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**Answer**: Option A

**Explanation**:

Ramesh alone finished \(\frac{1}{2}\) of the work in 10 days.

Remaining \(\frac{1}{2}\) was finished by Ramesh and Dinesh together in 2 days.

Therefore, they both together can finish the complete job in 4 days.

**Q2. Ram decided to plough a farmland in 50 days. He employed 50 men in the beginning and 50 more after 35 days and completed the construction in stipulated time. If he had not employed the additional men, how many days behind schedule would it have been finished?**

**Answer**: Option A

**Explanation**:

Given that, 50 men employed for 35 days and 50 more men employed for 15 days so that the work could be finished in 50 days.

Thus, the total work = 50*35 + (50 + 50) 15 = 3250

Suppose, it takes x days to finish the whole work if additional men were not employed.

So, we have an equation here.

50*x = 3250

x = 65 days

Therefore, it takes 15 days more than the stipulated time.

**Q3. 20 men can complete a piece of work in 15 days and 12 women can complete the same piece of work in 24 days. What is the ratio of amount of work done by 30 men in one day to the amount of work done by 16 women in 1 day ? **

**Answer**: Option B

**Explanation**:

M : W = \(\frac{30}{20*15}\) : \(\frac{16}{12*24}\)

= \(\frac{1}{10}\) : \(\frac{1}{18}\) = \(\frac{1}{5}\):\(\frac{1}{9}\)

Ratio = 9:5

**Q4. X alone can do a piece of work in 5 days. Y can do the same piece of work in 4 days. X and Y are assigned to do the work for Rs.5000.They complete the work in 2 days with the help of Z. How much is to be paid to Z ?**

**Answer**: Option D

**Explanation**:

Z’s one day work = \(\frac{1}{2}\) – [ \(\frac{1}{5}\) + \(\frac{1}{4}\)] = 10 – 4 + \(\frac{5}{20}\) = \(\frac{1}{20}\)

Ratio = \(\frac{1}{5}\):\(\frac{1}{4}\):\(\frac{1}{20}\) = \(\frac{4}{20}\) : \(\frac{5}{20}\) : \(\frac{1}{20}\) = 4:5:1

Z = 5000*\(\frac{1}{10}\) = 500

**Q5. 9 men and 12 women can complete the job in 12 days.In how many days can 3 men and 4 women finish the same job working together ?**

**Answer**: Option A

**Explanation**:

9m + 12w = \(\frac{1}{13}\)

3m + 4w = ?

3(3m + 4w) = \(\frac{1}{13}\)

3m + 4w = \(\frac{1}{36}\)

**Answer**: Option B

**Explanation**:

S+T = 23*(39-37) = 46

\(\frac{S}{T}\) = \(\frac{15}{8}\)

T = 16

**Q2. The ages of Four members of a family are in the year 2010 are ‘X’,’X+12’,’X+24’ and ‘X+36’. After some years Oldest among them was dead then average reduced by 3. After how many years from his death, the average age will same as in 2010?**

**Answer**: Option B

**Explanation**:

In 2010: 4x+\(\frac{72}{4}\) = x + 18

After death : 3x + 36 + \(\frac{3N}{3}\) = x + 18 – 3

N = 3 years

3x + 36 + \(\frac{3N}{3}\) = (x + 18)

N = 6 years

6-3 = 3 years from his death

**Q3. The average of Four numbers is 24.5. of the four numbers, the first is 1.5 times the second, the second is 1/3 rd of the third, and the third is 2 times the fourth number. Then what is smallest of all those numbers?**

**Answer**: Option C

**Explanation**:

First = 1.5x Second = x Third = 3x Fourth = 1.5x

average = 24.5 = \(\frac{(1.5x + x + 3x + 1.5x)}{4}\)

x = 14

**Q4. There are 459 students in a hostel. If the number of students increased by 36, the expenses of the mess increased by Rs .81 Per day while the average expenditure per head reduced by 1. Find the original expenditure of the mess?**

**Answer**: Option D

**Explanation**:

Total expenditure = 459x

36 students joined then total expenditure = 459x +81

average = 459x + \(\frac{81}{495}\) = x-1

x = 16

original expenditure = 16*459 = 7344

**Q5. The average cost 32 different Mobiles is Rs. 9000. Among them, Oppo which is the costliest is 70% higher price than the cheapest Mobile Lava. Excluding those both mobiles, the average of the Mobiles is Rs.8880. Then what is the cost of Oppo Mobile?**

**Answer**: Option D

**Explanation**:

L+O = 21600

O = L*\(\frac{170}{100}\)

O = 13600

**Answer**: Option B

**Explanation**: \(\frac{8-x}{12 –x}\) = \(\frac{12-x}{20 –x}\)

(8-x)(20-x) = (12 – x)(12 –x )

160 – 8x – 20x + \({x}^{2}\) = 144 – 12x – 12x + \({x}^{2}\)

4x = 16

x = 4

**2. The sum of square of 2 number is 1972 and the difference of their square is 620. Find the number ?**

**Answer**: Option A

**Explanation**:

\({x}^{2}\) + \({y}^{2}\)= 1972

\({x}^{2}\) – \({y}^{2}\) = 620

Solve ths

2\({x}^{2}\) = 2592

\({x}^{2}\) = 1296

x = 36

\({y}^{2}\) = 1972 – 1296

\({y}^{2}\) = 676

y = 26

**3. Product of one fourth of the number and 120% of another number is what % of the product of the original number ?**

**Answer**: Option D

\(\frac{x}{4}\) = one fourth of the no

\(\frac{120y}{100}\) = another no = \(\frac{6y}{5}\)

% = [(\(\frac{x}{4}\) * \(\frac{6y}{5}\))*100]x \(\frac{1}{xy}\)

= \(\frac{600}{20}\) = 30%

**4. If three numbers are added in pairs, the sums equal to 11,16 and 23.Find the three numbers**

**Answer**: Option B

**Explanation**:

X + y = 11

Y + z = 16

Z + x = 23

X + y + z = 25

X = 25 – 16 = 9

y = 25 – 23 = 2

z = 25 – 11 = 14

**5. What is the no of zeroes at the end of the product of the no from 1 to 100 ?**

**Answer**: Option C

**Explanation**:

\(\frac{100}{5}\) + \(\frac{100}{25}\) = 20 + 4 = 24