,537−319
  2. 87+46−109,87+46−109,87+46−109,87+46−109,87+46−109,87+46−109, 87 – 46 + 109, 87 −(46+109),(87−46)+109−(46+109),(87−46)+109

Solution:

  1. 319+537,319−537,−537+319,537−319319+537,319−537,−537+319,537−319
    Same value →319−537→319−537 and −537+319→−537+319→ Follow the commutative property of addition
  2. 87+46−109,87+46−109,87+46−109,87+46−109,87+46−109,87+46−109, 87−46+109,87−(46+109),(87−46)+10987−46+109,87−(46+109),(87−46)+109
    Same value →87−46+109→87−46+109 and (87−46)+109→(87−46)+109→ Follow the associative property of addition.
    Exactly same →87+46−109,87+46−109,87+46−109→87+46−109,87+46−109,87+46−109.

Q.57: Add brackets at appropriate places in the expressions such that they lead to the values indicated.

  1. 34−9+12=1334−9+12=13
  2. 56−14−8=3456−14−8=34
  3. −22−12+10+22=−22−22−12+10+22=−22

Solution:

  1. 34−9+12=1334−9+12=13
    34−(9+12)=1334−(9+12)=13
  2. 56−14−8=3456−14−8=34
    (56−14)−8=34(56−14)−8=34
  3. −22−12+10+22=−22−22−12+10+22=−22
    −22−(12+10)+22=−22−22−(12+10)+22=−22

Q.58: Using only reasoning of how terms change their values, fill the blanks to make the expressions on either side of the equality (=) equal.

  1. 423 + ________ = 419 + ________
  2. 207 – 68 = 210 – ________

Solution:

  1. 423 + ________= 419 + ________
    Since 419 is 4 less than 423, we must add 4 more to 419’s side to balance.
    ∴∴ 433 + 2 = 419 + 6. (6 is more than 2).
  2. 207 – 68 = 210 – ________
    Since 210 is 3 more than 207, subtract 3 more from right side to balance.
    ∴∴ 207 – 68 = 210 – 71 (71 is 3 more than 68)

Q.59: Using the numbers 2, 3 and 5, and the operators ‘+’ and ‘-’, and brackets, as necessary, generate expressions to give as many different values as possible. For example, 2 – 3 + 5 = 4 and 3 – (5 – 2) = 0.

Solution:

  1. 2+(3+5)=102+(3+5)=10
  2. 2−(3+5)=2−8=−62−(3+5)=2−8=−6
  3. (−2)−(3+5)=−2−8=−10(−2)−(3+5)=−2−8=−10
  4. (−2)+(3−5)=−2+(−2)=−4(−2)+(3−5)=−2+(−2)=−4
  5. 5+(3−2)=5+1=65+(3−2)=5+1=6

Different values obtained are 10,−10,6,−6,−410,−10,6,−6,−4.

Q.60: Whenever Jasoda has to subtract 9 from a number, she subtracts 10 and adds 1 to it. For example, 36 – 9 = 26 + 1.

  1. Do you think she always gets the correct answer? Why?
  2. Can you think of other similar strategies? Give some examples.

Solution:

  1. Yes, Jasoda always gets the correct answer.
    Subtracting 9 is the same as subtracting 10 and adding 1 back, because:
    −9=−10+1−9=−10+1
    So, 36−9=36−10+1=26+1=2736−9=36−10+1=26+1=27, which is correct.
  2. Examples of some similar strategies:-
    To subtract 19, subtract 20 and add 1:
    54−19=54−20+1=34+1=3554−19=54−20+1=34+1=35
    To subtract 8, subtract 10 and add 2 :
    43−8=43−10+2=33+2=3543−8=43−10+2=33+2=35
    To subtract 18, subtract 20 and add 2 :
    65−18=65−20+2=45+2=4765−18=65−20+2=45+2=47

Q.61: Consider the two expressions: (a) 73−14+173−14+1, (b) 73−14−173−14−1. For each of these expressions, identify the expressions from the following collection that are equal to it.

  1. 73−(14+1)73−(14+1)
  2. 73 – (14  -1)
  3. 73+(−14+1)73+(−14+1)
  4. 73+(−14−1)73+(−14−1)

Solution:

(a) 73−14+1=59+1=6073−14+1=59+1=60
(b) 73−14−1=59−1=5873−14−1=59−1=58

  1. 73−(14+1)=73−15=5873−(14+1)=73−15=58.
  2. 73−(14−1)=73−13=6073−(14−1)=73−13=60.
  3. 73+(−14+1)=73+(−13)=6073+(−14+1)=73+(−13)=60.
  4. 73+(−14−1)=73+(−15)=5873+(−14−1)=73+(−15)=58.

Therefore,
73−14+173−14+1 is equal to 73−(14−1)73−(14−1) and 73+(−14+1)73+(−14+1).
73−14−173−14−1 is equal to 73−(14+1)73−(14+1) and 73+(−14−1)73+(−14−1).

Q.62: Fill in the blanks with numbers, and boxes by signs, so that the expressions on both sides are equal.

  1. 3×(6+7)=3×6+3×73×(6+7)=3×6+3×7
  2. (8+3)×4=8×4+3×4(8+3)×4=8×4+3×4
  3. 3×(5+8)=3×5 ◻ 3×3×(5+8)=3×5 ◻ 3× ________
  4. (9+2)×4=9×4 ◻ 2×(9+2)×4=9×4 ◻ 2× ________
  5. 3 ×× (________ +4)=3+4)=3 ________ + ________
  6. (________ +6)×4=13×4++6)×4=13×4+ ________
  7. 3 ×× (________ + ________) =3×5+3×2=3×5+3×2
  8. (________ + ________) ×× ________ =2×4+3×4=2×4+3×4
  9. 5×(9−2)=5×9−5×5×(9−2)=5×9−5× ________
  10. (5−2)×7=5×7−2×(5−2)×7=5×7−2× ________
  11. 5×(8−3)=5×8 ◻ 5×5×(8−3)=5×8 ◻ 5× ________
  12. (8−3)×7=8×7 ◻ 3×7(8−3)×7=8×7 ◻ 3×7
  13. 5×(12−)=5×(12−)= ________  ◻ 5× ◻ 5× ________
  14.  (15-  (15-  ________  ) ×7= ) ×7= ________  ◻ 6×7 ◻ 6×7
  15. 5 ×× (________ – ________) =5×9−5×4=5×9−5×4
  16. (________ – ________) ×× ________ =17×7−9×7=17×7−9×7

Solution:

  1. 3×(5+8)=3×5±3×8⎯⎯3×(5+8)=3×5±3×8_
  2. (9+2)×4=9×4+⎯⎯⎯2×4⎯⎯(9+2)×4=9×4+_2×4_
  3. 3×(7⎯⎯+4)=3×7⎯⎯⎯⎯⎯+3×4⎯⎯⎯⎯⎯⎯⎯⎯3×(7_+4)=3×7_+3×4_
  4. (13⎯⎯⎯⎯+6)×4=13×4+6×4⎯⎯⎯⎯⎯⎯⎯⎯(13_+6)×4=13×4+6×4_
  5. 3×(5⎯⎯+2⎯⎯)=3×5+3×23×(5_+2_)=3×5+3×2
  6. (2⎯⎯+3⎯⎯)×4⎯⎯=2×4+3×4(2_+3_)×4_=2×4+3×4
  7. 5×(9−2)=5×9−5×2⎯⎯5×(9−2)=5×9−5×2_
  8. (5−2)×7=5×7−2×7⎯⎯(5−2)×7=5×7−2×7_
  9. 5×(8−3)=5×8−⎯⎯⎯5×3⎯⎯5×(8−3)=5×8−_5×3_
  10. (8−3)×7=8×7−⎯⎯⎯3×7(8−3)×7=8×7−_3×7
  11. 5×(12−3⎯⎯)=5×12⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯−⎯⎯⎯5×3⎯⎯5×(12−3_)=5×12_−_5×3_
  12. (15−6⎯⎯)×7=15×7⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯=6×7(15−6_)×7=15×7_=6×7
  13. 5×(9⎯⎯−4⎯⎯)=5×9−5×45×(9_−4_)=5×9−5×4
  14. (17⎯⎯⎯⎯−9⎯⎯)×7⎯⎯=17×7−9×7(17_−9_)×7_=17×7−9×7

Q.63: In the boxes below, fill ‘’ or ‘=’ after analysing the expressions on the LHS and RHS. Use reasoning and understanding of terms and brackets to figure this out and not by evaluating the expressions.

  1. (8−3)×29(8−3)×29 ________ (3−8)×29(3−8)×29
  2.  15+9×1815+9×18 _______ (15+9)×18(15+9)×18

Solution:

  1. (8−3)×29_(3−8)×29(8−3)×29_(3−8)×29
    Positive ×29(>)×29(>) Negative ×29×29
    ∴(8−3)×29(>)(3−8)×29.∴(8−3)×29(>)(3−8)×29.
  2. 15+9×18_(15+9)×1815+9×18_(15+9)×18
    Smaller value (multiplication first) (<) Bigger Value (addition first)
    ∴15+9×18(<)(15+9)×18∴15+9×18(<)(15+9)×18

Q.64: In the boxes below, fill ‘<’, ‘>’ or ‘=’ after analysing the expressions on the LHS and RHS. Use reasoning and understanding of terms and brackets to figure this out and not by evaluating the expressions.

  1. 23×(17−9)23×(17−9) ________ 23×17+23×923×17+23×9
  2. (34−28)×42(34−28)×42 ________ −34×42−28×42−34×42−28×42

Solution:

  1. 23×(17−9)_23×17+23×923×(17−9)_23×17+23×9
    Smaller value (involves subtraction of terms) (<) Bigger value (involves addition of terms)
    23×(17−9)(<)23×17+23×923×(17−9)(<)23×17+23×9
  2. (34−28)×42_34×42−28×42(34−28)×42_34×42−28×42
    This is a property called distributive property, so both are equal.
    (34−28)×42(=)34×42−28×42(34−28)×42(=)34×42−28×42

Q.65: Here is one way to make 14:2⎯⎯×(1⎯⎯+6⎯⎯)=1414:2_×(1_+6_)=14. Are there other ways of getting 14? Fill them out below:

  1. ________ ×× (________ + ________) = 14
  2. ________ ×× (________ + ________) = 14
  3. ________ ×× (________ + ________) = 14
  4. ________ ×× (________ + ________) = 14

Solution:

2×(3+4)=142×(3+4)=14
7×(2+0)=147×(2+0)=14
2×(5+2)=142×(5+2)=14
2×(0+7)=142×(0+7)=14

Q.66: Find out the sum of the numbers given in each picture below in at least two different ways. Describe how you solved it through expressions.

Solution:

  1. Sum =4×5+8×4=20+32=52=4×5+8×4=20+32=52.
    Also, Sum =4×(5+8)=4×13=52=4×(5+8)=4×13=52.
  2. Sum =5×8+6×8=40+48=88=5×8+6×8=40+48=88.
    Also, Sum =8×(5+6)=8×(11)=88=8×(5+6)=8×(11)=88.

Q.67: Read the situations given below. Write appropriate expressions for each of them and find their values.

  1. The district market in Begur operates on all seven days of a week. Rahim supplies 9 kg of mangoes each day from his orchard and Shyam supplies 11 kg of mangoes each day from his orchard to this market. Find the amount of mangoes supplied by them in a week to the local district market.
  2. Binu earns ₹20,000 per month. She spends ₹5,000 on rent, ₹5,000 on food, and ₹2,000 on other expenses every month. What is the amount Binu will save by the end of a year?
  3. During the daytime a snail climbs 3 cm up a post, and during the night while asleep, accidentally slips down by 2 cm. The post is 10 cm high, and a delicious treat is on its top. In how many days will the snail get the treat?

Solution:

  1. Mangoes supplied by Rahim per day =9 kg=9 kg
    Mangoes supplied by Shyam per day =11 kg=11 kg
    Total amount of mangoes supplied by both in a week =7×(9+11)kg=7×(9+11)kg
    =7×9+7×11=7×9+7×11
    =63+77=140 kg=63+77=140 kg
  2. Total salary =₹20,000=₹20,000
    Rent expenses = ₹5,000
    Food expenses = ₹5,000
    Other expenses = ₹2,000
    Total amount saved at the end of year =12×{20,000−(5000+5000+2000)}=12×{20,000−(5000+5000+2000)}
    =12×{20,000−12,000}=12×{20,000−12,000}
    =12×8,000=₹96,000.=12×8,000=₹96,000.
  3. Height of post =10 cm=10 cm
    Climb during the day =3 cm=3 cm
    Slip during the night =2 cm=2 cm
    Let, total days taken by the snail =y=y
    ∴10=y×(3−2)∴10=y×(3−2)
    10=y×110=y×1
    y=10y=10
    Therefore, the snail will take 10 days to reach the top.

Q.68: Melvin reads a two-page story every day except on Tuesdays and Saturdays. How many stories would he complete reading in 8 weeks? Which of the expressions below describes this scenario?

  1. 5×2×85×2×8
  2. (7−2)×8(7−2)×8
  3. 8×78×7
  4. 7×2×87×2×8
  5. 7×5−27×5−2
  6. (7+2)×8(7+2)×8
  7. 7×8−2×87×8−2×8
  8. (7−5)×8(7−5)×8

Solution:

Stories completed in 8 weeks are represented by the expressions given below:
(b) (7 – 2) × 8
(g) 7 × 8 – 2 × 8

Q.69: Find different ways of evaluating the following expressions:

  1. 1−2+3−4+5−6+7−8+9−101−2+3−4+5−6+7−8+9−10
  2. 1−1+1−1+1−1+1−1+1−11−1+1−1+1−1+1−1+1−1

Solution:

  1. Way 1: 1−2+3−4+5−6+7−8+9−101−2+3−4+5−6+7−8+9−10 =(1−2)+(3−4)+(5−6)+(7−8)+(9−10)=(1−2)+(3−4)+(5−6)+(7−8)+(9−10)
    =(−1)+(−1)+(−1)+(−1)+(−1)=(−5).=(−1)+(−1)+(−1)+(−1)+(−1)=(−5).
    Way 2: 1−2+3−4+5−6+7−8+9−101−2+3−4+5−6+7−8+9−10 =1+3+5+7+9−2−4−6−8−10=1+3+5+7+9−2−4−6−8−10
    =(1+3+5+7+9)−(2+4+6+8+10)=(1+3+5+7+9)−(2+4+6+8+10)
    =25−30=(−5)=25−30=(−5)
    Way 3: 1−2+3−4+5−6+7−8+9−101−2+3−4+5−6+7−8+9−10 =−1+3−4+5−6+7−8+9−10=−1+3−4+5−6+7−8+9−10
    =2−4+5−6+7−8+9−10=2−4+5−6+7−8+9−10
    =−2+5−6+7−8+9−10=−2+5−6+7−8+9−10
    =3−6+7−8+9−10=3−6+7−8+9−10
    =−3+7−8+9−10=−3+7−8+9−10
    =4−8+9−10=4−8+9−10
    =−4+9−10=−4+9−10
    =5−10=−5=5−10=−5
  2. Way 1: 1−1+1−1+1−1+1−1+1−11−1+1−1+1−1+1−1+1−1 =(1−1)+(1−1)+(1−1)+(1−1)+(1−1)=(1−1)+(1−1)+(1−1)+(1−1)+(1−1)
    =0+0+0+0+0=0=0+0+0+0+0=0
    Way 2 : 1−1+1−1+1−1+1−1+1−11−1+1−1+1−1+1−1+1−1 =1+1+1+1+1−1−1−1−1−1=1+1+1+1+1−1−1−1−1−1
    =(1+1+1+1+1)−(1+1+1+1+1)=(1+1+1+1+1)−(1+1+1+1+1)
    =5−5=0=5−5=0
    Way 3: 1−1+1−1+1−1+1−1+1−11−1+1−1+1−1+1−1+1−1 =0+1−1+1−1+1−1+1−1=0+1−1+1−1+1−1+1−1
    =1−1+1−1+1−1+1−1=1−1+1−1+1−1+1−1
    =0+1−1+1−1+1−1=0+1−1+1−1+1−1
    =1−1+1−1+1−1=1−1+1−1+1−1
    =0+1−1+1−1=0+1−1+1−1
    =1−1+1−1=1−1+1−1
    =0+1−1=0+1−1
    =1−1=0.=1−1=0.

Q.70: Compare the following pairs of expressions using ‘<’, ‘>’ or ‘=’ or by reasoning.

  1. 49−7+849−7+8 & ◻◻ & 49−7+849−7+8
  2. 83×42−1883×42−18 & ◻◻ & 83×40−1883×40−18
  3. 145−17×8145−17×8 & ◻◻ & 145−17×6145−17×6
  4. 23×48−3523×48−35 & ◻◻ & 23×(48−35)23×(48−35)
  5. (16−11)×12(16−11)×12 & ◻◻ & −11×12+16×12−11×12+16×12

Solution:

  1. 49−7+849−7+8 ________ 49−7+849−7+8
    Both sides are exactly the same.
    ∴49−7+8(=)49−7+8∴49−7+8(=)49−7+8
  2. 83×42−18_83×40−1883×42−18_83×40−18
    3486−18_3320−183486−18_3320−18
    3468(>)3302.3468(>)3302.
    ∴83×42−18(>)83×40−18∴83×42−18(>)83×40−18
  3. 145−17×8_145−17×6145−17×8_145−17×6
    145−136_145−102145−136_145−102
    9(<)439(<)43
    ∴145−17×8(<)145−17×6∴145−17×8(<)145−17×6
  4. 23×48−35_23×(48−35)23×48−35_23×(48−35)
    1104−35_23×131104−35_23×13
    1069(>)2991069(>)299
    ∴23×48−35(>)23×(48−35)∴23×48−35(>)23×(48−35)
  5. (16−11)×12−11×12+16×12(16−11)×12−11×12+16×12
    5×12−132+1925×12−132+192
    60(=)6060(=)60
    ∴(16−11)×12(=)−11×12+16×12∴(16−11)×12(=)−11×12+16×12

Q.71: Compare the following pairs of expressions using ‘’ or ‘=’ or by reasoning.

  1. (76−53)×88(76−53)×88 ◻◻ 88×(53−76)88×(53−76)
  2. 25×(42+16)25×(42+16) ◻◻ 25×(43+15)25×(43+15)
  3. 36×(28−16)36×(28−16) ◻◻ 35×(27−15)35×(27−15)

Solution:

  1. (76−53)×88_88×(53−76)(76−53)×88_88×(53−76)
    23×8888×−2323×8888×−23
    2024(>)−20242024(>)−2024
    ∴(76−53)×88(>)88×(53−76)∴(76−53)×88(>)88×(53−76)
  2. 25×(42+16)_25×(43+15)25×(42+16)_25×(43+15)
    25×58_25×5825×58_25×58
    1450(=)14501450(=)1450
    ∴25×(42+16)(=)25×(43+15)∴25×(42+16)(=)25×(43+15)
  3. 36×(28−16)_35×(27−15)36×(28−16)_35×(27−15)
    36×1235×1236×1235×12
    432(>)420432(>)420
    ∴36×(28−16)(>)35×(27−15)∴36×(28−16)(>)35×(27−15)

Q.72: Identify which of the following expressions are equal to the given expression without computation. You may rewrite the expressions using terms or removing brackets. There can be more than one expression which is equal to the given expression.
83 – 37 – 12

  1. 84 – 38 – 12
  2. 84 – (37 + 12)
  3. 83 – 38 – 13
  4. – 37 + 83 -12

Solution:

83−37−1283−37−12

  1. 84-38-12 contains different terms than the given expression. Therefore, it is not equal to the original expression.
  2. 84−(37+12)=84−37−1284−(37+12)=84−37−12 which is same as the given expression.
  3. 83-38-13 contains different terms than the given expression. Therefore, it is not equal to the original expression.
  4. −37+83−12=83−37−12−37+83−12=83−37−12 which is same as the given expression.
    ∴∴ (83−37−1283−37−12) is equal to (ii) and (iv).

Q.73: Identify which of the following expressions are equal to the given expression without computation. You may rewrite the expressions using terms or removing brackets. There can be more than one expression which is equal to the given expression.
93+37×44+7693+37×44+76

  1. 37+93×44+7637+93×44+76
  2. 93+37×76+4493+37×76+44
  3. (93+37)×(44+76)(93+37)×(44+76)
  4. 37×44+93+7637×44+93+76

Solution:

93+37×44+76=93+(37×44)+7693+37×44+76=93+(37×44)+76

  1. 37+93×44+76=37+(93×44)+7637+93×44+76=37+(93×44)+76. The multiplication uses different numbers than the given expression. Therefore, it is not equal to the original expression.
  2. 93+37×76+44=93+(37×76)+4493+37×76+44=93+(37×76)+44. The multiplication uses different numbers than the given expression. Therefore, it is not equal to the original expression.
  3. (93+37)×(44+76)(93+37)×(44+76). The grouping and operation order are different than the given expression. Therefore, it is not equal to the original expression.
  4. 37×44+93+76=(37×44)+93+7637×44+93+76=(37×44)+93+76 which can be rearranged as 93+(37×44)+7693+(37×44)+76. Therefore, it is equal to the original expression.
    ∴∴ (93+37×44+7693+37×44+76) is equal to (iv).

Q.74: Choose a number and create ten different expressions having that value.

Solution:

Number chosen: 7

  1. 3+(9−5)3+(9−5)
  2. (10−5)+2(10−5)+2
  3. 2×3+12×3+1
  4. (11−8)+(15−11)(11−8)+(15−11)
  5. (8×2)−9(8×2)−9
  6. 5+(6−4)5+(6−4)
  7. 2+(9−4)2+(9−4)
  8.  ii) (12×2)−17 ii) (12×2)−17
  9. 3+(8−4)3+(8−4)
  10. (7×2)−7(7×2)−7

myCBSEguide App

Test Generator

Create question paper PDF and online tests with your own name & logo in minutes.

Create Now
myCBSEguide App

Learn8 App

Practice unlimited questions for Entrance tests & government job exams at ₹99 only

Install Now