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# 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.

These 10 th Class CBSE Solutions of Most Important Maths Questions of 3 and 4 marks for 2021-22 board-II are designed as per the updated syllabus of CBSE, so you need not have any kinds of doubt in the selections of the questions. Don’t forget to make a comment and to subscribe us.

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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 nth  term is defined as 7n – 6 .

Solution.

The  nth 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 Sn}, where n = 50

Applying the formula for sum of the AP

$\fn_cm S_{n} =\frac{n}{2}\left [ 2a + \left ( n-1 \right )d \right ]$

$\fn_cm S_{50} =\frac{50}{2}\left [ 2\times 1 + \left ( 50-1 \right )7 \right ]$

= 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= 10 cm, r= 30 cm, h = 42 cm

The volume of bucket is

$=\frac{1}{3}\pi h\left ( r_{1}^{2}+r_{2}^{2}+r_{1}r_{2} \right )$

$=\frac{1}{3}\times \frac{22}{7}\times 42 \left ( 10^{2}+30^{2}+10\times 30 \right )$

The rate of milk is Rs 40 /liter

$= \frac{22}{21}\times 42 \left ( 100+900+300 \right )=44\times 1300=57200$

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(y2 – y3 ) + x2 (y3 -y) + x3(y1 -y2 ) ]

x=3, x2 =5 , x3=16, y=4, y2 =-2 , y3=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(x1 ,y) = (2+2)/2, (5+1)/2 = (2, 3)

P(x2 ,y) = (2+4)/2, (5+3)/2 = (3, 4)

P(x3 ,y) = (2+4)/2, (1+3)/2 = (3, 2)

Area of Δ = 1/2 [x(y2 – y3 ) + x2 (y3 -y) + x3(y1 -y2 ) ]

x=2, x2 =3 , x3=3, y=3, y2 = 4 , y3=2

Area of Δ = 1/2 [x(y2 – y3 ) + x2 (y3 -y) + x3(y1 -y2 ) ]

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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