SolveItClass 9 · NCERT

NCERT Solutions · Class 9 Mathematics Introduction to Linear Polynomials

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Exercise Set 2.2 1–7 (part 2 of 8)

  1. Exercise 1

    Find the value of the linear polynomial 5x−3\displaystyle 5 x-3 if:
    (i)
    x=0\displaystyle x=0
    (ii)
    x=−1\displaystyle x=-1
    (iii)
    x=2\displaystyle x=2

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    Substitute the given number for the variable, then simplify.
    "The value of the polynomial at \(\displaystyle x=a\)" means: replace every \(\displaystyle x\) by \(\displaystyle a\) and work out the resulting number. Keeping the substituted value in brackets stops sign mistakes.
    (i)
    \(\displaystyle x=0\):
    \[5(0)-3 = 0-3 = -3 \]
    (ii)
    \(\displaystyle x=-1\):
    \[5(-1)-3 = -5-3 = -8 \]
    Here \(\displaystyle 5 \times (-1) = -5\), and subtracting \(\displaystyle 3\) from \(\displaystyle -5\) moves further left on the number line, giving \(\displaystyle -8\).
    (iii)
    \(\displaystyle x=2\):
    \[5(2)-3 = 10-3 = 7 \]
    (i) \(\displaystyle -3\) (ii) \(\displaystyle -8\) (iii) \(\displaystyle 7\)

  2. Exercise 2

    Find the value of the quadratic polynomial 7s2−4s+6\displaystyle 7 s^{2}-4 s+6 if:
    (i)
    s=0\displaystyle s=0
    (ii)
    s=−3\displaystyle s=-3
    (iii)
    s=4\displaystyle s=4

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    Substitute, square first, then combine — the squared term is never negative.
    (i)
    \(\displaystyle s=0\):
    \[7(0)^{2}-4(0)+6 = 0-0+6 = 6 \]
    (ii)
    \(\displaystyle s=-3\): work out \(\displaystyle (-3)^{2}\) before multiplying, since \(\displaystyle (-3)^{2}=(-3)\times(-3)=9\), not \(\displaystyle -9\).
    \[7(-3)^{2}-4(-3)+6 = 7(9)+12+6 = 63+12+6 = 81 \]
    Note the middle term: \(\displaystyle -4 \times (-3) = +12\), a plus, because a negative times a negative is positive.
    (iii)
    \(\displaystyle s=4\):
    \[7(4)^{2}-4(4)+6 = 7(16)-16+6 = 112-16+6 = 102 \]
    (i) \(\displaystyle 6\) (ii) \(\displaystyle 81\) (iii) \(\displaystyle 102\)

  3. Exercise 3

    The present age of Salil's mother is three times Salil's present age. After 5\displaystyle 5 years, their ages will add up to 70\displaystyle 70 years. Find their present ages.

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    Name the unknown with one letter, write everything else in terms of it.Let Salil's present age be \(\displaystyle x\) years. His mother's present age is three times this, so it is \(\displaystyle 3x\) years.After $\displaystyle 5$ years each of them is $\displaystyle 5$ years older, so their ages then will be \(\displaystyle x+5\) and \(\displaystyle 3x+5\). We are told these add up to $\displaystyle 70$: \[(x+5)+(3x+5)=70 \]Collect the like terms on the left: \[4x+10=70 \]Subtract \(\displaystyle 10\) from both sides (whatever is done to one side must be done to the other, so the two sides stay equal): \[4x=60 \]Divide both sides by \(\displaystyle 4\): \[x=15 \]So Salil is \(\displaystyle 15\) and his mother is \(\displaystyle 3 \times 15 = 45\).Check. In $\displaystyle 5$ years they will be \(\displaystyle 20\) and \(\displaystyle 50\), and \(\displaystyle 20+50=70\), as required. Also \(\displaystyle 45 = 3 \times 15\), so the "three times" condition holds too.Salil is $\displaystyle 15$ years old and his mother is $\displaystyle 45$ years old.

  4. Exercise 4

    The difference between two positive integers is 63\displaystyle 63 . The ratio of the two integers is 2\displaystyle 2:5. Find the two integers.

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    A ratio lets you describe both numbers with a single unknown.The two integers are in the ratio \(\displaystyle 2:5\), which means they can be written as \(\displaystyle 2k\) and \(\displaystyle 5k\) for the same positive number \(\displaystyle k\) — the ratio only fixes their relative sizes, and \(\displaystyle k\) fixes the actual size.Since \(\displaystyle 5k > 2k\), the larger integer is \(\displaystyle 5k\) and the difference is \[5k-2k=3k \]We are told this difference is $\displaystyle 63$: \[3k=63 \quad\Longrightarrow\quad k=21 \]Therefore the integers are \[2k = 2 \times 21 = 42, \qquad 5k = 5 \times 21 = 105 \]Check. \(\displaystyle 105-42=63\), and \(\displaystyle 42:105\) simplifies (dividing both by $\displaystyle 21$) to \(\displaystyle 2:5\). Both are positive integers, as required.The two integers are $\displaystyle 42$ and 105.

  5. Exercise 5

    Ruby has 3\displaystyle 3 times as many two-rupee coins as she has five rupee-coins. If she has a total ₹88\displaystyle 88, how many coins does she have of each type?

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    Count the coins with one unknown, then add up their value, not their number.The number of two-rupee coins is described in terms of the five-rupee coins, so let the five-rupee coins be the unknown.Let Ruby have \(\displaystyle x\) five-rupee coins. Then she has \(\displaystyle 3x\) two-rupee coins.Their values in rupees are: \[\text{five-rupee coins: } 5 \times x = 5x, \qquad \text{two-rupee coins: } 2 \times 3x = 6x \]The total value is ₹$\displaystyle 88$: \[5x+6x=88 \] \[11x=88 \quad\Longrightarrow\quad x=8 \]So there are \(\displaystyle 8\) five-rupee coins and \(\displaystyle 3 \times 8 = 24\) two-rupee coins.Check. \(\displaystyle 8 \times 5 = 40\) rupees and \(\displaystyle 24 \times 2 = 48\) rupees, and \(\displaystyle 40+48=88\). Also \(\displaystyle 24 = 3 \times 8\), so the "$\displaystyle 3$ times as many" condition holds.Notice that there are more two-rupee coins ($\displaystyle 24$) than five-rupee coins ($\displaystyle 8$), yet the two-rupee coins are worth only a little more in total — counting coins and counting rupees are different things.Ruby has $\displaystyle 24$ two-rupee coins and $\displaystyle 8$ five-rupee coins ($\displaystyle 32$ coins in all).

  6. Exercise 6

    A farmer cuts a 300\displaystyle 300 feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces?

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    Let the shorter piece be the unknown, since the longer one is described in terms of it.Let the shorter piece be \(\displaystyle x\) feet long. The longer piece is four times as long, so it is \(\displaystyle 4x\) feet.The two pieces together make up the whole $\displaystyle 300$-foot fence: \[x+4x=300 \] \[5x=300 \quad\Longrightarrow\quad x=60 \]So the shorter piece is \(\displaystyle 60\) feet and the longer piece is \(\displaystyle 4 \times 60 = 240\) feet.Check. \(\displaystyle 60+240=300\) feet, and \(\displaystyle 240 \div 60 = 4\), so the longer piece really is four times the shorter one. The two pieces are indeed of different sizes, as the question says.The two pieces are $\displaystyle 60$ feet and $\displaystyle 240$ feet long.

  7. Exercise 7

    If the length of a rectangle is three more than twice its width and its perimeter is 24\displaystyle 24 cm, what are the dimensions of the rectangle?

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    Write both dimensions in terms of the width, then use the perimeter formula.Let the width of the rectangle be \(\displaystyle w\) cm. The length is "three more than twice its width", so the length is \(\displaystyle 2w+3\) cm.The perimeter of a rectangle is twice the sum of its length and width: \[P = 2(\text{length}+\text{width}) \] \[2\big((2w+3)+w\big)=24 \]Divide both sides by $\displaystyle 2$ first — it keeps the numbers small: \[(2w+3)+w=12 \] \[3w+3=12 \] \[3w=9 \quad\Longrightarrow\quad w=3 \]Then the length is \(\displaystyle 2(3)+3 = 6+3 = 9\) cm.Check. Length \(\displaystyle 9\) is three more than twice the width \(\displaystyle 2 \times 3 = 6\), and the perimeter is \(\displaystyle 2(9+3)=2 \times 12=24\) cm, as required.The rectangle is $\displaystyle 9$ cm long and $\displaystyle 3$ cm wide.