31. What is the value of(a,y) in (13^-20-a*13^y)=168*13^-22?
Answer:
Explanation:(13^-20-a*13^y)=168*13^-22
13^-20 - 168*13^-22 = a * 13^y
13^-22 (13^2 - 168) = a * 13^y
Thus,
a = 13^2 -168 = 169 - 168 = 1.
y = -22 (substitute powers of 13)
Therefore, a=1, y=-22.
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32. Ramesh has solved 108 questions in an examination. If he got only '0' marks, then how many questions were wrong when one mark is given for each one correct answer and 1/3 mark is subtracted on each wrong answer.
Answer:
Explanation:If Ramesh attempts 'x' questions correct and 'y' questions wrong, then
x + y=108 ---(i) &
x - (1/3)y=0 ---(ii)
On solving x=27, y=81
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33. Total num of pupiles in 3 grades of scholl is 333. The no of pupiles in grades 1 and 2 are un 3:5 ratio and 2 and 3 grades are 7:11 ratio .What is the strength of class that has highest no of pupils?
Answer:
Explanation:Ratio G1:G2=3:5 and G2:G3=7:11
So G1:G2:G3=21 : 35 : 55
Let the strength of three classes are 21x, 35x and 55x respectively, then 21x+35x+55x= 333
=> 111x= 333 or x=3
So strength of the class with highest number of pupils=55x=55*3= 165
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34. A train is approaching a tunnel which is AB in length.Ther is a cat inside the tunnel which is 3/8 distance from the starting point of the tunnel.When the train whistes the cat starts to run.The train catches the cat exactly at the entry point of the tunnel.If the cat runs towards the exit,the train catches tha cat exactly at the exit point.The speed of the train is greater than the speed of the cat in what order?
Answer:
Explanation:1.Time taken for x km by train = Time taken for 3/8 of AB by cat
2.Time taken for x+8 km by train = Time taken for 5/8 of AB by cat
eq1 => x/v1=(3/8)AB/v2
eq2 => (AB+x)/v1=(5/8)AB/v2 => x/v1=AB((5/8v2)-1/v1)
LHS of eq1 and eq 2 are same so RHS of 2 eqns are equal so :
(3/8)AB/v2 = AB((5/8v2)-1/v1)
(3/8)/v2=(5/8v2)-1/v1
1/v1=(2/8v2)
1/v1=1/4v2
v1=4v2
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35. Total Rs.700 are divided among 3 persons (A, B, and C). A gets 1/2 of B and B gets 1/2 of C. How much C have?
Answer:
Explanation:A gets 100 Rs.
B gets 200 Rs.
C gets 400 Rs.
A=B/2
B=C/2
A+B+C=700
(B/2)+B+(2B)=700
B=200
A=B/2=200/2=100
C=2B=2*200=400
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36. A is 25% bigger than B, B is 20% less than C, then the relation between A, B, C?
Answer:
Explanation:Assume C = 100. Then, B is 20%less than C
Therefore, 20*100/100 = 20 ;
B= 100-20 = 80
So, B =80
A = 25% more than B :
So, A = B + 25% of B= 80 + 20 = 100
So, the proportion are 100:80:100 = 5:4:5
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37. The average of 11 observations is 60. If the average of first five observations is 58 and that of the last five is 56, then the sixth observation is :
Answer:
Explanation:Let the 6th observation be x.
According to the question,
Total = 11 * 60 -----(i)
Total of first five = 58 * 5 ---(ii)
Total of Last five = 56 * 5 ---(iii)
Now, 58 * 5 + 56 * 5 + x = 660.
=> 114 * 5 + x = 660.
=> x = 660 - 570 = 90.
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38. A man can complete a job in 12 days, his wife will complete in 15 days. How many days, working together they will finish the job?
Answer:
Explanation:man's one day work=1/12
wife's one day work=1/15
both=(1/12)+(1/15)=9/60
reciprocal of work is days
so 60/9=6 2/3 days.
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Hexaware Technologies
39. If 35 people play hockey and 10 people play both football and hockey and if 10 people play hockey and cricket and 5 people play all three games then the number of people playing hockey alone is?
Answer:
Explanation:hockey =35 (given)
hockey , cricket , football = 5(given)
hockey , cricket = 10(given)
hockey , football = 10(given)
only hockey = 35 - 5(h,c,f)-5(h,c (10-5)) -5(h,f(10-5))=35-5-5-5=20
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Hexaware Technologies
40. What is the discount for 'buy 6 and get 1 for free'?
Answer:
Explanation:let 6 article be purchased for Rs 100.
.'. each article worth Rs.100/6.
so for 7 article it would be 100+100/6.
Discount% is equal to what % of (100+100/6) equal to 100,minus by 100%.
i.e.100=x of (100+100/6),
100=x% of 350/3.
x%=300/350.
x%=0.85714.
discount=(1-0.85714)*100.
=14.28% is the disount percentage on 116.6666.
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