CBSE Set Qa3 Q14 Sample Test Papers For Class 10th for students online
Q14) Flow Chart - Omitted being out of Syllabus. (Marks 2)
Q15) The median
of the following observations arranged in ascending order is 24. Find x.
11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 (Marks 2)
Ans15) n = 10 (even) Medium = 24
Medium lies in (5th + 6th items)/2
= Median = ((x + 2) + (x + 4))/2 = 24
= 2x + 6 = 48
= 2x = 48 - 6 = 42 = x = 21
Q16) Solve
graphically the following system of linear equations : 2x - y = 2; 4x - y = 8.
Also, find the co-ordinates of the points where the lines meet the axis of x.
(Marks 4)
Ans16)
| 2x - y = 2 y = -2 + 2x Let x = 1 y = -2 + 2(1) = 0 2x - y = 2
(1, 0), (2, 2), (3, 4) |
4x - y = 8 y = 4x - 8 Let x = 1 y = -8 + 4(1) = 4 4x - y = 8
(1, -4), (2, 0), (3, 4) |
By plotting the points and joining them, the lines intersect at 3, 4.
... x = 3, y = 4
The line 2x - y = 2 meets the x - axis at (1, 0).
The line 4x - y = 8 meets the x - axis at (2, 0).
Q17) Factorize. Omitted, being out of Syllabus. (Marks 4)
Q18) Students of a
class are made to stand in rows. If 4 students are extra in a row, there would
be 2 rows less. If 4 students are less in a row, there would be 4 more rows.
Find the number of students in the class. (Marks 4)
Ans18) Let the number of students = x
And the number of rows = n
Then x/n = y
= Number of students in each row
or x/y = n = x = ny ...(i)
As per the question in (i) case
x/(y + 4) = n - 2
= x = (n - 2)(y + 4)
= x = ny - 2y + 4n - 8
= ny = ny - 2y + 4n - 8 ...[... x = ny
= ny - ny + 2y - 4n + 8 = 0
= 2y - 4n + 8 = 0
= y - 2n + 4 = 0 ...(ii)
x/(y - 4) = n + 4
= x = (n + 4)(y - 4)
= ny = ny - 4n + 4y - 16
= ny - ny + 4n - 4y + 16 = 0
= 4n - 4y + 16 = 0 ...[Dividing by 4
= n - y + 4 = 0 ...(iii)
Subtracting (iii) from (ii), we get n = 8
Putting n = 8 in (ii), we get
y - 2 x 8 + 4 = 0 = y - 16 + 4 = 0
= y - 12 = 0 = y = 12
Putting n = 8 and y = 12 in (i)
We get x = 12 x 8 = 96
... Number of students in the class = 96
Q19)
In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its
average speed for the trip was reduced by 100 km/hour and time increased by 30
minutes. Find the original duration of flight. (Marks 4)
Ans19) Let the usual speed of aircraft = x km/hr
Reduced speed or aircraft = (x - 100) km/hr
Distance = 2800 km
According to the question
2800/(x - 100) - 2800/x = 30 minutes (... Time
=Distance/Speed)
= (2800x - 2800x + 280000)/x(x-100 )= 1/2 hr
= x2 - 100x = 560000
= x2 - 100x - 560000 = 0
= x2 - 800x + 700x - 560000 = 0
= x(x - 800) + 700(x - 800) = 0
= (x - 800)(x + 700) = 0
= x = 800 or x = -700
... Speed can never be -ve
... Speed = 800 km/hr
Time = D/S = 2800/800 = 7/2 = 3 ½ hours
Q20)
A solid is in the form of a cylinder with hemispherical ends. The total
height of the solid is 19 cm and the diameter of the cylinder is 7cm. Find the
volume and surface area of the solid. (Marks 4)
Ans20)
| Diameter of
cylinder = 7 cm ... Radius of cylinder = 3.5 cm ... Radius of each hemispherical end = 3.5 cm Height of solid = 19 cm Height of cylinder = 19 - (3.5 + 3.5) = 19 - 7 = 12 cm Volume of cylinder =  r2h= 22/7 x 7/2 x 7/2 x 12 = 462 cm3 Volume of each hemispherical end =2/3 r3= 2/3 x 22/7 x 7/2 x 7/2 x 7/2 = 539/6 cm3 Total volume of solid = 462 + 539/6 + 539/6 = (2772 + 539 + 539)/6 = 3850/6 = 641.67 cm3 Curved surface area of cylinder = 2 rh = 2 x 22/7 x 7/2 x 12 = 264cm2 Curved surface area of each hemispherical end = 2 r2= 2 x 22/7 x 7/2 x 7/2 = 77 cm2 Total surface area of solid = 264 + 77 + 77 = 418 cm2 |
 |
Q21)
In a right triangle ABC, right angled at C, at point D is
taken on AB such that CD is perpendicular to AB, Prove that 1/AC2 +
1/BC2 = 1/CD2 (Marks 4)
Ans21) Given : In 
ABD, 
C = 90o
and CD | AB
To prove : 1/AC2 + 1/BC2 = 1/CD2
Proof : ar ABC
= 1/2 x BC x AC ...(i) (Taking BC as base)
[Area of = 1/2 x base
x altitude]
Taking AB as base and ED as altitude
ar ABC = 1/2 x AB x
CD ...(ii) (Taking AB as base)
1/2 x BC x AC = 1/2 x AB x CD [ From (i) and (ii)]
= CD = (BC x AC)/AB
Squaring both side, we get
CD2 = (BC2 x AC2)/AB2
Taking reciprocal of both altitude
1/CD2 = AB2/(BC2 x AC2)
1/CD2 = (AC2 + BC2)/(BC2 x AC2)
[In rt ABC, by
Pythagoras' theorem AB2 = AC2 + BC2]
1/CD2 = AC2/(BC2 x AC2) + BC2/(BC2
x AC2)
1/CD2 = 1/BC2 + 1/AC2 = 1/AC2
+ 1/BC2 = 1/CD2
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