Quantitative Aptitude Test

Number of ways the first, second and third positions may be won is 12 × 11 × 10 = 1320. 

DAUGHTER has 8 letters with 3 vowels (A, U, E) and 5 consonants (D, G, H, T, R). The 4 odd positions are 1,3,5,7. Arrange 3 vowels in these 4 positions: 4P3 = 4x3x2 = 24 ways. Arrange 5 consonants in remaining 5 positions: 5! = 120 ways. Total = 24 x 120 = 2880. But checking the calculation: 3 vowels in 4 odd places means 4P3 = 24, and 5 consonants in 5 places = 5! = 120, giving 2880 total arrangements. The answer 576 suggests only 3 odd places or different counting.

For three numbers in G.P, let them be a/r, a, ar. Sum = a/r + a + ar = 19. Product = (a/r) x a x ar = a³ = 216, so a = 6. Substituting: 6/r + 6 + 6r = 19. Divide by 6: 1/r + 1 + r = 19/6. Multiply by r: 1 + r + r² = 19r/6. This gives r = 1.5 or 2/3. So numbers are 4, 6, 9 (with r = 1.5) or 9, 6, 4 (with r = 2/3). Option A matches one valid ordering.

For a 4-digit number to be greater than 5000, the first digit must be 5, 6, or 7 (3 choices). After choosing the first digit, we have 4 choices for the second digit, then 3 for the third, and 2 for the fourth, with no repetition allowed. Total = 3 × 4 × 3 × 2 = 72.

For a 4-digit number from digits {0,1,2,3,4} with no repetition, the first digit cannot be 0 (otherwise it's a 3-digit number). So first digit has 4 choices (1,2,3,4). Second digit has 4 remaining choices (including 0), third has 3 choices, fourth has 2 choices. Total = 4 × 4 × 3 × 2 = 96.

  The 4!=24 permutations of the four digits and consider their sum.2468
2486
2648
2684
2846
28648246
8264
8426
8462
8624
+8642We find that each column has: six 2's, six 4's, six 6's, six 8's.The total of each column is 62 + 64 + 66 + 68 = 120Hence, the addition has the form:000120
001200
012000
+120000

133320

This is a geometric progression with first term a = 1 paise and common ratio r = 2 (each day doubles the previous day's savings). For 14 days (two weeks), the sum is S = a(r^n - 1)/(r - 1) = 1(2^14 - 1)/(2 - 1) = 16383 paise = Rs. 163.83. The key is recognizing the doubling pattern.

This is an infinite geometric series with first term a = 1 and common ratio r = 2/3. The sum of an infinite GP is S = a/(1-r) when |r| < 1. Here, S = 1/(1 - 2/3) = 1/(1/3) = 3. The series converges because 2/3 < 1.

The word TRIANGLE has 8 distinct letters. If 'angle' must always be present as a contiguous unit, we treat it as one single entity. Then we arrange TRI (3 letters) + (angle) as 4 units, giving 4! = 24 arrangements. The letters within 'angle' stay in that fixed order.

For a GP with second term ar = 24 and fifth term ar^4 = 81, we divide to find r^3 = 81/24 = 27/8, so r = 3/2. Then a = 24/r = 24/(3/2) = 16. The series is 16, 16×(3/2), 16×(3/2)^2, ... = 16, 24, 36, 54, 81,...

Let the three numbers in G.P. be a/r, a, ar. Their sum is 70. When extremes are multiplied by 4 and mean by 5, we get 4a/r, 5a, 4ar. For these to be in A.P., the middle term must equal the average of first and third: 5a = (4a/r + 4ar)/2. Solving with sum=70 gives the numbers 10, 20, 40. Check: 10+20+40=70 and 4×10, 5×20, 4×40 = 40, 100, 160 which is an A.P. with common difference 60.

Let the AP be (a-d), a, (a+d). Sum = 3a = 21, so a = 7. After adding 1, 5, 15 respectively: (8-d), 12, (22+d). For GP: 12^2 = (8-d)(22+d). Solving d^2 + 14d - 32 = 0 gives d = 2 or d = -16. For d = 2: numbers are 5, 7, 9. For d = -16: numbers are 23, 7, -9. Both work, but only 5, 7, 9 appears in the options.

For two numbers x and y: AM = (x+y)/2 = 40, so x + y = 80. GM = √(xy) = 24, so xy = 576. We can find (x-y)^2 = (x+y)^2 - 4xy = 6400 - 2304 = 4096, so x-y = 64. Solving x + y = 80 and x - y = 64 gives x = 72, y = 8.

Let the GP be a, ar, ar^2, ar^3. Given a(1+r+r^2+r^3) = 60 and (a+ar^3)/2 = 18, so a+ar^3 = 36. Dividing the sum equation by this gives (1+r+r^2+r^3)/(1+r^3) = 5/3. This leads to 2r^3 - 3r^2 - 3r + 2 = 0, which has r = 2 as a solution. With r = 2, a(1+8) = 36 gives a = 4. The numbers are 4, 8, 16, 32.

The odd numbers between 200 and 300 are 201, 203, 205, ..., 299. This is an arithmetic progression with first term 201, last term 299, and 50 terms. Sum = n/2 × (first + last) = 50/2 × (201 + 299) = 25 × 500 = 12500.

Numbers between 500 and 1000 divisible by 13 are: 507 (13×39), 520 (13×40), ..., 988 (13×76). This is an AP with first term 507, last term 988, and 38 terms. Sum = 38/2 × (507 + 988) = 19 × 1495 = 28405.

Use inclusion-exclusion: Sum of multiples of 4 (10200) + Sum of multiples of 5 (8200) - Sum of multiples of 20 (2200, counted twice). Total = 10200 + 8200 - 2200 = 16200.

For AP 3, 5, 7, 9, ...: first term a=3, common difference d=2. Sum of n terms S_n = n/2[2a + (n-1)d] = n/2[6 + 2n - 2] = n(n+2). Therefore S_n + 1 = n² + 2n + 1 = (n+1)², which is a perfect square.

Numbers divisible by both 4 and 5 are divisible by LCM(4,5)=20. Multiples of 20 from 100 to 300 are: 100, 120, ..., 300. This is 11 terms. Sum = 11/2 × (100 + 300) = 11/2 × 400 = 2200.

Monthly installments form an AP: 100, 95, 90, ..., with sum 975. Let n be number of months. S_n = n/2[2a + (n-1)d] = n/2[200 + (n-1)(-5)] = n/2[205 - 5n] = 975. Solving: n(205 - 5n) = 1950, which gives n = 15.

Annual savings form an AP with sum 16500 over 10 years. S_10 = 10/2[2a + 9d] where d = 100. So 16500 = 5[2a + 900], giving 2a + 900 = 3300, therefore 2a = 2400 and a = 1200.

Using compound interest formula A = P(1 + r/100)^t: 8650 = P(1.1)^5. Since (1.1)^5 = 1.61051, P = 8650/1.61051 ≈ 5372.29. This exact value is not listed among options A, B, or C.

Population growth: 55 crore × 2.01 = 56.1 crore (not 5705, 6005, or 6700). The option values are incorrectly scaled - they appear to be in lakhs but the unit is unclear, making all given options wrong.