Mathematics Important Questions

Real Numbers – Important Questions:

Ques.1: Check whether 6n can end with the digit 0 or any natural number n.

Ans:  If any number ends with the digit 0, it should be divisible by 10. Which means, it will also be divisible by 2 and 5 as 10 = 2 × 5.

Therefore, Prime factorization of 6ⁿ = (2 ×3)ⁿ. It's interesting to note that 5 isn't included in the prime factorization of 6ⁿ. Hence, for any value of n, 6ⁿ will not be divisible by 5. Hence 6n cannot end with the digit “0” for any natural number” n”.

Ques.2: Prove that 3 + 2√5 is irrational.

Ans: Let 3 + 2√5 be a rational number. Then the co-primes x and y of the given rational number, where (y ≠ 0) will be-

  3 + 2√5 = x/y

Rearranging these, we get,

2√5 = (x/y) – 3

√5 = 1/2[(x/y) – 3]

Since x and y are integers, thus, 1/2[(x/y) – 3] is a rational number.

Therefore, √5 is also a rational number. But this confronts the fact that √5 is irrational.

Thus, our assumption that 3 + 2√5 is a rational number is wrong.

Hence, this proves that 3 + 2√5 is irrational.

Ques.3: Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Ans: Consider x to be any positive integer and y = 3.

By Euclid’s division algorithm;

x =3q + r (for some integer q ≥ 0 and r = 0, 1, 2 as r ≥ 0 and r < 3)

Therefore,

x = 3q, 3q + 1 and 3q + 2

If we take the square on both the sides, we get;

x² = (3q)² = 9q² = 3x3q²

Let 3q² = m

Therefore,

x2 = 3m ...(1)

x² = (3q + 1)²

= (3q)² + 12 + 2 × 3q × 1

= 9q² + 1 + 6q

= 3(3q² + 2q) + 1

Substitute, 3q²+2q = m, to get,

x² = 3m + 1 ... (2)

x² = (3q + 2)²

= (3q)² + 22 + 2 × 3q × 2

= 9q² + 4 + 12q

= 3(3q² + 4q + 1) + 1

Again, substitute, 3q² + 4q + 1 = m, to get,

x² = 3m + 1… (3)

Hence, from eq. 1, 2 and 3

We can conclude that the square of any positive integer is either of the form 3m or 3m+1 for some integer m.

Ques.4.: An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Ans: To find the maximum number of columns in which the army contingent can march.

First we need to get the HCF(616, 32) 

Using Euclid’s algorithm: -

616 = 32 × 19 + 8

32 = 8 × 4 + 0

The HCF (616, 32) is 8.

So, we can say that they can march in 8 columns each.

Ques.5: There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Ans: It can be observed that Ravi takes less time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi has completed 1 round of that circular path with respect to Sonia.

The total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.

LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36

Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.

Polynomials – Important Questions:

Ques.1: Find the value of ‘p’ in the polynomial x² + 3x + p, if one of the zeroes of the polynomial is 2.

Ans: As 2 is the zero of the polynomial. We know that if α is a zero of the polynomial p(x), then p(α) = 0

Substituting x = 2 in x² + 3x + p,

⇒ 22 + 3(2) + p = 0

⇒ 4 + 6 + p = 0

⇒ 10 + p = 0

⇒ p = -10

Ques.2: If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, then find the value of a and b.

Ans: Let the given polynomial be:

p(x) = x³ – 3x² + x + 1

Given,

The zeroes of the p(x) are a – b, a, and a + b.

Now, compare the given polynomial equation with general expression.

px³ + qx² + rx + s = x³ – 3x² + x + 1

Here, p = 1, q = -3, r = 1 and s = 1

For sum of zeroes:

Sum of zeroes will be = a – b + a + a + b

-q/p = 3a

Substitute the values q and p.

-(-3)/1 = 3a

a = 1

So, the zeroes are 1 – b, 1, 1 + b.

For the product of zeroes:

Product of zeroes = 1(1 – b)(1 + b)

-s/p = 1 – ????²

=> -1/1 = 1 – ????²

Or, ????² = 1 + 1 

=2

So, b = √2

Thus, 1 – √2, 1, 1 + √2 are the zeroes of equation ????³ − 3????² + ???? + 1

Ques.3: Find the quadratic polynomial if its zeroes are 0, √5.

Ans: A quadratic polynomial can be written using the sum and product of its zeroes as:

x² – (α + β)x + αβ

Where α and β are the roots of the polynomial.

Here, α = 0 and β = √5

So, the polynomial will be:

=x² – (0 + √5)x + 0(√5)

= x² – √5x

Ques.4: How many zeros does the polynomial (x – 3)² – 4 have? Also, find its zeroes.

Ans: Given polynomial is (x – 3)² – 4

Now, on expanding this expression.

We get:

= x² + 9 – 6x – 4

= x² – 6x + 5

As the polynomial has a degree of 2, the number of zeroes will be 2.

Now, solve x² – 6x + 5 = 0 to get the roots.

So, x² – x – 5x + 5 = 0

=> x(x – 1) -5(x – 1) = 0

=> (x – 1)(x – 5) = 0

x = 1, x = 5

So, the roots are 1 and 5.

Ques.5: Find the value of ‘x’ in the polynomial 2a² + 2xa + 5a + 10 if (a + x) is one of its factors.

Ans: Let f(a) = 2a² + 2xa + 5a + 10

Since, (a + x) is a factor of 2a² + 2xa + 5a + 10, f(-x) = 0

So, f(-x) = 2x² – 2x² – 5x + 10 = 0

-5x + 10 = 0

5x = 10

x = 10/5

Therefore, x = 2

Pair Of Linear Equations in Two Variables + Quadratic Equations – Important Questions:

Ques.1: Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Ans: Given, half the perimeter of a rectangular garden = 36 m

so, 2(l + b)/2 = 36

(l + b) = 36 …(1)

Given that:

The length is 4 m more than its width.

Let width = x

And length = x + 4

Substituting this in eq(1), we get;

x + x + 4 = 36

2x + 4 = 36

2x = 32

x = 16

Therefore, the width is 16 m and the length is 16 + 4 = 20 m.

Ques.2: Find two consecutive positive integers, sum of whose squares is 365.

Ans: Let us say, the two consecutive positive integers are x and x + 1.

Therefore, as per the given statement,

x² + (x + 1)2 = 365

⇒ x² + x² + 1 + 2x = 365

⇒ 2x² + 2x – 364 = 0

⇒ x² + x – 182 = 0

⇒ x² + 14x – 13x – 182 = 0

⇒ x(x + 14) -13(x + 14) = 0

⇒ (x + 14)(x – 13) = 0

Thus, either, x + 14 = 0 or x – 13 = 0,

⇒ x = – 14 or x = 13

Since, the integers are positive, so x can be 13, only.

So, x + 1 = 13 + 1 = 14

Therefore, the two consecutive positive integers will be 13 and 14.

Ques.3: The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Ans: Let us say, the shorter side of the rectangle is x m.

Then, larger side of the rectangle = (x + 30) m

Diagonal of the rectangle = √[x²+(x+30)²] 

Given that: 

The length of the diagonal = x + 60 m

⇒ x² + (x + 30)² = (x + 60)²

⇒ x² + x² + 900 + 60x = x² + 3600 + 120x

⇒ x² – 60x – 2700 = 0

⇒ x² – 90x + 30x – 2700 = 0

⇒ x(x – 90) + 30(x -90)

⇒ (x – 90)(x + 30) = 0

⇒ x = 90, -30

Ques.4: Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

Ans: Let us consider,

Speed of the boat in still water = x km/hr

Speed of the current = y km/hr

Now, speed of Ritu in downstream = x + y km/hr

Upstream = x – y km/hr

As per the question given,

2(x + y) = 20

Or x + y = 10 …(1)

And, 2(x – y) = 4

Or x – y = 2 …(2)

Adding both eq.1 and eq. 2

We get,

2x = 12

x = 6

Putting the value of x in eq.1

We get,

y = 4

Therefore, Speed of Ritu in still water = 6 km/h

And 

Speed of current = 4 km/hr

Ques.5: A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Ans: Let the fixed charge for the first three days = Rs. A and the charge for each extra day = Rs. B.

According to the question,

A + 4B = 27 ………. (i)

A + 2B = 21 ……… (ii)

When equation (ii) is subtracted from equation (i) we get,

2B = 6

B = 3 …………(iii)

Substituting B = 3 in equation (i) we get,

A + 12 = 27

A = 15

Hence, the fixed charge is Rs. 15.

And the Additional charge per day is Rs. 3.

Arithmetic Progression – Important Questions:

Ques.1: Write first four terms of the A.P. when the first term a and the common difference d are  (i) a = 10, d = 10

Ans: Consider the Arithmetic Progression series a₁, a₂, a₃, a₄, a₅ …

a₁ = a = 10

a₂ = a₁ + d = 10 + 10 = 20

a₃ = a₂ + d = 20 + 10 = 30

a₄ = a₃ + d = 30 + 10 = 40

a₅ = a₄ + d = 40 + 10 = 50

And so on…

Therefore, the A.P. series will be 10, 20, 30, 40, 50 …

And First four terms of this A.P. will be 10, 20, 30, and 40.

Ques.2: How many terms of the AP : 24, 21, 18, . . . must be taken so that their sum is 78?

Ans: Here, a = 24, d = 21 – 24 = –3 and S_n = 78. 

We need to find n. 

So,

S_n = n/2[2a+(n-1)d]

78 = n/2 [48+(n-1)(-3)]

= n/2 [51 – 3n]

3n² – 51n +156 = 0

n² – 17n + 52 = 0

(n-4) (n-13) = 0

n = 4 or 13

Both values of n can be considered. So, the number of terms can be either 4 or 13.

Ques.3: The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Ans: First term, a = 5

Last term, l = 45

Sum of the AP, Sn = 400

As we know, the sum of AP formula is;

Sn = n/2 (a + l)

400 = n/2 (5 + 45)

400 = n/2 (50)

Number of terms, n = 16

As we know, the last term of AP series can be written as:

Last term, l = a + (n − 1) d

45 = 5 + (16 − 1) d

40 = 15d

Therefore the common difference, d = 40/15 = 8/3

Ques.4: Find the sum of the first 22 terms of an AP in which d = 7 and 22nd term is 149.

Ans. Common difference, d = 7

22nd term, a22 = 149

We have to find the sum of first 22 term, S22

By the formula of nth term, 

an = a + (n − 1)d

a22 = a + (22 − 1)d

149 = a + 21 × 7

149 = a + 147

a = 2 = First term

Sum of nth term is given by the formula;

Sn = n/2 (a + an)

= 22/2 (2 + 149)

= 11 × 151

= 1661

Ques.5: If the sum of the first n terms of an AP is 4n − n2, what is the first term (that is S1)? What is the sum of the first two terms? What is the second term? Similarly find the 3rd, the 10th and the nth terms.

Ans: Sn = 4n − n2

First term, a = S1 = 4(1) − (1)2 = 4 − 1 = 3

Sum of first two terms = S2= 4(2) − (2)2 = 8 − 4 = 4

Second term, a2 = S2 − S1 = 4 − 3 = 1

Common difference, d = a2 − a = 1 − 3 = −2

Nth term is given by,

an = a + (n − 1)d

= 3 + (n − 1) (−2)

= 3 − 2n + 2

= 5 − 2n

Therefore, a3 = 5 − 2(3) = 5 − 6 = −1

a10 = 5 − 2(10) = 5 − 20 = −15

Hence, the sum of the first two terms is 4. The second term is 1.

The 3rd, 10th, and nth terms are −1, −15, and 5 − 2n respectively.

Triangle + Coordinate Geometry – Important Questions:

Ques.1: Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.

(i) 7 cm, 24 cm, 25 cm

Ans: (i) Given, sides of the triangle are 7 cm, 24 cm, and 25 cm.

Squaring the lengths of the sides of the, we will get 49, 576, and 625.

49 + 576 = 625

(7)2 + (24)2 = (25)2

Length of Hypotenuse = 25 cm

Therefore, the above equation satisfies Pythagoras theorem. Hence, it is a right-angled triangle.

Ques.2: Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear.

Ans:  A(2, 3)= (x1, y1)

B(4, k) = (x2, y2)

C(6, -3) = (x3, y3)

If the given points are collinear, the area of the triangle formed by them will be 0.

½ [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0

½ [2(k + 3) + 4(-3 -3) + 6(3 – k)] = 0

½ [2k + 6 – 24 + 18 – 6k] = 0

½ (-4k) = 0

4k = 0

k = 0

Ques.3: Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5)

Ans: Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).

Then, AP = BP

AP2 = BP2

Using distance formula,

(x – 7)2 + (y – 1)2 = (x – 3)2 + (y – 5)2

x2 – 14x + 49 + y2 – 2y + 1 = x2 – 6x + 9 + y2 – 10y + 25

x – y = 2

Hence, the relation between x and y is x – y = 2.

Ques.4: In the given figure, PS/SQ = PT/TR and ∠ PST = ∠ PRQ. Prove that PQR is an isosceles triangle. 

Ans: Given,

PS/SQ = PT/TR

We know that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Therefore, ST // QR

And ∠ PST = ∠ PQR (Corresponding angles) ……..(i)

∠ PST = ∠ PRQ (given( ………(ii)

From (i) and (ii),

∠ PRQ = ∠ PQR

Therefore, PQ = PR (sides opposite the equal angles)

Hence, PQR is an isosceles triangle.

Ques.5: If ΔABC ~ ΔQRP, ar (ΔABC) / ar (ΔPQR) =9/4 , AB = 18 cm and BC = 15 cm, then PR is equal to

(A) 10 cm (B) 12 cm (C) 20/3 cm (D) 8 cm

Ans: Given that ΔABC ~ ΔQRP.

ar (ΔABC) / ar (ΔQRP) =9/4

AB = 18 cm and BC = 15 cm

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

ar (ΔABC) / ar (ΔQRP) = BC2/RP2

9/4 = (15)2/RP2

RP2 = (4/9) ×225

PR2 = 100

Therefore, PR = 10 cm

Introduction to Trigonometry + Applications of Trigonometry – Important Questions:

Ques.1: If tan (A + B) =√3 and tan (A – B) =1/√3, 0° < A + B ≤ 90°; A > B, find A and B.

Ans: Since we know that:

Tan (A + B) = √3 & tan 60° = √3

So, we can write:

⇒ tan (A + B) = tan 60°

⇒(A + B) = 60° …… (i)

Now again

tan (A – B) = 1/√3

Since, tan 30° = 1/√3

So, we can write;

⇒ tan (A – B) = tan 30°

⇒(A – B) = 30° ….. (ii)

Adding the equation (i) and (ii), we get;

A + B + A – B = 60° + 30°

2A = 90°

A= 45°

To find the value of B-

Now, put the value of A in eq. (i) 

45° + B = 60°

B = 60° – 45°

B = 15°

Therefore A = 45° and B = 15°

Ques.2: Show that :

(i) tan 48° tan 23° tan 42° tan 67° = 1

Ans: To show: Tan 48° tan 23° tan 42° tan 67°= 1

We can also write the above given tan functions in terms of cot functions, such as;

tan 48° = tan (90° – 42°) = cot 42°

tan 23° = tan (90° – 67°) = cot 67°

Hence, substituting these values, we get

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°)

= 1 × 1 [since cot A. tan A = 1]

= 1

Ques.3: A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

 

Ans: Total height of the tree is the sum of AB and AC i.e. AB+AC

In right angled triangle ΔABC,

Using Cosine and tangent angles,

cos 30° = BC/AC

We know that, cos 30° = √3/2

√3/2 = 8/AC

AC = 16/√3 …(1)

Also,

tan 30° = AB/BC

1/√3 = AB/8

AB = 8/√3 ….(2)

From (1) and (2),

Total height of the tree = AB + AC = 16/√3 + 8/√3 = 24/√3 = 8√3 m.

Ques.4: A pole 6 m high casts a shadow 2√ 3 m long on the ground, then the Sun’s elevation is ?

Ans: Let BC be the pole and AB be its shadow.

In triangle ABC,

tan θ = BC/AB

= 6/2√ 3

= 3/√ 3

= (√ 3 × √ 3 )/ √ 3

= √ 3

tan θ = tan 60°

θ = 60°

Hence, the Sun’s elevation is 60°.

Ques.5: Prove : (1 + tan2A/1 + cot2A) = (1 – tan A/1 – cot A)2 = tan2A

Ans: LHS:

= 1+tan²A / 1+cot²A

Using the trigonometric identities we know that 1+tan²A= Sec²A and 1+cot²A= Cosec²A

= Sec²A/ Cosec²A

On taking the reciprocals we get

= Sin²A/Cos²A

= tan²A

RHS:

=(1-tanA)²/(1-cotA)²

Substituting the reciprocal value of tan A and cot A we get,

=(1-sinA/cosA)²/(1-cosA/sinA)²

=[(cosA-sinA)/cosA]²/ [(sinA-cos)/sinA)²

=(cosA-sinA)²×sin²A /Cos²A. /(sinA-cosA)²

=1×sin²A/Cos²A×1.

=tan2A

The values of LHS and RHS are the same.

Hence proved.

Circle & Area Related to Circles – Important Questions:

Ques.1: If the radius of a circle is 4.2 cm, compute its area and circumference.

Ans: Area of a circle = πr2

So, area = π(4.2)2 = 55.44 cm2

Circumference of a circle = 2πr

So, circumference = 2π(4.2) = 26.4 cm

Ques.2: Find the area of the sector of a circle with radius 4cm and of angle 30°. Also, find the area of the corresponding major sector.

Ans: Radius = r = 4 cm, θ=30°

Area of sector = [????/360]×????????2

= 30/360×3.14×(4)2

= 1/12×3.14×4×4

= 1/3×3.14×4

= 12.56/3 cm2

= 4.19 cm2

Area of major sector = ((360 − θ)/360)×????????2

= ((360 − 30))/360×3.14×(4)2

= 330/360×3.14×4×4

= 11/12×3.14×4×4

= 46.05 cm2

Ques.3: The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Ans: The radius of car’s wheel = 80/2 = 40 cm (as D = 80 cm)

So, the circumference of wheels = 2πr = 80 π cm

Now, in one revolution, the distance covered = circumference of the wheel = 80 π cm

It is given that the distance covered by the car in 1 hr = 66 km

Converting km into cm we get,

Distance covered by the car in 1hr = (66 × 105) cm

In 10 minutes, the distance covered will be = (66 × 105 × 10)/60 = 1100000 cm/s

∴ Distance covered by car = 11 × 105 cm

Now, the no. of revolutions of the wheels = (Distance covered by the car/Circumference of the wheels) 

= 11 × 105 /80 π 

= 4375 revolutions.

Ques. 5: Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. (All India- 2015, 4 Marks)

Ans: From the figure- 

OA=OC (Radii of circle)

Now OB=OC+BC

∴OB>OC (OC being radius and B any point on tangent)

⇒OA<OB

B is an arbitrary point on the tangent. 

Thus, OA is shorter than any other line segment joining O to any 

point on tangent.

Shortest distance of a point from a given line is the perpendicular distance from that line.

Hence, the tangent at any point of the circle is perpendicular to the radius.

Ques.5: Define Tangent and Secant.

Ans. A tangent is a line which meets the circle only at one point.

A secant is a line which meets the circle at two points while intersecting it. These two points are always distinct.

Surface Area and Volumes – Important Questions:

Ques. 1: A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm, and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference between the volumes of the cylinder and the toy. (Take π = 3.14)

Ans. Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere as shown in the above figure.

The radius BO of the hemisphere (as well as of the cone) =( ½) × 4 cm = 2 cm.

So, volume of the toy = (?) πr3 + (?) πr2h

= (?) × 3.14 × 23 + (?)× 3.14 × 22 × 2

= 25.12 cm3

Now, let the right circular cylinder EFGH circumscribe the given solid.

The radius of the base of the right circular cylinder = HP = BO = 2 cm, and its height is

EH = AO + OP = (2 + 2) cm = 4 cm

So, the volume required = volume of the right circular cylinder – the volume of the toy

= (3.14 × 22 × 4 – 25.12) cm3

= 25.12 cm3

Hence, the required difference between the two volumes = 25.12 cm3

Ques.2: Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Ans: For Sphere 1:

Radius (r1) = 6 cm

∴ Volume (V1) = (4/3) × π× r13

For Sphere 2:

Radius (r2) = 8 cm

∴ Volume (V2) = (4/3) × π × r23

For Sphere 3:

Radius (r3) = 10 cm

∴ Volume (V3) = (4/3) × π× r33

Also, let the radius of the resulting sphere be “r”

Now, Volume of resulting sphere = V1 + V2 + V3

(4/3)× π× r3 = (4/3)× π× r13 + (4/3)× π× r23 + (4/3) × π× r33

r3 = 63 + 83 + 103

r3 = 1728

r = 12 cm

Ques. 3: A canal is 300 cm wide and 120 cm deep. The water in the canal is flowing at a speed of 20 km/h. How much area will it irrigate in 20 minutes if 8 cm of standing water is desired?

Ans: The volume of water flows in the canal in one hour = width of the canal × depth

of the canal × speed of the canal water 

= 3 × 1.2 × 20 × 1000 m3

= 72000m3

In 20 minutes, the volume of water = (72000 × 20)/ 60 = 24000 m3

Area irrigated in 20 minutes, if 8 cm, i.e., 0.08 m standing water is required

=24000/0.08

=300000 m2

=30 hectares

Ques.4: A cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.

Ans: The volume of cone = (?) × π × 6 × 6 × 24 cm3

If r is the radius of the sphere, then its volume is (4/3) πr3.

Since the volume of clay in the form of the cone and the sphere remains the same.

We have (4/3) πr3 = (?) × π × 6 × 6 × 24 cm3

r3 = 3 × 3 × 24 = 33 × 23

So, r = 3 × 2 = 6

Therefore, the radius of the sphere is 6 cm.

Ques.5: 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.

Ans: (A) a cone and a cylinder

The Nib of a sharpened pencil = conical shape

The rest of the part of a sharpened pencil = cylindrical

Therefore, a pencil is a combination of a cylinder and a cone

Statistics & Probability – Important Questions:

Ques. 1: Find the mean of the 32 numbers, such that if the mean of 10 of them is 15 and the mean of 20 of them is 11. The last two numbers are 10.

Ans: The given mean of 10 numbers = 15

So, Mean of 10 numbers = sum of observations/ no. of observations

15 = sum of observations / 10

Sum of observations of 10 numbers = 150

Similarly, Mean of 20 numbers = sum of observations/ no. of observations

11 = sum of observations / 20

Sum of observations of 20 numbers = 220

Hence, Mean of 32 numbers = (sum 10 numbers + sum of 20 numbers +sum of last two numbers)/ no. of observations

Mean of 32 numbers = (150 +220 + 20 ) / 32 

= 390 /32 

= 12.188

Ques.2: If P(E) = 0.05, what is the probability of ‘not E’?

Ans. We know that,

P(E) + P(not E) = 1

It is given that, P(E) = 0.05

So, P(not E) = 1 – P(E)

P(not E) = 1 – 0.05

∴ P(not E) = 0.95

Ques. 3: 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determining the probability that the pen is taken out is a good one.

Ans: Numbers of pens = Numbers of defective pens + Numbers of good pens

∴ Total number of pens = 132 + 12 = 144 pens

P(E) = (Number of favourable outcomes) / (Total number of outcomes)

P(picking a good pen) = 132/144 = 11/12 = 0.916

Ques. 4: While checking the value of 20 observations, it was noted that 125 was wrongly noted as 25 while calculating the mean and then the mean was 60. Find the correct mean.

Ans: Let y be the sum of observation of 19 (20 – 1) numbers leaving 125,

So, y + 25 = 20 × 60 = 1200 (Mean = sum of observations/ no. of observations)

As we know,

x+25 = 20×60

=1200

Also

x+125 = 20×y

=20y

Next, Subtract 

125−25 = 20y−1200

20y = 1300

y = 65

Ques. 5: An aircraft has 120 passenger seats. The number of seats occupied during 100 flights are given in the following table :

Determine the mean number of seats occupied over the flights.

Ans:

∴ Assumed mean, a = 110

Class width, h = 4

And total observations, N = 100

Mean(x¯¯¯)=a+∑fidi∑fi

= 110 + (-8/100)

= 110 – 0.08

= 109.92

But we know that the seats cannot be in decimal.

Therefore, the number of seats = 109 (approx.).

Previous Years’ Questions

Ques 1: Prove that the product of any three consecutive positive integers is divisible by 6. (All India- 2016, 4 Marks)

Ans: 

Let three consecutive positive numbers are n, n + 1 and n + 2

Case 1: If n is even

This means n + 2 will also be even.

Hence n and n + 2 can be divisible by 2

Also, the product of n and (n + 2) will be divisible by 2.

Means, n(n + 2) will be divisible by 2.

This means that-

n(n + 2) (n + 1) will be divisible by 2 …(i)

Since, 

n, n + 1 and n + 2 are three consecutive numbers. So, the product of these three numbers i.e. n(n + 1) (n + 2) will be a multiple of 3.

This means that-

n(n + 1) (n + 2) will be divisible by 3. …(ii)

On solving Eq. (i) and (ii) we can say

n(n + 1) (n + 2) can be divisible by 2 and 3 both.

Hence, we can conclude that- 

n(n + 1) (n + 2) is divisible by 6.
 

Case 2: When n is odd.

Which means (n + 1) is even

Hence (n + 1) is divisible by 2. …(iii)

This concludes that n(n + 1) (n + 2) is an even number and can be divisible by 2.

Also we know that-

The product of three consecutive numbers is a multiple of 3.

n(n + 1)(n + 2) is divisible by 3. …(iv)

From Equation (iii) and (iv) we can say

n(n + 1) (n + 2) can be divisible by both 2 and 3.

Hence, we can conclude that-

n(n + 1)(n + 2) is divisible by 6.

Ques. 2: Find HCF of 378,180 and 420 by prime factorisation method. Is HCF x LCM of three numbers equal to the product of the three numbers? (All India- 2014, 4 Marks)

Ans: By Prime Factorization Method, the HCF of 378, 180 and 420 can be written as- 

378 = 2 x 33 x 7

180 = 22 x 32 x 5

420 = 22 x 3 x 5 x 7

So, the HCF (378, 180, 420) = 2 x 3 = 6.

No, the HCF (p, q, r) x LCM (p, q, r) ≠ p x q x r. 

Where p, q and r are positive integers.

Ques. 3: Explain whether the number 3 x 5 x 13 x 46 + 23 is a prime number or a composite number. (All India- 2016, 3 Marks)

Ans: We have 3 x 5 x 13 x 46 + 23 = 23 x (3 x 5 x 13 x 2) + 23

= 23 x (3 x 5 x 13 x 2) + 23 = 23 x 390 + 23 = 8,993

From the above equation, it is clear that the (3 x 5 x 13 x 2) + 23 is a multiple of 23. Therefore, (3 x 5 x 13 x 2) + 23 is not a prime number. 

Ques. 4: Three bells toll at intervals of 12 minutes, 15 minutes and 18 minutes respectively. If they start tolling together, after what time will they next toll together? (All India- 2015, 3 Marks)

Ans: LCM of 12, 15, 18 

= 2²x 3² x 5

=4 x 9 x 5 = 180

So, we can say that after 180 minutes the bells will ring together next time.

Ques. 5: If one zero of the quadratic polynomial f(x) = 4x2 – 8kx + 8x – 9 is negative of the other, then find the zeros of kx2 + 3kx + 2. (All India- 2015, 3 Marks)

Answer: 

Ques.6: Obtain all other zeroes of the polynomial x4 + 4x3-2x2-20x -15 if two of its zeroes are √5 and -√5 (All India- 2014, 3 Marks)

Answer: 

Ques.7: If the product of zeroes of the polynomial ax2 – 6x – 6 is 4, find the value of a. Find the sum of zeroes of the polynomial. (All India- 2014, 2 Marks)

Answer: 

Ques.8: Solve the following pair of linear equations for x and y:

141x + 93y = 189; 93x + 141y = 45 (All India-2013, 3 Marks)

Answer:

Ques.9: Solve for x and y: 27x + 31y = 85; 31x + 2 7y = 89 (All India- 2017, 3 Marks)

Answer: 

Ques.10: The owner of a taxi company decides to run all the taxis on CNG fuel instead of petrol/diesel. The taxi charges in the city comprises fixed charges together with the charge for the distance covered. For a journey of 13 km, the charge paid is ?129 and for a journey of 22 km, the charge paid is ?210.

What will a person have to pay for travelling a distance of 32 km? ( All India 2014, 5 Marks )

Answer: Let fixed charge be Rs x and the charge for the distance = Rs y per km

According to the Question,

For a journey of 13 km,

x + 13y = 129 ⇒ x = 129 – 13y …(i)

For a journey of 22 km, x + 22y = 210 …(ii)

⇒ 129 – 13y + 22y = 210 …[From (i)

⇒ 9y = 210 – 129 = 81

⇒ 9y = 81 ⇒ y = 9

From (i), x = 129 – 13(9)

= 129 – 117 = 12

∴ Fixed charge, x = ?12

∴ The charge for the distance, y = ?9 per km

To pay for travelling a distance of 32 km

= x + 32y = 12 + 32(9) = 12 + 288 = ?300

Ques.11: Find the value of k for which the equation x2 + k(2x + k – 1) + 2 = 0 has real and equal roots. (Delhi 2017, 2 Marks)

Answer: We have, x2 + k(2x + k – 1) + 2 = 0

x2 + 2kx + k2 – k + 2 = 0

Here a = 1, b = 2k, c = k2 – k + 2

D = 0 …[real and equal roots

∴ b2 – 4ac = 0

⇒ (2k)2 – 4 × 1(k2 – k + 2) = 0

⇒ 4k2 – 4 (k2 – k + 2) = 0

⇒ 4(k2 – k2 + k – 2) = 0 ⇒ 4(k – 2) = 0

⇒ k – 2 = 0 ⇒ k = 2

Ques. 12: If -5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, find the value of k. (All India 2016, 3 Marks)

Answer: We have, 2x2 + px – 15 =0

Since (-5) is a root of the given equation

∴ 2(-5)2 + p(-5) – 15 = 0

⇒ 2(25) – 5p – 15 = 0

⇒ 50 – 15 = 5p

⇒ 35 = 5p ⇒ p = 7 …(i)

Now, p(x2 + x) + k ⇒ px2 + px + k = 0

7x2 + 7x + k = 0 …From (i)

Here, a = 7, b = 7, c = k

D = 0 …(Roots are equal)

b2 – 4ac = 0

⇒ (7)2 – 4(7)k = 0 ⇒ 49 – 28k = 0

⇒ 49 = 28k 

∴ k = 49/28 = 7/4

Ques. 13: If the roots of the quadratic equation (a – b)x2 + (b – c)x + (c – a) = 0 are equal, prove that 2a = b + c. (All India 2016, 3 Marks)

Answer: Here-

 ‘a’ = a – b, ‘b’ = b – c, ‘c’ = c – a

D = 0 ….(Roots are equal)

b2 – 4ac = 0

⇒ (b – c)2 – 4(a – b)(c – a) = 0

⇒ b2 + c2 – 2bc – 4(ac – a2 – bc + ab) = 0

⇒ b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ab = 0

⇒ 4a2 + b22 + c2 – 4ab + 2bc – 4ac = 0

⇒ (-2a)2 + (b)2 + (c)22 + 2(-2a)(b) + 2(b)(c) + 2(c)(-2a) = 0

⇒ [(-2a) + (b) + (c)]2 = 0

? x2 + y2 + z2 + 2xy + 2yz + 2zx = (x + y + z)2

Taking square-root on both sides

-2a + b + c = 0

⇒ b + c = 2a 

⇒ 2a = b + c

Ques. 14: Which term of the A.P. 3, 14, 25, 36, … will be 99 more than its 25th term? (All India 2011, 3 Marks)

Answer:

Let the required term be nth term, i.e., an

Here, d = 14 – 3 = 11, a = 3

According to the Question, an = 99 + a25

∴ a + (n – 1) d = 99 + a + 24d

⇒ (n – 1) (11) = 99 + 24 (11)

(n – 1) (11) = 11 (9 + 24)

n – 1 = 33

n = 33 + 1 = 34

∴ 34th term is 99 more than its 25th term.

Ques. 15: The 14th term of an AP is twice its gth term. If its 6th term is -8, then find the sum of its first 20 terms. (All India 2015, 4 Marks)

Answer:

Let a = First term, d = Common difference

Given that

a14 = 2.a8

⇒ a + 13d = 2 (a + 7d) 

⇒ a + 13d = 2a + 14d

⇒ 1a – 2a = 14d – 13d

⇒ -1a = d ⇒ a = -d … (i)

a6 = -8

⇒ -8 = a + 5d

From eq 1

⇒ -d + 5d = -8

⇒ 4d = -8 ⇒ d = -2

Value of d put in equation (i), we get

a = -d ⇒ a=-(-2)

Now, a = 2, d = -2

Now, Sum of first 20 terms,

S20 = 20/2 [2 × 2 + (20 – 1)(-2)] 

S20 = 10[4 + 19(-2)]

S20 = 10[4 – 38] 

S20 = -340

Ques.16: In the given figure, AD = 3 cm, AE = 5 cm, BD = 4 cm, CE = 4 cm, CF = 2 cm, BF = 2.5 cm, then find the pair of parallel lines and hence their lengths. (All India 2015, 4 Marks) 

Answer: 

Ques. 17: If sides AB, BC and median AD of AABC are proportional to the corresponding sides PQ, QR and median PM of PQR, show that ?ABC ~ ?PQR. (All India 2017, 4 Marks)

Answer: 

Ques. 18: Find the value of: (All India 2013, 4 Marks)

Answer:

Ques. 19: Prove that the lengths of tangents drawn from an external point to a circle are equal. (All India 2016, 4 Marks)

Ans: Given that- A circle C (O,r), P is a point outside the circle and PA and PB are tangents to a circle.

To Prove- PA=PB

Construction: Draw OA, OB and OP 

Proof: Consider ?? OAP and ?? OBP 

∠OAP = ∠OBP = 90° …(i)

OA = OB = Radius …(ii)

OP is common …(iii)

From eq. (i), (ii) and (iii) 

We can say that, 

 ?? OAP ≅ ?? OBP 

Hence, It can be concluded that 

PA = PB 

Ques. 20: In a given figure, O is the centre of a circle of radius 5 cm. T is a point such that OT intersects a circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle. (All India 2016, 4 Marks)

Ans: 

Ques. 21: Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. (All India 2016, 4 Marks)

Ans: 

In the figure - 

OA=OC (Radii of circle)

Now OB=OC+BC

∴OB>OC (OC being radius and B any point on tangent)

⇒OA<OB

B is an arbitrary point on the tangent. 

Thus, OA is shorter than any other line segment joining O to any 

point on tangent.

Since, the shortest distance of a point from a given line is the perpendicular distance from that line.

Hence, the tangent at any point of the circle is perpendicular to the radius.

Ques. 22: From a solid cylinder of height 7 cm and base diameter 12 cm, a conical cavity of same height and same base diameter is hollowed out. Find the total surface area of the remaining solid. (Use π = 22/7) (All India 2017, 3 Marks)

Answer: 

Ques. 23: A hemispherical tank, full of water, is emptied by a pipe at the rate of 25/7 litres per sec. How much time will it take to empty half the tank if the diameter of the base of the tank is 3 m? (All India 2015, 4 Marks)

Answer: 

Ques. 24: Find the mean and median for the following data: (All India 2015, 3 Marks)

Answer: 

Ques. 25: Weekly income of 600 families is given below: (All India 2012, 3 Marks)

Answer: 

Ques. 26: The following table gives the daily income of 50 workers of a factory. Draw both types (“less than type” and “greater than type”) ogives. (All India 2015, 4 Marks)

Answer: 

Ques. 27: The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is ¼ The probability of selecting a blue ball at random from the same jar is ? . If the jar contains 10 orange balls, find the total number of balls in the jar. (All India 2015, 3 Marks)

Answer: 

Ques. 28: A number x is selected at random from the numbers 1, 4, 9, 16 and another number y is selected at random from the numbers 1, 2, 3, 4. Find the probability that the value of xy is more than 16. (All India 2016, 3 Marks)

Answer: x can be any one of 1, 4, 9, or 16. So total cases= 4 ways

y can be any one of 1, 2, 3, So total cases= 4 ways

Total number of cases of xy = 4 × 4 = 16 ways

Number of cases, where product is more than 16

(9, 2), (9, 3), (9, 4), (16, 2), (16, 3), (16, 4)

Which means total 6 ways

∴ Required probability = 6/16 = 3/8