← All topics💰 Change, Income & SavingsMaths
Ratio & Proportion · Topic 6 of 6

💰 Change, Income & Savings

2 exam question types, fully solved

Here a ratio is deliberately disturbed — the same number is added to each term, or income and expenditure are pushed up and down by percentages. Give every part one shared multiplier and the disturbance turns into a single equation.

Re-forming a ratio after a change

• To change a ratio by adding the same number to both terms, build the new ratio , set it equal to the target and cross-multiply for . (To subtract, use and ; a negative answer just means you add instead — report the size.)

• The “let the common multiple be ” technique: whenever a quantity is “in the ratio ”, rewrite the parts as and so one unknown controls everything, then form one equation from the extra fact given.

• Income, expenditure and savings all hang on one rule: .

• If income is in the ratio to expenditure, set income and expenditure ; then savings . A given savings amount fixes , and every other value follows.

• For two people, set incomes and use expenditure to write the expenditure ratio as an equation, then cross-multiply for .

• For percentage shifts, work in units: new savings new income new expenditure, then percent change .

📊Ratio ScalerEnter A and B to explore the ratio — add C for compound ratio
A
:
B
+
C (optional)
A
3
B
5
Simplified Ratio
3 : 5
3rd Proportional (x where A:B = B:x)
x = 8.3333
Mean Proportional (√A×B)
√15≈ 3.873
Proportion Checker — Is a:b = c:d ?
:=:

If income and expenditure , then savings .

📝Practice Questions

Q1What number must be added to each of 5, 9, 7 and 13 to make the four numbers proportional?

Q2A person has income : expenditure = 7:5. If income stays the same while expenditure rises by ₹500, savings fall to ₹1500. Find the original savings.

Q3Two persons have incomes in the ratio 5:6 and expenditures in the ratio 7:9. If each saves ₹3000, find the smaller income.

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Real exam questions — Change, Income & Savings

2 question types · 6 solved examples from real SSC papers

These problems take a ratio and disturb it — the same number is added to each term, or income and expenditure are pushed up and down. The cure is always the same: give every part the SAME multiplier, write the new ratio as an equation, and solve.

How to solve this type
Let the number added be xx and rebuild the ratio: aa and bb become a+xa+x and b+xb+x. Set the new ratio equal to the target and cross-multiply. For four numbers "in proportion" use a+xb+x=c+xd+x\dfrac{a+x}{b+x}=\dfrac{c+x}{d+x} — the x2x^2 terms cancel, leaving a quick linear equation. If the algebra gives a negative xx it simply means you SUBTRACT (or vice-versa); report the magnitude. Handy filter: four numbers are already proportional when a×d=b×ca\times d=b\times c.

A certain ratio is 5:7. What number must be added to BOTH terms to make the ratio 7:9?

A2B3C4D5

What number must be SUBTRACTED from each of 7, 12, 17 and 27 so that the results are in proportion?

A1B3C4D2

What number must be added to each of 4, 6, 8 and 12 to make them proportional?

A1B2C0D3