Chapter 2: Arithmetic Expressions (Expressions)

Take 36:

$30+6$, $40-4$, $4\times 9$, $6\times 6$, $72÷2$, $108÷3$, $50-14$, $18+18$, $3\times 12$, $100-64$, $2\times 18$, $360÷10$

E.g. 36 = 30 + 6 = 40 − 4 = 4 × 9 = 6 × 6 = 72 ÷ 2 = 18 + 18 = 100 − 64.

(a) $13+4=17=$ 11 $+6$

(b) $6\times 5=30$, so $22+$ 8 $=30$

(c) $64÷2=32$, so $8\times$ 4 $=32$

(d) $34-$ 9 $=25$

(a) 11 (b) 8 (c) 4 (d) 9

Values: (a) 48 (b) 47 (c) 60 (d) 55 (e) 40

$120÷3<67-20<67-19<5\times 11<35+25$

(e) < (b) < (a) < (d) < (c), i.e. 40 < 47 < 48 < 55 < 60.

(a) 245 is 1 less than 246, but 289 is 4 more than 285. Overall the left side is 3 more: $245+289$ > $246+285$.

(b) 273 is 1 more than 272, and 145 (taken away) is also 1 more than 144. The extra 1 is added and taken away: $273-145$ = $272-144$.

(c) 364 is 1 more than 363, but 587 is 2 less than 589. The left side is 1 less: $364+587$ < $363+589$.

(d) Both add 245; 124 is less than 129: $124+245$ < $129+245$.

(e) 213 is 1 less than 214, and we take away 77, which is 1 more than 76. The left side is 2 less: $213-77$ < $214-76$.

(a) > (b) = (c) < (d) < (e) <

Examples: $18-10=8$ and $18+(-10)=8$; $5-9=-4$ and $5+(-9)=-4$; $-3-4=-7$ and $-3+(-4)=-7$.

Token Model: a positive token (+1) and a negative token (−1) cancel to make zero. To take away 10 positive tokens from 18 positive tokens leaves 8. Adding 10 negative tokens to 18 positive tokens makes 10 zero pairs, again leaving 8. So "taking away 10" and "adding −10" always leave the same tokens.

Yes: e.g. 18 − 10 = 18 + (−10) = 8. In the Token Model, removing 10 positive tokens has the same result as adding 10 negative tokens, which cancel 10 positive ones.

4 + 15 + (−9): terms 4, 15, −9; 23 + (−2 × 4) + 16: terms 23, −2 × 4, 16; 28 + 19 + (−8): terms 28, 19, −8.

No. Examples:

• Swapping: $(-4)+(-2)=-6=(-2)+(-4)$; $7+(-10)=-3=(-10)+7$.
• Grouping: $(-7)+10+(-11)$: $(-7+10)+(-11)=3+(-11)=-8$ and $-7+(10+(-11))=-7+(-1)=-8$.
• Any order, four terms: $15+(-8)+5+12=24$ and $12+5+15+(-8)=24$.

Token Model: adding terms means putting all their tokens together in one heap. The final heap (and so the count left after cancelling zero pairs) is the same whichever heap we pour in first or which two we combine first.

Swapping, grouping or reordering terms never changes the sum, even with negative terms, because all the tokens end up in the same heap.

No. Since terms can be added in any order, she can add the missed number to her total:

$11749+9055=$ 20,804

No: 11749 + 9055 = 20,804.

$7\times 23+5=161+5=$ ₹166. Terms: $7\times 23$ and $5$.

7 × 23 + 5 = ₹166; terms 7 × 23 and 5.

For '5': 6 groups of 5 are formed and 3 children are left out: $6\times 5+3=33$.

• Call '4': $33=8\times 4+1$; terms $8\times 4$ and $1$.
• Call '7': $33=4\times 7+5$; terms $4\times 7$ and $5$.
• For a class of 40 with call '6': $40=6\times 6+4$; with call '9': $40=4\times 9+4$.

6 groups of 5 with 3 left over; for '4': 8 × 4 + 1; for '7': 4 × 7 + 5.

Terms of the first: $4\times 100$, $1\times 20$, $1\times 10$, $2\times 1$. Terms of the second: $8\times 50$, $1\times 10$, $4\times 5$, $2\times 1$.

More ways: $432=4\times 100+6\times 5+2\times 1$; $432=21\times 20+1\times 10+2\times 1$; $432=3\times 100+2\times 50+1\times 20+1\times 10+2\times 1$.

Terms are the products such as 4 × 100, 1 × 20, …; e.g. also 432 = 4 × 100 + 6 × 5 + 2 × 1.

(a) $28+(-7)+8=$ 29 (terms 28, −7, 8)

(b) $39+(-2\times 6)+11=39-12+11=$ 38 (terms 39, −2 × 6, 11)

(c) $40+(-10)+10+10=$ 50 (terms 40, −10, 10, 10)

(d) $48+(-10\times 2)+16÷2=48-20+8=$ 36 (terms 48, −10 × 2, 16 ÷ 2)

(e) $6\times 3+(-4\times 8\times 5)=18-160=$ −142 (terms 6 × 3, −4 × 8 × 5)

(a) 29 (b) 38 (c) 50 (d) 36 (e) −142

(a) A bus had 89 passengers. At a stop 21 got on and 10 got off. Passengers now: $89+21-10=$ 100.

(b) Meena bought 5 boxes of 12 pencils and gave away 6 pencils. Pencils left: $5\times 12-6=$ 54.

(c) Raju bought 4 notebooks at ₹9 each and 2 pens at ₹6 each. Total cost: $4\times 9+2\times 6=36+12=$ ₹48.

(a) 100 (b) 54 (c) 48, with stories like the ones above.

(a) $2\times 100+100÷2=200+50=$ 250 coins. Terms: $2\times 100$, $100÷2$.

(b) (i) $4\times 40+3\times 20=160+60=$ ₹220. Terms: $4\times 40$, $3\times 20$.

(ii) $2\times (3\times 40)=2\times 120=$ ₹240. Single term: $2\times (3\times 40)$.

(c) In the picture the window has a border at the top and the bottom, 6 grills and 7 gaps:

$2\times 3+6\times 2+7\times 5=6+12+35=$ 53 cm. Terms: $2\times 3$, $6\times 2$, $7\times 5$.

(a) 2 × 100 + 100 ÷ 2 = 250 (b) (i) 4 × 40 + 3 × 20 = ₹220 (ii) 2 × (3 × 40) = ₹240 (c) 2 × 3 + 6 × 2 + 7 × 5 = 53 cm

Column 1: $54+(-16)=38$ (given). $53+(-15)$: −15 is one more than −16, so the value is 1 more: 38.

Column 2: $52+(-16)$: 52 is one less than 53, so 36. $53+(-17)$: −17 is one less than −16, so 36.

Column 3: $-87+(-16)=$ −103.

• $-88+(-15)$: −88 is 1 less, −15 is 1 more, so the value is unchanged: −103.
• $-86+(-18)$: −86 is 1 more, −18 is 2 less, so 1 less overall: −104.
• $-97+(-26)$: −97 is 10 less and −26 is 10 less, so 20 less: −123.

38; 36; 36; −103; −103; −104; −123

(a) $24+(6-4)=24+6$ − 4

(b) $38+($9 − 4$)=38+9-4$

(c) $24-(6+4)=24$ − $6-4$

(d) $24-6-4=24-6$ − 4

(e) $27-(8+3)=27$ − $8$ − $3$

(f) $27-($8 − 3$)=27-8+3$

(a) − 4 (b) 9 − 4 (c) − (d) − 4 (e) −, − (f) 8 − 3

When a bracket follows a minus sign, every term inside changes sign; otherwise the terms stay as they are.

(a) $14+12+10$ (= 36)

(b) $14-12-10$ (= −8)

(c) $14+12-10$ (= 16)

(d) $14-12+10$ (= 12)

(e) $-14+12-10$ (= −12)

(f) $14+12+10$ (= 36)

(a) 14 + 12 + 10 (b) 14 − 12 − 10 (c) 14 + 12 − 10 (d) 14 − 12 + 10 (e) −14 + 12 − 10 (f) 14 + 12 + 10

(a) $6+10-2=14$ and $6+10-2=14$: equal.

(b) $16-(8-3)=16-8+3=11$, but $(16-8)-3=5$: not equal.

(c) $27-18-4=5$ and $27-18-4=5$: equal.

The two expressions are equal when, after removing the brackets, they have the same terms (with the same signs).

(a) 14 and 14, equal (b) 11 and 5, not equal (c) 5 and 5, equal; they are equal when they have the same terms after removing brackets.

(a) $319-537$ and $-537+319$ have the same terms (319 and −537), so they are equal. ($537-319$ has the opposite terms, and $319+537$ is different.)

(b) The three copies of $87+46-109$ are equal (terms 87, 46, −109). Also $87-46+109$ and $(87-46)+109$ are equal to each other (terms 87, −46, 109). $87-(46+109)=87-46-109$ is different from all of these.

(a) 319 − 537 = −537 + 319. (b) The three 87 + 46 − 109 are equal; 87 − 46 + 109 = (87 − 46) + 109.

(a) $34-(9+12)=34-21=13$

(b) $(56-14)-8=42-8=34$ (this is the usual order; the brackets just show it)

(c) $-22-(12+10)+22=-22-22+22=-22$

(a) 34 − (9 + 12) (b) (56 − 14) − 8 (c) −22 − (12 + 10) + 22

(a) 423 is 4 more than 419, so the blank on the right must be 4 more than the blank on the left. For example, $423+6=419+10$ (both 429), or $423+419=419+423$.

(b) 210 is 3 more than 207, so we must take away 3 more: $207-68=210-$ 71.

(a) e.g. 423 + 6 = 419 + 10 (right blank 4 more than the left) (b) 71

Each expression is $\pm 2\pm 3\pm 5$ with at least one plus, so these 6 values (10, 6, 4, 0, −4, −6) are all that are possible.

10, 6, 4, 0, −4 and −6 (e.g. 2 + 3 + 5, 3 + 5 − 2, 2 − 3 + 5, 2 + 3 − 5, 3 − (2 + 5), 2 − (3 + 5)).

(a) Yes, always. $9=10-1$, so $n-9=n-(10-1)=n-10+1$. Taking away 10 takes away 1 too many, so she adds it back.

(b) Examples:

• Subtract 8: subtract 10, add 2. $53-8=43+2=45$
• Subtract 99: subtract 100, add 1. $425-99=325+1=326$
• Add 9: add 10, subtract 1. $47+9=57-1=56$
• Subtract 19: subtract 20, add 1. $64-19=44+1=45$

(a) Yes, because n − 9 = n − (10 − 1) = n − 10 + 1. (b) E.g. subtract 99 as −100 + 1; add 9 as +10 − 1.

Remove the brackets: (a) $73-14-1$ (b) $73-14+1$ (c) $73-14+1$ (d) $73-14-1$

• $73-14+1$ equals (b) and (c).
• $73-14-1$ equals (a) and (d).

73 − 14 + 1 = (b) and (c); 73 − 14 − 1 = (a) and (d).

$3\times (43+24)=3\times 43+3\times 24=129+72=$ ₹201

3 × (43 + 24) = ₹201

$5\times 4+3$ has two terms, $5\times 4$ and 3, so it is $20+3=23$: "3 more than 5 fours". But $5\times (4+3)=5\times 7=35$: "5 sevens". In the second, the 3 is also multiplied by 5.

Yes, $5\times (4+3)=5\times (3+4)=(3+4)\times 5=35$, since $4+3=3+4$ and swapping the numbers in a product does not change it.

5 × 4 + 3 = 23 but 5 × (4 + 3) = 35, since in the second the 3 is multiplied too; yes, all three are 35.

$97\times 25=100\times 25-3\times 25=2500-75=$ 2,425

(a) $95\times 8=(100-5)\times 8=800-40=$ 760

(b) $104\times 15=(100+4)\times 15=1500+60=$ 1,560

(c) $49\times 50=(50-1)\times 50=2500-50=$ 2,450

Yes, it is quicker. It works best when one number is close to a "round" number such as 10, 50, 100 or 1000: e.g. $998\times 6=6000-12=5988$, $102\times 45=4500+90=4590$, $59\times 7=420-7=413$.

2,425; (a) 760 (b) 1,560 (c) 2,450; it is quickest when one number is near 10, 50, 100, 1000, …

(a) $3\times (6+7)=3\times 6+3\times 7$ (already complete)

(b) $(8+3)\times 4=8\times 4+3\times 4$ (already complete)

(c) $3\times (5+8)=3\times 5$ + $3\times$ 8

(d) $(9+2)\times 4=9\times 4$ + $2\times$ 4

(e) $3\times ($10$+4)=3\times$ 10 $+$ 3 × 4 (any number can replace 10)

(f) $($13$+6)\times 4=13\times 4+$ 6 × 4

(g) $3\times ($5 + 2$)=3\times 5+3\times 2$

(h) $($2 + 3$)\times$ 4 $=2\times 4+3\times 4$

(i) $5\times (9-2)=5\times 9-5\times$ 2

(j) $(5-2)\times 7=5\times 7-2\times$ 7

(k) $5\times (8-3)=5\times 8$ − $5\times$ 3

(l) $(8-3)\times 7=8\times 7$ − $3\times 7$

(m) $5\times (12-$3$)=$ 5 × 12 − $5\times$ 3 (any number can replace 3)

(n) $(15-$6$)\times 7=$ 15 × 7 − $6\times 7$

(o) $5\times ($9 − 4$)=5\times 9-5\times 4$

(p) $($17 − 9$)\times$ 7 $=17\times 7-9\times 7$

(c) +, 8 (d) +, 4 (e) 10; 3 × 10 + 3 × 4 (f) 13; 6 × 4 (g) 5 + 2 (h) 2 + 3, 4 (i) 2 (j) 7 (k) −, 3 (l) − (m) 3; 5 × 12 − 5 × 3 (n) 6; 15 × 7 − (o) 9 − 4 (p) 17 − 9, 7

(a) $8-3$ is positive and $3-8$ is negative, so the left is positive and the right negative: >.

(b) On the left only 9 is multiplied by 18; on the right 15 is also multiplied by 18: <.

(c) The left is $23\times 17-23\times 9$; the right adds $23\times 9$ instead: <.

(d) By the distributive property, they are the same: =.

(a) > (b) < (c) < (d) =

(a) $2\times (3+4)=14$

(b) $2\times (2+5)=14$

(c) $7\times (1+1)=14$

(d) $1\times (8+6)=14$

(Also $2\times (0+7)$, $1\times (10+4)$, $14\times (0+1)$, …)

E.g. 2 × (3 + 4), 2 × (2 + 5), 7 × (1 + 1), 1 × (8 + 6).

Picture I (five 4s and four 8s):

• Way 1: count each number: $5\times 4+4\times 8=20+32=$ 52
• Way 2: row by row: $(4+8+4)+(8+4+8)+(4+8+4)=2\times 16+20=52$
• Way 3: 4 + 8 pairs: $4\times (4+8)+4=48+4=52$

Picture II (eight 5s and eight 6s):

• Way 1: $8\times 5+8\times 6=40+48=$ 88
• Way 2: pair each 5 with a 6: $8\times (5+6)=8\times 11=88$
• Way 3: each row has two 5s and two 6s: $4\times (5+6+6+5)=4\times 22=88$

Picture I: 52 (e.g. 5 × 4 + 4 × 8); Picture II: 88 (e.g. 8 × (5 + 6)).

(a) $7\times (9+11)=7\times 20=$ 140 kg

(b) Monthly saving $=20000-(5000+5000+2000)=8000$. Yearly saving $=12\times (20000-(5000+5000+2000))=12\times 8000=$ ₹96,000

(c) Each full day and night, the snail goes up $3-2=1$ cm. After 7 days and nights it is at $7\times (3-2)=7$ cm. On the 8th day it climbs 3 cm: $7+3=10$ cm, and it reaches the top before it can slip back. So it gets the treat on the 8th day.

(a) 7 × (9 + 11) = 140 kg (b) 12 × (20000 − (5000 + 5000 + 2000)) = ₹96,000 (c) On the 8th day: 7 × (3 − 2) + 3 = 10 cm.

He reads one story on $7-2=5$ days a week, so in 8 weeks: $(7-2)\times 8=$ 40 stories.

The matching expressions are (b) (7 − 2) × 8 and (g) 7 × 8 − 2 × 8 (both 40). (Expression (a), $5\times 2\times 8=80$, gives the number of pages read, not stories.)

40 stories; expressions (b) and (g). [(a) gives the 80 pages.]

(a) Way 1, in pairs: $(1-2)+(3-4)+(5-6)+(7-8)+(9-10)=5\times (-1)=$ −5

Way 2, positive and negative terms separately: $(1+3+5+7+9)-(2+4+6+8+10)=25-30=-5$

Way 3: $1+(-2+3)+(-4+5)+(-6+7)+(-8+9)-10=1+4-10=-5$

(b) Way 1, in pairs: $(1-1)+(1-1)+(1-1)+(1-1)+(1-1)=$ 0

Way 2: five 1s and five −1s: $5-5=0$

(a) −5 (b) 0, e.g. by pairing terms or by adding positive and negative terms separately.

(a) The two expressions are identical: =.

(b) 83 × 42 is more than 83 × 40, and both take away 18: >.

(c) The left takes away more (17 × 8 > 17 × 6): <.

(d) The left takes away only 35 from 23 × 48; the right takes away 23 × 35: >.

(e) $(16-11)\times 12=16\times 12-11\times 12$, the same terms as the right: =.

(f) The left is positive (76 > 53), the right negative: >.

(g) $42+16=58=43+15$: =.

(h) $28-16=12=27-15$, and 36 > 35: >.

(a) = (b) > (c) < (d) > (e) = (f) > (g) = (h) >

(a) (i) 84 is 1 more than 83 and we take away 1 more (38 instead of 37): equal. (ii) $84-37-12$ is 1 more: not equal. (iii) takes away 2 more: not equal. (iv) same terms 83, −37, −12 in another order: equal.

(b) The terms are 93, 37 × 44 and 76. Only (iv) $37\times 44+93+76$ has the same terms. (i), (ii) and (iii) have different terms.

(a) (i) and (iv) (b) (iv)

Take 50:

1. $25+25$
2. $100-50$
3. $5\times 10$
4. $200÷4$
5. $7\times 7+1$
6. $60-2\times 5$
7. $2\times (20+5)$
8. $(100+50)÷3$
9. $8\times 6+2$
10. $1+2+3+4+5+6+7+8+9+5$

E.g. 50 = 25 + 25 = 5 × 10 = 200 ÷ 4 = 7 × 7 + 1 = 2 × (20 + 5) = (100 + 50) ÷ 3 = 8 × 6 + 2 …

*A computer search shows that 11, 13, 14, 18 and 19 cannot be made with only +, −, ×, ÷, brackets and joined 4s (like 44). For these we need a little extra: $\sqrt{4}=2$ or the decimal $0.4$.

See the table; e.g. 1 = 44 ÷ 44, 7 = 44 ÷ 4 − 4, 20 = 4 × (4 + 4 ÷ 4). 11, 13, 14, 18, 19 need √4 or 0.4.

Every whole number from −10 to 10 can be made. One way for each:

All 21 values from −10 to 10 are possible (see the table).

$0+1+2+3+4+5+6+7+8\times 9=28+72=$ 100

Another: $5\times (4+6)+9\times 8-7\times 3-2+1+0=50+72-21-2+1=100$

0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 × 9 = 100