Differentiation Test 10

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Differentiation Test 10

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Weekly Quiz Competition

Question 1
1 / -0

Find the next number in the series.
$$3, 6, 9, 12, 15,....$$

A
$$17$$

B
$$18$$

C
$$19$$

D
$$20$$

Solution

The next number is $$18$$.
Since the numbers are multiple of $$3$$.
$$3, 6, 9, 12, 15, \underline {18 }$$.
Question 2
1 / -0

Given three vertices of a triangle whose coordinates are A (1, 1), B (3, -3) and (5, -3). Find the area of the triangle.

A
3 square units

B
4 square units

C
5 square units

D
6 square units

Solution

We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
The three vertices of the triangle are $$A(1,1), B(3,-3), C(5,-3)$$
Area of triangle $$=\dfrac{|1(-3-(-3)+3(-3-1)+5(1-(-3))|}{2}$$
$$=\dfrac{|1(0)+3(-4)+4(4)|}{2}$$
$$=\dfrac{|-12+20|}{2}$$
$$=\dfrac{8}{2}$$
$$=4$$ square units.
Question 3
1 / -0

The area of triangle whose vertices are $$A (-3, -1), B(5, 3)$$ and $$C(2, -8)$$ is ____ $$\text{ sq. units}$$.

A
$$34\text{ sq. units}$$

B
$$36\text{ sq. units}$$

C
$$38\text{ sq. units}$$

D
$$32\text{ sq. units}$$

Solution

We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$

The given vertices of the triangle are $$A(-3,-1), B(5,3)$$ and $$C(2,-8)$$.
So, by using the above formula,
$$\begin{aligned}{}\text{Area of the triangle} &= \frac{1}{2} {[ - 3(3 - ( - 8)) + 5( - 8 - ( - 1)) + 2( - 1 - 3)]}\\ &= \frac{1}{2}{[ - 33 - 35 - 8]}\\ &= \frac{{[ - 76]}}{2}\\& = \frac{{76}}{2}\quad\quad\quad\quad\dots[\text{Area can never be negative so, we ignore negative sign}]\\& = 38\text{ sq. units}\end{aligned}$$

So, the area of the triangle is equal to $$38\text{ sq. units}$$.
Question 4
1 / -0

If $$y$$ is expressed in terms of a variable $$x$$ as $$y = f(x)$$, then $$y$$ is called

A
Explicit function

B
Implicit function

C
Linear function

D
Identity function

Solution

If $$y$$ is expressed in terms of a variable $$x$$ as $$y=f(x)$$, then $$y$$ is called explicit function.
Question 5
1 / -0

If there are five consecutive integer in a series and the first integer is $$1$$, what is the value of the last consecutive integer?

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

Let the five consecutive integers be $$x, x + 1, x + 2, x + 3$$ and $$x + 4$$.
Therefore, the value of first integer is $$1$$ i.e., $$x = 1$$
So, the last integer is $$x + 4$$
The value of last integer $$= 5$$.
Question 6
1 / -0

$$10,$$ __$$, 15, 15, 20, 20, 25, 25,...$$. What number should fill the blank?

A
7

B
8

C
10

D
15

Solution

Here addition with repetition series, each number in the series repeats itself, and then increases by $$5$$ to get the next number.
The next number is $$10$$.
Question 7
1 / -0

The average IQ of $$4$$ people is $$110$$. If three of these people each have an IQ of $$105$$, what is the IQ of the fourth person?

A
110

B
115

C
120

D
125

Solution

Average IQ of $$4$$ people $$=110$$
$$\therefore$$ total IQ of $$4$$ people $$=$$ $$110\times 4=440$$
Average IQ of $$3$$ people $$=105$$
$$\therefore $$ total IQ of $$3$$ people $$=$$ $$105\times 3=315$$
Then IQ of forth person $$=$$ $$440-315=125$$
Question 8
1 / -0

A garrison of '$$n$$' men had enough food to last for $$30$$ days. After $$10$$ days, $$50$$ more men joined them. If the food now lasted for $$16$$ days, what is the value of $$n$$?

A
$$200$$

B
$$240$$

C
$$280$$

D
$$320$$

Solution

After $$10$$ days, the food for n men is there for $$20$$ days.
This food can be eaten by $$(n + 50)$$ men in $$16$$ days.
$$\therefore 20n = 16(n + 50)$$
$$\therefore n = 200$$
Question 9
1 / -0

A bag contains Rs. $$112$$ in the form of $$1$$-rupee, $$50$$-paise and $$10$$-paise coins in the ratio $$3 : 8 : 10$$. What is the number of $$50$$-paise coins?

A
$$112$$

B
$$128$$

C
$$96$$

D
$$24$$

Solution

Here bag contains Rs. $$112$$ in the form of $$1$$-rupee, $$50$$-paise and $$10$$-paise.
Coin ratio is $$ 3 : 8 : 10$$.
Value ratio $$= 3 \times 1: 8\times \dfrac {1}{2} : 10 \times \dfrac {1}{10}$$
$$= 3 : 4 : 1$$
Number of $$50$$ paise coins $$= \dfrac {4}{8} \times$$ Rs. $$112 =$$ Rs. $$56$$
Therefore, number of $$50$$ paise coins $$= 2\times 56 = 112$$.
Question 10
1 / -0

The sum of two numbers is $$80$$. If the larger number exceeds four times the smaller by $$5$$, what is the smaller number?

A
$$5$$

B
$$15$$

C
$$20$$

D
$$25$$

Solution

Given, sum of two numbers is $$80$$.
Let the smaller number be $$x$$
Thus the larger will be $$80 - x$$.
Also given, larger number exceeds four times the smaller number by $$5$$.
Therefore, $$ 80 - x = 4x + 5$$
$$\Rightarrow 5x = 75$$
$$\Rightarrow x = 15$$
Thus the smaller number is $$15$$.

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Differentiation Test 10

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Differentiation Test 10

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SHARING IS CARING

Question 1

$$17$$

$$18$$

$$19$$

$$20$$

Solution

Question 2

3 square units

4 square units

5 square units

6 square units

Solution

Question 3

$$34\text{ sq. units}$$

$$36\text{ sq. units}$$

$$38\text{ sq. units}$$

$$32\text{ sq. units}$$

Solution

Question 4

Explicit function

Implicit function

Linear function

Identity function

Solution

Question 5

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 6

7

8

10

15

Solution

Question 7

110

115

120

125

Solution

Question 8

$$200$$

$$240$$

$$280$$

$$320$$

Solution

Question 9

$$112$$

$$128$$

$$96$$

$$24$$

Solution

Question 10

$$5$$

$$15$$

$$20$$

$$25$$

Solution

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