Home
Class 12
MATHS
Let m be a positive integer and let the ...

Let m be a positive integer and let the lines ` 13x + 11y = 700 ` and ` y= mx - 1 ` intersect in a point whose coordinates are integer. Then m equals to :

A

4

B

5

C

6

D

7

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the value of \( m \) such that the lines \( 13x + 11y = 700 \) and \( y = mx - 1 \) intersect at integer coordinates. ### Step 1: Rewrite the equations We start with the two equations: 1. \( 13x + 11y = 700 \) 2. \( y = mx - 1 \) We can rearrange the first equation to express \( y \) in terms of \( x \): \[ 11y = 700 - 13x \] \[ y = \frac{700 - 13x}{11} \] ### Step 2: Substitute for \( y \) Now, we substitute \( y \) from the second equation into the rearranged first equation: \[ \frac{700 - 13x}{11} = mx - 1 \] ### Step 3: Clear the fraction To eliminate the fraction, we multiply both sides by 11: \[ 700 - 13x = 11(mx - 1) \] \[ 700 - 13x = 11mx - 11 \] ### Step 4: Rearrange the equation Now, we rearrange the equation to isolate terms involving \( x \): \[ 700 + 11 = 11mx + 13x \] \[ 711 = 11mx + 13x \] \[ 711 = x(11m + 13) \] ### Step 5: Solve for \( x \) Now we can solve for \( x \): \[ x = \frac{711}{11m + 13} \] ### Step 6: Ensure \( x \) is an integer For \( x \) to be an integer, \( 11m + 13 \) must be a divisor of 711. We need to find the divisors of 711. ### Step 7: Find the divisors of 711 The prime factorization of 711 is: \[ 711 = 3 \times 237 = 3 \times 3 \times 79 = 3^2 \times 79 \] The divisors of 711 are: \( 1, 3, 9, 79, 237, 711 \). ### Step 8: Set up the equation We set up the equation: \[ 11m + 13 = d \quad \text{(where \( d \) is a divisor of 711)} \] Rearranging gives: \[ 11m = d - 13 \] \[ m = \frac{d - 13}{11} \] ### Step 9: Check each divisor We will check which divisors yield a positive integer \( m \): 1. For \( d = 1 \): \[ m = \frac{1 - 13}{11} = \frac{-12}{11} \quad \text{(not positive)} \] 2. For \( d = 3 \): \[ m = \frac{3 - 13}{11} = \frac{-10}{11} \quad \text{(not positive)} \] 3. For \( d = 9 \): \[ m = \frac{9 - 13}{11} = \frac{-4}{11} \quad \text{(not positive)} \] 4. For \( d = 79 \): \[ m = \frac{79 - 13}{11} = \frac{66}{11} = 6 \quad \text{(positive integer)} \] 5. For \( d = 237 \): \[ m = \frac{237 - 13}{11} = \frac{224}{11} \quad \text{(not an integer)} \] 6. For \( d = 711 \): \[ m = \frac{711 - 13}{11} = \frac{698}{11} \quad \text{(not an integer)} \] ### Conclusion The only positive integer value of \( m \) that satisfies the condition is: \[ \boxed{6} \]
Promotional Banner

Topper's Solved these Questions

  • STRAIGHT LINES

    VIKAS GUPTA (BLACK BOOK) ENGLISH|Exercise Exercise-2 : One or More than One Answer is/are Correct|12 Videos
  • STRAIGHT LINES

    VIKAS GUPTA (BLACK BOOK) ENGLISH|Exercise Exercise-3 : Comprehension Type Problems|4 Videos
  • SOLUTION OF TRIANGLES

    VIKAS GUPTA (BLACK BOOK) ENGLISH|Exercise Exercise-5 : Subjective Type Problems|11 Videos
  • TRIGONOMETRIC EQUATIONS

    VIKAS GUPTA (BLACK BOOK) ENGLISH|Exercise Exercise-5 : Subjective Type Problems|9 Videos

Similar Questions

Explore conceptually related problems

Let P = { x: x is a positive integer , x lt 6 }

The graph of|y-1|=|x+1| forms an X. The two branches of the X intersect at a point whose coordinates are

Let the images of the point A(2, 3) about the lines y=x and y=mx are P and Q respectively. If the line PQ passes through the origin, then m is equal to

The number of integral point inside the triangle made by the line 3x + 4y - 12 =0 with the coordinate axes which are equidistant from at least two sides is/are : ( an integral point is a point both of whose coordinates are integers. )

Let S be the circle in the x y -plane defined by the equation x^2+y^2=4. (For Ques. No 15 and 16) Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N . Then, the mid-point of the line segment M N must lie on the curve (x+y)^2=3x y (b) x^(2//3)+y^(2//3)=2^(4//3) (c) x^2+y^2=2x y (d) x^2+y^2=x^2y^2

The y-intercept of the line through the two points whose coordinates are (5,-2) and (1,3) is

If the line y=mx meets the lines x+2y-1=0 and 2x-y+3=0 at the same point, then m is equal to

Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m. [Hint: Let x be any positive integer then it is of the form 3q , 3q+1 or 3q+2 Now square each of these and sho

If the straight lines (x-1)/(k)=(y-2)/(2)=(z-3)/(3) and (x-2)/(3)=(y-3)/(k)=(z-1)/(2) intersect at a point, then the integer k is equal to