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Suppose that number is xnowx+12= 160/xx^2+12x-160= 0

x^2+(20-8)x-160= 0

x^2+20x-8x-160= 0

x(x+20)-8(x+20)= 0

x= 8 x= -20x= 8 Answer

Let the natural number = c

If the number increased by 12 = x + 12

Reciprocal of the number = 1/x

According to the problem given,

x + 12 = 160 times of reciprocal of x

x + 12 = 160/ xx( x + 12 ) = 160x^2 + 12x - 160 = 0x^2 + 20x - 8x - 160 = 0x( x + 20) - 8( x + 20)= 0( x + 20 ) ( x - 8 ) = 0x + 20 = 0 or x - 8 = 0x = - 20 or x = 8But x is a natural number.

Therefore,Required number = x = 8

let the natural numbe be " n "

when it is increased by 12

it will be n +12

and when it will be n+12

it will equal to 160 times of its reciprocal

reciprocal of n would be 1/n

and now According to given condition

n+12 = 160 X 1/n

n+12 = 160/n

n^2 + 12n = 160

solve the quadratic equation by factorization

n^2 + 12n - 160 = 0

n^2 + 20n - 8n - 160 = 0

n(n+20) - 8(n +20) = 0

(n+20) ( n - 8) = 0

n = - 20

or

n = 8

n = -20 is not not possible because value of natural numbers always positive

so the required number n = 8

(1)40(2)-7(3)-57(4)-6(5)0(6)-20(7)-45

1.402.-7 3.-574.-6 5.06.-20 7.-45

=35/6*45/7+11/2=225/6+11/2=450+66/12=516/12=43

Let the number be x.80/100×x=3000x=3000×10/8 =3000×5/4=750×5 =3750

106 × 106 - 94 × 94

106² - 94²

(106 - 94) ( 106 + 94)

12 × 200

2400

7x-3= 187x= 21x=3 will be the answer

sin60°cos30° + cos60° sin30°=sin(60°+30°)=sin90°=1

Identity usedsin(A+B)=sinAcosB+cosAsinB

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By comparison2x= 6,x= 33y= 15,y= 54z= 20,z= 55w= 35w= 7

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4.7-2.3>0.3a2.4>0.3a8>aOption- B✓✓please like the solution 👍 ✔️

the value of x would be 10 degreeas 80+x= 90x= 10

4sin^2 A = 3sin^2 A=3/4sinA=√3/2A =arc sin(√3/2) =60°1+cosA=1+cos60°=1+(1/2)=3/2

Given:Pass percentage=94%Thus fail percentage=100-94=6%Also given 78 students failedThat is 6× x/100=78Here x is total students 6x=7800x=1300Therefore total candidates appeared is 1300

let the total number of candidates participated - xif 94 % candidates passedthen 6 % candidate failed

if number of failed students = 78 thanATQ6x/100 = 78

6x = 7800

x = 1300

Total number of boys + girls appeared in the examination = 3200 students

Total number of boys = 2000

Boys passed = 40 %

Boys failed = 100 - 40 = 60 %

Total number of boys failed = (60*2000)/100

= 1200 boys failed out of 2000 boys.

Total number of girls = 1200

Girls passed = 60 %

Girls failed = 100 - 60 = 40 %

Total number of girls failed = (40*1200)/100

= 480 girls failed out of 1200 girls

Total number of boys + girls failed = 1200+480 = 1680 Students

Percentage of failed students = (1680*100)/3200

= 52.5 %

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let no of candidates be Xhence 5% failed as 95% passedhence5/100*X=92X=92*100/5=18401840 students appeared for the examination.

there is a deduction of 5

i) 10,5,0,-5,-10,-15

(a) 28 6 8 14 4 2 6 8

(b) 8 1 6 3 5 7 4 9 2

(c) 4 8 12 16

(d) 1 4 9 16 25

=35÷ (9-2) (3+2) =35÷(7) (5) =35÷35=1

Let the cost price of watch = CPLet selling price of watch = SP

Given CP*(8/100) = 76 CP = 76*25/2 = 38*25 = 950

For 12% profitSP = CP(1 + 12/100) = CP(1+3/25) = 950(28/25) = 1064

Therefore for 12% profit selling price should be Rs 1064

Let the CP of each pen be Rs.x.

Then,

CP of 12 pens=Rs.12

SP of 12 pens=CP of 15 pnes=Rs.15

now,gain=SP-CP=Rs.(15-12)=Rs.3

gain%=gain/cp*100

=(3/12*100)%=25%

Hence,the gain=25%.

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(x+8)(x-8) = 0 x+8=0 or x-8=0 x = -8 or 8

As in power one Value c = 0 there both power of 2 and 5 will be 0 . Therefore thier value will be 1.

So, answer is 1

this is from the pair of linear equations in two variables chapter sir

Given equations are 2x+4y=10 and 3x+6y=12

now in these two equations takea1 =2 and a2 = 3b1 = 4 and b2 = 6c1 =10 and c2 =12

by a1/a2 =2/3b1 /b2 = 4/6 =2/3c1/c2 = 10/12 =5/6

when a1/a2 = b1/b2 ≠c1/c2the two equations have no solution