Spider Chart Questions and Answers Quiz-2

The average monthly income for 2008 was 30,000.
Thus yearly income becomes 30,000*12= Rs 3,60,000
Assume income in January 2008 = x
And increment is 1500 every month. Thus february = x+1500
Thus it forms a series x+(x+1500)+(x+3000)+......+(x+11th term of AP) = yearly income
And the AP formed is 1500,3000,......, 11th term. => Solving the AP we get 11th term = 16500
Sum of this AP (with 1st term =1500, 11th term = 16500 and d=1500) = 99000
Thus the yearly income become 12x+99000 which must equal 360000
On solving, x = 21,750

2013 Income = 20% more than 2012 = 20% more of 40K = 40,000 + 8000 = Rs. 48,000
2013 Expenditure = 20% more than 2012 = 20% more of 25K = 25000+5000 = Rs 30,000

Monthly saving of 2013 = 18,000
Monthly saving of 2008 = 15,000
% increase = (18,000-15000)/15000 X 100 = (3000/15000) X 100 = 20%

On day1
Total work is LCM of time taken by All.
TW= 192 And
Efficiency is TW/Time taken
R= 192/32 = 6 , S= 192/16 = 12 , T= 192/64=3, U = 192/48=4 , V = 192/32=6
NOW ½ of the work is completed By R,S and T in 192×1/2×/(6+12+3) = 32/7
And remaining work 192/2 = 96 is completed by U and V on rotation.
Starting from U On 1st day work by U = 4
On 2nd day work by V= 6
In 2-days, total work done is = 4+6=10
In 18-day, total work done is = 10*9=90
On 19th day work by U is = 4
And rest work (96-90-4 = 2) is now completed by V in 2/6 days
So total work required is = 32/7 + 19 + 2/6 = 23.9 days or 24 days approx.

Time taken by R & S is 32 and 16 Respectively
And the ratio of efficiency is = reverse ratio of time
(Note TW=32 LCM of Time taken by them)
On Monday efficiency of R&S is (1:2) and sum of efficiency is 3.
On Tuesday if total work is same the efficiency is 32/12 & 32/36 and sum of efficiency is 32/9
Now the ratio is = 3:32/9 or 27:32

On Tuesday time taken by R, T & U to complete the work is 12, 24 and 60 respectively.
Total work is (LCM of time taken) – 120
Now their efficiency is (TW/time) is 10, 5 and 2 respectively work done alternatively,
then total work done by them in 3 days is = 10+5+2=17 (3-day work 17)
now they have to complete 17 times of total work 120 = 120*17
now the time require to finish the job is = 120*3 =360 days

On Monday time taken by R and S is 32 and 16 respectively
Total work (LCM of time taken) = 32
Efficiency of R & S on Monday (total work /time) R = 32/32 = 1 & S = 32/16 =2
Time taken to finish work on Monday is = 32/3
On Tuesday time taken by S and T is 36 and 24 respectively
Total work (LCM of time taken) =72
Efficiency of S & T on Monday (total work /time) S =72/36 =2 & T= 72/24 =3
Time taken to finish work on Tuesday is = 72/5
NOW % more/less = (32/3 – 72/5) ÷ 72/5 = 700/27%

On Tuesday times taken by T, U and V are 24, 60, and 48 respectively.
Total work is (LCM of time taken – 240)
The efficiency of T, U, and V is (TW/Time) T- 240/24=10, U- 240/60=4, V- 240/48=5
Now they have to complete the total work in 10 days
That means total efficiency required is = 240/10 = 24
But the combined Efficiency of T, U, and V is 10+4+5=19
So, T has to increase his efficiency by 5 in order to achieve total efficiency of 24
So % es increase in efficiency of T is 5/10*100= 50%

Total question 70
6 marks question = 5 out of which 2 wrong – marks = 3*6 – (6*2)*1/4 = 15marks
3marks question = 3*5=15 marks
2marks question = 2*10=20 marks
1 marks Question = 1*20=20 marks
0.5 marks question (12 wrong) = (0.5*18) – (12*0.5*1/4) =7.5
Total marks scored is = 77.5
And % is 77.5/100 *100 = 77.5%

Total 6&3 marks question from all subject = 25+25=50
Total 1&2 marks question from all subject = 50+40=90
% = (40/90) X 100 = 400/9%.
Thus Answer is ‘A’ option.

Total attempt is = 30
10 question of 1 marks,
16 question of 0.5 marks ,
and rest 4 question in 3:1 (3question of 6 marks and 1 question of 3 marks)
Total score is = 10+8+18+3=39
More marks than passing = 39-30 = 9 Marks

Minimum passing marks is =33
Minimum no of question that a person has to do is 5 question of 6 marks question and 1quesion of 3 marks
Minimum no of question = 6