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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. |
We have a basket full of apples, oranges and mangoes.If 50% are apples, 30% are oranges, then what percentare mangoes?3. |
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Answer» Apples= 50%Oranges= 30%Mangoes= 100-50-30= 20% were mangoes |
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| 2. |
We have a basket full of apples, oranges and mangoes.If 50% are apples, 30% are oranges, then what per centare mangoes?3. |
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| 3. |
5. In a retaill market, fruit vendors were selling mangoes kept in packing boxes. Thesevendorsboxes contained varying number of mangoes. The following was the distribution ofmangoes according to the number of boxes.Number of mangoes 50-52 53-55 56-58 59-61 62 -64Number of boxes1511013511525Find the mean number of mangoes kept in a packing box. Which method of findingthe mean did you choose? |
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| 4. |
.In a retail market, fruit vendors were selling mes. IILC.uaccordine yng number of mangoes. The following was the distribution ofangoes kept in packing bof boxes.62-64mangoes 50-52 53-55 56-58 59-61Number of boxes1352515110Find the mean number of mangoes kept in a packing box. Which method of findingthe mean did you choose? |
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| 5. |
A school hall measures 20 m in length and 15 m in breadth. Find its area |
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Answer» The length is=20m The breadth is=15m=20×15=The area is=300=300Answer Area is length*breadthhence area=20*15=300m^2 Area is length × breadthhence,Area = 20× 15 = 300 area= breadth × lengtharea=20×15area= 300sq.m area equals l×w. 15×20=300m^2 the correct answer of the given question is 300sq.m a hall is 20m long and 15m widearea of rectangle =lxb20x15=300m 300 m is the answer |
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| 6. |
A PersonSells itbuysat toPencil atfor llifindfour & to andProfit to |
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Answer» 1/10 x100= 10 hope this will help you 10 is the answer hope its helpful for u 1/10*100=10the answer is correct |
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| 7. |
OA Personsells itbuys Pencil atat To for it.IFindfour & to andProfut to |
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Answer» and the answer of this is 1/10 x100 = 10 sory I don't know ......... 10 is right answer......... |
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| 8. |
ic hulilbers.2)In the year 2010 in the village, there were4000 people who were literate. Every yearthe number ofliterate people increases by400. How many people will be literate inthe year 2020? |
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Answer» Rate = 400 per yearor 10 percentageso it is just like a question of simple interesthere P= 4000time = 10 rate = 400= 4000*10*10/100= after 10 yerastotal literacy will be 4000+4000= 8000 |
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| 9. |
2. A school had 1000 students in tstudents increases by 20%,culdgesDa 1000 students in the year 2016. If every year the number ofases by 20%, then what will be the strength of the school in theyear 2018? |
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Answer» 1200 students will be the strength of the school in the year 2018 1200 students will be the strength of the school in the year 2018 1400 students will be the strength of the school in the year 2018 |
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| 10. |
15. The population of an Indian state is 8 crore. If it increases by 2% every year, find the population of |the state after one year. |
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Answer» where is the answere |
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| 11. |
38. A circular piece of thin wire is converted into a squareof side 6.25cm. If there is no loss or gain in its length,find the radius of the circular wire.6.25 cmAm |
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Answer» Perimeter will be same so2πr= 4*6.252πr= 25r= 25/2πr= 3.98cm Perimeter will be same so2πr= 4*6.252πr= 25r= 25/2πr= 3.98cm |
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| 12. |
हरण- दिए गए आव्यूह में प्रश्नवाचक चिह्न (?)| पर कौनसी संख्या आएगी?X9X3+५4|3|2| 8 | 32 |5 |3|1| 9 | 247| 3 | 3 | 7 | 70|2|9|4|12| ? | |
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Answer» 60 is correct sorry. . ..... 84 (2 × 9 × 4) + 12 = 84 set 1(4 × 3 × 2) + 8 = 32set 2(5 × 3 × 1) + 9 = 24set 3(7 × 3 × 3) + 7 = 70 the answer is the 70 |
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| 13. |
A circular piece of thin wire is converted into a rhombus of side 11 cm. Find thediameter of the circular piece. |
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| 14. |
2. A circular piece of thin wire is converted into a rhombus of side 11 cm. Find thediameter of the circular piece. |
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Answer» Circumference of circular wire= Perimeter of square wire2πr=4*side2*(22/7)*r=4*11(22/7)*r=22r=7cmdiameter=2r=14cm |
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| 15. |
6 207) + 6(10 12) + 15= (..t2..) + 1537 |
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| 16. |
A person sells his computer for Rs. 18000making a profit of 20%. How much did thecomputer cost him ? |
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Answer» Cost * (100 + 20 /100) = 18000=> Cost = 18000*5 / 6=> Cost = Rs. 15000 |
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| 17. |
1. Cost of 42 computers is * 1199940. Find the cost of1 computer. |
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Answer» cost of 42 computers = 11,99,940 rs .cost of 1 computer = 11,99,940 ÷ 42 = 28,570 rs. 28570₹ is the answer of the following 28570 rs. is correct answer. cost of computer=11,99,940no. of computer=4211,99,940 cost of 42 computer = 1199940cost of one computer= 1199940÷42=Rs 28570 |
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| 18. |
BSE sP 2013xifa and β are the zeroes of the quadratic polynomial f(x) = 3x2-5-2, find theHOTSt if α and β are the zeroes of the quadratic polynomial f(x) 4x2-4x + 1, find thevalue of +value of + . |
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| 19. |
The marked price of a computer is 22,000. After allowing a 10% discountdealer still makes a profit of 20%. Find the cost price of a computer.discount of |
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Answer» Profit = selling price - cost price let the cost price be cThe selling price: = 10% discount= discounted price = 100%(marked price) - discount (10%) = 90% = 90%/100% × 22000 = 0.9× 22000 = rs. 19800This discounted price is the selling price. Therefore profit = 19800 - C % profit = profit/ cost price×100% = 19800-c/ c×100% = 20% = 100(19800-c)/c = 20 = 100(19800-c)/c = 20 = 1980000 - 100c = 20c = 120c = 1980000 = c = 1980000/120 c = Rs. 16500 |
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| 20. |
(ap/f α and pare the zeroes ofthe quadratic polynomial f(x)-kx2 +4x + 4 such that α, p, 24, find the.ue of k |
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Answer» Solution :- α and β are the zeros of the given polynomial Kx² + 4x + 4 = 0so, product of zeros = αβ = constant/coefficient of x² = 4/Ksum of zeros = α + β = -coefficient of x/Coefficient of x² = -4/k Now, α² + β² = 24 ⇒(α + β)² - 2αβ = 24 ⇒(-4/k)² - 2(4/k) = 24 ⇒16/K² - 8/k = 24 ⇒ 2 - k = 3k² ⇒3k² + k -2 = 0 ⇒ 3k² + 6k - k - 2 = 0⇒3k(k + 2) - 1(k +2) = 0⇒(3k -1)(k +2) = 0 Hence, k = 1/3 and -2 |
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| 21. |
+ k such that22. If α and β are the zeros of the polynomial f(x)value ofa A4.+x-2, find the23 I |
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Answer» Like my answer if you find it useful! |
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| 22. |
Evaluate: sin 35° - sin 55° — c0S 35° - cos 55°. |
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| 23. |
4-2x3 t 3x2-9x + 3a-7. When divided by x + 1 leavesf the polynomial f (x)the remainder 20, then find the value of a. |
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| 24. |
\sin 35^{\circ} \sin 55^{\circ}-\cos 35^{\circ} \cos 55^{\circ}=0 |
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| 25. |
8१80 -guis * 0500 +0 पा न |
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| 26. |
sin(35^circ)*cos(55^circ) %2B sin(55^circ)*cos(35^circ) |
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Answer» sin35 cos55 +cos35sin55/ cosec10^2-tan80^2= sin(90-35)cos55 + cos(90-35)sin55/ cosec^2(90-10) - tan80^2=cos55 cos55+sin55sin55/cosec^2(90-10)-tan80^2= cos55^2+ cos55^2/1/ sin80^2- sin80^2/ cos80^2=( coss55^2+ sin55^2)/ cos80^2- sin80^2/ sin80=1/cos80^2+ sin80^2/sin80^2= sin88^2/ cos88^2+sin80^2/+1=tan80^2+1 |
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| 27. |
Use Euclid's algorithm to find the HCF of 408 and 1032. |
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Answer» If the HCF of 408 and 1032 can be written as By Euclid 's division algorithm, 1032 = 408×2 + 216 408 = 216×1 + 192 216 = 192×1 + 24 192 = 24×8 + 0 Since the remainder becomes 0 here, so HCF of 408 and 1032 is 24 |
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| 28. |
\operatorname { sin } 35 ^ { \circ } \operatorname { sin } 55 ^ { \circ } - \operatorname { cos } 35 ^ { \circ } \operatorname { cos } 55 ^ { \circ } = 0 |
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Answer» Since cos (a+b) = - (sin a sinb - cos a cos b) so cos (35+55) = -(sin35 sin 55- cos 35 cos 55) so = - cos 90= 0 ѕιη35°×¢σѕ(90°-35°)-- ¢σѕ35°×ѕιη(90°-35°)=0ѕιη35°×1/ѕιη35°-- ¢σѕ35°×1/¢σѕ35°=01-1=00=0 |
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| 29. |
भुजाओं 0? और २0 को क्रमश: बिन्दुओं 5 और पं=135° R ok £POT=110° &, /1 LPRQ T3-ASPA135° |
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Answer» Given,∠SPR =135° and ∠PQT =110° A.T.Q ∠SPR+∠QPR = 180° (by Linear pair axiom.) ⇒ 135°+∠QPR =180° ⇒ ∠QPR = 180° - 135° ⇒ ∠QPR = 45° ∠PQT+∠PQR = 180° (by Linear pair axiom.) ⇒ 110°+∠PQR =180° ⇒ ∠PQR = 70° Now,∠PQR+∠QPR+ ∠PRQ = 180° (Sum of the interior angles of thetriangle.)⇒70°+ 45° + ∠PRQ =180°⇒ 115°+ ∠PRQ =180° ⇒ ∠PRQ = 65° Like my answer if you find it useful! English mein kyo nahi pasand aa rahi |
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| 30. |
in 35°\? _ 1००8 55cos 55° sin 35°2) - 2०08 60 का मान है |
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Answer» (sin35/cos55)²+(cos55/sin35)²-2cos60 =[cos(90-35)/cos55]²+[cos55cos(90-35)]²-2cos60 =(cos55/cos 55)²+(cos55/cos 55)²-2cos60 =1+1-2×1\2=1+1-1=1 |
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| 31. |
( \frac { \operatorname { sin } 35 ^ { \circ } } { \operatorname { cos } 55 ^ { \circ } } ) ^ { 2 } + ( \frac { \operatorname { cos } 55 ^ { \circ } } { \operatorname { sin } 35 ^ { \circ } } ) - 2 \operatorname { cos } 60 ^ { \circ } |
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Answer» sin35=sin(90-55)=cos55cos55=cos(90-35)=sin35cos60=1/2so (cos55/cos55)^2+(sin35/sin35)^2-2(1/2)=1+1-1=1 Thanks |
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| 32. |
Calculate Arithmetic Mean, Median and Mode for thefollowing data.lessClass| less | less less less less less less less lessthan than than than than than than than than than10 20 30 40 50 60 70 80 90 100MarksNo. oftudentsNo Of5151527 | 422742686888 100 1 06 110 11588 100 106 110 115 |
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| 33. |
271women were examined in'a hospital by a doctor and the number of heartbeats perwere recorded and summarised as follows. Find the mean heartbeats per minutefor these women, choosing a suitable method.Number of heartbeats 65-68per minuteNumber of women68-7171-74 4-777-30 30-3 6-s. In a retail market, fruit vendors were selling mangoes kept in packing boxes. Thescboxes contained varying number of mangoes. The following was the distribution ofmangoes according to the number of boxes.Number of mangoes 50-52 53-55 56-58 59-61 62135 11S215110Number of boxesFind the mean number of mangoes kept in a packing box. Which method of findingthe mean did you choose?nai of 25 houscholds in a lecality |
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| 34. |
if the hcf of 408 and 1032 is expressed in the form of 1032m-408×5. find m |
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Answer» HCF(408, 1032) = 24 1032m - 408*5 = 24 1032m = 24 + 2040 1032m = 2064 m = 2 |
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| 35. |
e HCF of 408 and 1032 is expressible in the form 1032 m- 408 5, find m |
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| 36. |
8064 7-1152Exercise 34Divide and find the quotient:5, 336 ÷ 39, 534 ÷213. 927 ÷97, 6312 ÷ 32. 80+46. 408 ÷10. 432 314. 714618. 9303+7 |
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Answer» 1)202)205)1126)1029)26714)119 |
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| 37. |
408 is a straight line. Find x andalso determine the measure of each angle |
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Answer» Concept : Sum of angles in staright line is 180°.a) 6x + 4x = 180°10x = 180°x = 180°/10x = 18°So angles are 108°, 72° b) x/2 + x/3 = 180°5x/6 = 180°x = (180° × 6)/5x = 216°So angles are , 216/2 = 108° , 216°/3 = 72° |
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| 38. |
Find the value of x or (*) in following question(i)2** 8+ 95167(ii)68xx408 |
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Answer» here in this question *=4 okkkkkkkkkkkk |
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| 39. |
The HCF of 408 and 1032 is expressible in the form 1032m-2040. Find the value of m.Also, find the LCM of 408 and 1032.14, |
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| 40. |
(#0500 0 (2 G0+408- 3B |
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| 41. |
11. For which value(s) on, do the pair of linear equations λχ +y = λ2 and x + Ay = 1 have(i) no solution? (i) infinitely many solutions? (ii) a unique solution? |
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| 42. |
3x = 1-12 , then find the value of | x-Solution: |
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| 43. |
tf - 2 is a solution of Ksthen find the value of |
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Answer» put x =✓2 in the given equationKx²+✓2x-4 = 0so K (✓2)² + ✓2 × ✓2 -4 = 0 2K -2 =0 K = 1 |
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| 44. |
price. Puu le malleu PICCOThe marked price of a computer is 22,000. After allowing a 10% discount,dealer still makes a profit of 20%. Find the cost price of a computer. |
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Answer» The market price of the computer is Rs 22000 After 10% discount, its cost be = Rs [22000 - (22000 × 10%)] = Rs [22000 - (22000 × 10/100)] = Rs (22000 - 2200) = Rs 19800 Let us take the cost price of the computer be Rs x The dealer made a profit of 20% Then, its selling price was = Rs [x + (x × 20%)] = Rs [x + (x × 20/100)] = Rs (x + x/5) = Rs (6x/5) By the given condition, 6x/5 = 19800 → x = 19800 × 5/6 → x = 16500 ∴ the cost price of the.computer is Rs 16500 the cost price of the computer is Rs 16500 the cost price of the computer is Rs 16500 The cost price of the computer is Rs 16500 the cost price is 16500 the cost price of computer is Rs 16500 |
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| 45. |
5.complete revuIULIUMIS!The ratio of the radii of two circles is 3:7.Calculate ratio of their circumferences. |
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| 46. |
to rastig45.The diameter of the wheel of a vehicle is 5 m.It makes 7 revolutions per 9 seconds. What isthe speed of vehicle in Km/h?(a) 34 km/h(c) 54 km/h(d) 45 km/h(b) 44 km/h Sud |
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Answer» Given,Diameter of wheel = 5 mRadius of wheel r = 5/2 = 2.5 m Circumference of wheel= 2*pi*r= 2*22/7*5/2= 22/7 * 5 Distance covered by wheel in 7 revolution = 7*circumference= 7*22/7*5= 110 m Speed = Distance /time = 110/9 m/sec = 110/9 * 18/5 = 22*2 = 44 km/hr (b) is correct option |
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| 47. |
p1cc (3) поxi) The cost price (c) of an article when 22% profitis made after selling it for 's'. |
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| 48. |
15. How long a train will take to travel a distance of200 km with a speed of 60 km h1 ?3h 20 min |
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Answer» time= distance/speed =200km / (60km/hr) =10/3 hr |
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| 49. |
Ans. clockwise.tal42. A circular coil is kept horizontally on a table. If a vertical magnet isfalling freely over the coil, then what will be its accelerationAns. Less than g |
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Answer» When the magnet approaches the coil the current induced in the coil will repel the approaching magnet, so acceleration of the magnet is less than that of acceleration due to gravity. When the magnet has passed through the coil then it is attracted by the coil in accordance with Lenz's law hence its acceleration is again less thang. |
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| 50. |
Exponents and Power95(m) The speed of light is 300000000 m/sec(iv) The distance between the moon and the earth is 384467000 m(app)(v) The charge of an electron is 0.00000000000000000016 coulombs(vi) Thickness of a piece of paper is 0.0016 cm(vii) The diameter of a wire on a computer chip is 0.000005 cmIn a pack, there are 5 books, each of thickness 20 mm and 5 paper sheets each of thickness0.016mm. What is the total thickness of the pack.Rakesh solved some problems of exponents in the following way. Do you agree with thesolutions? If not why? Justify your argument.(1) xxx2x6cm)=x*() () = x2 = x(iv) x ² = d(0)3x = 1 / 1 |
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Answer» x^-3 ×x^-2 = x^-6;;; ii) x^3/ x^2=x^3-2= x^1 |
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