3 Marks Question

Question 13 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.
(i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly?
(ii) Mohini scores ( -5 ) marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?

Answer

Given, marks for every correct answer $=+3$
Marks for every incorrect answer $=-2$
(i) Total marks scored by Radhika $=20$
Marks scored for correct answer $=12 \times 3=36$
$\therefore$ Marks given for incorrect answers
$=$ Total marks - Marks given for correct answers
$=20-36=-16$
$\therefore$ Number of incorrect answers
$
\begin{array}{l}
=\frac{\text { Total marks for incorrect answers }}{\text { Marks for every incorrect answer }} \\
=\frac{(-16)}{(-2)}=\frac{-16}{-2}=8
\end{array}
$
Hence, Radhika attempted 8 questions incorrectly.
(ii) Marks obtained by Mohini $=-5$
Marks obtained for 7 correct answers $=7 \times 3=21$
$\therefore$ Marks obtained for incorrect answers
$
=-5-21=-26
$
$\therefore$ Number of incorrect answers
$
\begin{array}{l}
=\frac{\text { Total marks obtained for incorrect answers }}{\text { Marks for every incorrect answer }} \\
=\frac{-26}{-2}=13
\end{array}
$

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Question 23 Marks

The temperature at 12 noon was $10^{\circ} C$ above zero. If it decreases at the rate of $2^{\circ} C / h$ untll mid-night, at what time would the temperature be $8^{\circ} C$ below zero? What would be the temperature at mid-night?

Answer

Given, temperature at 12 noon $=+10^{\circ} C$
Rate of change in temperature $=-2^{\circ} C / h$
Total time from 12 noon to mid-night $=12 h$
Change in temperature in $12 h=12^{\circ} C \times(-2)=-24^{\circ} C$
Temperature at mid-night $=10^{\circ} C +\left(-24^{\circ}\right) C$
$
=(10-24)^{\circ} C=-14^{\circ} C
$
Now, temperature difference between $10^{\circ} C$ and $-8^{\circ} C$
$
\begin{array}{l}
=10-(-8)=10+8 \\
=18^{\circ} C \\
=\frac{18}{2}=9
\end{array}
$
So, temperature change of $18^{\circ} C$ will take place in 9 h from 12 noon.
Thus, the temperature $8^{\circ} C$ below $0^{\circ}\left(-8^{\circ} C \right)$ would be at 9 pm .

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Question 33 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

(i) We know that $3 \div1=3$, so $3 \div(-1)=-3$
On comparing with $a \div b=-3$, we get
$ a=3 \text { and } b=-1 $
Hence, the required pair of integers is $(3,-1)$.
(ii) $(12,-4)$.
(iii) $(9,-3)$.
(iv) $(-12,4)$.
(v) $(15,-5)$.

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Question 43 Marks

In a quiz, Team A scored -40, 10, 0 and Team B scored 10, 0, -40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?

Answer

Given, Ist round score of Team A = -40,
IInd round score of Team A = 10
and IIIrd round score of Team $A=0$
$\therefore$ Total score of Team $A=$ Sum of all rounds
$
\begin{array}{l}
=-40+10+0 \\
=-40+10=-30
\end{array}
$
Similarly, Ist round score of Team $B=10$,
IInd round score of Team $B=0$
and IIIrd round score of Team $B=-40$
$\therefore$ Total score of Team $B= Sum$ of all rounds
$
\begin{array}{l}
=10+0+(-40) \\
=10-40=-30
\end{array}
$
$\because$ Total score of Team $A=$ Total score of Team $B=-30$
So, the score of both teams are equal.
Yes, we can add integers in any order.

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