**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, and 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**- 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?**- 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 Triangle CAB 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

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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**Twice the speed of a boat downstream is equal to thrice the speed upstream. The ratio of its speed in still water to the speed of current is**- 1 : 5
- 1 : 3
- 5 : 1
- 2 : 3

Answer : Option C.

Let the boat speed in still water be b.

Let the stream speed be x.

2(b+ x) = 3(b-x)

5x=b

b/x=5/1**A person has a chemical of Rs. 25 per litre. In what ratio should water be mixed with that chemical so that after selling the mixture at Rs. 20/litre he may get a profit of 25%?**- 13 : 16
- 12 : 15
- 9 : 16
- 19 : 22

Answer : Option C.

This can be solved using allegation.

What is required at the end of mixing is a price of 20/1.25 = 16.

So the allegation would look like this –

Hence the ratio would be (25 – 16) : 16 = 9 : 16

Hence required ratio of Water : Chemical is 9:16.**The difference between the simple interest and compound interest on a certain sum of money for 2 years at 15% p. a. is Rs. 45. Find the sum.**- Rs. 2700
- Rs. 2500
- Rs. 2000
- None of these

Answer : Option C.

Since we know that the interest rate is 0.15, and knowing that the difference between two years of compound interest is nothing but interest on interest, we can find the first year’s interest as –

45/0.15 = 300.

Now if the interest is 300 at the end of one year, then the principal is 300 / 0.15 = 2,000.**How many terms are there in an A.P. whose first and fifth terms are -14 and 2, respectively, and the sum of terms is 40?**- 15
- 10
- 5
- 20

Answer : Option B.

Now the common difference of this AP is 16/4 = 4.

The sum of an AP is n/2 {2a + (n – 1)d}

Substituting we get, 40 = n/2 {2×-14 + (n – 1)4}

The best way to solve this is by plugging options. Put in n = 10 and get the RHS as 40.