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The product of two numbers is 1900 and their LCM is 100, then the HCF of the numbers is:
If a number when divided by 71 gives 37 as quotient and 42 as remainder, then the number is:
Here, divisor = 71, quotient = 37 and remainder = 42.
∴ Dividend = Divisor × Quotient + Remainder
= (71 × 37) + 42
= 2669.
The decimal expansion of the rational number will terminate after:
The number 0.121212 ...... is the form of p/q will be equal:
Let x =0.121212.... ...(i)
100x =12.1212.... ...(ii)
On subtracting eqn. (i) from (ii)
If the sum of the zeroes of the quadratic polynomial 5x^{2} – px + 8 is 4, then the value of k is:
Let the zeroes of the given polynomial be a and b, then sum of the zeroes = 4
i.e p/5 = 4
⇒ p = 20
If the polynomial is p(x) = x^{3}  3x^{2} + x + 1, then the value of p( 3) is:
We have,
p(x) = x^{3} – 3x^{2} + x + 1
Then, p(– 3) = (3)^{3} – 3(– 3)^{2} + (– 3)
+ 1
= – 27 – 27 – 3 + 1
= – 56.
If α, β and γ are the zeroes of the polynomial 2x^{3} + x^{2} – 13x + 6, then abg is equal to:
Let the polynomial be f(x), then
The distance of the point P(– 4, 3) from the origin is:
Here, two points (–4, 3) and
(0, 0) are given, then
x_{1} = – 4, y_{1} = 3 and x_{2} = 0, y_{2} = 0
∴ Distance between them
If is the mid point of the line segment joining the points A(–7, 3) and B(– 3, 7), then the value of a is:
Then, comparing on both sides, we get
a/3 = – 5
⇒ a = – 15.
The point divide the line segment joining the points A(3, 5) and (– 3, – 2) is the ratio:
Let the required ratio be k : 1, then the coordinates of P are
⇒ –15k + 15 = 3k + 3 and –10k + 25 = 11k + 11
⇒ –18k = –12 and –21k = –14
⇒ k = 2/3 and k = 2/3
Hence the point P divides AB in the ratio
2 : 3.
The distance between the points
Distance between
In a given ABC, DE  BC and AB/DB = 3/5. If AC = 5.6 cm, then AE = .
Let AE = x cm,
then EC = (5.6 – x) cm
and DE  BC (given)
⇒ 3(5.6 – x) = 5x
⇒ 16.8 – 3x = 5x
⇒ 8x = 16.8
⇒ x = 2.1
If sin A = 1/√2, then the value of tan A + cot A is:
∵ sin A = 1/√2,
∴ sin A = sin 45°
⇒ A = 45°
Now, tan A + cot A = tan 45° + cot 45°
= 1 + 1
= 2.
If x = 0, then the value of
We have,
x = 0
or cos x = cos 0
or cos x = 1
The value of sin^{2} 30° – cos^{2} 30° is:
=  2/4
=  1/2.
The circumference of a circle of radius 4 cm is: (Use π = 3.14)
Given, radius (r) = 4 cm
∴ Circumference, (C) = 2πr
= 2 × 3.14 × 4 cm
= 25. 12 cm.
Aarti had bought a circular pot wherein area of crosssection is mentioned as 38.5 cm^{2}. She wantedto calculate the radius of the circle and observed its value.
Area of a circle = 38.5 cm^{2}
πr^{2} = 38.5
If a piece of wire 30 cm long is bent into the form of an arc of a circle, subtending an angle of 60° at its centre, then radius of the circle is :
Length of arc = 30 cm
So, 2 x 5 – 5p + 2 = 0
⇒ 10 + 2 – 5p = 0
⇒ 5p = 12
⇒ p = 2.4
Which of the following can not be the probability of an event ?
Probability of an event can not be more than one or negative 9/8 > 1.
A box contains 80 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears a prime number less than 30, is:
Total number outcomes = 80
i.e., n(S) = 80
Prime number less than 30 = 2, 3, 5, 7, 11,
13, 17, 19, 23, 29
i.e., n(E) = 10
∴ Required probability,
What number must be added to numerator and denominator of 2/7 to make fraction equal to 2/3 ?
Let x be added to numerator and denominator of 2/7, so
The difference of LCM and HCF of 28 and 42 is:
28 = 2^{2} × 7
and 42 = 2 x 3 x 7
∴ HCF (28, 42) = 2 x 7 = 14
and LCM (28, 42) = 2^{2} x 3 x 7 = 84
Now, difference = 84 – 14 = 70.
If the sum of ages of father and son is 40 years and its difference is 20 years then age of father will be:
Let father's age be x years and son's age be y, then
x + y = 40 ...(i)
and x – y = 20
On adding eq. (i) and (ii), we get
2x = 60
⇒ x = 30.
If 2 sin 5θ = √3, then the value of q is:
The system of equations 2x + ay = 1 and 5x – 7y = 9 are given.
If the system has a unique solution, then all real values except :
∵ The system of equations has unique
solution
i.e., all real values except  14/5
If the radius of a circle with centre O is 3.5 cm and two radii OA and OB are drawn at right angles toeach other, then the area of minor segment is :
Here,
r = 3.5 cm, θ = 90°
∴ Area of minor segment
If the area of a sector of the circle of radius 18 cm is 72π cm^{2}, then the angle q is :
Here, r = 18 cm and area of sector = 72 π
cm^{2}
The zeroes of the quadratic polynomial 3x^{2} + 2x – 1 = 0 are :
Given, 3x^{2} + 2x – 1 = 0
⇒ 3x^{2} + 3x – x – 1 = 0
⇒ 3x(x + 1) – 1 (x + 1) = 0
⇒ (x + 1) (3x – 1) = 0
⇒ x = – 1 and 1/3
In the given figure, if PQ = 8 cm and PR = 6 cm, then the radius of the circle is ..... cm, where O is thecentre of circle.
Since ∠P in a semicircle, so
∠P = 90°
∴ In DRPQ, RQ^{2} = PR^{2} + PQ^{2}
(Using Pythagoras theorem)
= (6)^{2} + (8)^{2}
= 100
∴ RQ = 10 cm
∴ Radius = 1/2RQ
= 5 cm
In the given figure, If DE  BC, AD = 2 cm, BD = 2.5 cm, AE = 3.2 cm and DE = 4 cm, then BC = 9 cmFind AC.
∵ DE  BC (given)
∴ ∠ADE = ∠ABC (corresponding angles)
and ∠AED = ∠ACB (corresponding angles)
∴ DADE ~ DABC (By AA similarity)
If cot (A + B – C) = √3 and cos (B + C – A) = 1/2, then the value of B is :
cot (A + B + C) = √3 = cot 30°
⇒ A + B – C = 30° ...(i)
and cos (B + C – A) = 1/2
⇒ cos (B + C – A)= cos 60°
⇒ B + C – A = 60° ...(ii)
Adding eq. (i) and (ii), we get
2B = 90°
⇒ B = 45°
If sin θ = 3/4, then cot θ =
If cos θ = 4/5, then tan θ =
If two positive integers p and q are written as p = a^{3}b^{4} and q = a^{4}b^{4}, where a and b are prime numbers,then LCM (p, q) × HCF (p, q) is :
p = a^{3}b^{4}
and q = a^{4}b^{4}
Then LCM (p, q) = a^{4} b^{4}
and HCF (p, q) = a^{3} b^{4}
∴ LCM (p, q) x HCF (p, q)
= a^{4} b^{4} x a^{3} b^{4}
= a^{7} b^{8}
In evening, during family time, Radha observed the dimensions of room. If the length, breadth andheight of a room are 24 m, 18 m and 9 m respectively, then the length of the longest rod that canmeasure the dimensions of the room exactly is :
Length = 24 m, breadth = 18 m and height = 9 m
Since, the length of the longest rod is equal to HCF (24, 18, 9), i.e., 24 = 2^{3} × 3
18 = 2 × 3^{2}
and 9 = 3^{2}
Then, HCF (24, 18 and 9) = 3
Thus, the longest rod that can measure the dimensions of the room exactly = 3 m.
The value of is:
If DABC ~ DQRP and AC = 18 cm then the value of QR is :
∵ DABC ~ DQRP, then
Rational number 7/125 will terminate after ........ decimal places.
which terminates after 3 places.
The difference of (3 + 2 √3) and (3 − 2 √3) is :
Difference = (3 + 2√3) −(3 − 2√3)
= 3 + 2√3 − 3 + 2√3
= 4√3 ,
which is irrational.
In the given figure, MN  QR. The length of PN is :
In DPQR, MN  QR
Case Study
Places A and B are 80 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 4 hours. If they travel towards each other, they meet in 1 hour.
Assuming that the speed of first car and second car be u km/h and v km/h respectively.
What is the relative speed of both cars while they are travelling in the same direction ?
Relative speed of both cars while they are travelling in same direction
= (50 – 35) km/hr = 15 km/h.
Case Study
Places A and B are 80 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 4 hours. If they travel towards each other, they meet in 1 hour.
What is the relative speed of both cars while they are travelling towards each other ?
Relative speed of both cars while they are travelling in opposite directions i.e., travelling towards each other
= (50 + 35) km/hr = 85 km/h.
Case Study
Places A and B are 80 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 4 hours. If they travel towards each other, they meet in 1 hour.
What is the actual speed of one car?
Let the speeds of first car and second car be u km/hr and v km/hr respectively.
According to the given information.
4 (u – v) = 80
i.e,. u – v = 20 ...(i)
and u + v = 80 ...(ii)
Solving eqs. (i) and (ii), we get u = 50 km/hr.
Case Study
Places A and B are 80 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 4 hours. If they travel towards each other, they meet in 1 hour.
What is the actual speed of other car?
Let the speeds of first car
and second car be u km/hr and v km/hr
respectively.
According to the given information.
4 (u – v) = 80
i.e,. u – v = 20 ...(i)
and u + v = 80 ...(ii)
Solving eqs. (i) and (ii), we get u = 50
km/hr and v = 30 km/hr.
Case Study
Places A and B are 80 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 4 hours. If they travel towards each other, they meet in 1 hour.
The given problem is based on which mathematical concept
The given problem is based
on pair of linear equations.
Case Study
The diagram show the plans for a sun room. It will be built onto the wall of a house. The four walls of
the sun room are square clear glass panels. The roof is made using,
Refer to Top View, find the midpoint of the segment joining the points J(6, 17) and I(9, 16).
Midpoint of J(6, 17) and I (9, 16) is
Case Study
The diagram show the plans for a sun room. It will be built onto the wall of a house. The four walls of
the sun room are square clear glass panels. The roof is made using,
Refer to front View, the distance of the point P from the yaxis is:
The distance of the point P from the Yaxis = 4.
Case Study
The diagram show the plans for a sun room. It will be built onto the wall of a house. The four walls of
the sun room are square clear glass panels. The roof is made using,
Refer to front view, the distance between the points A and S is
A's coordinates = (1, 8)
S's coordinates = (15, 8)
Case Study
The diagram show the plans for a sun room. It will be built onto the wall of a house. The four walls of
the sun room are square clear glass panels. The roof is made using,
Refer to front view, find the coordinates of the point which divides the line segment joining the points
A and B in the ratio 1 : 3 internally.
The coordinates of A = (1, 8)
The coordinates of B = (4, 10)
Also, m = 1 and n = 3
Then,
Case Study
The diagram show the plans for a sun room. It will be built onto the wall of a house. The four walls of
the sun room are square clear glass panels. The roof is made using,
Refer to front view, if a point (x, y) is equidistant from the Q(9, 8) and S(17, 8), then
Let point be P(x, y)
PQ^{2} = PS^{2}
or, (x – 9)^{2} + (y – 8)^{2} = (x – 17)^{2} + (y – 8)^{2}
or, x – 13 = 0
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