[administration]

The sum of lowest 2 weight = 94 and the sum of highest 2 weights =104

Taking Option a) 40
It has to be the lowest weight.
so,  2nd lowest = 54
but sum of highest two = 104

So,  One of them will be less than or equal to 52 (not possible as both of them should be greater than 54….greater than the 2nd lowest)
b) 45
It has to be the lowest weight.
2nd lowest = 49
Highest = 101 – 49 = 52
=> 3rd lowest = 104 – 52 = 52

c) 48
It has to be the 2nd lowest
=> Lowest = 46
=> Highest = 101 – 48 = 53
=> 3rd lowest = 104 – 53 = 51
But 51 + 48 = 99 (not possible, as only possibilities are 94, 97, 101, 104)

d) 53
We will get the above values (so not possible)

So, 45 is the only possibility

[administration]

(x² + 4x + 5) / (x + 2) = (x2 + 4x +4 + 1) / (x + 2) = [(x+2)² +1]/(x+2) = (X+2) +1/(x+2)

 

Since y + 1/y >=2 or <=-2 for y=/= 0

so (x+2) +1/(x+2)  >=2 or <=-2 for x=/= -2 Hence -1 is the correct answer

[administration]

The correct answers are … Q1)  b i.e. 500 Q2) d i.e. 3360 Q3) c i.e. 8400 Q4) d i.e. 34 🙂 @cattack

[administration]

Yup @magnumopus22605 shadab… this one is also correct … 🙂 Ans is 34… 2 more to go 🙂

[administration]

ok right answer @magnumopus22605 shadab :)…. carry on guys/gals Q2, Q3, Q4 … go , go , go …

[administration]

So let me give a green signal to the Race with these 4 questions 🙂

 

Q1. A candidate must secure 40% of the total aggregate to pass an examination in which each question has equal marks. He attempts only 8 questions out of every 10 questions. On 75% of his answers, he secures only 26% of the marks and on the remaining of the answers, 100% of the marks. Thus he gets 22 marks, too few to pass. The maximum number of marks is
a) 425  b) 500  c) 380  d) 620

 

Q2.   In how many ways we can form committee of 7 members out of 6 boys nd 4 girls if at least one of the president and the secretary is female?

a) 768   b) 812 c) 1680 d) 3360  e) None

 

Q3. One plans to prepare schedule for 7 day period in which he shall play exactly one sport out of A,B,C,D on each day,number of ways in which a schedule can be prepared such that each sport is played on atleast one day in the 7 day period is ?
a) 4200  b) 6300   c) 8400   d) 53760   e) None

 

Q4.  All 4 digit no. using digits 1,2,3,4,5 once are taken and their product is found How many zeroz at the end will such a product have ?
a)18  b) 24  c) 30 d)34

[administration]

a⁴– 62a² + 1 = 0

Dividing by a² we get
=> a² + 1/a² = 62
=> (a + 1/a) = 8

a³ + 1/a³ = (a + 1/a)(a² – 1 + 1/a²) = 8x(62-1) = 488 Answer

 

[administration]

We need to minimize E/s

E/s = s^2 – 20s + 124 = (s – 10)^2 + 24
So, s = 10

=> E/s = 24

=> d = 72*10^5/24 = 300 km

 

soln by Hemant Yadav

[administration]

Let there are 100 houses
so number of items required = 300
number of item missing in whole village = 67+83+73=223
19 houses not have any 1 of these so
->100-19=81 houses should not have 2 or 3 item in there house
-> 223-19=204 item missing in this 81 house
let X number of house have 2 item missing so 81-X houses have 3 item missing
so-> 2X + 3(81-X)=204
X=39
81-x= 42 houses not have any of these item ..and 39 have 2 items missing

[administration]

g(k) = g(k-1)+f(k) = [g(k-2)+f(k-1)] +f(k)= [g(k-3)+f(k-2)]+f(k-1)+f(k) = [g(4)+f(5)] + f(6)+….+f(k)

 

so g(k)- g(4) = f(k)+ …+f(5);

g(10)-g(8) = f(10)+f(9)

f(1)=1, f(2)=2,

f(3)=6, f(4)=10,

f(5)=15, f(6)=21,

f(7)=28,f(8)=36,

f(9)=45, f(10)=55.

so g(10)-g(8) = 100,

 

now f(5)+f(6) +… f(k)= 100 so k =8.

[administration]

yeah acc to options it shud be 64 !! going by last 2-digits will be very lengthy

see 36! have 8 zeros and 24! will have 4

so wen we subtract fr last 2 non zerodigit we’ll have
xxx00- last 2 non zero digit of 24!

now last non zerodigit of 24! is 6
so ans has to be in form of x4 using options 64!! (but yeah risky!! as none of dese is dere) u cn find last two digits also bt a bit lengthy

answer by Yatish Jain

[administration]

putting x=3
9a+3b+c=0
from f(5)=-3f(2)
25a+5b+c=-12a-6b-3c
–>37a+11b+4c=0
als o 36a+12b+4c=0 from f(3)=0 multiplied by 4
–>a-b=0
–>from f(3)=0 we have 12a=-c
–>ax^2+ax-12a=0
–>x^2+x-12=0
–>2nd root=-4

[administration]

x⁴ + 1/x⁴ = [x²+1/x²]^2 – 2 (x²)(1/x²) = [(x-1/x)² +2x(1/x)]² -2 = [2² +2]² -2 = 34

[administration]

For an AP, T_n=a+(n-1)d
so its clear that, Tn=a(mod d)
Considering above question, Tn=a(mod 9)
So if we find  the remainder of  (Tn/9), we get the first term
66541=4(mod 9)
4⁸⁷⁶⁸=(4³)²⁹²² x 4²=1×16(mod 9)=7(mod 9)
Hence a = 7 Ans.

[administration]

Let grandmother has x sons

so each son has x-1 brothers and hence x-1 grandsons.

so total number of grandsons = x(x-1)

 

also since grandmother’s age is in between 50-60 .

so 50< x(x-1) <60 which is satisfied by x=8  and no other natural number
Hence grandmother’s age = 8×7 = 56

[Author]

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