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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... 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... 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... 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,... 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... 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... 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... 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... 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... 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... View Article

In the below figure AB, CD and EF are three lines concurrent at O. Find the value of y.

Solution: Given that ∠BOF = 5y So ∠EOA = 5y (vertically opposite angles) ∠COE + ∠EOA + ∠AOD = 1800... View Article

The LCM of two prime numbers p and q (p > q) is 221. Find the value of 3p – q.

(a) 4 (b) 28 (c) 38 (d) 48 Solution: LCM of two prime numbers = product of the numbers 221... View Article

If the zeroes of the quadratic polynomial ax² + bx + c, where c ≠ 0, are equal, then

(a) c and a have opposite signs (b) c and b have opposite signs (c) c and a have the... View Article

What is the temperature of the Fahrenheit scale corresponding to 30 degree Celsius?

Solution: 30^{\circ}C=30^{\circ}*\frac{9}{5}+32 = 86^{\circ}F

AC power is transmitted from a power house at a high voltage as

a) Power cannot be transmitted at low voltage b) The rate of transmission is faster at high voltages c) A... View Article

A 12 ohm resistor and an inductance of 0.21 henry are connected in series to an AC source operating at 20volts, 50 cycles/second. The phase angle between the current and the source voltage is

Solution: \(\begin{array}{l}phase angle \phi = tan^{-1}\frac{X_{L}}{R} = tan^{-1}\frac{\omega L}{R}\end{array} \) \(\begin{array}{l}\phi = tan^{-1}\frac{2\pi*50*0.21}{12}\end{array} \) \(\begin{array}{l}\phi = tan^{-1}\frac{21\pi}{12}\end{array} \) \(\begin{array}{l}\phi =... View Article

A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. What is the normal reaction on A?

a) N_{A}=\frac{Wx}{d} b) N_{A}=\frac{W(d+x)}{d} c) N_{A}=\frac{W(d-x)}{d} d) N_{A}=\frac{Wd}{x} Solution: c) N_{A}=\frac{W(d-x)}{d}

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