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Marking Scheme

Class- X Session- 2021-22

PRE-BOARD 1-TERM 1

Subject- Mathematics (Standard)

SECTION A

QN Correct

Option HINTS/SOLUTION

MARKS

1 (b) If xy=180 and HCF (x, y) =3, then find the LCM (x, y).

Solution:

1

2 (a) For what value of k, the pair of equation 4x-3y = 9, 2x + k y =

11 has no solution.

Solution:

1

3 (b)

The areas of two similar triangles ∆ABC and ∆PQR are 25 cm2

and 49 cm2 respectively. If QR = 9.8 cm, find BC

Solution:

The ratio of the areas of two similar triangles is equal to the ratio

of the squares of their corresponding sides.

Here,

area of ∆ABC = 25 cm²,

area of ∆PQR = 49 cm²,

QR = 9.8 cm

BC = ?

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DELHI PUBLIC SCHOOL BANGALORE & MYSORE

2

Therefore,

4 (d)

If the sum and the product of zeros of the polynomial ax²-6x-c is

equal to 12 find the values of a and c

Solution:

1

5

(a)

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:

1

6 (d)

If ABC and DEF are two triangles and AB/DE = BC/DF, then the

two triangles are similar if:

Solution: If the sides of a triangle are respectively proportional to

the sides of another triangle, their corresponding angles are equal

∴ ∠𝐵 = ∠𝐷

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3

7 (b)

If tan α =5/12 find the value of sec α.

Solution:

1

8 (c)

Which of the following numbers is irrational?

Solution:

Here , which is a rational number

and , which is a rational number

and , which is a rational number

but , which is an irrational number

Hence is an irrational number.

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9 (a)

If the point (x, 4) lies on the circle whose center is at the origin

and diameter is 10, then find x

Solution:

1

10 (d)

If the distance between the points (4, p) and (1,0) is 5, then the value

of p is:

Solution:

1

4

11 (b)

The prime factorization of the number 140 is:

Solution:

Correct Answer: (b)

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12 (c)

The least number that is divisible by all the numbers from 1 to 5

is:

Solution:

Answer: (b) 60

Explanation: The least number will be LCM of 1, 2, 3, 4, 5.

Hence, LCM (1, 2, 3, 4, 5) = 2 x 2 x 3 x 5 = 60

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13 (b)

If tan α=√3 and tan β=𝟏

√𝟑, 0<𝛂, 𝛃< 𝟗𝟎°, then find the value

of cot(α+β).

Solution:

1

14 (a)

1

5

15 (d)

Solution:

1

16 (b)

ΔABC is an equilateral triangle whose sides measure 12 cm each.

Then, the length of its altitude AD is:

Solution:

The altitude of an equilateral triangle bisects its base. BD = DC =

6 cm

Now, ABD is right-angled at D.

Thus, by Pythagoras theorem,

= 144 - 36

= 108

cm.

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6

17 (b)

If triangles ABC and DEF are similar and AB=4 cm, DE=6 cm,

EF=9 cm and FD=12 cm, the perimeter of triangle is:

Solution:

Explanation: ABC ~ DEF

AB=4 cm, DE=6 cm, EF=9 cm and FD=12 cm

AB/DE = BC/EF = AC/DF

4/6 = BC/9 = AC/12

BC = (4.9)/6 = 6 cm

AC = (12.4)/6 = 8 cm

Perimeter = AB+BC+AC

= 4+6+8

=18 cm

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18 (a)

If tan θ =1

√3 , then evaluate

cosec ²θ− sec²θ

cosec ²θ+ sec²θ .

Solution:

1

19 (d)

Solve for x and y: 2x+3y=7 and 4x+3y=11

Solution:

1

7

20 (a)

Two dice are thrown simultaneously. what is the probability that the

sum of the two numbers appearing on the top is 13?

Solution:

For a dice, the maximum number that it has is 6.

For two dices thrown, the maximum possible number that can be

obtained is

6+6=12.

Thus, obtaining the sum 13 is an unlikely event to happen in the

throwing of the two dice.

So, the probability of an unlikely event is zero. Here too, the

probability is zero.

Final Answer

The probability is zero.

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SECTION B

21 (c)

Three farmers have 490 kg, 588 kg and 882 kg of wheat respectively.

Find the maximum capacity of a bag so that the wheat can be packed

in exact number of bags.

Solution

Explanation: [Hint. HCF of 490, 588, 882 = 98 kg]

To find the maximum capacity of the bag to pack all the weight in

exact number of bags, we need to find the HCF of these weights

490=2*5*7*7

588=2*2*3*7*7

882=2*3*3*7*7

common factors are-2 and 7(occurring twice)

so HCF=7*7*2=98

so the capacity of the bag is 98 kg

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22 (c)

The perimeters of two similar triangles are 25 cm and 15 cm

respectively. If one side of the first triangle is 9 cm, then find the

corresponding side of second triangle.

Solution

Assume ΔABC & ΔPQR to be the 2 triangles. ΔABC ≌ ΔPQR,

Then the ratio of perimeter= ratio of corresponding sides

⇒25/ 15 = 𝐴𝐵 /𝑃𝑄 ⇒ 25 /15 = 9 /𝑃𝑄 ⇒ 𝑃𝑄 = 15×9/ 25 = 5.4cm

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Ans: corresponding side= 5.4cm.

23 (b)

If tanθ+ cotθ= 2, find the value of tan2 θ+ cot2θ.

Solution

We have,

tanθ + cotθ = 2

(tanθ + cotθ)2 = 4 tan2θ + cot2θ + 2tanθcotθ = 4

tan2θ + cot2θ+ 2 = 4 ( tanθcotθ= 1)

tan2θ + cot2θ = 2

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24 (a)

If α and β are the roots of the equation x²+ 5 x + a=0 and

2α + 5β = −1, then find the value of a.

Solution:

1

25 (c)

The decimal expansion of the rational number 23/ (22. 5) will

terminate after

(a) one decimal place

(b) two decimal places

(c) three decimal places

(d) more than 3 decimal places

Answer: two decimal places

Explanation:

23/ (22 . 5) = (23 × 5)/ (22 . 52) = 115/ (10)2 = 115/100 = 1.15

Hence, 23/ (22 . 5) will terminate after two decimal places.

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26 (a)

The odds in favour of an event are 3: 5. Find the probability of

occurrence of this event.

Solution:

Let E be the event. Then the number of cases favourable to the

occurrence of E = 3 and against the occurrence of E = 5.

Therefore, the total number of outcomes = 8 (1/2 mark)

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9

Hence, P (E) = (1/2 mark)

27 (b)

A number is selected at random from first 50 natural numbers. Then

the probability that it is a multiple of 3 and 4 is:

Solution:

Let M be the event of getting a multiple of 3&4 from the

numbers 1to 50. Multiples of 3 & 4 up to 50 =12,24,36,48. So,

number of favorable outcomes = 4 Total no. of possible

outcomes = 50 ∴ P (E) = No of favorable outcomes/ Total no. of

possible outcomes = 4/50 = 2/25

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28 (b)

Solution:

1

29 (a)

If the point C (−1,2) divides the line segment AB in the ratio 3:4, where

the coordinates of A are (2,5). Find the coordinates of B.

Solution:

30 (c)

D and E are the midpoints of side AB and AC of a triangle ABC,

respectively and BC = 6 cm. If DE || BC, then the length (in cm) of

DE is:

Solution: 3

Explanation: By midpoint theorem,

DE=½ BC

DE = ½ of 6

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10

DE=3 cm

31 (d)

What is the distance of the point (-12,5) from the origin?

Solution:

Now,

The distance between the origin

and the point (x, y) = (-12,5)

=

=

=

=13 units.

Therefore,

The distance between the origin and the point (-12,5)

= 13 units.

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32 (b)

The diagonals of a rhombus are 16 cm and 12 cm, in length. The side

of rhombus in length is:

Solution: 10 cm

Explanation: Here, half of the diagonals of a rhombus are the sides of

the triangle and side of the rhombus is the hypotenuse.

By Pythagoras theorem,

(16/2)2+(12/2)2=side2

82+62=side2

64+36=side2

side=10 cm

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33 (a) Find the smallest positive rational number by which 1 / 7 should be

multiplied so that its decimal expansion terminates after 2 places of

decimal.

Solution:

The smallest number by which 1/7 should be multiplied so that its

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decimal expansion terminates after two decimal points is 7/100 as

1/7 x7/100 = 1/100 = 0.01

Ans: 7/100

34 (b)

In given figure, if ∠ACB = ∠CDA, AC = 8 cm and AD = 3cm, then

find BD.

Solution:

Given, AC = 8 cm, AD = 3cm and ∠ACB = ∠CDA

From figure, ∠CDA = 90°

∠ACB = ∠CDA = 90°

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12

35 (a)

Find the distance between the point (-8/5,2) and (2/5,2)

Solution:

1

36 (c)

A horse is tied to a peg at one corner of a square shaped grass field of

side 15 m by means of a 7 m long rope. The area of that part of the

field in which the horse can graze, is

Solution:

Here, the side of the field is 15 m.

But, the maximum length of the rope is 7 m.

So, thus forms a quarter circle.

Area of that quarter circle -

=> (90 / 360) π r ²

=> (1 / 4) × (22 / 7) × 7²

=> (77 / 2)

=> 38 .5 m²

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37 (d)

If the perimeter of the circle and square are equal, then the ratio of

their areas will be equal to:

Solution: 14:11

Explanation: Given,

The perimeter of circle = perimeter of the square

2πr = 4a

a=πr/2

Area of square = a2 = (πr/2)2

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A circle/A square = πr2/(πr/2)2

= 14/11

38 (b)

Find a quadratic polynomial whose one zero is √5 and the product of

its zeroes is -2√5.

Solution

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39 (c)

If the area of a circle is equal to sum of the areas of two circles

of diameters 10 cm and 24 cm, then the diameter of the larger circle (in

cm) is:

Solution

Radius of the first circle r1=210=5 cm.

Radius of the second circle r2=224=12 cm.

∴ Sum of their area = π r12+ πr2

2

=π52+π122

=25π+144π=169π.

Let the radius of the larger circle be r.

∴Area=πr2.

According to the question:

⇒πr2=169π

⇒r2=169

⇒r2=√169=13 cm

∴Radius of the large circle=13 cm.

Then, diameter of the larger circle=13×2=26 cm.

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40 (b)

The graphical representation of a pair of equations 4x + 3y – 1 = 5

and 12x + 9y = 15 will be

Solution

Answer parallel lines

Explanation:

Given pair of equations are 4x + 3y – 1 = 5 and 12x + 9y = 15.

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14

Comparing with the standard form,

a1 = 4, b1 = 3, c1 = -6

a2 = 12, b2 = 9, c2 = -15

a1/a2 = 4/12 = 1/3

b1/b2 = 3/9 = 1/3

c1/c2 = -6/-15 = 2/5

Thus, a1/a2 = b1/b2 ≠ c1/c2

Hence, the given pair of equations has no solution.

That means, the lines representing the given pair of equations are

parallel to each other.

SECTION C

41 (c) 𝑥2−24𝑥+128

𝑥2−8𝑥−16𝑥+128

(𝑥−16) (𝑥−8) X=8,16

1

42 (b) α + β + αβ= 8+16+128=152 1

43 (d) p (2) =2²- 24(2) +128 =4-48+128=84

1

44

(c) x²+x-2, then 1/α+ 1/β

zeroes are 1 and -2. So, 1

1 +

1

−2 =

−2+1

−2 =

−1

−2=

1

2

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45 (a) Given that the polynomial has sum of zeroes equal to the product

of zeroes.

So, -b/a = -2/k and c/a=3k/k

-2/k=3

3k=-2

K=-2/3.

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46 (a) d=[(x2–x1)2+(y2–y1)2] (½)

d= [(8−4)2+(5−-3)2] (½)

d= [(4)2+(8)2] (½)

d= [16+64] (½)

d= [80] (½)

d=4√5

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15

47 (c) Using the section formula, if a point (x, y) divides the line joining the

points (x1, y1) and (x2, y2) in the ratio m:n, then

(x, y) = (mx2+nx1/m+n, my2+ny1/m+n)

= (1x8+2x7/3, 1x5+2x3/3)

= (8+14/3, 5+6/3)

= (22/3, 11/3)

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48 (a) M.P= (x1+x2/2, y1+y2/2)

= (4+8/2, -3+5/2)

= (12/2, 2/2) = (6,1)

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49 (b) (7,3) =m1x8+m2x4/m1+m2

7(m1+m2) =8m1+4m2

7m1-8m1=4m2-7m2

m1=3m2

m1/m2=3/1 3:1

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50 (d) Scalene triangle

AB=√45 Units

BC=√5 units

AC=√80 units

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