KHOJ-2019 ANSWER KEY WITH SOLUTION CLASS - 9

UG – 1 & 2, Concorde Complex, Above OBC Bank. R.C. Dutt Road., Alkapuri Baroda. Ph. (0265) 6625979

KHOJ-2019

ANSWER KEY WITH SOLUTION CLASS - 9

PART - I PART - II PART - III

Q. No. Answer Q. No. Answer Q. No. Answers Q.

No. Answers

1 C 1 D 1 C 31 D 2 C 2 C 2 C 32 D 3 A 3 A 3 A 33 C 4 C 4 D 4 C 34 B 5 A 5 B 5 B 6 A 6 C 6 A 7 C 7 A 7 D 8 B 8 D 8 B 9 A 9 A 9 D PART - IV

10 B 10 A 10 C 1 B 11 D 11 C 11 B 2 A 12 B 12 B 12 D 3 C 13 A 13 B 13 C 4 B 14 B 14 D 14 C 5 C 15 D 15 A 15 A 6 A 16 D 16 C 16 D 7 B 17 A 17 A 17 B 8 B 18 C 18 D 18 A 9 C 19 D 19 B 19 C 10 D 20 C 20 A 20 C 11 B

21 D 21 C 12 B 22 C 22 B 13 B 23 A 23 D 14 A 24 B 24 C 15 D 25 C 25 A 16 A 26 D 26 D 27 B 27 C 28 A 28 C 29 B 29 C 30 B 30 D

DATE : 06/10/2019 KHOJ SOLUTIONS CLASS : 9

Page # 1UG–1 & 2, Concorde Complex, Above OBC Bank. R.C. Dutt Road., Alkapuri Baroda.IIT ASHRAM

PART - I

1. (c)According to the given information:

Green

Yellow

Red

Orange

Blue Black

It is clear from the above diagram that the colour of the side having question mark is green.2. (c)

Given that,33 11 9 28 4 5

after changing the sign 33 11 9 28 4 5

= 33 11 9 28 4 5

= 33 11 9 7 5

= 42 46 4 So, option (c) is correct answer.

3. (a)18 5 1 19 15 14R E A S O N

As, 2 2 2 2 2 2

16 7 25 21 13 16P G Y U M P

Similarly,4 9 18 5 3 20D I R E C T

2 2 2 2 2 2

B K P G A V2 11 16 7 1 22So, DIRECT will be coded as BKPGAV.

4. (c)

As, In 1st figure : 3 26 5 1 2,

In 2nd figure : 3 212 10 2 12 and

In 4th figure : 3 224 20 4 80

Similarly, In 3rd figure : 3 218 15 3 36

So, 36 will replace the question mark.



5. (a)

40

S-W

2836

52

29 km

So option (a)

6. (a)

po ki top ma Usha is playing cards

kop ja ki ma Asha is playing tennis

po sur kop Cards and tennis

So code for Asha is ja.

7. (c)

Teacher Woman Doctor

So option (c)

8. (b)2[ 6 (4 4) 1] 20 1 19 S 2[ 4 (1 7) 1] 9 1 8 H 2[ 8 (5 10) 1] 14 1 13 M

Similarly, 2[ 5 (5 2) 1] 15 1 14 N 9. (a) 13 represents non educated non working urban temales.10. (b) 4011. (d) 3 represents non-working rural male whose not educated.12. (b) (70 + 9) = 79 So, option (b) is the correct answer.13. (a) 50 + 1 + 60 = 111 So, option (a) is the correct answer.14. (b)

Father

Mother-in-lawHusband

Son

Father-in-law

Husband

Son-in-law

SonGirl



15. (d)

16. (d)

17. (a) 3 teachers are both players and Artists.

18. (c) 64:9::81: ?

( 64 8) : (8's next number 9)

::( 81 9):(9's next number)

64:9::81: 10

19. (d)

20. (c)



PART - II1. (d)

215.01 81.009 32

2 215.01 15 225 81.009 9 225 + 9 32 = 513 ANS.

2. (c)Suppose speed of person = x km/hTotal time taken = t hrs.distance travelled = xt km.Case-I Speed = x + 2

time = t - 2distance = (x + 2) (t - 2)xt = xt - 2x + 2t - 4t - x = 2 ......(1)

Case-II Speed = x + 6time = t - 4distance = (x + 6) (t - 4)xt = xt + 6t - 4x - 243t - 2x = 12 .......(2)

by (1) & (2)3t - 2x = 122t - 2x = 4- + -t = 8x = 8 - 2 = 6 distance = 8 6 = 48 km. ANS.

3. (a)Suppose radius of sphere = x cm

Volume of cylinder = 100 5.4 = 3540 cm

Volume of one sphere = 34 r3

So, 3415 r 5403

3 540 3r 27

4 15

r 3 diameter = 6 cm. ANS.



4. (d)

M 125x

x

y

y

135100

N

R

O

reflex 0MON 360 235 125

In 0ORM x x 100 180 x 40

In 0

0 45ORN y y 135 180 y2

x 1 y4 2

40 1 454 2 2

= 852 =

01214

ANS.

5. (b)P = 25600 r = 25% PA A = 28900 t = n years

25r %4

compounded quaterly

28900 = 25600 4n

2514 100

4n28900 1725600 16

2 4n17 17

16 16

4n = 2 1n2

year. ANS.

6. (c)Suppose MP = xdiscount by A = 30% of x = 0.3xdiscount by B = 20% of x = 0.2x

= 0.8x 10% = 0.08x Total = 0.2x + 0.08x = 0.28x

So, 0.3x - 0.28 x = 600

0.02x = 600 600 100x 30000Rs.

60x

ANS.

7. (a)no. of total alphabet on odd position = 13no. of vowels on odd position = 5

probability of odd vowels = 5

13



8. (d)

3 2 3 2x 3xy 14, y 3yx 13 3 3 3 2 2(x y) x y 3xy 3yx 14 + 13 = 27

3(x y) 27

x + y = 3 ........... (1)

3 3 3 2 2(x y) x y 3xy 3yx 14 13 1

(x y) 1 ............(2)

by (1) & (2)

x = 2, y = 1 2 2 2x y 2 1 5 . ANS.

9. (a)

2 2a 2b b 2a 5 2 2a 2a 1 b 2b 6 2 2a 2a 1 b 2b 1 5

2 2(a 1) (b 1) 5 2 2(a 1) (b 1) 5 (a b 2) (a b) 5

a & b are positive integer So a + b - 2 = 5, a - b = 1 hence b = 3. ANS.

10. (a)

In ADB

2 2 2 2 2AC AD CD 15 20 = 225 + 400 = 625 = 252

A

B CD15 15

30

20E

So, AC = 25

area of 10

1ABC 30 20 3002

area of 1ADC BE AC2

1300 BE 252

300 2BE 24 cm

25

ANS.

11. (c)

7 4 5 4 2 5 10 5 5p q 5

6

1 3 5 p q 56

On comparing we get, 1 1p , q6 2

1 1 8 6So, p q6 2 12 3 ANS.

12. (b)

5 5 5..... 7 7 7.....

3 3 3.....

Consider- y = 5 5 5.....



y 5y 2y 5y 2y 5y 0 y (y - 5) = 0

y = 0 or y = 5So, y = 5

Similarly x = 7 7 7..... x = 7

& z = 3 3 3.....

z = 3 So5 7 12 4

3 3

ANS.

13. (b)

In ABCD , AB CD and AB = CD

So ABCD is a parallelogram

Hence area of 1ACD2

as ABCD 21 66 33cm2

14. (d)Suppose Polynomial -

3 2p(x) ax bx cx d p(1) = a+b+c+d

1 a b c d ........................(1)

p(2) 8a 4b 2c d

2 8a 4b 2c d ........................(2)

p(3) 27a 9b 3c d

3 27a 9b 3c d .........................(3)

p(4) 64a 16b 4c d

5 64a 16b 4c d ...........................(4)

p(6) 216a 36b 6c d ............................(5)

(2) (1) | (3) (2)7a + 3b + c = 1 ............................(6)19a + 5b + c = 1 ....................... ....(7)

7 6 12a 2b 0 ............................(9)

4 3 37a 7b c 2 .............................(8)

8 7 18a 2b 1 ............................(10)

10 9 6a 1 a 1 /617b 1, c , d 16

So, p (6) = 16. ANS.



15. (a)(n + 20) + (n + 21) + ------------- + (n + 100)

= 81 (n + 60)= 92 (n + 60) So least value of n is 4.

16. (c)

For real 29 n 2 0

2n 2 9 3 n 2 3 5 n 1

So total integers -5, -4, -3, -2, -1, 0, 1. ANS.17. (a)

2 2 2a b1 1 2 ab

a b aba b a b

2

21 1 aba b a b

2

2 2a b ab 1 1

ab 2015ab a b

18. (d)

2 2 2 2a 5 b c c d b c d 9 0

for this each term must be zero.So, a = 5, b = c, c = d

b c d 9

c c d 9

3d 9

d 3

d c b 3

So, a b c b c d

5 3 3 3 3 3 11 9 99 . ANS.

19. (b)A

M

y x

2x 2y

B CD

EF

G

2 : 1

Let G be centroid, D, E, F be mid points of BC, CA, AB & M be mid point of FE.Let BE = 3x, CF = 3y & AD = 30 (given)



So AG = 20, GD = 10, AM = MG = 10 (M is midpoint of AG)BG = 2x, GE = x, CG = 2y, GF = y

Now, D is mid point of BC of BGC . So D is circumcentre of BGC .

Hence, BD = GD = CD = 10 So BC = 20

In BGC

2 24y 4x 400 2 2x y 100

In ABC

2 2 23 AB BC CA = 2 2 24 AD BF CF

(side median property)

2 2 23 AB BC CA = 2 24 900 9x 9y

= 4 900 900 = 600

4 18003

= 2400 2 2AB BC CA

24100

. ANS.

20. (a)Let no. be abc.

Since a b c is prime, we should have two of them as 1 each and third a prime no.

So a, b, c (2, 3, 5, 7)If we take 2, then

112, 121, 211 3 numbersfor each numbers there are 3 no.

So total no. = 12. ANS.

21. (d)The last digit of 2015n is 5 for all n for 2016n is 6. the no. must be divisible by 10, should have 0 in lastdigit.

n2015 5 (in last digit) n2016 6 (in last digit)

5 + 6 = 11 (in last digit) So n2017 should have 9 as last digit

So n should be 2 . ANS.



22. (c)Let the radius of outer circle be r.

Perimeter of circle = 2 r

but OQ BC r (diagonal of square BQCO]

P Q

RS

A

B

C

D

O

So perimeter of ABCD = 4r Ratio = 2 r4r 2

23. (a)

Clearly DEF is also equilateral .

Let 2a be the side of ABC .

In ACFA

B

C

D

E F

300

600

300

CAF 30º

AFC 60º

So, AFC 2 CAF

Hence, AF = 2CFAF = 2x

by Pythagorous theorem-

30º

60º

A

C F

2x2a

x

4x2 = 4a2 + x2

3x2 = 4a2

2ax3

2aAD

3

So, 6aDF

3

2

2

36aarea DEF 3 3area ABC 4a

ANS.



24. (b)So OA = 6, OP = 2

In APO -

2 2 2AP OA OP 36 4 32

A2 2 2 2

2

S1 S2

BO

Cx

xP

AP 4 2

Let PC = BC = x

In ABC -

2 2 2AC AB BC 2 2 24 2 x 8 x 2 232 x 8 2 x 64 x

8 2 x 32 32x

8 2 4 2x

2

x 2 2 So k = 2. ANS.

25. (c)

Suppose EAD x

So CED 180 140 x 40 x

DCE CED 40 x

A x

B

C

D140

E

BCA x

CBD CDB 2x

So at C -

0ACB BCD CDCE 180

0x 180 4x 40 x 180

x = 100 ANS.26. (d)

We need to find the two digit no. that divide 265 - 5 = 260.

260 2 2 5 13 , the two digit factors 10, 13, 20, 26, 52, 65

So, 6 such numbers. ANS.



27. (b)

4 3 2m 6m 11m 6m 1 = 2 2

21 6m m 11 6m

mm

=2 2

21 1m m 2 9 6 m

mm

=2

2 1 1m m 6 m 9m m

= 2

2 1m m 3m

= 22

22

m 3m 1m

m

= 22m 3m 1 So Square Root = 2m 3m 1

28. (a)Extend AD to G and EF to G.

AGF BEC

So BC = DF

So area of Rect. DBEF = area of DBCG . A B

CD E

FG

1

2

DBCG is A Parallelogram.

So area DBCG = 2 area BCD

area DBCG12 ABCD2

area DBCG 2 1 = 2 ANS.

29. (b) n = 560560560560563n = 560560560560560 + 3n = 8K + 3 (K = 70070070070070)

So, 2 2n 64K 16K 9 8 (8 K + 2K + 1) + 1 So when n2 is

divided by 8 Remainder is 1. ANS.30. (b)

AXY & BXC are similar

So,BC BXAY AX

AY AXBC BX

AY BC AX BX

BC BX

DY ABBC BX

DY BX BC AB

ORlet AD = BC = a

BX = AX = b AB = 2b

AXY BXC

hence AY = BC = a

Now BX DY b 2a 2ab

area of Rectangle ABCD = 2ab hence option B.



PART - III1.Sol. (c)

Initial relative velocity = v1 – v2. Final relative velocity = 0From v2 = u2 – 2as 0 = (v1 – v2)2 – 2 × a × s

2

1 2v vs

2a

If the distance between two cars is ‘s’ then collision will take place. To avoid collision d > s.

2

1 2v vd

2a

where d = actual initial distance between two cars.2.Sol. (c)

With respect to the first train, the second train is approaching with a velocity of (100 m/s + 150 m/s) = 250 m/s. To completely cross the first train the second train has to travel a distance of 2000 mrelative to first train.

D = 1000 + 1000 = 2000 m relative speed v = 150 + 100 = 250 m/s

Dis tance 2000mTime 8secVelocity 250m/s

3.Sol. (a)

FBD of 2 kg block

T = 2g sin 300

= 10 N4.Sol. (c)

According to newton’s III law, when we push water backwards, water apply force on us in forwarddirection



5.

Sol. (b) 1 1 2 2 v at a t a t 2 21 1

2 1 1 2

/ 2 1/ 4 2

a F mt m

t a F m m

6.Sol. (a)

W = F. s. cos

when 0º cos 1

mW F.s.

7.Sol. (d)

2

1 2

FrGm m

Unit of 2 2 3

3 2 12 2 2 2

kg mNm m mG m s kgkg s kg s kg

8.

Sol. (b) e 22GM 2 GMRv 2gR

R R

'e ev 2 g 4R 2 2gR 2v

9.Sol. (d)

Inside the earth.

dg ' g 1R

Linear (increasing)

Outside the earth, 2gg '

(l h/R)

exponential

(decreasing)



10.Sol. (c)

Here net driving force

= mg mgmg2 2

downward

Hence frction will act upward and its magnitude should be mg2

f

If the block ‘m’ in stationary the friction between m and wall should be static.

If limf < f

mg 1. N mg2 2

mg2

11.Sol. (b)

Oily road has less friction. Vehicles moves forward due to friction exerted by ground in forwarddirection when friction is less vehicle slide.

12.Sol. (d)

Relative density = density of objectdensity of water

Relative density no unit13.Sol. (c)

P = gh

pressure does not depends on area of the surface.14.Sol. (c)

Frequency = 54 9 Hz60 10

9V f 10 9m /s10

15.Sol. (a)

wavelength is distance between two particles in same phase16.Sol. (d) F = 2mv = 2000 N

F = 2000 N

2 4 2A 1cm 10 m 3

7 24

F 2 10P 2 10 N/mA 10

17.Sol. (b)



The feeling of weight comes from the experience of normal force acting on the feel of the person.

In stationary lift,W = mgWhen the lift moves downward with acceleration a,

So, 2mg W ma

2W m(g a)

Now, 2

WW

= mg

m g a = 32

g 3g a 2

2g = 3g – 3a3a = g

ga3

18. (a)Sol. Amalgam is alloy of mercury with another metal.19. (c)Sol. Non-metals are generally brittle.20. (c)Sol. The element which is mainly electroplated on car parts, bicycle handles to give shiny appearnce is

chromium.21.Sol. (c)

Phosphrous is stored in water.22. (b)Sol. Fog, Mist, Cloud are aersol.23.Sol. (d)

y is gas, z is liquid, x is solid, kinetic energy in increasing order X < Z < Y24.Sol. (c)

When Aluminium react with NaOH it produces sodium hydroxide.



25. (a)Sol. a deadly poisnous gas produce during in complete combustion is carbon monoxide.26. (d)Sol. Carbon is limiting reagent, 6 g of carbon will produce 22 g CO2

27. (c)The rate of evaporation decreases with increase in humidity.

28. (c)Mass of solution - 100 + 34.7 = 134.7g

Volume of solution = mass of solution

density 3134.7 103.61m

1.3

(m/v) % = Mass of solute 100

Volume of solution =

34.7 100 33.49%103.61

29. (c)Sol. Weight of oxygen in compound = 12 16 = 192 u

Total weight of compound = 2 27 + 3 32 + 12 16 = 342 uSo, mass % of oxygen

100 192 56.14%342

30. (d)

Sol. One mole of H2SO4 consist of 237 6.022 10 atoms.

31. (d)

Sol. molecular weight of 2 3 2 27 48 102Al O gram

102 gram consist of 2 3Al O 232 6.022 10 ions

0.051 gram 2 3Al O consist of

23

20.051 2 6.022 10

10

= 206.022 10 ions



32. (d)Sol. Ideal gas eqn PV = nRT

case (1) 17P. V RT28

eqn (1)

case (2) 27 14P. V RT

28

eqn (2)

Where V2 = 12 litDividing eqn (2) by (1)

1

21 RTP 12 287P V RT28

Solving this

1

12 3V

112V 4lit3

33. (c)

Sol. The temperature at which celsius & faranhite scales shows same reading is 040 C

34. (b)

Sol.39

19K Neutrons = 20

24

12Mg Neutrons = 12

ratio = 2012 = 5/3



PART - IV1. (b)

The above stsatement is correct explanation of Agriculture.2. (a)

Separating grains from chaff is called: winnowing3. (c)

The organism is called fungi and its reproduce by spore formation.4. (b)

P organism is Amoeba and D is hydra both perform asexual method of Reproduction. (Binnary fissionand budding)

5. (c)It is chloroplast which trap solar energy and convert it in chemical energy.

6. (a)It acts as semi permiable barrier to the outside of the cell. Allow flow of certain molecules in and outas needed.

7. (b)Less surface area of cell ensure more number of it so it can be present so many in a perticular area.

8. (b)Charles Darwin proposed the theory of evolution.

9. (c)It is meant to conserve both, the biodiversity and the culture of that area.

10. (d)The talking mechanism is performed by the birds to Attract the other bird.

11. (b)B+ is a biological technological concept a bacteria introduced in the plant to make it insect resistance.

12. (b)Scientist who discovered fermentation is Louis Pasteur

13. (b)In these statements only (b) is correct.

14. (a)Nitrogen is very essential for growth of the plant.

15. (d)Option D is correct.

16. (a)Malaria Is caused by Protozoa.

KHOJ-2019 ANSWER KEY WITH SOLUTION CLASS - 9

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