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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. |
Find ten rational numbers betweenala |
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| 2. |
S 20 m Rnd ala18 m25 mb. T10 m |
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Answer» area of the upper part ( SRTQ) which is a trapezium ,will be 1/2 *( 20+25)* 18 = 405 m² area of lower part (TQP) which is a triangle ,will be 1/2 * 25 * 10 = 125 m² So, the area will be (405+125) m² = 530 m² |
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| 3. |
o the bisectore d angle LB and Lc alatuangle ABC b meets at a point athen prove that the230C = 96 + I LA |
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| 4. |
4) 3show that the points Ala, btc), BC 6, cra), CCC, athare collinear |
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| 5. |
ala borrowedounded yearly. What amount will she pay at the end of 2 years and 4 montthe loan?26,400 from a Bank to buy a scooter at a rate of 15% ! |
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| 6. |
Find the area of a trapezium whose parallel sides are 25 cm, 13 cm anthe other sides are 15 cm each. |
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| 7. |
Find the area of a trapezium whose parallel sides are 25 cm, 13 cm anthe other sides are 15 cm each |
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| 8. |
. A shopkeeper bought achair for t 375 and sold it for t 400. Find the gĂĄin Percentage.2. Cost ofan item Ă :30 It was sold with a profit of 12%. Find the selling price. |
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Answer» 1. |
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| 9. |
12-A shopkeeper bought 200 kg wheat at Rs 8 per kg.He sold 150 kg wheat at Rs 10 per kg and theremaining quantity at Rs 6 per kg. Find his gainor loss per cent in the whole transaction. |
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| 10. |
Ayushree bought 6.75kg of biscuits at the rate of Rs.25.5 per kg and 0475kg ofnlumslat the rate of Rs. 165.6 per kg. She gives the shopkeeper a 500 rupee note. How much changeswill the shopkeeper return to her? |
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Answer» Biscuit price = 6.75x24.5 =172.125 Plum price=80rs Therefore total price=172.12+80 =252.12rs So shopkeeper give change to her =500-252.12 =247.88 rs |
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| 11. |
,1. A shopkeeper bought 3050 per ball. Atransaction?) cricket balls at the rate of 40 per ball. He sold 20 balls at the rate oft what rate should he sell the remaining/balls to have a profit of 40% in the whole |
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Answer» 30×40=1200rs 20×50=1000rs 40÷100×1200=480rs He should sold the remaning balls at rs 480 to have 40% profit wrong hai |
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| 12. |
7^(1/2)*8^(1/2) |
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| 13. |
(b) A man wants to buy 124 shares available at 66 (par value = 50).(1) How much should he invest?(1) If the dividend is 7.5%, what will be his annual income ?(ii) If he wants to increase income by 600, how many extra shares should he buy? |
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Answer» 1.The man will invest ₹8184.2.His annual income is ₹464 |
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| 14. |
( 5 ( 8 ^ { 1 / 3 } + 27 ^ { 13 } ) ) ^ { 1 / 2 } = |
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| 15. |
if f(x) = x² - 3x +4, find the value of x satisfying f(x) = f(2x +1). |
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Answer» here the value of x is =-1 or 2/3 Given:f (x) = x2 – 3x + 4Therefore,f (2x + 1) = (2x + 1)2 – 3(2x + 1) + 4 = 4x2 + 1 + 4x – 6x – 3 + 4 = 4x2 – 2x + 2Now,f (x) = f (2x + 1)⇒ x2 – 3x + 4 = 4x2 – 2x + 2⇒ 4x2 – x2 – 2x + 3x + 2 – 4 = 0⇒ 3x2 + x – 2 = 0⇒ 3x2 + 3x – 2x – 2 = 0⇒ 3x(x + 1) – 2(x +1) = 0⇒ (3x – 2)(x +1) = 0⇒ (x + 1) = 0 or ( 3x – 2) = 0⇒x=−1 or x=23⇒x=-1 or x=23Hence, x=−1,23x=-1,23. |
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| 16. |
\left. \begin{array} { l } { \operatorname { sin } 10 ^ { \circ } + \operatorname { sin } 20 ^ { \circ } + \operatorname { sin } 40 ^ { \circ } + \operatorname { sin } 50 ^ { \circ } = \operatorname { sin } 70 ^ { \circ } + \operatorname { sin } 80 ^ { \circ } } \\ { \operatorname { sin } ( 45 ^ { \circ } + A ) + \operatorname { sin } ( 45 ^ { \circ } - A ) = \sqrt { 2 } \operatorname { cos } A } \end{array} \right. |
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Answer» L.H.S. = 2sin15cos5+2sin45cos5 [ using sin C+sin D= 2sin C+D/2 cos C-D/2 for sin10+sin20 & sin40+sin50] = 2cos5 (sin15+sin45) = 2cos5 (2sin30cos15) [ using sin C+sin D= 2sin C+D/2 cos C-D/2 ] = 2cos5 (2 x 1/2 x cos15) = 2cos5 cos15 R.H.S. = sin70+sin80 = 2sin75cos5 [ using sin C+sin D= 2sin C+D/2 cos C-D/2 ]
sin75 = sin(90-15) = cos 15 L.H.S = 2cos5 cos15 R.H.S. = 2cos15 cos5 [ since, sin75 = cos15 ] |
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| 17. |
Find the values of a and b, if the function f is defined by-(f(x) =x2 + 3x + a , x 1bx +2, x>1is differentiable at χ=1 |
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| 18. |
\operatorname { sin } 10 ^ { \circ } + \operatorname { sin } 20 ^ { \circ } + \operatorname { sin } 40 ^ { \circ } + \operatorname { sin } 50 ^ { \circ } = \operatorname { sin } 70 ^ { \circ } + \operatorname { sin } 80 ^ { \circ } |
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Answer» L.H.S. = 2sin15cos5+2sin45cos5 [ using sin C+sin D= 2sin C+D/2 cos C-D/2 for sin10+sin20 & sin40+sin50]= 2cos5 (sin15+sin45)= 2cos5 (2sin30cos15) [ using sin C+sin D= 2sin C+D/2 cos C-D/2 ]= 2cos5 (2 x 1/2 x cos15)= 2cos5 cos15 R.H.S. = sin70+sin80= 2sin75cos5 [ using sin C+sin D= 2sin C+D/2 cos C-D/2 ] sin75 = sin(90-15) = cos 15 L.H.S = 2cos5 cos15R.H.S. = 2cos15 cos5 [ since, sin75 = cos15 ] |
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| 19. |
when 0 <= 5x-4,= 4x2-3x, when x > 12. Let f(x)1Discuss the differentiability of f(x) at x = 1 |
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| 20. |
47. If f(x)=x2-3x+7 evaluate f(2)-f(-1)=f().fanales of a parallelogra |
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| 21. |
3. Write the rational number for each point marked with a letterP Q R S T Ua.2222.221224MN-11-10 21-9--5-4-3-2-111 7 7 7 7 7 7 7 771-10798317323210this to |
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Answer» p=2 Q=5 R=6 S=9 T=10 u=13 a. p=2/4 , q=5/4 , r=6/4 , s=9/4 , t=10/4 , u=13/4 P=2, Q=5,R=6, S=9, T=10, U=13 |
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| 22. |
31. What is the multiplicative inverse of7977377 |
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Answer» A multiplicative inverse or reciprocal for a number x, denoted by 1/x or x−1, is a number which when multiplied by x yields the multiplicative identity 1. So multiplicative inverse of(-3/7)^-9 = (-3/7)9 (3) is correct option |
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| 23. |
2. A man allows 20% discount on the M.P. of his goods and still earns a profit of 4%. What isthe cost price of the goods for him, marked at 600? |
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| 24. |
4. Thewheel of a bullock cart has a diameter of 1.4m. How many rotations will thewheel complete as the cart travels 1.1km? |
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Answer» In one rotation distance covered= 2*pi*r= 2*22/7*.7= 44/10 = 22/5 Let wheel complete n rotations in 1.1 kmThenn*22/5 = 1.1*1000n = 1100*5/22 = 500/2n = 250 Thus bullock cart complete 250 rotations thanks ♡ |
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| 25. |
The wheel of a bullock cart has a diameter of 1.4 m. How many rotations will the wheel complete as the cart travels 1.1 km? |
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| 26. |
15.0 9If the circumference of a circle is 176 cm, find its radius.The radius of a circular garden is 56 m. What would it cost to put a 4-round faround this garden at a rate of 40 rupees per metre?The wheel of a bullock cart has a diameter of 1.4m. How many rotations wilwheel complete as the cart travels 1.1 km ?11. SSS98 |
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Answer» r=28 is the following question answer. |
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| 27. |
10) Simplify (3 + V2) (6+4 3)11) Simplify V4Ă 16 |
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| 28. |
car travels 60 km in 35 minutes. At the same rate,ow many kilometres will it travel in one hour? |
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Answer» Distance travelled by car = 60kmtime taken by car = 45min. = 35/60 hr = 7/12 hr.speed = distance/time = 60/(7/12)speed = 60*12/7 = 102.85km/hr. |
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| 29. |
Simplify3/3-22 |
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Answer» Bai ji Ada nahi |
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| 30. |
213 101 2) Simplify( 2 ) Simplify : 189% (20 x 294(3) Simplify: 2x -11x- (4x-3) 2x) -x4 |
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Answer» (1) (8^2)3 ÷ (2^10x2^6)/4 = (8)^6*4/(2)^16) = (2)^18*(2)^2 /(2)^16 = (2)^[20 - 16] = (2)^4 = 16 (2) 2x - [1 +{ x-(4x - 3) + 2x} - x} + 4= 2x - [4 - 2x] + 4= 2x - 4 + 2x + 4= 4x |
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| 31. |
A shopkeeper earns a profit of 1 on selling one pen and suffers a loss of 30 paise on selling one pencin a particular month, he incurs a loss ofき5. In that month, he sold 40 pens. How many pencils didell in that period? |
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| 32. |
A shopkeeper earns a profit of 10% after allowing a discount of 20% on the marked price. Find the costprice of the article whose marked price is 880 |
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| 33. |
5. A shopkeeper earns a profit of? 2 by selling a new register and incurs a loss of 1.20 per register of oldstock. In a particular day he incurs a loss of R 4, he sold 10 new registers. How many old registers did hesell on that particular day? |
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Answer» Let the old registers sold be xso10*2-1.2x= -4-1.2x= -24x=-24/1.2= 20 |
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| 34. |
A shopkeeper earns a profit of 12% on selling a-book at 10% discount on the printed price. The ratioof the cost price to the printed price of the book is:(a) 50: 61(c) 99 125(b) 45:56(d) 55:69 |
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Answer» Let the CP be 100.Hence, SP = 100 + 12% of 100 = 112.If the marked price be X, then90% of X = 112=>X = (112*100)/90 = Rs. 1120/9Hence,Required ratio = 100:1120/9 = 900:1120 = 45:56. |
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| 35. |
20. How much percent above the cost price should a shopkeeper mark his goods so thatafter allowing a discount of 20% he still earns a profit of 20%? |
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Answer» let the CP be Rs. 100 and MP be Rs. xgain % after discount = 20%gain = 20% of 100 = Rs. 20⇒SP of the article = CP + Gain = Rs. (100 + 20) = Rs. 120Discount % = 20%Discount = 20% of Marked Price= Rs.20/100 * x = Rs. 20x/100 = Rs. x/5as the question saysmarked Price - discount = SP=> x - x/5 = 120=> (5x - x)/5 = 120=> 4x/5 = 120=> x = 120 * 5/4 = Rs. 150marked Price = Rs. 150amount marked above the CP = MP - CP = Rs. (150 - 100) = Rs. 50∴% amount marked above the CP= Amount increased/CP * 100= 50/100 * 100= 50% |
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| 36. |
The marked price of a mobile pof 20% and earns a profithone 1s 8,000. A shopkeeper sells it at a discountof 25%. Find the cost price of the mobile phone |
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| 37. |
A man bought 360 ten-rupee shares paying 12% per annum. He sold them when theprice rose to 721 and invested the proceeds in five-rupee shares paying 41% perannum at 3.5 per share. Find the annual change in his income. |
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| 38. |
ten-rupee shares at R8. Find the changein his annual incor25 A man bought 360 ten-rupee shares paying 12% per annum. He sold them when thep s paying 4o% per annumq1 He sells out/price rose to21 and invested the proceeds in five-rupee shar?n at3.5 per share. Find the annual change in his income |
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Answer» 1st case Nominal value of 1 share = Rs. 10 Nominal value of 360 shares = Rs. 10 × 360 = Rs. 3,600 Market value of 1 share = Rs. 21 Market value of 360 shares = Rs. 21 × 360 = Rs. 7,560 Dividend% = 12% Dividend = 12% of Rs. 3600=12/100*3600=rs 4322nd caseNominal value of 1 share=rs 5market value of 1 share =rs 3.50no of shares purchased=7560/3.50=2160sharesNominal value of 2160 shares = Rs. 5 × 2160 = Rs. 10,800 Dividend% = 4.5% Dividend = 4.5% of Rs. 10,800 = (4.5/100) ×10,800= Rs. 486 Annual change in income = Rs. 486 – Rs. 432 = Rs. 54 increase |
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| 39. |
3 My sister had a five hundred rupee note in her purse. Shewent to market to purchase some fruits. She bought fruitsfor 185.75. How much money was leftover with her? |
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Answer» original money=500amount left=500-185.75 =314.25 |
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| 40. |
Find the sum of the following A.Ps(0 2, 7, 12, to 12 terms |
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| 41. |
In Figure 2, DE BC. Find the length of side AD, given that AE - 18 emBD = 7.2 cm and CE = 5.4 cm.1-8 cm7-2 cmcm |
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Answer» Applying BPTso hereAD/ DB= AE/ECx/7.2= 1.8/5.4x/7.2= 1/3x= 7.2/3= 2.4 cm |
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| 42. |
how many 20p coins make ₹1 ? |
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Answer» 1rs=100paise100/20=5So,5 20p coins make rs1 five coins of 20 p makes 1 rs ₹1=100pSo, 5 20p coin is needed 1 rs=100paise 100/20=5 5 20p coins make rs1 5 coin of 20p make 1 rupee 5 coins of 20p make 1 rupee |
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| 43. |
Find the value)12+12+/12nfinite terms. |
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| 44. |
. How many 25-paisa coins make a rupee? |
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Answer» 25 पैसे +25 पैसे =50 25 पैसे+25पैसे= 50 तो 4 coin |
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| 45. |
A shopkeeper bought a fan forRs. 200 and he sold it forRs. 180. What was his loss per cent? |
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Answer» Loss percentage = 200-180/200 = 20/200 = 1/10 = 10% |
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| 46. |
50 p coins make I rupee. |
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Answer» We know 1 rupee = 100 paise So, Two 50 p coins make 1 rupee |
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| 47. |
2) The sum of 36 and (-52)(ii) 16(i) -88(iv) 88(iii) -163) Ratio 3:4 is same as(ii) 6:8(i) 4:3(iii) 8:6(iv) 1:44) Which of the following gives negative integers?(1) 7 (2) |
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Answer» 2) 36 + (-52) = 36 - 52 = -163) 6:8 6:8 = 3:4 quest 2nd ans...3rd.,, nd 3rd quest.. ans... is 2nd 2) 36+(-52)》 36-52= -16 3) 3:4 is same as 6:8 because by dividing 6:8 by 2 we will get the same ratio(3:4) 2) Answer:(III) -16Explanation:36-52= -16 3)Answer: (ii)6:8Explanation6/8=3/4Thus3:4=6:8 |
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| 48. |
how to make angle of 22° using compass |
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Answer» Construct a right angle, Draw it's bisector using compass.Now, the the bisecyor of the new angles formed using a compass. |
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| 49. |
10. Write a pair of integers whose :(i) sum gives an integer smaller than both the integers.(ii) sum gives an integer smaller than only one of the integers.(iii) difference gives a negative integer.(iv) difference gives an integer greater than only one of the integers. |
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| 50. |
Write a pair of negative integers whose difference gives 10 |
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Answer» Let the two integer be -4,-14Now -4-(-14)=-4+14=10Thus the difference is 10The two integer can be -4 & -14 |
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