CBSE Maths Multiple Choice Questions with Answers for Class 10

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CBSE Maths Multiple Choice Questions with Answers for Class 10

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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.
NCERT Exemplar Problems Class 10 Maths - Real Numbers-1.1-1s

Question 2:
For some integer q, every odd integer is of the form
(a) q                            (b) q +1                   (c) 2g                        (d) 2q +1
Solution:
NCERT Exemplar Problems Class 10 Maths - Real Numbers-1.1-2

Question 3:
NCERT Exemplar Problems Class 10 Maths - Real Numbers-1.1-3
Solution:
NCERT Exemplar Problems Class 10 Maths - Real Numbers-1.1-3S

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:
NCERT Exemplar Problems Class 10 Maths - Real Numbers-1.1-4S

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]
NCERT Exemplar Problems Class 10 Maths - Real Numbers-1.1-5

Question 6:
If two positive integers a and b are written as a = x3y2 and b = xy3, wfiere x, y are prime numbers, then HCF (a, b) is
(a) xy                         (b) xy2                      (c)x3y3                          (d) xy2
Solution:
NCERT Exemplar Problems Class 10 Maths - Real Numbers-1.1-6s

Question 7:
If two positive integers p and q can be expressed’ as p=ab2 and q = a3b; where 0, b being prime numbers, then LCM (p, q) is equal to
(a) ab                       (b) a2b2                      (c) a3b2          (d) a3b3
Solution:
NCERT Exemplar Problems Class 10 Maths - Real Numbers-1.1-7s

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:
NCERT Exemplar Problems Class 10 Maths - Real Numbers-1.1-8s

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
NCERT Exemplar Problems Class 10 Maths - Real Numbers-1.1-9s

Question 10:
NCERT Exemplar Problems Class 10 Maths - Real Numbers-1.1-10
(a) one decimal place                               (b) two decimal places
(c) three decimal places                           (d) four decimal places
Solution:
NCERT Exemplar Problems Class 10 Maths - Real Numbers-1.1-10s


Pplynomials Class 10 Maths Multiple Choice Questions with Answers

Question 1:
ncert-exemplar-problems-class-10-maths-polynomials-EX-2.1-1
Solution:
(a) Given that, one of the zeroes of the quadratic polynomial say p(x) = (k- 1)x2 + kx + 1
ncert-exemplar-problems-class-10-maths-polynomials-EX-2.1-1S

Question 2:
A quadratic polynomial, whose zeroes are -3 and 4, is
(a) x2 – x + 12          (b)x2 + x + 12            (c)\frac { { x }^{ 2 } }{ 2 } -\frac { x }{ 2 } -6          (d)2x2 + 2x-24
Solution:
(c) Let ax2 + bx + c be a required polynomial whose zeroes are -3 and 4.
ncert-exemplar-problems-class-10-maths-polynomials-EX-2.1-2s-1
ncert-exemplar-problems-class-10-maths-polynomials-EX-2.1-2s-2
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 xz + (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) ={ x }^{ 2 }+\left( a+1 \right) x+ b
Given that, 2 and -3 are the zeroes of the quadratic polynomial p(x).
ncert-exemplar-problems-class-10-maths-polynomials-EX-2.1-2s-3s-1
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) = ax2 + bx + c be the required polynomial whose zeroes are -2 and 5.
ncert-exemplar-problems-class-10-maths-polynomials-EX-2.1-2s-4s-1
ncert-exemplar-problems-class-10-maths-polynomials-EX-2.1-2s-4s-2
Hence, the required number of polynomials are infinite i.e., more than 3.

Question 5:
If one of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is zero,
the product of then other two zeroes is
(a)\frac { -c }{ a }                (b)\frac { c }{ a }                        (c)0                            (d) \frac { -b }{ a }
Solution:
(b) Let p(x) =ax3 + bx2 + 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,
ncert-exemplar-problems-class-10-maths-polynomials-EX-2.1-5s

Question 6:
If one of the zeroes of the cubic polynomial x3 + ax2 + 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) = x3 + ax2 + 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,
ncert-exemplar-problems-class-10-maths-polynomials-EX-2.1-6s-1
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) = x2 + ax2 + bx + c i.e., (x + 1) is a factor of f{x).
Now, using division algorithm,
ncert-exemplar-problems-class-10-maths-polynomials-EX-2.1-6s-2
⇒x3 + ax2 + bx +c = (x + 1) x {x2 + (a – 1)x + (b – a + 1)> + (c – b + a -1)
⇒x3 + ax2 + bx + (b – a + 1) = (x + 1) {x2 + (a – 1)x + (b -a+ 1}}

Let a and p be the other two zeroes of the given polynomial, then
ncert-exemplar-problems-class-10-maths-polynomials-EX-2.1-6s-3

Question 7:
The zeroes of the quadratic polynomial x2 + 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) =x2 + 99x + 127.
On comparing p(x) with ax2 + bx + c, we get
a = 1, b = 99 and c = 127
ncert-exemplar-problems-class-10-maths-polynomials-EX-2.1-7s-1
Hence, both zeroes of the given quadratic polynomial p(x) are negative.
Alternate Method
ncert-exemplar-problems-class-10-maths-polynomials-EX-2.1-7s-2
the above condition.
So, both zeroes of the given quadratic polynomial are negative.

Question 8:
The zeroes of the quadratic polynomial x2 + 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) = x2 + kx + k, k≠0
On comparing p(x) with ax2 + bx + c, we get
ncert-exemplar-problems-class-10-maths-polynomials-EX-2.1-8s
Here, we see that
k(k − 4)> 0
⇒                       k ∈ (-∞, 0) u (4, ∞)
Now, we know that
In quadratic polynomial ax2 + 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 ax2+ 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 ax2 + bx + c, c ≠ 0 are equal. If coefficient of x2 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.
ncert-exemplar-problems-class-10-maths-polynomials-EX-2.1-9s
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 x2 + 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) = x2 + ax + b.
Put a = 0, then,                            p(x) = x2 + b = 0
⇒                                              x2 = -b
⇒                                              x = ± \pm \sqrt { -b }
[∴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) = x2 + ax+ b
and by given condition the zeroes area and – α.
Sum of the zeroes = α- α = a
=>a = 0
f(x) = x2 + b, which cannot be linear,
and product of zeroes = α .(- α) = b
⇒                         -α2 = 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?
ncert-exemplar-problems-class-10-maths-polynomials-EX-2.1-11
Solution:
(d) For any quadratic polynomial ax2 + bx + c, a≠0, the graph of the Corresponding equation y = ax2 + 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+ \frac { 10 }{ 3 } = 0         [dividing by 3]… (i)
and       2x-y+9=0                                                                …(ii)
Now, table for 2x – y +  \frac { 10 }{ 3 } = 0,
ncert-exemplar-problems-class-10-maths-pair-linear-equations-two-variables-EX-3.1-1s-1
ncert-exemplar-problems-class-10-maths-pair-linear-equations-two-variables-EX-3.1-1s-2
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:
ncert-exemplar-problems-class-10-maths-pair-linear-equations-two-variables-EX-3.1-2s

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:
ncert-exemplar-problems-class-10-maths-pair-linear-equations-two-variables-EX-3.1-3s

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.
ncert-exemplar-problems-class-10-maths-pair-linear-equations-two-variables-EX-3.1-4
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
ncert-exemplar-problems-class-10-maths-pair-linear-equations-two-variables-EX-3.1-5
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) \frac { 1 }{ 2 }                           (b) -\frac { 1 }{ 2 }                   (c) 2                         (d) -2
Solution:
(c) Condition for coincident lines is
ncert-exemplar-problems-class-10-maths-pair-linear-equations-two-variables-EX-3.1-6-1
ncert-exemplar-problems-class-10-maths-pair-linear-equations-two-variables-EX-3.1-6-2

Question 7:
If the lines given by 3x+ 2ky = 2 and 2x + 5y = 1 are parallel, then the value of is
(a) -\frac { 5 }{ 4 }                        (b)  \frac { 2 }{ 5 }
(c) \frac { 15 }{ 4 }                          (d) \frac { 3 }{ 2 }
Solution:
ncert-exemplar-problems-class-10-maths-pair-linear-equations-two-variables-EX-3.1-7

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
ncert-exemplar-problems-class-10-maths-pair-linear-equations-two-variables-EX-3.1-8
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
ncert-exemplar-problems-class-10-maths-pair-linear-equations-two-variables-EX-3.1-9s-1
ncert-exemplar-problems-class-10-maths-pair-linear-equations-two-variables-EX-3.1-9s-2

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) x2 + 2x + 1 = (4 – x)2 + 3
(b) – 2x2 = (5 – x) (2x – \frac { 2 }{ 5 }
(c) (k + 1)x2 + – \frac { 3 }{ 2 } x = 7, where k = -1
(d) x3 – x2 =(x -1)3
Solution:
ncert-exemplar-problems-class-10-maths-quadratic-equations-Ex-4.1-1s-1
ncert-exemplar-problems-class-10-maths-quadratic-equations-Ex-4.1-1s-2

Question 2:
Which of the following is not a quadratic equation?
(a) 2 (x -1)2 = 4x2 – 2x +1
(b) 2x – x2 = x2 + 5
(c) (-√2X +√3)2 = 3x2 – 5x
(d) (x2 + 2x)2 = x4 + 3 + 4x2
Solution:
ncert-exemplar-problems-class-10-maths-quadratic-equations-Ex-4.1-2s

Question 3:
Which of the following equations has 2 as a root?
(a) x2-4x + 5=0
(b) x2 + 3x-12 =0
(c) 2x2 – 7x + 6 = 0
(d) 3x2 – 6x – 2 = 0
Solution:
ncert-exemplar-problems-class-10-maths-quadratic-equations-Ex-4.1-3s

Question 4:
If \frac { 1 }{ 2 }  is a root of the equation x2 + kx –\frac { 5 }{ 4 }  = 0, then the value of k is
(a) 2         (b) -2                          (c) \frac { 1 }{ 4 }                              (d)\frac { 1 }{ 2 }
Solution:
ncert-exemplar-problems-class-10-maths-quadratic-equations-Ex-4.1-4s

Question 5:
Which of the following equations has the sum of its roots as 3?
(a) 2x2 – 3x + 6 = 0
(b) -x2 + 3x – 3 = 0
(c) √2x2 – \frac { 3 }{ \sqrt { 2 }}x + 1 = 0
(d) 3x2 – 3x + 3 = 0
Solution:
ncert-exemplar-problems-class-10-maths-quadratic-equations-Ex-4.1-5s-1
ncert-exemplar-problems-class-10-maths-quadratic-equations-Ex-4.1-5s-2

Question 6:
Value(s) of k for which the quadratic equation 2x2 -kx + k = 0 has equal roots is/are
(a) 0                           (b) 4                            (c) 8                          (d) 0, 8
Solution:
(d)
Given equation is 2x2 – kx + k- 0
On comparing with ax2 + bx + c = 0, we get
a = 2, b= – k and c = k
For equal roots, the discriminant must be zero.
ncert-exemplar-problems-class-10-maths-quadratic-equations-Ex-4.1-6s
Hence, the required values of k are 0 and 8.

Question 7:
Which constant must be added and subtracted to solve the quadratic
equation 9x2 +\frac { 3 }{ 4 }  x – √2= 0 by the method of completing the square?
(a) \frac { 1 }{ 8 }                   (b) \frac { 1 }{ 64 }                            (c) \frac { 1 }{ 4 }                  (d) \frac { 9 }{ 64 }
Solution:
ncert-exemplar-problems-class-10-maths-quadratic-equations-Ex-4.1-7s

Question 8:
The quadratic equation 2x2 – √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:
ncert-exemplar-problems-class-10-maths-quadratic-equations-Ex-4.1-8s

Question 9:
Which of the following equations has two distinct real roots?
(a)2x2-3√2x +\frac { 9 }{ 4 }  =0
(b) x2 + x – 5 =0
(c) x2 + 3x + 2√2 =0
(d)5x2-3x + 1=0
Solution:
ncert-exemplar-problems-class-10-maths-quadratic-equations-Ex-4.1-9s
ncert-exemplar-problems-class-10-maths-quadratic-equations-Ex-4.1-9s-1

Question 10:
Which of the following equations has no real roots?
(a) x2 – 4x + 3√2 =0
(b)x2+4x-3√2=0
(c) x2 – 4x – 3√2 = 0
(d) 3x2 + 4√3x + 4=0
Solution:
ncert-exemplar-problems-class-10-maths-quadratic-equations-Ex-4.1-10s-1
ncert-exemplar-problems-class-10-maths-quadratic-equations-Ex-4.1-10s-2

Question 11:
(x2 +1)2 – x2 = 0 has
(a) four real roots
(b) two real roots
(c) no real roots
(d) one real root
Solution:
ncert-exemplar-problems-class-10-maths-quadratic-equations-Ex-4.1-11s


Arithemetic Progressions Class 10 Maths Multiple Choice Questions with Answers

Question 1:
In an AP, if d = – 4, n = 7 and an = 4, then a is equal to
(a) 6                             (b) 7                        (c) 20                            (d) 28
Solution:
ncert-exemplar-problems-class-10-maths-arithmetic-progressions-Ex-5.1-1s

Question 2:
In an AP, if a = 3.5, d = 0 and n = 101, then an will be
(a) 0                             (b) 3.5                     (c) 103.5                       (d) 104.5
Solution:
(b) For an AP an = 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, a2 = -6, a3 = -2 and a4 = 2………
ncert-exemplar-problems-class-10-maths-arithmetic-progressions-Ex-5.1-3s
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,\frac { -5 }{ 2 } , 0,\frac { 5 }{ 2 }
(a)-20                        (b)  20                        (c) -30                         (d) 30
Solution:
Given AP,- 5,\frac { -5 }{ 2 } , 0,\frac { 5 }{ 2 }
ncert-exemplar-problems-class-10-maths-arithmetic-progressions-Ex-5.1-4s

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, a2 = 13 and a5 = 25
ncert-exemplar-problems-class-10-maths-arithmetic-progressions-Ex-5.1-7

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
ncert-exemplar-problems-class-10-maths-arithmetic-progressions-Ex-5.1-8
Hence, the 10th term of an AP is 210.

Question 9:
If the common difference of an AP is 5, then what is a18 – a13?
(a) 5                            (b) 20                       (c) 25                        (d) 30
Solution:
(c) Given, the common difference of AP i.e., d = 5
ncert-exemplar-problems-class-10-maths-arithmetic-progressions-Ex-5.1-9s

Question 10:
What is the common difference of an AP in which a18 – a14 = 32?
(a) 8                            (b) -8                        (c) – 4                        (d) 4
Solution:
ncert-exemplar-problems-class-10-maths-arithmetic-progressions-Ex-5.1-10s

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 d1 and d2, respectively.
Bycondition,                          d1 =d2 =d                                                                      …(i)
Let the first term of first AP (a1) = -1
and the first term of second AP (a2) = – 8
We know that, the nth term of an AP, T1 = a + (n – 1) d
∴4th term of first AP, T4 = a, + (4 – 1)d = – 1 + 3d .
and 4th term of second AP, T’4 = a2 + (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,
ncert-exemplar-problems-class-10-maths-arithmetic-progressions-Ex-5.1-12s

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
ncert-exemplar-problems-class-10-maths-arithmetic-progressions-Ex-5.1-13s
From Eq, (i), a4 = 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:
ncert-exemplar-problems-class-10-maths-arithmetic-progressions-Ex-5.1-15s

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,…
ncert-exemplar-problems-class-10-maths-arithmetic-progressions-Ex-5.1-16s

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:
ncert-exemplar-problems-class-10-maths-arithmetic-progressions-Ex-5.1-17s

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
ncert-exemplar-problems-class-10-maths-arithmetic-progressions-Ex-5.1-18s


Triangles Class 10 Maths Multiple Choice Questions with Answers

Question 1:
In figure, if ∠BAC =90° and AD⊥BC. Then,
 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-1Q
(a) BD.CD = BC²                         (b) AB.AC = BC²                        (c) BD.CD=AD²                          (d) AB.AC =AD²
Solution:
(c) In ΔADB and ΔADC,
 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-1S

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,
 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-2S

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:
 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-3S

Question 4:
If in two Δ PQR ,\frac { AB }{ QR } =\frac { BC }{ PR } =\frac { CA }{ PQ } ,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,\frac { AB }{ QR } =\frac { BC }{ PR } =\frac { CA }{ PQ }
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
 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-4s

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:

 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-5s

Question 6:
If in two Δ DEF and Δ PQR,∠D =∠Q and ∠R = ∠E,then which of the following is not true?
 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-6Q
Solution:
(b) Given,in ΔDEF,∠D =∠Q,∠R = ∠E
 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-6S

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
 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-7S
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:

 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-8Q
Solution:
 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-8S

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, AABC ~ ADFE, then ∠A = ∠D = 30°, ∠C ∠E = 50°
 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-9S

Question 10:
If in ΔABC and ΔDEF, \frac { AB }{ DE } =\frac { BC }{ FD } , 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,
 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-10S

Question 11:
 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-11Q
(a) 10 cm                  (b) 12 cm                 (c) \frac { 20 }{ 3 } cm                    (d) 8 cm
Solution:
(a) Given, Δ ABC ~Δ QRP, AB = 18cm and BC = 15cm
 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-11s-1
We know that, the ratio of area of two similar triangles is equal to the ratio of square of their corresponding sides.
 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-11s-2

Question 12:
If S is a point on side PQ of a Δ PQR such that PS = QS = RS, then
 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-12Q
Solution:
 Wordncert-exemplar-problems-class-10-maths-triangles-Ex-6.1-12s


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
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-1s
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 (x1, y1) and (x2, y2),
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-2s

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 (x1,y2)and (x2, y2)
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-3s

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 (x1,y1) and (x2, y2),
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-4s

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)
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-5s
Now, length of the diagonal AB = Distance between the points A(0, 3) and B(5, 0).
∴ Distance between the points (x,, y,) and (x2, y2),
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-5s-1
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
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-6s-1
Now,perimeter of ΔAOB=Sum of the length of all its sides = d(AO) + d(OB) + d(AB)
∴ Distance between the points (x1,y1) and (x2, y2),
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-6s-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≡(x1,y1),B≡(x2,y2) and C≡(x3, y3) are given by
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-7s
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),
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-8s
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(x1,y2) and B(x2, y2) internally in the ratio
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-9s

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)
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-10s
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≡(x4 ,y4) and L, M be the middle points of AC and BD, respectively,
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-11s-1
Since, ABCD is a parallelogram, therefore diagonals AC and BD will bisect each other. Hence, L and M are the same points.
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-11s-2
Hence, the fourth vertex of parallelogram is D s (x4, y4) 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 =  \frac { 1 }{ 3 } AB                              (b) AP = PB                            (c)PB =  \frac { 1 }{ 3 } AB                      (d)AP = \frac { 1 }{ 2 } 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:
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-12s
Hence, required condition is AP = \frac { AB }{ 2 }

Question 13:
If P(\frac { 1 }{ 2 } ,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(\frac { 1 }{ 2 } ,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
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-13s-1
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-13s-2
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.
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-14s-1
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.
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-14s-2
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 (x1, y1) and (x2, y2) is
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-14s-3
Also, we know that the perpendicular bisector of the line segment is perpendicular on the line segment.
Let slope of line segment is m2.
From Eq. (iii),
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-14s-4
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
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-15Q
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,
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-15s

Question 16:
If a circle drawn with origin as the centre passes through (\frac { 13 }{ 2 } ,0), then the point which does not lie in the interior of the circle is
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-16Q
Solution:
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-16s-1
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-16s-2

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.

ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-17s
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 ≡ (x1, y1) ≡ (a, b + c)
B ≡ (x2, y2) ≡ (b,c + a) and C = (x3, y3) ≡ (c, a + b)
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-18s
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
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-19s
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 = (x1,y1) =  (1,2),
B = (x2,y2) = (0,0) and C3 = (x3,y3)= (a, b).
ncert-exemplar-problems-class-10-maths-coordinate-geometry-Ex-7.1-20s
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 =\frac { 4 }{ 5 } , then the value of tan A is
(a) \frac {3 }{ 5 }
(b) \frac {3 }{ 4 }
(c) \frac {4 }{ 3 }
(d) \frac {5}{ 3 }
Solution:

ncert-exemplar-problems-class-10-maths-introduction-to-trigonometry-and-its-applications-Ex-8.1-1s

Question 2:
If sin A =\frac {1 }{ 2 }  then the value of cot A is
(a) √3
(b) \frac { 1 }{ \sqrt { 3 }}
(c) \frac {\sqrt { 3 }}{ 1}
(d) 1
Solution:

ncert-exemplar-problems-class-10-maths-introduction-to-trigonometry-and-its-applications-Ex-8.1-2s

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) \frac {3 }{ 2 }
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 =\frac {3 }{ 5 }, then cosθ is equal to
ncert-exemplar-problems-class-10-maths-introduction-to-trigonometry-and-its-applications-Ex-8.1-4Q
Solution:
ncert-exemplar-problems-class-10-maths-introduction-to-trigonometry-and-its-applications-Ex-8.1-4s

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:
ncert-exemplar-problems-class-10-maths-introduction-to-trigonometry-and-its-applications-Ex-8.1-5s

Question 6:
The value of (tanl° tan2° tan3°… tan89°) is
(a) 0                           (b) 1                           (c) 2                            (d) \frac {1 }{ 2 }
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°
ncert-exemplar-problems-class-10-maths-introduction-to-trigonometry-and-its-applications-Ex-8.1-6s

Question 7:
If cos 9α = sin α and 9α < 90° ,then the value of tan 5α is
(a) \frac { 1 }{ \sqrt { 3 }}
(b) \frac {\sqrt { 3 }}{ 1}
(c) 1
(d) 0
Solution:
ncert-exemplar-problems-class-10-maths-introduction-to-trigonometry-and-its-applications-Ex-8.1-7s

Question 8:
If ΔABC is right angled at C, then the value of cos (A + B) is
(a) 0
(b) 1
(c) \frac { 1 }{ 2 }
(d) \frac {\sqrt { 3 }}{ 2}
Solution:
ncert-exemplar-problems-class-10-maths-introduction-to-trigonometry-and-its-applications-Ex-8.1-8s

Question 9:
If sin A + sin2 A = 1, then the value of (cos2 A + cos4 A) is
(a) 1                      (b) \frac { 1 }{ 2 }                         (c) 2                                          (d) 3
Solution:
ncert-exemplar-problems-class-10-maths-introduction-to-trigonometry-and-its-applications-Ex-8.1-9s

Question 10:
If sinα = \frac { 1 }{ 2 }  and cosβ = \frac { 1 }{ 2 }  then the value of (α + β) is
(a) 0°                (b) 30°                        (c) 60°                     (d) 90°
Solution:
ncert-exemplar-problems-class-10-maths-introduction-to-trigonometry-and-its-applications-Ex-8.1-10s

Question 11:
The value of the expression
ncert-exemplar-problems-class-10-maths-introduction-to-trigonometry-and-its-applications-Ex-8.1-11Q
(a) 3                                 (b) 2                                        (c) 1                              (d) 0
Solution:
ncert-exemplar-problems-class-10-maths-introduction-to-trigonometry-and-its-applications-Ex-8.1-11s

Question 12:
ncert-exemplar-problems-class-10-maths-introduction-to-trigonometry-and-its-applications-Ex-8.1-12Q
(a) \frac { 2 }{ 3 }
(b) \frac { 1 }{ 3 }
(c) \frac { 1 }{ 2 }
(d) \frac { 3 }{ 4 }
Solution:
ncert-exemplar-problems-class-10-maths-introduction-to-trigonometry-and-its-applications-Ex-8.1-12s

Question 13:
If sinθ – cosθ= 0, then the value of (sin4 θ + cos4 θ) is
(a) 1
(b) \frac { 3 }{ 4 }
(c) \frac { 1 }{ 2 }
(d) \frac { 1 }{ 4 }
Solution:
ncert-exemplar-problems-class-10-maths-introduction-to-trigonometry-and-its-applications-Ex-8.1-13s

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:
ncert-exemplar-problems-class-10-maths-introduction-to-trigonometry-and-its-applications-Ex-8.1-15s


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 C1 and C2, whose radii are r1 = 4 cm and r2 = 5 cm. Now, we draw a chord AC of circle C2, which touches the circle C1 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]
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-1s

Question 2:
In figure, if ∠AOB = 125°, then ∠COD is equal to
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-2Q
(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.
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-2s

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
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-3Q
(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.
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-3s

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 cm2                  (b) 65 cm2                (c) 30 cm2                 (d) 32.5 cm2
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.
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-4s-1
Thus, quadrilateral POOR is formed.
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-4s-2

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.
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-5s

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
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-6Q
(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.
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-6s

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
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-7Q
(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.
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-7s

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
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-8Q
(a) 25°               (b) 30°                    (c) 40°                     (d) 50°
Solution:
(a) Given, PA and PB are tangent lines.
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-8s

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) \frac { 3 }{ 2 } √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°.
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-9s-1
Tangent at any point of a circle is perpendicular to the radius through the point of contact.
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-9s-2
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
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-10Q
(a) 20°               (b) 40°                    (c) 35°                     (d) 45°
Solution:
ncert-exemplar-problems-class-10-maths-circles-Ex-9.1-10s


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 A1 A2, A3,… are located at equal distances on the ray AY and the point B is joined to
(a) A12                       (b) A11                        (c) A12                         (d) A9
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 A11

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 A1, A2, A3,… and B1, B2, B3,… are located to equal distances on ray AY and BY, respectively. Then, the points joined are
(a) A5 and A6                     (b) A6 and B5             (c) A4 and B5              (d) A5 and B4
Solution:
(a) Given a line segment AB and we have to divide it in the ratio 5:6.
ncert-exemplar-problems-class-10-maths-constructions-Ex-10.1-3s
Steps of construction

  1. Draw a ray AX making an acute ∠BAX.
  2. Draw a ray BY parallel to AX by making ∠ABY equal to ∠BAX.
  3. Now, locate the points A1, A 2, A3, A 4 and A5 (m= 5) on AX and B1, B2, B3, B4, B5 and B6 (n = 6) such that all the points are at equal distance from each other.
  4. Join B6A5. 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 \frac {3}{7} 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 B1, B2, B3,… on BX at equal distances and next step is to join
(a) B10 to C               (b) B13 to C                      (c) B7 to C                   (d)Bto C
Solution:
(c) Here, we locate points B1, B2, B3, B4, B5, B6 and B7 on BX at equal distance and in next step join the last points is B7 to C.

Question 5:
To construct a triangle similar to a given ΔABC with its sides \frac {8}{5} 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 \frac {m}{n} 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 \frac {8}{5}
Hence,                                      \frac {m}{n}=\frac {8}{5}
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 R1 and R2 is equal to the area of a circle of radius R, then
ncert-exemplar-problems-class-10-maths-areas-related-circle - Ex-11.1-1Q
Solution:
(b) According to the given condition,
                                                          Area of circle =Area of first circle + Area of second circle
ncert-exemplar-problems-class-10-maths-areas-related-circle - Ex-11.1-1s

Question 2:
If the sum of the circumferences of two circles with radii R1 and R2 is equal to the circumference of a circle of radius R, then
(a) R1 + R2=R
(b) R1 + R2 > R
(c) R1 + R2 < R
(d) Nothing definite can be said about the relation among R1,R2 and R
Solution:
(a) According to the given condition,
Circumference of circle = Circumference of first circle + Circumference of second circle
ncert-exemplar-problems-class-10-maths-areas-related-circle - Ex-11.1-2s

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
ncert-exemplar-problems-class-10-maths-areas-related-circle - Ex-11.1-3s
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) rsqunits                  (b) \frac { 1 }{ 2 }  r2 sq units                  (c) 2r2 sq units                              (d) √2 r2 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.
ncert-exemplar-problems-class-10-maths-areas-related-circle - Ex-11.1-4s

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,
ncert-exemplar-problems-class-10-maths-areas-related-circle - Ex-11.1-5s

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
ncert-exemplar-problems-class-10-maths-areas-related-circle - Ex-11.1-6s

Question 7:
The area of the circle that can be inscribed in a square of side 6 cm is
(a) 36π cm2                 (b) 18π cm2                       (c) 12π cm2                            (d) 9π cm2
Solution:
ncert-exemplar-problems-class-10-maths-areas-related-circle - Ex-11.1-7s

Question 8:
The area of the square that can be inscribed in a circle of radius 8 cm is
(a) 256 cm2               (b) 128 cm2               (c)64√2 cm2             (d)64 cm2
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.
ncert-exemplar-problems-class-10-maths-areas-related-circle - Ex-11.1-8s

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 = πd1 = 36 π cm                                  [given, d1 = 36 cm]
and circumference of second circle = πd2 = 20 π cm                                 [given, d2 = 20 cm]
According to the given condition,
Circumference of circle = Circumference of first circle + Circumference of second circle
ncert-exemplar-problems-class-10-maths-areas-related-circle - Ex-11.1-9s

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 r1 = 24 cm and r2 = 7 cm
ncert-exemplar-problems-class-10-maths-areas-related-circle - Ex-11.1-10s


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
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-1s

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
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-2s

Question 3:
A plumbline (sahul) is the combination of (see figure)
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-3Q
(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)
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-3s

Question 4:
The shape of a glass (tumbler) (see figure) is usually in the form of
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-3s
(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
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-4s
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
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-5Q
(a) two cylinders                                         (b) a cone and a cylinder
(c) two cones and a cylinder                    (d) two cylinders and a cone
Solution:
(c)
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-5s

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:

ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-7s

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
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-8s-1
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-8s-2

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
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-9s
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 cm3
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-10s
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 \frac{1}{8}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:
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-11s

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 cm3
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-12s
∴                    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 cm2               (b) 4951 cm2                  (c) 4952 cm2                              (d) 4953 cm2
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)
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-13s

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 cm3                                  (b) 0.35 cm3                          (c) 0.34 cm3                          (d) 0.33 cm3
Solution:
(a) Given, diameter of cylinder = Diameter of hemisphere = 0.5 cm
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-14s-1
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-14s-2
Hence, the capacity of capsule is 0.36 cm3

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) 47πr2                      (b) 6πr2                     (c) 3πr2                    (d) 8πr2
Solution:
(a) Because curved surface area of a hemisphere is 2 w2 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 πr2 + 2 πr2 = 4πr2

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
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-18s-1
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-18s-2
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:
(d) Let the radii of the two spheres are r1 and r2, respectively.
ncert-exemplar-problems-class-10-maths-surface-areas-volumes-Ex-12.1-20s
Hence, the required ratio of their surface area is 16 : 9.


Statistics and Probability Class 10 Maths Multiple Choice Questions with Answers

Question 1:
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-1Q
(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, {d}_{i} = { x }_{ i } – a
i.e., {d}_{i}‘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:

ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-3Q
(a) 0                  (b) -1                           (c) 1                         (d) 2
Solution:
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-3s

Question 4:
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-4Q
Solution:
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-4s

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,
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-6Q
the sum of lower limits of the median class and modal class is
(a) 15                                (b) 25                                    (c) 30                         (d) 35
Solution:
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-6s
Now,\frac{N}{2} =\frac{66}{2}= 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
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-7Q
The upper limit of the median class is
(a) 7                                (b) 17.5                      (c) 18                                 (d) 18.5
Solution:
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-7s

Question 8:
For the following distribution,
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-8Q
the modal class is
(a) 10-20                             (b) 20-30            (c) 30-40                           (d) 30-40
Solution:
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-8s
Here,we see that  the highest frequency is 30. which lies in the interval 30-40.

Question 9:
consider the data.
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-9Q
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:
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-9s
Here,\frac{N}{2} =\frac{67}{2} = 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
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-10Q
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
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-11Q
the frequency of the class 30-40 is
(a) 3                          (b) 4                                 (c) 3                          (d) 4
Solution:
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-11s
Hence,frequency in the class interval 30-40 is 3

Question 12:
If an event cannot occur, then its probability is
(a) 1                          (b) \frac{3}{4}                          (c) \frac{1}{2}                 (d) 0
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) \frac{1}{2}                    (b) 0.1                   (c) 3                        (d) \frac{17}{16}
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 — \frac{1}{P}
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) \frac{3}{26}                        (b) \frac{3}{13}                                  (c)  \frac{2}{13}                                    (d)  \frac{1}{2}
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 = \frac{6}{52} =\frac{3}{26}

Question 19:
The probability that a non-leap your selected at random will contains 53 Sunday is           ‘
(a) \frac{1}{7}                        (b) \frac{2}{7}                                  (c)  \frac{3}{7}                                    (d)  \frac{5}{7}
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 = \frac{1}{7}

Question 20:
When a die is thrown, the probability of getting an odd number less than 3 is ,
(a) \frac{1}{6}                        (b) \frac{1}{3}                                  (c)  \frac{1}{2}                                    (d)  0
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 = \frac{1}{6}

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
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-22s
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.
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-23s
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) \frac{1}{5}                        (b) \frac{3}{5}                                  (c)  \frac{4}{5}                                    (d)  \frac{1}{3}
Solution:
(a) Number of total outcomes = 40
Multiples of 5 between 1 to 40 = 5,10,15,20,25. 30 35, 40
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-24s

Question 25:
Someone is asked to take a number from 1 to 100. The probability that it is a prime,is
(a) \frac{1}{5}                        (b) \frac{6}{25}                                  (c)  \frac{1}{4}                                    (d)  \frac{13}{50}
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.
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-25s

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) \frac{4}{23}                        (b) \frac{6}{23}                                  (c)  \frac{8}{23}                                (d)  \frac{17}{23}
Solution:
(b) Total number of students = 23
Number of students in house A, B and C = 4+ 8 + 5 = 17
ncert-exemplar-problems-class-10-maths-statistics-probability Ex -13.1-26s


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