Algebra MCQs for UPSC Prelims

Correct answer: D - 25

Let the price of a brown pair be b, so a black pair costs 3b. Let the original number of brown pairs be x. Original bill = 5 black + x brown = 5(3b) + x(b) = 15b + xb. After interchanging the counts, there are x black pairs and 5 brown pairs, so the wrong bill = x(3b) + 5(b) = 3xb + 5b. A 100% increase means the wrong bill is twice the original: 3xb + 5b = 2(15b + xb). Dividing by b: 3x + 5 = 30 + 2x, so x = 25.

Correct answer: A - Alok

Alok plus Bhupesh equals Chander plus Dinesh, and together they total 100, so each pair holds 50. Chander has half of Dinesh, so Chander plus Dinesh equals 1.5 times Dinesh equals 50, giving Dinesh = 33.33 and Chander = 16.67. Alok is 5 more than Dinesh, so Alok = 38.33, and Bhupesh = 50 - 38.33 = 11.67. The largest share belongs to Alok.

Correct answer: C - 3

Add column by column (units to thousands). Each letter is a distinct digit greater than 3, so digits come from {4,5,6,7,8,9}. Units: C + F must end in 2. Tens: B + 2 (+ carry) must end in 0. Thousands plus a leading carry gives the 1 in 15902. Try the consistent solution: C + F = 12 ends in 2 with carry 1; with B in tens, B + 2 + 1(carry) must end in 0, so B = 7 (7+2+1=10), carry 1. Then hundreds: 3 + E + 1(carry) ends in 9, so E = 5, no carry. Units choice giving C + F = 12 with distinct digits >3 not equal to 7 or 5: take C = 8, F = 4. Remaining unused large digits for A and D are 9 and 6. Thousands: A + D = 15 (since the total is 15902, the thousands column with no carry-in gives A + D = 15, producing 5 in thousands place and carry 1 to make the leading 1). 9 + 6 = 15 works. So {A, D} = {9, 6}, and the difference is 9 - 6 = 3.

Correct answer: A - 9

AB + CD = 1CE means (10A + B) + (10C + D) = 100 + 10C + E. The 10C terms cancel, leaving 10A + B + D = 100 + E. Since B and D are distinct digits, B + D is at most 9 + 8 = 17, so 10A must be at least 100 + E - 17, which is at least 83, forcing A = 9. (A = 8 would require B + D = 20 + E, at least 20, impossible; A at most 7 makes the left side at most 87, below 100.) With A = 9 the equation B + D = 10 + E has valid solutions, e.g. 96 + 25 = 121 with C = 2, E = 1 and digits 9, 6, 2, 5, 1 all distinct. Hence A = 9.

Correct answer: C - Only three

Let the integers be n - 1, n, n + 1. Sum = 3n and product = n(n^2 - 1). Setting 3n = n^3 - n gives n^3 - 4n = 0, so n(n - 2)(n + 2) = 0 and n = 0, 2 or -2. The triples are (-1, 0, 1), (1, 2, 3) and (-3, -2, -1); each indeed has sum equal to product (0, 6 and -6 respectively). So there are exactly three possibilities, option (c). Option (a) counts only the familiar (1, 2, 3) and forgets zero and negative integers, option (b) misses one of the remaining cases, and option (d) is plainly false.

Correct answer: A - z < x < y

From the first condition: 0.7x + y = 1.65y, so 0.7x = 0.65y and y = (70/65)x = (14/13)x, which is greater than x. From the second condition: 0.6x + z = 1.65z, so 0.6x = 0.65z and z = (60/65)x = (12/13)x, which is less than x. Therefore z < x < y, option (a). Options (b), (c) and (d) all contradict at least one of the derived relations y > x and z < x.

Correct answer: B - ₹360

Translate X's statement into equations. If Z gives X Rs. 40, all three are equal: X + 40 = Y = Z - 40, so Y = X + 40 and Z = X + 80. If Y gives X Rs. 40, Y is left with half of Z: Y - 40 = Z/2, i.e. (X + 40) - 40 = (X + 80)/2, giving 2X = X + 80, so X = 80. Then Y = 120 and Z = 160, and the total is 80 + 120 + 160 = Rs. 360. Verification: Y giving 40 leaves Y with 80, exactly half of Z's 160; Z giving 40 makes each of them 120. Option (a) Rs. 420, option (c) Rs. 300 and option (d) Rs. 270 do not satisfy both conditions simultaneously - the two conditions force the unique split 80/120/160.

Correct answer: A - 10 kg

Let the empty vessel weigh E and a full load of water weigh W. Full: E + W = 40. One-third filled: E + W/3 = 20. Subtract the second from the first: W - W/3 = 40 - 20, so (2/3)W = 20, giving W = 30 kg (full water). Then E = 40 - 30 = 10 kg. The empty vessel weighs 10 kg. The other options ignore that the difference of 20 kg corresponds to two-thirds of the water, not the full water.