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SEE Compulsory Math Question Paper With Solution

SEE 2077 (2021)

Write down the relationship between the area of the square and a triangle standing on the same base CD and between the same parallel lines CD and EF.

The marked price of a radio is Rs. 8000. If it is sold at a discount of 10%, how much a buyer needs to pay for the radio.

Find the compound interest compounded annually on Rs. 5000 at the rate of 12% p.a. for 2 years.

Given,

Principal (P) = Rs. 5000

Rate of interest (R) = 12% p.a.

Time (T) = 2years

To find Compound interest (C.I)

By formula,

\(C.I\;=\;p\left\{(1\;+\;\frac R{100})^T-1\right\}\)

= \(Rs.5000\left\{(1\;+\;\frac{12}{100})^2-1\right\}\)

=Rs. 5000 {(1 + 0.12)^{2} -1}

=Rs. 5000 (1.2544 - 1}

=Rs. 5000 x 0.2544

=Rs. 1272.

Therefore, Compound interest in 2 years (C.I.) = Rs.1272.

If the length of each of two equal sides of an isosceles triangle is 4cm and the base is \(\sqrt[4]2\). Find the area of the triangle.

Given,

In an isosceles triangle,

Length of each of two equal sides (a) = 4cm

and length of the base (b) = \(\sqrt[4]2\;cm\)

To find: Area of a triangle (A)

By formula,

Area of an isosceles triangle (A) = \(\frac B4\sqrt{4a^2-b^2}\)

= \(\frac{\sqrt[4]2}4\times\sqrt{4\times(4cm)^2-(\sqrt[4]2cm)^2}\)

= \(\sqrt2cm\times\sqrt{64cm^2-32cm^2}\)

= \(\sqrt2cm\times\sqrt{32cm^2}\)

= \(\sqrt2cm\times\sqrt[4]2cm\)

= 8cm^{2}

Therefore, the Area of the given triangle (A) = 8cm^{2}

The given figure is an isosceles triangular-based prism.

If AB = AC = 10cm, BC = 12cm and the height of the prism CC' = 30cm, find the volume of the prism

.

Given:

In a triangular based prism,

AB = AC = 10cm, BC = 12cm

and height of the prism CC' = 30cm

To find: Volume of the prism (V)

By formula,

Area of the isosceles triangular base = \(\frac b4\times\sqrt{4a^2-b^2}\)

= \(\frac{12cm}4\times\sqrt{4\times(10cm)^2-(12cm)^2}\)

= \(3cm\times\sqrt{400cm^2-144cm^2}\)

= \(3cm\times\sqrt{256cm^2}\)

= 3cm x 16cm

= 48cm^{2}

Again, by formula

Volume of prism (V) = Area of the base x height

= 48cm^{2} x 30cm

= 1440cm^{2}

The volume of a cylinder is 448π cu. Cm and height 7cm. Find its diameter.

Given,

The volume of a cylinder (V) = 448π cm^{3}

Height (h) = 7cm

To find: Diameter of the cylinder (d)

By formula,

or, Volume of a cylinder (V) = πr^{2} h

or, 448π cm^{3} = πr^{3} x 7cm

or, r^{2} = \(\frac{448\pi\;cm^3}{7\pi cm}\)

or, r^{2} = 64cm^{3}

or, r^{2} = (8cm)^{2}

or, r = 8cm

Therefore, diameter of the base (d) = 2r

= 2 x 8cm

= 16cm.

Rationalize the denominator of: \(\frac{\sqrt5+\sqrt4}{\sqrt5-\sqrt4}\)

Here,

\(\frac{\sqrt5+\sqrt4}{\sqrt5-\sqrt4}\)

= \(\frac{\sqrt5+2}{\sqrt5-2}\)

Multiplying both numerator and denominator by \(\sqrt5+2\)

\(\frac{\sqrt5+2}{\sqrt5-2}\times\frac{\displaystyle\sqrt5+2}{\displaystyle\sqrt5+2}\)

= \(\frac{(\sqrt5+2)^2}{(\sqrt5)^2-\;(2)^2}\)

= \(\frac{(\sqrt5)^2+2.\sqrt5.2+(2)^2}{5-4}\)

= \(\frac{5+4\sqrt5+4}1\)

= \(9+4\sqrt5\)

The age of a daughter is one-sixth of the age of her mother. If the difference between their ages is 35 years, find the mother's age.

In the given figure, M is the mid-point of AB and AD⊥BD. If BC = 6cm and AD = 8cm, find the area.

Given,

M is the midpoint of AB, AD⊥BD,

BC = 6cm and AD = 8cm.

To find, Area of triangle AMC

Here,

(i) Area of triangle ABC = \(\frac12\) BC x AD {Area of a triangle = \(\frac12\) base x height}

= \(\frac12\) x 6cm x 8cm

= 24cm^{2}

(ii) Area of triangle AMC = \(\frac12\) area of triangle ABC {Median CM bisects triangle ABC}

= \(\frac12\) x 24cm^{2}

Therefore area of triangle AMC = 12cm

In the given figure, ABCD is a parallelogram and ABCE is a cyclic quadrilateral. If ∠ADC=52°, find the measure of ∠AEC.

Given, ABCD is a parallelogram, ABCE is a cyclic quadrilateral and ADC = 52°

To find: The measure of AEC

Here,

(I) ABC = ADC {Being the opposite angles of a parallelogram}

52°

(II) ABC + AEC = 180° {Being opposite angles of a cyclic quardrilateral)

or, 52° + AEC = 180°

or, AEC = 180° - 52°

or, AEC = 128°

In the figure, L is the center of the circle, MN is the diameter and M is the point of contact. If MCL=55°, find the measure of CLN.

Given,

L is the centre of the circle, MN is the diameter,

M is the point of contact and MCL = 55°

To find, the measure of CLN

Here,

(I) LMC = 90° {Diameter drawn from the point of contact to a circle is perpendicular to the tangent of the circle}

(II) CLN = LMC + MCL {The exterior angle of a triangle is equal to the sum of the non-adjacent interior angles of the triangle}

or, CLN = 90° + 55°

=145°

The measure of CLN is 145°.

In triangle ABC, the sides a = 4cm, b = 6cm and c = 60°, find the area of the given triangle.

In a continuous series, mean=24, the number of terms=10, and ∑FX=200+a find the value of a.

A fair coin is tossed and at the same time a fair dice is rolled. What is the probability of getting head in coin and 5 in dice? Find it.

Let, the event of getting head in a coin be H and the event of getting 5 in a die be F.

Then, P(H) = \(\frac12\) and P(F) = \(\frac16\)

To find, P(H∩F)

Here, H and F are independent events.

Then, by formula

P(H∩F) = P(H) x P(F)

= \(\frac12\) x \(\frac16\)

= \(\frac1{12}\)

The probability of getting head in coin and getting 5 in dice is \(\frac1{12}\)

Two children were given birth by a married couple. Represent the probability of giving birth to at least a daughter in a tree diagram.

Let, the event of giving birth to a son be B.

And the event of giving birth to a girl be G.

Then, P(B) = \(\frac12\) and P(G) = \(\frac12\)

A probability tree diagram for giving birth to two children by a married couple is given below.

**Possible Outcomes**

To find, \(P(\overline{B\cap B)}\)

Here, \(P(\overline{B\cap B)}\) = P(B) x P(B) = \(\frac12\times\frac14=\frac14\)

and \(P(\overline{B\cap B)}\) = 1 - \(P(\overline{B\cap B)}\) = \(1-\frac14\)

= \(\frac34\)

Therefore, the probability of giving birth to at least a daughter is \(\frac34\)

In a survey of 400 people, 300 people drink only one drink out of tea and coffee, 50 people drink non of them. Find the number of people who drink both the drinks.

Let, the set of people who drink tea be T

And the set of people who drink coffee be C.

Then, n_{o} (T) + n_{o} (C) = 300,

\(n(\overline{T\cup C})\) = 50 and n(U) = 400

to find, n(T∪C)

By formula,

n_{o} (T) + n_{o} (C) + n(T∩C) + \(n(\overline{T\cup C})\) = n(U)

or, 300 + n(T∩C) + 50 = 400

or, n(T∩C) + 350 = 400

or, n(T∩C) = 50

Therefore, the number of people who drink both the drinks is 50.

A cycle was sold after allowing 20% discount on the marked price and levying 10% VAT. If the discount amount is Rs. 750, how much VAT amount was levied on the price of the cycle.

Given,

Discount percent = 20%

VAT percent = 10%

Amount of discount = Rs. 750

To find, Amount of VAT

Let, marked price (M.P) = X

By formula,

Amount of discount = discount % of M.P.

or, Rs. 750 = 20% of X

or, Rs. 750 = \(\frac{\displaystyle20}{100}\times X\)

or, Rs. 750 = \(\frac x5\)

or, x = Rs. 3750

Marked price (x) = Rs. 3750

Here,

Selling price = M.P. - amount of discount = Rs. 3750 - Rs. 750

= Rs. 3000

Again, by formula

Amount of VAT = VAT% of S.P

= 10% of Rs.3000

= \(\frac{10}{100}\) x Rs. 3000

= Rs. 300

Therefore, the amount of VAT is Rs. 300.

If the sum of two sides containing the right angle of a right-angled triangle is 7cm and its area is 6cm^{2} , find the perimeter of the triangle.

Let, the length of the sides containing the right angle be P and B and the hypotenuse be h.

Now, by the question

P + B = 7cm

And area of the triangle (A) = 6cm^{2}

To find: Perimeter of the triangle (P)

Here, P + B = 7cm

or, B = 7cm - P......(I)

By formula,

Area of a right angle triangle = \(\frac12\) p.b

or, 6cm^{2} = \(\frac12\) p (7cm - p)

or, 12cm^{2} = 7cm. p - p^{2}

or, p^{2} - (4cm + 3cm). p + 12cm^{2} = 0

or, p^{2} - 4cm. p - 3cm. p + 12cm^{2} = 0

or, p(p - 4cm) - 3cm(p - 4cm) = 0

or, (p - 4cm) (p-3cm) = 0

**Either, p - 4cm = 0**

**Therefore, p = 4cm**

**Or,**

**p -3cm = 0**

**Therefore, p = 3cm**

When p = 4cm, then b = 7cm - 4cm = 3cm

and when p = 3cm, then b = 7cm - 3cm = 4cm

Now, by pythagoras theorem

h= \(\sqrt{p^2+b^2}\)

=\(\sqrt{(4cm)^2+(3cm)^2}\)

= \(\sqrt{16cm^2+9cm^2}\)

= \(\sqrt{25cm^2}\)

= 5cm

**Therefore, perimeter of the gievn right angle triangle (P) = 3cm + 4cm + 5cm**

Find the HCF of: (1-x^{2}) (1-y^{2})+4xy, 1-2x+y-x^{2}y+x^{2}

Here,

**First expression = (1-x ^{2}) (1-y^{2})+4xy**

= 1 - y^{2} - x^{2} + x^{2} y^{2} + 4xy

= x^{2} y^{2} + 2xy + 1 - x^{2} + 2xy - y^{2}

= (xy)^{2} + 2.xy.1 + (1)^{2} - (x^{2} - 2xy + y^{2} )

= (xy + 1)^{2} - (x-y)^{2}

= {(xy + 1) + (x-y)} {(xy + 1) - (x-y)}

= (xy + 1 + x - y) (xy - x + y + 1)

**Second expression = 1-2x+y-x ^{2}y+x^{2}**

= x^{2} - 2x + 1 - x^{2} y + y

= (x - 1)^{2} - y(x^{2} - 1)

= (x - 1)^{2} - y(x + 1) (x - 1)

= (x - 1) (x - 1 - xy - y)

= (x - 1) (x-1 - xy - y)

= -(xy - x + y + 1) (x - 1)

Therefore, H.C.F = (xy - x + y + 1)

Solve: \(\sqrt x+\;\sqrt{x+13}=\frac{91}{\sqrt{x+13}}\)

Here,

\(\sqrt x+\;\sqrt{x+13}=\frac{91}{\sqrt{x+13}}\)

or, \(\sqrt x\times\sqrt{x+13}+(\sqrt{x+13})^2\;=\;91\)

or, \(\sqrt{x(x+13})+(x+13)=91\)

or, \(\sqrt{x^2+13x}+x+13=91\)

or, \(\sqrt{x^2+13x}=91-x-13\)

or, \(\sqrt{x^2+13x}=78-x\)

Squaring both sides, we get

\(\sqrt[{}]{x^2+13x})^2=(78-x)^2\)

or, \(x^2+13x=(78)^2-2x78.x+x^2\)

or,\(x^2+13x=6084-156x+x^2\)

or, \(x^2+13x+156x-x^2=6084\)

or, 169x = 6084

or, x = \(\frac{6084}{169}\)

or, x = 36

Therefore the value of is 36.

Prove that the triangles on the same base and between the same parallel lines are equal in area.

**Given,**

Triangle ABC and DBC are standing on the same base BC and between the same parallel lines AD and BC.

**To prove:** Area of triangle ABC = Area of triangle DBC

**Construction**: Through point C, CE is drawn parallel to BA which meets AD at E.

**Proof:**

Statements |
Reasons |

ABCE is a parallelogram. |
By construction. |

Triangle ABC = \(\frac12\) parallelogram ABCE. | Diagonal AC bisects parallelogram ABCE. |

Triangle DBC = \(\frac12\) parallelogram ABCE. | Both the triangle and parallelogram are standing on the same base and between the same parallels. |

Therefore, Triangle ABC = Triangle DBC | From Statements(2) and (3), Halves of the same parallelogram. |

**Proved**

Construct a Triangle in which AB = 5cm, BC=7cm and ABC=60º. Also, construct another triangle DBC having CD 5.2cm.

Verify experimentally that the relationship between the center angle URN and circumference angle USN standing on the same are UN of a circle.(Two circles having radii at least 3 cm are necessary).

**Statement:** The central angle URN is double of the circumference angle USN standing on the same are UN of a circle.

**Construction of figures:**

Two circles of different radii with center R are drawn. In each circle, central angle URN and circumference angle USN are drawn which are standing on the same arc UN.

To verify: URN = 2 USN

Method of verification:

By using a protractor, URN and USN are measured and the results are tabulated below.

Figure | URN | USN | 2 USN | Results |

I | 94° | 47° | 94° | URN = 2 USN |

II | 86° | 43° | 86° | URN = 2 USN |

**Conclusion: **Hence, the central angle of a circle is double the circumference angle standing on the same arc.

**Verified**

A 1.5 Meters tall person is standing in front of a 41.5m high tree. When observing the top of the tree an angle of 45° is formed with the eyes. Find the distance between the tree and the person.

Let, AB be the height of a man, CD be the height of a tree,

AE = BD be the distance between the man and the tree,

and CAE is the angle of elevation of the top of the tree.

Then, by the question

ED = AB = 1.5m

CD = 41.5m, CAE = 45° ,BD = ?

Here, CE = CD - ED = 41.5m - 1.5m = 40m

In right angled triangle CEA

Tan 45° = \(\frac{CE}{AE}\)

or, 1 = \(\frac{40m}{BD}\)

or, BD = 40m

Therefore, the distance between the man and the tree (BD) = 40m

Calculate the first quartile from the data given below.

Class | 0 - 5 | 5 - 10 | 10 - 15 | 15 - 20 | 20 - 25 | 25 - 30 |

Frequency | 2 | 4 | 14 | 15 | 6 | 4 |

Preparing a cumulative frequency table from the given data, we get.

Class (X) |
Frequency(F) |
Cumulative frequency (C.F) |

0 - 5 | 2 | 2 |

5 - 10 | 4 | 6 |

10 - 15 | 14 | 20 |

15 - 20 | 15 | 35 |

20 - 25 | 6 | 41 |

25 - 30 | 4 | 45 |

N = 45 |

By formula,

First quartile (Q_{1}) = The value of \((\frac N4)^{th}\) item

=The value of \((\frac{45}4)^{th}\) item

= The value of 11.25^{th} item.

Here, C.F just greater than 11.25 is 20 and it lies in the class (10 - 15).

Therefore, First quartile class = (10 - 15)

Again, by formula

First quartile (Q_{1}) = \(L\;+\;(\frac{{\displaystyle\frac N4}-c.f}f)\times i\)

Where, L = 10, \(\frac N4\) = 11.25, c.f = 6, f = 14, i = 5

Therefore, Q_{1} = 10 + \(10\;+\;(\frac{11.25\;-\;6}{14})\times5\)

= \(10+\frac{5.25}{14}\times5\)

= 10 + 1.88

First quartile (Q_{1}) = 11.88

The difference between compound interest compounded annually and simple interest of a certain sum of money at the rate of 10% per annum in 3 years is Rs. 3875. Find the sum.

Given,

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

Time (T) = 3 years

Compound interest - simple interest (CI - SI) = Rs. 3875

To find: Principal (P)

By formula,

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

= \(\frac{P\times3\times10}{100}\)

= 0.3 P

Again, by formula

Compound interest (CI) = \(P\left\{(1+\frac R{100})^t-1\right\}\)

= \(P\left\{(1+\frac{10}{100})^3-1\right\}\)

= P{(1 + 0.1)^{3} - 1}

= P{(1.1)^{3} - 1}

= P(1.331 - 1)

= 0.331P

Now, by the question

CI - SI = Rs. 3875

or, 0.331P - 0.3P = Rs. 3875

or, 0.0031p = Rs. 3875

or, P = \(\frac{Rs.3875}{0.031}\)

or, P = Rs. 125000

Therefore, the required principal (P) is Rs. 125000.

Calculate the total surface area of the given combined solid object.

Given,

In a combined solid of a cone and hemisphere,

Height of the cone (h) = 24cm

and diameter of the circular base (d) = 14cm

To find, Total surface area of the solid (TSA)

Here,

The radius of the base (r) = 7cm

and slant height of the cone = \(\sqrt{r^2+h^2}\)

= \(\sqrt{(7)^2+(24)^2}\)

= \(\sqrt{49cm^2+576cm^2}\)

= \(\sqrt{625cm^2}\)

= 25 cm

Now by formula,

C.S.A of a conical part = Πrl

=\(\frac{22}7\times7cm\times25cm\)

= 550cm^{2}

and C.S.A of hemispherical part = 2Πr^{2}

= \(2\times\frac{22}7\times(7cm)^2\)

= \(\frac{44}7\times49\)

= 308cm^{2}

Therefore, the total surface area of the combined solid = C.S.A of conical part + C.S.A of hemispherical; part

= 550cm^{2} + 308cm^{2}

= 858cm^{2}

Find the HCF and LCM of x^{2} + 2xy + y^{2} -z^{2} , y^{2} + 2yz + z^{2} - x^{2} and z^{2} + 2zx + x^{2} - y^{2} .

Here,

First expression = x^{2} + 2xy + y^{2} -z^{2}

= (x + y)^{2} - z^{2}

= (x + y + z) (x + y + z)

Second expression = y^{2} + 2yz + z^{2} - x^{2}

= (y + z)^{2} - x^{2}

= (x + y + z) (y + z - y)

Third expression = z^{2} + 2zx + x^{2} - y^{2}

= (z + x)^{2} - y^{2}

= (x + y + z) (x - y + z)

Therefore, HCF = (x + y + z)

And LCM = (x + y + z) (x + y - z) (y + z - x) (x - y + z)

In the given figure, AD//BE//GF and AB//DG//EF. If the areas of parallelogram ABCD and parallelogram CEFG are equal, then prove that DE//BG.

**Given:** AD//BE//GF, AB//DG//EF

and parallelogram ABCD = Parallelogram CEFG

**To prove:** DE//BG

**Construction:** BD and GE are joined.

**Proof:**

Sn |
Statements |
Reasons |

1 | Triangle BCD = 1/2 parallelogram ABCD | Diagonal of a parallelogram bisects it. |

2 | Triangle GCE = 1/2 parallelogram CEFG | Same reason as (1). |

3 | Parallelogram ABCD = Parallelogram CEFG | Given |

4 | Triangle BCD = Triangle GCE | From statements 1, 2, and 3, both are halves of the equal parallelogram |

5 | Triangle BCD + Triangle BCG = Triangle GCE + Triangle BCG | Adding triangle GCE on both sides of statement 4. |

6 | Triangle BDG = Triangle BEG | From statement 5, by whole part axiom. |

7 | Therefore, DE//BG | From statements 6, being triangles on the same like segment and towards the same side of it equal in area. |

**Proved**

- 2022-03-20 12:45
- 60 Mins
- 12 Enrolled

- 25 Full Marks
- 10 Pass Makrs
- 25 Questions

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