**Maths Subject Multiple Choice Questions MCQ** is very important for students who want to score good marks in their CBSE board examination. Students who Opted for Basic and Standard Mathematics can Practice These Questions to improve your score in Board Exams.

## CBSE Maths Multiple Choice Questions with Answers for Class 10

Practicing Maths MCQ for Class 10 CBSE is one of the best ways to prepare for CBSE class 10 board exam. There is no substitute for consistent practice whether one wants to understand a concept thoroughly or one wants to score better. By practicing more Science MCQ for Class 10 CBSE with answers, students can improve their speed and accuracy which can help them during their board exam.

**MCQ Questions for Class 10 Maths Real Numbers with Solutions****MCQ Questions for Class 10 Maths Polynomials with Solutions****MCQ Questions for Class 10 Maths Pair of Linear Equations in Two Variables with Answers****MCQ Questions for Class 10 Maths Quadratic Equations with Answers****MCQ Questions for Class 10 Maths Arithmetic Progressions with Answers****MCQ Questions for Class 10 Maths Coordinate Geometry with Answers****MCQ Questions for Class 10 Maths Introduction to Trigonometry with Answers****MCQ Questions for Class 10 Maths Application of Trigonometry with Answers****MCQ Questions for Class 10 Maths Circles with Answers****MCQ Questions for Class 10 Maths Geometrical Constructions with Answers****MCQ Questions for Class 10 Maths Areas Related to Circles with Answers****MCQ Questions for Class 10 Maths Surface Areas and Volumes with Answers****MCQ Questions for Class 10 Maths Statistics with Answers****MCQ Questions for Class 10 Maths Probability with Answers**

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### Real Numbers Class 10 Maths Multiple Choice Questions with Answers

**Question 1:
**For some integer m, every even integer is of the form

(a) m (b) m +1 (c) 2m +1 (d) 2m

**Solution:**

(c) We know that, even integers are 2, 4, 6,…

So, it can be written in the form of 2m.

**Question 2:
**For some integer q, every odd integer is of the form

(a) q (b) q +1 (c) 2g (d) 2q +1

**Solution:**

**Question 3:**

**Solution:**

**Question 4:
**If the HCF of 65 and 117 is expressible in the form 65m -117, then the value of m is

(a) 4 (b) 2 (c) 1 (d) 3

**Solution:**

**Question 5:
**The largest number which divides 70 and 125, leaving remainders respectively, is

(a) 13 (b) 65 (c) 875 (d) 1750

**Solution:**

(a) Since, 5 and 8 are the remainders of 70 and 125, respectively. Thus, after subtracting these remainders from the numbers, we have the numbers 65 = (70-5),

117 = (125 – 8), which is divisible by the required number.

Now, required number = HCF of 65,117 [for the largest number]

**Question 6:
**If two positive integers a and b are written as a = x

^{3}y

^{2}and b = xy

^{3}, wfiere x, y are prime numbers, then HCF (a, b) is

(a) xy (b) xy

^{2}(c)x

^{3}y

^{3}(d) xy

^{2 Solution: }

**Question 7:
**If two positive integers p and q can be expressed’ as p=ab

^{2}and q = a

^{3}b; where 0, b being prime numbers, then LCM (p, q) is equal to

(a) ab (b) a

^{2}b

^{2}(c) a

^{3}b

^{2}(d) a

^{3}b

^{3 Solution: }

**Question 8:
**The product of a non-zero rational and an irrational number is

(a) always irrational (b) always rational

(c) rational or irrational (d) one

**Solution:**

**Question 9:
**The least number that is divisible by all the numbers from 1 to 10 (both inclusive)

(a) 10 (b) 100 (c) 504 (d) 2520

**Solution:**

(d) Factors of 1 to 10 numbers

**Question 10:**

(a) one decimal place (b) two decimal places

(c) three decimal places (d) four decimal places

**Solution:**

### Pplynomials Class 10 Maths Multiple Choice Questions with Answers

**Question 1:**

**Solution:**

(a) Given that, one of the zeroes of the quadratic polynomial say p(x) = (k- 1)x^{2} + kx + 1

**Question 2:
**A quadratic polynomial, whose zeroes are -3 and 4, is

(a) x

^{2}– x + 12 (b)x

^{2}+ x + 12 (c) (d)2x

^{2}+ 2x-24

**Solution:**

(c) Let ax

^{2}+ bx + c be a required polynomial whose zeroes are -3 and 4.

We know that, if we multiply/divide any polynomial by any constant, then the zeroes of polynomial do not change.

**Alternate Method**

Let the zeroes of a quadratic polynomial are α = – 3 and β = 4.

Then, sum of zeroes =α + β = -3+4=1 and product of zeroes = αβ = (-3) (4) = -12

**Question 3:
**If the zeroes of the quadratic polynomial x

^{z}+ (a +1)* + b are 2 and -3, then

(a) a = -7, b = -1 (b) a = 5,b = -1

(c) a=2, b = -6 (d)a=0,b = -6

**Solution:**

(d) Let p{x) =+ b

Given that, 2 and -3 are the zeroes of the quadratic polynomial p(x).

required values are a = 0 and b = – 6.

**Question 4:
**The number of polynomials having zeroes as -2 and 5 is

(a) 1 (b) 2 (c) 3 (d) more than 3

**Solution:**

(d) Let p (x) = ax

^{2}+ bx + c be the required polynomial whose zeroes are -2 and 5.

Hence, the required number of polynomials are infinite i.e., more than 3.

**Question 5:**

If one of the zeroes of the cubic polynomial ax^{3} + bx^{2} + cx + d is zero,

the product of then other two zeroes is

(a) (b) (c)0 (d)

**Solution:**

(b) Let p(x) =ax^{3} + bx^{2} + cx + d

Given that, one of the zeroes of the cubic polynomial p(x) is zero.

Let α, β and γ are the zeroes of cubic polynomial p(x), where a = 0.

We know that,

**Question 6:
**If one of the zeroes of the cubic polynomial x

^{3}+ ax

^{2}+ bx + c is -1, then the product of the other two zeroes is

(a) b – a +1 (b) b – a -1 (c) a – b +1 (d) a – b -1

**Solution:**

(a) Let p(x) = x

^{3}+ ax

^{2}+ bx + c

Let a, p and y be the zeroes of the given cubic polynomial p(x).

∴ α = -1 [given]

and p(−1) = 0

⇒ (-1)

^{3}+ a(-1)

^{2}+ b(-1) + c = 0

⇒ -1 + a- b + c = 0

⇒ c = 1 -a + b …(i)

We know that,

αβγ = -c

⇒ (-1)βγ = −c [∴α = -1]

⇒ βγ = c

⇒ βγ = 1 -a + b [from Eq. (i)]

Hence, product of the other two roots is 1 -a + b.

**Alternate Method**

Since, -1 is one of the zeroes of the cubic polynomial f(x) = x

^{2}+ ax

^{2}+ bx + c i.e., (x + 1) is a factor of f{x).

Now, using division algorithm,

⇒x

^{3}+ ax

^{2}+ bx +c = (x + 1) x {x

^{2}+ (a – 1)x + (b – a + 1)> + (c – b + a -1)

⇒x

^{3}+ ax

^{2}+ bx + (b – a + 1) = (x + 1) {x

^{2}+ (a – 1)x + (b -a+ 1}}

Let a and p be the other two zeroes of the given polynomial, then

**Question 7:
**The zeroes of the quadratic polynomial x

^{2}+ 99x + 127 are

(a) both positive (b) both negative

(c) one positive and one negative (d) both equal

**Solution:**

(b) Let given quadratic polynomial be p(x) =x

^{2}+ 99x + 127.

On comparing p(x) with ax

^{2}+ bx + c, we get

a = 1, b = 99 and c = 127

Hence, both zeroes of the given quadratic polynomial p(x) are negative.

Alternate Method

the above condition.

So, both zeroes of the given quadratic polynomial are negative.

**Question 8:
**The zeroes of the quadratic polynomial x

^{2}+ kx + k where k ≠ 0,

(a) cannot both be positive (b) cannot both be negative

(c) are always unequal (d) are always equal

**Solution:**

(a)Let p(x) = x

^{2}+ kx + k, k≠0

On comparing p(x) with ax

^{2}+ bx + c, we get

Here, we see that

k(k − 4)> 0

⇒ k ∈ (-∞, 0) u (4, ∞)

Now, we know that

In quadratic polynomial ax

^{2}+ bx + c

If a > 0, b> 0, c> 0 or a< 0, b< 0,c< 0,

then the polynomial has always all negative zeroes.

and if a > 0, c < 0 or a < 0, c > 0, then the polynomial has always zeroes of opposite sign

Case I If k∈ (-∞, 0) i.e., k<0

⇒ a = 1>0, b,c = k<0

So, both zeroes are of opposite sign.

Case II If k∈ (4, ∞)i.e., k≥4

=> a = 1> 0, b,c>4

So, both zeroes are negative.

Hence, in any case zeroes of the given quadratic polynomial cannot both be positive.

**Question 9:
**If the zeroes of the quadratic polynomial ax

^{2}+ bx+ c, where c≠0, are equal, then

(a) c and a have opposite signs (b) c and b have opposite signs

(c) c and a have same signs (d) c and b have the same signs

**Solution:**

(c) The zeroes of the given quadratic polynomial ax

^{2}+ bx + c, c ≠ 0 are equal. If coefficient of x

^{2}and constant term have the same sign i.e., c and a have the same sign. While b i.e., coefficient of x can be positive/negative but not zero.

which is only possible when a and c have the same signs.

**Question 10:
**If one of the zeroes of a quadratic polynomial of the form x

^{2}+ ax + b is the negative of the other, then it

(a) has no linear term and the constant term is negative

(b) has no linear term and the constant term is positive

(c) can have a linear term but the constant term is negative

(d) can have a linear term but the constant term is positive

**Solution:**

(a) Let p(x) = x

^{2}+ ax + b.

Put a = 0, then, p(x) = x

^{2}+ b = 0

⇒ x

^{2}= -b

⇒ x = ±

[∴b < 0]

Hence, if one of the zeroes of quadratic polynomial p(x) is the negative of the other, then it has no linear term i.e., a = O and the constant term is negative i.e., b< 0.

**Alternate Method**

Let f(x) = x

^{2}+ ax+ b

and by given condition the zeroes area and – α.

Sum of the zeroes = α- α = a

=>a = 0

f(x) = x

^{2}+ b, which cannot be linear,

and product of zeroes = α .(- α) = b

⇒ -α

^{2}= b

which is possible when, b < 0.

Hence, it has no linear term and the constant term is negative.

**Question 11:
**Which of the following is not the graph of a quadratic polynomial?

**Solution:**

(d) For any quadratic polynomial ax

^{2}+ bx + c, a≠0, the graph of the Corresponding equation y = ax

^{2}+ bx + c has one of the two shapes either open upwards like u or open downwards like ∩ depending on whether a > 0 or a < 0. These curves are called parabolas. So, option (d) cannot be possible.

Also, the curve of a quadratic polynomial crosses the X-axis on at most two points but in option (d) the curve crosses the X-axis on the three points, so it does not represent the quadratic polynomial.

### Pair of Linear Equations in Two Variables Class 10 Maths Multiple Choice Questions with Answers

**Question 1:
**Graphically, the pair of equations

6x – 3y + 10 = 0

2x – y + 9 = 0

represents two lines which are

(a) intersecting at exactly one point

(b) intersecting exactly two points

(c) coincident

(d) parallel

**Solution:**

The given equations are

6x-3y+10 = 0

⇒ 2x-y+ = 0 [dividing by 3]… (i)

and 2x-y+9=0 …(ii)

Now, table for 2x – y + = 0,

Hence, the pair of equations represents two parallel lines.

**Question 2:
**The pair of equations x + 2y + 5 = 0 and – 3x – 6y +1 = 0 has

(a) a unique solution (b) exactly two solutions

(c) infinitely many solutions (d) no solution

**Solution:**

**Question 3:
**If a pair of linear equations is consistent, then the lines will be

(a) parallel (b) always coincident

(c) intersecting or coincident (d) always intersecting

**Solution:**

**Question 4:
**The pair of equations y = 0 and y = – 7 has

(a) one solution (b) two solutions

(c) infinitely many solutions (d) no solution

**Solution:**

**(d)**The given pair of equations are y = 0 and y = – 7.

By graphically, both lines are parallel and having no solution

**Question 5:
**The pair of equations x = a and y = b graphically represents lines which are

(a) parallel (b) intersecting at (b, a)

(c) coincident (d) intersecting at (a, b)

**Solution:**

**(d)**By graphically in every condition, if a, b>>0; a, b< 0, a>0, b< 0; a<0, b>0 but a = b≠ 0.

The pair of equations x = a and y = b graphically represents lines which are intersecting at (a, b).

If a, b > 0

Similarly, in all cases two lines intersect at (a, b).

**Question 6:
**For what value of k, do the equations 3x – y + 8 = 0 and 6x – ky = – 16

(a) (b) (c) 2 (d) -2

**Solution:**

**(c)**Condition for coincident lines is

**Question 7:
**If the lines given by 3x+ 2ky = 2 and 2x + 5y = 1 are parallel, then the value of

**k**is

(a) (b)

(c) (d)

**Solution:**

**Question 8:
**The value of c for which the pair of equations cx- y = 2 and 6x – 2y = 3

will have infinitely many solutions is

(a) 3 (b) – 3 (c)-12 (d) no value

**Solution:**

**(d)**Condition for infinitely many solutions

Since, c has different values.

Hence, for no value of c the pair of equations will have infinitely many solutions.

**Question 9:
**One equation of a pair of dependent linear equations is – 5x+ 7y – 2 = 0. The second equation can be

(a) 10x + 14y + 4=0 (b)-10x-14y + 4 =0

(c) -10x + 14y + 4 = 0 (d) 10x-14y + 4=0

**Solution:**

**(d)**Condition for dependent linear equations

**Question 10:
**A pair of linear equations which has a unique solution x – 2 and y = – 3 is

(a) x + y = 1 and 2x – 3y = – 5

(b) 2x+ 5y= -11 and 4x + 10y = -22

(c) 2x – y = 1 and 3x + 2y = 0

(d) x – 4y -14 = 0 and 5x – y -13 = 0

**Solution:**

**(b)**If x = 2, y = – 3 is a unique solution of any pair of equation, then these values must satisfy that pair of equations.

From option (b), LHS = 2x + 5y = 2(2) + 5(- 3) = 4 – 15 = – 11 = RHS

and LHS = 4x + 10y = 4(2) + 10(- 3)= 8 – 30 = – 22 = RHS

**Question 11:
**If x = a and y = b is the solution of the equations x- y = 2 and x + y = 4, then the values of a and b are, respectively

(a) 3 and 5 (b) 5 and 3 (c) 3 and 1 (d) – 1 and – 3

**Solution:**

**(c)**Since, x = a and y = b is the solution of the equations x – y = 2 and x+ y = 4, then these values will satisfy that equations

a-b= 2 ,..(i)

and a + b = 4 … (ii)

On adding Eqs. (i) and (ii), we get

2a = 6

a = 3 and b = 1

**Question 12:
**Aruna has only ₹ 1 and ₹ 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is ₹ 75, then the number of ₹ 1 and ₹ 2 coins are, respectively

(a) 35 and 15 (b) 35 and 20 (c) 15 and 35 (d) 25 and 25

**Solution:**

**(d)**Let number of ₹ 1 coins = x

and number of ₹ 2 coins = y

Now, by given conditions x+y=50 …(i)

Also, x×1+y×2=75

⇒ x + 2y = 75 …(ii)

On subtracting Eq. (i) from Eq. (ii), we get

(x + 2y) – (x + y) = 75 – 50

⇒ y = 25

When y = 25, then x = 25

**Question 13:
**The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages (in year) of the son and the father are, respectively

(a) 4 and 24 (b) 5 and 30

(c) 6 and 36 (d) 3 and 24

**Solution:**

(c)

Let x yr be the present age of father and y yr be the present age of son.

Four years hence, it has relation by given condition,

(x + 4) = 4(y + 4)

⇒ x-4y = 12 …(i)

and x = 6y …(ii)

On putting the value of x from Eq. (ii) in Eq. (i), we get

6y-4y=12

⇒ 2y = 12

⇒ y = 6

When y = 6, then x = 36

Hence, present age of father is 36 yr and age of son is 6 yr.

### Quadratic Equations Class 10 Maths Multiple Choice Questions with Answers

**Question 1:
**Which of the following is a quadratic equation?

(a) x

^{2}+ 2x + 1 = (4 – x)

^{2}+ 3

(b) – 2x

^{2}= (5 – x) (2x –

(c) (k + 1)x

^{2}+ – x = 7, where k = -1

(d) x

^{3}– x

^{2}=(x -1)

^{3 Solution: }

**Question 2:
**Which of the following is not a quadratic equation?

(a) 2 (x -1)

^{2}= 4x

^{2}– 2x +1

(b) 2x – x

^{2}= x

^{2}+ 5

(c) (-√2X +√3)

^{2}= 3x

^{2}– 5x

(d) (x

^{2}+ 2x)

^{2}= x

^{4}+ 3 + 4x

^{2 Solution: }

**Question 3:
**Which of the following equations has 2 as a root?

(a) x

^{2}-4x + 5=0

(b) x

^{2}+ 3x-12 =0

(c) 2x

^{2}– 7x + 6 = 0

(d) 3x

^{2}– 6x – 2 = 0

**Solution:**

**Question 4:
**If is a root of the equation x

^{2}+ kx – = 0, then the value of k is

(a) 2 (b) -2 (c) (d)

**Solution:**

**Question 5:
**Which of the following equations has the sum of its roots as 3?

(a) 2x

^{2}– 3x + 6 = 0

(b) -x

^{2}+ 3x – 3 = 0

(c) √2x

^{2}– x + 1 = 0

(d) 3x

^{2}– 3x + 3 = 0

**Solution:**

**Question 6:
**Value(s) of k for which the quadratic equation 2x

^{2}-kx + k = 0 has equal roots is/are

(a) 0 (b) 4 (c) 8 (d) 0, 8

**Solution:**

(d)

Given equation is 2x

^{2}– kx + k- 0

On comparing with ax

^{2}+ bx + c = 0, we get

a = 2, b= – k and c = k

For equal roots, the discriminant must be zero.

Hence, the required values of k are 0 and 8.

**Question 7:
**Which constant must be added and subtracted to solve the quadratic

equation 9x

^{2}+ x – √2= 0 by the method of completing the square?

(a) (b) (c) (d)

**Solution:**

**Question 8:
**The quadratic equation 2x

^{2}– √5x + 1 = 0 has

(a) two distinct real roots

(b) two equal real roots

(c) no real roots

(d) more than 2 real roots

**Solution:**

**Question 9:
**Which of the following equations has two distinct real roots?

(a)2x

^{2}-3√2x + =0

(b) x

^{2}+ x – 5 =0

(c) x

^{2}+ 3x + 2√2 =0

(d)5x

^{2}-3x + 1=0

**Solution:**

**Question 10:
**Which of the following equations has no real roots?

(a) x

^{2}– 4x + 3√2 =0

(b)x

^{2}+4x-3√2=0

(c) x

^{2}– 4x – 3√2 = 0

(d) 3x

^{2}+ 4√3x + 4=0

**Solution:**

**Question 11:
**(x

^{2}+1)

^{2}– x

^{2}= 0 has

(a) four real roots

(b) two real roots

(c) no real roots

(d) one real root

**Solution:**

### Arithemetic Progressions Class 10 Maths Multiple Choice Questions with Answers

**Question 1:
**In an AP, if d = – 4, n = 7 and a

_{n}= 4, then a is equal to

(a) 6 (b) 7 (c) 20 (d) 28

**Solution:**

**Question 2:
**In an AP, if a = 3.5, d = 0 and n = 101, then a

_{n}will be

(a) 0 (b) 3.5 (c) 103.5 (d) 104.5

**Solution:**

(b) For an AP a

_{n}= a + (n – 1)d= 3.5+ (101 – 1 )x 0 [by given conditions]

∴ = 3.5

**Question 3:
**The list of numbers – 10, – 6, – 2, 2,… is

(a) an AP with d = -16

(b) an AP with d = 4

(c) Fan AP with d = – 4

(d) not an AP

**Solution:**

**(b)**The given numbers are -10,-6,-2, 2…………………..

Here, a., = -10, a

_{2}= -6, a

_{3}= -2 and a

_{4}= 2………

Each successive term of given list has same difference i.e., 4.

So, the given list forms an AP with common difference, d=4.

**Question 4:
**The 11th term of an AP – 5,, 0,…

(a)-20 (b) 20 (c) -30 (d) 30

**Solution:**

Given AP,- 5,, 0,

**Question 5:
**The first four terms of an AP whose first term is – 2 and the common difference is-2 are

(a) -2,0,2, 4

(b) -2, 4, -8,16

(c) -2,-4,-6,-8

(d) -2, – 4, -8, -16

**Solution:**

(c) Let the first four terms of an AP are a, a + d, a + 2d and a + 3d.

Given, that first term, a = – 2 and common difference, d = – 2, then we have an AP as follows

-2, – 2 – 2, – 2 + 2(-2), – 2 + 3(-2)

= – 2, – 4, – 6, – 8

**Question 6:
**The 21st term of an AP whose first two terms are – 3 and 4, is

(a) 17 (b) 137 (c) 143 (d)-143

**Solution:**

(b) Given, first two terms of an AP are a = – 3 and a + d = 4.

⇒ – 3 + d = 4

**Question 7:
**If the 2nd term of an AP is 13 and 5th term is 25, what is its 7th term?

(a) 30 (b) 33 (c) 37 (d) 38

**Solution:**

(b) Given, a

_{2}= 13 and a

_{5}= 25

**Question 8:
**Which term of an AP : 21, 42, 63, 84,… is 210?

(a) 9th (b) 10th (c)11th (d) 12th

**Solution:**

(b) Let nth term of the given AP be 210

Hence, the 10th term of an AP is 210.

**Question 9:
**If the common difference of an AP is 5, then what is a

_{18}– a

_{13}?

(a) 5 (b) 20 (c) 25 (d) 30

**Solution:**

(c) Given, the common difference of AP i.e., d = 5

**Question 10:
**What is the common difference of an AP in which a

_{18}– a

_{14}= 32?

(a) 8 (b) -8 (c) – 4 (d) 4

**Solution:**

**Question 11:
**Two APs have the same common difference. The first term of one of these is – 1 and that of the other is – 8. The difference between their 4th terms is

(a) -1 (b) -8 (c) 7 (d) -9

**Solution:**

**(c)**Let the common difference of two APs are d

_{1}and d

_{2}, respectively.

Bycondition, d

_{1}=d

_{2}=d …(i)

Let the first term of first AP (a

_{1}) = -1

and the first term of second AP (a

_{2}) = – 8

We know that, the nth term of an AP, T

_{1}= a + (n – 1) d

∴4th term of first AP, T

_{4}= a, + (4 – 1)d = – 1 + 3d .

and 4th term of second AP, T’

_{4}= a

_{2}+ (4 – 1)d = – 8 + 3d

Now, the difference between their 4th terms is i.e.,

|T4 -T’4|= (-1 + 3d)-(-8+3d)

= -1+ 3d + 8 – 3d = 7

Hence, the required difference is 7.

**Question 12:
**If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be

(a) 7 (b) 11 (c) 18 (d) 0

**Solution:**

(d) According to the question,

**Question 13:
**The 4th term from the end of an AP – 11, -8,-5,…, 49 is

(a) 37 (b) 40 (c)43 (d) 58

**Solution:**

**(b)**We know that, the n th term of an AP from the end is

From Eq, (i), a

_{4}= 49 – (4 -1) 3 = 49 – 9 = 40

**Question 14:
**The famous mathematician associated with finding the sum of the first 100 natural

numbers is

(a)’Pythagoras

(b) Newton

(c) Gauss

(d) Euclid

**Solution:**

**(c)**Gauss is the famous mathematician associated with finding the sum of the first 100 natural numbers i,e., 1,2, 3……………..100.

**Question 15:
**If the first term of an AP is – 5 and the common difference is 2, then the

sum of the first 6 terms is

(a) 0 (b) 5 (c) 6 (d) 15

**Solution:**

**Question 16:
**The sum of first 16 terms of the AP 10, 6, 2, … is

(a)-320 (b) 320 (c)-352 (d)-400

**Solution:**

**(a)**Given, AP is 10, 6, 2,…

**Question 17:
**In an AP, if a = 1, an = 20 and Sn = 399, then n is equal to

(a) 19 (b) 21 (c) 38 (d) 42

**Solution:**

**Question 18:
**The sum of first five multiples of 3 is

(a) 45 (b) 55 (c) 65 (d) 75

**Solution:**

**(a)**The first five multiples of 3 are 3, 6, 9,12 and 15.

Here, first term, a = 3, common difference, d = 6-3 = 3 and number of terms, n = 5

### Triangles Class 10 Maths Multiple Choice Questions with Answers

**Question 1:
**In figure, if ∠BAC =90° and AD⊥BC. Then,

(a) BD.CD = BC² (b) AB.AC = BC² (c) BD.CD=AD² (d) AB.AC =AD²

**Solution:**

**(c)**

*In*ΔADB

*and*ΔADC,

**Question 2:
**If the lengths of the diagonals of rhombus are 16 cm and 12 cm. Then, the length of the sides of the rhombus is

(a) 9 cm (b) 10 cm (c) 8 cm (d) 20 cm

**Solution:**

**(b)**We know that, the diagonals of a rhombus are perpendicular bisector of each other.

Given, AC = 16 cm and BD = 12 cm [let]

∴ AO = 8cm, SO = 6cm

and ∠AOB = 90°

In right angled ∠AOB,

**Question 3:
**If ΔABC ~ ΔEDF and ΔABC is not similar to ΔDEF, then which of the following is not true?

(a) BC · EF = AC · FD (b) AB · EF = AC · DE

(c) BC · DE = AB · EF (d) BC · DE = AB · FD

**Solution:**

**Question 4:
**If in two Δ PQR ,,then

(a)Δ PQR~Δ CAB (b) Δ PQR ~ Δ ABC

(c)Δ CBA ~ Δ PQR (d) Δ BCA ~ Δ PQR

**Solution:**

**(a)**Given, in two Δ ABC and Δ PQR,

which shows that sides of one triangle are proportional to the side of the other triangle, then their corresponding angles are also equal, so by SSS similarity, triangles are similar.

i.e., Δ CAB ∼ Δ PQR

**Question 5:
**In figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to

(a) 50° (b) 30° (c) 60° (d) 100°

**Solution:**

**Question 6:
**If in two Δ DEF and Δ PQR,∠D =∠Q and ∠R = ∠E,then which of the following is not true?

**Solution:**

**(b)**Given,in ΔDEF,∠D =∠Q,∠R = ∠E

**Question 7:
**In Δ ABC and ΔDEF, ∠B = ∠E, ∠F = ∠C and AB = 30E Then, the two triangles are

(a) congruent but not similar (b) similar but not congruent

(c) neither congruent nor similar (d) congruent as well as similar

**Solution:**

**(b)**In ΔABC and ΔDEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE

We know that, if in two triangles corresponding two angles are same, then they are similar by AAA similarity criterion. Also, ΔA8C and ΔDEF do not satisfy any rule of congruency, (SAS, ASA, SSS), so both are not congruent.

**Question 8:**

**Solution:**

**Question 9:
**If ΔABC ~ΔDFE, ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and OF = 7.5 cm. Then, which of the following is true?

(a) DE =12 cm, ∠F =50° (b) DE = 12 cm, ∠F =100°

(c) EF = 12 cm, ∠D = 100° (d) EF = 12 cm,∠D =30°

**Solution:
**

**(b)**

*Given, A*ABC ~ ADFE,

*then*∠A = ∠D = 30°, ∠C

*=*∠E =

*50°*

**Question 10:
**If in ΔABC and ΔDEF, , then they will be similar, when

(a) ∠B = ∠E (b) ∠A = ∠D

(c)∠B = ∠D (d) ∠A = ∠F

**Solution:**

**(c)**Given, in ΔABC and ΔEDF,

**Question 11:**

(a) 10 cm (b) 12 cm (c) cm (d) 8 cm

**Solution:
**

**(a)**Given, Δ ABC ~Δ QRP, AB = 18cm and BC = 15cm

We know that, the ratio of area of two similar triangles is equal to the ratio of square of their corresponding sides.

**Question 12:
**If S is a point on side PQ of a Δ PQR such that PS = QS = RS, then

**Solution:**

### Coordinate Geometry Class 10 Maths Multiple Choice Questions with Answers

**Question 1:
**The distance of the point P(2, 3) from the X-axis is

(a) 2 (b) 3 (c) 1 (d) 5

**Solution:**

**(b)**We know that, if (x, y) is any point on the cartesian plane in first quadrant.

Then, x = Perpendicular distance from Y-axis

and y = Perpendicular distance from X-axis

Distance of the point P(2, 3) from the X-axis = Ordinate of a point P(2, 3)= 3.

**Question 2:
**The distance between the points A(0, 6) and 5(0,- 2) is

(a) 6 (b) 8 (c) 4 (d) 2

**Solution:**

**(b)**v Distance between the points (x

_{1}, y

_{1}) and (x

_{2}, y

_{2}),

**Question 3:
**The distance of the point P(- 6, 8) from the origin is

(a) 8 (b) 2√7 (c) 10 (d) 6

**Solution:**

**(c)**∴Distance between the points (x

_{1,}y

_{2})and (x

_{2}, y

_{2})

**Question 4:
**The distance between the points (0, 5) and (- 5, 0) is

(a) 5 (b) 5√2 (c)2√5 (d) 10

**Solution:**

**(b)**∴ Distance between the points (x

_{1},y

_{1}) and (x

_{2}, y

_{2}),

**Question 5:
**If AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0), then the length of its diagonal is

(a) 5 (b) 3 (c) √34 (d) 4

**Solution:**

**(c)**

Now, length of the diagonal AB = Distance between the points A(0, 3) and B(5, 0).

∴ Distance between the points (x,, y,) and (x

_{2}, y

_{2}),

Hence, the required length of its diagonal is √34.

**Question 6:
**The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is

(a) 5 (b) 12 (c)11 (d)7+√5

**Solution:**

**(b)**we Further, adding all the distance of a triangle to get the perimeter of a triangle.We plot the vertices of a triangle i.e., (0, 4), (0,0) and (3,0) on the paper shown as given below

Now,perimeter of ΔAOB=Sum of the length of all its sides = d(AO) + d(OB) + d(AB)

∴ Distance between the points (x

_{1},y

_{1}) and (x

_{2}, y

_{2}),

Hence, the required perimeter of triangle is 12.

**Question 7:
**The area of a triangle with vertices A(3,0), B(7, 0) and C(8, 4) is

(a) 14 (b) 28 (c) 8 (d) 6

**Solution:**

**(c)**Area of Δ ABC whose Vertices A≡(x

_{1},y

_{1}),B≡(x

_{2},y

_{2}) and C≡(x

_{3}, y

_{3}) are given by

Hence, the required area of AABC is 8.

**Question 8:
**The points (- 4, 0), (4, 0) and (0, 3) are the vertices of a

(a) right angled triangle (b) isosceles triangle

(c) equilateral triangle (d) scalene triangle

**Solution:**

(b) Let A(- 4, 0), B(4, 0), C(0, 3) are the given vertices.

Now, distance between A (-4, 0) and B (4, 0),

Hence, ΔABC is an isosceles triangle because an isosceles triangle has two sides equal.

**Question 9:
**The point which divides the line segment joining the points (7, – 6) and (3, 4) in ratio 1: 2 internally lies in the

(a) I quadrant (b) II quadrant (c) III quadrant (d) IV quadrant

**Solution:**

**(d)**If P(x, y) divides the line segment joining A(x

_{1},y

_{2}) and B(x

_{2}, y

_{2}) internally in the ratio

**Question 10:
**The point which lies on the perpendicular bisector of the line segment joining the points A(-2, – 5) and B(2, 5) is

(a) (0,0) (b) (0, 2) (c) (2, 0) (d)(-2,0)

**Solution:**

**(a)**We know that, the perpendicular bisector of the any line segment divides the^jjpe segment into two equal parts i.e., the perpendicular bisector of the line segment always passes through the mid-point of the line segment.

Mid-point of the line segment joining the points A (-2, -5) and S(2, 5)

Hence, (0, 0) is the required point lies on the perpendicular bisector of the lines segment.

**Question 11:
**The fourth vertex D of a parallelogram ABCD whose three vertices are A(- 2, 3), B(6, 7) and C(8, 3) is

(a) (0,1) (b) (0,-1) (c) (-1,0) (d) (1,0)

**Solution:**

**(b)**Let the fourth vertex of parallelogram, D≡(x

_{4 ,}y

_{4}) and L, M be the middle points of AC and BD, respectively,

Since, ABCD is a parallelogram, therefore diagonals AC and BD will bisect each other. Hence, L and M are the same points.

Hence, the fourth vertex of parallelogram is D s (x

_{4}, y

_{4}) s (0,-1).

**Question 12:
**If the point P(2,1) lies on the line segment joining points A(4, 2) and 6(8, 4), then

(a)AP = AB (b) AP = PB (c)PB = AB (d)AP = AB

**Solution:**

**(d)**Given that, the point P(2,1) lies on the line segment joining the points A(4,2) and 8(8, 4), which shows in the figure below:

Hence, required condition is AP =

**Question 13:
**If P(,4) is the mid-point of the line segment joining the points Q(- 6, 5) and fl(- 2, 3), then the value of a is

(a)-4 (b) -12 (c) 12 (d) -6

**Solution:**

**(b)**Given that, P(,4) is the mid-point of the line segment joining the points Q(-6, 5) and

R (-2, 3), which shows in the figure given below

Hence, the required value of a is -12.

**Question 14:
**The perpendicular bisector of the line segment joining the points A(1,5) and 8(4,6) cuts the y-axis at

(a) (0,13) (b) (0,-13) (c) (0,12) (d) (13,0)

**Solution:**

**(a)**Firstly, we plot the points of the line segment on the paper and join them.

We know that, the perpendicular bisector of the line segment AB bisect the segment AB, i.e., perpendicular bisector of line segment AB passes through the mid-point of AB.

Now, we draw a straight line on paper passes through the mid-point P. We see that the perpendicular bisector cuts the Y-axis at the point (0,13).

Hence, the required point is (0,13).

**Alternate Method**

We know that, the equation of line which passes through the points (x

_{1}, y

_{1}) and (x

_{2}, y

_{2}) is

Also, we know that the perpendicular bisector of the line segment is perpendicular on the line segment.

Let slope of line segment is m

_{2}.

From Eq. (iii),

So, the required point is (0,13).

**Question 15:
**The coordinates of the point which is equidistant from the three vertices of the ΔAOB as shown in the figure is

**Solution:**

**(a)**Let the coordinate of the point which is equidistant from the three vertices 0(0, 0), A(0,2y) and B(2x, 0) is P(h,k).

Then, PO = PA = PB

⇒ (PO)² = (PA)²= (PB)

^{2}… (i)

By distance formula,

**Question 16:
**If a circle drawn with origin as the centre passes through (,0), then the point which does not lie in the interior of the circle is

**Solution:**

**Question 17:
**A line intersects the y-axis and X-axis at the points P and Q, respectively. If (2, – 5) is the mid-point of PQ, then the coordinates of P and Q are, respectively ’

(a) (0,-5) and (2, 0) (b) (0, 10) and (- 4, 0)

(c) (0, 4) and (- 10, 0) (d) (0, – 10) and (4, 0)

**Solution:**

**(d)**Let the coordinates of P and 0 (0, y) and (x, 0), respectively.

So, the coordinates of P and Q are (0, -10) and (4, 0).

**Question 18:
**The area of a triangle with vertices (a, b + c) , (b, c + a) and (c, a + b) is

(a) (a + b + c)² (b) 0 (c) (a + b + c) (d) abc

**Solution:**

**(b)**Let the vertices of a triangle are, A ≡ (x

_{1}, y

_{1}) ≡ (a, b + c)

B ≡ (x

_{2}, y

_{2}) ≡ (b,c + a) and C = (x

_{3}, y

_{3}) ≡ (c, a + b)

Hence, the required area of triangle is. 0.

**Question 19:
**If the distance between the points (4, p) and (1, 0) is 5, then the value of pis

(a) 4 only (b) ±4 (c) – 4 only (d) 0

**Solution:**

**(b)**According to the question, the distance between the points (4, p) and (1, 0) = 5

Hence, the required value of p is ± 4,

**Question 20:
**If the points A(1, 2), B(0, 0) and C(a, b) are collinear, then

(a) a = b (b) a = 2b (c) 2a = b (d) a = – b

**Solution:**

**(c)**Let the given points are B = (x

_{1},y

_{1}) = (1,2),

B = (x

_{2},y

_{2}) = (0,0) and C3 = (x

_{3},y

_{3})= (a, b).

Hence, the required relation is 2a = b.

### Introduction to Trigonometry and Its Applications Class 10 Maths Multiple Choice Questions with Answers

**Question 1:
**If cos A =, then the value of tan A is

(a)

(b)

(c)

(d)

**Solution:**

**Question 2:
**If sin A = then the value of cot A is

(a) √3

(b)

(c)

(d) 1

**Solution:**

**Question 3:
**The value of the expression cosec (75° + 0) – sec (15° – 0) – tan (55° + 0) + cot (35° – 0) is

(a) -1 (b) 0 (c)1 (d)

**Solution:**

**(b)**Given, expression = cosec (75° + 0) – sec (15° – 0) – tan (55° + 0) + cot (35° – 0)

=cosec [90° – (15° – 0)] – sec (15° – 0)- tan (55° + 0) + cot (90° – (55° + 0)}

= sec (15° – 0) – sec (15° – 0) – tan (55° + 0) + tan (55° + 0)

[∴ cosec (90° – 0) = sec 0 and cot (90° – 0) = tan 0]

= 0

Hence, the required value of the given expression is 0.

**Question 4:
**If sinθ =, then cosθ is equal to

**Solution:**

**Question 5:
**If cos(α+ β) = 0, then sin (α – β) can be reduced to

(a) cos β (b) cos 2β (c) sin α (d) sin 2α

**Solution:**

**Question 6:
**The value of (tanl° tan2° tan3°… tan89°) is

(a) 0 (b) 1 (c) 2 (d)

**Solution:**

**(b)**tan1°-tan2°-tan3°… tan 89°

= tan1°-tan2°-tan3°… tan44° . tan 45° . tan 46°… tan 87°-tan 88°tan 89°

= tan 1°- tan2 °- tan 3°… tan 44° . (1)- tan (90° – 44°)… tan (90° – 3°)

tan (90° -2°)- tan (90° -1°) (∴ tan 45° = 1)

= tan1°-tan2°-tan3°…. tan44° (1) . cot 44°……. cot3°-cot2°-cot1°

**Question 7:
**If cos 9α = sin α and 9α < 90° ,then the value of tan 5α is

(a)

(b)

(c) 1

(d) 0

**Solution:**

**Question 8:
**If ΔABC is right angled at C, then the value of cos (A + B) is

(a) 0

(b) 1

(c)

(d)

**Solution:**

**Question 9:
**If sin A + sin

^{2}A = 1, then the value of (cos

^{2}A + cos

^{4}A) is

(a) 1 (b) (c) 2 (d) 3

**Solution:**

**Question 10:
**If sinα = and cosβ = then the value of (α + β) is

(a) 0° (b) 30° (c) 60° (d) 90°

**Solution:**

**Question 11:
**The value of the expression

(a) 3 (b) 2 (c) 1 (d) 0

**Solution:**

**Question 12:**

(a)

(b)

(c)

(d)

**Solution:**

**Question 13:
**If sinθ – cosθ= 0, then the value of (sin

^{4}θ + cos

^{4}θ) is

(a) 1

(b)

(c)

(d)

**Solution:**

**Question 14:
**sin (45° + θ) – cos (45° – θ) is equal to

(a) 2 cosθ (b) 0 (c) 2 sinθ (d) 1

**Solution:**

**(b)**sin(45° + θ) – cos(45° – θ) = cos[90°- (45° + θ)] – cos(45°- 6) [∴ cos(90° – θ) = sin0]

= cos (45° – 0) – cos (45° – 0)

= 0

**Question 15:
**If a pole 6 m high casts a shadow 2√3 m long on the ground, then the Sun’s elevation is

(a) 60° (b) 45° (c) 30° (d) 90°

**Solution:**

### Circles Class 10 Maths Multiple Choice Questions with Answers

**Question 1:
**If radii of two concentric circles are 4 cm and 5 cm, then length of each chord of one circle which is tangent to the other circle, is

(a) 3 cm (b) 6 cm (c) 9 cm (d) 1 cm

**Solution:**

**(b)**Let 0 be the centre of two concentric circles C

_{1}and C

_{2}, whose radii are r

_{1}= 4 cm and r

_{2}= 5 cm. Now, we draw a chord AC of circle C

_{2}, which touches the circle C

_{1}at B.

Also, join OB, which is perpendicular to AC. [Tangent at any point of circle is perpendicular to radius throughly the point of contact]

**Question 2:
**In figure, if ∠AOB = 125°, then ∠COD is equal to

(a) 62.5° (b) 45° (c) 35° (d) 55°

**Solution:**

**(d)**We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

**Question 3:
**In figure, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to

(a) 45° (b) 60° (c) 50° (d) 55°

**Solution:**

**(c)**In figure, AOC is a diameter of the circle. We know that, diameter subtends an angle 90° at the circle.

**Question 4:
**From a point P which is at a distance of 13 cm from the centre 0 of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle is drawn. Then, the area of the quadrilateral PQOR is

(a) 60 cm

^{2}(b) 65 cm

^{2}(c) 30 cm

^{2}(d) 32.5 cm

^{2 }

**Solution:**

**(a)**Firstly, draw a circle of radius 5 cm having centre O. P is a point at a distance of 13 cm from O. A pair of tangents PQ and PR are drawn.

Thus, quadrilateral POOR is formed.

**Question 5:
**At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A, is

(a) 4 cm (b) 5 cm

(c) 6 cm (d) 8 cm

**Solution:**

**(d)**First, draw a circle of radius 5 cm having centre 0. A tangent XY is drawn at point A.

**Question 6:
**In figure, AT is a tangent to the circle with centre 0 such that OT = 4 cm and ∠OTA = 30°. Then, AT is equal to

(a) 4 cm (b) 2 cm (c) 2

*√3 cm*(d) 4√3 cm

**Solution:**

**(c)**Join OA

We know that, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

**Question 7:
**In figure, if 0 is the centre of a circle, PQ is a chord and the tangent PR at P , ; makes an angle of 50° with PQ, then ∠POQ is equal to

(a) 100° (b) 80° (c) 90° (d) 75°

**Solution:**

**(a)**Given, ∠QPR = 50°

We know that, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

**Question 8:
**In figure, if PA and PB are tangents to the circle with centre 0 such that ∠APB = 50°, then ∠OAB is equal to

(a) 25° (b) 30° (c) 40° (d) 50°

**Solution:**

**(a)**Given, PA and PB are tangent lines.

**Question 9:
**If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm, then the length of each tangent is

(a) √3 cm (b) 6 cm (c) 3 cm (d) 3 √3 cm

**Solution:**

(d) Let P be an external point and a pair of tangents is drawn from point P and angle between these two tangents is 60°.

Tangent at any point of a circle is perpendicular to the radius through the point of contact.

Hence, the length of each tangent is 3√3 cm.

**Question 10:
**In figure, if PQR is the tangent to a circle at Q whose centre is 0, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to

(a) 20° (b) 40° (c) 35° (d) 45°

**Solution:**

### Constructions Class 10 Maths Multiple Choice Questions with Answers

**Question 1:
**To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn, so that ∠BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is

(a) 8 (b) 10 (c) 11 (d) 12

**Solution:**

**(d)**We know that, to divide a line segment AB in the ratio m: n, first draw a ray AX which makes an acute angle ∠BAX, then marked m + n points at equal distance.

Here, m = 5, n = 7

So, minimum number of these points = m+n = 5 + 7 = 12.

**Question 2:
**To divide a line segment AB in the ratio 4 : 7, a ray AX is drawn first such that ∠BAX is an acute angle and then points A

_{1}A

_{2}, A

_{3},… are located at equal distances on the ray AY and the point B is joined to

(a) A

_{1}

_{2}(b) A

_{11}(c) A

_{12}(d) A

_{9 }

**Solution:**

**(b)**Here, minimum 4+7 = 11 points are located at equal distances on the ray AX, and then B is joined to last point is A

_{11}

**Question 3:
**To divide a line segment AB in the ratio 5 : 6, draw a ray AY such that ∠BAX is an acute angle, then draw a ray BY parallel to AY and the points A

_{1}, A

_{2}, A

_{3},… and B

_{1}, B

_{2}, B

_{3},… are located to equal distances on ray AY and BY, respectively. Then, the points joined are

(a) A

_{5}and A

_{6}(b) A

_{6}and B

_{5}(c) A

_{4}and B

_{5}(d) A

_{5}and B

_{4 }

**Solution:**

**(a)**Given a line segment AB and we have to divide it in the ratio 5:6.

**Steps of construction**

- Draw a ray AX making an acute ∠BAX.
- Draw a ray BY parallel to AX by making ∠ABY equal to ∠BAX.
- Now, locate the points A
_{1}, A_{2}, A_{3}, A_{4}and A_{5}(m= 5) on AX and B_{1}, B_{2}, B_{3}, B_{4}, B_{5}and B_{6}(n = 6) such that all the points are at equal distance from each other. - Join B
_{6}A_{5}. Let it intersect AB at a point C.

Then, AC:BC = 5:6

**Question 4:
**To construct a triangle similar to a given ΔABC with its sides of the corresponding sides of ΔABC, first draw a ray BX such that ∠CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then, locate points B

_{1}, B

_{2}, B

_{3},… on BX at equal distances and next step is to join

(a) B

_{10}to C (b) B

_{13 }to C (c) B

_{7}to C (d)B

_{4 }to C

**Solution:**

**(c)**Here, we locate points B

_{1}, B

_{2}, B

_{3}, B

_{4}, B

_{5}, B

_{6}and B

_{7}on BX at equal distance and in next step join the last points is B

_{7}to C.

**Question 5:
**To construct a triangle similar to a given ΔABC with its sides of the corresponding sides of ΔABC draw a ray BX such that ∠ CBX is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is

(a) 5 (b) 8 (c)13 (d) 3

**Solution:**

**(b)**To construct a triangle similar to a given triangle, with its sides of the corresponding sides of given triangle the minimum number of points to be located at equal distance is equal to the greater of m and n is

Hence, =

So, the minimum number of point to be located at equal distance on ray BX is 8.

**Question 6:
**To draw a pair of tangents to a circle which are inclined to each other at an angle of 60°, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be

(a) 135° (b) 90° (c) 60° (d) 120°

**Solution:**

**(d)**The angle between them should be 120° because in that case the figure formed by the intersection point of pair of tangent, the two end points of those-two radii tangents are drawn) and the centre of the circle is a quadrilateral.

From figure it is quadrilateral,

∠POQ + ∠PRQ = 180° [∴ sum of opposite angles are 180°]

60°+ θ = 180°

θ=120

Hence, the required angle between them is 120°.

### Areas Related to Circles Class 10 Maths Multiple Choice Questions with Answers

**Question 1:
**If the sum of the areas of two circles with radii R

_{1}and R

_{2}is equal to the area of a circle of radius R, then

**Solution:**

**(b)**According to the given condition,

^{ }Area of circle =Area of first circle + Area of second circle

**Question 2:
**If the sum of the circumferences of two circles with radii R

_{1}and R

_{2}is equal to the circumference of a circle of radius R, then

(a) R

_{1}+ R

_{2}=R

(b) R

_{1}+ R

_{2}> R

(c) R

_{1}+ R

_{2}< R

(d) Nothing definite can be said about the relation among R

_{1,}R

_{2}and R

**Solution:**

**(a)**According to the given condition,

Circumference of circle = Circumference of first circle + Circumference of second circle

**Question 3:
**If the circumference of a circle and the perimeter of a square are equal, then

(a) Area of the circle = Area of the square

(b) Area of the circle > Area of the square

(c) Area of the circle < Area of the square

(d) Nothing definite can be said about the relation between the areas of the circle and square

**Solution:**

**(b)**According to the given condition,

Circumference of a circle = Perimeter of square

Hence, Area of the circle > Area of the square.

**Question 4:
**Area of the largest triangle that can be inscribed in a semi-circle of radius r units is

(a) r

^{2 }squnits (b) r

^{2}sq units (c) 2r

^{2}sq units (d) √2 r

^{2}sq units

**Solution:**

**(a)**Take a point C on the circumference of the semi-circle and join it by the end points of diameter A and B.

**Question 5:
**If the perimeter of a circle is equal to that of a square, then the ratio of their areas is

(a) 22 :7 (b) 14:11 (c) 7:22 (d) 11:14

**Solution:**

**(b)**Let radius of circle be r and side of a square be a.

According to the given condition,

**Question 6:
**It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be

(a) 10 m (b)15m (c) 20 m (d) 24 m

**Solution:**

**(a)**Area of first circular park, whose diameter is 16 m

**Question 7:
**The area of the circle that can be inscribed in a square of side 6 cm is

(a) 36π cm

^{2}(b) 18π cm

^{2}(c) 12π cm

^{2}(d) 9π cm

^{2 }

**Solution:**

**Question 8:**

The area of the square that can be inscribed in a circle of radius 8 cm is

(a) 256 cm^{2} (b) 128 cm^{2} (c)64√2 cm^{2} (d)64 cm^{2
}**Solution:
**

**(b)**Given, radius of circle, r = OC = 8cm.

∴ Diameter of the circle = AC = 2 x OC = 2 x 8= 16 cm

which is equal to the diagonal of a square.

Let side of square be x.

**Question 9:
**The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm is

(a) 56 cm (b) 42 cm (c) 28 cm (d) 16 cm

**Solution:**

**(c)**∵ Circumference of first circle = 2 πr = πd

_{1}= 36 π cm [given, d

_{1}= 36 cm]

and circumference of second circle = πd

_{2}= 20 π cm [given, d

_{2}= 20 cm]

According to the given condition,

Circumference of circle = Circumference of first circle + Circumference of second circle

**Question 10:
**The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is

(a) 31 cm (b) 25 cm (c) 62 cm (d) 50 cm

**Solution:**

**(d)**Let r

_{1}= 24 cm and r

_{2}= 7 cm

### Surface Areas and Volumes Class 10 Maths Multiple Choice Questions with Answers

**Question 1:
**A cylindrical pencil sharpened at one edge is the combination of

(a) a cone and a cylinder

(b) frustum of a cone and a cylinder ‘

(c) a hemisphere and a cylinder

(d) two cylinders

**Solution:**

**(a)**Because the shape of sharpened pencil is

**Question 2:
**A surahi is the combination of

(a) a sphere and a cylinder (b) a hemisphere and a cylinder

(c) two hemispheres (d) a cylinder and a cone

**Solution:**

**(a)**Because the shape of surahi is

**Question 3:
**A plumbline (sahul) is the combination of (see figure)

(a) a cone and a cylinder

(b) a hemisphere and a cone

(c) frustum of a cone and a cylinder

(d) sphere and cylinder

**Solution:**

**(b)**

**Question 4:
**The shape of a glass (tumbler) (see figure) is usually in the form of

(a) a cone (b) frustum of a cone

(c) a cylinder (d) a sphere

**Solution:**

**(b)**We know that, the shape of frustum of a cone is

So, the given figure is usually in the form of frustum of a cone.

**Question 5:
**The shape of a gilli, in the gilli-danda game (see figure) is a combination of

(a) two cylinders (b) a cone and a cylinder

(c) two cones and a cylinder (d) two cylinders and a cone

**Solution:**

**(c)**

**Question 6:
**A shuttle cock used for playing badminton has the shape of the combination of

(a) a cylinder and a sphere (b) a cylinder and a hemisphere

(c) a sphere and a cone (d) frustum of a cone and a hemisphere

**Solution:**

(d) Because the shape of the shuttle cock is equal to sum of frustum of a cone and hemisphere.

**Question 7:
**A cone is cut through a plane parallel to its base and then the cone that is formed on one side of that plane is removed. The new part that is left over on the other side of the plane is called

(a) a frustum of a cone (b) cone (c) cylinder (d) sphere

**Solution:**

**Question 8:
**If a hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that – space of the cube remains unfilled. Then, the number of marbles that the cube can accomodate is

(a) 142244 (b) 142344 (c) 142444 (d) 142544

**Solution:**

**(a)**Given, edge of the cube = 22 cm

**Question 9:
**A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively is melted and recast into the form a cone of base diameter 8 cm. The height of the cone is

(a) 12 cm (b) 14 cm (c) 15 cm (d) 18 cm

**Solution:**

**(b)**Given, internal diameter of spherical shell = 4 cm

and external diameter of shell = 8 cm

Hence, the height of the cone is 14 cm.

**Question 10:
**If a solid piece of iron in the form of a cuboid of dimensions 49 cm x 33 cm x 24 cm, is moulded to form a solid sphere. Then, radius of the sphere is

(a) 21 cm (b) 23 cm (c) 25 cm (d)19cm

**Solution:**

**(a)**Given, dimensions of the cuboid = 49 cm x 33 cm x 24 cm

∴ Volume of the cuboid = 49 x 33 x 24 = 38808 cm

^{3}

Hence, the radius of the sphere is 21 cm.

**Question 11:
**A mason constructs a wall of dimensions 270 cmx 300 cm x 350 cm with the bricks each of size 22.5 cm x 11.25 cmx 8.75 cm and it is assumed that space is covered by the mortar. Then, the number of bricks used to construct the wall is

(a) 11100 (b) 11200 (c) 11000 (d) 11300

**Solution:**

**Question 12:
**Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is

(a) 4 cm (b) 3 cm (c) 2 cm (d) 6 cm

**Solution:**

**(c)**Given, diameter of the cylinder = 2 cm

∴ Radius = 1 cm and height of the cylinder = 16 cm [∵ diameter = 2 x radius]

∴ Volume of the cylinder = π x (1)

^{2}x 16 = 16 π cm

^{3}

∴ Diameter of each sphere, d=2r = 2×1=2 cm

Hence, the required diameter of each sphere is 2 cm.

**Question 13:
**The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm, respectively. The curved surface area of the bucket is

(a) 4950 cm

^{2}(b) 4951 cm

^{2}(c) 4952 cm

^{2}(d) 4953 cm

^{2 }

**Solution:**

**(a)**Given, the radius of the top of the bucket, R = 28 cm

and the radius of the bottom of the bucket, r = 7 cm

Slant height of the bucket, l= 45 cm

Since, bucket is in the form of frustum of a cone.

∴ Curved surface area of the bucket = π l (R + r) = π x 45 (28 + 7)

**Question 14:
**A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. The capacity of the capsule is

(a) 0.36 cm

^{3}(b) 0.35 cm

^{3}(c) 0.34 cm

^{3}(d) 0.33 cm

^{3 }

**Solution:**

**(a)**Given, diameter of cylinder = Diameter of hemisphere = 0.5 cm

Hence, the capacity of capsule is 0.36 cm

^{3}

**Question 15:
**If two solid hemispheres of same base radius r are joined together along their bases, then curved surface area of this new solid is

(a) 47πr

^{2}(b) 6πr

^{2}(c) 3πr

^{2}(d) 8πr

^{2 }

**Solution:**

**(a)**Because curved surface area of a hemisphere is 2 w

^{2}and here, we join two solid hemispheres along their bases of radius r, from which we get a solid sphere.

Hence, the curved surface area of new solid = 2 πr

^{2}+ 2 πr

^{2}= 4πr

^{2}

**Question 16:
**A right circular cylinder of radius r cm and height h cm (where, h>2r) just encloses a sphere of diameter

(a) r cm (b) 2r cm (c) h cm (d) 2h cm

**Solution:**

**(b)**Because the sphere encloses in the cylinder, therefore the diameter of sphere is equal to diameter of cylinder which is 2r cm.

**Question 17:
**During conversion of a solid from one shape to another, the volume of the new shape will

(a) increase (b) decrease

(c) remain unaltered (d) be doubled

**Solution:**

**(c) During conversion of a solid from one shape to another, the volume of the new shape will remain unaltered.**

**Question 18:
**The diameters of the two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. The capacity of the bucket is

(a) 32.7 L (b) 33.7 L (c) 34.7 L (d) 31.7 L

**Solution:**

**(a)**Given, diameter of one end of the bucket

Hence, the capacity of bucket is 32.7 L.

**Question 19:
**In a right circular cone, the cross-section made by a plane parallel to the base is a

(a) circle (b) frustum of a cone (c) sphere (d) hemisphere

**Solution:**

**(b)**We know that, if a cone is cut by a plane parallel to the base of the cone, then the portion between the plane and base is called the frustum of the cone.

**Question 20:
**If volumes of two spheres are in the ratio 64 : 27, then the ratio of their surface areas is

(a) 3: 4 (b) 4 : 3 (c) 9 : 16 (d) 16 : 9

Solution:

Solution:

**(d)**Let the radii of the two spheres are r

_{1}and r

_{2}, respectively.

Hence, the required ratio of their surface area is 16 : 9.

### Statistics and Probability Class 10 Maths Multiple Choice Questions with Answers

**Question 1:**

(a) lower limits of the classes (b) upper limits of the classes

(c) mid-points of the classes (d) frequencies of the class marks

**Solution:
**

**(c)**We know that, = – a

i.e., ‘s are the deviation from a of mid-points of the classes.

**Question 2:
**While computing mean of grouped data, we assume that the frequencies are

(a) evenly distributed over all the classes

(b) centred at the class marks of the classes

(c) centred at the upper limits of the classes

(d) centred at the lower limits of the classes

**Solution:**

**(b)**In computing the mean of grouped data, the frequencies are centred at the class marks of the classes.

**Question 3:
**

(a) 0 (b) -1 (c) 1 (d) 2

**Solution:**

**Question 4:**

**Solution:**

**Question 5:
**The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its

(a) mean (b) median (c) mode (d) All of these

**Solution:**

(b) Since, the intersection point of less than ogive and more than ogive gives the median on the abscissa.

**Question 6:
**For the following distribution,

the sum of lower limits of the median class and modal class is

(a) 15 (b) 25 (c) 30 (d) 35

**Solution:**

Now,= 33, which lies in the interval 10-15. Therefore, lower limit of the median class is 10. . ,

The highest frequency is 20, which lies in the interval 15-20. Therefore, lower limit of modal class is 15. Hence, required sum is 10 + 15 = 25.

**Question 7:
**Consider the following frequency distribution

The upper limit of the median class is

(a) 7 (b) 17.5 (c) 18 (d) 18.5

**Solution:**

**Question 8:
**For the following distribution,

the modal class is

(a) 10-20 (b) 20-30 (c) 30-40 (d) 30-40

**Solution:**

Here,we see that the highest frequency is 30. which lies in the interval 30-40.

**Question 9:
**consider the data.

The difference of the upper limit of the median class and the lower limit of the modal class is

(a) 0 (b) 19 (c) 20 (d) 38

**Solution:**

Here, = 33.5 which lies in the interval 125 -145.

Hence, upper limit of median class is 145.

Here, we see that the highest frequency is 20 which lies in 125-145. Hence, the lower limit of modal class is 125.

Required difference = Upper limit of median class – Lower limit of modal class

= 145-125 = 20

**Question 10:
**The times (in seconds) taken by 150 atheletes to run a 110 m hurdle race are tabulated below

The number of atheletes who completed the race in less than 14.6 s is

(a) 11 (b) 71 (c) 82 (d) 130

**Solution:**

**(c)**The number of atheletes who completed the race in less than 14.6

= 2 + 4+ 5+71 =82

**Question 11:
**Consider the following distribution

the frequency of the class 30-40 is

(a) 3 (b) 4 (c) 3 (d) 4

**Solution:**

Hence,frequency in the class interval 30-40 is 3

**Question 12:
**If an event cannot occur, then its probability is

(a) 1 (b) (c) (d) 0

Solution:

Solution:

**(d)**The event which cannot occur is said to be impossible event and probability of impossible event is zero.

**Question 13:
**Which of the following cannot be the probability of an event?

(a) (b) 0.1 (c) 3 (d)

**Solution:**

**(d)**Since, probability of an event always lies between 0 and 1.

**Question 14:
**An event is very unlikely to happen. Its probability is closest to

(a) 0.0001 (b) 0.001 (c) 0.01 (d) 0.1

**Solution:**

**(a)**The probability of an event which is very unlikely to happen is closest to zero and from the given options 0.0001 is closest to zero.

**Question 15:
**If the probability of an event is P, then the probability of its completmentry event will be

(a) P -1 (b) P (c) 1 – P (d) 1 —

**Solution:**

**(c)**Since, probability of an event + probability of its complementry event = 1

So, probability of its complementry event = 1 – Probability of an event = 1 – P

**Question 16:
**The probability expressed as a percentage of a particular occurrence can never be

(a) less than 100 (b) less than 0

(c) greater than 1 (d) anything but a whole number

**Solution:**

**(b)**We know that, the probability expressed as a percentage always lie between 0 and 100. So, it cannot be less than 0.

**Question 17:
**If P (A) denotes the probability of an event A, then

(a) P(A) < 0 (b) P(A) > 1 (c) 0 ≤ P(A) ≤ 1 (d) -1 ≤ P(A) ≤ 1

**Solution:**

**(c)**Since, probability of an event always lies between 0 and 1.

**Question 18:
**If a card is selected from a deck of 52 cards, then the probability of its being a red face card is

(a) (b) (c) (d)

**Solution:**

(c) In a deck of 52 cards, there are 12 face cards i.e.,6 red and 6 black cards.

So, probability of getting a red face card =

**Question 19:
**The probability that a non-leap your selected at random will contains 53 Sunday is ‘

(a) (b) (c) (d)

**Solution:**

**(a)**A non-leap year has 365 days and therefore 52 weeks and 1 day. This 1 day may be Sunday or Monday or Tuesday or Wednesday or Thursday or Friday or Saturday. Thus, out of 7 possibilities, 1 favourable event is the event that the one day is Sunday.

∴ Required probability =

**Question 20:
**When a die is thrown, the probability of getting an odd number less than 3 is ,

(a) (b) (c) (d) 0

Solution:

Solution:

**(a)**When a die-is thrown,then total number of outcomes = 6 Odd number less than 3 is 1 only.

Number of possible outcomes = 1

Required probability =

**Question 21:
**A card is drawn from a deck of 52 cards. The event E is that card is not an ace of hearts. The number of outcomes favourable to E is

(a) 4 (b) 13 (c) 48 (d) 51

**Solution:**

**(d)**In a deck of 52 cards, there are 13 cards of heart and 1 is ace of heart.

Hence, the number of outcomes favourable to E = 51

**Question 22:
**The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot is

(a) 7 (b) 14 (c) 21 (d) 28

**Solution:**

**(b)**Here, total number of eggs = 400

Probability of getting a bad egg = 0.035

**Question 23:**

A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, then how many tickets has she bought?

(a) 40 (b) 240 (c) 480 (d) 750

**Solution:**

(c) Given, total number of sold tickets = 6000

Let she bought x tickets.

Hence, she bought 480 tickets.

**Question 24:
**One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is

(a) (b) (c) (d)

**Solution:**

**(a)**Number of total outcomes = 40

Multiples of 5 between 1 to 40 = 5,10,15,20,25. 30 35, 40

**Question 25:
**Someone is asked to take a number from 1 to 100. The probability that it is a prime,is

(a) (b) (c) (d)

**Solution:**

**(c)**Total numbers of outcomes = 100

So, the prime numbers between 1 to 100 are 2, 3, 5, 7,11,13,17,19, 23, 29, 31,37, 41. 43, 47, 53, 56, 61, 67, 71, 73, 79, 83, 89 and 97.

**Question 26:
**A school has five houses A, B, C, D and E. A class has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D and rest from house E. A single student is selected at random to be the class monitor. The probability that the selected student is not from A, B and C is

(a) (b) (c) (d)

**Solution:**

**(b)**Total number of students = 23

Number of students in house A, B and C = 4+ 8 + 5 = 17

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Subject – Science |

Class 10 | NCERT Solutions for Class 10 Maths | NCERT Solutions for Class 10 Science |