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ICSE Class 9 Master Editorial Notes

Problems on Simultaneous Linear Equations

Complete ICSE Class 9 Word Problems (M.L. Aggarwal Reference) + Exhaustive 100/100 Examination Kit

Topic 1: Two-Digit Number & Digit Problems

Place Value Formulation
Topic 1: Two-Digit Number Practice Kit
Problem 1.1 (Four Times Sum & Reversed Digits) 3 MARKS / ICSE

A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Step-by-Step Solution:
1. $10x + y = 4(x + y) \implies 6x = 3y \implies y = 2x$.
2. $(10x + y) + 18 = 10y + x \implies 9y - 9x = 18 \implies y - x = 2$.
3. Substitute $y = 2x \implies 2x - x = 2 \implies \mathbf{x = 2}, \mathbf{y = 4}$. $$\text{Original Number} = 10(2) + 4 = \mathbf{24}$$
x=2, y=4 Number = 24
Fig 6.1: Digit Solution 1
Problem 1.2 (Seven Times Number Equals Four Times Reversed) 4 MARKS / ICSE

Seven times a two-digit number is equal to 4 times the number obtained by reversing the order of its digits. If the difference between the digits is 3, find the number.

Step-by-Step Solution:
1. $7(10x + y) = 4(10y + x) \implies 70x + 7y = 40y + 4x \implies 66x = 33y \implies y = 2x$.
2. Digits differ by 3 $\implies y - x = 3 \implies 2x - x = 3 \implies \mathbf{x = 3}, \mathbf{y = 6}$. $$\text{Original Number} = 10(3) + 6 = \mathbf{36}$$
x=3, y=6 Number = 36
Fig 6.2: Digit Solution 2

Topic 2: Fraction Problems (Numerator & Denominator Algebra)

Fraction Formulation Rules
Topic 2: Fraction Problems Practice Kit
Problem 2.1 (Numerator & Denominator Alterations) 3 MARKS / ICSE

A fraction becomes $\frac{1}{3}$ when 1 is subtracted from numerator, and $\frac{1}{4}$ when 8 is added to denominator. Find the fraction.

Step-by-Step Solution:
1. $\frac{x - 1}{y} = \frac{1}{3} \implies 3x - y = 3$.
2. $\frac{x}{y + 8} = \frac{1}{4} \implies 4x - y = 8$.
3. Subtract: $x = 5 \implies y = 12$. $$\text{Fraction} = \mathbf{\frac{5}{12}}$$
5 12
Fig 6.3: Fraction 5/12
Problem 2.2 (Dual Addition Fraction) 4 MARKS / ICSE

A fraction becomes $\frac{9}{11}$ if 2 is added to both numerator and denominator. If 3 is added to both, it becomes $\frac{5}{6}$. Find the fraction.

Step-by-Step Solution:
1. $\frac{x+2}{y+2} = \frac{9}{11} \implies 11x + 22 = 9y + 18 \implies 11x - 9y = -4$.
2. $\frac{x+3}{y+3} = \frac{5}{6} \implies 6x + 18 = 5y + 15 \implies 6x - 5y = -3$.
3. Solve system: Multiply 1st by 5, 2nd by 9 $\implies \mathbf{x = 7}, \mathbf{y = 9}$. $$\text{Fraction} = \mathbf{\frac{7}{9}}$$
7 9
Fig 6.4: Fraction 7/9

Topic 3: Age Related Word Problems

Topic 3: Age Related Word Problems Practice Kit
Problem 3.1 (Past & Future Comparison) 4 MARKS / ICSE

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. Find their present ages.

Step-by-Step Solution:
1. $(x - 5) = 3(y - 5) \implies x - 3y = -10$.
2. $(x + 10) = 2(y + 10) \implies x - 2y = 10$.
3. Subtract: $y = 20\text{ yrs} \implies x = 50\text{ yrs}$. $$\text{Nuri} = \mathbf{50\text{ years}}, \qquad \text{Sonu} = \mathbf{20\text{ years}}$$
Nuri = 50 yrs Sonu = 20 yrs
Fig 6.5: Age Solution 1
Problem 3.2 (Father & Two Children Age Sum) 4 MARKS / ICSE

The age of a father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his two children. Find the age of the father.

Step-by-Step Solution:
Let Father's age $= x$, sum of 2 children's ages $= y$.
1. $x = 2y \implies y = x/2$.
2. After 20 years: Father becomes $(x + 20)$, each child gains 20 years $\implies$ sum becomes $(y + 40)$.
3. $(x + 20) = (y + 40) \implies x - y = 20 \implies x - x/2 = 20 \implies x/2 = 20 \implies \mathbf{x = 40}$. $$\text{Father's Age} = \mathbf{40\text{ years}}$$
2 Children Sum Father = 40 yrs
Fig 6.6: Father Age

Topic 4: Speed, Distance, Time & Upstream/Downstream Boats

Upstream & Downstream Master Formula
Topic 4: Speed & Boat Motion Practice Kit
Problem 4.1 (Classic River Boat Upstream/Downstream) 5 MARKS / ICSE

A boat goes $30\text{ km}$ upstream and $44\text{ km}$ downstream in $10\text{ hours}$. In $13\text{ hours}$, it can go $40\text{ km}$ upstream and $55\text{ km}$ downstream. Find speed of stream and boat in still water.

Step-by-Step Solution:
Let $u = 1/(x-y), v = 1/(x+y) \implies 30u + 44v = 10$ and $40u + 55v = 13$.
Solve system: $u = 1/5 \implies x - y = 5$, and $v = 1/11 \implies x + y = 11$.
Add: $2x = 16 \implies \mathbf{x = 8\text{ km/h}}, \mathbf{y = 3\text{ km/h}}$. $$\text{Boat} = \mathbf{8\text{ km/h}}, \qquad \text{Stream} = \mathbf{3\text{ km/h}}$$
Boat & Stream 8 km/h & 3 km/h
Fig 6.7: River Boat Result
Problem 4.2 (Highway Two Cars Meeting) 4 MARKS / ICSE

Places A and B are $100\text{ km}$ apart on a highway. One car starts from A and another from B. If they travel in same direction, they meet in 5 hours. If towards each other, they meet in 1 hour. Find speeds.

Step-by-Step Solution:
1. Same direction: $5(x - y) = 100 \implies x - y = 20$.
2. Opposite direction: $1(x + y) = 100 \implies x + y = 100$.
3. Add: $2x = 120 \implies \mathbf{x = 60\text{ km/h}}, \mathbf{y = 40\text{ km/h}}$. $$\text{Car A} = \mathbf{60\text{ km/h}}, \qquad \text{Car B} = \mathbf{40\text{ km/h}}$$
Highway Cars 60 km/h & 40 km/h
Fig 6.8: Two Cars Speed

Topic 5: Commercial Charges, Work-Time & Cost Problems

Topic 5: Commercial & Work-Time Practice Kit
Problem 5.1 (City Taxi Fixed Charge) 3 MARKS / ICSE

A taxi charge in a city consists of a fixed charge together with distance charge. For $10\text{ km}$, charge is Rs. $105$; for $15\text{ km}$, charge is Rs. $155$. Find fixed charge and rate per km.

Step-by-Step Solution:
1. $x + 10y = 105$ and $x + 15y = 155$.
2. Subtract: $5y = 50 \implies \mathbf{y = \text{Rs. } 10/\text{km}}$.
3. Substitute: $x + 100 = 105 \implies \mathbf{x = \text{Rs. } 5}$. $$\text{Fixed} = \mathbf{\text{Rs. } 5}, \qquad \text{Rate/km} = \mathbf{\text{Rs. } 10}$$
Taxi Rates Rs. 5 & Rs. 10/km
Fig 6.9: Taxi Charges
Problem 5.2 (Work & Time: Women & Men) 4 MARKS / ICSE

2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find time taken by 1 woman alone and 1 man alone.

Step-by-Step Solution:
Let 1 woman take $x$ days, 1 man take $y$ days.
1. $\frac{2}{x} + \frac{5}{y} = \frac{1}{4} \implies 2u + 5v = 1/4$.
2. $\frac{3}{x} + \frac{6}{y} = \frac{1}{3} \implies 3u + 6v = 1/3$.
3. Solve system: $u = 1/18 \implies \mathbf{x = 18\text{ days}}, v = 1/36 \implies \mathbf{y = 36\text{ days}}$. $$\text{Woman} = \mathbf{18\text{ days}}, \qquad \text{Man} = \mathbf{36\text{ days}}$$
Work & Time 18 days & 36 days
Fig 6.10: Work & Time

6. ICSE Board Verbal Translation Master Guide

Verbal to Algebraic Translation Reference
ICSE English Problem Statement Algebraic Formulation
A 2-digit number with tens digit $x$ & units digit $y$ $10x + y$
Reversed 2-digit number $10y + x$
Fraction numerator $x$ & denominator $y$ $x/y$
$n$ years ago (past age) $x - n$
$m$ years hence / after $m$ years (future age) $x + m$
Boat Upstream Speed (still speed $x$, stream $y$) $x - y$
Boat Downstream Speed (still speed $x$, stream $y$) $x + y$
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