National Mathematics Day

Srinivasa Ramanujan, (born December 22, 1887, Erode, India—died April 26, 1920, Kumbakonam), Indian mathematician whose contributions to the theory of numbers include pioneering discoveries of the properties of the partition function.

When he was 15 years old, he obtained a copy of George Shoobridge Carr’s Synopsis of Elementary Results in Pure and Applied Mathematics, 2 vol. (1880–86). This collection of thousands of theorems, many presented with only the briefest of proofs and with no material newer than 1860, aroused his genius. Having verified the results in Carr’s book, Ramanujan went beyond it, developing his own theorems and ideas. In 1903 he secured a scholarship to the University of Madras but lost it the following year because he neglected all other studies in pursuit of mathematics.

Ramanujan continued his work, without employment and living in the poorest circumstances. After marrying in 1909 he began a search for permanent employment that culminated in an interview with a government official, Ramachandra Rao. Impressed by Ramanujan’s mathematical prowess, Rao supported his research for a time, but Ramanujan, unwilling to exist on charity, obtained a clerical post with the Madras Port Trust.

Ramanujan’s knowledge of mathematics (most of which he had worked out for himself) was startling. Although he was almost completely unaware of modern developments in mathematics, his mastery of continued fractions was unequaled by any living mathematician. He worked out the Riemann series, the elliptic integrals, hypergeometric series, the functional equations of the zeta function, and his own theory of divergent series, in which he found a value for the sum of such series using a technique he invented that came to be called Ramanujan summation. On the other hand, he knew nothing of doubly periodic functions, the classical theory of quadratic forms, or Cauchy’s theorem, and he had only the most nebulous idea of what constitutes a mathematical proof. Though brilliant, many of his theorems on the theory of prime numbers were wrong.

RAMANUJAN'S NUMBER 

1729 is the natural number following 1728 and preceding 1730. It is a taxicab number, and is variously known as Ramanujan's number and the Ramanujan-Hardy number, after an anecdote of the British mathematician G. H. Hardy when he visited Indian mathematician Srinivasa Ramanujan in hospital. He related their conversation.

I remember once going to see him when he was ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavourable omen. "No," he replied, "it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways."

The two different ways are:

1729 = (1)^3 + (12)3 = (9)^3 + (10)^3

67 BPSC Math solution(Answer Key)

Q1. A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours. What will be the true time when the clock indicates 1 p.m. on the following day? 

Q9. The perimeter of a rhombus is 52 m and its shorter diagonal is 10 m. The length of the longer diagonal is

Solution

We have,

Parameter of a rhombus$=52m$Its diagonal$=10m$

Let the length of the other diagonal$=Xm$

 $Now,$

$Parameter\; of\; rhombus=4\times side$

52=4×side ⇒side=13m

Therefore, each side of rhombus $=13m$

Then, the length of other diagonal$=2\sqrt{{13}^2-{\left(\frac{10}{2}\right)}^2}$

x=2×144​

x=2×12

$x=24m$

We know that,

Area of rhombus$=\frac{1}{2}\times$ product of diagonals$=\frac{1}{2}\times 10\times 24$=5×24=120m.2     

Q10. In a 100 m race, A runs at 8 kmph. If A gives Ba start of 4 m and still beats him by 15 seconds, what is the speed of B?

Solution 

Important questions of class 8

Q1. Construct a rectangle whose diagonal is 5 cm and the angle between the diagonal is 50°.

Solution:

Construction:

Step I: Draw AC = 5 cm.

Step II: Draw the right bisector of AC at O.

Step III: Draw an angle of 50° at O and product both sides.

Step IV: Draw two arcs with centre O and of the same radius 2.5 cm to cut at B and D.

Step V: Join AB, BC, CD and DA.

Thus, ABCD is the required rectangle.

Q2. Construct a quadrilateral ABCD in which BC = 4 cm, ∠B = 60°, ∠C = 135°, AB = 5 cm and ∠A = 90°.

Solution.

Construction:

Step I: Draw AB = 5 cm.

Step II: Draw the angle of 60° at B and cut BC = 4 cm.

Step III: Draw an angle of 135° at C and angle of 90° at A which meet each other at D.

Thus, ABCD is the required quadrilateral.

Q3. Is it possible to construct a quadrilateral ABCD in which AB = 5 cm, BC = 7.5 cm, ZA = 80°, B = 140° and C=145°? If not, give reason.

Solution:

No, it is not possible to construct a quadrilateral ABCD with the given measurements.

Angle (A + B + C)= 80° +140° +145° =365° is greater than 360°.

The sum of all the four angles is 360°. quadrilateral cannot be constructed.

Question 4.

In the parallelogram given alongside if m∠Q = 110°, find all the other angles.

Solution:

Given m∠Q = 110°

Then m∠S = 110° (Opposite angles are equal)

Since ∠P and ∠Q are supplementary.

Then m∠P + m∠Q = 180°

⇒ m∠P + 110° = 180°

⇒ m∠P = 180° – 110° = 70°

⇒ m∠P = m∠R = 70° (Opposite angles)

Hence m∠P = 70, m∠R = 70°

and m∠S = 110°

Q5. Write true and false against each of the given statements.

(a) Diagonals of a rhombus are equal.

(b) Diagonals of rectangles are equal.

(c) Kite is a parallelogram.

(d) Sum of the interior angles of a triangle is 180°.

(e) A trapezium is a parallelogram.

(f) Sum of all the exterior angles of a polygon is 360°.

(g) Diagonals of a rectangle are perpendicular to each other.

(h) Triangle is possible with angles 60°, 80° and 100°.

(i) In a parallelogram, the opposite sides are equal.

Solution:

(a) False

(b) True

(c) False

(d) True

(e) False

(f) True

(g) False

(h) False

(i) True

Q6. If AM and CN are perpendiculars on the disgonal BD of a parallelogram ABCD,Is △AMD≅△CNB?Give reason.

Solution

In triangles AMD and CNB

AD=BC(opposite sides of a parallelogram)

∠AMB=∠CNB=90∘

 

∠ADM=∠NBC since AD∥BC and BD is the transversal.

So, △AMD≅△CNB by AAS

Q7.Verify whether a polyhedron can have 10 faces, 20 edges and 15 vertices.

Solution:

We have

Number of faces F = 10

Number of edges E = 20

and number of vertices V = 15

Euler’s formula:

V + F – E = 2

⇒ 15 + 10 – 20 = 2

⇒ 5 ≠ 2

Hence, it is not possible to have a polyhedron satisfying the above data

Q8. Draw the nets of the following polyhedrons.

(i) Cuboid

(ii) Triangular prism with a base equilateral triangle.

(iii) Square pyramid.

Solution:

i.

ii.

iii.

Q9. Name the solids that have:

(i) 4 faces

(ii) 8 triangular faces

(iii) 6 faces

(iv) 1 curved surface

(v) 5 faces and 5 vertices

(vi) 6 rectangular faces and 2 hexagonal faces

Solution:

(i) Tetrahedron

(ii) Regular octahedron

(iii) Cube and cuboid

(iv) Cylinder

(v) Square and a rectangular pyramid

(vi) Hexagonal prism

Q10. The scale of a map is given as 1 : 50,000. Two villages are 5 cm apart on the map. Find the actual distance between them.

Solution

Let the map distance be x cm and actual distance be y.

1 : 50,000 =x:y

Q11. 15 men can build a wall in 42 hours, how many workers will be required for the same work in 30 hours?

Solution:

Let the required number of workers be x.

The number of workers, faster will they do the work.

So, the two quantities are inversely proportional

x1y1 = x2y2

⇒ 42 × 15 = 30 × x

⇒ x = 21

Hence, the required number of men = 21

Q12. The cost of 5 metres of cloth is ₹ 210. Tabulate the cost of 2, 4, 10 and 13 metres of cloth of the same type.

Solution:

Let the length of the cloth be x m and its cost be ₹ y. We have the following table.

Q12.Mohit deposited a sum of ₹ 12000 in a Bank at a certain rate of interest for 2 years and earns an interest of ₹ 900. How much interest would be earned for a deposit of ₹ 15000 for the same period and at the same rate of interest?

Solution:

Let the required amount of interest be ₹ x.

Important and tricky question on mathematics

the water in a rectangular reservoir having a base 80 m by 60 m is 6.5 m deep . in what time be emptied by a pipe of which the crosssection is a square of side 20 cm if the water runs though the pipe at the rate of 15 km per hour

Solution The reservoir is in the shape of a cuboid. 

Volume of water in the reservoir in =

    80×60×213​m3
=31200m3

Volume of water flown in 1  hour = Speed × Area of  the cross section 

=15×1000×10020​×10020​=600m3

Time taken to empty 31200m3=60031200​=52 hours