**A worker is kept on a contract for 100 days to make some toys. On any of these 100 days he does not make more than 20 toys. If on any day, he makes more than 12 toys, then he makes at most 6 toys each on the next two days. What is the maximum possible number of toys that he can make over the period of 100 days?**- 1109
- 1208
- 1100
- 1076
- None of these

Answer: Option B.

If the worker makes more than 12 toys on any day then in three days period he can make a maximum of 20 + 6 + 6 = 32 toys.

On the other hand he could have made 36 toys over this span by making 12 toys each day.

So to achieve the maximum he must not make more than 12 toys on any day except possibly the last day.

So, maximum number of toys he could have made = 99 × 12 + 20 = 1208**From a list of four comics, four friends discuss their favourite comics. At least 2 friends vote for Mandrake, not more than 3 vote for Mammaji, 1 votes for Alibaba and 2 vote for Tom & Jerry. If two friends have exactly voted for 2 different comics each, and 2 friends for exactly 3 different comics each, then how many votes did Mandrake get?**- 2
- 3
- 4
- Cant say
- None of these

Answer: Option C.

There are a total of 10 votes, of which exactly 3 go to Alibaba and Tom & Jerry together.,

The remaining 7 are distributed between Mandrake and Mammaj

Maximum votes that Mandrake can get is 4 (since there are only 4 comics) and maximum votes that Mammaji can get are 3.**From a bag containing 242 balls, one ball weighs 19.9 grams and all the other weigh 19.5 grams each. Using a simple balance where balls can be kept on either pan, what is the minimum weighs required to identify the defective ball?**- 3
- 4
- 5
- 7
- none of the above

Answer: Option C.

By dividing the total balls in three parts and putting two parts on the simple balance we can find which of the three parts has ball of different weight. Again divide this part into further three parts and put two parts on the simple balance. From here again we can find the part having ball with different weight. Again this part can be divided into three parts and so on. So 35 = 243 i.e up to 243 balls can be checked within 5 weighings.**DIRECTIONS for the question 4-5: Read the information given below and answer the question that follows**.

Five friends, viz. Shan, Monu, Jai, Karan and Bunty are living in five different cities named Karanpur, Jaipur, Vizanagar, Barnala and Patiala, not necessarily in that order.

Their salaries are 7 Lacs, 8 Lacs, 9 Lacs, 11 Lacs, 13 Lacs (INR per annum), in no particular order. Further, the following information is given about them:

I. Karan, who does not live in Barnala, earns a salary that is a prime number multiple of 1 Lac.

II. Monu made a call to one of his four mentioned friends who lives in Patiala and who earns a perfect square multiple of 1 Lac in salary.

III. Jai's salary is 1 Lac more than the average salary of Karan and Shan

IV. Monu lives in the city, which has the shortest name amongst the above cities.**If Karan lives in Vizanagar, then what is the average salary of the persons living in Barnala and Karanpur?**- 9
- 10
- 12
- Data insufficient
- None of these

Answer: Option D.

From the given information, we can summarize the data in the following table:

Karan Jai Shan Bunty Monu

Salaries:7 or 13 11 13 or 7 9 8

Cities: K/V B/K/V B/K/V Patiala Jaipur

Where ‘K’, ‘V’ and ‘B’ stands for ‘Karanpur’, ‘Vizanagar’ and ‘Barnala’ respectively.

If Karan, lives in Vizanagar, then Jai and Shan must be staying at Karanpur and Barnala, not necessarily in that order. Their average salary in any case will be Rs. 12 lakhs or 9 lakhs. So, the data is insufficient.**Who stays in Patiala?**- Shan
- Monu
- Bunty
- Jai

Answer: Option C.

Monu called a friend, who gets Rs. 9 lakh as his salary is a perfect square multiple of 100000 and stays in Patiala.

Bunty stays in Patiala.Impetus Preparation Links- Sample Aptitude Questions of Impetus
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**If Monu and Jai live in cities with names starting with consecutive alphabets, then who lives in Vizanagar?**- Shan
- Monu
- Bunty
- Karan

Answer: Option D.

Monu lives in Jaipur, so Jai must be living at Karanpur.

Since Karan is not staying at Barnala, he must be staying at Vizanagar.**A square and a regular hexagon have the same area. Find the ratio of the perimeter of the square to the perimeter of the hexagon.**- √3 : 2
- 1 : 2
^{4}√3 :^{4}√4^{4}√4 :^{4}√3

Answer: Option D.Area of regular hexagon of side a = 6 x √3/4 a^{2}Area of square of side b = b^{2}Now according to the question,6 x √3/4 a^{2 }= b^{2 }=>b^{2 }/ a^{2 }= 3√3/2 b/a = 3^{3/4}/√2.Required Ratio = 4b/6a= 4 x 3^{3/4 }/ 6 x √2 = √2 /^{4}√3 = (4/3)^{1/4.}**If x + y = 1, then what is the value of (x^3 + y^3 + 3xy)?**- 1
- 2
- 9
- 0
- None of these

Answer: Option A.

From a point P, the tangents PQ and PT are drawn to a circle with centre O and radius 2 units. From the centre O, OA and OB are drawn parallel to PQ and PT respectively. The length of the chord TQ is 2 units. Find the measure of the ∠AOB.- 30°
- 90°
- 60°
- 120°
- None of these

Answer: Option D.

Since, OQ = TQ = 2 units, therefore Triangle OTQ is equilateral. TDQ = 60° Since, PQ is tangent to the circle, so OQP = 90Deg Since, PQ is parallel to OA so AOQ = 90Deg For the same reason BOT = 90° AOB = 360° - (Angle TOQ +Angle AOQ +Angle BOT) = 120°**Last Sunday, every customer who visited the CENTRA MALL was given a gift coupon, on every purchase worth Rs. 1000, with a unique six-digit code written on it. Each code was such that-**

(i). The first digit was non-zero.

(ii). All the six digits were distinct.

(iii). The 1st and the 6th digits added up to 9 and so do the 2 nd and 5 th digits, and also the 3rd and 4th digits.

A gift was given to a customer who had two coupons with codes such that the numbers formed using the first three digits of each code were the reverse of each other.

The number of coupons distributed could not have been more than- 504
- 729
- 432
- 648
- None of The above

Answer: Option C.

The six-digit number on the coupon will look like

x y z 9 – z 9 – y 9 – x

Once we select the first, second and third digits of the number, the remaining three digits get fixed. The first digit can be chosen out of 1, 2, 3, …9 in 9 ways.

Then, the second digit can be selected in (10 – 2) = 8 ways and the third digit in (10 – 4) = 6 ways.

Hence, the maximum possible number of such six digit number is 9 × 8 × 6 = 432.