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Applied Maths 2 Marks Questions

Numbers and Quantification · CBSE STD 12 Science (CBSE - English Medium). 68 questions.

Questions · Page 2 of 2

2 Marks Questions

Question 512 Marks
Two runners A and B complete a 100 metres race in 36seconds and 48 seconds respectively, by how
many metres will A defeat B?
Answer
The distance covered by $B$, after $A$ has finished the race i.e. distance covered by $B$ in
$
(48-36=12 s)
$
Distance $=(100 / 48) \times 12=25$ meters
$A$ beats $B$ by 25 metres.
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Question 522 Marks
Priya can run 1 km in 3 min 10 sec and Preeti can cover the same distance in 3 min 20 sec. By what
distance can Priya beat Preeti?
Answer
Priya beats Preeti by $(200-190) sec =10 sec$
Distance covered by Preeti in $10 sec =\frac{1000}{200} \times 10=50 m$
Therefore, Priya beats Preeti by 50 meters.
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Question 532 Marks
In a 1000 metres race. A, B, and C get the gold, silver, and bronze medals, respectively. If A beats
B by 100 metres and B beats C by 100 metres, then by how many metres does A beat C?
Answer
$A$ beats $B$ by 100 metres, means $A$ travels 1000 metres in the same time in which $B$ travels 900 metres.
$B$ beats $C$ by 100 metres, means $B$ travels 1000 metres in the same time in which $C$ travels 900 metres.
$A: B=10: 9$
$B: C=10: 9$
$\Rightarrow$ The ratio of $A: B: C=100: 90: 81$
$A$ travels 100 metres and in the same time $C$ travels 81 metres.
Therefore when $A$ travels 1000 metres, in the same time $C$ travels 810 metres.
Thus $A$ beats $C$ by 190 metres.
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Question 542 Marks
A person can row a boat 5 km an hour in still water. It takes him thrice as long to row upstream as to row downstream. Find the rate at which the stream is flowing.
Answer
Let the rate at which the stream is flowing be $x km / hr$ and let the distance covered by the boat be y km .According to the question
$
\begin{array}{l}
\frac{3 y}{5+x}=\frac{y}{5-x} \\
\Rightarrow 3(5-x)=5+x \\
\Rightarrow 15-3 x=5+x \\
\Rightarrow 4 x=10 \\
\therefore x=2.5
\end{array}
$
The stream is flowing at the rate of $2.5 km / hr$.
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Question 552 Marks
A man rows 15 km upstream and 25 km downstream in 5 hours each time. What is the speed of the current?
Answer
Let the speed of the boat in still water be $x km / hr$ and the speed of the current be $y km / hr$ then $(x+y) \times 5=25$ and $(x-y) \times 5=15$
$($ Distance $=$ Speed $\times$ Time $)$
and
$
\Rightarrow x+y=5
$
On solving we get, $x=4$ and $y=1$
Thus, speed of the current is $1 km / hr$.
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Question 562 Marks
If X = 7 and Y = 3, then verify that X mod Y = (X +kY) mod Y for k = 6
Answer
For given $X=7$ and $Y=3$
$X \bmod Y=7 \bmod 3=1$
Also, $(X+k Y) \bmod Y=(7+6 \times 3) \bmod 3=25 \bmod 3= 1$
As $25=8 \times 3+1$
Therefore, $7 \bmod 3=25 \bmod 3=1$
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Question 612 Marks
Find $\mu(12)$ and $\mu(35)$
Answer
As $12=2^2 \times 3$
thus $\mu(12)=0$
and $35=5 \times 7$
therefore $\mu(35)=(-1)^2=1$
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Question 642 Marks
Find $\varphi(12)$
$
12=2^2 \times 3
$
Answer
$\varphi(12)=12\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)$ $=12 \times \frac{1}{2} \times \frac{2}{3}=4$
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Question 652 Marks
Find $\varphi(n)$ for $n =3$
Answer
If n = 3,
as, gcd (1,3) = 1 and gcd (2,3) = 1 (gcd = greatest common divisor)
Therefore, numbers coprime to 3 are 1 and 2
Hence $\varphi$(3) = 2
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Question 662 Marks
Find $5^6(\bmod 4)$
Answer
$
\begin{aligned}
5 & \equiv 1(\bmod 4)=1 \\
5^6 & \equiv 1^6(\bmod 4) \quad(\text { Using property } 5)
\end{aligned}
$
Hence, $5^6(\bmod 4)=1$
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Question 672 Marks
If $a=31, b=21, m=5$, then check $a a^{\circ} b(\bmod m)$.
Answer
a – b = 31 – 21 =10 which is divisible by 5
and a mod m = b mod m
31 mod 5 = 21 mod 5 = 1
31 = 21 (mod 5)
Hence, a =b (mod m) is true
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Question 682 Marks
Verify a  b (mod m), if a = 41, b = 21 and m = 5
Answer
a - b = 41 - 21 = 20, which is divisible by 5
a mod m = 41 mod 5 = 1
b mod m = 21 mod 5 = 1
therefore, 41 = 21 (mod 5)
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2 Marks Questions - Page 2 - Applied Maths STD 12 Science Questions - Vidyadip