Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 6 · Chapter companion part
Translate numbers, transfers, ages, digits, fractions, rectangles and prices into two equations, then interpret and verify the result.
Class 8 English Version: part C of the chapter 6 companion. Eight teaching sections, 30 written tasks and a notebook challenge. Selected companion coverage, not complete textbook coverage; use all four parts. Numeric checks cover only the requested value; full equations, graphs, conditions and explanations need guardian or teacher review.
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If two numbers total 84 and the larger exceeds the smaller by 18, let x be the larger and y the smaller. Write x+y=84 and x−y=18. Adding gives x=51, then y=33. Twice a number means 2x; twice the same number and three times that same number still use one variable, giving 5x, not a new independent variable.
For two shares in the ratio 3:2 with total Tk 250, use x=3k,y=2k and 5k=250. Thus k=50 and the shares are Tk 150,Tk 100. Equivalently use x+y=250 and 2x=3y. For a ratio of ages, compare the ages at the same time. A ratio is not a difference in years or taka.
Let A and B start with x and y counters. If A gives 4 to B, the new counts are x−4 and y+4. The condition B then has twice A becomes y+4=2(x−4). In a separate alternative where B gives 2 to A, equal counts give x+2=y−2. Solving gives x=16,y=20. These are separate scenarios from the same starting counts, not two transfers performed in sequence.
Let present ages be F and C. Five years ago a 3:1 ratio means F−5=3(C−5); fifteen years from now a 2:1 ratio means F+15=2(C+15). Solve to obtain C=25,F=65. Their difference stays 40 years; their ratio changes. Past ages in a stated ratio must be meaningful and its denominator nonzero.
If a is the tens digit and b the units digit, the number is 10a+b, with 1≤a≤9 and 0≤b≤9. Its reversed numeric value is 10b+a. A difference of 27 gives a−b=3. If the digit sum is 11 and the tens digit is larger, a=7,b=4 and the number is 74. Without specifying which digit is larger, an absolute digit difference can permit reversed answers; the source exercise 6.2(11) needs that distinction.
Let n/d be a written fraction. If adding 8 to its numerator gives 2 and subtracting 2 from its denominator gives 1, write (n+8)/d=2 and n/(d−2)=1, with d≠0,2. These imply n−2d=−8 and n−d=−2, so d=6,n=4. The requested written fraction is 4/6. Its value is 2/3, but changing the numerator of 2/3 by 8 is a different operation.
For a rectangle of perimeter 68 m whose length is 10 m greater than width, write 2L+2W=68 and L−W=10. Thus L+W=34, L=22 m,W=12 m. Check 2(22+12)=68. For a second condition L=1.5W or L=2W+5, translate that full relationship; perimeter is not area.
Suppose 3 notebooks and 2 pencils cost Tk 76, while 2 notebooks and 5 pencils cost Tk 69. Let n,p be the unit prices in taka: 3n+2p=76 and 2n+5p=69. Eliminate to find n=22,p=5. Verify both basket totals. A price is not a basket total, and count constraints may exclude an algebraic result that is negative or not a whole number.
For the two basket equations, 5(3n+2p)=380 and 2(2n+5p)=138. Subtraction yields n=22; substitution yields p=5. The checks concern fictional unit prices in taka.
A digit difference without naming its direction can describe reversed numbers. Preserve and explain that ambiguity rather than inventing an unstated condition.
A fraction value can have many equivalent forms. An instruction to change its numerator or denominator needs the original written terms.
Keep definitions, time and units consistent.
| Situation | First condition | Second condition |
|---|---|---|
| Total 84, difference 18 | x+y=84 | x−y=18, x≥y |
| Tk 250, ratio 3:2 | x+y=250 | 2x=3y |
| Rectangle P=68 m, extra length 10 m | 2L+2W=68 | L−W=10 |
| Three notebooks/two pencils; two/five | 3n+2p=76 | 2n+5p=69 |
These are invented classroom quantities, not real prices or measurements. Each situation is solved using its own two unknowns.
Both scenarios start from x counters for A and y for B.
A gives 4: y+4=2(x−4) B gives 2: x+2=y−2 y=x+4; x=16; y=20
Performing the two alternatives successively changes the story and creates the wrong equations.
Let F,C be the present ages in years.
F−5=3(C−5); F+15=2(C+15) F−3C=−10; F−2C=15 C=25; F=65
The age difference stays 40 years; the ratio does not stay fixed.
Changes apply to the original numerator n and denominator d.
| Condition/check | Original n=4,d=6 | Reduced terms n=2,d=3 |
|---|---|---|
| Original value | 4/6=2/3 | 2/3 |
| Add 8 to numerator | 12/6=2 | 10/3≠2 |
| Subtract 2 from denominator | 4/4=1 | 2/1≠1 |
Equal fraction values do not make their written terms interchangeable in a problem that changes those terms. Original d must avoid 0 and 2.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original counter-count model. Cards organise values rather than represent areas or actual possessions.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original proportional rectangle, not a surveyed site. Metre labels give the mathematical dimensions.
Show your working on paper. Check your answer, then choose Next.
51
Adding x+y=84 and x−y=18 gives 2x=102.
33
84−51=33, and 51−33=18.
150
Five equal ratio parts give Tk 50 each; three parts are Tk 150.
60
Six parts total 360, so one part is 60.
16
The second gives y=x+4; substitute into the first to obtain x=16.
20
y=x+4=20; the original total is 36.
25
F−3C=−10 and F−2C=15 give C=25.
65
F=2×25+15=65; past ages 60,20 have ratio 3:1.
74
a+b=11 and a−b=3 give a=7,b=4.
3
Divide by 9; a and b are digits in their defined places.
6
Subtract n−2d=−8 from n−d=−2 to get d=6.
4
n=d−2=4. Verify 12/6=2 and 4/4=1.
12
L+W=34 and L−W=10 give 2W=24.
22
Five times the first minus twice the second gives 11n=242. This is one notebook’s price, not a basket total.
5
3×22+2p=76 gives p=5; 44+25=69.
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Let x,y be the first and second numbers. Use 3x+2y=34,2x−y=4. Then y=2x−4; 7x=42, x=6,y=8. Check 18+16=34 and 12−8=4.
Translate the order of subtraction explicitly.
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Let x,y be the shares. x+y=420 and 3x=4y give y=180,x=240. Both shares sum to 420 and their ratio reduces to 4:3.
Give units and verify both conditions.
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Starting at 16,20, the first alternative gives 12,24 and B has twice A. The second alternative gives 18,18. Each alternative preserves total 36 and starts from the original counts.
Do not apply the alternatives successively.
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F−5=3(C−5) gives F−3C=−10; F+15=2(C+15) gives F−2C=15. Hence C=25,F=65. Past ages 60,20 have ratio 3:1; future ages 80,40 have ratio 2:1.
Shift both people and check the same time in each ratio.
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Let F=4C and (F−4)+(C−4)=52. Then F+C=60, C=12,F=48. Four years ago the ages were 8 and 44, totalling 52.
Do not subtract four from the combined age only once.
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a=7,b=4, so the number is 74 and the reversed value 47. Their difference is 27=9×3.
Use 10a+b, not a+b, for the number itself.
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The digit values are 7 and 3, giving both 73 and 37 as two-digit numbers. Add a condition such as tens digit exceeds units digit to select 73. Source exercise 6.2(11) needs this ordering clarification.
Do not automatically mark one reversal wrong when both satisfy the wording.
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10a+b−18=10b+a gives a−b=2. Also a+b+7=3a gives 2a−b=7. Subtract to get a=5,b=3, number 53. Check 53−18=35 and 5+3+7=15.
State the place definitions and check the digit range.
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The original terms are n=4,d=6, so the written fraction is 4/6, with value 2/3. The operations attach to the original numerator and denominator. Applying +8 to the reduced numerator 2 gives 10/3, not 2.
Keep d≠0,2; distinguish a representation from its reduced value.
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n+d=16,d−n=6 give d=11,n=5. The proper fraction is 5/11; both original conditions hold.
Clarify which difference is positive instead of assuming n−d=6.
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L+W=50 and L=1.5W give 2.5W=50, W=20 m,L=30 m. The perimeter is 2(30+20)=100 m.
Differentiate dimensions, perimeter and area.
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L+W=50,L=2W+5. Thus 3W+5=50,W=15 m,L=35 m. Check perimeter 100 m and 35=2×15+5.
A labelled graph can also show the intersection (35,15).
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Multiply by 5 and 2: 15n+10p=380 and 4n+10p=138. Subtract: 11n=242, n=22,p=5. Verify 66+10=76 and 44+25=69.
The model prices are invented classroom values, not current market prices.
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Whole notebooks require non-negative integer counts. Recheck translation, quantities and arithmetic. If the equations were copied correctly and still give 2.5, the stated whole-item situation has no feasible solution as described.
Distinguish a mathematical solution from a feasible count.
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Both phrases name the same quantity t, so 3t+2t=85. Thus t=17. Writing independent x and y would discard the fact that the number is the same.
Translate identity of quantities, not just the number of phrases.
Make four original two-condition stories: one transfer, one age, one digit and one dimension or basket problem.
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This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 6, printed pages 101–120. It does not replace the complete chapter or its practical teaching.
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