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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 11851. |
What is the formula of baking soda? |
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Answer» Nahco3 NaHCo3 nahco3 |
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| 11852. |
How to solve the equation x²+x-306=0 |
| Answer» Use middle spillt term method or quadratic formula | |
| 11853. |
Show that pp will a remainder when divided by 8 if p isan odd positive integer |
| Answer» Since p is an odd positive integer.Therefore, p = 2q\xa0+ 1 or 2q + 3 for some integer q.case 1: p = 2q + 1squaring both sides, we get,p2= (2q+1){tex}^2{/tex}=4q2\xa0+ 4q + 1= 4\xa0q(q+ 1)\xa0+ 1 Since we know that q(q+1) is even\xa0therefore , q(q+1)=2m{tex}\\implies {/tex}p{tex}^2 {/tex}\xa0={tex}4\\times2m+1{/tex}\xa0p{tex}^2 {/tex}\xa0=8m+1{tex}\\Rightarrow{/tex}\xa0p2\xa0leaves remainder 1 when divided by 8.case 2: p = 2q +\xa03p{tex}^2{/tex}=(2q+3){tex}^2{/tex}{tex}p^2\xa0= 4q^2\xa0+ 12q\xa0+ 9{/tex}{tex}p^2\xa0= 4q^2\xa0+ 12q\xa0+ 8 + 1{/tex}= 4q(q+3)+8+1\xa0={tex}4\\times2n+8+1{/tex}(since q(q+3) is even number, so q(q+3)=2n\xa0= 8n+8+1\xa0=8(n+1)+1 =8m+1, where m=n+1{tex}{/tex}{tex}\\therefore{/tex}p2\xa0leaves a remainder 1 when divided by 8. | |
| 11854. |
S and T are points on sides PR and QR ofΔ PQR such that ∠ P = ∠ RTS. Show thatΔ RPQ ~ Δ RTS. |
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Answer» F In triangle PQR and STR P=T givenR=R commontherefore, triangle RPQ~RTS |
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| 11855. |
3jgfjf |
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| 11856. |
solve using BODMAS[5 (2×3)+3-{(18-24)}÷[13+(-4)×2-3×(-7)]] |
| Answer» 3/2 | |
| 11857. |
What is collinear |
| Answer» When three or more points lie on a straight line. These points are all collinear\xa0Please note ( Two points are always in a line) | |
| 11858. |
Under root 5× under root 5 |
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| 11859. |
If n is an odd integer, then show that n2-1 is divsible by 8 |
| Answer» Let n = 4q + 1 (an odd integer){tex}\\therefore \\quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \\quad \\text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}{tex}= 16{q^2} + 8q{/tex}{tex}= 8 \\left( 2 q ^ { 2 } + q \\right){/tex}= 8m, which is divisible by 8. | |
| 11860. |
If tan A=cotB, prove that A+B=90° |
| Answer» As we know cot(90-A)=tanASo cot (90-A)=cotB90°-A=B90°=A+BHence proved | |
| 11861. |
Formulae of trigonometry. |
| Answer» See in last page of chapter or guide page no 318 | |
| 11862. |
Is - 156 is a term of ap 17, 14, 11, 8 |
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Answer» No No |
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| 11863. |
Prove 3 is irrational no |
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Answer» Let us assume that |
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| 11864. |
x^3-3x^2-3x+5 ➗ x^2-x-1 |
| Answer» X-2 is the quotient and -4x+7 is the remainder | |
| 11865. |
Four wall formula |
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Answer» 2(l+b)h 2(lb+bh+hl) |
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| 11866. |
cot +cosec-1%cot-cosec+1 |
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| 11867. |
(X have power 0) then x0=1 |
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| 11868. |
Is we find the area if pentagon |
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| 11869. |
Find two numbers whose sum is 27 and product is 183 |
| Answer» Let the two numbers be x, 27 - x.According to the question x (27 - x) =182{tex}\\implies{/tex}27x - x2\xa0-182 = 0{tex}\\implies{/tex}x2 - 27x + 182 = 0{tex}\\implies{/tex}\xa0x2 - 13x -14x + 182 = 0{tex}\\implies{/tex}x (x-13) - 14 (x-13) = 0{tex}\\implies{/tex}(x-13) (x-14) = 0Either x-13 = 0 or x - 14 = 0{tex}\\implies{/tex}x = 13, 14Hence, the required numbers are 13, 14 | |
| 11870. |
EF is parallel to qr why |
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| 11871. |
If sin theta + 2cos theta = 1 prove that 2sin theta - costheta = 2 |
| Answer» (sinθ + 2cosθ)2 = 12(sinθ + 2cosθ)2 + (2sinθ – cosθ)2 = 1 + (2sinθ – cosθ)2\xa05sin2θ + 5cos2θ = 1 + (2sinθ – cosθ)25 - 1 = (2sinθ – cosθ)2root4 = 2sinθ – cosθor, 2sinθ – cosθ = 2. | |
| 11872. |
5 (6-7? |
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| 11873. |
2x+3y=9 |
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| 11874. |
Which term f the sequence 114, 109, 104 .... Is the first negative term |
| Answer» 24th term | |
| 11875. |
Tips to perform best in exam |
| Answer» There are no shortcut to achieve success just be postive. And dont take tension be mently good. | |
| 11876. |
Fot |
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| 11877. |
Can i get my video class in English |
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| 11878. |
Paper pattern of 2017-18 |
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| 11879. |
y(y+2)(y+5)(y+3)-72=0 |
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| 11880. |
solve the quadratic eq. 4x^2 + 4x - (a^2 - b^2) by the method of FACTORISATION ONLY..N |
| Answer» We have,4x2 - 2 (a2 + b2) x + a2b2 = 0{tex}{/tex}{tex}\\Rightarrow 4x^2-(2a^2+2b^2)x+a^2b^2=0{/tex}{tex}\\Rightarrow{/tex}4x2 - 2a2x - 2b2x + a2b2 = 0{tex}{/tex}{tex}\\Rightarrow{/tex}\xa02x(2x - a2) - b2(2x - a2) = 0{tex} \\Rightarrow{/tex}\xa0(2 x - a2) ( 2x - b2) = 0{tex} \\Rightarrow{/tex}\xa0(2x - a2) = 0 or, (2x - b2) = 0{tex} \\Rightarrow \\quad x = \\frac { a ^ { 2 } } { 2 } \\text { or, } x = \\frac { b ^ { 2 } } { 2 }{/tex} | |
| 11881. |
Find roots :- 2x2-7x+3=0 |
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| 11882. |
Repeated questions of math SA 1 class X |
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| 11883. |
How many polynomials are there having 4 and -2 as zeros ? |
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| 11884. |
Derivation of the median formula for ungroup data |
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| 11885. |
Give tanA=4÷3 , find the other triangls ratiou of angel A |
| Answer» SinA = 4/5CosA = 3/5CotA = 3/4SecA = 5/3CosecA = 5/4 | |
| 11886. |
Prove that root 3 +root 5 is a rational number |
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| 11887. |
What happens when carbon mono oxide is added to calcium carbonate |
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Answer» R R |
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| 11888. |
If a,b,c are in AP,then show that a (b+c)/bc,b (a+c)/ca,c (a+b)/ab |
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| 11889. |
2/x + 2/y and 3/x+ 2/y=0 |
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| 11890. |
1+1 |
| Answer» 2 | |
| 11891. |
ax^2+bx+x=0 ,a not equal to 0 solve by quadratic formula |
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| 11892. |
A number multply by 3and added by 6 divide by 3 minus by the number |
| Answer» 50 | |
| 11893. |
2×2 |
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Answer» 4 but this is not a good question 4 |
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| 11894. |
Prove that the product of two consecutive integer is divided by 2. |
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| 11895. |
If n is an odd integer then show that n in a square minus one is divisible by 8. |
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| 11896. |
Prove 100-100 when divided by 100-100 gives 2 or 1/2 |
| Answer» 100-100÷100-100= 10×10 _10×10 ÷ 10( 10 - 10 )= (10+10) (10-10)÷ 10 (10-10)10-10 will be cancelled10+10 ÷1020÷10= 2 | |
| 11897. |
STN and PQR are similar triangle such t |
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| 11898. |
Triangle STN and PQR a |
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| 11899. |
Write all the trigonometric ratios in terms of secA |
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| 11900. |
Sum of prime numbers up to 19 |
| Answer» Prime numbers upto 19 are = 2,3,5,7,11,13,17,19Sum = 77 | |