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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. |
A 15 m long ladder reached a window 12m highfrom the ground on placing it against a wall atadistance a. Find the distance of the foot of theladder from the wall.3.15m12m |
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| 2. |
The brimary system uses powers ofAL 16[C] 810D]102 |
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Answer» Answer:D)2Binary representation, just because it only uses two digits has an interesting interpretation. Binary representation of a number is a sum of powers of 2 |
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| 3. |
10. In the adjoining figure, ABCD is aquadrilateral in which diag. BD 14 cm. IfAL L BD and CM 1 BD such that AL 8 cmand CM = 6 cm, find the area of quad. ABCD |
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| 4. |
4. IfA=1012|,then show that 13A|-27/Al |
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| 5. |
10. The solution of the pair of equations +4 and-2 isof the pair of equations-+y=4 and a-y=2 isalal(a) x 3a, yb(c) x--3a, y =-b(b) x Ba, y = b-3a, y b |
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| 6. |
nt che sum o al saral oinibers lying herwceem 10) and 1on, which aemhiples of |
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Answer» 36-31+26-128÷8×3+7×2+6 |
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| 7. |
(9) Fr afFinda p. third term = 8 f seventh term = 20nineth term. |
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Answer» Yes this is correct answer ye wrong answer he dont follow let 1st term =a , difference= d3rd term=8a+(3-1)d=8a+2d= 8----(1)a+6d= 20---(2)by subtracting -4d= -12d= 12/4=3a+2d=8a= 8-6=29th term= a +(n-1)d = 2+(9-1)×3 = 2+8×3= 2+24=26 |
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| 8. |
(7) Prove thatcos 20° cos 40 cos 60 .cos 8016 |
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Answer» cos20°cos40°cos60°cos80°=(1/2)(2cos20°cos40°)(1/2)cos(80°){cos60°=1/2; and multiplying and dividing by 2}=(1/4){cos(20°+40°)+cos(40°-20°)}cos80°=(1/4){cos60°+cos20°}cos80°=(1/4){(1/2)cos80°+cos20°cos80°}=(1/4)(1/2)cos80°+(1/4)cos20°cos80°=(1/8)cos80°+(1/4)(1/2){2cos20°cos80°}=(1/8)cos80°+(1/8){cos (80+20)+cos(80-20)}=(1/8)cos80°+(1/8){cos100°+cos60°}=(1/8)cos80°+(1/8)cos100°+(1/8)(1/2)=(1/8){cos80°+cos100°}+1/16=(1/8){2cos((100°+80°)/2)cos((100°-80°)/2)}+1/16=(1/8){2cos90°cos10°}+1/16=1/16 you not seen problem carefully |
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| 9. |
a letter of English alphabet is chosen at random determine the probability that the chosen letter is constant |
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| 10. |
\operatorname { cos } 100 ^ { \circ } + \operatorname { cos } 20 ^ { \circ } = \operatorname { cos } 40 ^ { \circ } |
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| 11. |
90210Short-Answer Questions5. A letter of English alphabet is chosen at random.isprobability that the chosen letter is a consonant. |
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| 12. |
\begin{array} { l } { \text { Prove that: } } \\ { \text { (i) } \quad \cos 20 ^ { \circ } + \cos 100 ^ { \circ } + \cos 140 ^ { \circ } = 0 } \end{array} |
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| 13. |
\operatorname { cos } 20 ^ { \circ } + \operatorname { cos } 100 ^ { \circ } + \operatorname { cos } 140 ^ { \circ } = 0 |
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Answer» Cos20+cos100=2cos(100+20)/2cos(100-20)/2 =2cos(60)cos(40) =2*1/2*cos(40) =cos(40)cos20+cos100+cos140=cos40 + cos140 =2cos(40+140)/2cos(40-140)/2 =2cos90cos50 =2*0*cos50 =0 |
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| 14. |
20 ^ { \circ } + \operatorname { cos } 100 ^ { \circ } + \operatorname { cos } 140 ^ { \circ } = 0 |
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Answer» Cos20+cos100=2cos(100+20)/2cos(100-20)/2 =2cos(60)cos(40) =2*1/2*cos(40) =cos(40)cos20+cos100+cos140=cos40 + cos140 =2cos(40+140)/2cos(40-140)/2 =2cos90cos50 =2*0*cos50 =0 |
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| 15. |
atleast.Q8) A letter is chosen at random from the letters of the message HEY BOY I KEEP TRY« lA/hat is the nrnhahility that the chosen letter is a consonant? |
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Answer» hey boy i keep trying total letters: 17total consonants: 11 probability of a consonant selection: 11/17 |
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| 16. |
a toy is sold in rupees 72 in the loss of 10%. find the cp |
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Answer» SP = 72 let CP be x sox - 10% of x = 72x - 10x/100 = 72x - x/10 = 729x/10 = 72x = (72×10)/9x = 80So, cost price is ₹80 |
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| 17. |
A letter is chosen at randorm from the letters of the word "ENTERTAINMENT". Findthe probability that the chosen letter is a vowel or T. (repetition of letters is allowed) |
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Answer» for vowelno of vowels in the word is 5hence probability=5/13 |
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| 18. |
nd12, A flask was sold at72 incurring a loss of 10%.What was its cost price? |
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| 19. |
A letter is chosen from the word PENCIL', what is the probabilitythat the letter chosen is a consonant? |
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| 20. |
ll 23, 33, at9-The monthly income of vikas was Rs 12800, He got 15% incrementin his salary find;a) How much increment he got?b) what is his present salary ? |
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Answer» monthly income=Rs 12800a) increement=15% of income =0.15 × 12800 =Rs1920b) present salary=12800+1920 =14720 |
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| 21. |
30. Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment ofRs. 200 each year. In which year his income reaches Rs. 7000?eteen in the |
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Answer» Yearly income of Subba Rao is following : 5000, 5200, 5400, 5600............... Here, a = 5000, d = 200, Tn= 7000We know that, Tn= a + (n – 1) d⇒7000 = 5000 + (n – 1) 200⇒7000 = 5000 + 200n – 200⇒7000 = 4800 + 200n⇒200n = 2200⇒n = 11 Hence, in the 11th year, his income reached by Rs. 7000. |
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| 22. |
The salary of a man was increased by 8%. If his present salary is Rs 19440, what was his salarybefore increment ? |
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Answer» x + 8x/100 = 19440 108x = 1944000 x = 18000 rps |
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| 23. |
03 <90°, find the value ofcos θ1 + cos θ2 + cos θ3 |
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| 24. |
0. Find the product of cos 30° cos 60 cos 90 |
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Answer» Cos30.cos60.cos901\2.root 3 by 2 by 0so whole answer is 0 |
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| 25. |
find Cos 20 + Cos 140 + COS 100 |
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| 26. |
Find the value of the following:13Ďcos-1(cos16)cos cos |
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| 27. |
(ii) C.P. =* 375 and S.P. = * 360(iv) C.P. = 810 and Loss =* 72 |
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Answer» loss =cp -sp 375 -36015 cp= 810loss =72sp = cp-loss = 810-72 = 738 Loss =CP - SP =75 - 460 = Rs 15 CP = 810SP = 72SP = CP - Loss = 810 - 72 = Rs 738 |
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| 28. |
810. Find the weea of a triangle whose vertices are 38)(-2,1), (7-1). |
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Answer» if you like plz like (3, 8), (-2, 1), (7, -1), formula 1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]; 1/2{3(1-(-1)-2(-1-8)+7(8-1)}= 1/2{6+18+49}=1/2(73)=36.5 = Area triangle |
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| 29. |
Initial Demand (QFall in Demand (ÎNew Demand (QPrice Elasticity of1.5Ă10 80 |
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Answer» 1.5 = 4/10 × P/80 1.5 = 1/10 × P/20 P = 200 × 1.5 P = 300 If you find this answer helpful then like it. |
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| 30. |
4 men or 7 children's can do a work in 29 days how many days to go the work finished 12 men and 8 childrens |
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Answer» days=nab/ bc+ad; n=29; a=4, b=7; c=12; d=8;; 4 men can do the work 1 men does in 1 day=1/29×4; amount of work 12 men do in 1 day=12/29×4=3/29; 7 boys can do work in 29 days, = amount of work 1 boy does in 1 day = 1/29 × 7; amount of work 8 boys=8/29×7=8/203; work 12 men and 8 children do in 1 day=3/29+8/203=21+8/203= 1/7; 12men and 8 boys the same work in 7 days 4M=7C=29 days, 12M+8C=?, we know that4M=7Cso12M=21Cmeans21C+8C=? dayswe know7C=29daysso7x29=(21+8) ×? 7x29=29×? ? =7daysyou can take any symbol in the place of ? as:x, y, z7 days is answer. 7 days is the answer of the following |
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| 31. |
8 The demand for milk is given by:PriceDemand: 100 50 0Find the linear demand function and its slope. |
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| 32. |
g ey“620 रु० है । TR WH W0. दो रेडियो का क्रयमूल्य \20% T जार.o 5 को 20% हानि पर बेचने पर विक्रयमूल्य, रूप्तन हो हे,e क्रयमूल्य कया होगा ?अलग-अलग ही (2) 200 रु० , १20 रू० |
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| 33. |
4.5 g of an alloy of copper and zinc contains 3.5 g of copper. Whthere be in 18.9 g of the alloy?ch letters can we bu |
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| 34. |
1/6 of the length of a stick is 5cm. What is the length of the stick? |
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Answer» length of stick will be 30 cm |
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| 35. |
a 2At any point t on the curvesubtangent and subnormal. (March-2017)#3.x=a(1+sint),y=a(l-cos), find the lengths of angent, normal |
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| 36. |
6. Write the value of (5(810+2710) |
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| 37. |
810 110oD |
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Answer» _________840)110000 (130.952 - 840 _______ 2600 - 2520 ________ 800 - 0 ________ 8000 - 7560 ________ 4400 - 4200 ________ 2000 - 1680 _________ 320 Therefore quotient is 130.952 & remainder is 320 |
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| 38. |
B.In the adjoining figure, AEDC is a rectangle with ED 14 cm,CD 10 cm and ABC is a semi-circle drawn on AC as diameterWhat is the perimeter (in cm) of figure ABCDE? (Use Ď =(C.B.S.E. 2011C) |
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| 39. |
B. A circle has radius 12 cm. What is the lendlongest stick that can be placed inside this cathat the two ends of the stick lie on the circleat is the length of theinside this circle such |
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Answer» diameter is the longest line in the circle radius=diameter/2 thandiameter=radius ×2 12×2 24 cm the length of the stick that can placed in the circle which meets the both ends is=24cm 24 cm is the right answer 24 is the right answer 24cm ie the diameter 24 is the right answer to 24 cm is the right answer 24 cm is the right answer 24 is the right answer 24 cm. is correct answer. diameter of the circle=the longest stick in the circle diameter of circle=2*radius=2*12=24 24 cm. is correct answer. |
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| 40. |
A stick assumed to be 1m long measured the side of a square field The area thus calculated was 625m sq Later on it was discovered that the actual length of the stick is 90cm Find the actual area of the sq field |
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Answer» Please try not to post the same question multiple times. Your doubt will definitely be answered. Consider stick as 1m. Area of square 625m^2. Sides of square 25m. No. of measurement per side =25 (hence measured 25m) Now correct length of stick is 90cm instead of 100cm No. Of measurement remains same=25 Corrected side length of square = 25 x 0.90m=22.5m Corrected Area of square = 22. 5 ^2 =506.25m^2 |
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| 41. |
om कुक 218 s ( Wh 13 klubie ME DRyRIDwh 0=9 -7/ 4 R LINY AP 2By | kAwie T ____EEE____M EEE 212 |
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Answer» ax²+bx + c ka विविक्तकर = b² - 4ac 3x² + 2√5x - 5 = 0 विविक्तकर = (2√5)² - 4 × (3) × (-5) = 20 + 60 = 80 > 0 To mul vastavik hai aur asamaanniya hai. Like kar ke protsahan badhaiye... |
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| 42. |
(1) The king, queen and jack of clubs are removed from a deck of 52 playing cards andthen well shuffled. One card is drawn at random from the remaining cards. Find theprobability of getting (i) a king (ii) the 10 of hearts. |
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| 43. |
-Wh |
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Answer» (3x + 20) + (4x - 36) = 180 [Linear pair] 7x - 16 = 180 7x = 196 x = 28 You so good |
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| 44. |
1M wh g ks L-_u - |
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Answer» 5/18*(-72/7)5/18*(-10.28)0.27*(-10.28)-2.85 |
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| 45. |
(3) GST system was introduced in our country from..(A) 31st March 2017 (B) 1st April 2017(C) 1st January 2017 (D) 1st July 2017 |
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Answer» GST was introduced in India on1st July 2017 by our new financial minister Arun Jaitley by passing the GST bill in Lok sabha. |
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| 46. |
9. Find the percentage of the letter 'M' in the word MATHEMATICS. |
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Answer» total letter = 11no. of time M occur = 2% of M is (2/11) × 100 = 18.18% Ur answer is wrong sorry!. the answer is 18 whole 1/2 |
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| 47. |
Find the percentage of the letter 'M' in the word MATHEMATICS. |
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Answer» 2/11×100=200/11=18.18% |
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| 48. |
A letter is chosen from the word MATHEMATICS. What is the probability that it is a consonant? |
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Answer» total number of letters=11consonent=7probability =7/11 |
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| 49. |
T bl el धु& छह 2 छा el BIE bl 2 (7 —17) (67+39) 1 Libk b lalbply °S |
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Answer» (4x + 2y) ( 4x - 2y) = (4x)² - (2y)² = 16x² - 4y² Jab x = 2 aur y = 1 ho, (4x + 2y) ( 4x - 2y) = 16 × (2)² - 4 × 1² = 16 × 4 - 4 = 64 - 4 = 60 |
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| 50. |
youvelROHL NOBwave Ronal phe810 MUCHIVES 111 m a day!7. A candidate won an election by beating his nearest rival by 24541 votes. If thecandidate received 63198 votes, how many votes did his rival get?79he |
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Answer» elections 63198-25541= 38657 63198-25541=38657 elections |
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