# class 10th math mcq question-class 10th math mcq test

class 10th math mcq question-class 10th math mcq test

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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. 1
2. 2
3. 3
4. 4

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.

1. -3
2.  3
3.  4
4. -4

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

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.

1. 18
2. 15
3. 40
4. 26

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.

1. 408
2. 308
3. 480
4. 564

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 =

$\frac{n}{2}\left[2a+\left(n–1\right)d\right]\phantom{\rule{0ex}{0ex}}\frac{16}{2}\left[2×3+\left(16–1\right)3\right]$

=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.

1. $\frac{1}{85}$

2. 3/20
3. 1/15
4. 1/7

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) =

$\frac{15}{100}=\frac{3}{20}$

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

1. 61.67
2. 81.71
3. 63.67
4. 65.76

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  =

$l+h\left[\frac{\frac{n}{2}–cf}{f}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}60+20\left[\frac{25–24}{12}\right]\phantom{\rule{0ex}{0ex}}60+20×\frac{1}{12}\phantom{\rule{0ex}{0ex}}61.67$

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

1. 22 m
2. 21 m
3. 19 m
4. 23 m

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.

1.      4.464
2.       3.464
3.      1.464
4.     2.464

Solution:

= 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)

1. (0, 4)
2. (0, 7)
3. (0, 3)
4. (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. 1/3
2. 2/3
3. 1/4
4. 1/6

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

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.

1. 3 cm
2. 4cm
3. 2cm
4. 5cm

Correct option – (C)

Solution – Let the radius be ‘r’ cm.

OA = OC = r cm

∠D = 90°

DS = 12cm

DC = ‘r’ cm

:-   CS = (12 – r ) cm

CS = SP {tangents}

SP = (12 – r) cm ——— 1

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°

1. 4/3
2. 2/3
3. -6/7
4. -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

1. 60m
2. 120m
3. 80m
4. 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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