# Calsoft Aptitude Questions

Posted on :15-04-2016

Q1. 7 cannibals of XYZ island, decide to throw a party. As you may be aware, cannibals are guys who eat human beings. The senior among them – Father Cannibal decides that any 6 of them will eat up one cannibal, then out of the remaining six – five of them will eat up one cannibal and so on till one is left. What is the time until one cannibal is left, if it takes one cannibal 3 hours to eat up one cannibal independently?

A) 7 hrs 11 min

B) 6 hrs 12 min

C) 7 hrs 21 min

D) 18 hrs 16 min

ANS: C

Explanation:

At the beginning 6 cannibals will eat one, so time required will be 180/6 = 30 min.
Then out of the remaining six – five will devour one, so time required will be 180/5 = 36 min.
Thus the time until one cannibal is left will be = (180/6 + 180/5 + 180/4 + 180/3 + 180/2 + 180/1) min
= (30 + 36 + 45 + 60 + 90 + 180) min
= 441 min
= 7 hrs 21 min

Q2. Three articles are purchased for Rs. 1050, each with a different cost. The first article was sold at a loss of 20%, the second at 1/3rd gain and the third at 60% gain. Later he found that their SPs were same. What was his net gain/loss?

A) 14.28% gain

B) 13% loss

C) 12% loss

D) 11.11% gain

ANS: A

Explanation:

Let us assume that their CPs are x, y & z respectively.
According to the given condition 0.8x = 1.33y = 1.6z
⇒ (80/100)x = 400y/(3 × 100) = (160/100)z
⇒ x : y = 5 : 3 & y : z = 6 : 5
Thus x : y : z = 10 : 6 : 5
Hence CPs of the articles are x = (10/21) × 1050 = 500,
y = (6/21) × 1050 = 300 &
z = (5/21) × 1050 = 250.
SP of the article with CP Rs. x is 0.8x = 0.8 × 500 = 400.
Since SPs are same, the total SP will be 400 × 3 = 1200.
Hence the gain % = (SP – CP)/CP × 100 = (1200 – 1050)/1050 × 100 = 14.28%.

Q3. In the figure below, line DE is parallel to line AB. If CD = 3 and AD = 6, which of the following must be true?

I. ΔCDE ≈ ΔCAB
II. (area of ΔCDE)/(area of ΔCAB) = ( CA/DA )2
III. If AB = 4, then DE = 2

A) I and II only

B) I and III only

C) II and III only

D) I, II and III

ANS: A

Explanation:

Since the lines are parallel, ΔCDE ≈ ΔCAB and hence statements I and II are true. However, the ratio CD : CA = 3:9 ⇒ If AB = 4, then DE = 4/3. So statement III is false

Q4. In a game of tennis, A gives B 21 points and gives C 25 points. B gives C 10 points. How many points make the game?

A) 50 points

B) 45 points

C) 35 points

D) 30 points

ANS: C

Explanation:

Suppose p points make the game. Hence the points played by A, B and C will be,

Therefore, (p – 21) × (p – 10) = p × (p – 25)
=> p2 – 31p + 210 = p2 – 25p
=> 6p = 210
=> p = 35.
Hence 35 points make the game.

Q5. What is the value of a if x3 + 3x2 + ax + b leaves the same remainder when divided by (x – 2) and (x + 1)?

A) 18

B) 3

C) -6

D) Cannot be determined

ANS: C

Explanation:

Suppose the remainder is R.
Substituting x = 2 and x = –1, we get R = 8 + 12 + 2a + b = –1 + 3 – a + b
=> 3a = –18
=> a = –6

Q6. What is the range of values of k if (1 + 2k)x2 – 10x + k – 2 = 0 has real roots?

A) –3 ≤ k ≤ 4.5

B) –1.5 ≤ k ≤ 9

C) k ≥ 4.5, k ≤ –3

D) k ≥ 3, k ≤ –9

ANS: A

Explanation:

Let Boys in class = B
Since the given expression has real roots, we know that (–10)2 – 4(1 + 2k)(k – 2) ≥ 0
100 – 8k2 + 12k + 8 ≥ 0
8k2 – 12k – 108 ≤ 0
2k2 – 3k – 27 ≤ 0
–3 ≤ k ≤ 9/2

Q7. A square, S1, circumscribes the circum circle of an equilateral triangle of side 10 cm. A square, S2, is inscribed in the in circle of the triangle. What is the ratio of the area of S1 to the area of S2?

A) 4:1

B) 32:1

C) 8:1

D) 2:1

ANS: C

Explanation:

The height of the equilateral triangle is 53 cm.
Since the height is also the median, we know that the circum-radius is 2/3 × 53 = 103/3 and the in-radius is 1/3 × 53 = 53/3.
The diameter of the circumcircle is the side of square S1. So the area of S1 is (2 × 103/3)2 = 1200/9.
The diameter of the in-circle is the diagonal of square S2. So the area of S2 is ½ × (2 × 53/3)2 = 300/18.
Thus the ratio of areas S1 : S2 is 1200/9 : 300/18 = 8 : 1.

Q8. The sum of the first n terms of an AP is Sn = 4n2 – 2n. Three terms of this series, T2, Tmand T32 are consecutive terms in GP. Find m

A) 7

B) 10

C) 16

D) 5

ANS: A

Explanation:

From the given information, S1 = 4 × 12 – 2 × 1 = 2 => T1 = 2. Now, 2 = 4 × 22 – 2 × 2 = 12.
Since S2 = T1 + T2 and T1 = 2, we get T2 = 10. So, the 1st term of the AP is 2 and the common difference is 8. From this, we get T32 = 2 + 31 × 8 = 250.
Since T2, Tm and T32 are consecutive terms in GP, we know that Tm / T2 = T32 / Tm ⇒ Tm2 = T2 × T32 = 2500 ⇒ Tm = 50.
Since Tm is a term in the AP, we have 50 = 2 + 8(m – 1) ⇒ 48 = 8(m – 1) ⇒ m = 7.

Q9. Three casks of equal capacities contain three liquids A, B & C in the ratio 1 : 2 : 3, 3 : 4 : 5 & 5 : 6 : 7 respectively. The mixtures from these casks are taken in the ratio 1 : 2 : 3 and poured into a 4th cask with the same capacity as that of the three casks and the cask is completely filled. What is the ratio of the liquids A, B and C in the resulting mixture?

A) 25:36:47

B) 16:21:26

C) 3:4:5

D) 1:2:3

ANS: C

Explanation:

⇒(1 + 2 + 3) = 6, (3 + 4 + 5) = 12 & (5 + 6 + 7) = 18. Common multiple of (6, 12, 18) = 36. So let us fix the capacities of the four casks as 36 litres each.

Since the mixtures are taken in the ratio 1:2:3, 6litres, 12 litres and 18 litres mixture are drawn from the three casks respectively.

Hence the ratio of the liquids in the resulting mixture is 9 : 12 : 15 = 3 : 4 : 5

Q10. In the figure below, the radii of the circles with centres at A and B are 5 cm and 3cm respectively. If ∠FAG = ∠DBC = 900 what is the perimeter of CEG?

A) 38.63cm

B) 30.14cm

C) 42.18cm

D) 27.31cm

ANS: A

Explanation:

Since, OQ = TQ = 2 units, therefore triangle OTQ is equilateral.
In ΔFAG, since AF = AG and ∠FAG = 900, we know that ∠FGA = 450. Similarly, ∠DCB = 450. So, we can conclude that ΔCEG is an isosceles right triangle with hypotenuse CG = 16 cm. From this, we can calculate EG = CE = 8√2 cm. Thus the perimeter of ΔCEG = 8√2 + 8√2 + 16 = 16√2 +16 ≈ 38.63cm

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