SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Linear Inequalities

40 questions · 40 still being checked

EXERCISE 5.1 21–26 (part 3 of 5)

  1. Exercise 21

    Ravi obtained 70\displaystyle 70 and 75\displaystyle 75 marks in first two unit test. Find the minimum marks he should get in the third test to have an average of at least 60\displaystyle 60 marks.

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    NCERT’s answer
    $\displaystyle 35$
    Forming the inequality. Let \(\displaystyle x\) be the marks Ravi gets in the third test. His average over the three tests is \(\displaystyle \dfrac{70+75+x}{3}\), and "at least $\displaystyle 60$" means this average must be \(\displaystyle \geq 60\).\[\frac{70+75+x}{3} \geq 60\]Multiply both sides by \(\displaystyle 3\) (positive, so the sense does not change):\[145 + x \geq 180 \quad\Longrightarrow\quad x \geq 35\]Since a test score of \(\displaystyle 35\) is attainable, the smallest admissible value is \(\displaystyle 35\).Ravi must score at least $\displaystyle 35$ marks in the third test; the minimum required is 35.

  2. Exercise 22

    To receive Grade 'A' in a course, one must obtain an average of 90\displaystyle 90 marks or more in five examinations (each of 100\displaystyle 100 marks). If Sunita's marks in first four examinations are 87\displaystyle 87, 92\displaystyle 92, 94\displaystyle 94 and 95\displaystyle 95, find minimum marks that Sunita must obtain in fifth examination to get grade 'A' in the course.

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    NCERT’s answer
    $\displaystyle 82$
    Forming the inequality. Let \(\displaystyle x\) be Sunita's marks in the fifth examination. The first four give\[87 + 92 + 94 + 95 = 368.\]Grade 'A' needs an average of \(\displaystyle 90\) or more over the five examinations:\[\frac{368 + x}{5} \geq 90\] \[368 + x \geq 450 \quad\Longrightarrow\quad x \geq 82\]Each paper is out of \(\displaystyle 100\), so \(\displaystyle 82 \leq x \leq 100\) is achievable.Sunita must obtain at least $\displaystyle 82$ marks in the fifth examination.

  3. Exercise 23

    Find all pairs of consecutive odd positive integers both of which are smaller than 10\displaystyle 10 such that their sum is more than 11.

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    $\displaystyle (5,7),(7,9)$
    Translating the conditions into inequalities. Let \(\displaystyle x\) be the smaller of the two integers. Consecutive odd integers differ by \(\displaystyle 2\), so the pair is \(\displaystyle x\) and \(\displaystyle x+2\), with \(\displaystyle x\) an odd positive integer.Both smaller than $\displaystyle 10$ — the larger one decides:\[x + 2 < 10 \quad\Longrightarrow\quad x < 8\]Their sum is more than $\displaystyle 11$:\[x + (x+2) > 11 \quad\Longrightarrow\quad 2x > 9 \quad\Longrightarrow\quad x > 4.5\]Both must hold, so \(\displaystyle 4.5 < x < 8\). The odd integers in this range are \(\displaystyle x = 5\) and \(\displaystyle x = 7\).Verify: \(\displaystyle 5+7 = 12 > 11\) and \(\displaystyle 7 < 10\); \(\displaystyle 7+9 = 16 > 11\) and \(\displaystyle 9 < 10\). Both pairs qualify.The required pairs are \(\displaystyle (5,\,7)\) and \(\displaystyle (7,\,9)\).

  4. Exercise 24

    Find all pairs of consecutive even positive integers, both of which are larger than 5\displaystyle 5 such that their sum is less than 23.

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    NCERT’s answer
    $\displaystyle (6,8)$, $\displaystyle (8,10)$, $\displaystyle (10,12)$
    Translating the conditions into inequalities. Let \(\displaystyle x\) be the smaller integer. Consecutive even integers differ by \(\displaystyle 2\), so the pair is \(\displaystyle x\) and \(\displaystyle x+2\), with \(\displaystyle x\) an even positive integer.Both larger than $\displaystyle 5$ — the smaller one decides:\[x > 5\]Their sum is less than $\displaystyle 23$:\[x + (x+2) < 23 \quad\Longrightarrow\quad 2x < 21 \quad\Longrightarrow\quad x < 10.5\]Both must hold, so \(\displaystyle 5 < x < 10.5\). The even integers in this range are \(\displaystyle x = 6,\ 8,\ 10\).Verify the sums: \(\displaystyle 6+8=14\), \(\displaystyle 8+10=18\), \(\displaystyle 10+12=22\) — each is less than \(\displaystyle 23\), and every integer involved exceeds \(\displaystyle 5\).The required pairs are \(\displaystyle (6,\,8)\), \(\displaystyle (8,\,10)\) and \(\displaystyle (10,\,12)\).

  5. Exercise 25

    The longest side of a triangle is 3\displaystyle 3 times the shortest side and the third side is 2\displaystyle 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61\displaystyle 61 cm, find the minimum length of the shortest side.

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    NCERT’s answer
    $\displaystyle 9$ cm
    Naming the sides in terms of one unknown. Let the shortest side be \(\displaystyle x\) cm. Then
    longest side \(\displaystyle = 3x\) cm,
    third side \(\displaystyle = (3x - 2)\) cm ($\displaystyle 2$ cm shorter than the longest).
    Perimeter at least $\displaystyle 61$ cm:\[x + 3x + (3x - 2) \geq 61\] \[7x - 2 \geq 61\] \[7x \geq 63 \quad\Longrightarrow\quad x \geq 9\]Consistency check at \(\displaystyle x = 9\): the sides are \(\displaystyle 9,\ 27,\ 25\) cm — \(\displaystyle 9\) is indeed the shortest and \(\displaystyle 27\) the longest, the perimeter is \(\displaystyle 61\) cm, and \(\displaystyle 9 + 25 > 27\) so such a triangle exists.The minimum length of the shortest side is $\displaystyle 9$ cm.

  6. Exercise 26

    A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5cm longer than the second? [Hint: If x\displaystyle x is the length of the shortest board, then x,(x+3)\displaystyle x,(x+3) and 2x\displaystyle 2 x are the lengths of the second and third piece, respectively. Thus, x+(x+3)+2x≤91\displaystyle x+(x+3)+2 x \leq 91 and 2x≥(x+3)+5\displaystyle 2 x \geq(x+3)+5 ].

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    NCERT’s answer
    Greater than or equal to 8cm but less than or equal to 22cm
    Two conditions, one unknown. Let the shortest piece be \(\displaystyle x\) cm. Then the second piece is \(\displaystyle (x+3)\) cm and the third is \(\displaystyle 2x\) cm.The three pieces come out of a $\displaystyle 91$ cm board:\[x + (x+3) + 2x \leq 91\] \[4x + 3 \leq 91 \quad\Longrightarrow\quad 4x \leq 88 \quad\Longrightarrow\quad x \leq 22\]The third piece is at least $\displaystyle 5$ cm longer than the second:\[2x \geq (x+3) + 5\] \[2x \geq x + 8 \quad\Longrightarrow\quad x \geq 8\]A solution must satisfy both, so we take the intersection:\[8 \leq x \leq 22\]Check the ends: at \(\displaystyle x=8\) the pieces are \(\displaystyle 8,\ 11,\ 16\) (total \(\displaystyle 35 \leq 91\), and \(\displaystyle 16 - 11 = 5\)); at \(\displaystyle x=22\) they are \(\displaystyle 22,\ 25,\ 44\) (total \(\displaystyle 91\), and \(\displaystyle 44 - 25 = 19 \geq 5\)).The shortest board may be of any length from $\displaystyle 8$ cm to $\displaystyle 22$ cm, i.e. \(\displaystyle 8 \leq x \leq 22\).