Mathematics (Standard) Test 1

Least composite number is 4 and the least prime number is 2.

LCM (4,2) : HCF (4,2)

= 4:2

= 2:1

For lines to coincide:

\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

So , \(\frac{5}{15}=\frac{7}{21}=\frac{-3}{-k}\)

i.e. k = 9

By Pythagoras theorem

The required distance

\(=\sqrt{(200)^2+(150)^2}\)

\(=\sqrt{40000+22500}\)

\(=\sqrt{62500}\)

= 250

So the distance of the girl from the starting point is 250m

Area of the Rhombus \(=\frac{1}{2}d_1d_2\)

\(=\frac{1}{2}\times24\times32\)

\(=384\;cm^2.\)

Using Pythagoras theorem

\(side^2=(\frac{1}{2}d_1)^2+(\frac{1}{2}d_2)^2\)

\(=12^2+16^2\)

\(=144+256\)

\( =400\)

Side = 20cm

Area of the Rhombus

= base \(\times\) altitude

\(384=20\times\) altitude

So altitude

\(=\frac{384}{20}\)

= 19.2cm

Possible outcomes are (HH), (HT), (TH), (TT)

Favorable outcomes(at the most one head) are (HT), (TH), (TT)

So probability of getting at the most one head  \(=\frac{3}{4}\)

Ratio of altitudes = Ratio of sides for similar triangles

So,

\(AM:PN = AB:PQ = 2:3\)

\(2sin^2\beta - cos^2\beta = 2, \)

Then

 \(2\;sin^2\beta-(1-sin^2\beta)=2\)

\(\Rightarrow3\;sin^2\beta=3\)

or

\(\Rightarrow sin^2\beta =1\)

\(\beta\) is \(90^{\circ}\)

Since it has a terminating decimal expansion,

so prime factors of the denominator will be 2,5.

Lines \(x=a\) is a line parallel to y axis and \(y=b\) is a line parallel to x axis.

So they will intersect.

Distance of point A(-5,6) from the origin(0,0) is

\(=\sqrt{(0+5)^2+(0-6)^2}\)

\(=\sqrt{25+36}\)

\(=\sqrt{61}\) units

\(a^2=\frac{23}{25},\) then

\(a=\frac{\sqrt{23}}{5},\) which is irrational

LCM \(\times\) HCF = Product of two numbers

\(36\times 2=18\times x\)

\(x = 4 \)

\(tan\;A=\sqrt3=tan\;60^{\circ}\) so

\(\angle A=60^{\circ},\)

Hence \(\angle C=30^{\circ}.\)

So \(cos \;A \;cos\; C- sin\; A sin\; C\)

\(=(\frac{1}{2})\times(\frac{\sqrt3}{2})-(\frac{\sqrt3}{2})\)\(\times(\frac{1}{2})\)

 = 0 

\(1x +1x +2x =180^{\circ}\)

\(x=45^{\circ}.\)

\(\angle A,\angle B\) and \(\angle C\) are \(45^{\circ},45^{\circ}\) and \(90^{\circ}\) resp.

\(\frac{sec\;A}{cosec\;B}-\frac{tan\;A}{cot\;B}\)

\(=\frac{sec\;45}{cosec\;45}-\frac{tan\;45}{cot\;45}\)

\(=\frac{\sqrt2}{\sqrt2}-\frac{1}{1}\)

\(= 1-1\)

\(= 0\)

Number of revolution \(=\frac{total\;distance}{circumference}\)

\(=\frac{176}{2\times\frac{22}{7}\times0.7}\)

\(= 40\)

\(\frac{perimeter\;of\;\Delta ABC}{perimeter\;of\;\Delta DEF}=\frac{BC}{EF}\)

\(=\frac{7.5}{perimeter\;of\;\Delta DEF}=\frac{2}{4}\)

So perimeter of \(\Delta DEF\)

= 15cm

Dividing both numerator and denominator by \(cos\;\beta\)

\(\frac{4\;sin\beta-3\;cos\;\beta}{4\;sin\;\beta+3\;cos\;\beta}=\) \(\frac{4\;tan\;\beta-3}{4\;tan\;\beta+3}\)

\(=\frac{3-3}{3+3}\)

=  0

\(-2(–5x + 7y = 2)\) gives

\(10x - 14y = -4\).

Now \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=-2\)

Number of Possible outcomes are 26 .

Favorable outcomes are

M, A, T, H, E, I, C, S

probability  = \(\frac{8}{26}\)

\(=\frac{4}{13}\)

Since HCF = 81, two numbers can be taken as \(81x\) and \( 81y,\)

ATQ

\(81x + 81y = 1215\)

or \(x+y = 15\)

which gives four co prime pairs

1,14

2,13

4,11

7, 8

\(tan\;\alpha +cot\;\alpha=2\) gives

\(\alpha=45^{\circ}.\)

So 

\(tan\;\alpha=cot\;\alpha=1\)

\(=tan^{20}\alpha+cot^{20}\alpha\)

\(=1+1\)

= 2

Adding the two given equations we get :

\(348x + 348y = 1740. \)

So

\(x +y =5\)

LCM of two prime numbers = product of the numbers

221 \(=13\times17.\)

So p = 17 and q = 13

\(\therefore\) 3p - q = 51 - 13 = 38