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Class 10 Maths Term 2 Important Questions
Central Board of Secondary Education is all set to conduct CBSE Term 2 Maths exam on 5th May 2022. As the exam dates are approaching the students must solve CBSE Class 10 Term 2 Maths Important Questions given on this page to pass the class 10th maths exam with flying colors. We have also given the solutions here, if students get stuck in any question they can check the solution. Go through the entire article and bookmark this page to get all the latest updates from the Central Board of Secondary Education.
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Maths Class 10 Term 2 Important Questions: 2 Marks
1. Determine the factors of the equation by using the factored form method.
3x2 – 10x + 8 = 0
Answer:
=>3x2 – 10x + 8 = 0
3x2 – 6x – 4x + 8 = 0
3x(x – 2) – 4(x – 2) = 0
x = 2, 4/3
2. What will be the approximate least and greatest value of “y” in the given equation:
(y-7)2=484?
Answer:
y-72=484
=> y -72 = 222
=>y = 22+7 and -22 +7
=> y = 29, -15
3. If the distance between points (5, 2) and (1, a) is 5, then find the value of a, If (a < 0).
Answer:
Distance = {(5-1)2+2-a2}
=>5= {(4)2+2-a2}
=>25 = 16+2-a2
=>25-16 = 2-a2
=>9=2-a2
As value of a is negative.
=>2-a=3
=>a = 2-3 = -1
OR
A circle with its center at O (x,11) has a radius of 26 units. Point P (3,-13) lies on the circle.Also, x> 10. Find x.
Answer:
As O is at (x, 11) and p (3, – 13)
=> OP = x-32+(11--132)
=> OP = (x-3)^2+24^2
=>OP = (x-3)^2+576
=>262 = x-32+576
=>676 = x-32 +576
=>x-32= 100
=>(x-3) = 10
=>x = 13 (As x > 10)
4. What will be the sum of the first 21 terms of the AP 4, 7, 10, …..?
Answer:
AP = 4, 7, 10, ……
Here, a = 4
d = 7-4 = 3
Number of terms = n = 21
So, sum of first 21 terms of AP will be Sn =n2 (2a+(n-1)d)
=>Sn =212 * [2*4 +(21-1)*3]
=>Sn =212 * [8+60]
=>Sn =212 * 68
=>Sn = 21*34 =714
5. Mohan bought a toy which is packed in box to gift his Sohan on his birthday. The dimension of the box is 6cm x 6cm x 6cm. What size of gift paper is required to wrap the gift?
Answer:
The dimension of box is 6cm x 6cm x 6cm
Curved surface area of cube = 6 x a2
= 6x 62
= 216 m2
6. Find the mode of the following items.
1,2,2,5,6,6,4,2,4,0,4,4,6,5,3
Answer:
On arranging the items in ascending order, we get:
0,1,2,2,2,3,4,4,4,4,5,5,6,6,6
As we can see 4 occurs the maximum number of times.
Therefore, the mode of the given items = 4
OR
A bag contains a red ball, a blue ball, a yellow ball and an orange ball, all the balls being of the same size. Priti takes out a ball from the bag without looking into it. What is the probability that she takes out the
(i) yellow ball? (ii) orange ball?
Answer:
Priti takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them.
Let be the event ‘the ball taken out is yellow ‘and be the event ‘the ball taken out is orange”
Now, the number of possible outcomes .
(i) The number of outcomes favourable to the event .
So,
Similarly, (ii)
7. If the mid point of line AB is (7, 1) determine the coordinates of B if A (9.5, -6.5).
Answer:
Given, (x1, y1) = (9.5, -6.5)
B = (x2, y2)
According to the formula we can find the midpoint (x, y):
(x, y) = [x1 + x22,y1 + y22]
(7, 1) = [9.5+ x22,-6.5+y22]
=> 9.5+x2=14
=> x2 = 5.5
And, y2 = -6.5+y2 = 2
=> y2=8.5
8. If the value of 13Secθ = 85, then what will be the value of Cosecθ * tanθ?
Answer:
OR
In the figure given below, ∠PTS = ∠QPR, PS || QR, PS : QR = 7 : 9 and PR = 36 cm, then find the length of TR.
Answer:-
∠PTS = ∠QPR
PS || QR
PS : QR = 8 :10
PR = 36 cm
Rule of similarity AA(Angle-Angle)
∆TSP and ∆PQR are similar.
Now, PS/QR = PT/PR
⇒ 7/9 = PT/36
⇒ PT = 28 cm
Now, TR = PR – PT
= 36 – 28 = 8 cm
9. A box contains lemon and banana, if there are 13 bananas in the box and the probability of selecting 2 bananas from the box is 1/10, then find the number of lemons in the box?
Answer:
The number of bananas in the box =13
Let the number of lemons in the box =X
13*12X+13X+12=110
156X2+13X+12X+156=110
1560=X2+25X+156
X2+25X-1404=0
X2-27X+52X-1404=0
X(X-27)+52(X-27)=0
X=27, -52 (Neglecting the negative values)
Number of lemons in the box =27
10. The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify?
Number of wickets | 20 – 60 | 60 – 100 | 100 – 150 | 150 – 250 | 250 – 350 | 350 – 450 |
Number of
bowlers |
7 | 5 | 16 | 12 | 2 | 3 |
Answer:
Here, the class size varies, and the xi,s are large. Let us still apply the step deviation method with a = 200 and h = 20. Then, we obtain the data as in Table
Number of wickets taken | Number of bowlers
( fi ) |
xi | di = xi – 200 | ui = di20 | ui fi |
20 – 60 | 7 | 40 | –160 | –8 | –56 |
60 – 100 | 5 | 80 | –120 | –6 | –30 |
100 – 150 | 16 | 125 | –75 | –3.75 | –60 |
150 – 250 | 12 | 200 | 0 | 0 | 0 |
250 – 350 | 2 | 300 | 100 | 5 | 10 |
350 – 450 | 3 | 400 | 200 | 10 | 30 |
Total | 45 | -106 |
So,
Therefore,
This tells us that, on an average, the number of wickets taken by these 45 bowlers in one-day cricket is 152.89.
11. Find the roots of the following quadratic equation by completing the square method. 14 x2+23 x-30=0.
Answer;
x2 + 13x – 17 = 0
Comparing to general equation:
ax2 +bx +c = 0
=> a = 1, b = 13, c = -17
=> Roots = [-b ± √(b² – 4ac)]/2a
So,
12. What is the length of the radius for the given figure?
Answer:
Here, B is the point of tangency.
So,
Subtract both sides by 2500,
=> r2+3900 = r2+100r
Cancelling r2
=> 3900 = 100r
=> r = 39 ft
OR
What is the radius of the circle given below?
Answer:-
Here, AB is a tangent and BC is the radius
∴ BC ⊥ AB
Now, in triangle CBA,
⇒ CA2= AB2 + CB2
⇒ 302 = 262 + OT2
⇒ 900 – 676 = OT2
⇒ 224 = OT2
⇒ 14.96 cm = OT
13. If the value of 25Sinθ = 24, then what will be the value of Sinθ * Cosθ * Cotθ?
Answer:
As, sin θ = Perpendicular/Hypotenuse
25sin θ = 24
=>sin θ = 24/25
=> Perpendicular = 24, Hypotenuse = 25
\[=>Base=\sqrt{625-576}=7\]
cos θ = Base/Hypotenuse = 7/25
Cotθ = Base/Perpendicular = 7/24
=>Sinθ * Cosθ * Cotθ
=>[24/25 * 7/25 ] * 7/24
=>49/625
14. A juice seller was serving his customers using glasses as shown in fig. below. The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 10 cm, find the apparent capacity of the glass and its actual capacity. (Use π = 3.14.)
Answer:
Since the inner diameter of the glass and height ,
the apparent capacity of the glass
But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass.
i.e., it is less by
So, the actual capacity of the glass =
15. Present weight of Saket is 100 kg. He has decided to take a weight loss challenge, during his challenge he can only reduce 2.5 kg in each month. After how many months his weight will be 70 kg?
Answer:
As here. a = 100
d = – 2.5 (As his weight is decreasing by 2.5 kg in each month).
So, an = 70
As, an = a + (n-1)d
=> 70 = 100 + (n -1) * (-2.5)
=> 70 = 100 -2.5n +2.5
=> 2.5 n = 102.5 -70
=> 2.5 = 32.5
=> n=32.52.5 = 13
16. Determine the factors of the equation by using the factored form method.
y2-23y+112=0
Answer:
=>y2-23y+112=0
y2-16y-7y+112=0
y(y-16)-7(y-16)=0
y=+16, +7
OR
Find the roots of the following equations:
Answer:
17. A tangent PQ at a point P of a circle of radius 4 cm meets a line through the centre O at a point Q so that OQ = 5 cm. Length PQ is
Answer:
Triangle OPQ is a right – angled triangle
Therefore,
OQ2 = PQ2 + OP2
PQ2 = OQ2 – OP2
PQ2 = OQ2– OP2
PQ2 = 52– 32
PQ2 = 25-9
PQ2 = 16
PQ =4 cm
18. A cube has a volume of 64 cm3 then, what is the length of the cube?
Answer:
Volume of cube = a3
=> 64 = a3
=> a = 364
=> a = 4 cm.
19. The daily minimum steps climbed by a man during a week were as under:
Monday | Tuesday | Wednesday | Thursday | Friday | Saturday |
30 | 35 | 24 | 26 | 32 | 28 |
Find the mean of the steps.
Answer:
Number of steps climbed in a week: 30,35,24,26,32,28.
So, we get,
Mean = sum of observation (steps) / total no of observations
=30 + 35 + 24 + 26 + 32 + 286
=1756
= 29.17
OR
A coin and a dice is tossed together. What will be the probability of getting a tail and a prime number?
Answer:
Since the probability of getting tail = 1/2
As when dice is thrown,
We get numbers from 1 to 6 i.e. 1, 2, 3, 4, 5, 6
So, prime numbers are = 2, 3, 5 = 3
The probability of getting a prime number = 3/6 = 1/2
So, the probability of getting tail and a prime number will be = 1/2 *1/2 =1/4
20. Find two numbers whose sum is 27 and product is 182.
Answer:
Let the number is x
Then equation will be
=> x2+ 27x + 182 =0
=> x2 + 14x + 13x + 182 =0
=> x(x+14) + 13 (x+14) = 0
=> (x+14) (x+13)
=> x = 14,13
CBSE Class 10 Term 2 Maths Important Questions: 3 Marks
1. Which is correct sigma notation of the AP:101, 97, 93, 89, ….?
Answer:
AP is 101, 97, 93, 89 ,….
Here, a = 101 and other terms are continuously decreasing with a common difference of 4
So, we can write “a” i.e. 101 as (105 – 4*1) = (105 – 4*n)
So, correct sigma notation will be,
OR
Present weight of Saket is 100 kg. He has decided to take a weight loss challenge, during his challenge he can only reduce 2.5 kg in each month. After how many months his weight will be 70 kg?
Answer:
As here, a = first term = 3 kg
d = common difference = 2.5 (As his weight is increasing by 2.55 kg in every year)
So, after 15 years, means value of n = 15
Weight of baby will be:
= a15 = a + (n-1)d
= 3+ (15-1)×2.5
=>3+ 14×2.5
=>3 + 35 = 38 kg.
2. In triangle ABC, ∠C = 90°, BC = 24 cm and AC = 7 cm. Determine the value of SinA * CotB.
Answer:
As here,
Hypotenuse2= 242 + 72
=> Hypotenuse = 25 cm
=> SinA = opposite side/ hypotenuse = 24/25
=> TanB =opposite side/adjacent = 7/24
=>SinA * CotB = 24/25 * 7/24 = 7/25
3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.
Answer:
fix and calculate the value of and
Calculate the unknown frequency
Mean
f = 402
f = 20
4. A box contains 4 white and 8 yellow balls. Two balls are taken out at random, without replacement. If the first ball is white, then what is the probability that the second ball is also white?
Answer:
Number of white balls = 4, Number of yellow balls = 8
Number of balls taken out at random = 2
Total balls = 4 + 8 = 12
Required probability = 4/12 * 3/11 = 1/11
5. What will be the length of AD and BC of the given figure?
Answer:-
Given that,
ABC is a triangle,
D is a point on the side BC of ΔABC
=> ∠ADC=∠BAC
By using properties of similar triangle,
=> AB/AD = BC/AC = AC/DC
=> 8/AD = 6/3
=> AD = 4cm
Now, BC/6 = 6/3
=> BC = 12cm
6. In the given figure, if ∠PQR = 30°, find the value of ∠POR.
Answer:
Here, OP ⊥ QP and OR ⊥ QR
∴ ∠OSR = ∠ORQ = 90°
In triangle OPQR, ∠POR + ∠OPQ + ∠PQR + ∠ORQ = 360°
∠POR + 90° + 30° + 90° = 360°
∠POR = 360° – 210° = 150°
OR
In the given figure, PQ and PR are tangents to the circle with centre O such that ∠QPR = 45°, then ∠QOR is equal to:
Answer:
In quadrilateral PQOR
∠PQO + ∠QOR + ∠ORP + ∠QPR = 360°
⇒ 90° + ∠QOR + 90° + 45° = 360°
⇒ ∠QOR = 360° – 225° = 135°
7. If the value of Cosecθ = 26/24, then what will be the value of Sinθ + 2Cosθ?
Answer:
As
8. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find The volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Answer:
Answer:
Height of cylinder
Radius of cylinder
Height of cone
Radius of cone
Radius of hemisphere
But
Volume of water left in the cylinder
8. If Mahesh were younger by 5 years than what he really is. The square of his age would have been 11 more than five times of his present age. What is his present age?
Answer:
Let x be the present age of Mahesh
Then,
=> x-52 = 5x+11
=> x2-10x+25 =5x+11
=> x2-15x+14=0
=>x2-14x-x+14 =0
=>x(x-14)-1(x-14)=0
So,
x=14 or x=1
As age can’t be negative so, now age of Mahesh is 14.
Class 10 Term 2 Maths Important Questions: 4 Marks
1. Find the midpoint of a line whose endpoints are (11, 13) and (9, 11).
Answer:
Given, (x1, y1) = (11, 13)
(x2, y2) = (9, 11)
According to the formula we can find the midpoint (x, y):
(x, y) = [x1 + x22,y1 + y22]
(x, y) = [11 + 92,13 + 112]
= (10, 12)
2. Disha has two dairy milk and 3 perk chocolates. What will be the probability of taking two chocolates, one of dairy milk and one of perk?
Answer
Dairy milk = 2
=> Probability of taking one dairy milk P(A) = 1/2
Perk = 3
=> Probability of taking one perk P(B) = 1/3
=> P(A and B)=P(A∩B)=P(A)⋅P(B) = 1/3*1/2 = 1/6
OR
Find the value of y from the following observations if these are already arranged in ascending order. The Median is 63.
20, 24, 42, y , y + 2, 73, 75, 80, 99
Answer:
As the number of observations made is odd, so the median will be the middle term, i.e. y + 2.
Therefore,
y + 2 = 63
y = 63 – 2 = 61
3. What will be the approximate least and greatest value of “x” in the given equation?
x+42 -15 =0?
Answer:
x+42 -15 =0
=>x+42 = 15
=>(x+4)^2 = 152
=>x = √15 -4 and √15+4
As, √15 = 3.8729
=> x = 3.873 – 4 and 3.873 + 4 => -0.127 and 7.873
OR
The earning for the month of January of Mr. Tiwari is Rs. 1000. If his earning gets increased by Rs. 125 every month, then what will be his earning in the month of August?
Answer:
As here, a = first term = 1000
d = common difference = 125 (As earning is increased by Rs. 125 in every month)
So, for August, value of n will be = 7
=> Earning of August will be
= a7 = a + (n-1)d
= 1000+ (7-1)×125
=>1000+ 6×125
=>1000 + 750 = Rs. 1750
4. A cylinder shaper container has a height of 5m and its radius is 2m. How much water can a container store?
Answer:
Volume of cylinder
= πr2h
= π×2×2×5
= 62.83 m3 ≃63 m3
The container can contain 63 m3 of water .
5. The weight of a new born baby is 3 kg after every year his weight is increasing by 2.5 kg. What will be his weight when he will be 16 years old?
As here, a = first term = 3 kg
d = common difference = 2.5 (As his weight is increasing by 2.55 kg in every year)
So, after 15 years, means age when he will be 16 years old,
So, value of n = 15
Weight of baby will be:
= a15 = a + (n-1)d
= 3+ (15-1)×2.5
=>3+ 14×2.5
=>3 + 35 = 38 kg.
6. Draw the line segment PQ = 5 cm. From the point P draws a line segment PS = 6cm making an angle of 600 with PQ. Draw a perpendicular bisector of PS.
Answer:
- If the value of Sinθ = 63/65, then what will be the value of Cosθ * Tanθ?
As, sin θ = Perpendicular/Hypotenuse = 63/65
=> Perpendicular =63 , Hypotenuse = 65
=> Base = 652–632 = 16
=> Cos θ = Base/ Hypotenuse = 16/65
tan θ = Perpendicular/Base = 63/16
=>Cos θ * Tan θ = 16/65 * 63/16 = 65/63
7. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number.
Answer:
Get the values of a and h
a = 57
h = 53 – 50 = 3
Calculate the mean
Mean (x) = a + h × ∑fiui∑fi
x = 57 + 3×25400
x = 57 + 316
x = 57 + 0.19
x = 57.19
8. A bag contains 3 red, 5 yellow and 4 blue balls. If three balls are drawn at random, then find the probability that balls are yellow?
Answer:
Required probability = 5C3/12C3
= 5 * 4 * 3/12 * 11 * 10
= 1/22
OR
Yugali is playing a cricket match and a game of football; the probability of winning cricket is 0.4 while the probability of winning football is 0.6. What will be the probability that she will win both matches assuming that the result of the matches is independent?
Answer:
Probability of winning cricket = 0.4
Probability of winning football = 0.6
Probability of winning cricket and football = 0.4*0.6 = 0.24
9. Find the roots of the following quadratic equation by completing the square method. 14 x2+23 x-30=0 .
Answer:
To convert the coefficient of the first term in 14x2, multiply the equation by 14.
Result is,
Take the square root of both sides of the equation.
Result is, 14 x+232=472
The roots of the equation are given by,
10. In the given figure, S is the point of tangency. What will be the radius of the circle T?
Answer:-
Given that,
S is the point of tangency,
So
Subtract both sides by 1296,
OR
What is the value of x for the given figure?
Answer:-
Here, US is a tangent to point R at S,
UV is a tangent to point R at V
Tangents are congruent,
Therefore,
SV = VU
11. If the value of 17Sinθ = 8, then what will be the value of (2Secθ + 4Tanθ)/ Cosec θsinθ?
Answer:
As, sin θ = Perpendicular/Hypotenuse
17sin θ = 8
=>sin θ = 8/17
=> Perpendicular = 8, Hypotenuse = 17
=> sec θ = Hypotenuse/Base = 13/12
tan θ = Perpendicular/Base = 5/12
Cosecθ = 17/8
(2Secθ + 4Tanθ)/ Cosec θsinθ
=>136+531
=>136+53
=>13+106 =236
12. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 16 cm by 10 cm by 3.7 cm. The radius of each of the depressions is 0.6 cm and the depth is 1.5 cm. Find the volume of the wood in the entire stand.
Answer:
The dimensions of the cuboid are16 cm by 10 cm by 3.7 cm.
So, Volume of the cuboid is = l × b × h = 16 × 10 × 3.7 = 592 cm3
Radius of the conical depression (r) = 0.6 cm
Height of the conical depression (h) = 1.5 cm
As the pen stand made of wood is in the shape of a cuboid with four conical depressions,
Volume of wood
= volume of cuboid – 4 (volume of cone)
= 592 – (4 × 13 × 227 × 0.6 × 0.6 × 1.5)
= 592 – 2.26
= 589.74 cm3
So, the Volume of wood = 589.74 cm3
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Q. Is NCERT enough for Class 10 Term 2 Mathematics?
Yes, NCERT is more than enough for Class 10 Term 2 Mathematics. The students can practice the CBSE Class 10 Term 2 Mathematics Important Questions given on this page.
Q. Where can I find CBSE Class 10 Term 2 Mathematics Important Questions?
You can find the CBSE Class 10 Term 2 Mathematics Important Questions here. We have given the CBSE Class 10 Term 2 Mathematics Important Questions based on the latest CBSE Term 2 Exam pattern on this page.
Q. When CBSE will conduct the Class 10 Term 2 Exam for Mathematics?
The CBSE will conduct the Class 10 Term 2 Exam for Mathematics on the 5th of May 2022.