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# Arithmetic Aptitude :: Simplification

Home > Arithmetic Aptitude > Simplification > General Questions

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31. In a normal 5-day work week, each of 12 employees produces 300 items per day-last week, 1/3 of the employees were absent from work for 2/5 of the days in the week. how many total items were produced last week?

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Explanation:

1 employee produces 300 items per day
1/3 of the employees were absent for 2/5 days of last week.
=> 4 employees were absent for 2 days.
Items produced by them in 3 days = 300*3*4 = 3600
Items produced by other employees = 8*5*300 =12000
Total items produced = 12000+3600 = 15600

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Tags:  IBM

32. If a certain computer is capable of printing 4900 monthly credit card bills per hour, while a new model is capable of printing at a rate of 6600 per hour, the old model will take approximately how much longer than the new model to print 10000 bills?

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Explanation:

old model's 1 hr work = 4900
new model's 1 hr work = 6600
old model' time taken to print 10000 cards
for 1 hr -- 4900 cards
x -- 10000
x= 10000/4900 = 100/49
new model' time taken to print 10000 cards
for 1 hr -- 6600 cards
y -- 10000
y= 10000/6600 = 100/66
Difference is
100/49 - 100/66 = 1700 /(49*66) = 850/(49*33) = 0.525 hrs => 31 mins
Old model ll take 31 moreminutes to ptint 10000 bills

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Tags:  IBM

33. An Office used three types of envelopes: small,medium and large. They used 80 small envelopes. The no.of Medium envelopes used was 5 times, the no.of small envelopes plus 5/20 of the number of envelopes. The no.of.large envelopes was equals to the no.of small envelopes plus 5/20 of the no.of medium envelope. Altogether, How many envelope did they use?

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Explanation:

Lets the total no. of envelopes = x
No. of small envelopes = 80
No. of medium envelopes = [5*80+5/20*x] = [400+5/20*x]
No. of large envelopes = [80 +5/20(400+5/20*x)] = [180+25/400*x]
Sum of small, medium and large envelopes = x
=> 80 + 400+5/20*x + 180+25/400*x = x
=> 660 + 5/16*x = x
=> 11/16*x = 660
=> x = 960

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Tags:  IBM

34. Find the approximate value of the following equation.
6.23% of 258.43 - ? + 3.11% of 127 = 13.87

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Explanation:

[(6.23/100)*258.43]-'X'+[(3.11/100)*127]=13.87
16.100189-X+3.9497=13.87
X=6.179889

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Tags:  Accenture

35. The age of the grand father is the sum of his three grandsons.The second is 2 year younger than first one and the third one is 2 year younger than the second one. Then what will be the age of the grandfather?

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Explanation:

Lets assume second grandson = x year.
So, first grandson age = (x+2) year ,
third grandson = (x-2) year
So, x+2+x+x-2=3x
i.e grandfather age is 3 times as older as his second grandson

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Tags:  Accenture

36. If x(a) = a^2+2a-1, then x(8)-x(5) =

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Explanation:

x(8) = 64 + 16 - 1 = 79
x(5) = 25 + 10 - 1 = 34
So, x(8)-x(5) = 79 - 34 = 45

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Tags:  Accenture

37. If 3/p = 6 and 3/q = 15 then p - q = ?

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Explanation:

Given, p=1/2,q=1/5
So, p-q = (3/6 - 3/15) = (1/2 - 1/5) = 3/10

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Tags:  Accenture

38. Which of the following values of 'n' satisfies the in-equality n2 - 24n + 143 < 0?

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Explanation:

n^2 - 24n + 143 < 0
=> (n-11)(n-13) < 0
=> 11 < n < 13

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Tags:  Accenture

39. Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q's age?

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Explanation:

As per question R > Q > T
So, R - Q = Q - T and R + T = 50
By solving above 2 condition.
=> Q = 25
As the difference between R & Q and Q & T is same
So it is 25 years

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Tags:  Accenture

40. A box contains 90 bolts each of 100 gm and 100 bolts each of 150 gm. If the entire box weighs 35.5 kg., then the weight of the empty box is :

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