Oswal Practice Papers CBSE Class 10 Mathematics Solutions (Practice Paper - 7)

Section-A 

1. (a) (4, 4)

Explanation :    

Given : A(2, 3), B(4, 7) and C(6, 2).

We know that,

$$\text{Centroid (G) =}\\\bigg(\frac{x_{1} + x_{2} +x_{3}}{3},\frac{y_{1} + y_{2} + y_{3}}{3}\bigg)\\=\bigg(\frac{2+4+6}{3},\frac{3+7+2}{3}\bigg)$$

= (4, 4)

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2. (b) 2 : 3

Explanation :    

Let P be the point that divides the line segment joining (3, – 1) and (8, 9) in ratio k : 1.

Then, coordinates of point P is given by using internal section formula,

$$\text{P(x,y)} =\bigg[\frac{8k+3}{k+1},\frac{9k-1}{k+1}\bigg]$$

Since, point P lies on line x – y – 2 = 0, so P must satisfy the above equation.

$$\text{Putting}\space x =\frac{8k+3}{k+1}, y =\frac{9k-1}{k+1}$$

in equation x – y – 2 = 0, we get

$$\frac{8k+3}{k+1}-\bigg(\frac{9k-1}{k+1}\bigg)-2 = 0\\\Rarr\space\frac{8k+3-9k+1-2k-2}{k+1} = 0\\\Rarr\space\frac{-3k+2}{k+1}= 0\\\Rarr\space -3k + 2 = 0\\\Rarr\space k =\frac{2}{3}\\\Rarr\space k:1=\frac{2}{3}:1\\\Rarr\space 2 : 3$$

Hence, the required ratio is 2 : 3.

3. (d) 70°

Explanation :    

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Let each base angle of isosceles triangle be x.

∴Vertical angle of an isosceles triangle = x + 15°

We know that,

∠A + ∠B + ∠C = 180°

⇒ x + 15° + x + x = 180°

⇒ 3x = 165°

⇒ x = 55°

∴ x + 15 = 55 + 15

= 70°.

4. (d) 6 cm

Explanation :    

In ΔABC, if AE is the angle bisector of ∠A, then according to the angle bisector theorem,

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$$\frac{\text{AB}}{\text{AC}} = \frac{\text{BE}}{\text{EC}}\\\frac{9}{\text{AC}}=\frac{3.6}{2.4}\\\text{AC} =\frac{9×2.4}{3.6}= 6\space\text{cm.}$$

5. (b) 45°

Explanation :    

∠AOD = 135°

∠BOC = 180° – 135° = 45°

Since, angle subtended at the centre by a pair of opposite sides are supplementary.

$$\textbf{6.\space}(b)\space\frac{2\pi \text{R}\theta}{360\degree}$$

Explanation :    

Length of arc of a sector =

$$\frac{\theta}{360\degree}×2\pi\text{R}=\frac{2\pi \text{R}\theta}{360\degree}.$$

7. (c) 9

Explanation :    

f(x) = x2 – x – (2k + 2)

Substituting x for – 4, we have f(– 4) = (– 4)2 – (– 4) – (2k + 2)

= 16 + 4 – 2k – 2

= 18 – 2k = 0

Thus, 18 – 2k = 0

or 2k = 18 or k = 9.

8. (c) cos 90°

Explanation :    

Given, ∆ABC is right angled at C.

i.e., ∠C = 90°

Then by triangle angle sum property

∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠B + 90° = 180°

⇒ ∠A + ∠B = 180° – 90°

⇒ ∠A + ∠B = 90°

Taking cos on both the sides, we get

cos (A + B) = cos 90°

cos (A + B) = 0

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[ cos 90° = 0]

$$\textbf{9.\space}(b)\space\frac{2}{3}$$

Explanation :    

$$\text{Given :\space}\text{tan A}=\frac{1}{\sqrt{5}}\\\Rarr\space\text{tan}^{2}\text{A} =\frac{1}{5}\\\text{...(i)}\\\Rarr\space\frac{1}{\text{cot}^{2}\text{A}}=\frac{1}{5}\\\bigg[\because\space\text{tan A}=\frac{1}{\text{cot A}}\bigg]\\\Rarr\space \text{cot}^{2}\text{A} = 5\space\text{...(ii)}$$

$$\text{Now,}\space\frac{\text{cosec}^{2}A -\text{sec}^{2}A}{\text{cosec}^{2}A + \text{sec}^{2}A}\\=\frac{1 + \text{cot}^{2}A-(1 +\text{tan}^{2}A)}{1 +\text{cot}^{2}\text{A} + 1 + \text{tan}^{2}\text{A}}\\\begin{bmatrix}\because\space\text{cosec}^{2}\text{A} = 1 + \text{cot}^{2}A\\\text{and sec}^{2}\text{A} = 1 + \text{tan}^{2}A\end{bmatrix}\\=\frac{\text{cot}^{2}\text{A} -\text{tan}^{2}\text{A}}{2 +\text{cot}^{2}\text{A} +\text{tan}^{2}\text{A}}\\=\frac{5 -\frac{1}{5}}{2 + 5 +\frac{1}{5}}$$

[From (i) and (ii)]

$$=\frac{\frac{25-1}{5}}{\frac{35+1}{5}}=\frac{24}{36}=\frac{2}{3}$$

$$\therefore\space\frac{\text{cosec}^{2}\text{A} - \text{sec}^{2}\text{A}}{\text{cosec}^{2}\text{A} + \text{sec}^{2}\text{A}}=\frac{2}{3}$$

10. (c) 9

Explanation :    

We have

9 sec2 A – 9 tan2 A = 9(sec2 A – tan2 A)

= 9(1 + tan2 A – tan2 A)

= 9 × 1   [ sec2 θ = 1 + tan2 θ]

= 9

11. (a) 6

Explanation :    

Given quadratic equation is,

kx(x – 2) + 6 = 0

⇒ kx2 – 2kx + 6 = 0

On comparing the above equation with ax2 + bx + c = 0, we get

a = k, b = – 2k, c = 6

For equal roots,

D = b2 – 4ac = 0

⇒ (– 2k)2 – 4 × k × 6 = 0

⇒ 4k2 – 24k = 0

⇒ 4k (k – 6) = 0

⇒ k = 0 (not possible) or k = 6

∴ k = 6

12. (b) 2 cm

Explanation :    

An incircle is drawn with centre O which touches the sides of the triangle ABC at P, Q and R. OP, OQ and OR are radii and AB, BC and CA are the tangents to the circle.

OP ⊥ AB, OQ ⊥ BC and OR ⊥ CA.

OPBQ is a square. ( ∠B = 90°)

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Let r be the radius of the circle

∴ PB = BQ = r

AR = AP = 8 – r

CQ = CR = 6 – r

AC = AR + CR

⇒ 10 = 8 – r + 6 – r

⇒ 10 = 14 – 2r

⇒ 2r = 14 – 10 = 4

⇒ r = 2

∴ Radius of the incircle = 2 cm.

13. (a) 5, 1

Explanation :    

It has infinite many solution

$$\text{So,\space}\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\\\Rarr\space\frac{2}{\text{m+n}} =\frac{3}{\text{2m-n}}=\frac{11}{33}\\\Rarr\space\frac{2}{m+n} =\frac{3}{2m-n} =\frac{1}{3}\\\Rarr\space\frac{2}{m+n} = \frac{1}{3}\\\Rarr\space m + n=6\\\text{...(i)}\\\text{and}\space\text{2m-n} =9\\\text{...(ii)}$$

on adding 3m = 15

⇒ m = 5

Putting m in equation (i), 5 + n = 6

⇒ n = 1.

14. (c) – 7, 10

Explanation :    

Given quadratic polynomial is x2 + 7x + 10

Here a = 1, b = 7 and c = 10

We know,

$$\text{sum of roots}\space(\alpha + \Beta)=\frac{-b}{a}\\\alpha +\beta =\frac{\normalsize-7}{1} =-7\\\text{Product of roots (ab) =}\frac{c}{a}\\\alpha\beta=\frac{10}{1}= 10$$

15. (d) – 1

Explanation :    

Given, f(y) = 2y2 + 7y + 5

$$\text{Here}\space\alpha +\beta =-\frac{7}{2}\\\bigg(\because\space\alpha +\beta=\frac{-b}{a}\bigg)\\\alpha\beta =\frac{5}{2}\\\bigg(\because\space\alpha + \beta=\frac{c}{a}\bigg)\\\therefore\space(\alpha+ \beta) +(\alpha\beta)=-\frac{7}{2}+\frac{5}{2}=-1.$$

16. (d) 4

Explanation :    

Given, the distance of the ladder AC from the base of the wall AB is 2 m.

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$$\text{So,\space cos 60\degree}=\frac{\text{BC}}{\text{AC}}=\frac{2}{x}\\\Rarr\space\frac{1}{2} = \frac{2}{x}\\\Rarr\space\text{x = 2×2}$$

The height of the ladder = 4 m.

$$\textbf{17.}\space\text{(C)}\frac{4}{45}$$

Explanation :    

Total number of possible outcomes = 90

Thus, the favourable outcomes are 2, 3, 5, 7, 11, 13, 17, 19.

Hence, the total number of favourable outcomes = 8

$$\therefore\space\text{P(E)} =\frac{8}{90}=\frac{4}{45}$$

$$\textbf{18.\space}\text{(c)}\space\frac{364}{365}$$

Explanation :    

If

Total number of possible outcomes = 365 × 365

Total number of favourable outcomes = 365 × (365 – 1)

= 365 × 364

$$\therefore\space\text{P(E)} =\frac{365×364}{365×365} =\frac{364}{365}$$

19. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Explanation :    

Maximum frequency = 23

Hence, modal class is 12 – 15

Now,

$$\text{Mode} =l +\frac{(f_{k}-f_{k-1})}{2f_{k}-f_{k-1}-f_{k+1}}×h\\= 12 +\bigg(\frac{23-21}{46 - 21 -10}\bigg)×3\\= 12 +\frac{6}{46-31}\\= 12 +\frac{6}{15}= 12.4 $$

20. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Explanation :    

For reason, as per the empirical formula

The sum of all the probabilities of all possible outcomes of experiment is 1.

P(Event) + P (not an Event) = 1

So, reason is true.

For assertion,

As per the description

$$\text{P(boundary) or P(E) =}\\\frac{9}{45} =\frac{1}{5}\\\text{So P(no boundary) or P(not E)}\\= 1 -\frac{1}{5} =\frac{4}{5}$$

So, assertion is also true and Reason is the correct explanation of the assertion.

Section-B 

21. On comparing the given equations with standard form of pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, we get

$$\text{(a)}\space\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\\\text{as}\frac{3}{1}=\frac{3}{1}=\frac{\normalsize-9}{\normalsize-3},\space\text{consistent}\\\text{(b)}\space\frac{a_{1}}{a_{2}}\neq\frac{b_{1}}{b_{2}}\space\\\text{as}\space\frac{4}{\normalsize-2}\neq =-1,\text{consistent}\\\text{(c)}\space\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\text{as}\\\frac{5}{10}=\frac{1}{2}=\frac{10}{20},\space\text{consistent}\\\text{(d)\space}\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq\frac{c_{1}}{c_{2}}\text{as}\\\frac{\normalsize-2}{\normalsize-4}=\frac{1}{2}\neq\frac{3}{10},\text{inconsistent}$$

(i) So, three pair of linear equations are consistent.

(ii) Fourth pair of linear equations is inconsistent.

22. Let P be the required point on Y-axis.

∴ P = (0, y)

Now, PA = PB  [Given]

$$\therefore\space\sqrt{(0+4)^{2} + (y-3)^{2}}\\=\sqrt{(0-6)^{2} + (y-5)^{2}}\\\text{[Using distance formula]}$$

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On squaring both sides,

(4)2 + (y – 3)2 = (6)2 + (y – 5)2

⇒ 16 + y2 – 6y + 9 = 36 + y2 – 10y + 25

⇒ – 6y + 25 = 61 – 10y

⇒ – 6y + 10y = 61 – 25

⇒ 4y = 36

⇒ y = 9

∴ Coordinates of the point P = (0, 9) Ans.

OR

Given : A = (3, – 4), B = (1, 2), 

$$\text{P}=(p,-2)\text{and Q}=\bigg(\frac{5}{3},q\bigg)$$

Now, AP = PQ = QB

Thus AP : PB = 1 : 2 and AQ : QB = 2 : 1

$$\therefore\space p =\frac{1 + 3(2)}{1 + 2} =\frac{7}{3}\\\text{and}\space q =\frac{2(2) + 1(\normalsize-4)}{1 + 2} = 0\\\therefore\space p=\frac{7}{3}\space\text{and q = 0}.$$

Ans.

23. Total number of cards = 52

∴ n(S) = 52

(i) Number of black king or red queen= 2 + 2 = 4

∴ n(A) = 4

$$\text{P(A)} =\frac{\text{n(A)}}{\text{n(S)}}\\=\frac{4}{52}=\frac{1}{13}.$$

Ans.

(ii) Number of non-face cards = Total cards – Number of face cards

= 52 – 12 = 40

∴ n(B) = 40

$$\text{P(B)}=\frac{\text{n(B)}}{\text{n(S)}}\\=\frac{40}{52}=\frac{10}{13}.\\\textbf{Ans.}$$

24. Given sequence is 114, 109, 104, ........

Here, first term (a) = 114

Common difference (d) = 109 – 114 = – 5

Let nth term be the first negative term of the given A.P.

∴ an < 0

⇒ a + (n – 1)d < 0

⇒ 114 + (n – 1) (– 5) < 0

⇒ 114 – 5n + 5 < 0

⇒ 119 – 5n < 0

⇒ 119 < 5n

$$\Rarr\space n\gt\frac{119}{5}\\\Rarr\space n\gt 23.8$$

∴ 24th term is the first negative term of the given A.P. Ans.

OR

Given : Sn = n (4n + 1)

So, Sn – 1 = (n – 1){4(n – 1) + 1}

= (n – 1) (4n – 4 + 1)

= (n – 1) (4n – 3)

Now, nth term of A.P. is given as,

Tn = Sn – Sn – 1

= n(4n + 1) – (n – 1) (4n – 3)

= 4n2 + n – (4n2 – 3n – 4n + 3)

= 4n2 + n – 4n2 + 7n – 3

= 8n – 3

∴ T6 = 8 (6) – 3

= 48 – 3 = 45 Ans.

25. Let the coordinates of point P be (2y, y). Since, P is equidistant from Q and R

∴ PQ = PR

$$\Rarr\space\sqrt{(2y-2)^{2} + (y + 5)^{2}}\\=\sqrt{(2y+3)^{2} + (y-6)^{2}}\\\Rarr\space (2y-2)^{2} +(y +5)^{2}\\=(2y+3)^{2} +(y-6)^{2}\\\Rarr\space 4y^{2} +4 =8y +y^{2} +25+10 y$$

4y2 + 9 + 12y + y2 + 36 – 12y

⇒ 2y + 29 = 45

⇒ 2y = 45 – 29

$$\Rarr\space y =\frac{16}{2} = 8$$

Hence, the coordinates of point P are (16, 8). Ans.

Section-C 

26. Let us suppose that

$$3 -\sqrt{5}\space\text{is a rational number.}\\\text{Then, it can be represented in}\\\space\frac{p}{q}\text{form}\space i.e.,\\ 3-\sqrt{5}=\frac{p}{q}$$

[where p and q are real numbers and q ≠ 0]

$$\Rarr\space 3 -\frac{p}{q} =\sqrt{5}\\\Rarr\space \frac{3q-p}{q} =\sqrt{5}$$

p and q are real numbers.

$$\text{Therefore,}\space\frac{3q-p}{q}$$

should also be a real number.

But we already know that

$$\sqrt{5}\space\text{is an irrational number.}$$

So, our assumption is wrong.

$$\text{Hence, 3}-\sqrt{5}\space\text{is an irrational number.}\\\textbf{Hence Proved.}$$

27. Let, HCF be ‘H’

then LCM = 14 H

Sum of LCM and HCF is 750.

∴ 14 H + H = 750

⇒ 15 H = 750

$$\Rarr\space \text{H} =\frac{750}{15}$$

⇒ H = 50

∴ LCM = 14 H

= 14 × 50

= 700

We know,

Product of LCM and HCF= Product of two numbers

Let other number be y. 700 × 50 = 250 × y

$$y =\frac{700×50}{250}$$

y = 140

Hence, the other number is 140. Ans.

28. Given : The quadratic polynomial is f(x) = x2 – p(x + 1) – c = x2 – px – p – c And α and β are the zeros of the given polynomial.

We know that,

Sum of the zeros =

$$-\frac{\text{Coefficient of x}}{\text{Coefficient of x}^{2}}\\\text{Product of the zeros =}\\\frac{\text{Constant term}}{\text{Coefficient of x}^{2}}\\\therefore\space\alpha +\beta =-\frac{(-p)}{1}\\\alpha + \beta = p\\\text{and}\space \alpha\beta=\frac{(-p-c)}{1}\\\alpha\beta =-(p+c)\\\text{Now\space L.H.S} =(\beta + 1)(\alpha + 1)\\=\alpha\beta + (\alpha +\beta) +1$$

= – (p + c) + p + 1

= – p – c + p + 1

= 1 – c

= R.H.S. Hence Proved.

29. Let, a be the first term and d be the common difference of the given A.P. Then,

$$a_{m}=\frac{1}{n}\\a +(m-1)d=\frac{1}{n}\\\text{...(i)}\\\text{and}\space a_{n}=\frac{1}{m}\\ a + (n-1)d=\frac{1}{m}\\\text{...(ii)}$$

Subtracting equation (ii) from equation (i), we get

$$(m-n)d=\frac{1}{n}-\frac{1}{m}\\\text{(m-n)d}=\frac{m-n}{mn}\\ d=\frac{1}{mn}\\\text{Putting\space d}=\frac{1}{mn}\space\\\text{in equation (i), we get}\\\text{a + (m - 1)}\frac{1}{mn}=\frac{1}{n}\\a + \frac{1}{n}-\frac{1}{mn}=\frac{1}{n}\\a =\frac{1}{mn}$$

$$\text{S}_{mn}=\frac{mn}{2}\bigg[2a +(mn-1)d\bigg]\\\text{S}_{mn}=\frac{mn}{2}\begin{Bmatrix}\frac{2}{mn}+(mn-1)\frac{1}{mn}\end{Bmatrix}\\\text{S}_{mn}=\frac{1}{2}(mn+1)\\\textbf{Hence Proved.}$$

30. In ∆ABC, we have

DE || BC

$$\text{So,}\space\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}\\\text{[Thales’ theorem]}\\\Rarr\space\frac{4x-3}{3x-1} = \frac{8x-7}{5x-3}$$

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⇒ (4x – 3)(5x – 3) = (8x – 7)(3x – 1)

⇒ 20x2 – 15x – 12x + 9 = 24x2 – 21x – 8x + 7

⇒ 4x2 – 2x – 2 = 0

⇒ 2x2 – x – 1 = 0

⇒ 2x2 – 2x + x – 1 = 0

⇒ 2x(x – 1) + 1(x – 1) = 0

⇒ (x – 1)(2x + 1) = 0

⇒ x = 1 [as length cannot be negative]

So, the value of x = 1. Ans.

OR

Let EO || DC meet AD at E.

So, in ∆ADC, EO || DC

$$\text{So,\space}\frac{\text{AO}}{\text{OC}}=\frac{\text{AE}}{\text{ED}}\\\text{[Thales’ theorem] …(i)}\\\text{But}\space\frac{\text{AO}}{\text{OC}}=\frac{\text{BO}}{\text{OD}}\\\text{[Given}]\space...\text{(ii)}$$

By (i) and (ii),

$$\text{So,\space}\frac{\text{BO}}{\text{OD}} = \frac{\text{AE}}{\text{ED}}\\\text{Hence},\space \frac{\text{BO}}{\text{OD}}=\frac{\text{AE}}{\text{ED}}\\\text{and BO, OD, AE and ED}$$

are segments of ∆DAB.

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So, EO || AB [By converse of Thales’ theorem]

But EO || DC

Thus, AB || DC

As there are only two parallel sides in this quadrilateral, it is a trapezium.

Hence Proved.

31. Let the line QR intersects the bigger circle at S.

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Join PS.

O is the mid-point of PQ.

[PQ is a diameter of the bigger circle]

QR is a tangent to the smaller circle and OR is a radius through the point of  contact R.

∴ OR ⊥ QR ⇒ OR ⊥ QS

Since, OR is perpendicular to a chord QS of the bigger circle.

∴ QR = RS

[Perpendicular from the centre to a chord bisects the chord]

⇒ R is the mid-point of QS.

∴ In ΔQSP, O is the mid-point of PQ and R is the mid-point of QS.

$$\therefore\space \text{OR} =\frac{1}{2}\text{PS}$$

[segment joining the mid-points of any two sides of a triangle is half of the third side]

⇒ PS = 2OR = 2 × 8 cm = 16 cm

In right ΔOQR, OR2 + QR2= OQ2

⇒ 82 + QR2 = 132

⇒ 64 + QR2 = 169

⇒ QR2 = 169 – 64 = 105

$$\Rarr\space \text{QR} =\sqrt{105}\\\therefore\space\text{RS = QR=}\sqrt{105}\\\text{In}\space \Delta\text{PRS},\\\space \text{PR}^{2} = \text{RS}^{2} +\text{PS}^{2}\\=(\sqrt{105})^{2} + 16^{2}$$

⇒ PR = 19 cm. Ans.

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Let P be the external point from which tangents PA and PB are drawn on the circle with centre O.

Thus, OB is perpendicular to PB and OA is perpendicular to PA.

Hence, ∠OBP = ∠OAP = 90°

In ΔPOB and ΔPOA, OA = OB

[Radii of the circle]

OP = OP [Common side]

∠OBP = ∠OAP [Each 90°]

∴ ΔPOB @ ΔPOA [S.A.S.]

So, PA = PB [By C.P.C.T.]

Hence Proved.

Section-D

32. Total outcomes = {1, 2, 3, 4, 5, 6, 7, 8}

⇒ n (S) = 8

(i) Let A be the event that arrow will point at 8.

A = {8}

⇒ n (A) = 1

$$\because\space\text{P(A)}=\frac{1}{8}\space\textbf{Ans.}$$

(ii) Let B be the event that arrow will point at an odd number.

∴ B = {1, 3, 5, 7}
 
⇒ n(B) = 4

$$\therefore\space\text{P(B)}=\frac{4}{8}=\frac{1}{2}\space\textbf{Ans.}$$

(iii) Let C be the event that arrow will point at a prime number.

C = {2, 3, 5, 7}

⇒ n (C) = 4

$$\because\space\text{P(C)} =\frac{4}{8}=\frac{1}{2}\\\textbf{Ans.}$$

OR

Total number of coins = (100 + 50 + 20 + 10) = 180

(i) Possible number of outcomes of a 50 p coin falling out = 100

$$\text{Hence,\space}\text{P(E)}=\frac{100}{180}=\frac{5}{9}$$

Ans.

(ii) Possible number of outcomes of a coin falling out having value more than ₹1 = (20 + 10) = 30

$$\text{Hence,\space P(E)}=\frac{30}{180}=\frac{1}{6}\\\textbf{Ans.}$$

(iii) Possible number of outcomes of a coin falling out having value less than ₹5 = (100 + 50 + 20) = 170

$$\text{Hence,}\space\text{P(E)}=\frac{170}{180}=\frac{17}{18}\\\textbf{Ans.}$$

(iv) Possible number of outcomes of a coin falling out having value either ₹1 or ₹2 = (50 + 20) = 70

$$\text{Hence,}\space \text{P(E)}=\frac{\text{70}}{\text{180}}=\frac{7}{18}\\\textbf{Ans.}$$

33. Given, ΔABC ~ ΔPQR

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$$\Rarr\space\frac{\text{AB}}{\text{PQ}} =\frac{\text{BC}}{\text{QR}}=\frac{\text{AC}}{\text{PR}}$$

(From the side-ratio property of similar triangles)

⇒ ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ...(i)

BC = 2BD and QR = 2QM

(P and M are the mid-points of BC and QR)

$$\Rarr\space\frac{\text{AB}}{\text{PQ}}=\frac{2\text{BD}}{2\text{QM}}=\frac{\text{AC}}{\text{PR}}\\\text{...(ii)}$$

Now in ΔABD and ΔPQM

$$\frac{\text{AB}}{\text{PQ}}=\frac{\text{BD}}{\text{QM}}\\\text{...[\text{from} (ii)]}\\\angle B =\angle Q\\\text{...[from (i)]}\\\Rarr\space\Delta\text{ABD}∼\Delta\text{PQM}$$

(By SAS similarity criterion)

$$\therefore\space\frac{\text{AB}}{\text{PQ}} = \frac{\text{AD}}{\text{PM}}\\\textbf{Hence proved.}$$

34. Let ACB be the given arc subtending an angle of 60° at the centre.

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Here, r = 14 cm and θ = 60°.

Area of the minor segment ACBA = (Area of the sector OACBO) – (Area of ∆OAB)

$$=\frac{\pi r^{2}\theta}{360\degree} -\frac{1}{2}r^{2}\text{sin}\theta\\\=\frac{22}{7}×14×14×\\\frac{60\degree}{360\degree}-\frac{1}{2}×14×14×\text{sin}\space 60\degree\\=\frac{308}{3}-7×14×\frac{\sqrt{3}}{2}\\=\frac{308}{3}-49\sqrt{3} = 17.89\text{cm}^{2}$$

Area of the major segment BDAB = Area of circle – Area of minor segment ACBA

= πr2 – 17.89

$$=\frac{22}{7}×14×14-17.89$$

= 616 – 17.89 = 598.11 ≈ 598 cm2. Ans.

35. Let XOX′ and YOY′ be the X-axis and Y-axis respectively.

Now, 3x – 2y + 2 = 0 ⇒ 3x = 2y – 2

$$\Rarr\space x =\frac{2(y-1)}{3}$$

If y = – 2, x = – 2

If y = 1, x = 0

If y = 4, x = 2

Therefore,

x – 2 0 2
y – 2 1 4

Thus, plotting the points P(– 2, – 2), Q(0, 1) and R(2, 4) on the graph paper, we get the graph of 3x – 2y + 2 = 0, which is represented by PR.

$$\text{Now,\space}\frac{3}{2}x - y + 3 = 0\\\Rarr\space 3x-2y+6 =0\\\Rarr\space y =\frac{3(x+2)}{2}$$

If x = 0, y = 3

If x = – 2, y = 0

If x = – 4, y = – 3

Therefore,

x 0 -2 – 4
y 3 0 – 3

Thus, plotting the points S(0, 3), T(– 2, 0) and U(– 4, – 3) on the graph paper, we get the graph of

$$\frac{3}{2}x - y + 3 = 0,$$

which is represented by SU.

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Since, the lines do not intersect each other at any point, so the pair of equations is inconsistent. Also the lines representing the graphs of the equations, meet the Y-axis at points (0, 1) and (0, 3). Ans.

OR

Let XOX′ and YOY′ be the X-axis and Y-axis respectively.

Now, x – y + 1 = 0 ⇒ x = y – 1

If y = 1, x = 0

If y = 0, x = – 1

If y = 3, x = 2

Therefore,

x 0 -1 2
y 1 0 3

Thus, plotting the points P(0, 1), Q(– 1, 0) and R(2, 3) on the graph paper, we get the graph of x – y + 1 = 0, which is represented by PR.

Now, 3x + 2y – 12 = 0 ⇒ 2y = 12 – 3x

$$\Rarr\space y =\frac{3(4-x)}{2}$$

If x = 4, y = 0

If x = 2, y = 3

If x = 0, y = 6

Therefore,

x 4 2 0
y 0 3 6

Thus, plotting the points S(4, 0), R(2, 3) and T(0, 6) on the graph paper, we get the graph of 3x + 2y – 12 = 0, which is represented by ST.

7paper_18

The solution of the pair of equations is given by (2, 3).

Hence, the coordinates of the triangle formed by these two lines and the X-axis are Q(– 1, 0), R(2, 3) and S(4, 0).

$$\text{Area} =\frac{1}{2}×\text{Base × Height}$$

Now, Base = 4 – (– 1)

= 5

Height = 3

$$\text{Thus\space Area}=\frac{1}{2}×5×3$$

= 7.5 sq. units. Ans.

Section-E

36. (i) Radius of a dome, r = 2.5 m

The dome is hemispherical in shape.

Then, Cloth material required = 2 × Surface area of hemisphere

= 2 × 2πr2

$$= 4×\frac{22}{7}×2.5×2.5$$

= 78.57 m2 Ans.

(ii) Height of each pillar, h = 7 m

Radius of base, r = 1.4 m

Lateral surface area or curved surface area of 2 pillars = 2 × 2πrh

$$= 4×\frac{22}{7}×1.4×7$$

= 123.2 m2 Ans.

OR

Volume of 2 hemispheres of radius 1 cm

$$= 2×\frac{2}{3}\pi r^{3}=\frac{4}{3}\pi(1)^{3}\\=\frac{4}{3}\pi\text{cm}^{3}$$

Volume of a sphere of radius 2 cm =

$$\frac{4}{3}\pi(2)^{3}=\frac{32}{3}\pi\space\text{cm}^{3}\\\text{Then,\space Required ratio =}\frac{\frac{4}{3}\pi}{\frac{32}{3}\pi}\\=\frac{1}{8}\space = 1:8\space\textbf{Ans.}$$

(iii) Radius of hemisphere, r = 3.5 m

Then, volume of a hemisphere,

$$\text{V}=\frac{2}{3}\pi r^{3}\\=\frac{2}{3}×\frac{22}{7}×(3.5)^{3}$$

= 89.83 m3 Ans.

37. (i) Let be distance between building and tower, CB = DE = x

AE = h

$$\text{In} \Delta \text{ADE,}\\\space\text{tan\space} 30\degree = \frac{\text{AE}}{\text{DE}}=\frac{\text{h}}{\text{x}}\\=\frac{1}{\sqrt{3}}=\frac{h}{x}\\x =\sqrt{3}h\\\text{In} ∆\text{ACB,}\\\text{tan\space} 60\degree=\frac{\text{AB}}{\text{CB}}\\=\frac{15+h}{x}\\\sqrt{3} =\frac{15 + h}{\sqrt{3}h}$$

[Using (i)]

⇒ 3h = 15 + h

⇒ 2h = 15

$$h =\frac{15}{2} = 7.5\space\text{m}$$

Height of tower = AB = BE + EA

= 15 + 7.5 = 22.5 m Ans.

OR

We have,

$$\text{tan}\space\theta + \frac{1}{\text{tan}\space\theta} = 2\\\Rarr\space\bigg(\text{tan}\space\theta +\frac{1}{\text{tan}\space\theta}\bigg)^{2}=(2)^{2}\\\Rarr\space\text{tan}^{2}\theta + \frac{1}{\text{tan}^{2}\theta} +\\2×\text{tan}\space\theta×\frac{1}{\text{tan}\space\theta} = 4\\\Rarr\space\text{tan}^{2}\theta + \frac{1}{\text{tan}^{2}\theta} \\= 4-2\\\Rarr\space \text{tan}^{2}\theta +\frac{1}{\text{tan}^{2}\theta} = 2$$

(ii) In ∆ADE,

$$\text{tan}\space 30\degree =\frac{\text{AE}}{\text{DE}}=\frac{h}{x}\\\frac{1}{\sqrt{3}}=\frac{7.5}{x}$$

x = 7.5 × 1.73

= 12.975 m

Ans.

(iii) In ∆ADE

$$\text{sin}\space 30\degree =\frac{\text{AE}}{\text{AD}}\\(\because\space\text{Length of wire = AD})\\\frac{1}{2}=\frac{7.5}{\text{AD}}$$

D = 7.5 × 2 = 15 m Ans.

38. (i) Let AB denote the lamp-post and CD be the girl after walking for 4 seconds away from the lamp-post.

Now, her distance from the base of the lamp

BD = 1.2 m × 4 = 4.8 m Ans.

(ii) ∆ABE ~ ∆CDE

$$\frac{\text{BE}}{\text{DE}}=\frac{\text{AB}}{\text{CD}}\\\text{(Let DE = x)}\\\Rarr\space\frac{4.8 +x}{x}=\frac{3.6}{0.9}$$

⇒ 4.8 + x = 4x

⇒ x = 1.6 m Ans.

So, the shadow of the girl after walking for 4 seconds.

OR

$$\frac{\text{AE}}{\text{CE}}=\frac{\text{BE}}{\text{DE}}\\=\frac{4.8 +1.6}{1.6}=\frac{6.4}{1.6} = 4$$

⇒ AE = 4CE

⇒ AC + CE = 4CE

⇒ AC = 3CE

$$\Rarr\space\frac{\text{AC}}{\text{CE}}=\frac{3}{1}\space\textbf{Ans.}$$

(iii) Since ratio of the area of two similar triangles is equal to the ratio of the square of their corresponding sides, Ratio of areas of similar triangles = (9)2 : (16)2 = 81 : 256.

CBSE Practice Paper Mathematics Class 10

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