**up board class 10 math question paper 12/04/2022**

**All Question With Answers in Hindi 2022**

**Answers no -1**

(a)-ii

(b)-ii

(c)-i

(d)-i

(f)- iv

**Question No2(a)**

Solution

2x^{2}+4x+3=0

∴a=2,b=4,c=3

Thus, discriminant of this equation is given by,

D=b^{2}−4ac

∴D=(4)^{2}−4(2)(3)

∴D=16−24

∴D=−8<0

Thus, roots of the equation are imaginary

**Question No2(b)**

Solution

Given: 15cotA=8

cotA= 8/15 = 1/tanA

= AC/BC

By pythagoras theorem:

AB^{2} =BC^{2}+AC^{2}

AB^{2}=15^{2}+8^{2}

AB=289

AB=17

sinA= perpendicular/hypotenuse

= 15/17

secA= hypotenuse/base

= 17/8

**Question No2(c)**

Solution

A_{2}/A_{1}=(Side_{2}/Side_{1})^{2}

x^{2}/y^{2} =A_{1}/A_{2}

x^{2}/y^{2} =289/ 121

x/y =√289 /√121

x/y = 17/11

**Question No2(d)**

Solution

relation between mean, median and mode is

mode= 3 median – 2 mean

(mean = 16 mode= 13)

13 = 3 median – 2x 16

13= 3 median-32

13 + 32= 3xmedian

45 = 3x median

45/3 = median

median = 15

**Question No3(a)**

Then, there exist positive integers a and b such that

3= a/b

Now,

3= a/b

3= a^{2}/b^{2}

3b^{2}=a^{2}

3 divides a^{2} [∵3 divides 3b^{2}]

3 divides a…(i)

a=3c for some integer c

a^{2}=9c^{2}

3b^{2}=9c^{2}[∵a^{2}=3b^{2}]

b^{2}=3c^{2}

3 divides b^{2} [∵3 divides 3c^{2}]

3 divides b…(ii)

From (i) and (ii), we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.

**Question no 3 (b)**

The value of a is 4.

ax + 2y = 2

8x + ay = 4

For infinite solutions,

a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

Therefore,

a/8 = 2/a = 2/4

a/8 = 2/4, 2/a = 2/4, a/8 = 2/a

Thus,

a = 4, a = 4, a = 4,-4

Since for infinite solutions all equations should be satisfied simultaneously, therefore a = 4.

**Question No3(c)**

Solution

We have,

PS/SQ= PT/TR

ST∣∣QR [By using the converse of Basic Proportionality Theorem]

∠PST=∠PQR [Corresponding angles]

∠PRQ=∠PQR [∵∠PST=∠PRQ (Given)]

PQ=PR [∵ Sides opposite to equal angles are equal]

△PQR is isosceles.

**Question no 3(d)**

Given,

Heightof frustum, h=40 cm

Radius of larger circular end, r1=20 cm

Radius of smaller circular end, r2=11 cm

∴ Slant height, l^{2}=h^{2}+(r_{1}−r_{2})^{2}

=40^{2}+(20−11)^{2}

=1681

=41cm

Hence, the slant height =41 cm

**Question 4(a)**

HCF(1032,272)= 8

Start with the larger integer, that is 1032, Apply the division Lemma to 1032 and 272, we get

1032 = 272 × 3 + 216

Since , the remainder 216 ≠ 0.

We apply the division Lemma to 272 and 216, we get

272 = 216 × 1 + 56

Remainder 56≠0

216 = 56 × 3 + 48

Remainder 48≠0

56 = 48 × 1 + 8

Remainder ≠ 0

48 = 8 × 6 + 0

The remainder has now become zero ,so our procedure stops.

Since , the divisor at this stage is 8 , the HCF of 1032 is 272 is 8

**Question No4(b)**

In △ADC and △BAC

∠ADC=∠BAC (Given)

∠C is Common

∴ by AA Criterion of Similarity, △ADC ∼ △BAC

⇒ AD/BA= DC/AC= AC/BC

⇒ DC/AC= AC/BC

∴ CA 2 =CB.CD

**Question No4(c)**

Steps of Construction :

1- Draw a line segment AB = 5 cm.

2- Draw any ray AX making an acute angle down ward with AB.

3- Mark the points A1,A2,A3,……………A10 on AX such that AA1 = A1A2 = …..= A9A10.

4- Join BA10.

5-Through the point A draw a line parallel to BA10. To meet AB at P Hence AP : PB = 3

**Question no4 (d)**

Solution:

We will use the basic concepts of trigonometric ratios to solve the problem.

Consider ΔABC as shown below where angle B is a right angle.

cot θ = side adjacent to θ / side opposite to θ = AB/BC = 7/8

Let AB = 7 and BC = 8,

By applying Pythagoras theorem in Δ ABC, we get

AC2 = AB^{2} + BC^{2}

= (7)^{2} + (8)^{2}

= 49 + 64

= 113

AC = √113

= √113

Therefore, sin θ = side opposite to θ / hypotenuse

= BC/AC

= 8/√113

cos θ = side adjacent to θ / hypotenuse

= AB/AC

= 7/√113

(1 + sin θ) (1 – sin θ) / (1 + cos θ) (1 – cos θ)

= 1 – sin2θ / 1- cos2θ [Since, (a + b)(a – b)= (a^{2 }– b^{2})]

= [1 – (8/√113)^{2}] / [1 – (7/√113)^{2}]

= (1 – 64/113) / (1 – 49/113)

= (49/113) / (64/113)

= 49/64

**Question NO5 (a)**

Let the two consecutive positive integers be x and x+1

Then,

x^{2}+(x+1)^{2}=365

x^{2}+x^{2}+2x+1=365

2x^{2} +2x−364=0

x^{2}+x−182=0

Using the quadratic formula, we get

x= [−1±√(1+728)]/2

⇒ (−1±27)/2

⇒x=13 and x=−14

But x is given to be a positive integer. ∴x=−14

Hence, the two consecutive positive integers are 13 and 14.

**Question NO 5(b)**

Solution

Given: (x_{1},y1 )=(−6,10)

(x_{2},y_{2})=(3,−8)

(x,y)=(−4,6)

Using the section formula,we get,

A(-6,10), B(3,-8), C(-4,6)

x= (m_{1}x_{2}+m_{2}x_{1})/m_{1}+m_{2}

−4=( 3m_{1}−6m_{2})/ m_{1}+m_{2}

−4(m_{1}+m_{2}) =3m_{1}−6m_{2}

−4m_{1}−4m_{2}=3m_{1}−6m_{2}

−7m_{1}=−2m_{2}

m_{1}/m_{2} = 2/7

Ratio is 2 : 7.

Hence, the ratio is 2 : 7.

**Question No5 (c)**

Solution :- As a metallic sphere is melted and recast into the shape of a cylinder, their volume must be the same.

Volume of the sphere = Volume of the cylinde

Let us find the volume of the sphere and cylinder by using formulae;

Volume of the sphere = 4/3πr3 where r is the radius of the sphere

Volume of the cylinder = πr2h where r and h are radius and height of the cylinder respectively

Radius of the sphere, r₁ = 4.2 cm

Radius of the cylinder, r₂ = 6 cm

Let the height of the cylinder be h.

Volume of sphere = Volume of cylinder

4/3πr₁3 = πr₂2h

(4/3)r₁3 = r₂2h

h = 4r₁3 / 3r₂2

= (4 × 4.2 cm × 4.2 cm × 4.2 cm) / (3 × 6 cm × 6 cm)

= 2.744 cm

Hence, the height of the cylinder so formed will be 2.744 cm.

**Question NO6 (a)**

Solution

We have,

Let the large number is =x

Square of smaller number is =8x

Now, According to given question,

x^{2}−8x=180

x^{2}−8x−180=0

x^{2}−(18−10)x−180=0

x^{2}−18x+10x−180=0

x(x−18)+10(x−18)=0

(x−18)(x+10)=0

x−18=0,x+10=0

x=18,x=−10

x=−10,18

Either

x=−10 and x=18

x=18 is true (Positive value)

Now, square of small number=8x=18×8=144

√144=12

**Question No 6 (b)**

Side of a square, a=14cm

Radius of each circle, r= 7/2cm

Area of the shaded portion = Area of a square – 4× Area of a circle=a^{2}−4(πr^{2})

=(14×14−4× 22× 7 × 7 )/28

=196−154

∴ Area of the shaded portion =42cm ^{2}

**Question No6 (c)**

Solution

(1)Draw two concentric circle C_{1}and C_{2}with common center 0 and radius 4cm and 6c

(2) Take a point P on the outer circle C_{2} and join OP.

(3) Draw the bisector of OP which bisect OP at M’.

(4) Taking M’ as center and OM’ as radius draw a dotted circle whichcut the inner circle C_{1 }at two point M and P.

(5) Join PM and PP’. Thus, PM and PP’ are required tangent.

On measuring PM and PP’.

PM=PP′=4.4cm

By calculation:

InΔOMP,∠PMO=90 0

PM^{2}=OP^{2}−OM^{2 } (by pythagoras theorem)

PM^{2}=(6)^{2} −(4)^{2}

=36−16 = 20

PM^{2} =20cm

PM= 20

=4.4cm

Hence,

the length of the tangent is 4.4cm

**Question No 6 (d)**

Height (in cm) | f | C.F |

below 140 | 4 | 4 |

140 – 145 | 7 | 11 |

145 – 150 | 18 | 29 |

150 – 155 | 11 | 40 |

155 – 160 | 6 | 46 |

160 – 165 | 5 | 51 |

N=51⇒N/2 =51/2=25.5

As 29 is just greater than 25.5, therefore median class is 145-150.

Here, l= lower limit of median class =145

C=C.F. of the class preceding the median class =11

h= higher limit – lower limit =150−145=5

f= frequency of median class =18

∴median=145+ (25.5−11×5)/18=145+4.03=149.03

**Question No7(a)**

let a= 1/(x+y) and b= 1/(x−y)

10a+2b=4 …….1

15a−5b=−2 ……..2

Multiply 1 with 3 and 2 with 2 we get

30a+6b=12 ……3

30a−10b=−4 ……4

subtracting (3) and (4) we get

16b=16

b=1

Substituting b=1 in 1 we get

10a+2×1=4

10a=2

a= 1/5

As

1/(x+y)=a= 1/5

x+y=5…………eq5

and 1/(x−y)=b=1

x−y=1………..eq6

subtracting eq5 and eq6 we get

2x=6

x=3

putting x=3 in eq 5 we get

3+y=5

y=2

hence x=3 and y=2

or

**Question No7 (a)**

Let Rehman’s present age =x yrs.

3 yrs ago, Rehman’s age =x−3 yrs

5 yrs hence, his age will be x+5 yrs

1/(x−3)+ 1/(x+5)= 1/3

(x+5+x−3)/(x+5)(x−3)= 1/3

x^{2}+2x−15=6x+6

x^{2}−4x−21=0

x^{2} −7x+3x−21=0

x(x−7)+3(x−7)=0

(x+3)(x−7)=0

x=−3,7

Rehman’s present age is 7 yrs.

**Question No7(b)**

Let AB and CD be the multi-storied building and the building respectively.

Let the height of the multi-storied building be hm and the distance between the two building be xm

AE=CD=8m [ Given ]

BE=AB−AE=(h−8)m and

AC=DE=xm [ Given ]

Now, in △ACB,

⇒ tan45^{o}=AB/ AC

⇒ 1= h/x

∴ x=h —- ( 1 )

In △BDE,

⇒ tan30^{o}= BE/ED

⇒ 1/√3 = (h−8)/x

∴ x=√3(h−8) —— ( 2 )

⇒ h= √3h−8 √3

⇒ √3 h−h=8 √3

⇒ h( √3 −1)=8 √3

⇒ h= 8 √3/(√3 −1)

⇒ h=[8 √3×8 (√3+1)]/3-1

∴ h=(12+4 √3 )m

∴ x=(12+4 √3)m [ From ( 1 ) ]

∴ The height of the multi-storied building and the distance between the two buildings is (12+4 √3 )m

Or

**Question No7(b)****Solution**

In ΔBAP

tan30 o = AB/AP

⇒ 1/√3= 10/x

x=10 √3 m

Distance of building

Now,

In ΔPAD

tan45 o= AD/AP

1= AD/AP

⇒x=AD=DB+BA

⇒10 √3=h+10

h=10 √3−10=10( √3−1)

h=10(1.732−1)

h=7.32m

Length of flag.