Post on 27-Dec-2021
1
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 = ?
1
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
1
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.
1
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)
1
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
1
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.
1
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
1
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.
1
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
1
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
1
8
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
1
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.
1
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)
1
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
1
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
1
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.
1
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
1
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
1
11
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°
1
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
1
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
1
13
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
1
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.
1
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.
1
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
1
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.
1
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
1
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)
1
48 (a) M.P= (x1+x2/2, y1+y2/2)
= (4+8/2, -3+5/2)
= (12/2, 2/2) = (6,1)
1
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
1
50 (d) Scalene triangle
AB=√45 Units
BC=√5 units
AC=√80 units
1