Miscellaneous  

Page 2 / 8
   RSS

0

112 Answers
0

Hi sir, please help

Capture

IMG 20181003 202625

 

hi richa 

PFA solution

thankyou

0

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 ) 

 

0

9oct

IMG 20181010 015504

0

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) 

0

BA5EC078 2399 45A1 B616 C340A62A1EB5

2754efb1 ed25 4a4e 89f9 5a164b2437c5

0

6B1C643B 1176 41DD 9A0F 69520DC8FED8

solution for 7,8,10

PFA the solution : 

Q. 7

Algebra 4.  7th

10th. 

Algbra 5 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 . 

0

1733773C D82B 424C 95EF 66E1DAD23663
D93033F7 7F83 426C B995 1094137E4184
F070C06F E7B5 4593 A46E 4265DA39D2B7

 

solution for questions 13,24,25,26 please. 

24th 

Algebra 4   24th Question.

Solution to the 26th Question : 

Algebra 4 . Q. 26

13th

1539176796716666929332

0

index

@utkarsh-garg25

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.

0

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. 

0

2a1d9df9 1e8e 42cc bfe3 9634feb88e1b

Q41

PFA the solution : 

Capture

0

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)

0

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

0

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 ?

1539766640442899564389

0

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

 

0

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. ]. 

Page 2 / 8
Share:

Please Login or Register