Q 1 – If one zero of the quadratic polynomial x2 + 3x + k is 2, then the value of k is
(a) 10
(b) – 10
(c) – 7
(d) – 2
(b) – 10
Q 2 – The 2 digit number which becomes th of itself when its digits are reversed. The difference in the digits of the number being 1, then the two digits number is
(a) 45
(b) 54
(c) 36
(d) None of these
(b) 54
Q 3 – In a number of two digits, unit’s digit is twice the tens digit. If 36 be added to the number, the digits are reversed. The number is
(a) 36
(b) 63
(c) 48
(d) 84
(c) 48
Q 4 – If the common difference of an AP is 5, then what is a18 – a13?
(a) 5
(b) 20
(c) 25
(d) 30
(c) 25
Q 5 – Two poles of height 6 m and 11 m stand vertically upright on a plane ground. If the distance between their foot is 12 m, then distance between their tops is
(a) 12 m
(b) 14 m
(c) 13 m
(d) 11 m
(c) 13 m
Q 6 – If a circular grass lawn of 35 m in radius has a path 7 m wide running around it on the outside, then the area of the path is
(a) 1450m2
(b) 1576m2
(c) 1694m2
(d) 3368m2
(c) 1694m2
Q 7 – A card is drawn from a deck of 52 cards. The event E is that card is not an ace of hearts. The number of outcomes favorable to E is
(a) 4
(b) 13
(c) 48
(d) 51
(d) 51
Q 8 – In an AP, if d =- 4, n = 7 and an = 4, then a is equal to
(a) 6
(b) 7
(c) 20
(d) 28
(d) 28
Q 9 – The 4th term from the end of an AP -11, -8, -5, ….., 49 is
(a) 37
(b) 40
(c) 43
(d) 58
(b) 40
Q 10 – For finding the popular size of readymade garments, which central tendency is used?
(a) Mean
(b) Median
(c) Mode
(d) Both Mean and mode
(c) Mode
Q 11 – In figure, on a circle of radius 7 cm, tangent PT is drawn from a point P such that PT = 24 cm. If O is the centre of the circle, then the length of PR is
(a) 30 cm
(b) 28 cm
(c) 32 cm
(d) 25 cm
(d) 25cm
Q 12 – Which term of an AP, 21, 42, 63, 84, … is 210?
(a) 9th
(b) 10th
(c) 11th
(d) 12th
(b) 10th
Q 13 – If the radius of the sphere is increased by 100%, the volume of the corresponding sphere is increased by
(a) 200%
(b) 500%
(c) 700%
(d) 800%
(c) 700%
Q 14 – The median and mode respectively of a frequency distribution are 26 and 29, Then its mean is
(a) 27.5
(b) 24.5
(c) 28.4
(d) 25.8
(b) 24.5
Q 15 – In the adjoining figure, OABC is a square of side 7 cm. OAC is a quadrant of a circle with O as centre. The area of the shaded region is
(a) 10. 5 cm2
(b) 38.5 cm2
(c) 49 cm2
(d) 11.5 cm2
(a) 10. 5 cm2
Q 16 – Assertion : The value of y is 6, for which the distance between the points P (2, –3 ) and Q (10, y) is 10
Reason : Distance between two given points A(x1, y1) and B(x2, y2) is given,
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
(d) Assertion (A) is false but reason (R) is true.
Q 17 – Assertion : 34.12345 is a terminating decimal fraction.
Reason : Denominator of 34.12345, when expressed in the form 
is of the form 2m × 5n ,where m and n are non-negative integers.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
Q 18 – Assertion : The HCF of two numbers is 5 and their product is 150, then their LCM is 30.
Reason : For any two positive integers a and b HCF (a,b) + LCM (a,b) = a b
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true
(c) Assertion (A) is true but reason (R) is false.
Q 19 – If the point P (k ,0) divides the line segment joining the points A( 2, -2) and B (- 7, 4) in the ratio 1 : 2, then the value of k is
(a) 1
(b) 2
(c) – 2
(d) – 1
(d) – 1
Q 20 – The P (A) denotes the probability of an event A, then
(a) P (A) < 0
(b) P (A) > 1
(c) 0 ≤ P (A) ≤ 1
(d) – 1≤ P(A) ≤ 1
(b) P (A) > 1
Q 21 – Sides of a right triangular field are 25 m, 24 m and 7 m. At the three corners of the field, a cow, a buffalo and a horse are tied separately with ropes of 3.5 m each to graze in the field. Find the area of the field that cannot be grazed by these animals.
We know that, area of triangle is
In above figure, base b = 7 and height h = 24
Therefore, area of ∆ABC is
∴ A = 84 m2 ………(1)
Now, sum of three angles of a triangle is 180⁰.
If we join portions of three circles with radii 3.5 we get sector of angle 180⁰ that is semicircle.
Area of this semicircle of radius 3.5 m is
∴ A’ = 19.25 m2
Therefore, area of field that cannot be grazed by animals
= area of triangular field – sum of areas of 3 sectors
= 84 – 19.25
= 64.75 m2
Therefore, area of field that cannot be grazed by animals is 64.75 m2.
Q 22 – Three bells toll at intervals of 9, 12, 15 minutes respectively. If they start tolling together, after what time will they next toll together?
The required answer is LCM of 9,12 and 15 minutes.
Finding prime factor of given of number we have,
9 = 3 × 3 = 32
12 = 2 × 2 × 3 = 22 × 3
15 = 3×5
LCM(9,12,15) = 22 × 33 × 5
= 150 minutes
The bells will toll next together after 180 minutes.
Q 23 – Aftab tells his daughter, ‘7 years ago, I was seven times as old as you were then. Also, 3 years from now, I shall be three times as old as you will be.’ Represent this situation algebraically and graphically.
Consider Aftab’s age as x and his daughter’s age as y.
Then , seven years ago,
Aftab’s age = x − 7
His daughter’s age =y −7
According to the question,
x − 7 = 7 (y − 7)
x – 7 = 7y − 49
x − 7y = − 49 + 7
x − 7y = − 42 …(i)
After three years,
Aftab’s age = x + 3
His daughter’s age = y + 3
According to the question,
x + 3 = 3 ( y + 3 )
x + 3 = 3y + 9
x − 3y = 9 − 3
x − 3y = 6 …(ii)
Representing equation (i) and (ii) geometrically, we plot these equations by finding points on the lines representing these two equations
x − 7y = − 42 ⟹ x = 7y − 42
x =7y − 42 | 42 | 35 | 49 |
y | 12 | 11 | 13 |
x − 3y = 6⟹ x = 3y + 6
x=3y + 6 | 42 | 36 | 48 |
y | 12 | 10 | 14 |
From the graph we can see that two lines will intersect at a point.
x −7y = − 42
x − 3y = 6
On subtracting the two equations, we get
4y = 48
or
y = 12
Substituting value of y in (2),
x − 36 = 6
∴ x = 42
Q 24 – From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle. Prove that OT is the right bisector of line segment PQ.
To Prove : OT is the right bisector of line segment PQ.
Construction : Join OP & OQ Proof : In ΔPTR and ΔQTR
In ΔOPT and ΔOQT
∠OPT = ∠OQT = 90°
OP = OQ (radius)
OT = OT (Common)
ΔOPT ≅ ΔOQT (By RHS congruence)
∠PTR = ∠QTR (cpct)
TP = TQ (Tangents are equal)
TR = TR (Common)
∠PTR = ∠QTR (OT bisects ∠PTQ)
ΔPTR ≅ ΔQTR (By SAS congruency)
PR = QR
∠PRT = ∠QRT
But ∠PRT+ ∠QRT = 180° (as PQ is line segment)
∠PRT = ∠QRT = 90°
Therefore TR or OT is the right bisector of line segment PQ
Q 25 – Find the mean of the following distribution :
Let a = 62.5 be assumed
Q 26 – A horse is placed for grazing inside a rectangular field 70 m by 52 m. It is tethered to one corner by a rope 21 m long. On how much area can it graze ? How much area is left unglazed ?
= 346.5 m2
Total area of the field = 70 × 52 = 3640 m2
Area left ungrazed = Total area of the field – Area of the grazed field =3640 −346.5 = 3296.5 m2
Q 27 – Find the ratio in which the segment joining the points(1,-3) and (4,5) is divided by x -axis? Also find the coordinates of this point on x -axis.
Let the point on X-axis be P (x,0) since the y-coordinate is always zero on x – axis.
Given A(1, -3) and B(4,5)
Let P divides the line joining the given points (1, -3) and (4, 5) in the ratio m : 1
The coordinates of P = (mx2 + x1 / m +1 , my2 + y1 / m + 1)
i.e. (x, 0 ) = ( 4m +1/ m+1 , 5m -3/ m+1)
By comparing both the sides, we get
5m -3/ m+1 = 0
5m-3 = 0
=> 5m = 3
m = 3/5
m:1 = 3/5 :1
= 3:5
Therefore, P divides the line segment joining two points in the ratio 3:5 internally.
Q 28 – Write the smallest number which is divisible by both 306 and 657.
we have to find smallest number which is divisible by 306 and 657
indirectly question ask us to find LCM of 306 and 657.
prime factors of 306 = 2 × 3² × 17
prime factors of 657 = 3 × 3 × 73
so, lowest common multiple or LCM of 306 and 657 = 2 × 3² × 17 × 73
= 2 × 9 × 17 × 73
= 18 × 17 × 73
= 306 × 73
= 22338
hence, smallest number is 22338 which is divisible by 306 and 657.
Q 29 – The person standing on the bank of river observes that the angle of elevation of the top of a tree standing on opposite bank is 60 . When he moves 30 m away from the bank, he finds the angle of elevation to be 30 . Find the height of tree and width of the river.
(i) the height of the tree.
(ii) the width of the river, correct to 2d.p
. Let CD=h be the height of the tree and BC=x be the breadth of the river.
From the figure ∠DAC = 30∘ and ∠DBC = 60∘
∴ Height of the tree = 34.64 m and width of the river = 20m
Q 30 – A hemispherical depression is cut from one face of a cubical block, such that diameter 1 of hemisphere is equal to the edge of cube. Find the surface area of the remaining solid.
Consider the diagram shown below.
It is given that a hemisphere of radius 1/2 is cut out from the top face of the cuboidal wooden block.
Therefore, surface area of the remaining solid
= surface area of the cuboidal box whose each edge is of length l − Area of the top of the hemispherical part + curved surface area of the hemispherical part
Q 31 – If sin θ + Cos θ = √2 prove that tan θ + cot θ = 2
sin θ + cos θ = √2
now square on both side
= ( sin θ + cos θ )² = √2²
= (sin² θ + cos² θ )+ 2 sin θ cos θ = 2
= 1+ 2sin θ cos θ = 2
=> sin θ cos θ = 1/2
now
tanθ + cot θ =sin θ/cos θ + cos θ/sin θ
=( sin² θ +cos² θ) / sin θ cos θ
= 1 / (1/2) = 2
Q 32 – Prove that tangent drawn at any point of a circle perpendicular to the radius through the point contact.
Referring to the figure:
OA = OC (Radii of circle)
Now OB = OC + BC
∴ OB > OC (OC being radius and B any point on tangent)
⇒ OA < OB
B is an arbitrary point on the tangent.
Thus, OA is shorter than any other line segment joining O to any point on tangent.
Shortest distance of a point from a given line is the perpendicular distance from that line.
Hence, the tangent at any point of circle is perpendicular to the radius.
Q 33 – Amit, standing on a horizontal plane, find a bird flying at a distance of 200 m from him at an elevation of 30 . Deepak standing on the roof of a 50 m high building, find the angle of elevation of the same bird to be 45 . Amit and Deepak are on opposite sides of the bird. Find the distance of the bird from Deepak.
Let Amit is standing at bottom of the building at B and observing the bird at C with angle of elevation 30o.
It is assumed that given distance 200m is BC, the distance along line of sight
We have,
If Deepak is seeing the same bird at top of the building with an angle of elevation 45o, we have,
Q 34 – The weight of two spheres of same metal are 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the new sphere.
Weight of two spheres of same metal are 1 kg and 7 kg and radius of smaller sphere = 3 cm
Thus 1 kg occupies 36πcm3 space also when both sphere are melted the weight of re-casted sphere is equal to sum of individual weights of sphere
⇒ weight of re-casted sphere = 1 kg + 7 kg = 8 kg
Let the radius of re-casted sphere be r.
Then weight of re-casted sphere occupies volume of re-casted sphere space
Hence the required diameter of sphere is 12 cm.
Q 35 – Water is flowing through a cylindrical pipe, of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m/s. Determine the rise in level of water in the tank in half an hour.
Volume of water flowing through pipe in 1 secs
=πr 2 H = π ×1 2 × 0.4 × 100 cm3
Volume of water flowing through pipe in 30mins=30×60secs
=π × 12 × 0.4 × 100 × 30 × 60
Volume of cylindrical tank in 30mins
Π × 402 × h = πr2 h
⇒ π × 402 × h = π ×12 × 0.4 × 100 × 30 × 60
Q 36 – John and Priya went for a small picnic. After having their lunch Priya insisted to travel in a motor boat. The speed of the motor boat was 20 km/hr. Priya being a Mathematics student wanted to know the speed of the current. So she noted the time for upstream and downstream.
She found that for covering the distance of 15 km the boat took 1 hour more for upstream than downstream.
(i) Let speed of the current be x km/hr. then speed of the motorboat in upstream will be
(ii) What is the relation between speed distance and time?
(iii) Write the correct quadratic equation for the speed of the current ?
.(i) In this case speed of the motor boat in upstream will be (20 – x) km/hr
(ii) distance = (speed) / time
(iii) As per question,
15 (20 + x) = 15(20 – x) + (20 – x) (20 + x)
15x = – 15x + (202 – x2)
30x = – x2 + 400
x2 + 30x – 400 = 0
x2 + 30x – 400 = 0
x2 + 40x – 10x – 400 = 0
x(x + 40) – 10x (x + 400 = 0
(x + 40) (x – 10) = 0
x = 10, – 40
Here x = 10 is only possible.
Q 37 – Rohan is very intelligent in maths. He always try to relate the concept of maths in daily life. One day he is walking away from the base of a lamp post at a speed of 1 m/s. Lamp is 4.5 m above the ground.
(i) If after 2 second, length of shadow is 1 meter, what is the height of Rohan ?
(ii) What is the minimum time after which his shadow will become larger than his original height?
OR
What is the distance of Rohan from pole at this point ?
(iii) What will be the length of h is shadow after 4 seconds?
As per Question statement we make the diagram at following.
At t = 2 seconds, BD = 2 1= 2m
DE = 1m
Since, ABE CDE
(ii) At point where shadow is equal to her height
CD = DE = 1.5m
4.5 = BD + 1.5
BD = 4.5 – 1.5 = 3m
Time to reach at BD t = 3/1 = 3sec
Or
As calculated in part (ii) we have BD = 3m
(iii) After 4 sec. BD = 1 ×4 = 4
3DE = 4 + DE ⇒ DE = 2m .