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
181
The geometric mean between -4 and -9 is:
Answer:
-6
The geometric mean between two numbers of the same sign is a number with that same sign. If a and b are negative, GM = -sqrt(ab). Here, GM = -sqrt(-4 * -9) = -sqrt(36) = -6.
182
Find the sum of the series 2 + 22 + 222 + ... to n terms.
Answer:
(2/81)[10(10^n - 1) - 9n]
Let S = 2 + 22 + 222 + ... n terms. S = 2(1 + 11 + 111 + ... n terms). Multiply and divide by 9: S = (2/9)(9 + 99 + 999 + ... n terms). This can be rewritten as (2/9)[(10-1) + (100-1) + (1000-1) + ...]. Grouping the powers of 10 gives a GP: (2/9)[(10 + 100 + 1000 + ... n terms) - (1 + 1 + ... n terms)] = (2/9) * [ 10(10^n - 1)/(10-1) - n ] = (2/9) * [ 10(10^n - 1)/9 - n ] = (2/81)[10(10^n - 1) - 9n].
183
Which term of the progression 18, -12, 8, ... is 512/729?
Answer:
9th
This is a GP with a = 18 and r = -12/18 = -2/3. We need to find n when a_n = 512/729. a_n = ar^(n-1) => 512/729 = 18 * (-2/3)^(n-1). Dividing by 18: 512 / (729 * 18) = 256 / (729 * 9) = 256 / 6561 = (-2/3)^(n-1). Notice that 256 = 2^8 and 6561 = 3^8, so (-2/3)^8 = (-2/3)^(n-1). Therefore, n - 1 = 8, meaning n = 9.
184
The 4th term of a GP is square of its second term, and the first term is -3. Determine its 7th term.
Answer:
-2187
We are given a_4 = (a_2)^2 and a = -3. Using the formula a_n = ar^(n-1), we have ar^3 = (ar)^2 = a^2 r^2. Substituting a = -3: -3r^3 = 9r^2. Since r cannot be 0, divide by r^2 to get -3r = 9, so r = -3. We need the 7th term: a_7 = ar^6 = (-3)(-3)^6 = (-3)(729) = -2187.
185
If the sum of an infinite geometric progression is 4 and the sum of the squares of its terms is 48/5, find the first term and the common ratio.
Answer:
a=3, r=1/4
Sum S = a/(1-r) = 4, so a = 4(1-r). The squares of the terms (a^2, a^2 r^2, ...) also form an infinite GP with first term a^2 and ratio r^2. Its sum is a^2 / (1-r^2) = 48/5. Substitute a = 4(1-r): [16(1-r)^2] / [(1-r)(1+r)] = 48/5. This simplifies to 16(1-r) / (1+r) = 48/5. Dividing by 16 gives (1-r)/(1+r) = 3/5. Cross-multiplying: 5 - 5r = 3 + 3r => 8r = -2 => r = -1/4. Oh wait, my math is wrong: 5 - 5r = 3 + 3r => 2 = 8r => r = 1/4. Let's recheck. (1-r)/(1+r) = 3/5 => 5-5r = 3+3r => 8r = 2 => r=1/4. Then a = 4(1 - 1/4) = 3. This is option b. Let me recheck the math carefully. If a=6, r=-1/2, Sum = 6/(1 - (-1/2)) = 6 / 1.5 = 4. Sum of squares = 36 / (1 - 1/4) = 36 / 0.75 = 48. 48 is not 48/5. If option b: a=3, r=1/4, Sum = 3/(3/4) = 4. Squares = 9 / (1 - 1/16) = 9 / (15/16) = 144/15 = 48/5! Option b is correct. I will fix the intended correct option.
186
A man saves Rs. 100 in the first month, Rs. 150 in the second month, Rs. 200 in the third month, and so on. In how many months will his total savings reach Rs. 3000?
Answer:
10 months
The savings form an AP: 100, 150, 200... where a = 100, d = 50. We need to find n when S_n = 3000. Using S_n = (n/2)[2a + (n-1)d], we get 3000 = (n/2)[200 + (n-1)50]. Multiplying by 2: 6000 = n[200 + 50n - 50] = n[150 + 50n] = 150n + 50n^2. Divide by 50: n^2 + 3n - 120 = 0. Factoring gives (n+12)(n-10) = 0. Since n must be positive, n = 10.
187
What is the 20th term from the end of the AP: 3, 8, 13, ..., 253?
Answer:
158
To find a term from the end, we can reverse the AP. The new AP is 253, 248, 243, ..., 3. Here, the first term a' = 253, and the common difference d' = -5. We need the 20th term of this new AP. a_20 = a' + 19d' = 253 + 19(-5) = 253 - 95 = 158.
188
If the p-th term of an AP is q and the q-th term is p, what is the n-th term?
Answer:
p + q - n
We are given a_p = a + (p-1)d = q and a_q = a + (q-1)d = p. Subtracting the second from the first gives (p-q)d = q-p, so d = -1. Substituting d back into the first equation: a - p + 1 = q, so a = p + q - 1. The nth term is a_n = a + (n-1)d = (p + q - 1) + (n-1)(-1) = p + q - 1 - n + 1 = p + q - n.
189
Find the sum of all natural numbers between 100 and 200 which are multiples of 3.
Answer:
4950
The first multiple of 3 after 100 is 102. The last multiple of 3 before 200 is 198. This is an AP with a=102, d=3, and l=198. To find n: 198 = 102 + (n-1)3 => 96 = (n-1)3 => n-1 = 32 => n=33. Sum = (n/2)(a + l) = (33/2)(102 + 198) = (33/2)(300) = 33 * 150 = 4950.
190
Determine the value of x such that 8x+4, 6x-2, and 2x+7 form an AP.
Answer:
7.5
For an AP, the difference between consecutive terms is equal. So, (6x-2) - (8x+4) = (2x+7) - (6x-2). This simplifies to -2x - 6 = -4x + 9. Add 4x and add 6 to both sides: 2x = 15. Therefore, x = 15/2 = 7.5.