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Chapter 2: Arithmetic Expressions Class 8th Mathematics (Ganita Prakash) NCERT Solution

July 11, 2026  •  Admin ETeam  •  Class 7, Class 7 NCERT Solution
Arithmetic Expressions — Full Step-wise Solutions | EduGrown
Ganita Prakash · Grade 7 · Chapter 2

Arithmetic Expressions — Full Step-wise Solutions

Every in-text question, Example, Math Talk and Figure-it-Out from the chapter — worked out step by step with LaTeX, terms marked out like the textbook, and diagrams for the story problems.

Terms = parts split by + Commutative: a+b=b+a Associative: (a+b)+c=a+(b+c) Distributive: a×(b+c)=a×b+a×c
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📘 In-text Questions ✍️ Exercise (Figure It Out) 🧩 Expression Engineer
2.1

Simple Expressions

An arithmetic expression (like \(13+2\)) has a value it evaluates to. We can compare two expressions the same way we compare numbers, using \(=\), \(<\), \(>\).
Q
Choose your favourite number and write as many expressions as you can having that value.
Solution Let’s choose 24. Some expressions with value 24:
  1. \(12 + 12 = 24\)
  2. \(4 \times 6 = 24\)
  3. \(48 \div 2 = 24\)
  4. \(34 – 10 = 24\)
  5. \(20 + 4 = 24\)
Try this with your own favourite number!
Ex.2
Which is greater? \(1023+125\) or \(1022+128\)? (Marble story — Raja vs Joy)
Raja: 1023 + 125
1022125
Joy: 1022 + 128
1022125
Raja starts 1 marble ahead, but Joy gains 3 more today — net, Joy ends up 2 marbles ahead.
Step-wise Solution
  1. Raja started with 1 more marble than Joy (1023 vs 1022).
  2. Today Joy received 128, which is 3 more than Raja’s 125.
  3. Net effect: Joy gains \(3 – 1 = 2\) more marbles than Raja overall.
\(1023 + 125 < 1022 + 128\) — Joy has more, without computing either sum fully.
Ex.3
Which is greater? \(113-25\) or \(112-24\)? (Marble story — losing marbles)
Step-wise Solution
  1. Raja had 1 marble more than Joy to start (113 vs 112).
  2. Raja also lost 1 marble more than Joy (25 vs 24).
  3. The extra marble Raja had is exactly cancelled by the extra marble he lost.
\(113 – 25 = 112 – 24\) — they end up equal.
FQ1
Fill in the blanks to make both sides equal: (a) 13+4 = ___+6 (b) 22+___ = 6×5 (c) 8×___ = 64÷2 (d) 34−___ = 25
Step-wise Solution
  1. (a) \(13+4=17\), so \(17 = \_\_ + 6 \Rightarrow \_\_=11\)
  2. (b) \(6\times5=30\), so \(22+\_\_=30 \Rightarrow \_\_=8\)
  3. (c) \(64\div2=32\), so \(8\times\_\_=32 \Rightarrow \_\_=4\)
  4. (d) \(34-\_\_=25 \Rightarrow \_\_=9\)
(a) 11   (b) 8   (c) 4   (d) 9
FQ2
Arrange in ascending order of value: (a) 67−19 (b) 67−20 (c) 35+25 (d) 5×11 (e) 120÷3
Step-wise Solution
  1. (a) \(67-19=48\)
  2. (b) \(67-20=47\)
  3. (c) \(35+25=60\)
  4. (d) \(5\times11=55\)
  5. (e) \(120\div3=40\)
Ascending order: 120÷3 (40) < 67−20 (47) < 67−19 (48) < 5×11 (55) < 35+25 (60)
Q
Use ‘>’, ‘<‘ or ‘=’ without complicated calculations: (a) 245+289 __ 246+285 (b) 273−145 __ 272−144 (c) 364+587 __ 363+589 (d) 124+245 __ 129+245 (e) 213−77 __ 214−76
Step-wise Reasoning
  1. (a) LHS’s first number is 1 less (245 vs 246) but second is 4 more (289 vs 285): net +3 → LHS is bigger. >
  2. (b) LHS: 273 is 1 more, but subtracting 145 which is 1 more than 144 — the two changes cancel. =
  3. (c) LHS: 364 is 1 more than 363, but 587 is 2 less than 589 — net −1. <
  4. (d) Same second term (245) both sides; compare first terms only: 124 < 129. <
  5. (e) LHS: 213 is 1 less than 214, and subtracting 77 (which is 1 more than 76) — both effects reduce LHS further. <
(a) >   (b) =   (c) <   (d) <   (e) <
2.2

Reading and Evaluating Complex Expressions

Without brackets, an expression like \(30+5\times4\) can be misread. Brackets and the idea of terms (parts split by a ‘+’ sign) remove the confusion.
Ex.4/5
Brackets in Expressions: \(30+(5\times4)\), and Irfan’s change: \(100-(15+56)\)
Step-wise Solution
  1. Marbles: \(30+(5\times4)=30+20=\mathbf{50}\)
  2. Irfan’s change: total spent \(=15+56=71\); change \(=100-(15+56)=100-71=\mathbf{29}\)
Brackets tell us exactly which operation to do first — removing ambiguity.
Q
Check that replacing subtraction by addition of the inverse does not change the value — try examples.
Step-wise Solution Take 18 and 10: \(18-10=8\) and \(18+(-10)=8\) — same value. ✓
Subtracting a number always equals adding its (negative) inverse.
FQ
Complete the table — write each expression as a sum of terms, then list its terms.
Step-wise Solution
13 − 2 + 6 = 13+−2+6 → Terms: 13, −2, 6
5 + 6×3 = 5+6×3 → Terms: 5, 6×3
4 + 15 − 9 = 4+15+−9 → Terms: 4, 15, −9
23 − 2×4 + 16 = 23+−2×4+16 → Terms: 23, −2×4, 16
28 + 19 − 8 = 28+19+−8 → Terms: 28, 19, −8
Q
Does changing the order in which terms are added give different values? (3-term example: (−7)+10+(−11))
Step-wise Solution
  1. Group first two: \(((-7)+10)+(-11) = 3 + (-11) = -8\)
  2. Group last two: \((-7)+(10+(-11)) = (-7)+(-1) = -8\)
Same result both ways: −8. Grouping/order doesn’t matter.
Q
Manasa added a long list and got 11,749, but forgot the fourth number 9,055. Does she have to start over?
Step-wise Solution
  1. By the associative property, she can simply add the missing number to her existing total.
  2. \(11{,}749 + 9{,}055 = 20{,}804\)
No — just add 9,055 to 11,749 to get 20,804. No need to restart!
Ex.7
4 dosas at ₹23 each + ₹5 tip. If friends increase to 7 (tip stays ₹5), what’s the new expression and total?
Step-wise Solution
  1. Expression: \(7 \times 23 + 5\)
  2. \(7\times23 = 161\)
  3. \(161+5 = 166\)
Total = ₹166. Terms: 7×23 and 5.
Ex.8
“Fire in the mountain” game: 33 students, teacher calls ‘5’, Ruby writes \(6\times5+3\). What would Ruby write if teacher calls ‘4’ or ‘7’?
Step-wise Solution
  1. For group size 4: \(33 \div 4 = 8\) remainder \(1\) → \(8\times4+1\)
  2. For group size 7: \(33 \div 7 = 4\) remainder \(5\) → \(4\times7+5\)
Group of 4: \(8\times4+1\) (Terms: 8×4, 1)  |  Group of 7: \(4\times7+5\) (Terms: 4×7, 5)
Ex.10
Kannan pays ₹432 using coins/notes. Identify terms in the two given expressions, and find more ways.
Step-wise Solution
  1. Way 1: \(432 = 4\times100 + 1\times20 + 1\times10 + 2\times1\) — Terms: 4×100, 1×20, 1×10, 2×1
  2. Way 2: \(432 = 8\times50 + 1\times10 + 4\times5 + 2\times1\) — Terms: 8×50, 1×10, 4×5, 2×1
  3. Another way: \(432 = 40\times10 + 3\times10 + 2\times1\) — Terms: 40×10, 3×10, 2×1
Many combinations of notes/coins can sum to ₹432 — try building your own!
Ex.11
Which arrangement matches \(5\times2+3\)? What’s the expression for the right-hand arrangement (yellow/blue squares)?
Step-wise Solution
  1. \(5\times2+3 = 10+3=13\) — this is “3 more than 5×2” → matches the left arrangement (5 pairs + 3 extra).
  2. Right arrangement (2 identical columns of 5+3 each) needs brackets: \(2\times(5+3)\)
  3. This equals \(5+3+5+3\), or equivalently \(5\times2+3\times2\)
Right arrangement’s expression: \(2\times(5+3) = 16\)
FQ1
Find the values, writing the terms in each case: (a) 28−7+8 (b) 39−2×6+11 (c) 40−10+10+10 (d) 48−10×2+16÷2 (e) 6×3−4×8×5
Step-wise Solution
  1. (a) Terms: 28, −7, 8 → \(28-7+8 = \mathbf{29}\)
  2. (b) Terms: 39, −2×6, 11 → \(39-12+11 = \mathbf{38}\)
  3. (c) Terms: 40, −10, 10, 10 → \(40-10+10+10 = \mathbf{50}\)
  4. (d) Terms: 48, −10×2, 16÷2 → \(48-20+8 = \mathbf{36}\)
  5. (e) Terms: 6×3, −4×8×5 → \(18-160 = \mathbf{-142}\)
FQ2
Write a story/situation for each expression and find its value: (a) 89+21−10 (b) 5×12−6 (c) 4×9+2×6
Step-wise Solution
  1. (a) A library had 89 books; the librarian bought 21 more; then 10 were borrowed. \(89+21-10=\mathbf{100}\)
  2. (b) 5 packs of 12 pencils each, but 6 pencils were defective and removed. \(5\times12-6=60-6=\mathbf{54}\)
  3. (c) 4 tables with 9 chairs each, plus 2 benches seating 6 each. \(4\times9+2\times6=36+12=\mathbf{48}\)
FQ3
(a) Princesses’ gold coins (b) Metro tickets (c) Window height — write expressions and evaluate.
Border  3 cm
Grill    2 cm
Gap     5 cm
(pattern repeats: Grill, Gap × several rows)
Window measurements — border, grill bars, and gaps between them
Step-wise Solution
  1. (a) Elsa doubled her 100 coins: \(2\times100\). Anna kept half of her 100: \(\dfrac{100}{2}\).
    Expression: \(2\times100 + \dfrac{100}{2}\). Terms: 2×100, 100/2.
    Value \(= 200+50=\mathbf{250}\) gold coins.
  2. (b)(i) 4 adults + 3 children: \(4\times40+3\times20\). Terms: 4×40, 3×20.
    \(=160+60=\mathbf{₹220}\)
  3. (b)(ii) Two groups of 3 adults each: \(2\times(3\times40)\).
    \(=2\times120=\mathbf{₹240}\)
  4. (c) Window: Expression \(=7\times5+6\times2+2\times3\). Terms: 7×5, 6×2, 2×3.
    \(=35+12+6=\mathbf{53\ cm}\)
Ex.12–14
Removing Brackets — I: \(200-(40+3)\), \(500-(250-100)\), and Hira’s coins \(28+(35-10)\)
Step-wise Solution
  1. \(200-(40+3) = 200-40-3 = \mathbf{157}\) — bracket preceded by ‘−’ → both signs inside flip.
  2. \(500-(250-100) = 500-250+100 = \mathbf{350}\) — the ‘−100’ inside flips to ‘+100’.
  3. \(28+(35-10) = 28+35-10 = \mathbf{53}\) — bracket preceded by ‘+’ → signs inside stay the same.
Rule: a ‘−’ before a bracket flips every sign inside; a ‘+’ before a bracket keeps them unchanged.
Tinker
Complete the “Tinker the Terms I” grid — reasoning about how small changes affect the sum.
Step-wise Solution
  1. Base: \(53+(-16)=37\)
  2. \(54+(-16)=38\) (54 is 1 more than 53 → sum is 1 more)
  3. \(52+(-16)=36\) (52 is 1 less than 53 → sum is 1 less)
  4. \(53+(-15)=38\) (−15 is 1 more than −16 → sum is 1 more)
  5. \(53+(-17)=36\) (−17 is 1 less than −16 → sum is 1 less)
  6. \((-87)+(-16)=\mathbf{-103}\)
  7. \((-88)+(-15)=\mathbf{-103}\) (−88 is 1 less than −87, −15 is 1 more than −16 — changes cancel)
  8. \((-86)+(-18)=\mathbf{-104}\) (−86 is 1 more than −87, but −18 is 2 less than −16 — net 1 less than −103)
  9. \((-97)+(-26)=\mathbf{-123}\) (−97 is 10 less than −87, −26 is 10 less than −16 — total 20 less than −103)
FQ1
Fill blanks with numbers/operation signs so both sides are equal (a)–(f).
Step-wise Solution
  1. (a) \(24+(6-4)=24+6-4\) → box = −, blank = 4
  2. (b) \(38+(9-4)=38+9-4\) → blanks = 9, −, 4
  3. (c) \(24-(6+4)=24-6-4\) → box = −
  4. (d) \(24-6-4=24-6-4\) → box = −, blank = 4
  5. (e) \(27-(8+3)=27-8-3\) → blanks = −, −
  6. (f) \(27-(8-3)=27-8+3\) → blanks = 8, −, 3
FQ2
Remove the brackets: (a) 14+(12+10) (b) 14−(12+10) (c) 14+(12−10) (d) 14−(12−10) (e) −14+(12−10) (f) 14−(−12−10)
Step-wise Solution
  1. (a) \(14+12+10\)
  2. (b) \(14-12-10\)
  3. (c) \(14+12-10\)
  4. (d) \(14-12+10\)
  5. (e) \(-14+12-10\)
  6. (f) \(14+12+10\)
Note: (a) and (f) give the same expression (both equal 36) — a ‘−’ before ‘(−12−10)’ flips both negatives to positive.
FQ3
Find values; guess if equal first: (a) (6+10)−2 & 6+(10−2) (b) 16−(8−3) & (16−8)−3 (c) 27−(18+4) & 27+(−18−4)
Step-wise Solution
  1. (a) \((6+10)-2=14\); \(6+(10-2)=6+8=14\) → Equal (both = 14)
  2. (b) \(16-(8-3)=16-5=11\); \((16-8)-3=8-3=5\) → Not equal (11 ≠ 5)
  3. (c) \(27-(18+4)=27-22=5\); \(27+(-18-4)=27-22=5\) → Equal (both = 5)
Two expressions are equal exactly when removing brackets gives the identical set of terms.
FQ4
Identify which expressions in each set have the same value (without evaluating — use terms).
Step-wise Solution
  1. (a) 319+537, 319−537, −537+319, 537−319 → 319−537 and −537+319 share the same terms (319, −537), so they’re equal.
  2. (b) Among 87+46−109 (×3 identical), 87−46+109, 87−(46+109), (87−46)+109 → the three identical copies of 87+46−109 are equal to each other; separately, 87−46+109 and (87−46)+109 share terms (87,−46,109 rearranged) and are equal to each other.
FQ5
Add brackets so the expressions equal the given values: (a) 34−9+12=13 (b) 56−14−8=34 (c) −22−12+10+22=−22
Step-wise Solution
  1. (a) \(34-(9+12)=34-21=\mathbf{13}\) ✓
  2. (b) \((56-14)-8=42-8=\mathbf{34}\) ✓
  3. (c) \(-22-(12+10)+22 = -22-22+22=\mathbf{-22}\) ✓
FQ6
Using reasoning only, fill blanks: (a) 423+___=419+___ (b) 207−68=210−___
Step-wise Solution
  1. (a) 419 is 4 less than 423, so the other term must be 4 more to balance: \(423+419=419+\mathbf{423}\)
  2. (b) 210 is 3 more than 207, so we must subtract 3 more too: \(207-68=210-\mathbf{71}\)
FQ7
Using 2, 3, 5 with ‘+’, ‘−’ and brackets, generate as many different values as possible.
Step-wise Solution
  1. \(3-2-5=-4\)
  2. \(5+2+3=10\)
  3. \(5+2-3=4\)
  4. \(2-(3+5)=-6\)
Try more combinations — many different values are reachable!
FQ8
Jasoda subtracts 9 by subtracting 10 and adding 1 (e.g. 36−9=26+1). Does this always work? Other strategies?
Step-wise Solution
  1. Since \(-9 = -10+1\): \(36-9 = 36-(10-1) = 36-10+1 = 26+1\) ✓ — Yes, always correct.
  2. Similar trick for subtracting 8: subtract 10, add 2 (since \(-8=-10+2\)). E.g., \(55-8=55-10+2=47\)
FQ9
For 73−14+1 and 73−14−1, identify equal expressions from: (a) 73−(14+1) (b) 73−(14−1) (c) 73+(−14+1) (d) 73+(−14−1)
Step-wise Solution
  1. 73−14+1 has terms {73, −14, +1}. Expression (b) 73−(14−1)=73−14+1 matches (terms {73,−14,+1}); expression (c) 73+(−14+1) also matches directly.
  2. 73−14−1 has terms {73, −14, −1}. Expression (a) 73−(14+1)=73−14−1 matches; expression (d) 73+(−14−1) also matches directly.
73−14+1 = (b) and (c)  |  73−14−1 = (a) and (d)
Ex.15/16
Removing Brackets — II: Lhamo & Norbu’s bill \(2\times(43+24)\); Scouts & Guides \(4\times5+3\times5=(4+3)\times5\)
4 rows of 5 scouts (green) + 3 rows of 5 guides (yellow)
Step-wise Solution
  1. Cutlets: Individually: \(2\times43+24\) means “24 more than double 43” — wrong for 2 shared orders.
    Correctly, together they pay double of \((43+24)\): \(2\times(43+24) = 2\times43+2\times24 = 86+48=\mathbf{134}\)
  2. Scouts/Guides: \(4\times5+3\times5 = (4+3)\times5\)
    \(=20+15=35\)   or   \(7\times5=35\) — same answer either way.
This is the distributive property: multiple × sum = sum of multiples.
Q
If Sangmu joins Lhamo & Norbu ordering the same items, what’s the new expression?
Solution\(3\times(43+24) = 3\times67 = \mathbf{201}\)
Q
Why is \(5\times4+3 \ne 5\times(4+3)\)? Is \(5\times(4+3)=5\times(3+4)=(3+4)\times5\)?
Step-wise Solution
  1. \(5\times4+3 = 20+3 = 23\)
  2. \(5\times(4+3) = 5\times7 = 35\)
  3. \(23 \ne 35\) — because without brackets, only the “5×4” is a single term; the “+3” stands alone, not multiplied.
  4. \(5\times(3+4)=5\times7=35\) and \((3+4)\times5=7\times5=35\) — yes, all three are equal (35), since addition and multiplication are both commutative.
Ex.17/18
Given \(53\times18=954\), find \(63\times18\). Also evaluate \(97\times25\) using the distributive shortcut.
Step-wise Solution
  1. \(63\times18 = (53+10)\times18 = 53\times18+10\times18 = 954+180 = \mathbf{1134}\)
  2. \(97\times25 = (100-3)\times25 = 100\times25-3\times25 = 2500-75 = \mathbf{2425}\)
Q
Use this method for: (a) 95×8 (b) 104×15 (c) 49×50. Is it quicker? What other products work like this?
Step-wise Solution
  1. (a) \(95\times8=(100-5)\times8=100\times8-5\times8=800-40=\mathbf{760}\)
  2. (b) \(104\times15=(100+4)\times15=100\times15+4\times15=1500+60=\mathbf{1560}\)
  3. (c) \(49\times50=(50-1)\times50=50\times50-50\times1=2500-50=\mathbf{2450}\)
Yes, this is much quicker than standard long multiplication! It works best whenever one factor is close to a round number like 10, 50, 100, or 1000 — e.g. \(98\times7=(100-2)\times7\), \(103\times9=(100+3)\times9\).
FQ1
Fill in the blanks with numbers and operation signs (a)–(p) — the distributive property.
Step-wise Solution
  • (a) 3×(6+7) = 3×6+3×7 ✓ given
  • (b) (8+3)×4 = 8×4+3×4 ✓ given
  • (c) 3×(5+8) = 3×5 + 3×8
  • (d) (9+2)×4 = 9×4 + 2×4
  • (e) 3×(10+4) = 3×10 + 3×4
  • (f) (13+6)×4 = 13×4 + 6×4
  • (g) 3×(5+2) = 3×5+3×2
  • (h) (2+3)×4 = 2×4+3×4
  • (i) 5×(9−2) = 5×9−5×2
  • (j) (5−2)×7 = 5×7−2×7
  • (k) 5×(8−3) = 5×8 − 5×3
  • (l) (8−3)×7 = 8×7 − 3×7
  • (m) 5×(12−3) = 5×12 − 5×3
  • (n) (15−6)×7 = 15×7 − 6×7
  • (o) 5×(9−4) = 5×9−5×4
  • (p) (17−9)×7 = 17×7−9×7
FQ2
Fill ‘<‘, ‘>’ or ‘=’ by reasoning (not evaluating): (a)–(d)
Step-wise Solution
  1. (a) \((8-3)\times29\) vs \((3-8)\times29\): since \(8-3 > 3-8\) (positive vs negative), and both multiplied by the same positive 29 → >
  2. (b) \(15+9\times18\) vs \((15+9)\times18\): LHS adds only 15 extra to \(9\times18\); RHS adds \(15\times18\) extra — RHS is much bigger → <
  3. (c) \(23\times(17-9)\) vs \(23\times17+23\times9\): LHS distributes to \(23\times17-23\times9\), which has a minus, while RHS has a plus — RHS is bigger → <
  4. (d) \((34-28)\times42\) vs \(34\times42-28\times42\): these are the same by the distributive property → =
FQ3
Find other ways (besides 2×(1+6)=14) to make 14 using ___×(___+___).
Step-wise Solution
  1. \(2\times(5+2)=2\times7=14\)
  2. \(2\times(3+4)=2\times7=14\)
  3. \(7\times(1+1)=7\times2=14\)
  4. \(2\times(6+1)=2\times7=14\)
FQ4
Find the sum of numbers in each picture (yellow squares/blue circles grid, and red/blue circle grid) in two different ways.
Step-wise Solution
  1. Picture I (3×3 grid alternating 4 and 8): Way 1: \((5\times4)+(4\times8)=20+32=\mathbf{52}\)
    Way 2: \(2\times(4+8+4)+(8+4+8)=2\times16+20=32+20=\mathbf{52}\)
  2. Picture II (4×4 grid of 5s and 6s): Way 1: \((8\times5)+(8\times6)=40+48=\mathbf{88}\)
    Way 2: \(8\times(5+6)=8\times11=\mathbf{88}\)
Exercise

Final Figure It Out — Real-world Situations

The chapter’s concluding exercise (pages 42–44) — translating word problems into expressions and comparing them.
1a
Begur market: Rahim supplies 9 kg mangoes/day, Shyam supplies 11 kg/day, all 7 days a week. Total mangoes supplied?
Step-wise Solution Expression: \(7\times(9+11)\)   \(=7\times20=\mathbf{140\ kg}\)
1b
Binu earns ₹20,000/month, spends ₹5,000 rent + ₹5,000 food + ₹2,000 other, every month. Savings by year end?
Step-wise Solution
  1. Monthly savings \(= 20{,}000 – (5000+5000+2000) = 20{,}000-12{,}000=8{,}000\)
  2. Expression: \(12\times20{,}000 – 12\times(5000+5000+2000)\)
  3. \(=2{,}40{,}000 – 1{,}44{,}000\)
Savings by year end = ₹96,000
1c
A snail climbs 3 cm by day, slips 2 cm at night. Post is 10 cm high. When does it reach the treat?
Step-wise Solution
  1. Net gain per full day-night cycle \(= 3-2 = 1\) cm
  2. After 7 full cycles: climbed 7 cm (still short of 10 cm)
  3. On the 8th day, the snail climbs 3 cm during daytime before any slip: \(7\times(3-2)+3 = 7+3=10\) cm — reaches exactly the top!
The snail reaches the treat on Day 8 (during the daytime climb, before it can slip back).
2
Melvin reads a 2-page story daily except Tuesdays & Saturdays. How many stories in 8 weeks? Which expression matches?
Step-wise Solution
  1. Reading days per week \(= 7-2 = 5\)
  2. Stories in 8 weeks \(= 5\times8 = 40\), or equivalently \(7\times8-2\times8 = 56-16=40\)
Matching expressions: (b) (7−2)×8 and (g) 7×8−2×8 — both equal 40.
3
Find different ways of evaluating: (a) 1−2+3−4+5−6+7−8+9−10 (b) 1−1+1−1+1−1+1−1+1−1
Step-wise Solution
  1. (a) Way 1 — group odds and evens: \((1+3+5+7+9)+(-2-4-6-8-10)=25-30=\mathbf{-5}\)
  2. (a) Way 2 — pair consecutively: \((1-2)+(3-4)+(5-6)+(7-8)+(9-10)=(-1)\times5=\mathbf{-5}\)
  3. (b) Way 1 — pair consecutively: \((1-1)+(1-1)+(1-1)+(1-1)+(1-1)=0+0+0+0+0=\mathbf{0}\)
  4. (b) Way 2 — group all 1’s and all −1’s: \((1+1+1+1+1)+(-1-1-1-1-1)=5-5=\mathbf{0}\)
4
Compare using ‘<‘, ‘>’ or ‘=’: (a)–(h)
Step-wise Solution
  1. (a) 49−7+8 vs 49−7+8: identical expressions → =
  2. (b) 83×42−18 vs 83×40−18: 42>40, same subtraction → >
  3. (c) 145−17×8 vs 145−17×6: subtracting more (17×8) makes LHS smaller → <
  4. (d) 23×48−35 vs 23×(48−35): distributing RHS gives 23×48−23×35, which subtracts far more than 35 → LHS >
  5. (e) (16−11)×12 vs −11×12+16×12: RHS is exactly the distributed form of LHS → =
  6. (f) (76−53)×88 vs 88×(53−76): (76−53) is positive, (53−76) is negative — opposite signs → >
  7. (g) 25×(42+16) vs 25×(43+15): both inner sums equal 58 → =
  8. (h) 36×(28−16)=36×12 vs 35×(27−15)=35×12: same second factor (12), but 36>35 → >
5
Identify (without computing) which expressions equal (a) 83−37−12 and (b) 93+37×44+76.
Step-wise Solution
  1. (a) Terms of 83−37−12 are {83,−37,−12}. (i) 84−38−12 changes both 83→84 and 37→38 by +1 each, which cancels → equal. (iv) −37+83−12 is the same terms reordered → equal.
  2. (b) Terms of 93+37×44+76 are {93, 37×44, 76}. Only (iv) 37×44+93+76 has the exact same terms reordered → equal.
(a) equals (i) and (iv)  |  (b) equals (iv) only
6
Choose a number and create ten different expressions having that value.
Step-wise Solution Let’s choose 26:
  1. \(10+16=26\)
  2. \(30-4=26\)
  3. \(13\times2=26\)
  4. \(\dfrac{52}{2}=26\)
  5. \(5+(3\times7)=26\)
  6. \((5\times5)+1=26\)
  7. \(20+6=26\)
  8. \(50-24=26\)
  9. \(2\times13=26\)
  10. \(27-1=26\)
Try this exercise with your own favourite number!
Puzzle Time

Expression Engineer!

Using a fixed set of digits, the four operations, and brackets, build expressions that reach as many target values as possible.
Q
Using four 4’s, create expressions to get all values from 1 to 20.
Step-wise Solution (sample values)
  • 1: \(4\div4+4-4=1\)
  • 2: \(4\div4+4\div4=2\)
  • 3: \((4+4+4)\div4=3\)
  • 4: \(4+4\times(4-4)=4\)
  • 5: \((4\times4+4)\div4=5\)
  • 6: \(4+(4+4)\div4=6\)
  • 8: \(4+4+4-4=8\)
  • 10: \((44-4)\div4=10\)
  • 12: \(4\times4-4\div4… \) → \(4+4+4\div4\times… \) try \(4\times4-4+4\div4\)? Best: \(4+4+4\times1=12\) using \(4\div4=1\): \(4+4+4\times(4\div4)=12\)
  • 16: \(4\times4\times(4\div4)=16\)
  • 20: \(4\times4+4\div4\times… \) → \(4\times(4+4\div4)=20\)
  • 0: \(4-4+4-4=0\)
  • Keep exploring the rest (7, 9, 11, 13–15, 17–19)!
This is an open-ended explore-and-check puzzle — the key trick is that \(4\div4=1\) and \(4-4=0\) let you “build” any small number.
Q
Using 1, 2, 3, 4, 5 exactly once in any order, get as many values as possible between −10 and +10.
Step-wise Solution (sample values)
  • \(1-2-3-4+5=-3\)
  • \(1+2+3-4-5=-3\) (another way to −3)
  • \(1-2+3+4-5=1\)
  • \(-1+2-3+4+5=7\)
  • \(1+2-3+4+5=9\)
  • \(1+2+3+4-5=5\)
Try rearranging signs and order systematically to see how many distinct values in [−10, 10] you can reach!
Q
Using digits 0 to 9 exactly once, make an expression with value 100.
Step-wise Solution One classic solution: \(70+24+3+9-6+1-8+5+2 \) — needs verification of digit use. A cleaner known solution:
\(1+2+3+4+5+6+7+8\times9=1+2+3+4+5+6+7+72=100\) ✓ (uses each digit 1–9 once; if 0 must also be used, append \(+0\): \(1+2+3+4+5+6+7+8\times9+0=100\))
EduGrown · Ganita Prakash Grade 7 · Chapter 2 — Arithmetic Expressions · Full step-wise solutions

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