Wipro previous placement questions - 8

Wipro previous placement questions - 8

1. 12 members were present at a board meeting. Each member shake hands with all of the other members before and after the meeting .How many hand shakes were there?

a) 130

b)134

c)132

d)135

Answer: C

Explanation:

In order to have a hand shake there must be two members.  Therefore to select 2 out of 12 we have ¹²C₂.

This happens twice that is before and after the meeting

Answer = ¹²C₂ ×2 = 132

2. An emergency vehicle travels 10 miles at a speed of 50 miles per hour.  How fast  must the vehicle travel on return trip if the round trip travel time is to  be 20 minutes?

a) 75 mph

b) 70 mph

c) 65 mph

d) 80 mph

Answer: a

Explanation:

Going trip time = t = ds${\displaystyle \frac{d}{s}}$ = 1050×60${\displaystyle \frac{10}{50}}\times 60$ = 12 minutes.

Given total time = 20 minutes

Return trip time = Total time – Going trip time = 20 – 12 = 8 minutes = 860${\displaystyle \frac{8}{60}}$ hours.

As the distance is 10 miles during even return trip,

The return speed = dt${\displaystyle \frac{d}{t}}$ = 10860${\displaystyle \frac{10}{{\displaystyle \frac{8}{60}}}}$ = 10×608${\displaystyle \frac{10\times 60}{8}}$ = 75 mph

3. Mary and John can do a piece of work in 24 day; John and Vino in 30 days;Vino and Mary in 40 days. If Mary, John and Vino work together they will complete work  in ?

a) 10 days

b) 20 days

c) 47 days

d) 45 days

Answer: b

Explanation:

Given that

Mary and John take 24 days; i.e., (Mary + John)'s 1 day's work = 124${\displaystyle \frac{1}{24}}$

John and Vino take 30 days; i.e., (John + Vino)'s 1 day's work = 130${\displaystyle \frac{1}{30}}$
Vino and Mary take 40 days; i.e., (Vino + Mary)'s 1 day's work = 140${\displaystyle \frac{1}{40}}$

Adding above 3 equations, we get,

[(Mary + John) + (John + Vino) + (John + Vino)]'s 1 day's work = 124+130+140${\displaystyle \frac{1}{24}}+{\displaystyle \frac{1}{30}}+{\displaystyle \frac{1}{40}}$

2((Mary + John + Vino)'s 1 days work = 124+130+140${\displaystyle \frac{1}{24}}+{\displaystyle \frac{1}{30}}+{\displaystyle \frac{1}{40}}$

2(Mary + John + Vino)'s 1 days work =  5+4+3120${\displaystyle \frac{5+4+3}{120}}$ = 110${\displaystyle \frac{1}{10}}$

Therefore, (Mary + John + Vino)'s 1 days work =  120${\displaystyle \frac{1}{20}}$

i.e., Mary, John and Vino together can complete the work in 20 days.

4. My friend collects antique stamps. she purchased two, but found that she needed to raise money urgently. So she sold them for Rs. 800 each. On one she made 20% and on the other she lost 20%. How much did she gain or lose in the entire transaction?

a) she lost Rs 500.67

b) she lost Rs 666.67

c) she gain Rs 666.67

d) she gain Rs 500.67

Answer: b

Explanation:
Selling prices were given. Assume that on the first stamp she made profit and on the second stamp she made loss. 
So cost prices of the both stamps = 800120%+80080%${\displaystyle \frac{800}{120\mathrm{\%}}}+{\displaystyle \frac{800}{80\mathrm{\%}}}$ = 1666.66
So She incurred a loss of 66.66 rupees.

5. If the sum of n terms of two series in A.P. are in the ratio (5n + 4) : (9n + 6) then find the ratio of their 13th terms.

a. 129231${\displaystyle \frac{129}{231}}$

b. 12${\displaystyle \frac{1}{2}}$

c. 2315${\displaystyle \frac{23}{15}}$

d. None of the above

Answer: a

Explanation:

Formula for sum of n terms in AP  = Sn$S_n$ =  n2${\displaystyle \frac{n}{2}}$(2a + (n – 1)d ]
5n + 4 ⇒ 5(n – 1) + 9 ⇒ [2(92)+(n−1)5]$\left[2\left({\displaystyle \frac{9}{2}}\right)+(n-1)5\right]$

Common difference (d) = 5, First term (a) = 92${\displaystyle \frac{9}{2}}$

Similarly

Second series given 9n + 6 ⇒$\Rightarrow$ 9(n – 1) + 15 ⇒ [2(152)+(n−1)9]$\left[2\left({\displaystyle \frac{15}{2}}\right)+(n-1)9\right]$

Common difference (d) = 9, First term (a) = 152${\displaystyle \frac{15}{2}}$

So

13th term for first series is = a + 12d = 1292${\displaystyle \frac{129}{2}}$

13th  terms for second series is = a + 12d = 2312${\displaystyle \frac{231}{2}}$

Ratio = 129231${\displaystyle \frac{129}{231}}$

6. A team P of 20 engineers can complete work or task in 32 days. Another team Q of 16 engineers can complete same task in 30 days.Then the ratio of working capacity of 1 member of P to the 1 member of Q is

a. 3 : 2

b. 4 : 3

c. 2 : 5

d. 3 : 5

Answer: b

Explanation:
Let the capacity of an engineer in P = x units, and in Q = y units. 

Working capacity of  P = x × 32 × 20 

Working capacity of Q = y × 16 × 30 
As the total work is same, we equate the above equations.
⇒ x × 32 × 20 = y × 16 × 30
⇒ xy=16×3032×20=34${\displaystyle \frac{x}{y}}={\displaystyle \frac{16\times 30}{32\times 20}}={\displaystyle \frac{3}{4}}$

7. Ravi's salary was reduced by 25%.  Percentage increase to be effected to bring salary to original level = 

a. 20%

b. 25%

c. 33 1/3 %

d. None of the above

Answer: b

Explanation:

Let the Salary be 100.

Salary was reduced by 25%.  So present salary = 75. 

Percentage has to be increased in order to get Original level = 2575×100${\displaystyle \frac{25}{75}}\times 100$ = 33.33 %

8. An ore contains 25% of an alloy that has 90% iron. Other than this, in remaining 75% of the ore ,there is no iron. How many kgs of the ore are needed to obtain 60 kg. of pure iron.?

a.250

b.275

c.300

d.166.66

Answer: d

Explanation:
Let us take 100x kgs of ore. Now it contains 25x kgs of alloy and it contains 90% (25x) kgs of iron.
90% (25x) = 60 kg ⇒ x = 60×10090×125$60\times {\displaystyle \frac{100}{90}}\times {\displaystyle \frac{1}{25}}$ = 83${\displaystyle \frac{8}{3}}$
So iron ore required = 100 × 83${\displaystyle \frac{8}{3}}$ = 266.66

9. Find the day of the week on 16th july, 1776.

a. Sunday

b. Monday

c. Tuesday

d. Wednesday
Answer: c
Explanation:
Split the year 1775 + 16th july 1776
Till 1600 years no odd days. 
1601 to 1700 = 5 odd days
1701 to 1775 = 75 + 18 = 93 = 2 odd days.  (75 years has 93 odd days (∵$\because$ 18 leap + 57 non leap years)
upto 15th july 1776 = 31 + 29 + 31 + 30 + 31 + 30 + 15 = 197  = 1 odd day. 
Total odd days = 5 + 2 + 1 = 8 = 1 odd day. 
So one day after Monday. That is Tuesday. 

10. The radius of a sphere is increased by 50%. The increase in surface area of the sphere is :

a. 100%
b. 125%
c. 150%
d. 200%

Answer: b

Explanation:

Take radius 100. Then surface area is 4 × π$\pi$ × 100 × 100.
After increase radius by 50% the radius becomes100 + 50% of 100 = 150

Then new surface area is 4 ×  π$\pi$ × 150 × 150
Then put the values into formula of percentage = 4π1502−4π10024π1002×100${\displaystyle \frac{4\pi {150}^2-4\pi {100}^2}{4\pi {100}^2}}\times 100$ = 125%

11. On 8th Dec, 2007 Saturday falls. What day of the week was it on 8th Dec, 2006?
A.Sunday
B.Thursday
C.Tuesday
D. Friday
Answer: d
Explanation:
8th Dec 2007 – 8th Dec 2006 = 52 weeks 1 day.  So Dec 8th 2006 is behind one day = Friday

12. Which one of the following option is the closest in the meaning to the word given below..
MITIGATE
a) Diminish
b) Divulge
c) Dedicate
d) Denote

Answer: a

Explanation:
Mitigate means to make something less severe.  Divulge means reveal. Denote means indicate. Diminish means to make or to cause something to become less in size, importance etc.

13. On increasing the price of T.V. sets by 30%, their sale decreases by 20%. What is the effect on the revenue receipts of the shop ?
a. 4% increase
b. 4% decrease
c. 8% increase
d. 8% decrease
Answer: a
Explanation:
Let the price be = Rs.100, and number of units sold = 100
Then, sale value = Rs.(100 × 100) = Rs.10000
New sale value = Rs.(130 × 80) = Rs.10400
Increase% =  40010000${\displaystyle \frac{400}{10000}}$ × 100 = 4%

14. In an examination, 35% of total students failed in Hindi, 45% failed in English and 20% in both. The percentage of these who passed in both the subjects is :
a. 10%
b. 20%
c. 30%
d. 40%
Answer: d
Explanation:
Formula n(a∪b) = n(a) + n(b) − n(a∩b)
Fail in Hindi or English = 35 + 45 – 20 = 60
Therefore students who passed = 100 – 60 = 40. 

15. Find the angle between the minute hand and hour hand of a click when the time is 7.20?
Answer: 100 degrees
Explanation:
Use formula θ=∣∣∣30h−112m∣∣∣$\theta =\left|30h-{\displaystyle \frac{11}{2}}m\right|$
θ=∣∣∣30×7−112×20∣∣∣=1000$\theta =\left|30\times 7-{\displaystyle \frac{11}{2}}\times 20\right|={100}^0$

16. How will you measure height of building when you are at the top of the building? And if you have stone with you.
Answer:
Explanation:

Throw stone from top and note the time,t.
Find height using formula 
s = u × t + 0.5  × g × t²

Where u = 0