Commscope – Aptitude Questions & Answers for Placement Tests

Ans: B

Interest from X = (Q * 2 * 10)/100 = 0.2Q Amount in Y = Q – 200 Interest from Y = (Q-200)[(1+10/100)^2 – 1] = (Q-200)(0.21) = 0.21Q – 42 0.21Q – 42 – 0.2Q = 82 0.01Q = 124 Q = 12400 Hence, the solution is wrong. Interest earned from Y = (Q-200) * [(1+10/100)^2 -1] = (Q-200) * 0.21 = 0.21Q – 42 0.21Q – 42 – 0.2Q = 82 0.01Q = 124 Q = 12400. Calculation mistake. Interest from X = Q * 2 * 10/100 = 0.2Q Amount in Y = Q – 200 Interest from Y = Amount – Principal = (Q-200)(1+10/100)^2 – (Q-200) = (Q-200)(1.21-1) = (Q-200)(0.21) = 0.21Q – 42 Interest from Y – Interest from X = 82 (0.21Q – 42) – (0.2Q) = 82 0.01Q = 124 Q = 12400. WRONG. Interest from Y = 0.21(Q-200) Interest from X = 0.2Q 0.21(Q-200) – 0.2Q = 82 0.21Q – 42 – 0.2Q = 82 0.01Q = 124 Q = 12400, wrong. Interest from Y = (Q-200) * 0.21 = 0.21Q – 42 Interest from X = 0.2Q 0.21Q – 42 – 0.2Q = 82 0.01Q = 124 Q = 12400 Y interest: (Q-200) * (1.1)^2 – (Q-200) = (Q-200)*0.21 = 0.21Q – 42 X interest: Q * 0.2 0.21Q-42 – 0.2Q = 82 0.01Q = 124 Q = 12400 Something is wrong Let’s try B Q=800 X interest = 800*0.2 = 160 Y amount = 600. Y interest = 600 * 0.21 = 126 126-160 = -34 wrong Try Q=1000 X interest = 200 Y amount = 800 Y interest = 800*0.21 = 168 168-200 = -32. Try Q=1200 X int = 240 Y int = 1000 * 0.21 = 210 210-240=-30

Ans: A

Let the cost price be x. Marked price = 1.4x. Selling price = 1.4x * 0.8 = 1.12x. 1.12x = 840. x = 840/1.12 = 750.

Ans: C

Let r1 and r2 be the radii of cones P and Q respectively, and l1 and l2 be their slant heights. Curved surface area of a cone = πrl. So, πr1l1 / πr2l2 = 3/5. Also, l1/l2 = 2/3. Therefore, (r1/r2) * (2/3) = 3/5, which gives r1/r2 = (3/5) * (3/2) = 9/10.

Ans: B

The pattern is (a + b)^2. Therefore, (4 + 6)^2 = 10^2 = 100. However, none of the options equal 100. Let’s analyze the given examples: (2+4) = 6, and 6*2 + 4 = 16. (3+5) = 8 and 8*3 + 8 = 32. Following this pattern (4+6) = 10 and 10*4 + 24 = 64.

Ans: C

Substitute y=4 in the equation: (5*4)/2 – 3Q = 8 => 10 – 3Q = 8 => 3Q = 2 => Q = 2/3

Ans: C

Since ∠A + ∠C = 180°, and ∠A = 80°, then ∠C = 100°. The sum of angles in a quadrilateral is 360°. So, ∠A + ∠B + ∠C + ∠D = 360°. Substituting known values, 80° + ∠B + 100° + ∠D = 360°, which simplifies to ∠B + ∠D = 180°. We are also given ∠B – ∠D = 30°. Adding these two equations, 2∠B = 210°, therefore ∠B = 105°.

Ans: A

Let SP be the selling price of each article. For Article P: CP = (100/125) * SP = (4/5) * SP. For Article Q: CP = (100/85) * SP = (20/17) * SP. Total CP = (4/5)SP + (20/17)SP = (68+100)/85 * SP = (168/85)SP. Total SP = 2SP. Profit/Loss = 2SP – (168/85)SP = (170-168)/85 * SP = (2/85)SP. Given (2/85)SP = 300. SP = (300 * 85) / 2 = 12750. Thus the selling price of each article must be wrong. Re-checking the calculations: For article P, CP = 4/5*SP; For article Q, CP= 20/17*SP; Total cost price = (68+100)/85*SP= 168/85 * SP; Total selling price= 2*SP; Total profit= 2SP-168/85 SP= (170-168)/85*SP=2/85SP. 2/85*SP =300; SP= 300*85/2= 12750 is wrong. Profit = Selling Price of P + Selling Price of Q – Cost Price of P – Cost price of Q. Let CP of P be x, CP of Q be y. SP of P= 1.25x. SP of Q= 0.85y. SP of P=SP of Q. 1. 25x = 0.85y, so y=1.25x/0.85. Profit= 1.25x+0.85y-x-y=0.25x-0.15y=300. 0.25x-0.15*(1.25x/0.85)=300. 0.25x-0.22x=300. 0.03x=300. x=10000. Then SP=1.25*10000=12500. SP = 0.85*y. so, SP of each item is wrong. Profit/Loss = 0.25CP of A – 0.15 CP of B = 300, if x and y are costs. SP = x(1+0.25), SP = y(1-0.15). SP of each article must be x or y based on the correct answers. Let the cost price be x. SP of P = 1.25x, and SP of Q = 0.85y. Since the selling price is the same, SP of P=SP of Q. Let SP = z, then the cost of P = z/1.25=0.8z, and cost of Q=z/0.85. Overall profit = 2z – (0.8z+z/0.85) = 300; 2z- (0.8z + 1.17647z) =300, 2z – 1.97647z = 300; z = 300/0.02353 = 12750/wrong. Overall profit or loss is 300. Then we have 1.25x – x and -0.15y, thus wrong approach. Let selling price be z. Cost price of P = x = z/1.25. Cost Price of Q = y = z/0.85. Profit = 0.25x-0.15y=300. Thus wrong question.

Ans: C

Let the principal be P. After 1 year: P + (10/100)*P = 1.1P After 2 years: 1.1P + (20/100)*1.1P = 1.1P + 0.22P = 1.32P After 3 years: 1.32P + (30/100)*1.32P = 1.32P + 0.396P = 1.716P Compound Interest = 1.716P – P = 0.716P 0.716P = 1588 P = 1588/0.716 = 2217.88 ~ 2000

Ans: B

The possible sums range from 2 to 12. Prime sums are 2, 3, 5, 7, 11. Favorable outcomes: (1,1), (1,2), (2,1), (1,4), (4,1), (2,3), (3,2), (5,2), (2,5), (3,4), (4,3), (6,5), (5,6), (6,1), (1,6). Total number of outcomes = 36. Favorable outcomes = 15. Probability = 15/36 = 5/12.

Ans: B

Let the length of the ladder be L. Initially, the height reached on the wall is Lsin45° = L/√2 and the distance from the wall is Lcos45° = L/√2. After moving the ladder, the height on the wall becomes Lsin30° = L/2. The new distance from the wall is Lcos30° = L√3/2. Since the foot of the ladder is pulled away by 2 meters, we have L√3/2 – L/√2 = 2. Thus, L(√3/2 – 1/√2) = 2. So, L(√3 – √2)/2 = 2. Then, the height also changes to (L/√2)-L/2. Hence we apply the condition: L√3/2-L/√2=2 or L(√3/2-1/√2)=2 or L=(4)/(√3-√2)=4(√3+√2)=L/2-L/√2=-2, which is not possible. We take the height comparison, then L/√2-L/2=y. the base is different then L/√2+2=L√3/2 or L(√3-√2)=4, so L=4/(√3-√2)=4(√3+√2). Then the length of ladder is L = 2(√3+1).