16/10/2018 10:33 pm
Algebra
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A quadratic polynomial P(x) passes through three points A(2, 5), B(3, -6) and C(10, 5) when plotted on Cartesian plane. What is the minimum value of the polynomial?
-141/7
-20
-139/7
-138/7
aniket prajapati 16/10/2018 11:09 pm
let a polynomial be
y=ax2 - bx + c
now we are given with three points put these points and we get 3 different equations
5=4a- 2b +c................(1)
-6=9a-3b +c..............(2)
5=100a -10b +c ...................(3)
solve them and find values of a,b,c
after solving we get a=11/7 ,b=132/7 ,c=255/7
minimum value of any quadratic polynomial is= -(b2-4ac)/4a
put values and we get answer -141/7
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17/10/2018 12:51 am
The sum of first 2013 terms of a geometric series is 300 and sum of first 4026 terms of same series is 480. Find the sum of first 6039 terms of same series.
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20/10/2018 3:31 pm
In an AP with 35 terms, the sum of first 4 terms is 122 and the sum of last 4 terms is 286. Find the sum of all the terms of the progression.
TG Team 20/10/2018 3:42 pm
Sum of first four terms + sum of last four = 4 ( sum of first + last term ) = 408
Sum of 35 terms : 35/2 [ first term + last term ]
35/2 × 102 = 1785
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20/10/2018 3:52 pm
In a Garden red flowers are n% of total flowers if we plant some more trees having 20 red flowers only, then red flowers becomes (n + 10)% of total flowers. How many values are possible for n?
TG Team 20/10/2018 7:05 pm
Total number of flowers : T
(Tn/100) + 20 = (T + 20)( n + 10)/100
Tn + 2000 = Tn + 10T + 20n + 200
1oT + 20n = 1800
10T = 1800 - 20n
T = 180 - 2n
n can take 9 values from 0 to 80.
[ 0 , 10 , 20 , ......., 80. ].
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20/10/2018 5:27 pm
A positive whole numbers m less than 100 is represented in base 2 notation , base 3 notation, and base 5 notation . It is found that in all three cases the last digit is 1, while in exactly 2 out of 3 cases the leading digit is 1 . Then M equals?
Utkarsh Garg 20/10/2018 6:10 pm
Since in all three cases the last digit is 1, the number should give remainder 1 when divided individually by 2,3,5 . So the no. may be 31 or 91 . Now 31 in base 2,3 and 5 give first digit as 1 in all the 3 cases while 91 gives exactly two out of the three cases the leading digit as 1
aniket prajapati 20/10/2018 6:17 pm
In precise it is simply saying that out of 4 option which no. when converted to base 2,3,5 will have last digit 1 for all three base(2,3,5) conversion and any two conversion will have leading digit as 1. let us see for 31(base 2 conversion)-11111 31(base 3 conversion)-1011 31(base 5 convertion0-111 here we see that all there conversion is having last as well as leading digit as 1 so second condition of having any two with leading 1 is not satisfied try with others. 63(base 5 conversion)---(223) it do not satisfy 1st condition so this will be incorrect. 75(base 5 conversion)---(300) it do not satisfy 1st condition so this will be incorrect. 91(base 5 conversion)---(331) it satisfy 1st condition. 91(base 3 conversion)---(10101) it satisfy 1st condition. 91(base 2 conversion)---(1011011) it satisfy 1st condition. so 91 is satisfying second condition also of having any 2 leading digit as 1.
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20/10/2018 7:00 pm
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20/10/2018 7:02 pm
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22/10/2018 12:05 pm
In a sale, out of 30 buyers, 10 buyers bought 5 or fewer items, 22 buyers bought fewer than 8 items and 14 buyers bought at least 7 items. Find the numbers of buyers who bought exactly 6 items.
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22/10/2018 2:30 pm
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22/10/2018 10:51 pm
sin36 sin72 sin108 sin144=?
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23/10/2018 11:20 am
please provide the solution for all the ques.
TG Team 23/10/2018 1:05 pm
1.
_ _ _ _ 0 _ _ _ _ :
for first four digits there are 9C4 ways , for last four digits there are 5C4 ways.
=> 9C4 x 5C4 = 630 numbers
_ _ _ _ 1 _ _ _ _ :
For first four digits select 4 digits from { 2, 3, 4, 5, 6, 7, 8, 9, } in 8C4 ways and last four in 4C4 ways.
=> 8C4 x 4C4 = 70 numbers
Total : 630 + 70 => 700 numbers .
aniket prajapati 23/10/2018 4:22 pm
2nd
let AB=a, BC=b, CA=c
now a+b+c=2
and we have to find the max value of ab+bc which can also be written as b(a+c)
from given equation a+c=2-b
b(a+c)= b(2-b)
=2b-b2
for any quadratic eq min value is -D/4a but in this condition since term of b2 is -ve graph will be inverted curve so min value will become maximum value
-D/4a= -4/-4=1 hence option 1st
TG Team 23/10/2018 6:31 pm
From 112 to 118 => sum of the product of the digits will be (1 x 1 x 2 )+ (1 x 1 x 4) , (1 x 1 x 6) , ......, (1x 1 x 8)
From 122 to 118 =>sum of the product of the digits will be (1 x 2 x 2) + (1 x 2 x 4) , (1 x 2 x 6) ,....., (1 x 2 x 8).
From 133 to 118 => sum of the product of the digits will be ( 1 x 3 x 2 ) + ( 1 x 3 x 4 ) + ........+ ( 1 x 3 x 8)
from 112 to 198
sum of all products : (2 + 4 + 6 + 8) x ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 ) i.e 20 x 45
and for 212 to 298 sum of all products 20 x 45 x 2 , for 312 to 398 => 20 x 45 x 3 and so on ...
Hence sum of all such products : 20 x 45 x ( 1 + 2 + 3 + ....+ 9) = 40500 .
Last three digits : 500
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23/10/2018 7:08 pm
Hello sir , kindly share the solution to this problem -
A retailer has 'n' stones of distinct integral weights (in kg) by which he can measure from 1 kg to 11 kg keeping them on either side of the weighing machine. Find the minimum value of 'n'.
aniket prajapati 23/10/2018 8:28 pm
This post was modified 6 years ago by aniket prajapati
1 kg weight is must
then for 2 kg weight we can have =1+1 or 2 0r 3-1
but 3-1 is best of them because then we can get 3 and 3+1=4 also from them
so for 1 to 4 we get two weights
for 5kg we can have 5kg or 4+1 or 7-3 or8-3 or 9-3
we can take either of the cases 8-3 or 9-3 to get all the weights from 5 to 11
6= 8-(3-1) , 7=8-1, 9=8+1 ,10=8+(3-1) ,11=8+3
so we used 1kg ,3kg ,8 or 9 kg
no. of weight used =3 (answer)
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23/10/2018 9:11 pm
N is a smallest such positive integer such that when It’s left most digit is removed, then remaining number is 1/37th of N. Find sum of digits of N.
13
14
15
16
TG Team 23/10/2018 9:44 pm
(abcde.....)/37 = bcde....
abcde.... = 37 × bcde....
a × 10ⁿ + bcde..... = 37bcde
a × 10ⁿ = 36bcde.....
Smallest value of a which makes LHS a multiple of 36 is 9.
So 9 × 10ⁿ = 36 bcde...
bcde... = 900/36
bcd...= 27
abcd..... = 927
Sum of the digits : 9 +2 + 7 = 16.
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23/10/2018 9:11 pm
n is a positive integer such that: 1 + 2 + 22 + 23 + … + 2n = 22n – 49. Find n.
6
5
4
3
aniket prajapati 23/10/2018 10:30 pm
its a G.P
sum of n terms of G.P =(arn-1)/(r-1)
here a=1 ,r=2
22n – 49= 2n -1
2n (3)=48
2n =16
n=4
