UG – 1 & 2, Concorde Complex, Above OBC Bank. R.C. Dutt Road., Alkapuri Baroda. Ph. (0265) 6625979 KHOJ-2019 ANSWER KEY WITH SOLUTION CLASS - 10 PART - I PART - II PART - III Q. No. Answer Q. No. Answer Q. No. Answers Q. No. Answers 1 C 1 C 1 A 31 A 2 C 2 B 2 B 32 B 3 C 3 D 3 A 33 A 4 A 4 A 4 D 34 A 5 A 5 D 5 A 6 A 6 D 6 C 7 A 7 D 7 C 8 D 8 A 8 A 9 B 9 C 9 C PART - IV 10 B 10 D 10 C 1 C 11 B 11 D 11 C 2 C 12 A 12 C 12 D 3 D 13 D 13 B 13 D 4 A 14 D 14 C 14 C 5 B 15 B 15 C 15 A 6 D 16 B 16 D 16 B 7 D 17 A 17 C 17 C 8 B 18 B 18 C 18 B 9 A 19 A 19 C 19 A 10 C 20 B 20 D 20 D 11 B 21 C 21 B 12 B 22 B 22 A 13 C 23 A 23 D 14 B 24 A 24 C 15 C 25 D 25 D 16 A 26 D 26 D 27 D 27 C 28 A 28 B 29 B 29 C 30 A 30 D
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KHOJ-2019 ANSWER KEY WITH SOLUTION CLASS - 10...UG – 1 & 2, Concorde Complex, Above OBC Bank. R.C. Dutt Road., Alkapuri Baroda. Ph. (0265) 6625979 KHOJ-2019 ANSWER KEY WITH SOLUTION
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1 C 1 C 1 A 31 A 2 C 2 B 2 B 32 B 3 C 3 D 3 A 33 A 4 A 4 A 4 D 34 A 5 A 5 D 5 A 6 A 6 D 6 C 7 A 7 D 7 C 8 D 8 A 8 A 9 B 9 C 9 C PART - IV
10 B 10 D 10 C 1 C 11 B 11 D 11 C 2 C 12 A 12 C 12 D 3 D 13 D 13 B 13 D 4 A 14 D 14 C 14 C 5 B 15 B 15 C 15 A 6 D 16 B 16 D 16 B 7 D 17 A 17 C 17 C 8 B 18 B 18 C 18 B 9 A 19 A 19 C 19 A 10 C 20 B 20 D 20 D 11 B
21 C 21 B 12 B 22 B 22 A 13 C 23 A 23 D 14 B 24 A 24 C 15 C 25 D 25 D 16 A 26 D 26 D 27 D 27 C 28 A 28 B 29 B 29 C 30 A 30 D
Feb. 29, 2020 = 5So, no. of odd days in 4 years = 5
To celebrate her birthday on Monday no. of odd days must be zero. By observing pattern,So, after 28 years her birthday would fall on (i.e. on 29 Feb. 2044) Monday.
Since she will live till 2099.So, no. of birthday on Monday.
2016 + 28 = 2044 2044 + 28 = 2072So in 2044 & 2072 her birthday will come on Monday, twice.
10. (b)Watch gain 5 sec. in 3 min.
So in 1 min. 5 sec.3
in 60 min. 5 603
100 sec.
from 7 : 00 am to 4:00 pmTotal hours = 9 hrs.
In 9 hrs watch will gain = 9 100 = 900 sec. = 15 min.
So, till 4:00 pm it will gain 15 min.hence it will show 4:15 pm when actual time is 4:00 pm.
11. (b)From the figure it is clear that the opposite faces are-
So when dice is closed either option (b) or (c) is correct (because opposite surface cannot cometogether).But when we move in anti clock wise direction (as shown)
should come in anticlock wise direction of
So only option (b) is correct.12. (a)
no. of cubes having no surface colored = 3(n 2) no. of cubes which are on surface of blank surface= 8 + 10 = 18 cubes
13. (d)no. of cubes have atleast red color = 16 + 16 = 32
14. (d)no. of cubes have atleast blue color = 16 + 16 = 32
1.(c) 3 3 3x 6 6 6 ... so 3x 6 x . Cube both sides to get 3x x 6 so 3x x 6 0 and
2x 2 x 2x 3 0. So x 2 . The other roots are complex.
2.(b) Let a 1991, so we have
21 11 11 1 1
aa aa a
a a a
1 1 a a
3.(d) By repeated application of the Pythagorean theorem (see fig.),we have:
2 2 2 41 BD BC CD
2 2 2 5 BF BD DF
2 2 2 54 BE BF EF2 2 2 45 AE BE AB .
So 45 3 5 AE .
A3 4
67
5B
C
DE FF
4.(a) Let s be the side length of the square (see fig.). We have 090 m BAM m BMA and m BMA0180 m AMN m NMC , so m BAM m NMC and BAM CMN , with ratio of similitude
5.(d) Let P, Q be the midpoints of the longer sides of the rectangle. Let R, S be the points where thesemicircles meet. Let T be the point where PQ meets RS.
5 cm
P
Q
R S
4 cm
T
Each of the semicircles has radius 5 cm. Therefore PR, PS, QR and QS all have length 5 cm. ThereforePSQR is a rhombus. Hence the diagonals PQ and RS bisect each other at right angles. It follows thatPT and QT each have length 4cm. Let the common length of RT and ST be x cm.
We now apply Pythagoras’ Theorem to the right-angled triangle PTR. This gives 2 2 24 5 , x and
hence 2 2 25 4 25 16 9 x . Therefore x = 3.
It follows that both RT and ST have length 3 cm. Hence the length of RS is 6 cm. Therefore the widthof the overlap of the two semicircles is 6 cm.
6.(d) To avoid a lot of complicated arithmetic, we exploit the facts that 201.72.017100
and 101610.16100
.
Then we take out the common factor 201.7100 . This gives
7. (d)Sol. Let I, J, K and L be the vertices of the sqaure. Let M be the midpoint of JK, let N be the point where the
diagonal IK meets LM. Let the line through N parallel to LK meet IL at R and JK at S. Let T be the footof the perpendicular from N to LK. Let the square have side length s.
I J
KL
MNR S
TIn the triangles INL and KNM, the opposite angles INL and KNM are equal. Also, as IL is parallelto JK, the alternate angles LIN and MKN are equal. Therefore the triangles INL and KNM are
The most important function of villi in the small intestine is to provide increased surface area for absorptionof digested food.
11. (b)The final product of digestion of carbohydrates and proteins are glucose and amino acids respectively
12. (b)
In a closed circulatory system, blood is completely enclosed with in vessels
13. (c) (ii) and (iv)Left ventricle pumps oxygenated blood to different body parts while right ventricle pumps deoxygenatedblood to lungs and Right atrium receives deoxygenated blood from different parts of the body while leftventricle pumps oxygenated blood to different parts of the body
14. (b)
Wavelength of visible light is 400 - 700 nm
15. (c)Nitrogen is used in the synthesis of proteins?
16. (a)
If the tip of a seeding is cut off, growth as well as bending ceases because it hampers perception of lightstimulus