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Total marks awarded to boy for correct answer = 12×5=60Marks deducted for wrong answer=8×2=16Total score = 60-16=44

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1/6 is the right answer

1/6 is the right answer

let X be the no of questions he got right and y be the no of of questions he got wrong.x+y=60(Total no. of questions)2x-y=90(2 marks for right questions and -1 for the wrong ones)Adding both the equations,we get x=50Therefore,y=10.

IN TRIANGLE AOB, AO=BO=6cmTHEREFORE,∠OAB=∠OBA=y(say)IN TRIANGLE AOB BY ANGLE SUM PROPERTY60°+y+y=180°⇒180°-60°=2y⇒120°=2y⇒y=60°THEREFORE AS∠A=∠B=∠C=60°THEREFORE TRIANGLE AOB IS AN EQUILATERAL TRIANGLE⇒LENGTH OF CHORD AB =6cm.

if 13 is the lenght of the chordwhen the PQ is the perpendicular to AB then the QB is 13×1/2=6.5 cm

22 cm is the correct answer

wrong hai bro

(√6)²ⁿ/5 = 6ⁿ/5 = 1/5*(5+1)ⁿ , The last term of the expansion of (5+1)ⁿ = 1ⁿ = 1 .. other than that all will be divisible by 5

so, the last term will be 1/5 and that will be the fractional part..

option C

numbers in the blanks are following:1) 252) 243) 164) 35) 10

aap kaun si class mein

the correct answer is option B

(2-2^1/2)^1/2 is the answer.

56000000000000000000

(-50/7) × (14/3) (-100/3)

Given ΔABC, extend BA to D such that AD = AC.Now in ΔDBC∠ADC = ∠ACD [Angles opposite to equal sides are equal]Hence ∠BCD > ∠ BDCThat is BD > BC [The side opposite to the larger (greater) angle is longer]Þ AB + AD > BC∴ AB + AC > BC [Since AD = AC)Thus sum of two sides of a triangle is always greater than third side.

x12–y12=x6 × 2–y6 × 2= (x6)2– (y6)2

= (x6–y6) (x6+y6)

= ((x3)2– (y3)2) ((x2)3+ (y2)3)

= (x3–y3) (x3+y3) (x2+y2) ((x)2–x2y2+ (y2)2)

= (x–y) (x2+xy+y2) (x+y) (x2–xy+y2)(x2+y2) (x4–x2y2+y4)

do the same for z^12

simran Singh your solution is wrong.

how?

extend it with z^12

x^12–y^12-z^12=x^6 × 2–y^6 × 2 -z^6×2= (x^6×2)– (y^6×2) -(z^6×2)

= (x6–y6) (x6+y6) -(z^6)

= ((x3)2– (y3)2) ((x2)3+ (y2)3)

= (x3–y3) (x3+y3) (x2+y2) ((x)2–x2y2+ (y2)2)

= (x–y) (x2+xy+y2) (x+y) (x2–xy+y2)(x2+y2) (x4–x2y2+y4)

Tan15 × tan75 × tan 105 ×tan 165tan (90-75) ×tan75 ×tan (270-165)×tan165cot75× tan75×cot165 ×tan1651×11

for this first find total number of set of beads and then subtract from wholehence4/17+3/17+5/17=12/17subtract from wholehence1-12/17=17-12/17=5/17

thanks for the answer

please like my answer if you find it useful

Subsets of:-i) {a} = { }, { a}ii) {a,b}= { }, {a},{b},{a,b}iii) {1,2,3}= { }, { 1}, {2},{3}, {1,2},{1,3}, {2,3},{1,2,3}

2's power has repetitive sequence of last digit that is 2,4,8,6,2,4,8,6,...so 2^2003's last digit is 8so fractional part of 2^2003/17=8/17

Let us suppose the no. of correct answers is x and wrong answers be yThen -x + y = 120and5x - 2y = 348 according to the given two conditions

solving eq. 1x = 120 - yputting this value in eq. 25 (120-y) - 2y = 348600 - 5y -2y = 348-7y = 348 -600 = -252y = -252 / -7 = 36so, x = 120 - 36 = 84so, he answered 84 correct and 36 wrong answers.

let he answered x correct and y incorrect answers

x+y=120....... (1)

and

5x-2y=348....... (2)

multiplying eq.1 by 2 and adding to eq.2,we get

7x=588x=588/7x=84

put in eq. 184 + y=120y=120-84y=36

5/3 - root(2) + 2/3 + root(2)

= 5*(3+root(2)) + 2*(3-root(2))/ (3-root(2))*(3+root(2))

= (15 + 5root(2) + 6 - 2root(2))/ ( 9 - 2)

= (21 + 3root(2))/7

= 3 + 3/7 root(2)

Let theta = AThen,x = acosecAx/a = cosecA..... (1)

y = bcotAy/b = cotA...........(2)

Squaring eq(1) and eq(2), x^2/a^2 = cosec^2 A.... (3)y^2/b^2 = cot^2 A......... (4)

Subtract eq(3) from eq(4)

x^2/a^2 - y^2/b^2 = cosec^2 A - cot^2 A= 1

Hence proved

He please tell me simplest answer of this question

Let us assume that √2 + √3 is a rational number

Thus,√2 + √3 = (a/b) where a and b are co-prime numbers

On squaring both sides , we get2 + 3 + 2√6 = (a²/b²)

So,5 + 2√6 = (a²/b²) a rational no.

So, 2√6 = (a²/b²) – 5

Since, 2√6 is an irrational no. and (a²/b²) – 5 is a rational no.

So, our contradiction is wrong.So, (√2 + √3) is an irrational no.

Common difference is 5Let first term is at18=a+17dt13=a+12d

t18-t13=a+17d-a-12d=5d=5(5)=25 Option c is correct

Thank u soooo much

multiply equation 2 by 2X+2y=136x+2y=28subtract the two equations-5x=-15hence X=3put in equation 13+2y=132y=10y=5

answer

I have already posted the answer

i) Take out a^2b^2 common you get

a^2b^2( sec^x - tan^2x)

= a^2b^2 because ( sec^x - tan^2x) =1

Given :a(11) = 38a(16) = 73Let first term be a and common difference be d.So,We know a(n) = a + (n-1) d

38 = a + 10d .....(1)73 = a + 15d ......(2)

eq(2) - eq(1)35 = 5dd = 7So, a = 38 - 70 = -32Therefore,a(31) = a + 30d = -32 + 210 = 178

Good reply sister

thank-you much, can you give me more solutions