835, 734, 642, 751, 853, 981, 532
Three man’s on day work = 1/6+1/7+1/8=73/168
three man’s can complete the work in 1/73/168 days
= 168/73 days
Now if they works together fro the alternet days they will
complete the worksin 2*168/73 days
(If three mans working for the alternate days then work
completion time will be doubbled)
1(1/7)hr
(p*r*t)/100 = I ………………..(p*3*10)/100 = 840 …..p = 84000/30 = 2800
1261
copy cat
( a ) 10 metres
(3/4)*(L/Sp)=30
L/Sq=75
(3/4)*(Sq/Sp)=30/75
Sq/Sp=8/15
Sp/Sq=15/8
Answer is 45
First we need to subtract those reminders from the respective numbers, then we have to find the hcf of two numbers(numbers got from the subtraction) then you will get the answer.
So,
After subtraction you will get
3026-11 = 3015
5053-13 = 5040
HCF of these two numbers
5 | 3015 5040
3 | 603 1008
3 | 201 336
| 67 112
We can’t find a common diviser since 67 is a prime number
So the HCF = 5 * 3 * 3
= 45
5*3*3 = 45
34
Dear friends i ofcourse confused like you when i come
across to solve this problem. Really friends its very
simple if you understand the question clearly. First thing
is what is mean by “as many as” means its called ‘idiom and
phrase’ in english and it means “the same number of”. now
read the question “how many pairs of letters in STAINLESS
which has same number of letters between them in the word
as they have in english alphabet”.
In the alphabetical order, A-Z can be numbered as 1-26.
In A(INL)E which is same as in the alphabetical order A
(BCD)E. In both the cases E is in the Fourth position. so
we got one pair.
And in ST, there are no letters between them in the word
stainless. In alphabetical orer from A-Z also there is no
letters between them..so we got the second pair…
In STAINLESS it has two pairs ST and AE
x/y =5/7…………………………..(1)
(x-25)/(y-25)=35/39
59x=35y+600
dividing b.t.s by y and substituting (x/y) value from equation (1)
y=84
Now, substitue y value in equation (1)
x = 60
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835,sum(digits)= 16
734 = 14
642=12
751= 13,
853=16,
981=18,
532=10
answer(c)751