**The average age of a man and his son is 28 years. The ratio of their ages is 3 : 1 respectively. What is the man's age?**- 30 years
- 38 years
- 44 years
- 42 years
- None of these

Answer : Option D.

Total sum of man's age & his son's age =28 A— 2 = 56

Now, the Ratio of their ages is 3 : 1.

Therefore, Man's age = (3/4) A— 56 = 42

So, the correct answer is option D.**A car manufacturing plant manufactures 96 dozen cars in eight days. How many dozen cars will the plant manufacture in 17 day ?**- 210
- 224
- 204
- 209
- None of these

Answer : Option C.

The plant manufactures 96 dozen cars in 8 days

that means, It manufactures 96/8 = 12 dozen cars/day

So, In 17 days, it will manufacture 17 A— 12 = 204 dozen cars

Therefore, the answer is option C.

**A & B together can complete a piece of work in 16 days, B alone can complete the same work in 24 days. In how many days can A alone complete the same work ?**- 34 days
- 50 days
- 48 days
- 42 days
- None of these

Answer : Option C.

Let A alone can complete the work in x days and B alone can complete the work in 24days.

Therefore, according to the given conditions,

1/x + 1/24 = 1/16

1/x = 1/48

So, A's one day work is 1/48 which means that A will complete the entire work in 48days.

Therefore, the answer is option C.**Find the average of the following set of scores**

142, 93, 102, 206, 115, 98- 122
- 106
- 138
- 117
- None of these

Answer : Option E.

Adding all, we get 756

So, average = 756/6 = 126

So, the correct answer is option E (none of these).**The average of four consecutive odd numbers P, Q, R and S respectively (in increasing order) is 104. What is the sum of P & S?**- 204
- 208
- 206
- 212
- None of these

Answer : Option B.

Let the four consecutive odd numbers P, Q, R and S be X, X+2, X+4 and X+6 respectively.

So, average = X+3 =104

=> X = 101

So P = 101 & S = 107

And Sum of S and P will be 208.

Hence, the answer is option B.Siemens Preparation Links- Sample Aptitude Questions of Siemens
- Careers in Siemens
- Test Pattern & Selection Procedure of Siemens
- Free Mock Siemens Placement Paper
- All About Siemens
- Technical Questions asked in Siemens

**If tan α = n tan β and sin α = m sin β, then cos**^{2}α is- (m
^{2}- 1)/(n^{2}- 1) - (m
^{2}+ 1)/(n^{2}- 1) - m
^{2}/(n^{2}+ 1) - m
^{2}/n^{2}

Answer : Option A.- (m
**If cos Θ + sin Θ = √2cos Θ, then cos Θ - sin Θ is**- -√2 sinΘ
- √2 sinΘ
- √2 tanΘ
- -√2 cosΘ

Answer : Option B.

cosΘ + sinΘ = √2 cosΘ on squaring both sides,

cos^{2}Θ + sin^{2}Θ + 2 cosΘ. sinΘ =2cos^{2}Θ

=> cos^{2}Θ - sin^{2}Θ = 2sinΘ, cosΘ

=> (cosΘ + sinΘ) (cosΘ - sinΘ)

= 2 sinΘ, cosΘ

=> (2 sinΘcosΘ)/ √2 cosΘ = √2 sinΘ**If cos**^{4}Θ - sin^{4}Θ = 2/3 , then the value of 1 â€“ 2 sin^{2}Θ is- 2/3
- 1/3
- 4/3
- 0

Answer : Option A.

cos^{4}Θ - sin^{4}Θ = 2/3

=> cos^{2}Θ + sin^{2}Θ (cos^{2}Θ - sin^{2}Θ) = 2/3

=> 1- sin^{2}Θ - sin^{2}Θ = 2/3

=> 1- 2sin^{2}Θ = 2/3**Number of students institute A & B were in the ration of 7: 15 respectively in 2012. In 2013 the no. of students in institute A increased by 25% and the no. of institute B institute B increased by 26%, then what was the respective ratio between no. of students in institutes A & B respectively in 2013?**- 25:56
- 24:55
- 24:53
- 25:53
- 25:54

Answer : Option E.

Let the students in institute A and B 7x,15x

In 2013

Students in A = 5/4(7x)=35x/4

Students in B = 126/100(15x)

Required. Ratio = (35x)100/4(126)(15x) = 25/54

Hence , Option 5**An employer pays Rs.26/- for each day a worker works and forfeits Rs. 7/- for each day he is idle. All the end of 56 days. If the worker got Rs. 829/- for how many days did the worker remains idle?**- 21
- 15
- 19
- 13
- 17

Answer : Option C.

Say he works for x days

26x-7 (56-x)=829

33x=1221

x =37

As he works for 37 days so he is idle for 19 days

Hence, Option 3