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Maths 5 Marks Questions

Integers · Jharkhand Board (JAC) STD 7 (Jharkhand - English Medium). 3 questions.

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Question 15 Marks
In a class test $(+ 3)$ marks are given for every correct answer and $(–2)$ marks are given for every incorrect answer and no marks for not attempting any question. Radhika scored $20$ marks. If she has got $12$ correct answers, how many questions has she attempted incorrectly?
Answer
Given that,
Marks given for every correct answer $= +3$
And, Marks given for every wrong answer $= -2$
Also, it is given that:
Marks obtained by Radhika $= 20$
Correct answer $= 12$
Hence,
Marks obtained for correct answers $= 12 \times 3 = 36$
Therefore,
Marks obtained for incorrect answers $=$ Total marks – Marks obtained for correct answers
$= 20 – 36 = - 16$
As, marks obtained for 1 wrong answer $= - 2$
Hence, Number of incorrect answers $= \frac{-16}{-2} = 8$
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Question 25 Marks
The temperature at $12$ noon was $10^\circ C$ above zero. If it decreases at the rate of $2^\circ C$ per hour until midnight, at which time would the temperature be $8^\circ C$ degrees below zero? What would temperature at midnight ?
Answer
Temperature at $12$ noon $= +10^\circ C$
Temperature at $1 P.M. = 10^\circ C – 2^\circ C = 8^\circ C$
Temperature at $2 P.M. = 8^\circ C – 2^\circ C = 6^\circ C$
Temperature at $3 P.M. = 6^\circ C – 2^\circ C = 4^\circ C$
Temperature at $4 P.M. = 4^\circ C – 2^\circ C = 2^\circ C$
Temperature at $5 P.M. = 2^\circ C – 2^\circ C = 0^\circ C$
Temperature at $6 P.M. = 0^\circ C – 2^\circ C = –2^\circ C$
Temperature at $7 P.M. = –2^\circ C – 2^\circ C = –4^\circ C$
Temperature at $8 P.M. = –4^\circ C – 2^\circ C = –6^\circ C$
Temperature at $9 P.M. = –6^\circ C – 2^\circ C = –8^\circ C$
So, the temperature was $8^\circ C$ below zero at $9 P.M.$
Temperature at $10 P.M. = –8^\circ C – 2^\circ C = –10^\circ C$
Temperature at $11 P.M. = –10^\circ C – 2^\circ C = –12^\circ C$
Temperature at $12 P.M. = –12^\circ C – 2^\circ C = –14^\circ C$
So, the temperature at midnight was $12$ degrees below zero.
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Question 35 Marks
Write five pairs of integers $(a, b)$ such that a $\div b = –3$. One such pair is $(6, –2)$ because $6 \div (–2) = (–3)$
Answer
Five pairs of integers $(a, b)$ such that a $\div b = –3$ are:
$1. [9, (–3)]$
$2. [12, (–4)]$
$3. [15, (–5)]$
$4. [–9, 3]$
$5. [–12, 4]$
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