10 th Class CBSE Solutions of Most Important Maths Questions of 3 and 4 marks
Here,10 th Class CBSE Solutions of Most Important Maths Questions of 3 and 4 marks for 2022-23 board part 2 published by Future Study Point are presented for the students of 10 th class who are going to appear in the 2022-23 board exam of CBSE.
All the questions in CBSE class 10 maths most important questions and answers for 2022-23 board Part 2 are designed by the expert of the subject and their solutions are readily explained in an ordered way.
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10 th Class CBSE Solutions of Most Important Maths Questions of 3 and 4 marks
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Q1.Find the sum of 50 terms of an AP whose n^{th } term is defined as 7n – 6 .
Solution.
The n^{th }term is defined as = 7n – 6
First term of the AP is = 7 × 1 – 6 = 1
Second term = 7 × 2 – 6 = 8
Third term = 7 × 3- 6 = 15
Hence the AP is 1, 8, 15,22…….
Where first term,a = 1, common difference, d = 8 -1 = 7
We have to find the sum of n term, let it is S_{n}}, where n = 50
Applying the formula for sum of the AP
= 25 (2 +49×7) = 25 (2 + 343) = 25 ×345 = 8625
Q2.A bucket is in the form of a frustum of a cone whose height is 42 cm and the radii of its circular ends are 30 cm and 10 cm. Find the amount of milk (in liters) which this bucket can hold. If the milkman sells the milk at the rate of Rs. 40 per liter, what amount he will get from the sale?
Solution.
.
r_{1 }= 10 cm, r_{2 }= 30 cm, h = 42 cm
The volume of bucket is
The rate of milk is Rs 40 /liter
Hence the volume of the milk= 57200 cm² = 57200/1000 =57.2 liter
Therefore the amount he will get from the sale of milk = 40 ×57.2 = Rs 2288
Q3.If the coordinates of two points are A(3,4), B (5,-2) and a point P (x,5) is such that PA = PB, then find the area of ΔPAB.
Solution.
We are given that
PA = PB
√[(x -3)² + (5 – 4)²] = √[(x -5)² + (5 + 2)²]
Squaring both sides
(x -3)² + (5 – 4)² = (x -5)² + (5 + 2)²
x ² + 9 – 6x + 1= x ² + 25 -10x+49
10x -6x = 25 + 49 -10 = 64
4x = 64
x = 16
Therefore the co-ordinates of P are (16,5)
Area of Δ = 1/2 [x_{1 }(y_{2 }– y_{3 }) + x_{2 }(y_{3 }-y_{1 }) + x_{3}(y_{1 }-y_{2 }) ]
x_{1 }=3, x_{2 }=5 , x_{3}=16, y_{1 }=4, y_{2 }=-2 , y_{3}=5
Area of ΔPAB = 1/2 [3_{ }(-2_{ }– 5) + 5_{ }(5_{ }-4) + 16(4+2_{ }) ]
= 1/2[ -21 + 5 + 96] = 1/2(80) = 40
Therefore area of ΔPAB = 40 sq.unit
Q4.A solid metallic cylinder of diameter 12 cm and height 15 cm is melted and recast into toys in the shape of a cone of radius 3 cm and height 9 cm. Find the number of toys so formed.
Solution.
The diameter of A solid metallic cylinder =12 cm
The radius of the solid metallic cylinder will be(R)= 6 cm
Height of the solid metallic cylinder(H) = 15 cm
Radii of the recasted cones(r) =3 cm
Height of recasted cones(h) = 9 cm
Let the number of toys formed = N
Therefore
The volume of cylinder = N × Volume of cone
N = Volume of cylinder/Volume of the cone = πR²H/(1/3)πr²h = 3R²H/r²h =(3×6²×15)/(3²×9) = 1620/81 = 20
Therefore the number of toys so formed = 20
10 th Class CBSE Solutions of Most Important Maths Questions of 3 and 4 marks
Q5. A and B working together can do a work in 6 days. If A takes 5 days less than B to finish the work, in how many days B alone can do the work.
Solution.
In 6 days A and B do = 1 piece of work
In 1 day A and B will do = (1/6) Part of the work
Let B is capable to finish that work in = x days
B will do part of that work in 1 day = 1/x
An alone can do that work in = (x–5) days
An alone do that piece off work in one day = 1/(x -5)
One day’s work of A + One day’s work of B = (A and B)’s one day work
1/(x-5) + 1/x = 1/6
(x +x -5)/x(x -5) = 1/6
(2x – 5)/(x²- 5x) = 1/6
12 x – 30 = x² – 5x
x² – 17x + 30 = 0
x ² – 15x -2x +30 = 0
x(x – 15) – 2 ( x – 15) = 0
( x -15)( x – 2) = 0
x = 15, x = 2
Neglecting 2 because 2 < 6
Therefore B alone can do that work in 15 days.
Q5.A hollow metallic sphere of external and internal diameters 8 cm and 4 cm respectively is melted to form a solid cone of base diameter 8 cm. Find the height of the cone.
Solution. The inner radius of the sphere = 4/2 = 2 cm
The outer radius of the sphere = 8/2 = 4cm
Volume of the sphere = (4/3) π r³
The volume of the metals the sphere is made up of = Outer volume of the sphere – Inner volume of the sphere
=(4/3) π 4³ – (4/3) π 2³
= (4π/3)×64 – (4π/3) ×8
= (4π/3) (64 – 8) = (4π/3) ×56
Therefore, the volume of sphere = Volume of the cone
Diameter of the cone is = 8 cm
The radius of the cone is = 8/2 = 4 cm
Volume of the cone = (1/3) πr²h
Since sphere is recasted into the cone ,therefore
Volume of the cone = Volume of the sphere
(1/3) π4²h = (4π/3) ×56
h = (4×56)/16 = 14
Therefore the height of the cone = 14 cm
Q6. Prove that the parallelogram circumscribing a circle is a rhombus.
Solution.
Given → Parallelogram ABCD circumscribing a circle with center O which is touching it at P,Q,R and S points.
To Prove → The given parallelogram ABCD is a rhombus
Proof → AB = DC……….(i)(Opposite side of the parallelogram)
AD = BC……….(ii)(Opposite side of the parallelogram)
AP = AS ………………(iii) [Tangents drown on a circle from a same external point are equal]
BP = BQ (—do—)………(iv)
DR = DS (—do—)………(v)
CR = CQ (—do—)……….(vi)
HINT: Write the tangents corresponding to opposite sides same side as written above.
Adding (iii) to (vi) all equation we get
AP + BP + DR + CR = AS + BQ + DS + CQ
Arranging these line segments in order to get the side of paralellogram
(AP + BP) + (DR + CR) = (AS + DS )+ (BQ + CQ)
AB + DC = AD + BC
From equation number (i) and (ii)
DC + DC = AD + AD
2DC = 2AD
DC = AD
It is clear that AB = BC = DC= AD
All sides of the given parallelogram are equal therefore it is a rhombus.
Q7. Find the area of the triangle formed by joining the mid-points of sides of triangle whose vertices (2,1), (4,3) and (2,5).
Solution. Let the Δ formed PQR after joining the midpoints of the sides of ABC
We are given a Δ ABC such that the coordinates of each vertex are A(2,1),B(4,3) and C(2,5)
The vertices of the ΔPQR are calculated as follow
P(x_{1} ,y_{1 }) = (2+2)/2, (5+1)/2 = (2, 3)
P(x_{2} ,y_{2 }) = (2+4)/2, (5+3)/2 = (3, 4)
P(x_{3} ,y_{3 }) = (2+4)/2, (1+3)/2 = (3, 2)
Area of Δ = 1/2 [x_{1 }(y_{2 }– y_{3 }) + x_{2 }(y_{3 }-y_{1 }) + x_{3}(y_{1 }-y_{2 }) ]
x_{1 }=2, x_{2 }=3 , x_{3}=3, y_{1 }=3, y_{2 }= 4 , y_{3}=2
Area of Δ = 1/2 [x_{1 }(y_{2 }– y_{3 }) + x_{2 }(y_{3 }-y_{1 }) + x_{3}(y_{1 }-y_{2 }) ]
Area of ΔPAB = 1/2 [2_{ }(4_{ }– 2) + 3_{ }(2_{ }-3) + 3(3-4_{ }) ]
= 1/2[ 4 -3 -3] = 1/2(-2) = -1
Neglecting its negative sign because area can not be negative
Therefore area of ΔPQR = 1 square unit
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