**The base of a rectangle is seven times the height. If the perimeter is 32 meters, what is the area?**- 28 square meters
- 24 square meters
- 16 square meters
- 14 square meters

Answer : Option A.

Let b be the base and h be the height, b=7h, perimeter=2(h+b)=2(h+7h)=2(8h)=32,h=2 and b=14, so area = 14 X 2= 28 sq.mtrs.

**How many faces does a cube have?**- 4
- 6
- 8
- 12

Answer : Option B.

A cube is 3 dimensional figure with 6 faces.**Find the least number which must be subtracted from 6156 to make it a perfect square.**- 62
- 68
- 72
- 76

Answer : Option C.

72 is left as a remainder which must be subtracted from it to make it a perect square**DIRECTIONS for the question 4 to 5 : Questions are based on the information given below. Choose the most appropriate answer.**

Prabhat was married on 10th Dec 1955, his friend Robert was married on 21st March 1956 and their third friend Amarjit singh got married on 30th April 1956. If 10th Dec 1955 was Saturday:**On which day of the week was Amarjit Singh married?**- Tuesday
- Friday
- Monday
- Wednesday

Answer : Option C.

If 10th Dec 1955 was a Saturday then uptil 1st Jan 1956 there is one odd day so it will be one day after saturday i.e it is a Sunday, and similarly odd days from 1st Jan 1956 to 1st April 1956 is zero so it will also be Sunday, so is 8th April, 15th April, 22nd and 29th. Therefore 30th April is a Monday. Hence 3rd Option. Note: year 1956 is a leap year.**On which day of the week was Robert married?**- Monday
- Tuesday
- Wednesday
- Friday

Answer : Option C.

If 10th Dec 1955 was a Saturday then uptil 1st Jan 1956 there is one odd day so it will be one day after saturday i.e it is a Sunday, and similarly odd days from 1st Jan 1956 to 1st March 1956 is 4 so it will be 4th day from Sunday i.e. Thursday, so is 8th, 15th, 22nd and so 21st is a Wednesday. Note : year 1956 is a leap year.Hewlett-Packard Preparation Links- Hewlett-Packard Reasoning Questions
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**DIRECTIONS for the question 6 to 7 : Complete the series by choosing the appropriate option.****A, E, I, B, F, ___?___**- C
- D
- G
- J

Answer : Option D.

The alphabets are in a pair of 3 and there is a gap of 3 alphabets i.e. (+4) between two consecutive alphabets. Therefore, F + 4 = J.**16, 15, 36, 35, 64, ____? ____**- 50
- 61
- 62
- 63

Answer : Option D.

Every pair shows a difference of 1, so for 64 the no, would be 63.**DIRECTIONS-for the question 8-10 : The questions are based on the information given below. Choose the most appropriate option.**

Tara has decided to- tutor biology for the summer and is trying to plan her schedule. She can meet with up to two students a day, Monday 'through Friday, and has three students, Anu, Bhupesh, and Chirag. She needs to arrange her schedule according to the following constraints:

I. Each student meets with Tara twice a week.

II. No student meets with Tara twice in the same day.

III. Anu refuses to meet with Tara on Fridays.

IV. Chirag will never meet with Tara the day after Bhupesh has a lesson.

V. Anu and Chirag meet with Tara on the same day exactly once a week.

VI. If Anu meets with Tara on Monday, so does Bhupesh.**If Anu meets with Tara on Monday and Wednesday, on what day-must Chirag meet with her?**- Monday
- Tuesday
- Wednesday
- Thursday

Answer : Option C.Monday Tuesday Wednesday Thursday Friday Anu Yes Yes No Bhupesh Yes Chirag No No Yes **If Bhupesh is the only student to meet with Tara on a Wednesday, on what days must Anu meet with her?**- Monday and Tuesday.
- Monday and Wednesday.
- Monday and Thursday.
- Tuesday and Thursday.

Answer : Option D.Monday Tuesday Wednesday Thursday Friday Anu Bhupesh Chirag **If Anu is the only student to meet with Tara on Tuesday and no one meets with her on Friday, who meets with her on Wednesday?**- Chirag
- Anu and Chirag
- Bhupesh
- Anu and Bhupesh

Answer : Option B.Monday Tuesday Wednesday Thursday Friday Anu Bhupesh Chirag