Hi sir, please help
hi richa
PFA solution
thankyou
Consider the sequence of numbers - 1,22,333,4444,..... where each natural number 'n' repeats 'n'times. The 2015th number in the sequence will be?
2015th number or 2015th digit ?
Number
If the question asks about 2015th number, then it will simply be 201520152015..... ( 2015 written 2015 times )
Let D(N) denote the difference of sum of all even digits of N from sum of all odd digits of N. For example D(5678) = (5 + 7) – (6 + 8) = -2 and D(786) = (7) – (6 + 8) = -7.
Evaluate D(1) + D(2) + D(3) + …..+ D(9999).
0
4000
10000
20000
0000 , 0001 , 0002 , 0003 , ........, 9999
Number of digits used : 4 x 10000 = 40000
Each digit used : 40000/10 = 4000times
D(1) + D(2) + D(3) + …..+ D(9999)= ( Sum of all odd digits used ) - (Sum of all even digits used )
= ( 1+ 3 + 5 + 7 + 9) x 4000 - ( 0 + 2 + 4 + 6 + 8) x 4000 = 4000 x 5 = 20000.
Option (D)
solution for 7,8,10
PFA the solution :
Q. 7
10th.
Alternative Approach for 7th Question :
Put x = 2 and check whether u is 8.
Put x = 4 and check whether u is 4.
Put x = 16 and check whether u is 2 or not .
Only x = 4 satisfies the equation .
solution for questions 13,24,25,26 please.
24th
Solution to the 26th Question :
13th
Total yearly investment of Jatinder = (20000*4)+(16000*5)+(10000*3)=190000. For this investment, he is getting 19000 profit. Hence, for Ravinder to get Rs 19200 profit, his total investment should be 192000 for the year. Therefore, his initial investment should be 192000/12=16000.
FInd the odd term
13, 21, 34, 64, 89, 144
13 , 21 , (13 + 21) , ( 21 + 34) , ( 34 + 55) , ( 55 + 89) ... and so on .
Hence , 64.
Q41
PFA the solution :
If a, b, c are roots of x³ - 7x² - 6x + 5 = 0, find the value of (a + b)(b + c)(c + a).
a+b+c=7 .................(1)
ab+bc+ca= -6 .................(2)
abc= -5
(a+b)(b+c)(c+a)= (ab+ac+b^2 +bc)(c+a)
putting values from 1 and 2
(b² -6)(7-b)
7b²-b³ +6b-42
-(b³ -7b² -6b +42).................(3)
since b is a root of given polynomial therefore b satisfies given equation
b³ - 7b² - 6b+ 5 = 0
b³ - 7b² - 6b= -5
put in (3)
= -(-5+42)
= -37 (answer)
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
-141/7 is the answer?
yes. please share the solution
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
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.
736 ?
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.
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
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?
Answer is 50 ?
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. ].
