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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.
| 9601. |
The hcf of 135 and 225 |
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Answer» 45 45 45 45 45 |
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| 9602. |
Find the HCF and LCM of 12,15,18,27. |
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Answer» Hcf and lcm solutions Mein kal hi solve ki answers bhi aya thanks app loge na bhi help ki meri. Thanku LCM =540, ,HCF =3 HCF - 3LCM - 3^3 × 2^2 × 5 |
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| 9603. |
For what value of k will the equation 3X+ y=1 and (2k-1)X+(k-1) y= 2k+1 have no solution |
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Answer» 3/2k-1=1/k-1 3k-3=2k-13k-2k=-1+3K=2 K = 2 K is not equal to 2 k=2 |
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| 9604. |
The value of m,n respectively , if 108 = 2^m × 33 × 5^n are :A) 2,0B)3,1C)0,1D)2,2 |
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Answer» 108=2^2×3^3×5^0 So,M=2 and 5=0 Hence option A is correct answer 108 = 2²×3³×5⁰ therefore m=2 and n=0. Hence option A is correct |
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| 9605. |
What is an algebraic expressions |
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Answer» ? ur in class 10th and asking this silly questions ?.Will j be passing boards this year?Really verrey much stupid and dumb? Algebraic expressions are the expressions which consist of unknown values / numbers expressed as variables i.e. x,y,z and many more ....Also these are equal to zero |
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| 9606. |
If A(4, 9), B(2, 3) and C(6, 5) are the vertices of ABC, then the length of median through C is |
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Answer» Under root 10 unit 10 unit 10 unit is answer |
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| 9607. |
3 dice are thrown together the probability of getting the same no. On the all the dice is |
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Answer» 1/4 1/4 |
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| 9608. |
In figure,we have AB//CD//EF . If AB=6cm,CD=x cm,EF=10cm andDE =ycm.find x and y |
| Answer» Bhai figure kis ka h? | |
| 9609. |
What is the centroid of a triangle |
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Answer» X¹+X²+X³ Y¹+Y²+Y³------- \' ------- 3. 3 The point where the medians of three side intersect each other |
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| 9610. |
If of tan^2 A=1+2tan^B then value of cos^2A/cos^2B |
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Answer» Heeloe Nhi malum |
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| 9611. |
Mcqs type s questions |
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| 9612. |
sin^2 + sin^2 A tan^2 A |
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Answer» sin²A + sin²A*tan²A= sin²A(1+tan²A)= sin²A*sec²A= (sin²A)/(cos²A)= tan²A 0 |
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| 9613. |
How to do lcm and hcf |
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Answer» Do throw semple paper of Dapeeak By your own mind ?? |
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| 9614. |
The positions of A,B respectively |
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| 9615. |
In fig, 9.22,da |
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| 9616. |
Sin theta=3/1waht is the value of cos theta |
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Answer» It is not possible for sinA to be greater than +1 ; ( -1 It is not possible for sinA to be greater than +1 ; ( -1 Sin=P/H i.e. P=3 and H=1 Now,we have to find B=? By PGT (1)^2 - (3)^2 = (B)^2 1-9 = (B)^2 B= -√8 Since, Cos=B/H = -√8/1 Root of 8/1 |
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| 9617. |
4x2+6x+3 |
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Answer» Here ,D = b²-4.a.c = 36-(4.4.3) = -12 . Therefore , it has no real roots . 11+6x |
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| 9618. |
I want Chennai sahodaya complex class 10 maths science social English and Hindi papers 2021-2022 |
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| 9619. |
In an equilateral triangle PQR if PS QR then PS² = |
| Answer» Question wrong hai.. | |
| 9620. |
prepare a chart that contain all the formula related in first term |
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Answer» ???diy Maths*Polynomial Sum of zeros (alpha+beta)= -b/a Products of zeroes(alpha×beta)=c/a*Linear equations in two variable - if A1/a2=b1/b2=C1/C2 then in this condition graph will coincide nd will have infintly many solutions.-if A1/a2=b1/be€(not equal to)C1/c2 then grpah will become parallel line nd will have no solution.-if A1/a2 is not equal to b1/b2 then graph will intersect nd will have exactly one solution.* Areas related to circle -Area of circle=πr2- Circumference of circle=2πr- Area of sector=thetha /360°×πr2 or- Area of sector= 1/2 × r × l - Area of segment=πr^2theta/360°-1/2r2sin thetha-Area of ring π(R2-r2)or π(R+r)(R-r)*Coordinate geometry- for a point PQ in graph PQ=√(x2-x1)^2+(y2-y1)^2#TrignometrySin = 1/cosecCos =1/secTan=1/cot Tan = sin/ cosCot=cos/sin*IdentitiesSine^2theta +cos^2thetha =1 Sorry if I have written any formula wrong ....nd hope it helps!! |
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| 9621. |
if HCF (72,120)=24 then LCM(72,120)is |
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Answer» 360 360 360 3y0 360 |
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| 9622. |
If xsi A=5 and 7cosecA=6secA, then the value of x is |
| Answer» 7cosecA = 6 secA Therefore 7/6 = secA/cosecAOr sinA/cosA = 7/6Or tanA = 7/6Now draw a triangle with base = 6k and perpendicular = 7k apply Pythagoras theorem to get hypotenuse = (√85)k therefore sinA = 7k/(√85)k = 7/√85 therefore xsinA = 5 implies7x/√85 = 5 therefore x = (5√85)/7 | |
| 9623. |
135 and 234 LCM |
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Answer» Answer is 3,510 ANSWER IS 3510 |
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| 9624. |
The solution of 7^x+y=343 and 343^x-y=7 isPls answer ?\u200d♀️?\u200d♀️ |
| Answer» X = 5/3 and Y = 4/3 | |
| 9625. |
Sin theta +cos theta =? |
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Answer» Ratio of opposite side of hypotenuse Reduction Sin theta +cos theta is sin 45 + cos 45= 1/root 2 +1/root 2 = 2/root 2 = root 2 1+2sincos in root 1_2 |
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| 9626. |
If one zero of the polynomial 3x²-8x+2k+1 is seven times the other find the value of k ? |
| Answer» Polynomial:- 3x²-8x+2k+1Let one zero be α Therefore second zero is 7αNow sum of zeroes= α+7α= -b/a=> 8α = -(-8)/3=> 8α = 8/3=> α = 1/3 —1Now product of zeroes =>α × 7α = c/a=>7α² =(2k+1)/3Putting value of α from —1=>7×(1/3)² = (2k+1)/3=>7×1/9= (2k+1)/3=>7/9=(2k+1)/3=>21=18k+9=>12=18k=> k = 12/18=> k= 2/3 | |
| 9627. |
Line formed by joining points (-1,1) and (5,7) is divided by a line x+y=4 in the ratio |
| Answer» Hence, the line x + y = 4 divides the line joining the points (-1, 1) and (5, 7) in the ratio of 1:2 internally | |
| 9628. |
How many solutions does the pair of equations x+ y =1 and x+ y = -5 have ? |
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Answer» No solution No solution This ques is actually wrongAs If x and y are repeated then the sum of both can\'t differ no 3 |
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| 9629. |
The sum of 3 numbers in AP is 54 and the product of the extremes is 275 find the number |
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| 9630. |
3.3 |
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| 9631. |
If 2x+y=23 and 4x-y=19 then the value of (5y-3x) and (y/x-2) are: |
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Answer» By elimination method we have: 2x + y = 23 ...?➡️ 4x - y = 19 ...?➡️➡️ 6x = 42 ....?➡️➡️➡️From eq.3 we have x = 7On substituting value of x in any of the eq: (let be in eq.2)2x + y = 23➡️ {We know, x=7}2(7)+y = 23➡️14+y = 23➡️y = 23-14➡️y = 9➡️➡️➡️Now, we have values of both x and y {x=7; y=9}➡️➡️➡️Final ques➡️Find value of (5y-3x) and (y/x-2)➡️➡️➡️Part-1➡️5y-3x {x=7; y=9}➡️5(9)-3(7)➡️45-21➡️Ans [24]➡️➡️➡️➡️Part-2➡️y/x -2 {x=7; y=9)➡️9/7 -2➡️9/7 -14/7➡️Ans [-5/7]➡️ By elimination method we have: 2x + y = 23 ...?➡️ 4x - y = 19 ...?➡️➡️ 6x = 42 ....?➡️➡️➡️➡️➡️➡️From eq.? we have x = 7➡️➡️➡️➡️➡️➡️➡️On substituting value of x in any of the eq: (let be in eq.?)➡️➡️➡️➡️➡️➡️2x + y = 23 {We know, x=7}2(7)+y = 2314+y = 23y = 23-14y = 9➡️➡️➡️➡️➡️➡️Now, we have values of both x and y {x=7; y=9}➡️➡️➡️➡️➡️➡️Final quesFind value of (5y-3x) and (y/x-2)➡️➡️➡️➡️➡️➡️Part-15y-3x {x=7; y=9}5(9)-3(7)45-21Ans [24]➡️➡️➡️➡️➡️➡️Part-2y/x -2 {x=7; y=9)9/7 -29/7 -14/7Ans [-5/7] By elimination method we have: 2x + y = 23 ...1 4x - y = 19 ...2➡️ 6x = 42 ....3From eq.3 we have x = 7On substituting value of x in any of the eq: (let be in eq.2)2x + y = 23 {We know, x=7}2(7)+y = 2314+y = 23y = 23-14y = 9Now, we have values of both x and y {x=7; y=9}Final quesFind value of (5y-3x) and (y/x-2)Part-15y-3x {x=7; y=9}5(9)-3(7)45-21Ans [24]Part-2y/x -2 {x=7; y=9)9/7 -29/7 -14/7Ans [-5/7] |
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| 9632. |
If f(x)=x²-px+q is a quadratic polynomial then the sum of zeroes of the quadratic polynomial is |
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Answer» Sum of zeros is -(-p)/1=p/1 Sum of zeroes is -(-p)/1 = p/1 Sum zeroes is -p/1 |
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| 9633. |
√34=√45 |
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| 9634. |
What is the squre root of 100001001 |
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Answer» 10,000.05004......... Tell pls |
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| 9635. |
Find the quadrant of a circle whose circumference is 22 cm |
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Answer» 77/8 cm² Solution:Let the radius of the circle be \'r\' Thus, the circumference C = 2πrTherefore, r = C/2πA quadrant of a circle means one of the four equal parts. Therefore, the area of a quadrant = 1/4 × Area of a circle = 1/4 × πr2Circumference (C) = 22 cmTherefore, radius (r) = C/2π= 22/(2 × 22/7)= (22 × 7)/(2 × 22)= 7/2 cmTherefore, the area of a quadrant = 1/4 × πr2= 1/4 × 22/7 × 7/2 × 7/2= 77/8 cm2 77/8 |
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| 9636. |
class,ten,cbse,sample,pepar |
| Answer» ਠੀਕ ਹੈ ਜੀ ? | |
| 9637. |
If cos + sin =5/2 then cos^2 + sin ^2 |
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Answer» (Cos+sin)^2=5/2^2Cos^2+2cos.sin+sin^2=25/4Cos^2+sin^2=25/4-2Cos^2+sin^2=17/4 cos^2+sin^2=1 Please help I want answers of this chapter Chapter 6 triangle |
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| 9638. |
Find the common difference of the AP 10,8,6,4,2..... |
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Answer» Answer is 3 -2 is the common difference of the given AP 2 I think. |
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| 9639. |
The distance of the point P(2,3)from x axis is |
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Answer» 3 X axis is 3 units 3 units 3 units Distance from x axis is 3 |
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| 9640. |
How to right root17 in form of xroot y? |
| Answer» ਹਾਂ ਜੀ | |
| 9641. |
Prove that 15 +17^3 is an irrational number. |
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Answer» 15+17^3 is not a irrational number. Ohk ✌ Bro is term me prove waale question nhi aenge No Bo |
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| 9642. |
hcf of 126 and 35 is H. if H is expressed as h=126*a+35*b then prove that a*b/h=-2 |
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| 9643. |
A rational number between 3÷4 and 4÷5 |
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Answer» There are infinite no. Between 3/4 and 4/5 ... And one of them can be 0.775 What\'s question term first examination 31/40 19/25 |
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| 9644. |
Prove that/5 is a irrational |
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Answer» If that is a root 5 then it is a irrational no. Sorry it was rational I should repeat that you can solve or prove it by taking its contradiction as it is a rational no. You can prove it by taking its contradiction as it is an irrational no. Hope it will help you? Is that root |
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| 9645. |
Prove that √sec square Q + cosec square Q =tanQ + cot Q |
| Answer» Sec²∅ =tan²∅+1Cosec²∅= Cot²∅+1After adding both, we getSec²∅+Cosec²∅=tan²∅+Cot²∅+2Sec²∅+Cosec²∅=tan²∅+Cot²∅+ 2tan∅.cot∅Sec²∅+cosec²∅=( tan∅+cot∅) ²So we can write √(sec²∅+cosec²∅) = tan∅+cot∅ Hence proved | |
| 9646. |
Answer the following question what is the length of a broken part |
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Answer» Plz ask clear question The total length-the remaining length |
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| 9647. |
Case study base question.... Chap 12 Find the radius of the circle..... |
| Answer» Please ask the full question Not ".... " | |
| 9648. |
If the circumference of two circles are in the ratio 4:5 what is the ratio of their radii? |
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Answer» 4x and 5 x 5x and 5x |
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| 9649. |
If LCM is divided by HCF of the number 2^3×5 and 2^2×5^2 , then the number is |
| Answer» HCF can\'t be more than LCM???? Bt...in this question HCF is more than LCM | |
| 9650. |
X2 - 2x - 8Solution |
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Answer» You can do it by middle term spilting you will get (x-4)(x+2) x^2-2x-8=x^2-4x+2x-8=x(x-4) +2(x-4) =(x-4) (x+2) |
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