NTSE MAT Class 8 Mathematics Questions and Answers Set 04

Question. If \( x + \frac{1}{x} = \sqrt{3} \) then \( x^3 + \frac{1}{x^3} \) is
(a) 3
(b) 3r3
(c) r3
(d) 0
Answer: (d) 0
\( x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x}) \)
\( = (\sqrt{3})^3 - 3\sqrt{3} = (\sqrt{3})^3 - (\sqrt{3})^3 = 0 \)

Question. 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
Answer: (a) 15
Number = x, Successor = x + 1
\( \frac{1}{3} \) rd of the successor number = \( \frac{x}{3} \)
\( \frac{1}{4} \) th of the successor number = \( \frac{x + 1}{4} \)
As per question \( \frac{x}{3} = \frac{x + 1}{4} + 1 \)
X = 15

Question. If \( 2^x = 8^{y + 1} \) & \( 9^y = 3^{x - 9} \) then y in
(a) 6
(b) 3
(c) 4
(d) 9
Answer: (a) 6
\( 2^x = (2)^{3(y + 1)} \)
X = 3y + 3 (i)
\( (3)^{2y} = 3^{(x - 9)} \)
2y = x – 9 or x = 2y + 9 (ii)
from equation (i) & (ii) 3y + 3 = 2y + 9
3y – 2y = 9 – 3 = 6
y = 6

Question. The sum of two numbers is 24 & the sum of their reciprocal is \( \frac{1}{5} \), find their product
(a) 80
(b) 100
(c) 60
(d) 40
Answer: (a) 80
x + y = 24 (i)
\( \frac{x}{y} = \frac{1}{5} \) or y = 5x (ii)
from equation (i) x + 5x = 24 or x = 4
& y = 5x = 5 \times 4 = 20
Their product is = 20 \times 4 = 80

Question. \( (1 - \frac{1}{2})(1 - \frac{1}{3})(1 - \frac{1}{4}) \dots (1 - \frac{1}{n}) = ? \)
(a) \( \frac{1}{n} \)
(b) \( \frac{2x - 1}{n} \)
(c) \( n(\frac{n + 1}{n}) \)
(d) None of these
Answer: (a) \( \frac{1}{n} \)

Question. 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 \( \Delta \) ABC to that of \( \Delta \) PQR.
(a) 16:81
(b) 32:92
(c) 33:94
(d) None of these
Answer: (a) 16:81
Now from fig. \( \frac{Area \text{ of } ABC}{Area \text{ of } PQR} = \frac{AD^2}{PS^2} = \frac{4^2}{9^2} = \frac{16}{81} \)

Question. 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
Answer: (a) 10, 40
Let father, age = x & son's age = y
as per the problem x - 5 = 7(y - 5) ....(i) & after 5 year
F.A = (Present ag(e) + 5 = x + 5
S.A = (Present ag(e) + 5 = y + 5
as per the question x + 5 = 3 (y + 5) (i)
= x – 3y = 10 (ii)
from equation (i) and (ii) on solving
x = 40 & y = 10.

Question. If \( \alpha \) & \( \beta \) be the root of the equation \( x^2 - px + 9 \)
(a) \( p^2 - 2q \)
(b) \( p^2 + 2q \)
(c) \( p^2 - q^2 \)
(d) None of these
Answer: (a) \( p^2 - 2q \)
\( \alpha + \beta = \frac{p}{1} = p \)
\( \alpha\beta = \frac{9}{1} = 9 \)
\( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha\beta \)
\( = (– p)^2 - 2q \)
\( = p^2 - 2q \)

Question. The value of \( (\frac{x^a}{x^b})^{a + b} \times (\frac{x^b}{x^c})^{b + c} \times (\frac{x^c}{x^a})^{c + a} = ? \)
(a) 1
(b) 0
(c) \( x^{abc} \)
(d) None of these
Answer: (a) 1
\( x^{(a - b)(a + b)} \times x^{(b - c)(b + c)} \times x^{(c - a)(c + a)} \)
\( (x)^{a^2 - b^2 + b^2 - c^2 + c^2 - a^2} = x^0 = 1 \)

Question. IF x + y = 12, the maximum value of the product of xy is
(a) 26
(b) 36
(c) 30
(d) None of these
Answer: (b) 36

Question. Divide 50 into two parts x & y so that the sum of their reciprocals is \( \frac{1}{12} \) and the parts are
(a) 30, 20
(b) 20, 30
(c) 20, 40
(d) 40, 20
Answer: (a) 30, 20
As per question x + y = 50 (i)
\( \frac{1}{x} + \frac{1}{y} = \frac{1}{12} \)
or \( \frac{x + y}{xy} = \frac{1}{12} \)
xy = 12 (x + y)
= 12 \times 50 = 600 (ii)
or \( x - y = \sqrt{(x + y)^2 - 4xy} \)
\( = \sqrt{50^2 - 4 \times 600} \)
\( = \sqrt{2500 - 2400} = \sqrt{100} = 10 \)
Solving x + y = 50
x – y = 10
2x = 60 or x = 30 & y = 20

Question. A man buys mangoes paying one variety Rs. 320 to 240 & another variety of 640 to 400. He mixes & sells them at 16 mangoes for Rs. 30. Find the percentage of profit?
Answer: 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 = \( \frac{30}{16} \times 640 = 1200/- \)
Profit = 1200 – 960 = 240
So percentage of profit = \( \frac{240}{960} \times 100 = 25 \% \)

Question. 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
Answer: 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 1 minute =
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 =

Question. 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
Answer: 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 =

Question. 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
Answer: Consider rupee, 50 paise & 10 paise respectively are 3:4:20
Hence, the value of 10 paise coins is =
So the 10 paise coins are =

Question. The area of a square inscribed inside a circle of a radius is
(a) \( 2r^2 \)
(b) \( r^2 \)
(c) \( 1r^2 \)
(d) None of these
Answer: (a) \( 2r^2 \)
Let AB = x
& OA = r & diagional AC = 2r
\( \therefore \) Area of square = \( a^2 \)
A square is a rhombus of equal diagional
So \( x^2 = \)

Question. 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
Answer: (a) 120

Question. 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
Answer: (a) 120
Let distance between station A and Station B is x
Average speed = \( \frac{\text{Total distance}}{\text{total time taken}} \)
\( = \frac{2x}{\frac{x}{100} + \frac{x}{150}} = 120 \text{ km/hr} \)

Question. 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) \( 10\frac{4}{5} \)
(b) \( 5\frac{4}{5} \)
(c) \( 6\frac{4}{15} \)
(d) None of these
Answer: Angle made by minute hand at center in 60 minute = \( 360^{\circ} \)
Angle made by minute hand at center in 1 minute = \( 360/60 = 6^{\circ} \)
\( \theta = 6^{\circ} \)
r = 14 cm
Area = \( \frac{\theta}{360} \times \pi r^2 = \frac{6}{360} \times \frac{22}{7} \times 14 \times 14 = 10\frac{4}{15} \)

Question. In fig. TAS is a tangent to the circle with center at O at a point A if \( \angle OBA = 32^{\circ} \), find the value of x and y.
(a) \( 40^{\circ} \)
(b) \( 58^{\circ} \)
(c) \( 32^{\circ} \)
(d) None of these
Answer: (b) \( 58^{\circ} \)
O is the center
OA = OB (Radii)
In AOB \( \Delta \),
\( \angle OAB = \angle OBA = 32^{\circ} \)
A is the point of contact of tangent.
\( \angle OAS = 90^{\circ} \) or \( \angle OAB + \angle BAG = 90^{\circ} \)
= 32 + y = \( 90^{\circ} \) or y = \( 58^{\circ} \)

Question. Find the mean, mode and median 133, 73, 89, 108, 94, 140, 94, 85, 100, 120
Answer: Arranging the data in increasing order,
73, 85, 89, 94, 94, 100, 108, 120, 133, 140
n = 10 So median = \( \frac{1}{2}(\frac{n}{2} \& \frac{n}{2} + 1) \)
\( \frac{n}{2} = \frac{10}{2} = 5 \)
\( \frac{n}{2} + 1 = 5 + 1 = 6 \)
5th term = 94
6th term = 100
Median = \( \frac{94 + 100}{2} = \frac{194}{2} = 97 \)

Question. 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
Answer: (a) 72
Volume of hemi – spherical bowl = \( 2/3 \pi r^3 \)
= \( 2/3 \pi \times 18^3 \)
Volume of right circular cylinder = \( \pi r^2 h = \pi \times 3^2 \times 6 \)
Where r = 3 and h = 6
Now number of bottles required to supply the bowl
\( = \frac{2/3 \pi \times 18^3}{\pi \times 3^2 \times 6} = 72 \)

Question. The value of \( \frac{\cos \theta}{\sin(90 + \theta)} + \frac{\sin \theta}{\sin(180 + \theta)} + \frac{\cos(90 + \theta)}{\tan \theta} \) Is equal to
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (a) 1

Question. Which figure has the greatest area
(a) Triangle
(b) Rectangular
(c) Hexagon
(d) Circular
Answer: (c) Hexagon

Question. \( \sin^2(90 - \theta) + \cos^2(90 - \theta) = ? \)
(a) 1
(b) 0
(c) \( \sin^2 \theta - \cos^2 \theta \)
(d) None of these
Answer: (a) 1

Question. If \( \cos\theta + \sin\theta = \sqrt{2}\cos\theta \), then value of \( \cos\theta - \sin\theta = ? \)
(a) \( \sqrt{2} \sin\theta \)
(b) 0
(c) \( \sqrt{2} \cos\theta \)
(d) \( 2 \sin\theta \)
Answer: (a) \( \sqrt{2} \sin\theta \)
Squaring both sides and simplifying, we get
\( \cos\theta - \sin\theta = \sqrt{2}\sin\theta \)

Question. 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
Answer: Let total number of books = x
Cost per book = 80
As per our question, we get
\( (x + 4)(80/x – 1) = 80 \)
\( 80x – x^2 + 320 – 4x = 80x \)
\( x^2 + 4x - 320 = 0 \)
\( x = \frac{-4 \pm \sqrt{16 + 1280}}{2} = -20, 16 \)
So number of books = 16

Question. If \( \frac{P}{9} = 3 + \frac{1}{4 + \frac{1}{1 + \frac{1}{5}}} \) then find P/9.
(a) 93/29
(b) 47/15
(c) 101/49
(d) 55/47
Answer: (a) 93/29

Question. If (x, y) are complex numbers then \( \sqrt{x^2 + y^2} \) 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.
Answer: (a) are always equal