Mathematics/General Ability MCQs
Topic Notes: Mathematics/General Ability
MCQs and preparation resources for competitive exams, covering important concepts, past papers, and detailed explanations.
Plato
- Biography: Ancient Greek philosopher (427–347 BCE), student of Socrates and teacher of Aristotle, founder of the Academy in Athens.
- Important Ideas:
- Theory of Forms
- Philosopher-King
- Ideal State
141
How many terms are identical in the two arithmetic progressions 2, 4, 6, 8, ... up to 100 terms and 3, 6, 9, ... up to 80 terms?
Answer:
33
The first AP consists of multiples of 2, its last term is 2*100=200. The second AP consists of multiples of 3, its last term is 3*80=240. The identical terms are numbers that are multiples of both 2 and 3, meaning they are multiples of LCM(2,3) = 6. We need to find the number of multiples of 6 that are less than or equal to the smaller maximum value (200). 200 / 6 = 33.33... So, there are 33 identical terms.
142
If the sum of n terms of an AP is 3n^2 + 2n, find its 15th term.
Answer:
89
The nth term is a_n = S_n - S_{n-1}. So a_15 = S_15 - S_14. S_15 = 3(15^2) + 2(15) = 3(225) + 30 = 675 + 30 = 705. S_14 = 3(14^2) + 2(14) = 3(196) + 28 = 588 + 28 = 616. Therefore, a_15 = 705 - 616 = 89. Alternatively, a_n = (3n^2+2n) - (3(n-1)^2+2(n-1)) = 6n - 1. For n=15, 6(15) - 1 = 89.
143
Find the sum of the first 50 even natural numbers.
Answer:
2550
The sum of the first n even numbers is n(n+1). For n=50, the sum is 50(51) = 2550.
144
If the lengths of the sides of a right-angled triangle are in AP, the ratio of its sides is:
Answer:
3:4:5
Let the sides be a-d, a, a+d. Since it's a right triangle, the longest side is the hypotenuse. By Pythagorean theorem: (a-d)^2 + a^2 = (a+d)^2. Expanding gives a^2 - 2ad + d^2 + a^2 = a^2 + 2ad + d^2. Simplifying yields a^2 = 4ad. Since length cannot be zero, a = 4d. The sides are 4d-d, 4d, 4d+d, which simplifies to 3d, 4d, 5d. The ratio is 3:4:5.
145
In an AP, the sum of the first 10 terms is -150 and the sum of the next 10 terms is -550. Find the AP.
Answer:
3, -1, -5, ...
S_10 = 10/2 * (2a + 9d) = 5(2a + 9d) = -150 => 2a + 9d = -30. The sum of the next 10 terms is S_20 - S_10 = -550, so S_20 = -550 + (-150) = -700. Using formula for S_20: 20/2 * (2a + 19d) = 10(2a + 19d) = -700 => 2a + 19d = -70. Subtract the two equations: 10d = -40 => d = -4. Substitute d back: 2a + 9(-4) = -30 => 2a - 36 = -30 => 2a = 6 => a = 3. The AP is 3, -1, -5, ...
146
Find the sum of the series 1*3 + 3*5 + 5*7 + ... to n terms.
Answer:
n(4n^2 + 6n - 1)/3
The nth term is a_n = (2n-1)(2n+1) = 4n^2 - 1. We need to sum this from 1 to n: Sum(4n^2 - 1) = 4*Sum(n^2) - Sum(1) = 4*n(n+1)(2n+1)/6 - n = n [ 2(2n^2+3n+1)/3 - 1 ] = n [ (4n^2+6n+2-3)/3 ] = n(4n^2 + 6n - 1)/3.
147
The sum of the first n terms of the series 1 + 3 + 6 + 10 + 15 + ... is:
Answer:
n(n+1)(n+2)/6
The terms are triangular numbers, where the nth term is given by n(n+1)/2. We need to find the sum of these terms: Sum[n(n+1)/2] = (1/2)Sum(n^2+n) = (1/2)[n(n+1)(2n+1)/6 + n(n+1)/2] = (1/2)[n(n+1)/2 * ((2n+1)/3 + 1)] = (1/2)[n(n+1)/2 * (2n+4)/3] = n(n+1)(n+2)/6.
148
If the sum of an infinite GP is 9 and its second term is 2, find the first term.
Answer:
3 or 6
We have S = a/(1-r) = 9, so a = 9(1-r). The second term is ar = 2, so a = 2/r. Equating the two gives 9(1-r) = 2/r, leading to the quadratic 9r - 9r^2 = 2, or 9r^2 - 9r + 2 = 0. Factoring gives (3r-1)(3r-2)=0. So r = 1/3 or r = 2/3. If r = 1/3, a = 2/(1/3) = 6. If r = 2/3, a = 2/(2/3) = 3. Both 3 and 6 are possible first terms.
149
If 1/x, 1/y, 1/z are in AP, then x, y, z are in:
Answer:
HP
By definition, a sequence is a Harmonic Progression (HP) if the reciprocals of its terms form an Arithmetic Progression (AP). Thus, x, y, z form an HP.
150
The harmonic mean of 2 and 8 is:
Answer:
3.2
The formula for Harmonic Mean is 2ab / (a+b). For a=2 and b=8, HM = (2 * 2 * 8) / (2 + 8) = 32 / 10 = 3.2.