see_model_question_set_2078_20210118204838

SEE Model Set

Compulsory Mathematics

Time: 3 hrs. Maximum Marks: 100

Candidates are required to answer in their own words as far as practicable. Credit shall be

given to originality in expression, creativity and neatness in hand, not to rote learning.

Attempt all the questions

Group- A 61 = 6

1. a) If the initial price of a computer is Rs X and annual rate of depreciation is D%,
b) what will be the value of the computer after K years? Write it.
What is the volume of a cricket ball having radius R cm?
2. a) What does ‘n’ denote in the surd n√a?
b) If L = lower limit of median class, N = sum of frequencies, f = frequency of
median class, c.f. = cumulative frequency of pre-median class and h= width of
3. a) median class, write the formula for calculating the median.
If ABC and BCD with equal areas stand on the same base BC and lie between
the lines AD and BC. Write down the relation between the lines AD and BC.

b) In the figure, write down the name of the cyclic quadrilateral.

Group- B 172= 34

4. a) One European Euro has buying rate NRs. 113.48 and selling rate NRs. 114.12.
How much profit does a bank make in buying and selling 900 Euro? Find.

b) The population of a rural municipality in the beginning of B.S. 2077 was 8,800
and the population the place has been increasing 3% by birth and 2% by
immigration every year since B.S. 2077. What will be the population of the place
at the end of B.S. 2078?

5. a) A hemispherical ‘Damaha’ has diameter 28 cm. How much plastic
b) is required to cover it tightly? Find.
c) The internal radius of an ice-cream cone is 3.5 cm and height is 9 cm. How many
cubic centimetres of ice-cream can it hold? Find.
Find the area of rectangular surfaces of the triangular prism given
alongside.

8x−2x+1
6. a) Evaluate: 4x.2x−1−2x

b) Simplify: 3√56a7b4 ÷ 189a4b7

7. a) Test whether the equation √2x − 1 = −3 has a unique solution or not.
b)
Simplify: m + n
c) mn−n2 mn−m2
8. a)
Simplify: a2+3ax−4x2 − 2ax
a2−16x2 2a2−8ax

In the given figure, if AB = 15 cm, BE = 14 cm and A D

AE = 13 cm, what is the area of parallelogram E
ABCD? Find it. C

B

b) In the given figure, O is the centre of circle. If

BCD = 1260, find the measures of BAD and ABD.

c) In the figure, O is centre of circle and TAN is the tangent to the
circle. If OM = 3cm and MN = 2 cm, find the length of AN.

9. a) In the given  PQR, PS = SQ. If QR = 8cm, SR = 9cm and
QRS = 300, find the area of SQR and PQR.

b) In a continuous data, if fm = 20a2 + 60a – 100 and f = a2 + 3a – 5, find the
mean.

10. a) Find the probability of occurring squared number or cubed number while
b)
drawing a flash card from the set of cards numbered from 3 to 32.

A bag contains 1 yellow, 1 black and 1 green ball of same shape and size. Two
balls are drawn at randomly one after another without replacement; show the
probabilities of all possible outcomes in a tree-diagram. Also, write the sample
space.

Group- C 104= 40

11. In a class of 80 students, 30 students liked cricket but not badminton and 35
liked badminton but not cricket. If the number of students who liked both the
games is twice the number of students who did not like both the games, find the
number of students who liked at most one games by drawing a Venn-diagram.

12. The price tagged on the laptop is Rs 78,000. If a customer paid Rs 74,919 for the

laptop with 13% value added tax after certain discount, how much discount did

he get? Also, find the discount percentage.

13. An umbrella is made by stitching 10 triangular pieces of cloth of two different

colours, each piece measuring 15 cm, 41 cm and 28 cm. How much cloth is

required for the umbrella? Find the total cost of cloth for the umbrella at the rate

of 50 paisa per square centimetre.

14. Find the H.C. F of 8x3 - 1, 4x3 – 2x2 – x – 1 and 16x4 + 4x2 + 1

15. A year hence, a mother will be 5 times as old as her son. Two years ago, she was

three times as old as her son will be four year hence. Find their present ages.

16. Prove that the rectangle PQRS and parallelogram AQRB standing on the same

base QR and lying between the same parallels PB and QR are equal in area.

17. Construct a quadrilateral ABCD in which AB = BC = 5.5 cm, CD = DA = 4.5 cm

and A = 600. Also construct ADE equal in area to the quadrilateral ABCD.

18. Explore experimentally the relationship between the inscribed angle and the

central angle of a circle standing on the same arc. (Two circles with radii at least

3 cm are necessary)

19. A man of height 5.3 ft observes a tree 42 ft high situated in front of him and finds

the angle of elevation of the top of the tree to be 300. How far is the man from the

foot of the tree? Find it

20. Calculate the third quartile from the data given below.

Marks obtained 10-25 10-40 10-55 10-70 10-85 10-100

No. of students 5 8 14 20 32 40

Group- D 45 = 20

21. NBL slightly changes its policy and decides to pay half-yearly interest at the
following rates for the deposits in the Fixed Account.

Deposit Time Rate of interest p.a. Minimum balance required
Up to 6 Months 6.75% Rs 25,000
Above 6 Months up to 1 Year 7.25% Rs 25,000
Above 1 Year up to 5 Years 7.5% Rs 25,000

Mr. Pandey gets the above information. He borrows a loan of Rs 2,00,000 from
Mr. Shakya at the rate of 5% simple interest for 2 years and immediately

deposits the same sum for the same duration of time in his Fixed Account of
NBL.
(i) Calculate the amount that Mr. Pandey accumulated in 2 years.
(ii) Calculate the amount that Mr. Pandey had to pay to Mr. Shakya?
(iii) How much profit did Mr. Pandey make at the end of 2 years?
22. A cylindrical tin of 40cm high and base radius 14 cm is
completely filled with the cement. For the cement work of
wall, a mason pours the cement on the ground from the tin
and finds a conical heap of height 30 cm.
(i) Calculate the base radius of the heap of the cement.
(ii) Find the surface area of heap.
(iii) If 1 cm3 equals to 2.5 g, find the mass of the cement.
23. The area of a window is 4x square meter. The curtain of size 3 m by
2x m is used in the window. If the area of the curtain is 2 square
meters more than the area of window, find the possible areas of the
window.
24. A farmer has a piece of field ABCDE in pentagonal shape. If he divides his land
into three triangular pieces ABC, ACD and ADE such that BE// CD, BC//AD and
AC//DE for vegetable farming. Show that the triangular pieces ABC and ADE of
the filed are equal in area.

The End

Marking Scheme

Group-A (6  1 = 6)

1. a) DK 1
b) Rs X (1 − 100) 1
4
1
R3 cm3 1
3 1
1
2. a) Order of surd
1
b) Md = L + N − c.f. × h 1
3. a) 2 f 1
1
AD//BC 1
1
b) BCDE 1
1
Group-B (17  2 = 34)
1
4. a) C. P. of 900 Euro in NRs = NRs 1,02,132 1
b) 1
S. P. of 900 Euro in NRs = NRs 1,02,708
5. a) 1
b) Profit = NRs 576
c)
PT = P (1 + R T = 8800 (1 + 5 2
6. a)
100 ) 100 )

= 9,702

TSA = 3r2 = 3 × 22 × (14cm)2
7

= 1,848cm2

Volume of cone = 1 = 1 × 22 × (3.5 cm)2 × 9 cm
r2h
3 37

= 115.5 cu. Cm

By Pythagoras theorem, base (b) = 3.5cm
Perimeter of base (Pb) = 12 cm + 12.5 + 3.5 = 28 cm

Area of rectangular surfaces = 28  20 = 560 cm2

23x−2x×2
22x×22x−2x

2x(22x−2)
2x(222x−1 ) = 2

b) 3 8a3 1

7. a) √ 27b3 1
b)
2a 1
c) 3b 1
1
8. a) √2x − 1 = −3 or, 2x – 1 = 9 x = 5
b) 1
c) Checking, 3 = -3 (Not true) 1
mn
9. a) 1
b) n(m − n) − m(m − n) 1
1
10. a) m2 − n2 m + n 1
b) mn(m − n) = mn 1
1
(a + 4x)(a − x) 2ax 1
(a + 4x)(a − 4x) − 2a(a − 4x) 1
1
a − x − x a − 2x
a − 4x = a − 4x 1

Area of  ABE =√21(21 − 13)(21 − 14)(21 − 15) = 84cm2 1

Area of parallelogram ABCD = 2×84 cm2 = 168cm2 1
1
BAD + 1260 = 1800 BAD = 540
1+1
ABD + 540 = 900 ABD = 360

OA = OM = 3cm , ON = 5cm

Using Pythagoras theorem, AN = 4cm

1

Area of SQR =  8cm 9cm sin300 = 18cm2

2

Area of PQR = 2×SQR = 2×18cm2 = 36cm2

∑fm 20a2 + 60a – 100
Mean = ∑f = a2 + 3a – 5

X̅ 20(a2+ 3a – 5) = 20
= a2 + 3a – 5

n (S) = (32 – 3) + 1 = 30, n (Sq) = 4 , n (Cu) = 2

P(Sq or Cu) = 4 + 2 = 1
30 30 5

Correct tree-diagram with probabilities and sample space

Group-C (10  4 = 40)

11. Suppose, n (C̅̅̅∪̅̅̅B̅) = x and n (C  B) = 2x

Presentation in Venn-diagram 1
From Venn-diagram or using formula, n (C̅̅̅∪̅̅̅B̅) = 5 1

n (C  B) = 10 1
no (C) + no (B) + n (C̅̅̅∪̅̅̅B̅) = 30+35 + 5 =70 1
OR, n (̅C̅̅∩̅̅̅B̅) = 80 − 10 = 70

12. Let the S.P. after discount be Rs x

Then, S.P. with VAT = x + 13% of x 1

Rs 74,919= 1.13 x x = Rs 66,300 1

Discount = M.P. – S.P. = Rs 78,000 – Rs 66,300 = Rs 11,700 1
Rs 11700 1

Discount percentage = Rs 78000 × 100% = 15%

(Give full marks for other relevant methods)

13. Area of each piece of cloth =√42(42 − 15)(42 − 41)(42 − 28) 1

= 126cm2 1

Area of 10 pieces of cloth =10× 126cm2 = 1260 cm2 1

Cost of cloth for umbrella =1260 ×Rs 0.50 = Rs 630 1

14. 1st expression = 8x3 - 1 = (2x – 1) (4x2 + 2x + 1) 1

2nd expression = 4x3 – 2x2 – x – 1

= x3 – 1 + 3x3 – 2x2 – x

= (x – 1) (x2 + x + 1) + x (3x2 – 2x – 1)

= (x – 1) (x2 + x + 1) + x (3x2 – 3x + x – 1)

= (x – 1) (x2 + x + 1) + x {(3x(x – 1) + 1 (x – 1)}

= (x – 1) (x2 + x + 1) + x (x – 1) (3x +1)

= (x – 1) (x2 + x + 1+3x2 + x)

= (x – 1) (4x2 + 2x + 1) 1

3rd expression = 16x4 + 4x2 + 1

= (4x2)2 + (1)2 + 4x2

=(4x2 + 1)2 – 4x2

=(4x2 +2x + 1)(4x2 – 2x + 1) 1

H.C.F. = 4x2 + 2x + 1 1

15. Let the present ages of mother and her son be x years and y

years respectively.

Case I : x + 1 = 5 (y + 1)

or, x = 5y + 4 … (i) 1

Case II: (x – 2) = 3 (y + 4)

or, x = 3y + 14 …(ii) 1

For solving (i) and (ii), y = 5 years, x = 29 years 1+1

16. Correct figure + description 1
1
PAQ SRB with reasons 1
1
PAQ + Trap. AQRS = SRB + Trap. AQRS 1
1
Rectangle PQRS = Parallelogram AQRB 1
1
17. Rough sketch 1
1+1
Construction of quadrilateral ABCD 1
1
Drawing DB//CE 1
1
Joining DE and conclusion 1

18. Correct figures 1
1
Table with correct measurements 1
1
Conclusion

19. Correct figure + Description

In right angled triangle, p = 42 ft – 5.3 ft = 36.7 ft

36.7

tan300 =

x

x = 63.56 ft.

20. Construction of correct cumulative frequency table

Q3 class = (70 – 85)

Q3 = 70 + 30 − 20 × 15
12

Q3 =82.5

Group-D (4 × 5 = 20)

21. (i) C.A. for the 1st 6 month = Rs 2,00,000 (1 + 6.75)1 1
1
200

= Rs 2,06,750
C.A. for the 2nd 6 month = Rs 2,06,750 (1 + 7.25)1

200

= Rs 2,14,244.69

C.A. for the 2nd year = Rs 2,14,244.69 (1 + 7.5 2×1

200 )

1

= Rs 2,30,614.32

Mr. Pandey accumulated Rs Rs 2,30,614.32 amount at the

end of 2 years. 1

(ii) S.I. at the rate of 5% = PTR = Rs 2,00,000×2×5 = Rs 20,000

100 100

The amount that Mr. Pandey had to pay to Mr. Shakya was

Rs 2,00,000 + Rs 20,000 = Rs 2,20,000 1
(ii) Profit = Rs 2,30,614.32 – Rs 2,20,000 = Rs 10, 614.32

22. (i) Volume of cement in the tin = πr2h = 22 × (14cm)2 × 40cm 1

7

= 24,640 cm3

Volume of conical heap = Volume of cement in the tin

or, 1 πr2h = 24,640 cm3 1

3

or, 1 × 22 × r2 × 30cm = 24,640 cm3

37

or, r = 28 cm

(ii) Slant height (l) = √h2 + r2 = √302 + 282 = 41.04 cm 1

22

Surface area of heap =πr = × 28 cm × 41.04 cm

7

= 3,611.52cm2 1

(iii) 1 cm3 = 2.5 g

Mass of cement = 24,640 × 2.5 g = 61,600 g = 61.6 kg 1

23. Area of a window = 4x sq. m

Area of a curtain = 3×2x sq. m 1

By question, 3×2x = 4x + 2

or, 3×2x = (2x)2 + 2 1

Let, 2x = a then 3a = a2 + 2

or, a2 – 3a + 2 = 0

or, a2 – 2a – a + 2 = 0

or, a (a – 2) – 1(a – 2)= 0 1

or, (a – 2) (a – 1) = 0

Either, a – 2 = 0 or, a = 2 … (i)

OR, a – 1 = 0 or, a = 1 … (ii) 1
From (i), a = 2 or, 2x = 2 x = 1
From (ii), a = 1 or, 2x = 1 or, 2x = 20 x = 0 1
When x = 1, the area of window = 41 sq. m = 4 sq. m 1
When x = 0, the area of window = 40 sq. m = 1 sq. m 1
24. Correct figure + Given + To prove
1
1
ABC= 2 BCDF 1
1
1
AED= 2 CDEG
 BCDF =  CDEG
ABC = AED

The End