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Applied Maths 2 Marks Questions

Numerical Applications · CBSE STD 12 Science (CBSE - English Medium). 15 questions.

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2 Marks Questions

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15 questions · timed · auto-graded

Question 12 Marks
Find the ratio of swimming speed of Raj in still water to speed of river, if ratio of time taken to go 10 km upstream to time taken to go 10 km downstream is $11: 5$?
Answer
Speed = $\frac{\text { Distance }}{\text { Time }}$
$\therefore$ $\frac{\text { Upstream time }}{\text { Downstream time }}=\frac{11}{5}$
$\therefore \frac {\text {Upstream Speed}}{\text {Downstream Speed}} = \frac {5}{11}$
$\therefore$ $\frac{X-Y}{X+Y}=\frac{5}{11}$
$\therefore$ $11X - 11Y = 5X + 5Y$
$\therefore$ $6X = 16Y$
$\therefore$ $\frac{X}{Y}=\frac{8}{3}$
$=\frac{\text { Swimming speed of Raj }}{\text { Speed of river }}$
$= 8 : 3$
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Question 22 Marks
Guddi speed of swimming in still water to that of river water is $7: 1.$ She swims 4.2 km up the river in just 14 min. How much time will Guddi take to swim 18.4 km down the river?
Answer
$X : Y = 7 : 1$
$\therefore$ $X = 7Y$
Speed = $\frac{\text { Distance }}{\text { Time }}$
$\therefore$ Upstream speed = $\frac{4.2~ \text {km}}{14~ \text{min}}$
$= 0.3$ km/min
Upstream speed $= X - Y$
$= 7Y - Y = 0.3$ km/min
$\therefore$ $Y = 0.05$ km/min
Downstream speed $= X + Y$
$= (7Y + Y) = 8Y$
$= 8 \times 0.05$
$= 0.4$ km/min
Downstream time = $\frac{\text { Distance }}{\text { Speed }}$
$=\frac{18.4~ \text{km}}{0.4~ \text{km/min} }$
Downstream time $=\frac{184}{4}$
$=46~ \text{minutes}$
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Question 32 Marks
Solve the linear inequality $|x - 2| >= 6$
Answer
We know that, $|x-a| \geq k$
$\Rightarrow-6 \leq x-2 \leq 6$
$\Rightarrow-4 \leq x<8$
$\Rightarrow x \in(-\infty,-4] \cup[8, \infty)$
Hence, the solution set of the inequation is $(-\infty, 4]$$\cup[8, \infty).$
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Question 42 Marks
In a 100 m race, A runs at 8 km/h. If A gives B a start of 4 m and still beats him by 15 seconds, what is the speed of B?
Answer
Time taken by A to cover 100 m $=\left(\frac{60 \times 60}{8000} \times 100\right) s$
= 45s
$\therefore$ B covers $(100 - 4)\text{m}$ $= 96~ \text{m}$ in $(45 + 15)s$
= 60s
$\therefore$ B's speed $=\left(\frac{96 \times 60 \times 60}{60 \times 1000}\right) km / h$
$= 5.76$ km/h
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Question 52 Marks
Two pipes A and B can fill a tank in 24 minutes and 32 minutes respectively. If both the pipes are opened simultaneously, after how much time B should be closed so that the tank is full in 18 minute?
Answer
Let B be closed after x minute.
Then, part filled by (A + B) in x min + part filled by A in $(18 - x)$ min = 1
$\therefore x\left(\frac{1}{24}+\frac{1}{32}\right)+(18-x) \times \frac{1}{24}=1$
$\Rightarrow$ $\frac{7 x}{96}+\frac{18-x}{24}=1$
$\Rightarrow$ $7x + 4(18 - x) = 96$
$\Rightarrow$ $x = 8$
Hence, B must be closed after 8 minutes.
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Question 62 Marks
There is a road beside a river. Two friends started from a place A, moved to a temple situated at another place B and then returned to A again. One of them moves on a cycle at a speed of $12$ km/h while the other sails on a boat at a speed of $10$ km/h. If the river flows at the speed of $4$ km/h which of the two friends will return to place A first ?
Answer
Here, cyclist moves both ways at a speed of $12$ km/h.
So, average speed of the cyclist $= 12$ km/h
The boat sailor moves at rate of $(10 \pm 4)$ i.e., 6 km/h
So, average speed of the boat sailor
$=\left(\frac{2 \times 14 \times 6}{14+6}\right) km / h$
$=\frac{42}{5} km / h =8.4 km / h$
Since, the average speed of the cyclist is greater, he will return to A first.
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Question 72 Marks
How many kg of wheat costing ₹ 8 per kg must be mixed with 36 kg wheat costing ₹ 5.40 per kg so that $20\%$ gain may be obtained by selling the mixture at ₹ 7.20 per kg?
Answer
S.P. of 1 kg mixture = ₹ 7.20, Gain = 20%
$\therefore$ C.P. of 1 kg mixture = ₹$\left(\frac{100}{120} \times 7.20\right)=$ ₹ 6
By the rule of alligation, we have:
Image
Wheat of 1st kind: Wheat of 2nd kind $= 60:200 $
$= 3:10$
Let x kg of wheat of 1st kind be mixed with 36 kg of wheat of 2nd kind.
Then, $3: 10 = x :36$ or $10x = 3\times36$ or $x = 10.8$ kg
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Question 82 Marks
Tap P alone fills a cistern in 2 hours; while tap Q alone fills the same cistern in 3 hours. A new tap R is attached to the bottom of the cistern which can empty the completely filled cistern in 6 hours. Sunny started all three taps together at 9 a.m. When will the tank be full?
Answer
Cistern filled or work done by Tap P in 1 hour $=\frac{1}{2}$
Cistern filled or work done by Tap Q in 1 hour $=\frac{1}{3}$
Tank emptied or work done by Tap R in 1 hour $=\frac{1}{6}$
Tank filed or work done by all three pipes in 1 hour $=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}$
So Tank gets completely filled in $\frac{3}{2}$ hours.
$=$ 1.5 hours $=$ 1 hr 30 min
So time will be 9.00 a.m. $+$ 1 hour 30 min $=$ 10.30 a.m.
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Question 92 Marks
Tea worth ₹ 126 per kg and ₹ 135 per kg are mixed with a third variety in the ratio $1:1:2.$ If the mixture is worth ₹ 153 per kg, then find the price of the third variety in per kg.
Answer
Since first and second varieties are mixed in equal proportions.
So, their average price $=$ ₹$\left(\frac{126+135}{2}\right)$
$=$ ₹ $130.50$
So, the mixture is formed by mixing two varieties, one at ₹ 130.50 per kg and the other at say ₹ x per kg in the ratio $2: 2,$ i.e., $1: 1.$ We have to find x.
By the rule of alligation, we have:
Image
$\therefore$ $\frac{x-153}{22.50}=1$
$\Rightarrow$ $x - 153 = 22.50$
$\Rightarrow$ $x = 175.50$ $\therefore$ ₹ $175.50/$kg
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Question 102 Marks
Let x and b are real numbers. If $b > 0$ and $|x| > b,$ then find interval for $x.$
Answer
Given, $|x| > b, b > 0$
$\Rightarrow$ $x < - b ~\text{or} ~x >b$
$\Rightarrow$ $x \in(-\infty,-b) \cup(b-, \infty)$
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Question 112 Marks
Solve for x(x is real number), $4x + 3 < 5x + 7.$
Answer
Given, $4x + 3 < 5x + 7$
Transposing $5x$ to LHS and $3$ to RHS, we get
$4x - 5x < 7 - 3$
оr, $- x < 4$
or, $x > - 4$
$\therefore$When $x$ is real, the solution is $(-4, \infty).$
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Question 122 Marks
In a game of 80 points, A can give 5 points to B and 15 points to C. Calculate the number of points that B can give to C in a game of 60 points?
Answer
$A: B=80: 75, A: C = 80: 65$
$\frac{B}{C}=\left(\frac{B}{A} \times \frac{A}{C}\right)$
$=\left(\frac{75}{80} \times \frac{80}{65}\right)=\frac{15}{13}$
$=\frac{60}{52}=60: 52$
$\therefore$ In a game of 60, B can give C 8 points.
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Question 132 Marks
An electric pump can fill a tank in 3 hours. Because 1 of a leak in the tank it took $3 \frac{1}{2}$ hours to fill the tank. If the tank is full, how much time will the leak take to empty it?
Answer
Work done by the leak in 1 hour $=\left[\frac{1}{3}-\frac{1}{\left(\frac{7}{2}\right)}\right]$
$=\left(\frac{1}{3}-\frac{2}{7}\right)=\frac{1}{21}$
$\therefore$ The leak will empty the tank in 21 hours.
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Question 142 Marks
A man can row upstream at 7 km/h and downstream at 10 km/h. Find man's rate in still water and rate of current.
Answer
Rate in still water $=\frac{1}{2}(10+7) km / h =8.5 km / h$
Rate of current $=\frac{1}{2}(10-7) km / h =1.5 km / h$
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2 Marks Questions - Applied Maths STD 12 Science Questions - Vidyadip