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Class-10th Combat- Maths
Type of Question: Concept
Question Variety: Text based
Difficulty Level: Easy
Expected Time to Solve (in seconds): 60
Topic: Number Systems
Concept:
Sub Concept:
Concept Field: N/A
Q.1 If LCM (954, 488) = 232776, find HCF (954, 488).
- 1
- 2
- 3
- 4
Answer: B
Solution
LCM(954, 488) = 232776
HCF =
Type of Question: Concept
Question Variety: Text based
Difficulty Level: Easy
Expected Time to Solve (in seconds): 60
Topic: Algebra
Concept: Pair of Linear Equations
Sub Concept:
Concept Field: N/A
Q.2 Find the value of k for which 2x + 3y =5, kx – 6y = 7 has no solution.
- -3
- 3
- 4
- -4
Answer = D
Solution:
Linear equation has no solution so,
Type of Question: Concept
Question Variety: Text based
Difficulty Level: Medium
Expected Time to Solve (in seconds): 60
Topic: Algebra
Concept: Quadratic Equations
Sub Concept:
Concept Field: N/A
Q3 Find the present age of Rahul, if the sum of Rahul’s and his son’s present age is 55 years and the product of their ages five years ago was 350.
- 18
- 15
- 40
- 26
Answer: C
Solution:
Let the Present of mother is = x years
Then the age of daughter is = 55 – x
According to Questions
(x – 5) (55 – x – 5 ) = 350
(x – 5) (50 – x) = 350
50x – x2 – 250 + 5x = 350
55x – x2 – 250 – 350 = 0
x2 – 55x + 600 = 0
x2 – 15x – 40x + 600 =0
x(x – 15) – 40(x – 15) = 0
(x – 15) (x – 40) = 0
X – 15 = 0, | x – 40 =0
x = 15 (Not Possible), | x = 40
so, Rahul’s age is 40 years
Type of Question: Concept
Question Variety: Text based
Difficulty Level: Medium
Expected Time to Solve (in seconds): 60
Topic: Algebra
Concept: Arithmetic Progression
Sub Concept: Sum of nth term
Concept Field: N/A
class 10th math mcq question-class 10th math mcq test
Q. 4Find the sum of first 17 terms of an AP whose third term is 9 and fifth term is 15.
- 408
- 308
- 480
- 564
Answer: A
Solution
a3 = 9
a + 2d = 9 ……..(i)
a5 = 15
a + 4d = 15 ……(ii)
from (i) and (ii)
2d = 6
d = 3
Now using d = 3 in equation (i)
We get
a + 2 (3) = 9
a + 6 = 9
a = 9 -6
a = 3
Now, using the formula Sn =
=408
Type of Question: Concept
Question Variety: Text based
Difficulty Level: Medium
Expected Time to Solve (in seconds): 60
Topic: Statistics & Probability
Concept: Probability
Sub Concept:
Concept Field: N/A
Q.5 A shopkeeper mixed 15 duplicate pencils with 85 original pencils and put all those in a sale for Rs. 2 each. If a customer purchased one pencil from that sale. Find the probability that drawns pencil is not original.
- 3/20
- 1/15
- 1/7
Answer: B
Solution:
Total number of duplicate pencils = 15
Total number of original pencils = 85
Total number of pencils = 15 + 85 = 100
Total number of total possible outcomes = 100
Total number of favourable outcomes (not original) = 15
P(E) =
Type of Question: Concept
Question Variety: Text based
Difficulty Level: Hard
Expected Time to Solve (in seconds): 60
Topic: Statistics & Probability
Concept: Statistics
Sub Concept:
Concept Field: N/A
Q.6 The following frequency distribution gives the pocket expenses of the students of class 10th in a specific locality. Find the median.
Class interval | Frequency |
0-20 | 6 |
20-40 | 8 |
40-60 | 10 |
60-80 | 12 |
80-100 | 6 |
100-120 | 5 |
120-140 | 3 |
- 61.67
- 81.71
- 63.67
- 65.76
Answer: A
Solution:
Class interval | Frequency | Cumulative Frequency |
0-20 | 8 | 8 |
20-40 | 6 | 14 |
40-60 | 10 | 24 = cf |
60-80 | 12 = f | 36 |
80-100 | 6 | 42 |
100-120 | 5 | 47 |
120-140 | 3 | 50 |
n/2 = 50
Cumulative frequency just greater than 25 is 36.
So, the median class is 60 – 80
Median =
Q.7
Type of Question: Concept
Question Variety: Text based
Difficulty Level: Hard
Expected Time to Solve (in seconds): 60
Topic: Mensuration
Concept: Areas Related to Circles
Sub Concept:
Concept Field: N/A
Find the length of the corresponding arc of the sector whose area is 154 m2 and the diameter is 28 m
- 22 m
- 21 m
- 19 m
- 23 m
Answer: A
Solution:
Given: Area of the sector = 154 m2
Type of Question: Concept
Question Variety: Text based
Difficulty Level: Very Hard
Expected Time to Solve (in seconds): 60
Topic: Mensuration
Concept: Surface Areas and Volumes
Sub Concept:
Concept Field: N/A
A conical cavity of same height and same diameter is hollowed out from the solid cylinder whose radius is 0.007 m and height is 2.4 cm. Find the volume of the remaining part of cylinder.
- 4.464
- 3.464
- 1.464
- 2.464
Answer: D
Solution:
Radius = 0.007m
= 0.007 x 100 cm
= 0.7cm
Height = 2.4 cm
Volume of remaining cylinder = volume of cylinder – volume of cone
Ques 9 –
Find the point on the Y axis which is equidistance from the points (-2,5) and (-2,9)
- (0, 4)
- (0, 7)
- (0, 3)
- (4. 0)
CORRECT OPTION: B
As the point is on Y Axis, let its coordinates be (0, y)
Now
2 + (5 – y)2 = 2 + (y – 9)2
2 + (5 – y)2 = 2 + (y – 9)2
4 + (5 – y)2 = 4 + (y – 9)2
4 + 25 + y2 – 10y = 4 + 81 + y2 – 18y
29 – 10y = 85 – 18y
– 10y + 18y = 85 – 29
8y = 56
y= 7
topic: coordinate geometry
time: 45 sec
type: easy
Ques. 10 –
In a triangle PQR, A and B are the mid-points on the sides PR and QR respectively, such that AB, If PA = 4x, QB = 4x – 1, RA = 2x + 1 and RB = 2x – 1, find the value of x.
- 1/3
- 2/3
- 1/4
- 1/6
Answer: D
Solution:
so AB Parralllel PQ
then
8x2 – 4x = 8x2 + 4x – 2x – 1
8x2 – 4x – 8x2 – 4x + 2x = – 1
– 4x – 2x = – 1
–x = 1/6
Difficulty level: Medium
Time: 45 sec
Topic: triangles
Ques 11 – In the given circle, diameter AC parallel to chord ED. If ∠EBC = 43°, find ∠DEC.
(A) 133
(B) 47
(C) 43
(D) 137
Answer: B
Solution:
∠AEC = 90 [Angles in semi-cricle]
∠A = ∠B = 43 [Angles in same segment]
∴ ∠ACE = 90- 43 [Angle Sum Property]
47
∴ ∠ACE = ∠DEC = 47 [Alternate Angles]
Difficulty Level: Medium
Time to
solve: 45 secs
Topic – Circles
class 10th math mcq question-class 10th math mcq test
Ques 12.
In the given diagram if DS= 12 cm, BD= 5 cm, & ∠D = 90° find the radius r.
- 3 cm
- 4cm
- 2cm
- 5cm
Correct option – (C)
Solution – Let the radius be ‘r’ cm.
OA = OC = r cm
∠D = 90°
:- ADCO is a square.
DS = 12cm
DC = ‘r’ cm
:- CS = (12 – r ) cm
CS = SP {tangents}
SP = (12 – r) cm ——— 1
Now, AD = ‘r’ cm
BD = 5 cm
: – BA = (5 – r)cm
AB = BP [tangents]
: – BP = (5 – r) cm ———— 2
In triangle BDS,
BD2 + DS2 = BS2
52 + 122 = BS2
: – BS2 = 25 + 144
BS = √169
= 13 cm
From – 1 and 2, we get.
BS = PS + BP
13 = (12 – r) + (5 – r)
13 = 12 – r + 5 – r
13 = 17 – 2r
: – 2r = 4
: – r = 2cm
Difficulty Level: Very Hard
Time to solve: 120 sec
Topic: Heights and distances.
class 10th math mcq question-class 10th math mcq test
Ques 13.
If sec ( A+ B)= 2 and sec (A-B) = 2/, Find the value of A & B, IF A > B and 00 < A +B < 900
A A= 300, B= 150
B A= 450, B= 150
C A= 600, B= 150
D None of the above
Correct option: B
Solution:
sec ( A+ B)= 2
sec ( A+ B)= sec 600
comparing both sides we get
A + B = 600 ——————– (1)
sec (A-B) = 2/
sec (A-B) = sec 300
comparing both sides we get
A – B = 300 ——————– (2)
Subtracting (2) from (1) we get
A + B = 600
A – B = 300
2B = 300
B = 150
Using B in equation (1) we get
A + 150 = 600
A = 450
Difficulty level: Easy
Time to solve: 45 sec
Topic: trigonometry
class 10th math mcq question-class 10th math mcq test
Ques 14 –
cos2 30° + sin2 45° + cot2 60° + sin2 30 – tan260°
- 4/3
- 2/3
- -6/7
- -7/6
Correct option – (D)
Solution –
= (√3/2)2 + (1/√2) 2 + (1/√3) 2 + (½)2 – (√3) 2
= ¾ + ½ + ⅓ + ¼ – 3
= 1 + ½ + ⅓ – 3
= 6 + 3 + 2 – 18
6
= 11 – 18
6
= – 7/ 6
Difficulty Level: Hard
Time to solve: 45 sec
Topic: trigonometry
class 10th math mcq question-class 10th math mcq test
Ques. 15 –
In the given diagram find the value of h
- 60m
- 120m
- 80m
- 95m
Correct option – B
Solution :
Let the height of tower be ‘h’ m
Let ED be ‘x’ m
: – AB = ED = ‘x’ m
Now, in triangle ECD,
Tanθ = EC
ED
Tan60° = h – 30
X
√3 = h – 30
X
: – x = h – 30 —————— 1
√3
In triangle DEB
Tanθ = EB/ED
Tan30° = 30/x
1/ √3 = 30/x
:- x = 30√3 ————– 2
From 1 and 2, we get,
h – 30 = 30√3
√3
M h – 30 = 30 x 3
h – 30 = 90
h = 120m
Difficulty Level: very Hard
Time to solve: 90 sec
Topic: heights and distances.
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