**Cost price of each of the articles A and B is Rs. X. Article A was sold at a profit of 10% and article B was sold at a profit of 30%. If the overall profit earned after selling both the articles is Rs. 136/-, what is the value of 'X'?**- Rs 340
- Rs 300
- Rs 360
- Rs 380
- Rs 320

Answer : Option A.

Given : 0.1x + 0.3x = 136 => x = 340**Population of a village increased by 5% from 2007 to 2008 and by 25% from 2005 to 2009. If the population of the village was 480 in 2007, what was its population in 2009?**- 640
- 610
- 630
- 620
- 650

Answer : Option C.

Population in 2007 = 480

In 2008 = 1.05 × 480 = 504

In 2009 = 1.25 × 504 = 630

**400 + 206 × 12 = x**- 2800
- 6666
- 4666
- 2400
- 2872

Answer : Option E.

400 + 206 × 12 = x

=> x = 400 + 2472 = 2872**430% of 25 + 75% of 430 = ?**- 340
- 860
- 516
- 86
- 630

Answer : Option A.

430% of 25 + 75% of 430 = x 430/100 ( 25 + 75 ) => x = 430**There is a natural number which becomes equal to the square of a natural number when 100 is added to it, and to the square of another natural number when 168 is added to it. Find the number.**- 189
- 69
- 156
- 224
- None of these

Answer : Option C.

Try by options. 3rd option is correct because 100 + 156 = 256 (square of 16) and 168 + 156 = 324 (square of 18).

No other option is satisfying the 2nd condition.Qualcomm Preparation Links- Sample Aptitude Questions of Qualcomm
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**In a rectangular auditorium, chairs are arranged in rows and columns. The number of chairs in each column is more than the number of chairs in each row by 5. If there are in all 300 chairs, find the number of chairs in each row and in each column.**- 25,20
- 30,10
- 23,18
- 25,12
- None of these

Answer : Option D.

Go by options. When there were 300 chairs, so product of chairs in rows and chairs in columns should be 300. Difference of 5 should be there between the seats in rows and columns. So no option out of 1st 3 is satisfying the conditions. So answer is 4th option and correct values are 20 and 15.**A person sets to cover a distance of 12 km in 45 minutes. If he covers of the distance in of time, then what is the speed in the remaining time?**- 16 km/hr
- 8 km/hr
- 12 km/hr
- 55 km/hr
- 10 km/hr

Answer : Option C.

Distance already covered = = 9 km, Time spent = min = 30 min Distance left = (12 – 9) km = 3 km, Time left = (45 – 30) min = 15 min ∴ Required speed = 3/(5/16) km/hr = 12 km/hr.**If a train 110 metres in length passes a man walking at the rate of 6 km/hr against it in 6 seconds, It will pass another man walking at the same speed in the same direction in time of**- 9 1/3sec
- 10 2/3sec
- 8 sec
- 6 sec
- 7 1/3 sec

Answer : Option E.

Let the speed of the train = x km/hr. Relative speed = (x + 6) km/hr = (x + 6)* 5/18 m/sec. ∴ (x + 6) * 5/18 * 6 = 110 ∴x = 60. ∴ Speed of train = 60 km/hr for 2nd person, Relative speed = (60 – 6) km/hr = 54 * 5/18 m/sec = 15 m/hr.∴Time taken to cross 2nd person = 110/15 = 22/3 = 7 1/3 sec.**The number of ways in which letters of the word PRAISE be arranged**- 720
- 610
- 360
- 210
- None of these

Answer : Option A.

Reqd. number of ways = 6! = 720.