Ages and Number Problems for CUET General Aptitude

Q: How do you solve an age problem quickly?

Take one person's present age as x, write every other age in terms of x, and move everyone the same number of years along the timeline before you write the equation.

Q: How do you handle a ratio of ages "then" and "now"?

Write the ages at the time of the first ratio as ak and bk, shift both by the same number of years, and set the new fraction equal to the second ratio. Cross-multiply and solve for k.

Q: How do you solve two-digit number and reversed-digit problems?

Write the number as 10a + b, with tens digit a from 1 to 9 and units digit b from 0 to 9; its reverse is 10b + a. The number minus its reverse is 9(a - b), and their sum is 11(a + b).

About 10 min to learn · 60 questions

Ages and number problems are one skill in two costumes: read a sentence, name one unknown, write one clean equation, and check the answer against the words. The mathematics rarely goes beyond Class 10 linear equations and the odd simple quadratic; what costs marks is misreading 'ago', 'hence', 'the larger number' or 'the sum'. This lesson gives the setups that work every time, with worked examples and the traps beside them.

Syllabus area: General Mental Ability, Numerical Ability

On this page · 7 sections

01The core idea
02Ages
03Number problems
04Exam shortcuts
05Common mistakes
06Quick revision
07Check your understanding

The core idea

Every question in this chapter is a sentence to translate. Name one unknown in words ("x = Anil's age now"), write every other quantity in terms of it, form one equation, and check the answer against the original sentence, not against your equation.

For ages, two facts do most of the work: everyone ages at the same rate, so the gap between two people never changes, and a group's total age grows by the number of people times the years passed.

Ages

How do you solve an age problem quickly?

Take one person's present age as $x$ , write every other age in terms of $x$ , and move everyone the same number of years along the timeline before you write the equation.

Person	$t$ years ago	Now	$t$ years from now
A	$x - t$	$x$	$x + t$
B	$y - t$	$y$	$y + t$

Exam shortcut

Let $x$ be the youngest person when the others are multiples of them: "mother is three times the son" is cleaner as son $x$ , mother $3x$ . And answer the time the question asks about; the age "5 years hence" is a distractor when the present age is wanted.

Why does the age difference between two people never change?

Because both get older by exactly the same amount each year. If Meera is 7 years older than her brother today, she was 7 years older ten years ago and will be 7 years older in forty years.

A given difference lets you write both ages with one variable, $x$ and $x + d$ .
"When A was born, B was 25" means B − A = 25, forever.
One person is twice the other's age at exactly one moment: when the younger's age equals the gap.

Example 1 · Twice the age

Harish is 26 years older than his daughter, who is 9. In how many years will he be exactly twice her age?

The gap is 26, so he is twice her age when she is 26.
She is 9 now, so that is $26 - 9 = 17$ years away.

Answer

17 years

Check: she will be 26 and he 52.

How do you handle a ratio of ages "then" and "now"?

Write the ages at the time of the first ratio as $ak$ and $bk$ , shift both by the same number of years, and set the new fraction equal to the second ratio. Cross-multiply and solve for $k$ .

Example 2 · Two ratios

The ages of Rekha and Tushar are in the ratio $5 : 3$ . In 8 years the ratio will be $7 : 5$ . Find their present ages.

Now: $5k$ and $3k$ . In 8 years: $5k + 8$ and $3k + 8$ .
$\tfrac{5k + 8}{3k + 8} = \tfrac{7}{5}$ , so $25k + 40 = 21k + 56$ .
$4k = 16$ , so $k = 4$ .

Answer

Rekha 20, Tushar 12

Check: in 8 years, $28 : 20 = 7 : 5$ .

Exam shortcut

If the ratios are at " $a$ years ago" and " $b$ years from now", each person jumps $a + b$ years between them. Start the variable at the earlier time and add $a + b$ in one go.

How do you solve age problems about a whole family?

Work with the total of the ages, and remember that the total grows by the number of people times the years passed. Four people each ageing 3 years raise the total by 12, not 3.

Example 3 · A child is born

Four years ago, the average age of a family of three was 22. A child has been born since, and today the average age of all four is 20. How old is the child?

Four years ago the total was $3 \times 22 = 66$ .
Each of the three is 4 years older now: $66 + 3 \times 4 = 78$ .
Today's total for four is $4 \times 20 = 80$ , so the child is $80 - 78 = 2$ .

Answer

2 years

Adding only 4 to the old total gives 70 and a child of 10.

What is the trick for "A is twice as old as B was when A was as old as B is now"?

Draw two columns, then and now, fill in what you know, and use the fact that the time gap is the same for both people. The riddle collapses into one linear equation.

Example 4 · The riddle

Kavita is 36. She is three times as old as Ramesh was when Kavita was as old as Ramesh is now. How old is Ramesh?

Let Ramesh be $r$ . Kavita was $r$ exactly $36 - r$ years ago.
Ramesh was then $r - (36 - r) = 2r - 36$ .
$36 = 3(2r - 36)$ , so $6r = 144$ and $r = 24$ .

Answer

Check: 12 years ago Kavita was 24 and Ramesh 12, and $36 = 3 \times 12$ .

Number problems

How do you find two numbers from their sum and difference?

The larger is half of the sum plus the difference, and the smaller is half of the sum minus the difference. There is no need for two equations.

Sum and difference

\text{Larger} = \frac{\text{sum} + \text{difference}}{2} \qquad \text{Smaller} = \frac{\text{sum} - \text{difference}}{2}

Exam shortcut

With a ratio $a : b$ and a known sum, one part is the sum over $a + b$ ; with a known difference $d$ , one part is $\tfrac{d}{b - a}$ .
A number increased by $\tfrac{2}{7}$ of itself is $\tfrac{9}{7}$ of it: if that is 45, the number is 35.

How do you solve consecutive integer problems?

Use the middle term. For an odd count, the middle term is the sum divided by the count. For an even count, that quotient sits halfway between the two middle terms.

Type	Write them as	Step
Consecutive integers	$n - 1, n, n + 1$	1
Consecutive odd or even	$n - 2, n, n + 2$	2
Consecutive multiples of 5	$n - 5, n, n + 5$	5

Example 5 · Five consecutive even numbers

The sum of five consecutive even numbers is 230. Find the largest.

Middle term: $230 \div 5 = 46$ .
The numbers are 42, 44, 46, 48 and 50.

Answer

Key idea

Writing around the middle tidies squares too: $(n - 1)^2 + n^2 + (n + 1)^2 = 3n^2 + 2$ . If three consecutive squares add to 149, then $3n^2 = 147$ , $n = 7$ , and the numbers are 6, 7 and 8.

How do you solve two-digit number and reversed-digit problems?

Write the number as $10a + b$ , with tens digit $a$ from 1 to 9 and units digit $b$ from 0 to 9; its reverse is $10b + a$ . The number minus its reverse is $9(a - b)$ , and their sum is $11(a + b)$ .

Example 6 · Reversing the digits

The digits of a two-digit number add to 12. Reversing them makes the number 18 smaller. Find the number.

$9(a - b) = 18$ , so $a - b = 2$ , and $a + b = 12$ .
$a = 7$ and $b = 5$ .

Answer

Check: $75 - 57 = 18$ .

Watch out

The reversed number is always among the options. Since the reverse is smaller here, the tens digit is the bigger one; decide that first. And when a number is $k$ times the sum of its digits, $10a + b = k(a + b)$ gives a fixed relation: with $k = 6$ , $4a = 5b$ , so the number is 54.

How do you solve reciprocal and sum-of-squares problems?

Turn the sentence into a quadratic, factorise, and keep the root the words allow: positive, natural, or a digit. These are Class 10 quadratics, and the numbers are chosen so they factorise.

$x + \tfrac{1}{x} = s$ gives $x^2 - sx + 1 = 0$ , whose roots are reciprocals of each other.
$\tfrac{1}{x} - \tfrac{1}{x + d} = \tfrac{d}{x(x + d)}$ .
$(a - b)^2 = (a + b)^2 - 4ab$ gives the difference straight from the sum and product.

Example 7 · Difference from sum and product

The sum of two numbers is 26 and their product is 165. Find their difference.

$(a - b)^2 = 26^2 - 4 \times 165 = 676 - 660 = 16$ .
So the difference is 4.

Answer

The numbers are 15 and 11.

Example 8 · A number and its reciprocal

A positive number exceeds its reciprocal by $\tfrac{8}{3}$ . Find it.

$x - \tfrac{1}{x} = \tfrac{8}{3}$ , so $3x^2 - 8x - 3 = 0$ .
$(3x + 1)(x - 3) = 0$ , so $x = 3$ or $x = -\tfrac{1}{3}$ .
The number is positive.

Answer

Check: $3 - \tfrac{1}{3} = \tfrac{8}{3}$ .

Exam shortcut

Check by putting the answer back into the sentence, not into your equation, which would only confirm a wrong setup. With four options, trying a middle option in the words is a fair method when the algebra feels slow.

Exam shortcuts

The gap is fixed. Twice the age happens when the younger is as old as the gap.
Group totals. They grow by people × years.
Sum and difference. Half of each combination gives the two numbers.
Consecutive numbers. Sum ÷ count is the middle.
Digits. Number − reverse $= 9(a - b)$ . Number + reverse $= 11(a + b)$ .

Common mistakes

Answering the wrong time. Underline "ago", "now" or "hence" in the question.
Answering for the wrong person.
Shifting a group total by one person's years.
Changing the age gap. How much older someone will be later is today's gap.
Counting gaps wrongly. Four children 2 years apart span 3 gaps, 6 years.
Dropping brackets. "9 less than a number, times 4" is $4(x - 9)$ .
Keeping a forbidden root, or flipping its sign.
Ticking the reversed number.
A ratio term as the number of parts. In $3 : 11$ with a difference of 24, the difference is 8 parts.
Stopping at the middle term when the smallest or largest is asked.

Quick revision

Ages

One unknown for the present age, shift everyone equally
The gap never changes
Ratios then and now: $ak$ , $bk$ , shift, cross-multiply
Family total: + people × years

Numbers

Larger = (sum + difference) ÷ 2
Consecutive: middle = sum ÷ count
Digits: $10a + b$ . Minus reverse $9(a - b)$ , plus reverse $11(a + b)$

Quadratics

$(a - b)^2 = (a + b)^2 - 4ab$
Keep the root the words allow
Check against the sentence

Ratios are worked in ratio and proportion, totals from averages in averages, and quadratics in linear and quadratic equations.

Next · 06Quick revision

Check your understanding

60 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

Problems on agesQ1 of 60Q1

Problems on ages

30 questions

Check your understanding

Problems on ages

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