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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. |
clAsSMAteDatePagefiang es AB ntOt are on the Sametach othe atE Sho thai AEXCEBEXDE2018-12 |
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| 2. |
)The area ofa trapezium is 192 sq. cm and the perpendicular distance betweenits parallel sides is 15 cm, What is the sum of its parallel side? |
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Answer» can you give me this ans with full method |
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| 3. |
12) One of the parallel sides of a trapezium is 150 cm long and its height is100 cm, If the area of trapezium is 13500 sq cm. What is the length of theother side? |
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Answer» Area of trapezium= 1/2(sum of length of parallel sides)*height Let length of other side is s 13500 = 1/2(150 + s)*100s + 150 = 270s = 270 - 150 = 120 Length of other side = 120 cm |
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| 4. |
In the figure, ABCD is a parallelogram with DE parallel to AB. If AB is 15 cm, AD = 10cm, and te area of the parallelogram is 120 sq cm, find AE. |
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| 5. |
A man starts a journey. Affer sometime he finds that of the journey is covered andstill then he has to cover 90 km. How long is the journey? |
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Answer» remaining=1-5/8 =3/8 so, 3/8=90 total=(90×8)/3 =720/3 =240km. ans 240 km is the correct answer of the given question 240km is the correct answer |
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| 6. |
p ^ { - 5 } q ^ { - 6 } \div \frac { 1 } { ( p ^ { - 2 } } q ^ { - 2 } ) ^ { 2 } |
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Answer» p^-5*q^-6 ÷ ( 1/p^-2* q^-2)²= p^-5×q^-6 × p^4× q^4= p^-1 × q^-2= 1/pq² |
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| 7. |
The sum of the digits of a two-digit number is 9. If9 is added to the number formed by reversingthe digits, then the result is thrice the original number. Find the original number24. |
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Answer» If you find this solution helpful, Please like it. |
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| 8. |
ACEgownfind the other biguemitic nationa. tank 4of angua |
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| 9. |
17. If x2 - 3x + 2 is one of the zero of x4 - px? + q thenthe values of p and q are1) 4 and 6 2) 3 and 5 3) 5 and 6 4) 5 and 4 |
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Answer» Since4x^2−3x+2is a factorx^4−Px^2+q=0 x=2,x=1x=2,x=1must be roots therefore16−4p+q=0 1−p+q=0 15−3p=0 −3p=−15 P=5P= 1−P+q=0 1−5+q=0 q=4 Thusp=5q=4Answer :5,4 Can you explain in video |
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| 10. |
Perimeterof rectangleArea ofPerimeterSquare 2of briangle |
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Answer» 2(l+b)(side)²sum of all sides is the correct answer of the given question p of rectangle 2(l+b) (side)2 2(l+b) side ×sidesum of all sidesis the answer 2(l+b)(side)2 is the right answer |
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| 11. |
Find the perimeter of the following figures:Perimeter- AB + BC+ CD + DA40 cm |
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| 12. |
\begin{array} { l } { \text { A man starts his work on } 12 \text { th December, } 1967 \text { and } } \\ { \text { finishes it in } 163 \text { days. On what date will he finish } } \\ { \text { the work? } } \end{array} |
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Answer» Man finished his work in 163 days totally. In the year 1967, we have 19 days remaining in the month December. Next year 1968 is a leap year. Number of days in 1968 => 163 - 19 = 144 days Therefore 31 + 29 + 31 + 30 + 23 = 144 The man finished his work on May 23rd 1968. No, you're answer is wrong.Ans is 22nd may |
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| 13. |
3. In the adjoining DPfigure, ABCD is aparallelogram inwhich ZA = 60°.If the bisectorsof ZA and 2B meet DC at P,prove that:(i) (APB = 90°(ii) AD = DP and PB = PC = BC(iii) DC=2AD |
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| 14. |
The area of a trapezium is 810 sq cm. Itsparallel sides are in the ratio 4:5. The heightof the trapezium is 10 cm. Find the length ofthe sides. |
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Answer» Areas of trapezium = 810 cm^2 height = 10 cm it's parallel side ratio 4:5 let the 1 ratio be 4xlet the 2 ratio be 5x area of trapezium = 2h (sum of parallel side )810 =2×10 ×(4x +5x) 4.5= x so it's parallel side 4x = 4× 4.5= 18and 2 parallel side 5x = 5 ×4.5=22.5 |
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| 15. |
: \tan ^{4} \theta+\tan ^{2} \theta=\sec ^{4} \theta-\sec ^{2} \theta |
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| 16. |
ABCD, two points P and Q aregonal BD such that DP BQIn parallelogramBDke 8.20). Show that:" uken onsee Fig. 8.20), ShoN) AQ CPAPCQ is a parallelogram |
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| 17. |
0.754 / 3 to 1 dp |
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Answer» 0.754/3 = 0.251333333.........So, according to 1 decimal place as indicated -0.754/3 = 0.2 0.754/3=0.2513333; = 0.2 0.2 is the correct answer 0.754/3=0.2513333; =0.2is the best answer |
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| 18. |
а +ь ь сPrestolfoe feb: b+c c a = 3abc-a3 -- 43 -3.eta a b |
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| 19. |
EXERCISE6.4Nine added thrice a whole number gives 45. Find the number. |
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Answer» The number is 12Solution: let the number be a Then 9+3a=453a=36a=12So the number is 12 Let the number be xThen 9 +3x=453x=45x=45-9=12therefore the number is 12 |
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| 20. |
(-a cos φ, b sin q6.Find θ, if sin 282 cos(0-189, where 2Θ is an acute angle |
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Answer» sin2 theta = cos(90- 2theta) your multification is wrong correct answer is 37 36 36 |
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| 21. |
1 6 5Q.2. Find the sum of the series 1,1,.. up to 16th term. |
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| 22. |
b) If tan θ + sin θp and tan θ-sin θq, then show that p2-q--4-1pq15. Express the rotios cag |
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Answer» Tanθ+sinθ=ptanθ-sinθ=q∴, p+q=tanθ+sinθ+tanθ-sinθ=2tanθp-q=tanθ+sinθ-tanθ+sinθ=2sinθpq=(tanθ+sinθ)(tanθ-sinθ) =tan²θ-sin²θ∴, p²-q²=(p+q)(p-q)=2tanθ.2sinθ=4sinθtanθ4√pq |
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| 23. |
36. समीकरण 6(89 + 9)) - 38 + 20 और6(bx - 80 न 30-28 को संतुष्ट करने के लिए» और ७ के मान है |
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| 24. |
An isosceles triangle has perimeter 0cm and each of the eqial sdes is 12 cm. Fithe anea of the tnangle |
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| 25. |
25 A manstarts repaying a loan as first instalment of 500. If he increases the instalmentby 25 every month, what amount will he pay in the 30th instalment?lr 1.75 |
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| 26. |
Area of the trapezium is 384 sq cm IHs parallel sdes are in the ratio 3 5 The distancebetween paralel sides is 12 cm.Find the length of the parallel sides |
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| 27. |
| 2. छ+* कन४न४ 2. 0) * 02 "20 ने |
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Answer» This identity is wrong.The correct identity is, (a+b)²=a²+2ab+b² plzz prove the correct identity |
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| 28. |
) हे + 02 +0 न 390८ डाएन (2 + 90 + दो “अं माने डिगाव का निधि (8 # ७ * 2)मद |
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Answer» a^3 + b^3 + c^3 - 3abc = (a + b + c )(a^2 + b^2 + c^2 -ab - ac -bc) Now it is given that : a^3 + b^3 + c^3 = 3abcSo, a^3 + b^3 + c^3 - 3abc = 0(a + b + c )(a^2 + b^2 + c^2 -ab - ac -bc) = 0 this means either(a^2 + b^2 + c^2 -ab - ac -bc) = 0 or (a + b + c ) = 0 (a^2 + b^2 + c^2 -ab - ac -bc) = 0 cannot be zero because:2a² + 2b² + 2c² -2ab - 2ac - 2bc = 0a² + b² -2ab + a² +b² +2c² - 2ac -2bc = 0(a-b)² + a² + c² -2ac + b² + c² -2bc = 0(a-b)² + (a-c)² + (b-c)² = 0 (a-b)² , (a-c)² ,(b-c)² >=0 As a≠b≠c , The given value cannot be zero. This means(a + b + c ) has to be zero |
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| 29. |
12 यदि 10? न 25, तो 107” बराबर हैe | L 15 छाका 55 (हर |
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Answer» 10²ʸ=25=>10²ʸ=5²=>10⁻ʸ=5⁻¹first option is correct |
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| 30. |
12. यदि 10^{2 y}-25 \text { at } 10^{-y}, तो 107 बराबर है--\begin{array}{lll}{\text { (A) } \frac{1}{5}} & {\text { (B) } \frac{1}{50}} & {\text { (C) } \frac{1}{625}}\end{array} |
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Answer» Thanks |
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| 31. |
(A) 11(B) 10(C) 20There are 361 doctors and nurses in a hospital. If the ratio of the doctorsto the nurses is 8 : 11, how many nurses are there in the hospital?4.(A) 152(B) 209(C) 57(D) 171 |
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Answer» let X be the common multiplehence8x+11x=361hence 19x=361X=19hence nurses will be 11x=11*19=209 |
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| 32. |
. A manufacturer of a line of patent medicines is preparing a productionplan on medicines A ลnd B. There are sufficient ingredients available tomake 20000 bottles of A and 40000 bottles of B but there are only 45000bottles into which either of the medicines can be put. Furthermore, it takes3 hours to prepare enough material to fill 1000 bottles of A and it takes 1hour to prepare enough material to fill 1000 bottles of B, and there are 66hours available for this operation. "The profit is 8 per bottle for and 7per bottle for BHow should the manufacturer schedule the production in order tomaximize hs profit? Also, find the maximum profit. |
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| 33. |
17. In a row of students, Sherin is 12from the left and Athira is 19thfrom the right. If they interchangetheir positions, Sherin becomes16th from the left. Then, what willbe the position of Athira from theright? |
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Answer» Sherin is at 12th position from left Athira is at 19th position from right When they interchange their positionThen Sherin is at 16th position from left Thus, 16th position from left = 19th position from right. Then total number of positions = 15 + 19= 34 Therefore Athira position from right = 34 - 11 = 23rd position |
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| 34. |
19. Rita had 2 300 she stand y.merey petebook eta of the remail deson stationary how much money isleft with her |
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Answer» I am not able to understand |
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| 35. |
() (x-6)+Bx-4)+(x-1) |
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Answer» x-6+3x-4+x-1=x+3x+x-6-4-1=5x-11 |
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| 36. |
9.In a row of boys, A is 20th from left and B is 16th from right, interchange their position, the Abecomes 30th from lett. How many boys are there in the row?46(h) AA |
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Answer» Total number of boys in row =30+16-1=45 is the correct answer. after interchange the position a is 16th from right and 30th from leftthen total number of boys=30+16-1=46-1=45 |
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| 37. |
17. Rita had 300. She spent of her money on notebooks andstotinery items. How much money is left with her?of the remainder |
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| 38. |
nte at the following theoremFiesdes of aTheg8.9 : The line segment joining the mid points ofto the shird side. |
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| 39. |
2.If (x -a) (x - b) = x? - bx, then the value of x is |
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Answer» (x-a)(x-b)=x^2-bxx^2-bx-ax+ab=x^2-bx-ax=-abx=b |
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| 40. |
Example 2: If B and Q areacute angles such that sin B sin Qthen prove that B ZQ |
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Answer» Answer.∴ AC = k × PR and AB = k × PQFrom right ΔACB, By Pythagoras theorem we haveAB2= AC2+ BC2⇒ (k × PR)2= (k × PQ)2+ BC2⇒ k2× PR2= k2× PQ2+ BC2⇒ BC2= k2× PR2– k2× PQ2 = k2[PR2– PQ2]BC= k root of PR^2-PQ^2From right ΔPRQ, by Pythagoras theorem we havePQ2= PR2+ QR2⇒ QR2= PQ2– PR2 Hence ΔACB ~ ΔPRQ (SSS similarity criterion)∴ ∠B = ∠Q |
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| 41. |
(निम्न वक्रों के किसी बिन्दु पर ध्रुवान्तर रेखा तथा स्पर्श रेखा के मध्य कोण ज्ञात कीजिए।(a) r = a (1-cos D) [Cardioid](b) 20 = 1+cose[Raj. B.Sc., 02पि |
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Answer» 1 is the correct answer by me |
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| 42. |
If two equations 5x +3y+7-0 andlx + my +9-0 are inconsistent, thenwhat is 1: m? |
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Answer» Given equations 5x + 3y + 7 = 0and lx + my + 9 = 0 are inconsistent Then,5/l = 3/m5/3 = l/mOrl/m = 5/3 Therefore,l:m = 5:3 |
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| 43. |
103SE 201556. Prove that: (cot e- cosec 02cos1+cos θ)ICBSE 20151 |
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| 44. |
In ∆ OPQ, right-angled at P,OP= 7 cm and OQ - PQ= 1 cm (see Fig. 8.12).Determine the values of sin Q and cos Q |
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| 45. |
(ü) cos AExample 5 : In Δ OPQ, right-angled at P,OP 7 cm and OQ - PQ 1 cm (see Fig. 8.12).Determine the values of sin Q and cos Q10. In A POR, right-asin P, cos P and taSolution: In Δ OPQ, we have11. State whether the0) The value(i) sec A(ii) cos A is t(iv) cot A ist(+PO OP+ PQ (Why?)leLe.,l + 2PQ72(Why?)(v) sin θ1.c.,PQ-24 cm and OQ1 + PQ7 cm25 cmc3 TrigonometFrom geometry, y24Fig.8.12 |
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Answer» Given:-OPQ is rt angled at P .OP=7cmOQ-PQ=1cmso, OQ = 1 + PQBY PYTHAGORUS THEOREM,OQ^2=OP^2+PQ^2(1+PQ)^2= 7^2 + PQ^21 + PQ^2+2PQ= 49 + PQ^22PQ= 48PQ=24cmOQ= 25cm |
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| 46. |
(Xİİİ) x-3-x-3-663 x+3x +3x-3 79X111(-3.3) |
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Answer» (x-3)/(x+3) - (x+3)/(x-3) = 6 6/7 = 48/7 (x*x-6x+9+x*x+6x+9)/(x*x-9) = 48/7 7(2x*x+18) = 48(x*x-9) 14x*x + 126 = 48x*x - 432 558 = 34x*x x*x = 558/34 = 279/17 = 16.411 x = 4.05 |
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| 47. |
xiii)Xv)।Xii) 3+2x = 1 - X<iv) 3 = 3 + 2xvi) 2t - 3 = (5t-2)| xviii) 2x+0.6x-6.6 = 0.4xxx) 0.18(5x-4) = 0.5x+0.8Xvixixনা দেখে সমীকরণ তৈরি করি ও গল্প লি |
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Answer» (1) -0. 666(2) 2(3) 4.8(4) 3(5) 3.8above all answer's are correct. |
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| 48. |
are there in JanuaryHow many Monday are thereeft for Dubai on 16th September 2016Rita left for Dubai on 1Which day was it? |
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Answer» Friday is the right answer Friday is the correct answer it was Friday is the best answer the correct answer is Friday friday is the correct answer the answer is friday Friday is the correct answer it will be correct answer Friday👍👍 Friday is the right answer Friday is the right answer Friday is correct answer. Friday is right answer |
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| 49. |
(xiii)of a day(xiy |
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| 50. |
19%part of a Kerosine oil is filed. If 6 bottle is re-moved and again we put 3 bottle in it, the tineft halif-filled Find the capacity of the tin ?A)20 B)30 C)40 D)50 |
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Answer» 30 bottles Initially tin is 3/5 filled 6 bottles are taken out and 3 bottles are filled again====> 3 bottles are taken out (3/5 of tin capacity - 3 bottles = 1/2 of tin capacity(3/5 - 1/2 ) of tin capacity = 3 bottles1/10 of tin capacity = 3 bottles===>tin capacity = 30 bottles |
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