Question 115

From given integers in Column I, match an integer of Column II, so that their product lies between - 19 and -6.
$\begin{matrix} Column I & Column II - 5 & 1 6 & - 1 - 7 & 3 8 & - 2 \end{matrix}$

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Solution

$- 5 \times 3 = (- 15)$ which lies between -19 and -6.
$6 \times (- 2) = (- 12)$ which lies between -19 and -6.
$- 7 \times 1 = (- 7)$ which lies between -19 and -6.
$8 \times (- 1) = (- 8)$ which lies between -19 and -6.

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Modulus Function

Standard IX Mathematics

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Question 115 From given integers in Column I, match an integer of Column II, so that their product lies between - 19 and -6. Column IColumn II−5 1 6−1−7 3 8−2

−5×3=(−15) which lies between -19 and -6. 6×(−2)=(−12) which lies between -19 and -6. −7×1=(−7) which lies between -19 and -6. 8×(−1)=(−8) which lies between -19 and -6.

Question 115

From given integers in Column I, match an integer of Column II, so that their product lies between - 19 and -6.
$\begin{matrix} Column I & Column II - 5 & 1 6 & - 1 - 7 & 3 8 & - 2 \end{matrix}$

$- 5 \times 3 = (- 15)$ which lies between -19 and -6.
$6 \times (- 2) = (- 12)$ which lies between -19 and -6.
$- 7 \times 1 = (- 7)$ which lies between -19 and -6.
$8 \times (- 1) = (- 8)$ which lies between -19 and -6.