CBSE Class 10 Mathematics Solved Question Paper 2017 · 2018-03-22 · CBSE Class 10 Mathematics...

CBSE Class 10 Mathematics Solved Question Paper 2017

Get SOLVED & UNSOLVED question papers, updated Syllabus, Sample papers and study material

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Solved Question Paper 2017

Class 10

Summative Assessment II

Subject - Mathematics

Time allowed: 3 hours Maximum Marks: 90

General Instructions:

(i) All questions are compulsory.

(ii) The question paper consists of 31 questions divided into four sections - A, B, C and D.

(iii) Section A contains 4 questions of 1 mark each. Section B contains 6 questions of 2 marks

each, Section C contains 10 questions of 3 marks each and Section D contains 11 questions of 4

marks each.

(iv) Use of calculators is not permitted.

SECTION A

Question 1. What is the common difference of an A.P. in which a21 – a7 = 84?

Solution.

Let first term of AP be a and d be its common difference.

Now, nth term of an AP is given as:

an = a + (n−1)d

a21=a + (21−1)d = a + 20d

and a7 = a + (7−1)d = a + 6d

According to the question:

a21 − a7 = 84

(a+20d) − (a+6d) = 84

a + 20d – a − 6d = 84

14d = 84



and much more…

d = 6

Thus, the required common difference of the AP is 6.

Question 2. If the angle between two tangents drawn from an external point P to a circle of

radius a and centre O, is 60°, then find the length of OP.

Solution.

Let PQ and PR be the two tangents drawn to the circle with radius a and centre O such

that ∠QPR = 60°.

In ∆OPQ and ∆OPR

OQ = OR = Radius

∠OQP = ∠ORP = 90° (Tangents drawn to a circle are perpendicular to radius at the

point of contact)

PQ = PR (Lengths of tangents drawn from an external point to the circle are

equal)

So, ∆OPQ ≌ ∆OPR (SAS Congruence Axiom)

∠ OPQ =∠ OPR = 30° (CPCT)

Now, in ∆OPQ,

o OQsin 30

OP

1[OQ = Radius = (given)]

2 OP

OP 2

aa

a

Thus, the length of OP is 2a.



and much more…

Question 3. If a tower 30 m high, casts a shadow 10 3 m long on the ground, then what is the

angle of elevation of the sun?

Solution.

Let AB be the 30 m high tower and BC = 10 3 m be the length of its shadow on ground.

Let the angle of elevation of the sun from the ground be .

Now, in ∆ABC,

o

o

tan

30tan

10 3

30 3tan

10 3

30 3tan 3

30

tan tan 60

60

AB

BC

Thus, the angle of elevation of the sun is 60°.

Question 4. The probability of selecting a rotten apple randomly from a heap of 900 apples is

0.18. What is the number of rotten apples in the heap?

Solution.



and much more…

We know that Probability of an event (E) = Number of favourable outcomes

Total number of outcomes

Let E be the event of selecting rotten apple.

Let n be the number of rotten apples in the heap.

P(E) = 900

n

⇒ 0.18 = 900

n

⇒ n =162

Thus, total rotten apples = 162.

SECTION B

Question 5. Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is

6 times the other.

Solution.

Given equation can be written as:

2 14 8 – 0x xp p

….(i)

Let the first root be α, then the second root will be 6α.

General equation is given as:

ax2 + bx + c = 0 ….(ii)

Now, sum of roots = b

a

α + 6α = 14

p [Comparing (i) and (ii)]

7α = 14

p



and much more…

α = 2

p

Also, product of roots = ca

α × 6α = 8

p

[From (i) and (ii)]

6α 2=

8

p

6

2

2

p

= 8

p

p = 3

Hence, the value of p is 3.

Question 6. Which term of the progression 1 1 3

20,19 ,18 ,17 ,...4 2 4

is the first negative term?

Solution.

The given sequence is 1 1 3

20,19 ,18 ,17 ,...4 2 4

Here, First term, a = 20

And common difference, d = 1

194

− 20 = 77 80 3

4 4



and much more…

Let the nth term of the AP be the first negative term.

0

1 0

320 1 0

4

83 30

4 4

83 3 0

3 83

227

3

28

Thus, the 28th term is the first negative term of the given AP.

na

a n d

n

n

n

n

n

n

Question 7. Prove that the tangents drawn at the end points of a chord of a circle make equal

angles with the chord.

Solution.

Let AB be a chord of a circle with centre O, and let AC and BC to the same circle.

Join OC to cut AB at D.

We have to prove that ∠CAD = ∠CBD.

As the line segment joining the centre to the external point from where tangents are drawn,

bisects the angle between two tangents.

So, ∠ACD = ∠BCD ....(i)

In ΔACD and ΔBCD, we have



and much more…

CA = CB [Tangents from an external point are equal]

∠ACD = ∠BCD [From (i)]

CD = CD [Common]

ΔACD ≅ ΔBCD [By SAS congruency rule]

∠CAD = ∠CBD [BY CPCT]

Question 8. A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD =

BC + DA.

Solution.

ABCD is a quadrilateral. Suppose a circle touches the sides AB, BC, CD and DA of the

quadrilateral ABCD at P, Q, R and S, respectively.

We know that the length of tangents drawn from an external point to a circle are equal.

DR = DS .....(i)

CR = CQ .....(ii)

BP = BQ .....(iii)

AP = AS .....(iv)

Adding (i), (ii), (iii) and (iv), we get

DR + CR + BP + AP = DS + CQ + BQ + AS

(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)

CD + AB = AD + BC

Or, AB + CD = BC + DA

Question 9. A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, –5) is

the mid-point of PQ, then find the coordinates of P and Q.

Solution.



and much more…

Suppose the line intersects the y-axis at P(0, y) and the x-axis at Q(x, 0).

It is given that (2, –5) is the mid-point of PQ.

Using mid-point formula, we have:

0 0, 2, 5

2 2

, 2, 52 2

2and 52 2

4, 10

x y

x y

x y

x y

Thus, the coordinates of P and Q are (0, −10) and (4, 0), respectively.

Question 10. If the distances of P(x, y) from A(5, 1) and B(–1, 5) are equal, then prove that 3x =

2y.

Solution.

Given: PA = PB

To Prove: 3x = 2y

Proof:

Since PA = PB

Thus, according to the distance formula,

2 2 2 2

2 2 2 2

2 2 2 2

5 1 1 5

5 1 1 5 Squaring both sides

10 25 2 1 2 1 10 25

10 2 2 10

8 12

3 2

Hence,3 2 .

x y x y

x y x y

x x y y x x y y

x y x y

y x

x y

x y

SECTION C

Question 11. If ad ≠ bc, then prove that the equation (a2 + b

2)x

2 + 2 (ac + bd)x + (c

2 + d

2) = 0

has no real roots.



and much more…

Solution.

The given equation is (a2 + b

2)x

2 + 2 (ac + bd)x + (c

2 + d

2) = 0

We know, D = b2−4ac

Thus,

D = [2(ac+bd)]2 − 4(a

2+b

2)(c

2+d

2)]

= [4(a2c

2+b

2d

2+ 2abcd)] − 4(a

2+b

2)(c

2+d

2)

= 4[(a2c

2+b

2d

2+ 2abcd) − (a

2c

2+a

2d

2+b

2c

2+b

2d

2)]

= 4[a2c

2+b

2d

2 + 2abcd−a

2c

2−a2d

2−b2c

2−b2d

2]

= 4[2abcd − b2c

2 − a2d

2]

= −4[a2d

2+b

2c

2−2abcd]

= −4[ad−bc]2

But we know that ad ≠ bc

Therefore, (ad−bc) ≠0

(ad−bc)2 > 0

−4(ad−bc)2 < 0

D < 0

Hence, the given equation has no real roots.

Question 12. The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 400.

Find the number of terms and the common difference of the A.P.

Solution.



and much more…

1Here, or 5, 45and 400

We know that th term of an AP is given as,

1

45 5 1

1 40 .... i

Now,sum of terms is given as:

S 2 12

400 2 5 12

80010 40 Using (i)

80016

50

P

n n

n

n

a a a S

n

a a n d

n d

n d

n

na n d

nn d

n

n

utting 16in equation (i), we get:

16 1 40

40 8

15 3

n

d

d

8Hence, the common difference of the A.P. is and number of terms are16.

3

Question 13. On a straight line passing through the foot of a tower, two points C and D are at

distances of 4 m and 16 m from the foot respectively. If the angles of elevation from C and D of

the top of the tower are complementary, then find the height of the tower.

Solution.

Let AB be the tower of height h m. Suppose the angle of elevation of the top of the tower from

point C on the ground be . Then the angle of elevation of the top of the tower from point D

is 90°− .

Here, BC = 4 m and BD = 16 m.



and much more…

o

In ΔABC,AB

tan .... iBC 4

InΔABD,AB

tan 90BD 16

cot ..... ii16

h

h

h

Multiplying (i) and (ii), we get:

2

2

tan cot4 16

164

64

64 8m

Thus, the height of the tower is 8 m.

h h

h

h

h

Question 14. A bag contains 15 white and some black balls. If the probability of drawing a black

ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag.

Solution.

Suppose number of black ball = x

It is given that White balls = 15



and much more…

According to question,

P(Black balls)=3 × P(White balls)

153

15 15

45

Hence, number of black balls in the bag is 45.

x

x x

x

Question 15. In what ratio does the point 24

,11

y

divide the line segment joining the points P(2,

–2) and Q(3, 7)? Also find the value of y.

Solution.

Let point R24

,11

y

divide the line segment joining the points P(2, –2) and Q(3, 7) in ratio k:1.

Thus using section formula, 2 1 2 1,mx nx my ny

x ym n m n

, we get the coordinates of R

24,

11y

as:

24 3 2 7 2,

11 1 1

24 1 11 3 2 , 1 7 2

24 24 33 22, 7 2

9 2

2

9

2Putting in 7 2, we get :

9

2 27 2

9 9

2 9 14 18

9 9

11 4

9 9

4

11

k ky

k k

k k y k k

k k yk y k

k

k

k yk y k

y y

y y

y

y



and much more…

Thus the given point24

,11

y

divides the the line segment in k:1 = 9

:1or 9:22

.

And value of y is 4

11

.

Question 16. Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a

semicircle of radius 4.5 cm are drawn in the given figure. Find the area of the shaded region.

Solution.

Given diagram can be represented as below:

Here, area of shaded region = Area of largest semicircle – Area of circle with centre R – Sum of

areas two of semicircles with cemtres as P and Q respectively + Area of semicircle with centre S.

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