Maths Mixed Test - Class 10

Let the two consecutive odd positive integers be x and x + 2.
x(x + 2) = 483 ^{$\Rightarrow$} x² + 2x - 483 = 0
x2 + 23 x - 21 x  - 483 = 0 ^{$\Rightarrow$} x (x + 23) - 21 (x + 23) = 0
(x + 23) (x - 21) = 0
x = -23 , x - 21 => x = 21.
Since integers are positive, -23 can't be the answer.
So, the two integers are 21 and 23.

Since the equation has equal roots,  D = 0.
b² - 4ac = 0
(k + 1)² - 4 (k + 4) (1) = 0
k² + 2k + 1 - 4k - 16 = 0
^{$\Rightarrow$} k² - 2k - 15 = 0
k² - 5k + 3k - 15 = 0
k(k - 5) +3(k - 5) = 0
^{$\Rightarrow$} (k - 5) (k + 3) = 0
k = 5, -3, so option (1) is the answer.

 a = 3 and d = 5 (18 - 13, 13 - 8, 8 - 3 = 5)
78 = a_n = a + (n - 1)d
78 = 3 + (n - 1)5
78 = 3 + 5n - 5
^{$\Rightarrow$} 78 = 5n - 2
^{$\Rightarrow$} 78 + 2 = 5n
80 = 5n ^{$\Rightarrow$} n = 16.
So, 78 is the 16^th term of the AP.

7 (a + 6d) = 11 (a + 10d) ^{$\Rightarrow$} 7a + 42d = 11a + 110d
7a - 11a = 110d - 42d ^{$\Rightarrow$} -4a = 68d
-a = 17d ^{$\Rightarrow$} a = -17d
Hence, a₁₈ = a + 17d 
a₁₈ = -17d + 17d = 0.

PQ = 10 units
^{$\sqrt{(10-2)^2 + (y+3)^2} = 10 \Rightarrow \sqrt{64+ y^2 + 6y + 9} = 10$}_. 
Squaring both sides: 64 + y² + 6y + 9 = 100 _{^{$\Rightarrow$}} y² + 6y - 27 = 0
y² + 9y - 3y - 27 = 0     _{^{$\Rightarrow$}} y(y + 9) - 3(y + 9) = 0
(y + 9) (y - 3) = 0  _{^{$\Rightarrow$}} y +9 = 0 _{^{$\Rightarrow$}} y= -9 and y -3 = 0 _{^{$\Rightarrow$}} y = 3 

By distance formula,
^{$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$}
$AB = \sqrt{(6-3)^2 +(7-4)^2} =\sqrt{9+9} = \sqrt{18}$_units
$AB = \sqrt{(9-6)^2 +(4-7)^2} =\sqrt{9+9} = \sqrt{18}$_units
$CD = \sqrt{(6-9)^2 +(1-4)^2} =\sqrt{9+9} = \sqrt{18}$_units
$CD = \sqrt{(3-6)^2 +(4-1)^2} =\sqrt{9+9} = \sqrt{18}$_units
AB = BC = CD = DA = ^{$\sqrt{18}$_units}

Also, AC = BD.
So, its a square.

Total number of outcomes = 52
There are total 12 face cards (king, queen and jack are face cards), out of which 6 are black colour and 6 are red colour.
Favourable outcomes = 6
Probability = 6/52 ^{_{$\Rightarrow$}} 3/26

Volume of sphere = 4/3$\pi$r³ = 4/3$\pi$ (10.5)³ cm³
Volume of cone of base radius 3.5 cm and height 3 cm = 1/3$\pi$ (3.5)² x 3 cm³
Number of cones formed =$\frac{Volume of sphere}{Volume of cone} \Rightarrow \frac{4/3\pi \times 10.5 \times 10.5}{{1/3\pi}\times 3.5 \times 3.5 \times 3}$     $\Rightarrow$ 126

Radius of base of cone is 7 cm and height is 8 cm.
Volume of cone = 1/3 ^_$\pi$ r² h = 1/3^_$\pi$ (7)² x 8 cm³
External radius of hollow sphere = 5 cm 
Let internal radius be r cm.
Volume of hollow sphere = 4/3 ^_$\pi$ (R³ - r³) cm³ = 4/3 ^_$\pi$ (5³ - r³) cm³
Since volume of material used for cone will be the same as the volume of material used to make hollow sphere,
1/3 ^_$\pi$ (7)² 8 = 4/3 ^_$\pi$ (125 - r³) ^{$\Rightarrow$} 98 = 125 - r ³
r³ = 125 - 98 = 27
r = 3 cm
So, internal diameter of the sphere will be 2r = 2 x 3 = 6 cm.

The polynomial will be of the form a((x - (1/4))(x +1))
On simplifying, and putting a = 4, we get the required polynomial as  4x² + 3x - 1
Hence, option (1 )is correct.

x + 2 = 0
So, x = -2

37x + 29y = 53...(i)29x + 37y =13...(ii)
Adding (i) and (ii), we get66x + 66y = 66^{$\Rightarrow$} x + y = 1 ... (iii)
Subtracting (ii) from (i), we get 8x - 8y = 40^{$\Rightarrow$} x- y = 5 ...(iv)
Adding (iii) and (iv), we get 2x = 6$\Rightarrow$x = 3.
Subtracting (iv) from (iii), we get 2y = -4^{$\Rightarrow$} y = - 2
Hence, the solution is x = 3 and y = -2.

kx + 2y = 3
3x + 6y = 10
Comparing above equations with 
a₁x + b₁y = c₁
a₂x + b₂y = c₂
We get
$\frac{a_1}{a_2} = \frac{k}{3}$ , $\frac{b_1}{b_2} = \frac{2}{6}$ , $\frac{c_1}{c_2} = \frac{3}{10}$
For no solution:
a₁/a₂  = b₁/b_{2 ^{$ \neq$}} c₁/c₂
$\Rightarrow$ k/3  = 2/6 ^{_{$ \neq$ 3}}/10
^{$\Rightarrow$} k = 1 ^{_{$ \neq$ 9}}/10
The given system of equations will have no solution if k  =  1.
 

Let the fixed charge for the first three days be Rs. x and the charge for each extra day be Rs. y.
As per the question,
                                                x + 4y = 27
                                                x + 2y  = 21
Subtracting (ii) from (i), we get
                        (x + 4y) - (x + 2y)  =  27 - 21
                                                2y  = 6 ^{_{$\Rightarrow$}} y = 3
Substituting this value of y in (i), we get
                                    x + 4 x 3 = 27
^{_{$\Rightarrow$}}                               x =  27- 12 =>  x = 15
The fixed charge for the first 3 days is Rs. 15 and the charge for each extra day is Rs. 3.

Since DE || BC,
    AE/ EC  =    AD/ DB$\Rightarrow$                x + 2 / x - 1  =     x / x - 2^{_{$\Rightarrow$}}               x² - 4     =    x (x - 1)                                                  
^{_{$\Rightarrow$}}              x² - 4     =     x² - x ^{_{$\Rightarrow$}} x  =  4.

 ^$\Delta$PQR is right-angled at Q.

Let                   QR =   x cm
                        PR  =  (25 - x) cm                      PR²  =    PQ²  + QR²                                    (25 - x)² =    (5)² + x^{2
_{$\Rightarrow$}}       625 + x² - 50x   = 25 + x^{2                                            
_{$\Rightarrow$}}                       50x    =   600
_{^{$\Rightarrow$}}                           x       =  12
Thus, in  ^$\Delta$PQR, PQ = 5 cm, QR  = 12 cm and PR  = 13 cm.
Therefore, sin P = QR/ PR  = 12 / 13, cos P = PQ / PR = 5 / 13 and tan P = QR/ PQ = 12/ 5.

cos (40° + x) = sin 30˚ = cos 60°
40° + x = 60° 
 x = 20°

Let AC be the road and poles AB and CD are standing opposite to each other.

AB = CD  = x m
Let    PC  =   y m
  AP  = (100 - y) m,
                 Angle  CPD  = 60° and angle APB = 30°
In ^_$\triangle$DCP,
DC/PC = tan 60°
x/y = tan 60° = ^{_{$\sqrt 3$}}
x = ^{$\sqrt 3$_y}.........(1)
In ^_$\triangle$ABP, AB/AP = tan 30°
x/(100-y) = 1/^{_{$\sqrt 3$}
$\sqrt 3$}x = 100-y ...... (2)
Substituting the value of x from (1) into (2), we get^{_{$\sqrt 3$}} x ^{$\sqrt 3$_y} = 100 - y
3y = 100 - y
y =25.
Putting the value of y in (1), we get
x = ^{$\sqrt 3$} x 25 = 25 (1.73) = 43.3 m
The height of pole is 43.3 m and distance of point P from the nearer of the 2 poles is 25 m.
 

Let AB represent the tree. Let D be a point from where the tree is broken.

Let BD = x m
Then, AD = CD = (12 - x) m
In right angle ^$\triangle$DBC,
sin 60° = x/(12-x)                                                                
_{^{$\sqrt 3$}}/2 = x/(12-x)
^{$12\sqrt 3 - \sqrt {3x} = 2x$}
(2 + ^{$\sqrt 3$}) x = 12^{$\sqrt 3$} 
x =^{$\frac{12\sqrt3}{2+ \sqrt3} \times \frac{2 -\sqrt3}{2- \sqrt3}$}                                       
x = 24^{$\sqrt 3$} -36 _{^{$\Rightarrow$}}
x = 24 (1.73) - 36
x = 41.568 - 36 = 5.57 m

Area of right ^$\triangle$ABC = 1/2 x AB x BC= 1/2 x (6 x 8)cm² = 24 cm²    
AC² = AB² + BC²                            
64 cm² + 36 cm² = 100 cm².
AC = 10 cm. 
Also, area (^$\triangle$ABC) = ar (^$\triangle$OBC) + ar (^$\triangle$OCA) + ar (^$\triangle$OAB)= 1/2 (BC x r) + 1/2 (AC x r) + 1/2 (AB x r)= 1/2 r (6 + 10 +8) cm = 12 r cm
12 r cm = 24 cm^{2 _{$\Rightarrow$}} r = 2 cm

Area of circle with radius 6 cm =  $\pi$(6)² = 36$\pi$ cm².
Area of circle with radius 8 cm = _^$\pi$ (8)² = 64_^$\pi$ cm². 
Total area of both circles = 36_^$\pi$ cm² + 64_^$\pi$cm² = 100_^$\pi$ cm².
 So, the radius of the new circle be R.
_^$\pi$R² = 100_^$\pi$cm²
R² = 100 cm² = R _{^{$\Rightarrow$}} 10 cm. 

Area of square = 21 x 21 = 441 cm²
Area of four quarter circular regions of radius 10.5 cm = 4 x [1/4 x 22/7 x 21 /2 x 21/2] cm²= 693/2 cm²
Area of remaining region = (441 - 693/2) cm² = 189/2 = 94.5 cm²

Mean ^{$\bar x = \frac{total f_i x_i}{total f_i}$}= 3620/140 = 25.86.

The modal class is 160 cm - 165 cm.
Mode = l + ((f₁ - f₀) / (2f₁ - f₀ - f₂)) x h
l = 160 cm, f₀ = 22, f₁ = 36 , f₁ = 36, f₂ = 23, h = 5 cm.
Plugging in the values in the formula for the mode, we get mode = 162.593 = 162.60 cm.

If two coins are tossed simultaneously, the outcomes are: HH, HT, TH, TT.
So, total number of outcomes = 4
At most one heads = no heads or one heads.
This is possible when {HT, TH, TT} is obtained.
Probability = Favourable outcomes/Total number of outcomes ^{$\Rightarrow$} 3/4