1

SundarSTEM
Admission Test for Class – 9
Year 2022 Phase-1
(Part-I)
Duration: 60 Minutes

Part-I contains 10 Fill in the blanks and 20 multiple choice questions.

For Questions 1 to 10, write answer in the given box.
For Questions 11 to 30, encircle the correct option.
Total marks for Part-I are: 100. Marks of each question are given in the [ ] at the end of
each question.
Attempt all the questions.

Please note that there will be a negative marking of 1 mark for each wrong
multiple choice answer.
Part-I has 21 sheets including this one. Last two sheets are left blank for working.
Please fill the below details before proceeding to test.
Test ID: _______________ Date of Birth (dd/mm/yyyy): _______________
Name: ________________________________________________________
Test Center: ____________________________________________________
Instructions:

 All answers should be exact. Never use approximations. For example:

o DO NOT approximate √𝟐 = 𝟏. 𝟒𝟏𝟒𝟐. Just leave it as √𝟐 in the formula.
𝟐𝟐
o DO NOT approximate 𝝅 ≈ 𝒐𝒓 𝝅 ≈ 𝟑. 𝟏𝟒. Just leave it as 𝝅.
𝟕
 Simplify your answers.
 Calculator is not allowed in the test. If a student is making long, time-consuming calculations, he is probably on the wrong track.
 The working of candidate will be used in assessment of his problem solving approach.
 2

1. 25 × 83 × 162 as a power of 4 is

[3]
Solution:

25 × 29 × 28 = 25+9+8 = 222 = (22 )11 = 411

𝟒𝟏𝟏

2. Let r be a positive real number such that 𝑟 2 + 7𝑟 = 35.

What is the value of (r + 3)(r + 4)?

[3]
Solution:
(𝑟 + 3)(𝑟 + 4) = 𝑟 2 + 7𝑟 + 12 = 35 + 12 = 47

47
 3

7
3. = 0.538461538461538461 … … … … … … …,
13
58 th digit to the right of the decimal point will be

[3]
Solution:
7
= 0.538461538461538461 … … … … …
13
Digits after decimal point are being repeated after every 6 digits so

When we divide 58 by 6 the remainder is 4 so the 4th digit in the groups of 6 is 4 so

the answer is 4

4
2 3
63 × √67
4. Evaluate: 3 6 =?
√6

[3]
Solution:
2 7
63 × 63 2 7 6
6 = 63+3−3 = 61 = 6
63

6
 4

5. The average age of 5 people in a room is 30 years. An 18- year-old

person leaves the room. The average age of the four remaining people is

[3]

Solution:
Sum of ages of 5 people = 5 x 30 = 150
Sum of 4 people when 18 year old man left = 150 − 18 = 132
132
Average of remaining 4 people = = 33 𝑦𝑒𝑎𝑟𝑠
4

33

(6.4)2 −(5.4)2
6. = [3]
8.92 +(8.9×2.2)+1.12

Solution:
(6.4 − 5.4)(6.4 + 5.4) 1 × 11.8 11.8 11.8
= = = = 0.118
8.92 + 2 × 8.9 × 1.1 + 1.12 (8.9 + 1.1)2 102 100

0.118
 5

7. Let , , , and be whole numbers. If 2𝑤 × 3𝑥 × 5𝑦 × 7𝑧 = 588, then

=

[3]
Solution:
588 = 22 × 31 × 72
comparing indices of 2,3,5 and 7
W=2, x=1, y=0, and z=2
So 2w + 3x + 5y + 7z = 2(2) + 3(1) + 5(0) + 7(2)=21

21

8. =
[3]

Solution:
(1 − 2) + (3 − 4) + (5 − 6) … … . (2015 − 2016) + (2017 − 2018) + 2019

= ((−1) + (−1) + (−1) + ⋯ … … … … … 1009 𝑡𝑖𝑚𝑒𝑠) + 2019

=−1009 + 2019 = 1010

1010
 6

9. The tens digit of 72011 is

[3]

Solution:
70 = 01, 71 = 07,
72 = 49, 73 = 343,
74 = 2401, 75 = 16807,
76 = 117649, 77 = 823543
78 = 5764801, 79 = 40353607
710 = 282475249, 711 = 1977326743
712 = 13841287201, 713 = 96889010407

From above it can be observed (except text in red) that whenever the
index is not multiple of 4 (in blue) except "71 "the tens digit is 4.
So 2011 is not a multiple of 4 so the “tens digit” in the number
72011 will be 4.

4
 7

1 1
10. If 𝑥, 𝑦 be non-zero real numbers such that (x + ) (y + ) = 7. The
y x
1 1
value of (x 2 + 2) (y 2 + 2 ) is
y x

[3]

Solution:
1 1
(x + ) (y + ) = 7.
y x
𝑥𝑦 + 1 𝑥𝑦 + 1
( )( )=7
𝑦 𝑥
(𝑥𝑦 + 1)2
=7
𝑥𝑦
𝑥 2 𝑦 2 + 1 + 2𝑥𝑦
=7
𝑥𝑦
𝑥 2 𝑦 2 + 1 2𝑥𝑦
+ =7
𝑥𝑦 𝑥𝑦
𝑥 2𝑦2 + 1
+2=7
𝑥𝑦
𝑥 2𝑦2 + 1
=5
𝑥𝑦
2
1 1 (𝑥 2 𝑦 2 + 1) (𝑥 2 𝑦 2 + 1 (𝑥 2 𝑦 2 + 1
(x + 2 ) (y 2 + 2 ) = (
2
) ( ) = ( ) = 52 = 25
y x 𝑦2 𝑥2 𝑥𝑦

25
 8

11. For what value of x is 125 × 55 = 5x + 5x + 5x + 5x + 5x

[3]

Solution: 125 × 55 = 5x + 5x + 5x + 5x + 5x
53 × 55 = 5 × 5 𝑥
58 = 51+𝑥 (In multiplication if bases are same, indices are added)
1+𝑥 =8
𝑥=7
a. 8

b.7
c. 9
d. 11
 9

12. In the diagram below, a diameter of each of the two smaller circles is a radius of the
larger circle. If the two smaller circles have a combined area of square unit, then what
is the area of the shaded region, in square units?

[4]

Solution:
Let radius of smaller circle = r units
Area of two smaller circles = 1 square unit

2𝜋𝑟 2 = 1
1
𝑟=√
2𝜋

1
𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑏𝑖𝑔𝑔𝑒𝑟 𝑐𝑖𝑟𝑐𝑙𝑒 = 2𝑟 = 2√
2𝜋
2
1
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑏𝑖𝑔𝑔𝑒𝑟 𝑐𝑖𝑟𝑐𝑙𝑒 = 𝜋 (2√ )
2𝜋
1
= 𝜋4 (2𝜋) = 2 square units
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠ℎ𝑎𝑑𝑒𝑑 𝑟𝑒𝑔𝑖𝑜𝑛 = 2 − 1 = 1square units
1
a. 4
1
b. 3
1
c. 2
d. 1
π
e.
2
 10

13. An ant walks over the lines in the figure given below and takes a
shortest route from A to B. How many routes are possible for the ant?
[3]

a. 7
b. 12
c. 18
d. 20
 11

14. In square ABCE, AF = 2FE and CD = 2DE. What is the ratio of the
area of to the area of square ABCE?

[4]

Solution:
Area of square ABCE=(3x)(3y)=9xy

Area of triangles ABF, FED and BCD=3xy+3xy+xy=7xy

Area of triangle BFD=Area of square ABCE - Area of triangles ABF, FED and BCD
Area of triangle BFD=9xy – 7xy=2xy
2𝑥𝑦 2
Required ratio=9𝑥𝑦 = 9

1
a.
6
2
b.
9
5
c.
18
1
d.
3
7
e.
20
 12

15. In the figure, the outer equilateral triangle has area16, the inner
equilateral triangle has area 1, and the three trapezoids are congruent.
What is the area of one of the trapezoids?

[3]

Solution:
Area of three trapezoids=Area of outer triangle – Area of inner triangle
=16 – 1 = 15
Area of one trapezoid = 15/3 =5

a. 3
b. 4
c. 5
d. 6
e. 7
 13

16. Two circles that share the same center have radii meters
and meters. A mouse runs along the path shown, starting at and ending
at . How many meters does the mouse run?

[4]
Solution:
1 1 1
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑏𝑦 𝑚𝑜𝑢𝑠𝑒 = × 2𝜋 × 20 + 10 + × 2𝜋 × 10 + 20 + × 𝜋 × 10 + 10 = 20𝜋 + 40
4 4 4

a. 10𝜋 + 20
b. 10𝜋 + 30
c. 10𝜋 + 40
d. 20𝜋 + 20
e. 20𝜋 + 40
 14

17. Let 𝐴𝐵𝐶 be an equilateral triangle and 𝐶𝐷𝐸𝐹 a square such that 𝐸 lies on
segment 𝐴𝐵 and 𝐹 on segment 𝐵𝐶. If the perimeter of the square equals 4,
what is the perimeter of triangle 𝐴𝐵𝐶?

Solution:
𝐹𝐶 = 𝐹𝐸 = 1
𝐼𝑛 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒𝑑 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐵𝐹𝐸, 𝑎𝑛𝑔𝑙𝑒 𝐵 = 60°, EF=1.
𝐸𝐹 1
𝑆𝑜 sin(60°) = =
𝐸𝐵 𝐸𝐵
𝐸𝐵 = 2
𝐵𝑦 𝑃𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑢𝑠 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 𝐵𝐹 = √22 − 12 = √3
𝐵𝐶 = 1 + √3
𝑆𝑜 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐴𝐵𝐶 = 3(1 + √3) = 3 + 3√3

]4]

a. √3
1
b. 3√2 +
2
c. 2 + √3
d. 3 + 3√3
 15

18. The -digit number is divisible by . What is the remainder

when this number is divided by ?

[3]
Solution:
As 2018Uis divisible by 9 so sum of digit must also be divisible by 9.
2+0+1+8+U=11+U, So U=7. When 20187 is divided by 8 the remainder is
3.

a. 1
b. 3
c. 5
d. 6
e. 7

19.How many positive divisors of 220 321 also divides28 325 ?

‫ کے بھی مقسم ہیں۔‬28 325 ‫ کے کتنے ایسے مثبت مقسم ہیں جو‬220 321

[4]
Solution:
LCM of 220 321 𝑎𝑛𝑑 28 325 = 28 321
No of divisors of 28 321 = (8 + 1) × (21 + 1) = 9 × 22 = 198.

a. 168
b. 176
c. 189
d. 198
 16

20. What is the smallest positive integer which when multiplied to 244 + 64
makes the product a perfect square?

[3]
Solution:
244 + 64 = (23 × 31 )4 + 26 = 26 (26 34 + 1) = 26 (5185)
26 𝑖𝑠 𝑎 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑠𝑞𝑢𝑎𝑟𝑒, 𝑠𝑜 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 5185 𝑤𝑖𝑡ℎ 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟
𝑡𝑜 𝑚𝑎𝑘𝑒 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑝𝑒𝑟𝑓𝑒𝑐𝑡 𝑠𝑞𝑢𝑟𝑒.

a. 1037
b. 2074
c. 5185
d. 10370

21. [3]

Using set notation, the shaded region in the Venn diagram is represented by

a. 𝐴 ∪ (𝐵 𝑈 𝐶)
b. 𝐴 𝑈 (𝐵 ∪ 𝐶)/
c. 𝐴 𝑈 (𝐵 ∩ 𝐶)/
d. 𝐴 ∪ (𝐵 ∩ 𝐶)
e. None of the above.
 17

22. A regular hexagon is filled with small circles of the same size, as
illustrated in the figure. The circles can be tangent, but they do not overlap.
Exactly four circles fit next to each other along the side of the hexagon.
What is the maximum number of circles that fit in the hexagon in this way?

[3]

Solution:
Bottom row to Top row = 4+5+6+7+6+5+4 = 37
a. 30
b. 37
c. 39
d. 41
 18

23. The dice in the figure on the right has a 6 on the back (opposite the
1), a 5 on the bottom (opposite the 2), and a 4 on the left side (opposite the
3). The dice is tilted along the side of a 4 × 4 grid, on the little squares,
until it lies on the little square at point B. This can be done along the side of
the
grid via A, or along the side of the grid via C. How can the dice be
positioned once it has arrived at B?

]4]
a.

b.

c.

d.

24. Muhammad wants to cut out the figure below, make the folds and then the
collages to obtain a cube. Which of the cubes below will he get?
 19

[4]

a.

b.

c.

d.
 20

25. The factors of 1 − (a2 + b2 ) + a2 b2 are

‫ کی تجزی کریں۔‬1 − (a2 + b2 ) + a2 b2

[3]
Solution:
1 − (a2 + b2 ) + a2 b2 = 1 − 𝑎2 − 𝑏 2 + 𝑎2 𝑏 2 = (1 − 𝑎2 ) − 𝑏 2 (1 − 𝑎2 ) = (1 − 𝑎2 )(1 − 𝑏 2 )
= (1 − 𝑎)(1 + 𝑎)(1 − 𝑏)(1 + 𝑏)

a. (1 + 𝑎)(1 − 𝑎)(1 + 𝑏 2 )
b. (1 + 𝑎2 )(1 + 𝑏 2 )
c. (1 + 𝑎2 )(1 − 𝑏)(1 + 𝑏)
d. (1 − 𝑎)(1 + 𝑎)(1 − 𝑏)(1 + 𝑏)

26.

In the figure, QX and RX are the bisectors of angles Q and R respectively of ∆PQR. If
XS ⊥ QR and XT ⊥ PQ, then ∆XTQ ≅ ∆XSQ by

[4]

a. SAS Property
b. RHS Property
c. AAS Property
d. ASA Property
 21

27. Danial took two of his favorite positive integers. He realized that the
product of these numbers is 7times greater compared to their sum. What is
the difference the Danial’s two favorite numbers?

[4]
Solution:
Let Daniyal thinks of x and y, where x and y both are positive integers.
𝑥𝑦 = 7(𝑥 + 𝑦)
→ 𝑥𝑦 = 7𝑥 + 7𝑦
→ 𝑥𝑦 − 7𝑥 − 7𝑦 + 49 = 49 (𝑎𝑑𝑑𝑖𝑛𝑔 49 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠)
𝑥(𝑦 − 7) − 7(𝑦 − 7) = 49
(𝑥 − 7)(𝑦 − 7) = 49
𝑇ℎ𝑒𝑛 (𝑥 − 7, 𝑦 − 7) = (1,49), (49,1), (7,7), (−1, −49), (−49, −1), (−7, −7)
(𝑥, 𝑦) = (8,56), (56,8), (14,14), (6, −42), (−42,6), (0,0)
𝑎𝑠 𝑥 𝑎𝑛𝑑 𝑎𝑟𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑠𝑜 𝑑𝑖𝑠𝑐𝑎𝑟𝑑(6, −42)𝑎𝑛𝑑 (−42,6)𝑎𝑛𝑑 (0,0).
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑖𝑡ℎ𝑒𝑟 (8,56), (56,8)𝑜𝑟 (14,14)
𝑠𝑜 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑐𝑎𝑛 𝑒𝑖𝑡ℎ𝑒𝑟 𝑏𝑒 0 𝑜𝑟 48
𝑠𝑜 𝑡ℎ𝑒 𝑎𝑛𝑠𝑤𝑒𝑟 𝑖𝑠 48.

a. 56
b. 52
c. 48
d. 44
 22

28.

In the given figure, AD = BC and AC = BD. Then ∆PAB is

[3]
Solution:
In triangles ADB and ACB:
AD=BC, AC=BD, and AB=AB. So by SSS property Triangle ADB and triangle ACB are
congruent. So angle PAB and angle PBA are equal and triangle PAB is an isosceles
triangle.
a. equilateral
b. right angled
c. scalene
d. isosceles
 23

29.

How many pairs (𝑥, 𝑦) of positive integers are there such that
𝑦 𝑦+2
1 ≤ 𝑥 ≤ y ≤ 20 and both and are integers?
𝑥 𝑥+2

[3]
𝐼𝑓 𝑦 = 𝑥, 𝑤𝑒 ℎ𝑎𝑣𝑒 20 𝑠𝑙𝑜𝑢𝑡𝑖𝑜𝑛𝑠.
𝐴𝑠𝑠𝑢𝑚𝑒 𝑡ℎ𝑎𝑡 𝑥 < 𝑦. 𝑇ℎ𝑒𝑛 𝑥 𝑑𝑖𝑣𝑖𝑑𝑒𝑠𝑦 − 𝑥 𝑎𝑛𝑑 𝑥 + 2 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑦 − 𝑥.
𝐼𝑓 𝑥 𝑖𝑠 𝑜𝑑𝑑 𝑡ℎ𝑒𝑛 𝑥, 𝑥 + 2 𝑎𝑟𝑒 𝑐𝑜𝑝𝑟𝑖𝑚𝑒. 𝑇ℎ𝑢𝑠, 𝑥(𝑥 + 2)𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑦 − 𝑥.
𝐻𝑒𝑛𝑐𝑒, 𝑥(𝑥 + 2 ≤ 𝑦 − 𝑥. 𝑇ℎ𝑖𝑠 𝑠ℎ𝑜𝑤𝑠 𝑡ℎ𝑎𝑡 𝑦 ≥ 𝑥 2 + 3𝑥.
𝑀𝑜𝑟𝑒𝑜𝑣𝑒𝑟, 𝑠𝑖𝑛𝑐𝑒 𝑦 ≤ 20, 𝑖𝑡 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑡ℎ𝑎𝑡 𝑥 2 + 3𝑥 ≤ 20.
𝑆𝑖𝑛𝑐𝑒 𝑥 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑜𝑑𝑑, 𝑖𝑡 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑡ℎ𝑎𝑡 𝑥 = 1 , 3.
𝐹𝑜𝑟 𝑥 = 1, 𝑤𝑒 𝑠ℎ𝑜𝑙𝑢𝑑 ℎ𝑎𝑣𝑒 𝑦 + 2 𝑏𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 3.
𝐻𝑒𝑛𝑐𝑒, 𝑦 = 4 , 7 , 10 , 13 , 16 , 19.
𝑆𝑜, 𝑤𝑒 ℎ𝑎𝑣𝑒 6 𝑚𝑜𝑟𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 ℎ𝑒𝑟𝑒.
𝐼𝑓 𝑥 = 3, 𝑡ℎ𝑒𝑛 𝑦 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 3 𝑎𝑛𝑑 𝑦 + 2 𝑏𝑦 5.
𝑆𝑖𝑛𝑐𝑒 𝑦 > 𝑥 𝑖𝑡 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑡ℎ𝑎𝑡 𝑦 = 18. 𝑆𝑜, 𝑤𝑒 ℎ𝑎𝑣𝑒 1 𝑚𝑜𝑟𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.
𝐹𝑖𝑛𝑎𝑙𝑙𝑦, 𝑖𝑓 𝑥 𝑖𝑠 𝑒𝑣𝑒𝑛, 𝑡ℎ𝑒𝑛 𝑏𝑜𝑡ℎ 𝑥, 𝑥 + 2 𝑎𝑟𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 2.
𝑥(𝑥 + 2)
𝐻𝑒𝑛𝑐𝑒, 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑦 − 𝑥.
2
𝐼𝑡 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑡ℎ𝑎𝑡 𝑦 = 18. 𝑆𝑜, 𝑖𝑡 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑡ℎ𝑎𝑡 𝑥(𝑥 + 2) ≤ 2(𝑦 − 𝑥).
𝑇ℎ𝑢𝑠, 𝑥 2 + 4𝑥 ≤ 2𝑦 ≤ 40.
𝑆𝑖𝑛𝑐𝑒 𝑥 𝑖𝑠 𝑒𝑣𝑒𝑛, 𝑤𝑒 𝑓𝑖𝑛𝑑 𝑡ℎ𝑎𝑡 𝑥 = 2 , 4.
𝐼𝑓 𝑥 = 2, 𝑡ℎ𝑒𝑛 𝑦 𝑖𝑠 𝑒𝑣𝑒𝑛 𝑎𝑛𝑑 𝑦 + 2 𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 4.
𝐻𝑒𝑛𝑐𝑒, 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑦 = 6 , 10 , 14 , 18.
𝑆𝑜 𝑤𝑒 ℎ𝑎𝑣𝑒 4 𝑚𝑜𝑟𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠.
𝐹𝑖𝑛𝑎𝑙𝑙𝑦, 𝑖𝑓 𝑥 = 4 𝑡ℎ𝑒𝑛 𝑦 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 4 𝑎𝑛𝑑 𝑦 + 2 𝑏𝑦 6.
𝐻𝑒𝑛𝑐𝑒, 𝑦 = 16. 1 𝑚𝑜𝑟𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.
. 𝑊𝑒 𝑡ℎ𝑒𝑛 ℎ𝑎𝑣𝑒 20 + 6 + 1 + 4 + 1 = 32 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑖𝑛 𝑡𝑜𝑡𝑎𝑙
a. 38
b. 36
c. 34
d. 32
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30. Let 𝑥, 𝑦 be real numbers such that [4]

𝑥 2 + 𝑦 2 + 3𝑥𝑦 = 14;
{
2𝑥 + 2𝑦 − 𝑥𝑦 = 10.
Which one is the positive value of 𝑥 + 𝑦?

Solution:
𝑥 2 + 𝑦 2 + 3𝑥𝑦 = 14
𝑥 2 + 𝑦 2 + 2𝑥𝑦 = 14 − 𝑥𝑦
(𝑥 + 𝑦)2 = 14 − 𝑥𝑦
2𝑥 + 2𝑦 = 10 + 𝑥𝑦
𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑖𝑛𝑔 𝑥𝑦 𝑏𝑦 𝑎𝑑𝑑𝑖𝑛𝑔 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
(𝑥 + 𝑦)2 + 2(𝑥 + 𝑦) − 24 = 0
(𝑥 + 𝑦)2 + 6(𝑥 + 𝑦) − 4(𝑥 + 𝑦) − 24 = 0
(𝑥 + 𝑦)(𝑥 + 𝑦 + 6) − 4(𝑥 + 𝑦 + 6) = 0
(𝑥 + 𝑦 + 6)(𝑥 + 𝑦 − 4) = 0
𝑥 + 𝑦 + 6 = 0 𝑜𝑟 𝑥 + 𝑦 − 4 = 0
𝑥 + 𝑦 = −6 𝑜𝑟 𝑥 + 𝑦 = 4
𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 + 𝑦 = 4

A. 2
B. 3
C. 4
D. 5
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(Rough work)
 26

(Rough work)