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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. |
Po न के Y Y(1) (sin 8+ cos 8)? =1+ 2sin fcos 6. |
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Answer» thanks |
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| 2. |
34*43 |
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Answer» the product of 34*43=1462 |
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| 3. |
nd inI5) |
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Answer» In cyclic quadrilateral sum of opposite angles is 180° Then,(2x + 5) + 3y = 1802x + 3y = 175.......(1) (4y + 20) + (x + 10) = 180x + 4y = 150.........(2) Multiply eq(2) by 22x + 8y = 300......(3) Subtract eq(1) from eq(3)5y = 125y = 125/5 = 25 x + 4*25 = 150x = 150 - 100 = 50 Therefore,angle A = (2x + 5) = 2*50 + 5= 105angle B = (4y + 20) = 4*25 + 20= 120angle C = 3y = 3*25 = 75angle D = (x + 10) = 50 + 10 = 60 |
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| 4. |
a father having 5 children at the gap of 3 years between them.their average . age of 50. find the age of youngest child? |
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Answer» 44 years is the following question answer 44 years is the correct answer |
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| 5. |
Find the po(5x+8)3xmolifth |
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Answer» 15x²-24 is the correct answer of the given question |
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| 6. |
)(iv) 6‘5 £ 34 43- |
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Answer» 5/6+3+3/4=5(4)+3(24)+3(6)/24=20+72+18/24=20+90/24=110/24=4.5833 5(4)+3(24)+3(6)/24=20+72+18/24=72+38/24=110/24=55/12 55/12 is the correct answer 55/12 is correct answer. 55/12 is correct answer in this questions 5/6+3+3/4÷ 10+36+9/12=55/12 This questions correct answer is 55/12 |
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| 7. |
Priove. cycticfomalteto grsI5 0 |
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Answer» Let ABCD be a cyclic quadrilateral such that its diagonals AC and BD are the diameters of the circle through the vertices A, B, C, and D. As, AC is a diameter and angle in a semi-circle is a right angle ⇒∠ADC = 900and∠ABC = 900 Similarly, BD is a diameter. ⇒∠BCD =900and∠BAD = 900 Thus, ABCD is a rectangle |
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| 8. |
31 34d) (92-43) Ă (17)2 |
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| 9. |
(i) 43 or 34 |
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Answer» 4×4×4 = 64 3×3×3×3 = 81 3×3×3×3 > 4×4×4 |
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| 10. |
the following three equations holdulthneonal forn and43-343x + Fyt S=08, pety 1 |
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Answer» 3x+7y+5=0; 4x-3y=8; 4(3x+7y+5=0)=12x+28y=-20; 3(4x-3y=8)= 12x-9y=24/37y=44; y=44-37=7; 3x+7y+5=3x+7(7)+5=3x+54; 3x=-54; x=-18; ; -px+y=1; -p(18)-7=1; -p(18)=8; -p=8/18=4/9; p=-4/9 |
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| 11. |
13.54887105 : 33 का मान है।5134 |
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| 12. |
B ० सरकार, gy v v Q.) Prove that 2.15151515........1s 2 rational numbse |
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Answer» Let 2.151515.... be xSo , x = 2.151515.... .. .....(1)multiply 100 on both sides100x = 215.1515...... (2)eq(2) - eq(1)99x = 213x = 213/99So , 2.151515.... can be represented as 213/99 which is rational |
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| 13. |
. In a class of 48 students, the number of girls is three-fifths the number of boys. Determine thenumber of boys in the class. |
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Answer» The number of boys be xNumber of girls= 3x/5Total number of students= 48 x+ 3x/5= 485x+3x/5= 488x/5= 48x= 48×5/8x= 30 boys. hit like if you find it useful |
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| 14. |
In class of 40 pupils the number of girls is three-fifths of the number of boys. Find thenumber of boys in the class. |
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Answer» let there be X boysgirls are 3/5th of boys=3/5xhence X+3/5x=405x+3x=5*408x=200X=25hence 25 boysso 15 girls. thanks |
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| 15. |
feose-sind = v/2sin, prove that cosθ + sin-y2smo |
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Answer» cosA - sinA = √2 sinA cosA = sinA + √2 sinA cosA = sinA (√2 + 1) sinA/cosA = 1/(√2 + 1) sinA/cosA = 1/(√2 + 1) * (√2 - 1)/(√2 - 1) sinA/cosA = (√2 - 1)/(2 - 1) sinA/cosA = √2 - 1 sinA = (√2 - 1)cosA sinA = √2 cosA - cosA sinA + cosA = √2 cosA |
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| 16. |
6.In class of 40 pupils the number of girls is three-fifths of the number of boys. Eind thenumber of boys in the class. |
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Answer» Thank you for asking the question!There is a class with 40 pupils, three fifth are the girls. so the number of boys will be calculated as follows.Suppose Total pupils= 40 Number of girls= 3\5 Number of boys= ?According to a equation: y + 3 y\5 = 40 8 y\5 = 40 = 40*5\8 = 25Hence the total number of boys are 25.Hope it helps. Let the number of boys in the class be x. ∴The number of girls in the class =3x/5 The total number of pupils in the class = 40. According to the given condition, 3x/5+x=40 ⇒3x + 5x = 200 (Multiplying throughout by 5) 8x = 200 ⇒x =200/8 (Transposing 8 to RHS) x = 25 ∴The number of boys in the class is 25. Total pupils= 40 Number of girls= 3\5 Number of boys= ? According to a equation: y + 3 y\5 = 40 8 y\5 = 40 y = 40*5\8 = 25 Hence the total number of boys are 25. 25 is the correct answer 😘😘😘😘😘😘😘😘😘😘😘😘😘😘😘😘 no of the boys = 25 is the best answer. 16 is right answer of this question |
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| 17. |
In a class of 48 students, the number of girls is three-fifths the number of boys. Thenumber of boys isa. 25b. 30c. 35d. 48 |
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Answer» x = 3y/5 x + y = 48 3y/5 + y = 48 8y = 48*5 y = 30 |
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| 18. |
*In class of 40 pupils the number of girls is three-fifths of the number of boys. Find thenumber of boys in the class.v |
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| 19. |
i) 13573225 ii) 196 ni3 38220 iii) 255 डे 867Jo fafimit ® nigw geads usT a9 |
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Answer» (i) 135 and 225Since 225 > 135, we apply the division lemma to 225 and 135 to obtain225 = 135 × 1 + 90Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain135 = 90 × 1 + 45We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain90 = 2 × 45 + 0Since the remainder is zero, the process stops.Since the divisor at this stage is 45,Therefore, the HCF of 135 and 225 is 45. (ii) 196 and 38220Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain38220 = 196 × 195 + 0Since the remainder is zero, the process stops.Since the divisor at this stage is 196,Therefore, HCF of 196 and 38220 is 196. (iii)255 and 867Since 867 > 255, we apply the division lemma to 867 and 255 to obtain867 = 255 × 3 + 102Since remainder 102 ? 0, we apply the division lemma to 255 and 102 to obtain255 = 102 × 2 + 51We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain102 = 51 × 2 + 0Since the remainder is zero, the process stops.Since the divisor at this stage is 51,so hcf of 255 and 867 is 51. Like my answer if you find it useful! |
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| 20. |
REVIEW EXERCISE1 After spending 45% of his money, a boy has 176 left with him. How much money did he have i2. If a man were to sell his TV set for 7200, he would have lost 25%. For what pricemust he sel ,gain 25%?If 60% of the students in a school are boys and the total number of girls in the school is 360 finumber of boys in the school.ased from 25000 to 24500. Find the percentage decreaserTaur much money did I get for it? |
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| 21. |
The average age of three boys is 15 yr. if their ages are in the ratio 3:5:7,then the age of youngest boy is1.182. 213. 94. 15 |
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| 22. |
by find values for aUSt B arb JTSJS-55 |
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| 23. |
the centre and radius of a circle is (3, 4) and 7 units respectively then what is the position of the point A (5,8) with respect to the circle |
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Answer» If you find this solution helpful, Please like it. |
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| 24. |
1. Calculate the H3u2. If the centre and radius of circle is (3, 4) and 7 units respectively, then what is the position o the poi5 8) with respect to circle?Frntand CE-x, then find the |
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Answer» Let P be (5,8), C be the center and R be the radius.Now,CP=√(3-5)² + (4-8)² [using distance formula] =√(-2)²+(-4)² =√4+16 =√20 =2√5 =4.47Since, CP < R.So, the point lies inside the circle!! |
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| 25. |
A hour digit numberstard uits 15.าง.net İLio dtylo assLthkte, kinas ust,it nbis tere495 |
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Answer» First two digit = 15 , next two digit =3 * 15=45therefore the four digit no. is 1545.multiplying 452 by 1545 ,we get698340 |
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| 26. |
34 x 43(a)3 x 32 |
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Answer» 81*64/96=5184/96=54 Answer 3³ ×4/2 81×2162 correct answer for this question Ok this is correct ans 54 is the correct answer 81×64/965184/96=54 is correct answer. 81×64/96=5184/96=54 is the best answer. 504 is the right answer =3⁴×4³/3×32=3³×4³/32=27×64/32=27×2=54 54 is the correct answer 81×64/96 5184/96= 54 is the right answer |
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| 27. |
Prove or disprove :Fundamental Theorem ofintegral calculus isapplicable to f(x) = (x + 1]on (1,5) · ([y] denotes thelargest integer notexceeding y). |
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Answer» Not, applicable because f is not continuous at [1,5] Disprove because applicable of F is not unintirrupted at 1,5 |
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| 28. |
What is the HCF of 33 x 5 and 32 x 52 ? |
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Answer» The H.C.F. is= 3^2×5=9×5=45 3^3×5 and 3^2×5^2; 27x5=135; 9x 25=198;; HCF=3×3×5=9×5=45 1st term = 3^3×52nd term= 3^2×5^2HCF = 3^2×5= 9×5=45in HCF lowest power is taken. Highest common factor is 45 45 is the answer of 9×5 45 is correct answer this question |
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| 29. |
(k, -1) (2,1) and (4,5) find the value of k,for which the given points are collinear |
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Answer» A =(k,-1) we have to apply area of triangle |
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| 30. |
x=3+/8 and y = 3 - v thenaz x = 3+vē ste y = 3- VE ta(1) - 34(2) 34(3) 12.18(4) -12V8 |
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| 31. |
1.Calculate the HCF of 33 x 5 and 32522 If the centre and radius of circle is(3, 4) and 7 units respectively, then what is the position of the pointA(5, 8) with respect to circle ?2 RE- 1 and CFx then find the |
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| 32. |
x+5=33 |
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Answer» Ans :- x+5=33 x=33-5 x=28 |
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| 33. |
(i) 32 x 34 x 38 |
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Answer» Ans :- 3^2 × 3^4 × 3^8 = 3^(2+4+8) = 3^14 |
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| 34. |
passing through the origin having slopes 3and 2.passing through the origin havinginclinations 60° and 120° with x-axis. |
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| 35. |
How many circles can be drawn to Pass through two given Points |
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Answer» infinite circles can be drawn through two given points. |
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| 36. |
73 x 34 x 5251 x 22 x 33 |
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| 37. |
t Ar the foot of a mowntain the elevation of its ssemmit iss: Ater ascending 1000m towards the mountain up a45"depe of 30° inclination, the elevation is found to be 60Find the heigh of the moumtain |
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| 38. |
Q2. At the foot of the mountain , the elevation of its suhmit is 45 . after ascending 1000mtowards the mountain up a slope of 30 inclination, the elevation is found to be 60.find theheight of the mountain. ANS (H 50003 +1 )m |
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| 39. |
26. P(xi, yi) and g(x2 , y2) are the given points. If PQ is parallel to Y-axis, then |
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| 40. |
\frac { \operatorname { cos } 10 ^ { \circ } + \operatorname { sin } 20 ^ { \circ } } { \operatorname { cos } 20 ^ { \circ } - \operatorname { sin } 10 ^ { \circ } } ? |
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Answer» (cos10° + sin20°) / (cos20° - sin10°) = tan60° * (cos60°/sin60°) * [(cos10° + sin20°) / (cos20° - sin10°)] = tan60° * [(2cos60° cos10° + 2cos60° sin20°) / (2sin60°cos20° - 2sin60° sin10°)] = √3 * [(cos70° + cos50° + sin80° - sin40°) / (sin80° + sin40° - cos50° + cos70°)] = √3 * [(cos70° + cos50° + sin80° - cos50°) / (sin80° + cos50° - cos50° + cos70°)]{sin40° = cos50° sin(90-A)=cosA} = √3 * [(cos70° + sin80°) / (sin80° + cos70°)]= √3 |
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| 41. |
\operatorname { sin } 80 ^ { \circ } \operatorname { cos } 20 ^ { \circ } - \operatorname { cos } 80 ^ { \circ } \operatorname { sin } 20 ^ { \circ } = \frac { \sqrt { 3 } } { 2 } |
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| 42. |
Calculate the HCF of 33 x 5 and 32 x 52 |
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Answer» The numbers are (3^3• 51) and (3^2• 5^2).HCF = 3^2• 51 [lowest powers of the factors] = 9 • 5 = 45LCM = 3^3• 5^2 [highest powers of the factors] = 27 • 25 = 675Therefore,HCF (3^3•5, 3^2•5^2) =45LCM (3^3•5, 3^2•5^2) =675 |
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| 43. |
\operatorname { cos } ( 50 ^ { \circ } + \theta ) \operatorname { cos } ( 20 ^ { \circ } + \theta ) + \operatorname { sin } ( 50 ^ { \circ } + \theta ) \operatorname { sin } ( 20 ^ { \circ } + \theta ) = \frac { \sqrt { 3 } } { 2 } |
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| 44. |
\frac { \operatorname { cos } 70 ^ { \circ } } { \operatorname { sin } 20 ^ { \circ } } + \frac { \operatorname { cos } 59 ^ { \circ } } { \operatorname { sin } 31 ^ { \circ } } - 8 \operatorname { sin } ^ { 2 } 30 ^ { \circ } |
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Answer» Sin(90°-α) = cosαcos(90°-α) = sinα so,cos70°/sin20° + cos59°/sin31° -8 sin²30°=[cos(90°-20)]/sin20° + [cos(90°-31°)]/sin31° - 8×(1/2)²=sin20°/sin20° + sin31°/sin31° - 2= 1+1- 2= 0 hand writing |
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| 45. |
\frac { \operatorname { cos } 70 ^ { \circ } } { \operatorname { sin } 20 ^ { \circ } } + \frac { \operatorname { cos } 59 ^ { \circ } } { \operatorname { sin } 31 ^ { \circ } } - 8 \operatorname { sin } ^ { 2 } 30 ^ { \circ } = 0 |
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Answer» sin(90-20)=cos70same wisecos59= sin31so 1+1-(8*1/4)=2-2=0 |
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| 46. |
7/10 %2B 3/5 %2B 5/8 |
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Answer» 3/5+5/8+7/10=L.C.M=40=24+25+28/40=77/40Ans-77/40 3/5+5/8+7/10=L. C.M =40=24+25+28/40=77/40 24+25+28------------------ =77 40 ------ answer 40 77/40 is right answer of question. 77/40 is right answer of question 24+25+28__________ =77/40 40 LCM is 40so24+25+28/40=77/40 |
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| 47. |
2/5 %2B 5/10 |
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Answer» 9/10 is right answer |
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| 48. |
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length ofminor arc of the chord, |
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Answer» Di. of the circle = 40 cmthen rad. = 20 cmnow chord = 20cmthe triangle formed is an equilateral triangle with each side = 20cmΘ = 60°now length of the arc is I = rΘ= 20(60°*п(pi) /180°)= 20π/3cm |
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| 49. |
8. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the lengthof minor arc of the chord. |
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Answer» Diameter=half of radiusso it will be 20 cm connect end points of chord and centre of the circleso every side will be equal so it will be a equilateral triangle and all side will be 60° therfore minar arc = 60°and major arc = 360-60=300° |
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| 50. |
8. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the lengthof minor arc of the chord |
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