#1
 Unregistered Guest User NTSE exam sample papers

I need the National Talent Search Examination sample paper, will you please provide me the same????? Similar Threads Thread Thread Starter Forum Replies Last Post Unregistered Question Papers 1 January 27th, 2018 04:39 PM Unregistered Question Papers 1 January 13th, 2018 03:46 PM Unregistered Question Papers 1 December 14th, 2017 12:12 PM Unregistered Question Papers 1 November 15th, 2017 10:53 AM Unregistered Question Papers 1 September 10th, 2016 09:46 AM Unregistered Question Papers 1 June 27th, 2016 01:21 PM Unregistered General Discussion 1 January 14th, 2016 10:58 AM Unregistered Question Papers 1 January 13th, 2016 05:15 PM Unregistered Question Papers 1 July 15th, 2014 03:17 PM Unregistered General Discussion 1 July 15th, 2014 10:28 AM Nitin Sharma General Discussion 0 June 26th, 2014 12:52 PM Anuj Bhola General Discussion 0 June 21st, 2014 09:50 AM Nitin Sharma General Discussion 0 June 20th, 2014 12:48 PM Garima Chauhan General Discussion 0 February 11th, 2013 09:41 AM Anuj Bhola 2013 0 January 30th, 2013 10:17 AM

#2
 Super Moderator Join Date: Mar 2012 Re: NTSE exam sample papers

As per your request here I am sharing the National Talent Search Examination sample paper:

PAPER II
MATHEMATICS

Q1. If 3
1 r
x
x = + then
3
3 1
x
x + is
(a) 3
(b) 3r3
(c) r3
(d) 0
Ans. ( ) ( ) ( )
0
3 3 3 3 3
1 3 1 1
3 3 3
3
3
3
=

x
x
x
x
x
x
Q2. One third of a number is greater then one fourth of its successor by 1, find the number
(a) 15
(b) 20
(c) 5
(d) 25
Ans. Number = x, Successor = x + 1
3
1
rd of the successor number = 3
x
th
4
1
of the successor number = 4
1 + x
As per question 1
4
1
3
+
+
= x x
X = 15
Q3. If 2x = 8y + 1 & 9y = 3x – 9 then y in
(a) 6
(b) 3
(c) 4
(d) 9
Ans. 2x = (2)3 (y + 1)
X = 3y + 1 (i)
(3) 2y + 3(x – 9)
2y = x – 9 or x = 2y + 9 (ii)
from equation (i) & (ii) 3y + 3 = 3y + 9
3y – 2y = 9 – 3 = 6
= 6
Q4. The sum of two numbers is 24 & the sum of their reciprocal is 5
1
, find their product
(a) 80 (b) 100
(c) 60 (d) 40
Ans. x + y = 24 (i)
5
1 =
y
x
or y = 5x (ii)
from equation (i) x + 5x = 24 or x = 4
& y = 5x = 5x = 5 × 4 = 20
Their product is = 20 × 4 = 80
Q5. ? 1 1
4
1 1
3
1 1
2
1 1 =

n
Κ Κ Κ Κ
(a) n
1
(b) n
x 1 2 −
(c) 


 +
n
n n 1
(d) None of these
Ans. (a)
Q6. In two similar triangle ABC & PQR, if their corresponding altitudes AD & PS are in ratio of
4:9, find the ratio of the Area of ∆ ABC to that of ∆ PQR.
(a) 16:81
(b) 32:92
(c) 33:94
(d) None of these
Ans. (a) Now from fig. 81
16
9
4
2
2
2
2
= = =
PS
PQR of Area
ABC of Area
Q7. Five year hence, father's age will be 3 times then the age of his son. Five years ago, father was 7
times as old as his son. Find their present age ?
(a) 10, 40
(b) 5, 50
(c) 3, 30
(d) None of these
Ans. Let father, age = x & son's age = y
as per the problem x = 7y ….(i) & after 5 year
F.A = (Present ag(e) + 5 = (x + 5) + 5 = x + 10
S.A = (Present ag(e) + 5 = (y + 5) = y + 10
as per the question x + 10 = 3 (y + 10) (i)
= x – 3y = 20 (ii)
from equation (i) and (ii) on solving
x = 40 & y = 10.
Q8. If α & β be the root of the equation x2 – px + 9
(a) p2 – 2q (b) p2 + 2q
(c) p2 – q2 (d) None of these
Ans. α + β = 1
p
= p
αβ = 1
9
= 9
α2β2 = (α + β)2 – 2 αβ
= (– p)2 – 2q
= p2 – 2q
Q9. The value of ? = 

b
(a) 1 (b) 0
(c) xabc (d) None of these
Ans. x(a – (b)(a + (b) × x(b – (c)(b + (c) × x(c – (a)(c + (a)
(x) a2 – b2 + b2 – c2 + c2 – a2 = x0 = 1
Q10. IF x + y = 12, the maximum value of the product of xy is
(a) 26 (b) 36
(c) 30 (d) None of these
Ans. (b)
Q11. Divide 50 into two parts x & y so that the sum of their reciprocals is 12
1
and the parts are
(a) 30, 20 (b) 20, 30
(c) 20, 40 (d) 40, 20
Ans. As per question x + y = 50 (i)
12
1 1 1 = +
y x
or 12
1 =
+
xy
y x
xy = 12 (x + Y)
= 12 × 50 = 600 (ii)
= 2400 2500 −
or x – y = ( ) xy y x 4 2 − +
= 502 – 4 × 600
= 2400 2500 −
= 10 100 =
Solving x + y = 50
x – y = 10
2x = 60 or x = 30 & y = 20
Q12. A man buys mangoes paying one variety Rs. 320 to 240 & another variety of 640 to 400. He
mixes & sells them at16 mangoes for Rs. 30. Find the percentage of profit?
C.P of 240 mangoes = Rs. 320
C.P of 640 mangoes = Rs. 640
C.P of 640 mangoes = Rs. 960
(on variety)
S.P pf 16 mangoes = Rs. 30
S.P pf 640 mangoes = − = × / 1200 640
16
30
Profit = 1200 – 960 = 240
So percentage of profit = V 25 100
960
240 = ×
Q13. Two taps A & B take 20 minutes & 30 minutes to fill a cistern independently. The cistern can
filled in 40 minutes with the taps A & B & the waste pipe are open altogether. If the taps are
closed, calculate the time taken by the discharging outlet to empty the full cistern.
(a) 10 minutes
(b) 15 minutes
(c) 20 minutes
(d) None of these
Ans. Let the volume of cistern = V
Volume of water filled by tap A in 1 minute =
Volume of water filled by tap B in 1 minute =
Taps (A + (B) together can fill in 1minute =
When the discharging outlet is open these taps can fill water in one minute =
The outlined empties the cistern in 1 minute =
So the time taken by the outlet to discharging the whole water volume v is =
Q14. The price of sugar has decreased by 20%, by what% are the consumption of the sugar be
increased in a house so that there is no decrease in the expenditure on the sugar
Ans. Let the sugar consumption was x kg
Total expenditure of sugar = wx
Decrease in price = 25%
So new cost of sugar = x
Now, let w1kg of sugar is consumed for the same total expenditure in wx. This wx = w1x
% increase in consumption =
Q15. Ram Babu deposits Rs. 280. Consisting of one rupee 50 paise & 10 paise coins which are in the
ratio of 3:4:20. The number of 10 paise coins is
(a) 400 (b) 300
(c) 200 (d) None of these
Ans. Consider rupee, 50 paise & 10 paise respectively are 3:
Hence, the value of 10 paise coins is =
So the 10 paise coins are =
Q16. A man borrows Rs. 2500 at 10% pa simple interest. He lends it in the same year & at the same
time at 15% pa for 2 years compound annually. Find the C.I ?
Q17. The area of a square inscribed inside a circle of a radius is
(a) 2r2 (b) r2
(c) 1r2 (d) None of these
Ans. Let AB = x
& OA = r & diagional AC = 2r
∴Area of square = a2
A square is a rhombus of equal diagional
So x2 =
Q18. The least number of square slab of side 1.25 which can be fitted in a varendah of 25 × 20 m is
(a) 320 (b) 340
(c) 280 (d) 200
Ans. The minimum number of slabs
Q19. While going for Station A to Station B a train traveled at a speed 100 km/h & 150 km/h during
return. The average speed of train
(a) 120
(b) 180
(c) 130
(d) 140
Q20. While going for station A to station B a train travelled at a speed 100 km/hr and 150 km/hr
during return. The average speed of train
(a) 120
(b) 180
(c) 130
(d) 140
Ans. Let distance between station A and Station B is x
Average speed =
hr km x x
x
taken time total
ce dis Total
/ 120
150 100
2
tan
=
+
Q21. The sum of length of minute hand of a clock is 14 cm. Find the area of swept by the minute
hand in one minute.
(a) 5
4 10 (b) 5
4 5
(c) 15
4 6 (d) None of these
Ans. Angle made by minute hand at center in 600 minute = 3600
Angle made by minute hand at center in 1 minute = 360/60
= 60
θ = 60
r = 14 cm
Area = 14 14
7
22
360
6
360
2 × × × = × r π
θ
= 15
4 10
Q22. In fig. TAS is a tangent to the circle with center at O at a point A if ∠OBA = 320, find the value
of x and y.
(a) 400
(b) 580
(c) 320
(d) None of these
Ans. O is the center
In AOB ∆,
∠OAB = ∠OBA = 320
A is the point of contact of tangent.
∠OAS = 900 or ∠OAB + ∠BAG = 900
= 32 + y = 900 or y = 580
Q23. Find the mean, mode and median
133, 73, 89, 108, 94, 140, 94, 85, 100, 120
Ans. Arranging the data in increasing order,
73, 85, 89, 94, 94, 100, 108, 120, 133, 140
n = 10 So median = 1
2
&
2
+ n n
= 5
2
10
2
= = n
= 6 1 5 1
2
= + = + n
5th term = 94
6th term = 100
Median = 97
2
194
2
100 94 = =
+
Q24. A hemi – spherical bowl of internal diameter 36 cm contains a liquid in a cylindrical bottles of
radius 3 cm and height 6 cm. How many bottled required
(a) 72
(b) 36
(c) 54
(d) None of these
Ans. Volume of hemi – spherical bowl = 2/3πr3
= 2/3π × 183
Volume of right circular cylinder = πr2h = π326
Where r = 3 and h = 6
Now number of bottles required to supply the bowl
= 72
6 3
18 3 / 2
2
3
=
× ×
×
π
π
Q25. The value of θ
θ
θ
θ
θ
θ
tan
) 90 cos(
) 180 sin(
sin
) 90 sin(
cos +
+
+
+
+
Is equal to
(a) 1 (b) 2
(c) 3 (d) 4
Ans. (a)
Q26. Which figure has the greatest area
(a) Triangle (b) Rectangular
(c) Hexagon (d) Circular
Ans. (c)
Q27. sin2 (90 – θ) + cos2 (90 – θ) = ?
(a) 1 (b) 0
(c) sin2 θ – cos2θ (d) None of these
Ans. (a)
Q28. If cosθ + sinθ = √2cosθ, then value of cosθ – sinθ = ?
(a) √2 sinθ (b) 0
(c) √2 cosθ (d) 2 sinθ
Ans. Squaring both sides and simplifying, we get
cosθ – sinθ = √2sinθ

Q29. A shop keeper buys a number of books for Rs 80. If he had to bought 4 more books for the
same amount, each book would have cost him Rs 1/ – less. How many books did he buy?
(a) 6 (b) 10
(c) 15 (d) 20
Ans. Let total number of books = x
Cost per book = 80
As per our question, we get
(x + 4)(80/x – 1) = 80
80x – x + 320 – 4x = 80x
x2 + 4x – 30 = 0
x = 16 , 20
2
1280 16 4 − =
+ ± −
So number of books = 16
Q30. If
5
1 1
1 4
1 3
9
+
+
+ = P
then find P/9.
(a) 93/29
(b) 47/15
(c) 101/49
(d) 55/47
Ans. 93/29

Q31. If (x, y) are complex numbers then 2 2 y x + is called its modulus. The modulli of a complex
number and its conjugate
(a) are always equal
(b) are always different
(c) are off and on equal
(d) None of these. Quick Reply

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