Mathematics Test

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Mathematics Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

What is the area of the region bounded by the lines y = x, y = 0 and x = 4?

A
4 sq units

B
8 sq units

C
12 sq units

D
16 sq unit

Solution
Question 2
1 / -0

Four points with position vectors $A=7i-4 j+7 k, B=1-6 j+10 k, c=-1-3 j+4 k$ and $D=5 i-$ $j+k$ form.

A
rhombus

B
parallelogram but not rhombus

C
rectangle

D
square

Solution

AB= position vector of B− position vector of A $=i - 6j + 10 k - (7i - 4j + 7 k) = (- 6, - 2, 3)$ $B C = (- 2, 3, - 6)$ $C D = (6, 2, - 3)$ $D A = (2, - 3, 6)$ All the vectors have same modulus =√(36+4+9)=7

Hence, the points enclose either a square or a rhombus.
But AB⋅BC=12−6−18=−12≠0
Hence, AB not perpendicular to BC.

∴ $A B C D i s a r h o m b u s .$
Question 3
1 / -0

The measure of each interior angle of a regular convex polygon is 156°. The number of sides of the polygon is?

A
10

B
8

C
15

D
12

Solution
Question 4
1 / -0

A
0

B
1

C
2

D
3

Solution
Question 5
1 / -0

Two circles touch internally. The sum of their areas is $116 \pi \mathrm{cm}^{2} \&$ distance between their centres is $6 \mathrm{cm}$. Find the radii of the circles.

A
10, 4

B
8, 6

C
12, 2

D
5, 2

Solution

Let $R \& r$ be the radii of the circles having centres at $0 \& 0^{\prime}$ respectively. Then sum of areas $=116 \pi \mathrm{cm}^{2}$ $\Rightarrow \pi R^{2}+\pi r^{2}=116 \pi$
$\Rightarrow R^{2}+r^{2}=116$
Distance between the centres $=6 \mathrm{cm}$
$\Rightarrow 00^{\prime}=6 \mathrm{cm}$
$\Rightarrow R-r-6$
Now $(R+r)^{2}+(R-r)^{2}=2\left(R^{2}+r^{2}\right)$
$(R+r)^{2}+36=2 \times 116 \quad[\text { using }(1) \&(2)]$
$(R+r)^{2}=(2 \times 116-36)$
$=196$
$R+r=14$
(3)
Solving
(2) & (3) we get $R=10 \& r=4$
Question 6
1 / -0

The coordinates of one end point of a diameter of a circle are (4, –1) and the coordinates of the centre of the circle are (1, –3). Find the coordinates of the other end of the diameter.

A
(–2, 3)

B
(0, 5)

C
(3,–6)

D
(–2, –5)

Solution

Let $A B$ be a diameter of the circle having its centre at $\mathrm{C}(1,-3)$ such that the coordinates of one end $\mathrm{A}$ are (4,-1) Let the coordinates of $\mathrm{B}$ be $(\mathrm{x}, \mathrm{y})$ since $C$ is the mid point of AB. Therefore, the coordinates of $\mathrm{C}$ are $\left(\frac{x+4}{2}, \frac{y-1}{2}\right)$

But the coordinates of $C$ are given to be (1,-3)
$\therefore \frac{x+4}{2}=1$ and $\frac{y-1}{2}=-3$
$\Rightarrow x+4=2 \quad$ and $y-1=-6$
$\Rightarrow x=-2$ and $y=-5$
Hence the coordinates of $\mathrm{B}$ are (-2,-5).
Question 7
1 / -0

A number x is selected from the numbers 1, 2, 3 and then a second number y is randomly selected from the numbers 1, 4, 9. What is the probability that the product xy of the two numbers will be less than 9?

A
1/8

B
5/8

C
3/9

D
5/9

Solution

Number x can be selected in three ways and corresponding to each such way there are three ways of selecting number y. Therefore, two numbers can be selected in 9 ways as listed below:

(1, 1) (1, 4) (1, 9), (2, 1), (2, 4), (2, 9), (3, 1), (3, 4), (3, 9)

So total number of elementary events = 9.

The product xy will be less than 9, if x & y are chosen in one of the following ways:

(1, 1), (1, 4), (2, 1), (2, 4), (3, 1)

∴ favourable number of elementary events = 5

Hence required probability = 5/9.
Question 8
1 / -0

If the number of sides of a regular polygon is ‘n’, then the number of lines of symmetry is equal to?

A
2n

B
n

C
n/2

D
n²

Solution

If we take a regular pentagon.

Regular pentagon has 5 sides

From figure we can see that there are exactly 5 lines of symmetry.
Question 9
1 / -0

The average age of 25 boys in a class decreases by 6 months when a new boy takes the place of a 20 years old boy. Find the age of new boy.

A
7 yr

B
7.5 yr

C
8 yr

D
8.5 yr

Solution

Total age decreased $=(\text { average age } \times \text { average decrement })$ $=25 \times \frac{1}{2}=12.5$
Thus, age of new boy $=20-12.5=7.5$ years
Question 10
1 / -0

The coordinates of a point on the parabola y² = 8x, whose focal distance is 4, is?

A
(2, 4)

B
(4, 2)

C
(–2, –4)

D
(4, –2)

Solution

Given $a+x=4$ $8 y^{2}=4 a x=8 x$
$\Rightarrow a=2$
$\Rightarrow 2+x=4$
$x=2$
On putting $x=2$ in $y^{2}=8 x,$ we get $y=\pm 4$ Hence coordinates of a point is (2,4) or (2,-4)