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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The cost price of 12 candles is equal to the sellingrice of 15 candles. Find the loss per cent? |
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Answer» I can't understand something |
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| 2. |
15+19*20*3*5+34 |
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Answer» 15+ 19*20*3*5+34= 15+ 5700+ 34= 15+ 5734= 5749 15+5700+345700+495749 |
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| 3. |
3.Give four rational numbers equivalent to:275 |
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| 4. |
= xil x <0.- Write four rational numbers equivalent to each of the following ratiam(1)(i) - |
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| 5. |
काekमद |
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Answer» Volume of a cuboid = l*b*h volume = 8.2×5.3×2.6 =112.996 m³ Like my answer if you find it useful! |
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| 6. |
find the product x²(x-3y²)-xy(y²-2xy)-x(y³-5x²) |
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Answer» x^3-3x^2y^2-xy^3+2x^2y^2-xy^3-5x^3 |
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| 7. |
1-12 +2xy-y*) รท(x-1) |
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| 8. |
factorize a^2-a(x+2y)+2xy |
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Answer» a²-a(x+2y) +2xy => a²-ax -2ay+2xy=> a(a-x) -2y(a-x)=> (a-x)(a-2y) |
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| 9. |
(a-1 )x^ + a^2xy + (a + 1 ) y^2 |
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Answer» (a - 1)x² + a²xy + (a + 1)y² = (a - 1)x² + (a² - 1 + 1)xy + (a + 1)y² = (a - 1)x² + {(a - 1)(a + 1) + 1}xy + (a + 1)y² = (a - 1)x² + (a - 1)(a + 1)xy + xy + (a + 1)y² = (a - 1)x{x + (a + 1)y} + y{x + (a + 1)y} = (x + ay + y)(ax - x + y) it is the answer plz help me to solve |
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| 10. |
x^2 + 2xy + y^2-a^2 + 2ab-b^2 |
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Answer» x*x + 2xy + y*y - a*a + 2ab - b*b = (x+y)(x+y) - (a-b)(a-b) = (x+y+a-b)(x+y-a+b) |
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| 11. |
a fas olfere for(c) afc cos A + sin A = V2.cos Acos A - sin A = 12.sin A |
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| 12. |
ERCISE 3.2(ii) 42+93 |
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Answer» 14.15 is correct answer. Your correct answer is 283/20 283/20 is the right answer plz like my answer 283/20 is the correct answer of the given question 4 3/4 + 9 2/5= 4×4+3 / 4 + 9×5+2 /5= 16+3 /4 + 45+2 / 5= 19 / 4 + 47 / 5 = 19×5 / 4×5 + 47×4 / 5×4= 95 / 20 + 188 / 20= 95+188 /20= 283 / 20=14.15 |
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| 13. |
ERCISE 3.2 |
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Answer» 14.15 is right answer. 283/20 is the correct answer of the given question |
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| 14. |
ERCISE 8.4The sum of two numbers is 70. Their difference is 16. Find the numbers |
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Answer» let the number be x and y case1 x+y = 70 x=70-y case 2x- y =16 70-y-y =16-2y =16-70-2y =-54 y = 27 now x=70-yx=70- 27x=43 thank you |
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| 15. |
g eke e 2h 2% LD Wk ek W0 S x WO ] x WA ST & 2h & BB8 12 Dullh 2iaj के हक सं ७ फू के तप ६1६6 2 1eD b b ey ) T |
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| 16. |
ek ladka ek din me 29 pencil khareedta he to weh ek saal me kitni kgareedega |
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| 17. |
Two adjacent sides of a rectangle are3a - b and 6b - a then its perimeter is |
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Answer» Perimeter of rectangle = 2(l+b)Taking 3a-b =l and 6b-a=bPerimeter= 2(3a-b+6b-a)= 2(2a+5b)= 4a+10b 2(3a-b+6b-a) 2(2a+5b)4a+10b |
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| 18. |
19. The two adjacent sides of a rectangle are 5x-3y and x"+ 2xy. Find the perimeter |
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| 19. |
19. The two adjacent sides of a rectangle are 5x2-3y and x+2xy, Find the perimeter. |
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| 20. |
Two adjacent sides of a rectangle are in the ratio 7:3. If its perimeter is 100cm,ind its area. |
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| 21. |
Fimol markt Price (mp) |
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| 22. |
and are marked by and respectively Whichter we marked tysDHIMO?and then markt |
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Answer» Question is not visible |
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| 23. |
ercise 6.2|Q=|||in the blanks.480 = 101900 = 101860 - 1003892 = 102300 = 1000Olify.00 = 10 + 42 =|||Q=_b. 9900 - 100 |
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Answer» 1st Q=48 R=02nd Q=190 R=03rd Q=186 R=04th Q=389 R=25th Q=2 R=300 1. Q=48 R=02. Q=190 R=03. Q=186 R=04. Q=389 R=25. Q=2 R=300answer. 1 q= 48 R= 02 q= 190 R = 03- q=186. R =0 |
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| 24. |
Qils pole 461ww |
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Answer» this is not any question 2/3m + 3/2m = 4m + 6m /6 = 10m/6 = 5m/3is the best answer (2/3m+3/2n)²4/9m²+9/4n²+2mn is the correct answer of the given question |
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| 25. |
Hari had invested some amount of money at compound interest. He received Rs. 2,420/-after 2 yearsand Rs.2,662/- after 3 years. What is the rate of interest'? |
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| 26. |
4. Find the missing number in the sequence: 3, 4,6.324. Find the missing number in the sequence: 3, 4, 6,9,a) 11b) 12c) 13d) 14 |
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Answer» Missing number-13 hereone number is being increased hereas 3+1= 44+2= 66+3= 99+4= 13 |
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| 27. |
Evaluate:\frac{\sin 18^{\circ}}{\cos 72^{\circ}} |
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Answer» it will be 1sin18° = cos72°sin(90-theta) = cos thetaso sin18°/sin18° = 1 is the answer |
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| 28. |
Q41. Find Missing number.20 |
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Answer» 19 10+4=14,14+6=20,20+8=28 ans |
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| 29. |
hd the missing number2x4524?11 |
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| 30. |
Simplify the following:1. 180 150 (3 x 60 2 x 25)) |
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| 31. |
16.sin18°cos72°(A)(C)B)-0-1(D)8 |
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Answer» Answer of a question is B Answer:B)1 Explanation: sin 18°/cos 72°=sin 18°/cos(90°-18°)=sin 18°/sin 18°{ Cos (90°-A)=sin A} =1 |
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| 32. |
Evaluate :\frac{\sin 18^{\circ}}{\cos 72^{\circ}} |
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Answer» sin18° = cos(90-72°)cos72°= sin (90-18°)so sin18°/sin18°= 1 |
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| 33. |
1. Evaluate:\frac{\sin 18^{\circ}}{\cos 72^{\circ}} |
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Answer» it will be 1 as sin(90-∅)⇒cos∅so sin18 will be equal to cos 72° |
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| 34. |
Mumbai Port TrustTemporaryFrom Date : 31/10/2019 To Date : 06/11/2019Permit No.: 0128024T00082Name : MANIKANDAN KAIPILLAI HARIANCompany: JYOTI TRADERSCategory: Misc. services suppliersDesignation : HELPERGender: MaleAge : 23Nationality: INDIANLocation: Dock and StreamHolder SignReceived Rs. 90.86Issuing Authority |
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Answer» yes I am going to Mumbai Is me question kahan hey ???🤔🤔🤔🤔🤔 I can't understand what's your question yes you go to mumbai happy jauranay |
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| 35. |
1. Find the value of,(i) log3 6561 |
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Answer» Log36561=Log(6561) ÷ Log(3) Log(6561)= 3.81697003776 Log(3)=0.47712125472 Log36561=3.81697003776 ÷ 0.47712125472 Ans=8 |
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| 36. |
If log,log2(log t)]-1, show that 6561 |
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| 37. |
The area of a square field is 6561 m². Find the length of each side. |
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Answer» Area of square =Side² So area of square is 6561m² Side²=6561side=√656181 m |
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| 38. |
13. 5x-1=74 |
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Answer» 5x-1=745x=74+15x =75 X=75/5=15 X=15 |
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| 39. |
\left. \begin array l \text Find the value of m \text if 3 ^ 2 m - 2 = 6561 \\ \text (1) 2 \\ \text (3) 4 \end array \right. |
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| 40. |
1: Convert in to the improper fractions8 \frac{1}{5} \text { (b) } 10 \frac{3}{5} \text { (c) } 120 \frac{7}{8} \text { (d) } 31 \frac{2}{7} \text { (e) } 18 \frac{3}{4} |
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| 41. |
seas 1506561Simplify. 65536 |
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| 42. |
18 a ^ { 2 } b - 45 a ^ { 3 } b ^ { 5 } \text { by } 9 a ^ { 2 } b |
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| 43. |
cos 6x=32cos6 x-48cos" x + 18 cos' x-1. |
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Answer» cos6x=cos[2(3x)]=2cos^2(3x)-1 cos2x=2cos^2x-1 =2[cos(3x)]^2-1 cos3x=4cos^3x-3cosx =2[4cos^3x-3cosx]^2-1 (a-b)^2=a^2+b^2-2ab =2[16cos^6x+9cos^2x-24cos^4x]-1 =32cos^6x-48cos^4x+18cos^2x-1 |
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| 44. |
\frac{\sin 18}{\cos 72} |
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Answer» sin18 =sin(90-72)=cos72so cos72/cos72 =1 |
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| 45. |
sin 18°® cos 72° |
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| 46. |
\operatorname { cos } 306 ^ { \circ } + \operatorname { cos } 234 ^ { \circ } + \operatorname { cos } 162 ^ { \circ } + \operatorname { cos } 18 ^ { \circ } = 0 |
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| 47. |
\frac{\sin 18^{\circ}}{\cos 72 !} |
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| 48. |
\frac { \operatorname { cos } 18 } { \operatorname { sin } 72 } |
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Answer» cos(18)=cos(90-18)=sin72hencecos18/sin72=1 |
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| 49. |
sin(18^circ)/cos(72^circ) |
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Answer» Sin18/ cos72= Sin( 90-18)/ Cos72= Cos 72/ Sin 72=1 1. is the correct answer of the given question. sin(90-18)/cos72=cos72/cos72=1 Sin(90-72)/cos72cos 72/cos 72=1 sin18°/cos72°sin(90-18)/cos72°cos72°/cos72°=1 |
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| 50. |
\left. \begin{array} { l } { \text { Find the value of } \sqrt { s ( s - a ) ( s - b ) ( s - c ) } } \\ { a = 15 , b = 18 , c = 12 , \text { and } s = \frac { 1 } { 2 } ( a + b + c ) } \end{array} \right. |
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