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Maths Mixed Test - Class 10

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The product of two consecutive odd positive integers is 483. The integers are

  1. 21 and 23

  2. 27 and 29

  3. 17 and 19

  4. 31 and 33


Correct Option: A
Explanation:

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.

(k + 4)x2 + (k + 1)x + 1 = 0 has equal roots if

  1. k = 5

  2. k = 3

  3. k = -5

  4. k > 0


Correct Option: A
Explanation:

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.

Which term of the AP: 3, 8, 13, 18, _______ is 78?

  1. 15th term

  2. 16th term

  3. 17th term

  4. 14th term


Correct Option: B
Explanation:

 a = 3 and d = 5 (18 - 13, 13 - 8, 8 - 3 = 5) 78 = an = 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 16th term of the AP.

If 7 times the 7th term of an AP is equal to 11 times the 11th term, then its 18th term will be

  1. 7

  2. 11

  3. 18

  4. 0


Correct Option: D
Explanation:

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.

Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.

  1. y = 3

  2. y = -9

  3. Both 1 and 2

  4. None of these


Correct Option: C
Explanation:

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 

Four friends A, B, C and D are sitting in a park at points: A (3 , 4), B (6, 7), C (9, 4) and D (6, 1). Their sitting positions form

  1. a rhombus

  2. a square

  3. neither a square nor a rhombus

  4. an undefined shape


Correct Option: B
Explanation:

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.

A card is drawn from a well-shuffled deck of playing cards. Find the probability of drawing a red face card.

  1. 3/13

  2. 4/13

  3. 2/13

  4. 3/26


Correct Option: D
Explanation:

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

A solid metallic sphere of radius 10.5 cm is melted and recast into a number of smaller right circular cones of base radius 3.5 cm and height 3 cm each. Find the number of cones formed?

  1. 125

  2. 126

  3. 130

  4. 120


Correct Option: B
Explanation:

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

A solid right circular cone of base diameter 14 cm and height 8 cm is melted to form a hollow sphere. If the external diameter of the sphere is 10 cm, find the internal diameter of the sphere.

  1. 3 cm

  2. 4 cm

  3. 6 cm

  4. 8 cm


Correct Option: C
Explanation:

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.

Which of the following quadratic polynomials has zeroes as 1/4 and - 1.

  1. 4x2 + 3x -1

  2. 4x2 - 3x - 1

  3. 4x2 - 3x + 1

  4. 4x2 + 3x + 1


Correct Option: A
Explanation:

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.

Which of the following options is the zero of the polynomial x + 2 ?

  1. -1

  2. 1

  3. 2

  4. -2


Correct Option: D
Explanation:

x + 2 = 0 So, x = -2

Find the respective values of x and y.

37x + 29y = 53 29x + 37y = 13

  1. 3 and - 2

  2. 2 and - 3

  3. 3 and 2

  4. -3 and - 2


Correct Option: A
Explanation:

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.

For what value(s) of k does the given system of equations have no solution?

kx + 2y = 3 3x + 6y = 10

  1. k = 3/10

  2. k = 1

  3. Either k = 3/10 or k = 1

  4. None of these


Correct Option: B
Explanation:

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$ \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.  

A library has a fixed charge for the first three days and an additional charge for each day afterwards. Nishu paid Rs. 27 for a book kept for seven days, while Sangeeta paid Rs. 21 for a book that she kept for five days. Find the fixed charge and the charge for each extra day, respectively.

  1. Rs. 3 and Rs. 15

  2. Rs. 2 and Rs. 12

  3. Rs. 12 and Rs. 2

  4. Rs. 15 and Rs. 3


Correct Option: D
Explanation:

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.

In the given figure, DE || BC. If AD = x, DB = x - 2, AE = x+ 2 and EC = x - 1, find the value of x.

  1. 2

  2. -2

  3. 4

  4. -4


Correct Option: C
Explanation:

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.

In $\Delta$PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Find the respective values of sin P, cos P and tan P.

  1. 12/13, 5/13 and 12/5

  2. 12/13, 12/5 and 5/13

  3. 5/13, 12/13 and 12/5

  4. 12/5, 12/13 and 5/13


Correct Option: A
Explanation:

 $\Delta$PQR is right-angled at Q.

Let                   QR =   x cm                         PR  =  (25 - x) cm                      PR2  =    PQ + QR2                                    (25 - x)2 =    (5)2 + x2 $\Rightarrow$       625 + x2 - 50x   = 25 + x2                                             $\Rightarrow$                       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.

If cos (40° + x) = sin 30°, find the value of x.

  1. 60°

  2. 20°

  3. 30°

  4. 90°


Correct Option: B
Explanation:

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

Two poles of equal heights are standing opposite to each other on either side of a 100 m wide road. From a point between them on the road, the angles of elevation of the top of the poles are 60o and 30o, respectively. Find the height of the poles and the distance of the point from the nearer of the two poles.

  1. 25 m and 43.3 m, respectively

  2. 43.3 m and 25 m, respectively

  3. 25 m and 25 m, respectively

  4. 43.3 m and 43.3 m, respectively


Correct Option: B
Explanation:

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.  

A 12 m high tree is broken into two parts. The top of the tree strikes the ground and makes an angle of 60° with the level ground. At what height from the ground did the tree break?

  1. 5.57 m

  2. 6 m

  3. 4.47 m

  4. 8 m


Correct Option: A
Explanation:

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

ABC is a right triangle in which angle B = 90°. The biggest circle possible is inscribed in the triangle. If AB = 8 cm and BC = 6 m, find the radius r of the incircle.

  1. 4 cm

  2. 1 cm

  3. 2 cm

  4. 0.5 cm


Correct Option: C
Explanation:

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 cm2 $\Rightarrow$ r = 2 cm

The radii of two circles are 6 cm and 8 cm. Find the radius of the circle with area equal to the sum of the areas of these circles.

  1. 20 cm

  2. 5 cm

  3. 2 cm

  4. 10 cm


Correct Option: D
Explanation:

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. 

From a square sheet of paper of side 21 cm, four quarter circular regions of radius 10.5 cm are removed. Find the area of remaining portion.

  1. 104.5 cm2

  2. 441 cm2

  3. 94.5 cm2

  4. 346.5 cm2


Correct Option: C
Explanation:

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) cm= 189/2 = 94.5 cm2

The following table shows the marks obtained by 140 students in an examination.

Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
Number of students 20 24 40 36 20

Find the mean marks.

  1. 25

  2. 25.86

  3. 31.20

  4. 24.38


Correct Option: B
Explanation:
Marks (classes) 0-10 10-20 20-30 30-40 40-50 Total
Class mark (xi) 5 15 25 35 45  
Frequency (fi) 20 24 40 36 20 140
fixi 100 360 1000 1260 900 3620

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

The table shows the heights of 100 students. Find the modal height.

           
Height (cm) 155-160 160-165 165-170 170-175 175-180 Total
No. of students 22 36 23 11 8 100
  1. 160 cm

  2. 162.5 cm

  3. 162.6 cm

  4. 165 cm


Correct Option: C
Explanation:

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, f= 23, h = 5 cm. Plugging in the values in the formula for the mode, we get mode = 162.593 = 162.60 cm.

Two coins are tossed simultaneously. Find the probability of getting at most one heads.

  1. 1/4

  2. 1/2

  3. 1

  4. 3/4


Correct Option: D
Explanation:

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

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