**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?**- 7 hrs 11 min
- 6 hrs 12 min
- 7 hrs 21 min
- 18 hrs 16 min

Answer: Option C

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.

Hence option 3.

**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?**- 14.28% gain
- 13% loss
- 12% loss
- 11.11% gain

Answer: Option A

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

**ΔCDE ≈ ΔCAB****(area ΔCDE / area ΔCAB) = (CD/CA)**^{2}**If AB = 4, then DE = 2**

- I and II only
- I and III only
- II and III only
- I, II and III

Answer: Option A

Since the lines are parallel, Triangle CDE is similar to TriangleCAB and hence statements I and II are true.

However, the ratio CD : CA = 3:9

So,If AB = 4, then DE = 4/3. So statement III is false Hence option A**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?**- 50
- 45
- 35
- 30

Answer: Option C

When B scored p -10 then C scored p - 25.

When B scores 1 then C scores (p-25)/(p-10)

So when B scores p points then C will score (p-25)/(p-10) × p

As per question (p-25)/(p-10) × p = p -10 . Solving this we get p = 35A B C p points (p-21) points (p-25) points p points (p-10) points **What is the value of a if x**^{3}+ 3x^{2}+ ax + b leaves the same remainder when divided by (x – 2) and (x + 1)?- 18
- 3
- -6
- Cannot be determined

Answer: Option C

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

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**What is the range of values of k if (1 + 2k)x**^{2}– 10x + k – 2 = 0 has real roots?- –3 ≤ k ≤ 4.5
- –1.5 ≤ k ≤ 9
- k ≥ 4.5, k ≤ –3
- k ≥ 3, k ≤ –9

Answer: Option A

Since the given expression has real roots, we know that (–10)^{2}– 4(1 + 2k)(k – 2) ≥ 0

100 – 8k^{2}+ 12k + 8 ≥ 0

8k^{2}– 12k – 108 ≤ 0

2k^{2}– 3k – 27 ≤ 0

–3 ≤ k ≤ 9/2**A square, S**_{1}, circumscribes the circum circle of an equilateral triangle of side 10 cm. A square, S_{2}, is inscribed in the in circle of the triangle. What is the ratio of the area of S_{1}to the area of S_{2}?- 4:1
- 32:1
- 8:1
- 2:1

Answer: Option C

The height of the equilateral triangle is 5√3 cm.

Since the height is also the median, we know that the circum-radius is 2/3 × 5√3 = 10√3/3 and the in-radius is 1/3 × 5√3 = 5√3/3.

The diameter of the circumcircle is the side of square S_{1}.

So the area of S_{1}is (2 × 10√3/3)^{2 }= 1200/9.

The diameter of the in-circle is the diagonal of square S_{2}.

So the area of S_{2}is ½ × (2 × 5√3/3)^{2}= 300/18.

Thus the ratio of areas S_{1}: S_{2 }is 1200/9 : 300/18 = 8 : 1**The sum of the first n terms of an AP is S**_{n}= 4n^{2}– 2n. Three terms of this series, T_{2}, T_{m}and T_{32}are consecutive terms in GP. Find m- 7
- 10
- 16
- 5

Answer: Option A

Explanation: From the given information, S_{1 }= 4 × 12 – 2 × 1 = 2 => T_{1 }= 2. Now, S_{2 }= 4 × 22 – 2 × 2 = 12.

Since S_{2 }= T_{1 }+ T_{2 }and T_{1 }= 2, we get T_{2 }= 10. So, the 1st term of the AP is 2 and the common difference is 8.

From this, we get T_{32 }= 2 + 31 × 8 = 250.

Since T_{2 }, T_{3 }and T_{32 }are consecutive terms in GP, we know that T_{m }/ T_{2 }= T_{32 }/ T_{m}

⇒ (T_{m })^{2 }= T_{2 }× T_{32 }= 2500

⇒ T_{m }= 50.**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?**- 25:36:47
- 16:21:26
- 3:4:5
- 1:2:3

Answer: Option C

(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 four casks as 36 liters each.Liquid A Liquid B Liquid C Cask 1(36 liters) 6 12 18 Cask 2(36 liters) 9 12 15 Cask 3(36 liters) 10 12 14 Liquid A Liquid B Liquid C Cask 1(6 liters) 1 2 3 Cask 2(12 liters) 3 4 5 Cask 3(18 liters) 5 6 7 Cask 4(36 liters) 9 12 15

Hence option C**A trader sells two bullocks for Rs. 8,400 each, neither losing nor gaining in total. If he sold one of the bullocks at a gain of 20%, the other is sold at a loss of**- 20 %
- 18%
- 14%
- 21 %
- None of these

Answer: Option A

A trader sells two bullocks for Rs. 8,400 each, neither losing nor gaining in total.

As he sold one bullocks at a gain of 20%, it means 120% of C.P = 8400

We get C.P of one bullock = 7000

So gain on one bullock = Rs 1400

Other bullocks is sold at lose and there is neither losing nor gaining in total

So loss on 2nd bullock = 1400/7000×100 = 20%