Calsoft – Aptitude Questions & Answers for Placement Tests

Ans: B

9 pirates eat 1 pirate: 2 hours. 8 pirates eat 1 pirate: 2 hours. 7 pirates eat 1 pirate: 2 hours. 6 pirates eat 1 pirate: 2 hours. 5 pirates eat 1 pirate: 2 hours. 4 pirates eat 1 pirate: 2 hours. 3 pirates eat 1 pirate: 2 hours. 2 pirates eat 1 pirate: 2 hours. Total time = 2*8 = 16 hours.

Ans: C

Let the cost price of the first item be x and the second item be y. Then x + y = 1500. Selling price of first item = 1.2x and selling price of second item = 0.9y. Since selling prices are same, 1.2x = 0.9y, which simplifies to y = (4/3)x. Substituting y in x + y = 1500, we get x + (4/3)x = 1500, (7/3)x = 1500, x = 4500/7. Then y = 1500 – 4500/7 = 6000/7. Difference in cost prices is y – x = (6000-4500)/7 = 1500/7 approximately 214. Hence 1.2x = 0.9y therefore x= 3y/4 therefore 3y/4 + y = 1500 i.e. 7y/4 = 1500 so y = 6000/7 and x= 4500/7 . The difference is 6000/7 – 4500/7 = 1500/7 ~ 214, close to 200

Ans: D

Since DE || BC, triangle ADE is similar to triangle ABC. The ratio of their sides is AD/AB = 4/(4+8) = 4/12 = 1/3. The ratio of their areas is (1/3)^2 = 1/9. Let the area of triangle ABC be x. Then 20/x = 1/9, so x = 180. The area of quadrilateral DBCE is the area of triangle ABC minus the area of triangle ADE, which is 180 – 20 = 160.

Ans: A

Total points of X and Y: 30 + 25 = 55. Team X gives 15 points to Z, Team Y gives 5 points to Z, total points to Z are 15 + 5 = 20. Total points in the game = 30 + 25 + 20 = 75.

Ans: A

If (x – 1) is a factor, the polynomial equals 0 when x = 1. Substituting x = 1, we get 2(1)^3 – 5(1)^2 + k(1) – 8 = 0. Simplifying, 2 – 5 + k – 8 = 0, so k – 11 = 0, and k = 11.

Ans: B

For no real roots, the discriminant (b² – 4ac) < 0. Here, a = m, b = 2m - 1, c = m. Therefore, (2m - 1)² - 4(m)(m) < 0. Simplifying, 4m² - 4m + 1 - 4m² < 0, which gives -4m + 1 < 0, so 4m > 1, and m > 1/4. However the original problem states that m is a coefficient of the quadratic equation, therefore, m cannot be equal to 0. So the correct answer is when the discriminant is <0, m should be > 1/4. Also, since the leading coefficient (a = m) shouldn’t be 0, we look for options which fits this scenario, which is m is greater than 1/8 and less than 0.

Ans: B

Let the side of the outer square be ‘s’. The area of the outer square is s^2 = 36, so s = 6 cm. The diameter of the circle is equal to the side of the outer square, so diameter = 6 cm and the radius is 3 cm. The diagonal of the inner square is equal to the diameter of the circle. Let the side of the inner square be ‘x’. By Pythagorean theorem, x^2 + x^2 = 6^2, so 2x^2 = 36, and x^2 = 18. The area of the inner square is x^2.

Ans: B

Sn = 3n2 + n Tn = Sn – S(n-1) = (3n2 + n) – (3(n-1)2 + (n-1)) = 6n – 2 T3 = 6(3) – 2 = 16 T15 = 6(15) – 2 = 88 Since T3, Tk, T15 are in GP, Tk2 = T3 * T15 Tk2 = 16 * 88 = 1408 Tk = 6k – 2 (6k – 2)2 = 1408 36k2 – 24k + 4 = 1408 This method is not correct, we use Tk2 = T3 * T15 to find k. So, (6k – 2)2 = 16 * 88. T3, Tk, T15 => 16, 6k-2, 88 So (6k-2)^2 = 16*88 = 1408, so 6k-2 = sqrt(1408) Since Tk2 = T3 * T15 or Tk = sqrt(T3*T15) Let the common ratio be r T3 * r = Tk, Tk * r = T15 So (6k – 2)^2 = (6(3) – 2) * (6(15) – 2) = 16 * 88 = 1408 Tk = sqrt(1408). Instead, we need to use the general term form Tn = a + (n-1)d S_n = n/2 (2a + (n-1)d) S_n = 3n^2 + n, S_1 = 4 = a S_2 = 3(4) + 2 = 14; a + a+d = 14 => 4 + 4+d = 14, d = 6 Tn = 4 + (n-1)6 = 6n – 2 T3 = 16, Tk = 6k – 2, T15 = 88 So, (6k – 2)^2 = 16 * 88 => doesn’t work So we use T(k) = (T3 + T15)/2 incorrect T3, Tk, T15 in GP => Tk/T3 = T15/Tk (6k-2)/16 = 88/(6k-2) (6k-2)^2 = 16*88=1408 We should use the formula for a geometric mean: Tk^2 = T3 * T15, 6k-2 = sqrt(T3*T15) is incorrect If Tk/T3 = T15/Tk, then the series is in GP so the ratio between terms is the same. Tk/16 = 88/Tk (6k-2)^2=16*88 T3, Tk, T15 in GP, so Tk/T3 = T15/Tk => Tk^2 = T3 * T15 => (6k-2)^2 = 16*88 Incorrect. The correct method is to use the GP property where if a, b, c are in GP, then b^2 = ac. Then, since T3, Tk, T15 are in GP, Tk^2 = T3 * T15. T3 = 6(3) – 2 = 16. T15 = 6(15) – 2 = 88. Tk = 6k-2. So (6k-2)^2 = 16 * 88 = 1408. This implies Tk is irrational which is not possible. T3, Tk, T15 are in GP => Tk^2=T3*T15 so, (6k-2)^2=16*88=1408; incorrect since k is not an integer T3=16, T15=88 Tk=sqrt(T3*T15) is not correct, we use Tk/T3 = T15/Tk Also incorrect to imply Tk= (T3+T15)/2 T3, Tk, T15 are in GP => Tk/T3 = T15/Tk => Tk^2 = T3*T15 So the only thing we know is that b^2 = ac. So, T3, Tk, T15 forms a GP. Tk^2 = T3*T15 Tk^2 = 16*88. Incorrect Then (6k-2)^2=16*88=1408 and this is not possible Use Tk/T3 = T15/Tk => (6k-2)/16 = 88/(6k-2) => (6k-2)^2 = 16*88 Tk/T3 = T15/Tk (6k-2)/16 = 88/(6k-2) (6k-2)^2 = 16*88 = 1408 not an integer T3 = 16, T15 = 88. Need to find k such that T3, Tk, T15 are in GP. If in GP, then Tk/T3 = T15/Tk. (6k-2)^2 = 16*88. Try (6k-2) / 16 = (88)/(6k-2) Tk^2 = 16*88 We cannot proceed by this way since k is not an integer here. If the common ratio is r, then T(k)/T3=r and T15/T(k)=r. Hence, Tk/T3 = T15/Tk or Tk^2= T3*T15. Tk^2= 16*88. This gives k as a non-integer number Let us try different approach and find common difference. T3= 16, T15= 88. This does not work. T3=16, T15=88 (6k-2)^2 = 16 * 88 T3 = 16. T15 = 88. In GP, common ratio is const, so: T(k)/16 = 88/T(k) T(k)^2 = 16 * 88 = 1408. Which leads to an non-integer k. T3=16, T15=88 So, if terms are in GP Tk^2 = T3 * T15 so (6k-2)^2 = 16 * 88 Thus the method seems wrong. Try a different approach, if T3, Tk, T15 are in GP, then: Tk / T3 = T15 / Tk => Tk^2 = T3 * T15 (6k – 2)^2 = 16 * 88 = 1408 Which results in an irrational number and since we seek k. So (6k-2)^2 = 16*88. The approach is incorrect. Tn=6n-2. T3=16, T15 = 88. 6k-2 The series has a common ratio. Thus the method is wrong.

Ans: A

Since DE is parallel to BC, triangle ADE is similar to triangle ABC. Therefore, AD/AB = AE/AC. We have AB = AD + DB = 4 + 6 = 10 cm. So, 4/10 = 5/AC. Thus, AC = 50/4 = 12.5 cm. Finally, EC = AC – AE = 12.5 – 5 = 7.5 cm.