The prime factorisation of 92 and 152 are:
Sol. (c); In ABC, DE || BC by BPT
Sol. (b); Since the given pair of linear equations has no solution.
Sol. (d); We know that,
Sol. (a); Let P(x, 1) be equidistant from A(0, 0) and B(2, 0)
Sol. (b); When two coins are tossed together, then the following outcomes are obtained
HH, HT, TH, TT
Exactly one head is obtained, if the events are:
Hence, P(exactly one head)
(c); Length of arc (l) = 22 cm
Sol. (b); Let and be two roots of the given polynomial.
Sol. (d); We know that, sum of the probability of happening event (E) and not happening event () is 1.
Sol. (d); Given that DEF ~ PQR and EF : QR = 3 : 2
Sol. (c); Given quadratic polynomial is
Sol.(d); Let be the length and be the breadth.
Sol. (b); Let P(4, 0) divides the line segment joining the points A(4, 6) and B(4, –8) in the ratio k : 1.
Total number of balls in the bag = 16 + 8 + 6 = 30
Number of blue balls in the bag = 6
Sol. (b); Given sin - cos = 0
Since, we know that sine and cosine have equal value at 45°.
Thus, = 45°.
Sol. (c); We know that, sum of the probability of happening and not happening of an event is 1.
(a); Outer radius (R) = 14 cm
Sol. (b); Let origin divides AB in the ratio k : 1.
Sol. (c); The coordinates of mid-point of line segment joining A(–8, 0) and B(8, 0) =
Since, given that perpendicular bisector AB passes through (0, k).
Thus, the value of k will be any real number as origin is the mid-point of AB.
Sol. (a); The graph of linear equations x = –5 and y = 6, intersect each other at (–5, 6).
Thus, the solution of the pair of linear equations x = –5 and y = 6 is (–5, 6).
Sol. (d); Radius of circle = 3 units
Sol. (b); The given pair of linear equations are
Sol. (c); HCF of two consecutive even numbers is 2.
Sol. (a); Let and β zeroes of x2 + 99x + 127.
Since, sum of zeroes is negative but product of zeroes is positive.
Thus, both the zeroes of the given quadratic polynomial are negative.
Sol. (c); The coordinates of mid-point of line segment joining the points (–3, 9) and (–6, –4)
(d); A rational number having denominator, whose prime factors are not in the form 2m × 5n, have non-terminating but repeating decimal expansion.
Sol. (a); We know that,
Product of two numbers is equal to the product of their HCF and LCM.
Thus, HCF x LCM = 50 x 20 = 1000.
We observe that for any natural number n, 6n always ends with 6.
Thus, for no natural number n, 6n ends with digit zero.
Sol. (b); (1 + tan2 A) (1 + sin A) (1 – sin A)
(Attempt any 4 questions from Q. No. 41 to 45)
Sukriti throws a ball upwards, from a rooftop which is 8 m high from ground level. The ball reaches to some maximum height and then returns and hit the ground.
If height of the ball at time t ( in sec) is represented by h (m), then equation of its path is given as
Based above information, answer the following:
Draw a perpendicular from maximum height on x-axis. It meet x-axis at (1, 0).
Thus, height of ball at time t = 1 is
Sol. (b) The polynomial represented by above graph is quadratic polynomial.
Sol. (c); Time taken by ball to reach maximum height is 1 sec.
Sol. (b) Since, given graph is of quadratic polynomial.
Thus, it has two zeros.
Sol. (b); Given polynomial,
On equating to zero.
Thus, zeroes of polynomial are –2, 4.
(Attempt any 4 questions from Q. No. 46 to 50)
Quilts are available in various colours and design. Geometric design include shapes like squares, triangles, rectangles, hexagons etc. One such design shown above. Two triangles are highlighted, .
Based on above information, answer the following questions:
Sol. (d); RHS is not suitable for to be similar to .
Sol. (a) In right
We know that, ratio of area of two similar triangles is equal to the ratio of square of corresponding sides.
It should be