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The numerical difference of the roots of x² – 7x – 9 = 0 is

(a) -2 (b) 16 (c) √85 (d) 2√85 Solution: Given x2 - 7x - 9 = 0 Let a and b be the roots. Sum of the roots = a+b = 7 Product... View Article

A number is increased by 40% and then decreased by 40%. Find the net increase or decrease percent.

(a) 16% (b) 20% (c) 10% (d) 32% Solution: Let the number be 100. 40% of 100 = 40 So the resulting number = 100+40 = 140 40% of 140 =... View Article

If (a²– b²) sin θ + 2ab cos θ = a² + b², find sin θ.

Solution: Given (a2 - b2) sin θ + 2ab cos θ = a2 + b2 [(a2 - b2) sin θ/(a2 + b2)] + 2ab/(a2 + b2) cos θ = 1 Put (a2 -... View Article

The value of 4(cos² 60⁰ cos² 45⁰– sin 30⁰) is

(a) 3/2 (b) -3/2 (c) -2 (d) none of these Solution: We know cos 60 = 1/2 cos 45 = 1/√2 sin 30 = 1/2 4(cos2 600 cos2 450- sin 300) =... View Article

AB is a line segment. AX and BY are two equal line segments drawn opposite sides of line AB such that AX || BY. If line segments AB and XY intersect each other at point P. Prove that

(a) ΔAPX ≅ ΔBPY (b) line segments AB and XY bisect each other at P. Solution: Given AX = BY AX parallel to BY XY cuts AB at P. In ∆AXP... View Article

Let g(1) = 1 and g(2) = 3 , if g(x) is described by the formula g(x) = ax- b, then the value of a and b is

Solution: g(x)= ax - b g(1) = 1 => a - b = 1 ..(i) g(2) = 3 2a - b = 3 ..(ii) (ii)-(i) => a = 2 Put a = 2 in (i) b = 1... View Article

The boys and girls in the college are in the ratio 3:2. If 20% of the boys and 25% of the girls are adults, what is the percentage of students who are not adults?

(a) 80% (b) 75% (c) 78% (d) 50% Solution: Let the number of students in the class=100 Let 3x be the number of boys and 2x be the number of... View Article

The edges of a triangular board are 3m, 4m and 5m. The cost of painting it at the rate of Rs.40 and 50 paisa per m²

Solution: Given sides of the triangular board are 3m, 4m and 5m. This is a right-angled triangle. Area of triangle = ½ ×b×h... View Article

In a ∆PQR, if QO and RO are internal angle bisectors of ∠Q and ∠R respectively and ∠QPR = 70⁰, then ∠QOR is

(a) 1100 (b) 1250 (c) 450 (d) 1300 Solution: Given ∠QPR =700 Since QO and RO are internal angle bisectors of ∠Q and ∠R... View Article

The HCF of all possible pairs among four numbers is 3, and the LCM of the four numbers is 198. What is the product of the numbers?

(a) 2673 (b) 32076 (c) 16038 (d) 5346 Solution: Given LCM = 198 Let the numbers be 3a, 3b, 3c, 3d. 3abcd = 198 =>... View Article

The value of tan 203⁰ +tan 22⁰ +tan 203⁰ tan 22⁰ is

(a) 1 (b) -1 (c) 2 (d) 0 Solution: tan 2250 = tan (203 + 22)0 => 1 = (tan 2030 + tan 220)/(1 - tan 2030 tan 220) 1 - tan 2030 tan 220 =... View Article

There are 4 numbers. The HCF of each pair is 7 and the LCM of all the numbers is 1470. What is the product of the 4 numbers?

(a) 504210 (b) 502410 (c) 504120 (d) 501420 Solution: Given HCF of each pair of numbers = 7 Let the numbers be 7a, 7b, 7c, 7d. LCM of the... View Article

For curve y = (x²/2) – 2x + 5, the rate of change of y is twice the rate of change of x, then the value of x is

Solution: Given y = (x2/2) - 2x + 5 dy/dt = 2 dx/dt dy/dt = x (dx/dt) - 2 dx/dt = (x-2) (dx/dt) 2 dx/dt = (x-2) (dx/dt) => x-2= 2 => x = 4... View Article

ln ΔABC, AD ⊥ BC and point D lies on BC such that 2 DB = 3 CD. Prove that 5AB² = 5 AC² + BC².

Solution: Given 2 DB = 3 CD DB/CD = 3/2 Let DB = 3x, CD = 2x BD+DC = 3x+2x => BC = 5x In ΔABD Using Pythagoras theorem AD2+DB2 = AB2... View Article

In the figure, find the area of the shaded region, enclosed between two concentric circles of radii 7 cm and 14 cm where ∠AOC = 40⁰. Consider π = 22/7.

Solution: Given radius of smaller circle = 7 cm Radius of larger circle = 14 cm ∠AOC = 400 Area of smaller circle = πr2 = (22/7)×72 = 154... View Article

The difference between the compound interest and simple interest on a certain sum of money at 10% per annum for 2 years is Rs.500. Find the sum when the interest is compounded annually.

Solution: Let P be the principal, r be the rate of interest and n be the number of years. Simple interest, S I = Pnr = P×2×0.1 =... View Article