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SEE Compulsory Math Question Paper With Solution

SEE 2077 (2021)

What is the relation between the areas of a parallelogram and a triangle standing on the same base and between the same parallel lines?

What is the price of an article costing Rs. 1400 after levying 13% VAT. Find it.

The present price of a motorcycle is Rs 225000. If its price is depreciated per year by 8%, after how many years the price of the motorcycle will be Rs 175204.8? Find it.

Given:

The present cost (C) = Rs. 225000

Rate of compound depreciation (R) = 10% p.a

Cost after depreciation (C_{T}) = Rs 175204.8

To find: Time (T)

By Formula,

\(C_T=c\left(1-\frac R{100}\right)^T\)

Or, \(175204.8=225000\left(1-\frac8{100}\right)^T\)

Or, \(\frac{175204.8}{225000}=\left(1-0.08\right)^T\)

Or, \(0.778688=\left(0.92\right)^T\)

Or, \(\left(0.92\right)^3=\left(0.92\right)^T\)

Or, 3 = T

Hence, the required time is 3 years.

If the area of an equilateral triangle is \(36\sqrt3\) . Find its perimeter.

Given,

Area of an equilateral triangle (A) = \(36\sqrt3\) cm^{2}

To find, Perimeter of the triangle (P)

By formula,

Area of an equilateral triangle (A) = \(\frac{\sqrt3}4a^2\)

or, \(36\sqrt3\) cm^{2 }= \(\frac{\sqrt3}4a^2\)

or, \(\frac{4\times36\sqrt3cm^2}{\sqrt3}=a^2\)

or, a^{2} = 144cm^{2}

or, a^{2} = (12cm)^{2}

or, a = 12

Now,

Perimeter of an equilateral triangle (P) = 3a

= 3 x 12

= 36cm

In the given solid triangular-based prism, if AB = 8cm, AC = 6cm, B'C' = 10cm, and area of the rectangular faces of the prism in 288cm^2, FInd the height of the prism CC'.

Given, In a triangular based prism,

AB = 8cm

Ac = 6cm

BC = B'C' = 10cm

and lateral surface area of the prism = 288cm^{2}

To find, Height of the prism (CC')

Here,

Perimeter of the base (P) = AB + BC + AC

=8cm + 10cm + 6cm

=24cm

Now, by the formula,

The lateral surface area of the prism = perimeter of the base x Height

or, 288cm^{2} 24cm x CC'

or, cc' = \(\frac{\displaystyle288cm^2}{24cm}\)

or, CC' = 12cm

Thereforem, height of the prism (CC') = 12cm.

If the total surface of a hemisphere is 7392 square cm, then find its radius.

Given, Total surface area of a hemisphere (TSA) = 7392cm^{2}

By formula,

Total surface area of a hemisphere (TSA) = ^{2}

or,7392cm^{2} = \(3\times\frac{22}7\times r^2\)

or, \(\frac{7\times7392cm^2}{66}=r^2\)

or, r^{2} = 784cm^{2}

or, r^{2} = (28cm)^{2}

or, r = 28cm

Therefore the radius of the hemisphere (R) = 28cm.

The cost of a pen and a pencil is Rs. 35. If the cost of a pen is Rs. 15 more than the cost of a pencil, find the cost of a pencil.

In the figure, PQRS is a rectangle in which PQ=3 PS=12cm. RS is extended up to the point M. What is the area of triangle PQM? Find it.

Given, PQRS is a rectangle

and PQ = 3 Ps = 12cm

To find, Area of triangle PQM

Here,

(I) 3PS = 12cm

or, PS = 12/3 = 4cm

(II) Area of rectangle PQRS = PQ x PS [Area of rectangle = length x breadth]

= 12cm x 4cm

= 48cm^{2}

(III) Area of triangle PQM = 1/2 Area of rectangle PQRS [Both are standing on the same base and between the same parallel lines.]

or, Area of Triangle PQM = 1/2 x 48cm^{2}

= 24cm^{2}

In the given figure, find the value of x and y.

Given,

O is the center of the circle

And BOD = 160°

To find: The value of x and y

Here,

(I) BAD = 1/2 BOD [Therefore, BAD is a circumference angle and BOD is a central angle standing on the same arc of a circle.]

or, x = (1/2) x 160°

or, x = 80°

(II) BAD + BCD = 180° [Therefore, being opposite angles of a cyclic quadrilateral]

or, 80° + y = 180°

or, y = 180° - 80°

or, y = 100°

Therefore, 80° and y = 100°

In the given figure, RT is a tangent and R is the point of contact. If QRT = 70° and PRT = 30°, find the measure of RPQ.

Given,

RT is a tangent, R is the point of contact,

QRT = 70° and PRT = 30°

To find, The measure of RPQ.

Here,

(I) QRS + QRT = 180° [Therefore, sum of the adjacent angles standing on a straight line]

or, QRS + 70° = 180°

or, QRS = 180° - 70°

=110°

(II) RPQ = QRS [Therefore, the angle made by a chord to the tangent of a circle at the point of contact is equal to the angle in the alternate segment.]

or, RPQ = 110°

In the given figure, PQ = PS and QR = SR. If QR = 27cm, PR = 30cm and PRQ = 30°, find the area of kite PQRS.

Given,

QR = 27cm, PR = 30cm and PRQ = 30°

To find, Area of kite PQRS

By formula,

Area of Triangle PQR = 1/2 .QR.PR. Sin 30°

= (1/2) x 27cm x 30cm x (1/2)

= 202.2cm^{2}

Here,

Area of triangle PQR = (1/2) area of kite PQRS [ Diagonal so formed by joining the vertices made by equal sides of kite bisects the kite.]

or, 202.5cm^{2} = 1/2 area of kite PQRS

or, Area of kite PQRS = 405cm^{2}

Find the class of the third quartile from the following data.

X | 0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 |

CF | 4 | 10 | 20 | 28 | 35 |

In the given data,

No. of terms (N) = 35

Here,

Third quartile (Q_{3}) = The value of \(\frac{3N^{th}}4\) item

= the value of \(\frac{(3\times35)^{th}}4\) item

= the value of 26.25^{th} item

Here, CF just greater than 26.25 is 28 and it lies in the class ( 60- 80 )

Therefore, Third quartile class (Q_{3}) = (60 - 80).

Find the probability of getting a free card or an ace of spade, when a card is drawn randomly from a well-shuffled pack of 52 cards.

Let, the event of getting a face card be F.

And the event of getting ace of spades be C.

Then, n(S) = 52, n(F) = 12 and n(C) = 1

To find, P(F∪C)

Here, F and C are mutually exclusive events.

Then, by formula

P(F∪C) = P(F) + P(C)

= \(\frac{n(F)}{n(S)}+\frac{n(C)}{n(S)}\)

= \(\frac{12}{52}+\frac1{52}\)

= \(\frac{12+1}{52}\)

= \(\frac{13}{52}\)

= \(\frac14\)

Therefore, the probability of getting a face card or an ace of spades is \(\frac14\).

There are 10 yellow and 12 green marbles of the same shape and size in a box. Two marbles are drawn from the box in succession without replacement. Show the probabilities of all outcomes in the tree diagram.

In a group of 50 students, 20 like only math and 15 like only science. If the number of students who do not like any of the two subjects is double the number of students who like both subjects, find the number of students who like at most one subject by using Venn-Diagram.

Let, the set of students who like maths be M.

and the set of students who like science be S.

Then, n(U) = 50, n_{o}(S) = 15 and \(n(\overline{M\cup S})=2n(M\cap S)\)

To find, \(n(\overline{M\cap S)}\)

Let, \(n(M\cap S)\) = x

Then, \(n(\overline{M\cup S)}\) = 2x

Now, representing the above information in a Venn diagram, we get

From the above Venn diagram,

n_{o} (M) + n_{o} (S) + \(n(M\cap S)\) + \(n(\overline{M\cup S})\) = n(U)

or, 20 + 15 + x + 2x = 50

or, 35 + 3x = 50

or, 3x = 50-35

or, x = 15/3

or, x = 5

Therefore, \(n(\overline{M\cap S)}\) = n(U) - \(n(M\cap S)\) = 50 - 5 = 45

Therefore, The number of students who like at most one subject is 45.

The marked price of a laptop is Rs.75000. Allowing 15% discount and including the same percentage of VAT the laptop is sold. Find the cost that has to pay by the customer and the VAT amount.

Given,

Marked price = Rs. 75000

Discount percent = 15%

VAT percent = 15%

To find, SP with VAT and amount of VAT

By formula,

Selling price (SP) = MP - discount % of MP

= Rs.75000-15% of Rs.75000

= Rs. 75000 - (15/100) x Rs. 75000

= Rs. 75000 - Rs. 11250

= Rs. 63750

Again,

Amount of VAT = VAT % of SP

= 15% of Rs.63750

= (15/100) x Rs. 63750

= Rs. 9562.50

An umbrella is made by stitching 12 pieces of triangular cloths. If the measure of each piece of cloth is 42cm, 42cm, and 21cm, how much cloth is required to make the umbrella? If per square meter the cost of the cloth is Rs. 1250, find the cost of cloth required to make the umbrella.

Given,

In an umbrella,

No. of triangular pieces of cloth = 12

Length of each of two equal sides of triangular pieces (a) = 42cm = 42m/100 = 0.42m

and the length of third side (b) = 21cm = 21/100 = 0.21

and the cost of per sq. meter of cloth (C) = Rs. 1250

To find: (I) Total area of cloth

(ii) Total cost of cloth

By formula,

Area of each isoceles triangle piece of cloth = \(\frac b4\sqrt{4a^2-b^2}\)

= \(\frac{0.21m}4\times\sqrt{4\times(0.42m)^2-(0.21m)^2}\)

= 0.0525m x \(\sqrt{0.7056m^2-0.0441m^2}\)

= 0.0525m x \(\sqrt{0.6615m^2}\)

= 0.0525m x 0.8133m

= 0.0427m^{2}

Total area of cloth (A) = Area of each piece x No. of pieces cloth

= 0.0427m^{2} x 12

= 0.5124m^{2}

Now,

The total cost of cloth (T) = Cost of per sq. meter of cloth x area

= Rs. 1250 x 0.5124

= Rs. 640.50

Total area of cloth (A) = 0.5124m^{2}

and total cost of cloth (T) = Rs.640.50

Find the HCF of 8m^{4} + 27mn^{3} , 8m^{3} n + 2m^{2} n^{2} - 15mn^{3} and 4m^{3} n - 9mn^{3} .

Here,

**First expression**

= 8m^{4} + 27mn^{3}

= m(8m^{3} + 27n^{3})

= m{(2m)^{3} + (3n)^{3}}

= m(2m + 3n) (4m^{2} - 6mn + 9n^{2})

**Second expression**

= 8m^{3}n + 2m^{2} n^{2} - 15mn^{3}

= mn(8m^{2} + 12mn - 10mn - 15n^{2})

= mn{4m(2m+3n)- 5n(2m + 3n)}

= mn(2m+3n) (4m-5n)

**Third expression**

= 4m^{3}n - 9mn^{3}

= mn(4m^{2} - 9n^{2})

= mn{(2m)^{2} - (3n)^{2}}

= mn(2m + 3n) (2m -3n)

∴ HCF of the given expressions is = m(2m+3n)

Solve: 9^{a} - 10 x 3^{a} + 9 = 0

Here,

9^{a} - 10 x 3^{a} + 9 = 0

or, (3^{a})^{2} - 10 x 3^{a} + 9 = 0

Let, 3^{a} = x, we get

X^{2} - 10x + 9 = 0

x^{2} - 9x -x + 9 = 0

or, x(x - 9) - 1(x - 9) = 0

or, (x - 9) (x - 1) = 0

Now, substituting the value of x, we get

(3^{a} - 9) (3^{a} - 1) = 0

Either, 3 or, 3 or, 3 or, a = 2 |
Or, 3 or, 3 or, 3 or, a = 0 |

Therefore a = 0 or 2.

Prove that the angle inscribed in a semi-circle is a right angle.

Given: O is the center of the circle, AB is diameter and ACB is an angle inscribed in a semi-circle.

To prove: ACB = 90°

S.N | Statements | Reasons |

1 | ACB = 1/2 AOB | ACB is a circumference angle and AOB is a central angle standing on the same arc as ADB. |

2 | AOB = 180° | A straight angle. |

3 | ACB = (1/2)*180 = 90° | From statements (1) and (2), by substitution axiom |

4 | Therefore, ACB is a right angle. | From statement 3. |

Construct a quadrilateral having AB = 4.5cm, Bc = 5.5cm, CD=5.7cm, DA=4.9cm and diagonal BD=5.9cm. Also construct a triangle DAE whose area is equal to given quarilateral.

Given,

In quadrilateral ABCD

AB = 4.5cm

BC = 5.5cm

CD = 5.7cm

DA = 4.9cm

and diagonal BD = 5.9cm

To Construct: Quadrilateral ABCD of given data and triangle DAE whose area is equal to the area of the given quadrilateral ABCD.

Hence,

Quadrilateral ABCD of given data and triangle DAE equal in area to the quadrilateral ABCD is constructed.

Verify experimentally that the opposite angles of a cyclic quadrilateral are supplementary.

**Statement:** The opposite angles of a cyclic quadrilateral are supplementary.

**Construction of figures:**

Two circles of different radii with center 'O' are drawn. In each circle, a cyclic quadrilateral ABCD is drawn.

**To verify: **

i) ABC + ADC = 180°

ii)BAD + BCD = 180°

**Method of verification:**

By using a protractor, ABC, ADC, BAD, and BCD are measured and the results are tabulated below.

Figure | ABC | ADC | ABC + ADC | BAD | BCD | BAD + BCD | Results |

(i) | 78° | 102° | 180° | 79° | 101° | 180° |
ABC+ADC=180° BAD+BCD=190° |

(ii) | 90° | 90° | 180° | 84° | 96° | 180° |
ABC+ADC=180° BAD+BCD=190° |

Hence, The opposite angles of a cyclic quadrilateral are supplementary.

**Verified**

The shadow of a tower is formed on the ground when the angle made by the sun's ray with the ground is 30°. If the height of the tower is 50m, find the length of the shadow of the tower.

Let, AB be the height of a tower, BC be the length of the shadow of the tower and ACB be the angle made by the sun's rays with the ground.

Then, by the question

AB = 50m, ACB = 30°, BC = ?

Here, in right-angled triangle ABC

Tan 30° = AB/BC

or, \(\frac1{\sqrt3}\) = \(\frac{50m}{BC}\)

or, BC = \(50\sqrt3m\)

or, BC = 5- x 1.732m

or, BC = 86.6m

Therefore, the length of the shadow of the tower is 86.6m.

Find the mean of the data given below.

Marks obtained | 5 - 15 | 15 - 25 | 25 - 35 | 35 - 45 | 45 - 55 |

No. of students | 3 | 5 | 9 | 3 | 2 |

Calculate the difference between half yearly compound interest and simple interest of Rs. 8000 for 1(1/2) years at the rate of 10% per annum.

Given,

Principal (P) = Rs.8000

Rate of interest (R) = 10% p.a

Time (T) = 1(1/2) years = 3/2 years

To find C. I compounded half-yearly- Simple interest.

By the formula,

Simple interest (SI) = \(\frac{P\times T\times R}{100}\)

\(\frac{8000\times{\displaystyle\frac32}\times10}{100}\)

= Rs. 120000/100

= Rs. 1200

Again by formula,

CI compounded half yearly = \(p\left\{(1+\frac R{200})^{2T}-1\right\}\)

= Rs. 8000 \(\left\{(1+\frac{10}{200})^{2\frac32}-1\right\}\)

= Rs. 8000 {(1.05)^{3} - 1}

= Rs. 8000 (1.15-1)

= Rs. 8000 x 0.15

= Rs. 1261

Therefore, the difference between C.I. compounded half-yearly and simple interest = Rs.1261 - Rs.1200

Rs.61. Ans

Given solid is made of cone and cylinder. The base area of the cylinder is 100 sq. cm and the height of the cylinder is 3cm. If the volume of the whole solid is 600 cubic cm, find the height of the whole solid.

Given,

In a combined solid,

Area of the circular base = 100cm^{2
}Height of the cylinder (h_{1}) = 3cm

Volume of the combined solid (V) = 600cm^{2}

To find: Height of the combined solid (h)

By formula,

The volume of the cylinder (V_{1}) = area of the base x height

= 100cm^{2} x 3cm

= 300cm^{3}

Therefore volume of the come (V_{2}) = combined volume - volume of cylinder

= 600cm^{3} - 300cm^{3}

= 300cm^{3}

Again, by formula

Volume of the cone (V_{2}) = 1/3 area of base x height

or, 300cm^{3} = (1/3) x 100cm^{2} x h_{2}

or, 900cm^{3} / 100cm^{2} = h_{2}

or, h_{2} = 9cm

Therefore, the volume of cone (h_{2}) = 9cm

Now, h = h_{1} + h_{2}

= 3cm + 9cm

= 12cm

∴ The height of the combined solid (h) = 12cm.

The area of a rectangular room is 45 meter square. If the length of the room is 3m less and breadth is 1m more, it will be a square room, find the length and the breadth of the room.

Let, length of the rectangular room (l) = x m

and breadth of the room (b) = y m

Here,

According to the first condition,

xy = 45 ..........(i)

Again, according to the second condition,

x - 3 = y + 1

x = y + 1 + 3

x = y + 4 ........(ii)

Putting the value of x from equation(ii) and equation (i), we get

(y + 4)y = 45

or, y^{2} + 4y - 45 = 0

or, y^{2} + (9-5)y - 45 = 0

or, y^{2} + 9y - 5y - 45 = 0

or, y(y + 9) - 5(y + 9) = 0

or, (y + 9) (y - 5) = 0

Either y + 9 = 0 ..........(iii)

Or, y - 5 = 0 ..........(iv)

From equation (iii)

y + 9 = 0

or, y = -9 which is impossible because length and breadth cannot be negative.

or, y = 5

Now, putting the value of y in equation (ii), we got,

x = 5 + 4

= 9

∴ Length of the room (l) = 9m and breadth of the room (b) = 5cm.

In the given figure, AD//BC. If the areas of the triangle ABE and Triangle ACF are equal, then prove that EF//AC.

**Given,**

AD//BC and area of

Triangle ABE = area of triangle ACF

**To prove:** EF//AC

**Construction:** EC is joined.

**Proof:**

S.N | Statements | Reasons |

1 | Triangle ABE = Triangle ACE | Being triangles on the same base and between the same parallel lines. |

2 | Triangle ABE = Triangle ACF | Given. |

3 | Triangle ACE = Triangle ACF | From statements (1) and (2), by equal quantity axiom. |

4 | ∴ EF//AC | From statement (3), being triangles on the same line segment and towards the same side of it equal in area. |

**Verified.**

- 2024-06-12 03:15
- 15 Mins
- 12 Enrolled

- 25 Full Marks
- 10 Pass Makrs
- 25 Questions

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