Sequences and Series Test 26

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Sequences and Series Test 26

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Question 1
1 / -0

With the help of match-sticks, Zalak prepared a pattern as shown below. When $97$ matchsticks are used, the serial number of the figure will be ...........

A
Figure 32

B
Figure 95

C
Figure 49

D
Figure 48

Solution

Pattern associated with the number of matchsticks:
Figure 1: No. of sticks $=3=3+0$
Figure 2: No. of sticks $=5=3+2$
Figure 3: No. of sticks $=7=3+2+2$
Figure 4: No. of sticks $=9=3+2+2+2$
----------------------------------------
The rule associated with it can be given as no. of sticks $=3+2(n-1)$ where $n$ is the number of the figure.
Now there are $97$ matchsticks
$\implies 3+2(n-1)=97$
$\implies 3+2n-2=97$
$\implies 2n=96$
$\implies n=48$
Hence, serial number of the figure having $97$ matchsticks is $48$ .
Question 2
1 / -0

Ten points lie in a plane so that no three of them are collinear. The number of lines passing through exactly two of these points are dividing the plane into two regions each containing four of the remaining points is

A
1

B
5

C
10

D
Dependent on the configuration of points.

Solution

You can clearly see from the attached figure
Five such lines are possible joining $AF,BG,CH,DI,EJ$
Question 3
1 / -0

Area of the triangle formed by the points $(0,0),(2,0)$ and $(0,2)$ is

A
$1$ sq.units

B
$2$ sq.units

C
$4$ sq.units

D
$8$ sq.units

Solution

Area of triangle having vertices $(x_1,y_1), (x_2,y_2)$ and $(x_3,y_3)$ is given by
Area $= \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ]$
Therefore,
$=\dfrac{1}{2}\left|0(0-2)+2(2-0)+0(0-0)\right|$

$=\dfrac{1}{2}\left|0+4+0\right|$

$=\dfrac{1}{2}\times 4=2\ sq.\ units$

Hence, this is the answer.
Question 4
1 / -0

Find the value of $a$ if area of the triangle is $17$ square units whose vertices are $(0,0), (4,a), (6,4)$ .

A
$a=-2$

B
$a=-3$

C
$a=3$

D
none of the above

Solution

vertices of the triangle are $A(0,0), B(4,a), C(6,4)$ .

Then area of triangle $ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$\Rightarrow 17=\dfrac{1}{2}[0(a-4)+4(4-0)+6(0-a)$

$\Rightarrow 34=16-6a$

$\Rightarrow a=-\dfrac{18}{6}=-3$
Question 5
1 / -0

Out of $7$ consonants and $4$ vowels, words are formed each having $3$ consonants and $2$ vowels. The number of such words that can be formed is

A
$210$

B
$25200$

C
$2520$

D
$302400$

Solution

Total no. of consonants is $7$ and of vowels is $4$
Out of $7$ consonants, $3$ are chosen to form a word. So, this can be done in ${ }^{7}C_{3}$ ways
Out of $4$ vowels, $2$ are chosen to form a word. So, that can be done in ${ }^{4}C_{2}$ ways
So, the no. of select $3$ consonants and $2$ vowels is $^{7}C_{3} \times ^{4}C_{2}$
Now, we can arrange the word containing $5$ letters in $5!$ ways.
Thus, the total no. of words formed each having $3$ consonants and $2$ vowels is
$^{7}C_{3} \times ^{4}C_{2}\times 5! = 25200$
Hence, the answer is $25200$ .
Question 6
1 / -0

Find the term independent of $x$ in ${ \left( \cfrac { 3 }{ 2 } { x }^{ 2 }-\cfrac { 1 }{ 3x } \right) }^{ 9 }$ .

A
$\dfrac {6}{15}$

B
$\dfrac {7}{18}$

C
$\dfrac {7}{8}$

D
$\dfrac {4}{9}$

Solution

We know that the general term in expansion of $(a+b)^n$ is given by $^nC_{r}a^{n-r}b^{r}$ where r ranges from $0$ to $9$ .
Here $a= \cfrac{3}{2}x^2, \ b= -\cfrac{1}{3x} \ \And \ n=9$
For any term to be independent of $x$ , coefficient of $x$ in $a^{n+1-r}b^{r-1}$ should be zero.
$\Rightarrow \ (x^2)^{9-r}(x^{-1})^{r}=x^0$
$\Rightarrow 18-2r-r=0$
$\Rightarrow r=6$
Therefore, the term independent of $x$ is $^9C_6 \left (\cfrac{3}{2}\right)^3 \left (-\cfrac{3}{3}\right)^6=\cfrac{7}{18}$ .
Question 7
1 / -0

Choose $3, 4, 5$ points other than vertices respectively on the sides $AB, BC$ and $CA$ of a $\triangle ABC$ . The number of triangles that can be formed by using only these points as vertices, is

A
$220$

B
$217$

C
$215$

D
$210$

E
$205$

Solution

Required number of triangles that can be formed by using only given points as vertices
$= ^{12}C_{3} - \left \{^{3}C_{3} + ^{4}C_{3} + ^{5}C_{3}\right \}$
$= \dfrac {12\times 11\times 10}{3\times 2\times 1} - \left \{1 + 4 + \dfrac {5\times 4}{2\times 1}\right \}$
$= 220 - (5 + 10)$
$= 220 - 15 = 205$
Question 8
1 / -0

If $f(x)=\left| \log { \left| x \right| } \right|$ , then

A
$f(x)$ is continuous and differentiable for all $x$ in its domain

B
$f(x)$ is continuous for all $x$ in its domain but not differentiable at $x=\pm 1$

C
$f(x)$ is neither continuous nor differentiable at $x=\pm 1$

D
None of the above

Solution

It is evident from the graph of $f(x)=\left| \log { \left| x \right| } \right| \quad$ that $f(x)$ is everywhere continuous but not differentiable at $x=\pm 1$
Question 9
1 / -0

Three bells commenced to toll at the same time and tolled at intervals of $20, 30, 40$ seconds respectively. If they toll together at $6$ am, then which of the following is the time at which they can toll together

A
$6:55$ am

B
$6:56$ am

C
$6:57$ am

D
$6:59$ am

Solution
Question 10
1 / -0

The first term of an AP is $148$ and the common difference is $-2$ . If the AM of first $n$ terms of the AP is $125$ , then the value of $n$ is

A
$18$

B
$24$

C
$30$

D
$36$

E
$48$

Solution

Given, $a=148, d=-2$
AM of $n$ terms $=125$
$\Rightarrow \dfrac { { a }_{ 1 }+{ a }_{ 2 }+\cdots +{ a }_{ n } }{ n } =125$
$\Rightarrow \dfrac { \dfrac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] }{ n } =125$
$\Rightarrow 2a+\left( n-1 \right) d=250$
$\Rightarrow 2\times 148+\left( n-1 \right) \left( -2 \right) =250$
$\Rightarrow 296-2n+2=250$
$\Rightarrow 298-2n=250$
$\Rightarrow n=24$

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Sequences and Series Test 26

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Sequences and Series Test 26

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Question 1

Figure 32

Figure 95

Figure 49

Figure 48

Solution

Question 2

1

5

10

Dependent on the configuration of points.

Solution

Question 3

111 sq.units

222 sq.units

444 sq.units

888 sq.units

Solution

Question 4

a=−2a=-2a=−2

a=−3a=-3a=−3

a=3a=3a=3

none of the above

Solution

Question 5

210210210

252002520025200

252025202520

302400302400302400

Solution

Question 6

615\dfrac {6}{15}156​

718\dfrac {7}{18}7 ​18

78\dfrac {7}{8}87​

49\dfrac {4}{9}94​

Solution

Question 7

220220220

217217217

215215215

210210210

205205205

Solution

Question 8

f(x)f(x)f(x) is continuous and differentiable for all xxx in its domain

f(x)f(x)f(x) is continuous for all xxx in its domain but not differentiable at x=±1x=\pm 1x=±1

f(x)f(x)f(x) is neither continuous nor differentiable at x=±1x=\pm 1x=±1

None of the above

Solution

Question 9

6:556:556:55am

6:566:566:56am

6:576:576:57am

6:596:596:59am

Solution

Question 10

181818

242424

303030

363636

484848

Solution

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$1$ sq.units

$2$ sq.units

$4$ sq.units

$8$ sq.units

$a=-2$

$a=-3$

$a=3$

$210$

$25200$

$2520$

$302400$

$\dfrac {6}{15}$

$\dfrac {7}{18}$

$\dfrac {7}{8}$

$\dfrac {4}{9}$

$220$

$217$

$215$

$210$

$205$

$f(x)$ is continuous and differentiable for all $x$ in its domain

$f(x)$ is continuous for all $x$ in its domain but not differentiable at $x=\pm 1$

$f(x)$ is neither continuous nor differentiable at $x=\pm 1$

$6:55$ am

$6:56$ am

$6:57$ am

$6:59$ am

$18$

$24$

$30$

$36$

$48$