JEE Main 2023 (25 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: A triangle is formed by X - axis, Y- axis and the line $3 x + 4 y = 60$ . Then the number of points $P (a, b)$ which lie strictly inside the triangle, where a is an integer and b is a multiple of a, is______.

Q: Why do we assume $k$ is a positive integer? Could $k$ be zero?

Great question! The problem states the point $(a, b)$ lies strictly inside the triangle. Since the triangle is bounded by the X-axis ($y=0$) and Y-axis ($x=0$), any point inside must have $x > 0$ and $y > 0$. If $k$ were zero, then $b = 0 \cdot a = 0$, which would place the point on the X-axis, not strictly inside. Thus, $k$ must be a positive integer.

Q: How do I know when to stop testing values for $k$?

You stop when the upper bound for $a$ becomes less than 1. Since $a$ must be an integer and $a > 0$, the smallest possible value for $a$ is 1. If $\frac{60}{3+4k} \le 1$, there are no positive integers $a$ that satisfy the condition. Solving $3+4k \ge 60$ gives $4k \ge 57$, or $k \ge 14.25$. So, $k=14$ is the last possible value.

Q: Is there a risk of double-counting points?

Not with this method! By fixing $k$ as the multiplier ($b=ka$), we define a unique relationship for each point. Every point $(a, b)$ has a unique ratio $b/a$. Since each $k$ represents a distinct ratio, each point $(a, b)$ is counted exactly once.

Q: Why can't we just count all integer points in the triangle and then filter them?

You absolutely could, but that is the 'brute force' path! Counting all integer points inside $3x+4y < 60$ involves summing up values for every $x$ from 1 to 19. It is prone to arithmetic errors. Our method of using the multiplier $k$ is much more elegant and structured.

Q: What if $a$ is not a divisor of $b$? Does the problem allow that?

The problem explicitly states '$b$ is a multiple of $a$'. This is a constraint we must satisfy. If a point $(a, b)$ exists where $b$ is not a multiple of $a$, it simply doesn't meet the criteria of the problem and is excluded from our count.

Enter Numerical Value:

Visualized Solution

Visualizing the Triangle

The triangle is bounded by the lines $x = 0$ (Y-axis), $y = 0$ (X-axis), and $3 x + 4 y = 60$ .
Intercepts of the line $3 x + 4 y = 60$ :
When $y = 0$ , $3 x = 60 ⟹ x = 20$ . Point is $(20, 0)$ .
When $x = 0$ , $4 y = 60 ⟹ y = 15$ . Point is $(0, 15)$ .

Defining Interior Points

For a point $P (a, b)$ to lie strictly inside the triangle:
1. $a > 0$ (Right of Y-axis)
2. $b > 0$ (Above X-axis)
3. $3 a + 4 b < 60$ (Below the line)

The Multiple Condition

Given: $b$ is a multiple of $a$ .
Let $b = k a$ , where $k \in {1, 2, 3, \dots}$
$k$ must be a positive integer because $a, b > 0$ .

Deriving the Inequality

Substitute $b = k a$ into $3 a + 4 b < 60$ :
$3 a + 4 (k a) < 60$
Factor out $a$ : $a (3 + 4 k) < 60$

Solving for $a$

Isolate $a$ : $a < \frac{60}{3 + 4 k}$
Since $a$ must be a positive integer ( $a \geq 1$ ), we check values of $k$ starting from $1$ .

Case $k = 1$

For $k = 1$ : $a < \frac{60}{3 + 4 ( 1 )} = \frac{60}{7} \approx 8.57$
Possible integers: $a \in {1, 2, 3, 4, 5, 6, 7, 8}$
Number of points = 8

Case $k = 2$

For $k = 2$ : $a < \frac{60}{3 + 4 ( 2 )} = \frac{60}{11} \approx 5.45$
Possible integers: $a \in {1, 2, 3, 4, 5}$
Number of points = 5

Cases $k = 3$ and $k = 4$

For $k = 3$ : $a < \frac{60}{15} = 4 ⟹ a \in {1, 2, 3}$ (3 points)
For $k = 4$ : $a < \frac{60}{19} \approx 3.15 ⟹ a \in {1, 2, 3}$ (3 points)

Cases $k = 5$ and $k = 6$

For $k = 5$ : $a < \frac{60}{23} \approx 2.6 ⟹ a \in {1, 2}$ (2 points)
For $k = 6$ : $a < \frac{60}{27} \approx 2.2 ⟹ a \in {1, 2}$ (2 points)

Cases $k = 7$ to $k = 14$

For $k = 7$ : $a < \frac{60}{31} \approx 1.93 ⟹ a = 1$
The condition $a \geq 1$ requires $\frac{60}{3 + 4 k} > 1 ⟹ 3 + 4 k < 60 ⟹ 4 k < 57 ⟹ k \leq 14$ .
For $k \in {7, 8, \dots, 14}$ , $a = 1$ is the only solution.
Total points = $8 \times 1 = 8$ points.

Final Summation

Total points $= 8 + 5 + 3 + 3 + 2 + 2 + 8$
Total points $= 31$
Final Answer: 31

00:00 / 00:00

The Sigma Insight: Various Forms of Equations of a Line

Solution Diagram

Analyzing the Setup

To find the number of integer-coordinate points

(a, b)

strictly inside the triangle formed by the origin

(0, 0)

, the X-intercept

(20, 0)

, and the Y-intercept

(0, 15)

, we must satisfy three primary constraints:

a > 0

b > 0

3 a + 4 b < 60

We are further constrained by the condition that

b

is a multiple of

a

. We can express this as

b = k a

, where

k

is a positive integer.

The Master Equation

Substituting

b = k a

into the inequality

3 a + 4 b < 60

, we obtain:

3 a + 4 (k a) < 60

a (3 + 4 k) < 60

a < \frac{60}{3 + 4 k}

Since

a

must be a positive integer (

a \geq 1

), the number of possible values for

a

for a fixed

k

is given by the largest integer strictly less than

\frac{60}{3 + 4 k}

Systematic Enumeration

We iterate through values of

k

to determine the number of valid integer values for

a

For $k = 1$ : $a < \frac{60}{7} \approx 8.57 \Rightarrow a \in {1, 2, \dots, 8}$ (8 points) For

k = 2

a < \frac{60}{11} \approx 5.45 \Rightarrow a \in {1, 2, 3, 4, 5}

(5 points) For $k = 3$ : $a < \frac{60}{15} = 4 \Rightarrow a \in {1, 2, 3}$ (3 points) For

k = 4

a < \frac{60}{19} \approx 3.15 \Rightarrow a \in {1, 2, 3}

(3 points) For $k = 5$ : $a < \frac{60}{23} \approx 2.60 \Rightarrow a \in {1, 2}$ (2 points) For

k = 6

a < \frac{60}{27} \approx 2.22 \Rightarrow a \in {1, 2}

(2 points)

For

k \geq 7

, the upper bound

\frac{60}{3 + 4 k}

becomes less than 2. Specifically, for

k \in {7, 8, 9, 10, 11, 12, 13, 14}

, the value of

a

must be 1.

For $k = 7$ to $k = 14$ : There are $14 - 7 + 1 = 8$ values of $k$ . Each yields $a = 1$ (8 points*).

k = 15

, then

a < \frac{60}{63} < 1

, which yields no positive integer solutions for

a

Final Calculation

Summing the points obtained from each case:

T o t a l = 8 + 5 + 3 + 3 + 2 + 2 + 8

T o t a l = 31

The total number of integer-coordinate points satisfying the given conditions is 31.

Common Traps & FAQs

Why do we assume

k

is a positive integer? Could

k

be zero?

Great question! The problem states the point

(a, b)

lies strictly inside the triangle. Since the triangle is bounded by the X-axis (

y = 0

) and Y-axis (

x = 0

), any point inside must have

x > 0

and

y > 0

. If

k

were zero, then

b = 0 \cdot a = 0

, which would place the point on the X-axis, not strictly inside. Thus,

k

must be a positive integer.

How do I know when to stop testing values for

k

You stop when the upper bound for

a

becomes less than 1. Since

a

must be an integer and

a > 0

, the smallest possible value for

a

is 1. If

\frac{60}{3 + 4 k} \leq 1

, there are no positive integers

a

that satisfy the condition. Solving

3 + 4 k \geq 60

gives

4 k \geq 57

, or

k \geq 14.25

. So,

k = 14

is the last possible value.

Is there a risk of double-counting points?

Not with this method! By fixing

k

as the multiplier (

b = k a

), we define a unique relationship for each point. Every point

(a, b)

has a unique ratio

b / a

. Since each

k

represents a distinct ratio, each point

(a, b)

is counted exactly once.

Why can't we just count all integer points in the triangle and then filter them?

You absolutely could, but that is the 'brute force' path! Counting all integer points inside

3 x + 4 y < 60

involves summing up values for every

x

from 1 to 19. It is prone to arithmetic errors. Our method of using the multiplier

k

is much more elegant and structured.

What if

a

is not a divisor of

b

? Does the problem allow that?

The problem explicitly states '

b

is a multiple of

a

'. This is a constraint we must satisfy. If a point

(a, b)

exists where

b

is not a multiple of

a

, it simply doesn't meet the criteria of the problem and is excluded from our count.

Animated Solution for Mathematics - Straight Lines: A triangle is formed by X - axis, Y- axis and the line $3 x + 4 y = 60$ . Then the number of points $P (a, b)$ which lie strictly inside the triangle, where a is an integer and b is a multiple of a, is______.

Enter Numerical Value:

Visualized Solution

Visualizing the Triangle

Defining Interior Points

The Multiple Condition

Deriving the Inequality

Solving for $a$

Case $k = 1$

Case $k = 2$

Cases $k = 3$ and $k = 4$

Cases $k = 5$ and $k = 6$

Cases $k = 7$ to $k = 14$

Final Summation

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Master Equation

Systematic Enumeration

Final Calculation

Similar Questions

The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices $(0, 0), (0, 41)$ and $(41, 0)$ is

The number of integral points (integral point means both the coordinates should be integer) exactly in the interior of the triangle with vertices $(0, 0), (0, 21)$ and $(21, 0)$ , is

If the sum of squares of all real values of $α$ , for which the lines $2 x - y + 3 = 0$ , $6 x + 3 y + 1 = 0$ and $α x + 2 y - 2 = 0$ do not form a triangle is $p$ , then the greatest integer less than or equal to $p$ is

The number of integer values of $m$ , for which the x-coordinate of the point of intersection of the lines $3 x + 4 y = 9$ and $y = m x + 1$ is also an integer, is

The number of integral values of m so that the abscissa of point of intersection of lines $3 x + 4 y = 9$ and $y = m x + 1$ is also an integer, is :

Let two straight lines drawn from the origin $O$ intersect the line $3 x + 4 y = 12$ at the points $P$ and $Q$ such that $△ O P Q$ is an isosceles triangle and $∠ P O Q = 9 0^{\circ}$ . If $l = O P^{2} + P Q^{2} + Q O^{2}$ , then the greatest integer less than or equal to $l$ is :

All points lying inside the triangle formed by the points $(1, 3), (5, 0)$ and $(- 1, 2)$ satisfy

Let the points $(\frac{11}{2}, α)$ lie on or inside the triangle with sides $x + y = 11$ , $x + 2 y = 16$ and $2 x + 3 y = 29$ Then the product of the smallest and the largest values of a is equal to:

Determine all values of $α$ for which the point $(α, α^{2})$ lies inside the triangle formed by the lines $2 x + 3 y - 1 = 0$ , $x + 2 y - 3 = 0$ , $5 x - 6 y - 1 = 0$ .

Enter Numerical Value:

Visualized Solution

Visualizing the Triangle

Defining Interior Points

The Multiple Condition

Deriving the Inequality

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=1

Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=2

Cases .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=3 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=4

Cases .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=5 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=6

Cases .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=7 to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=14

Final Summation

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Master Equation

Systematic Enumeration

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices (0,0),(0,41) and (41,0) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of integral points (integral point means both the coordinates should be integer) exactly in the interior of the triangle with vertices (0,0),(0,21) and (21,0), is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of integer values of m, for which the x-coordinate of the point of intersection of the lines 3x+4y=9 and y=mx+1 is also an integer, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of integral values of m so that the abscissa of point of intersection of lines 3x+4y=9 and y=mx+1 is also an integer, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } All points lying inside the triangle formed by the points (1,3),(5,0) and (−1,2) satisfy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the points (211​,α) lie on or inside the triangle with sides x+y=11, x+2y=16 and 2x+3y=29 Then the product of the smallest and the largest values of a is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Determine all values of α for which the point (α,α2) lies inside the triangle formed by the lines 2x+3y−1=0, x+2y−3=0, 5x−6y−1=0.

Solving for $a$

Case $k = 1$

Case $k = 2$

Cases $k = 3$ and $k = 4$

Cases $k = 5$ and $k = 6$

Cases $k = 7$ to $k = 14$

The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices $(0, 0), (0, 41)$ and $(41, 0)$ is

The number of integral points (integral point means both the coordinates should be integer) exactly in the interior of the triangle with vertices $(0, 0), (0, 21)$ and $(21, 0)$ , is

The number of integer values of $m$ , for which the x-coordinate of the point of intersection of the lines $3 x + 4 y = 9$ and $y = m x + 1$ is also an integer, is

The number of integral values of m so that the abscissa of point of intersection of lines $3 x + 4 y = 9$ and $y = m x + 1$ is also an integer, is :

All points lying inside the triangle formed by the points $(1, 3), (5, 0)$ and $(- 1, 2)$ satisfy

Let the points $(\frac{11}{2}, α)$ lie on or inside the triangle with sides $x + y = 11$ , $x + 2 y = 16$ and $2 x + 3 y = 29$ Then the product of the smallest and the largest values of a is equal to:

Determine all values of $α$ for which the point $(α, α^{2})$ lies inside the triangle formed by the lines $2 x + 3 y - 1 = 0$ , $x + 2 y - 3 = 0$ , $5 x - 6 y - 1 = 0$ .