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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. |
In figure 1.73, ZABC DCB 909AB 6, DC 8A(Δ ABC)then A A DCB)Fig. 1.73 |
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Answer» these two triangles are congruent to each otherand we know that if sides of two similar triangles are in the ratioa:b, their areas are in the proportiona^2:b^2ratio of side= 6/8= 3/4so area would be (3/4)^29/16 |
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| 2. |
How many liters of water flow outa pipe having an area of crosssection of 5 cm² in one minute, ifthe speed of water in the pipe is 30cm/sec.? |
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Answer» the right answers is 9 litres of water may flow through the pipe |
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| 3. |
. In the given figure, triangles ABC and DCB are right-angled at A and D respectively and ACProve that ΔABC DCB.DB. |
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| 4. |
If a, a1, a2,,a10 , b are in Đ.Đ. and a, g1,9between910, b are in G.P. and h is the H.Ma and bthen+ag++5sas+as91910929995.96is101530 |
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| 5. |
in the given figure, name the collinear points : |
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Answer» Collinear points are points which lie on the same line Therefore,Collinear points are (A, B), (A,C), (A,D), (B,C,D) |
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| 6. |
Express 1947 into product of prime factors. |
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| 7. |
A can do 3/8 part of a work in 9 hrs. B can do 1/4 part of the same work in 4 hrs.Both work together and complete the work in:48(a) hrs35(c) hrs(d) 5 hrs(h) ag hrs48 |
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Answer» A can do 3/8 part work in 9 hrs so total work in 24hB can do 1/4 part work in 4 hrs so total work in 16hif both work together then 1/24+1/16=(16+24)/(24×16)=10/96so time needed is 9.6h |
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| 8. |
अंकों का योग =+) = 4+ 5 = 9|75 हो, तो निम्नलिखित में उसका कौन-केसी संख्या कास्सा 45 है ?(C)|Solution : (D)(RRB भोपाल T.C., 2005या का 5-75 |
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Answer» D) 3/8 beacuse 8 into 6 +3 =45 |
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| 9. |
By water harvesting, we mean:(a) Conservation of water(c) Water flow1.(b) Protection of land(d) Effect on rainfall. |
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Answer» conservation of water is called water harvesting.Wateringharvestingmeans capturing rainwater, where it falls and capture the runoff from, catchment and streams etc. |
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| 10. |
Answerthefollowingquestions.(i) What are the Temperate Grasslands of North America called? |
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Answer» it is called prairies |
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| 11. |
k at the fgure alongside ande the tollowing:Collinear pointsine segmentsnesthe different names of the line |
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Answer» Rays: OM, OR, OP |
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| 12. |
Aug) XU+ o -x)ueyT ueyâ x uey |
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| 13. |
27gBBM2351-H223-AUG-18 |
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Answer» please send the meaningful question |
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| 14. |
IF3 +Find the value oFc tOc |
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Answer» Given x = 3 + 2√2 ⇒x^2 = 17 + 12√2 ⇒1/x2 = 1/(17 + 12√2) On rationalizing the denominator, we get 1/x2 = (17 – 12√2) ∴ (x2 +1/x2) = (17 + 12√2) + (17 – 12√2) = 34 I think the wright answer is 10+√2 . My answer is correct so click the like option please |
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| 15. |
कि नि कीही हि ८2 b0, L 3T 0०0, 4 322.206८८ 1, ॥ ( किए am £ 2 En /'3 A 3 e ‘ |
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| 16. |
1.5 = 0.04gy हो, तोका मान क्या होगा?।(d) इनमें से कोई नहीं/RRB ASM/Goods Guard 2007] |
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Answer» (C) 1.5 x = 0.04yx = 0.04/1.5yx = 4/15*10y = 2y/75 Now, y - x/ y + x Put value of x = y - 2y/75/ y + 2y/75 = (75 - 2)y/ (75 + 2)y = 73/77 (c) is correct option |
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| 17. |
38. 1.5 बराबर है--7 54 (0,4rard 2004] 3 ®) 3 कि[RRB TC/CC 2006] |
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Answer» the answer is 4/3 of above question |
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| 18. |
[RRB 2016] |
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Answer» 17+7×27÷9-3=17+7×3-3=17+21-3=17+18=35 |
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| 19. |
| 0.27 यदि दो पूर्णाकों के वर्गमूलों का योग 1/14 + 843 है, तो इन- दो पूर्णाकों के वर्गों का योग क्या है?RRB ग्रुप डी परीक्षा, 2018 |
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Answer» gfysjwjyyd gfysjwjyyd gyjwjsthsnehne .......................... 772 is the right answer 37060 is the answer... 388 is the answer.... √14+8√3=17.5980638473 is correct answer 388 is the right answer of the following |
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| 20. |
कशॉप कैलकुलेशन एण्ड साइंस*7777,.. किस साधारण भिन्न के बराबर .77===e 2437il‘o |
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Answer» maan le ki0.4777777 x haix = 0.477777...... (1)10 se yogfal kare10x = 4.777777...... (2)(1) ko 100 se yogfal kare 100x = 47.777777..... (3) (3) - (2)90x = 43x = 43/90 ke barabar hai |
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| 21. |
1322+7777 |
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Answer» 1322+7777=9099 is the answer |
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| 22. |
8+0-7777+8 ? |
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Answer» 8+0-7777+8=-7777+16=-7761 |
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| 23. |
[RRB JE 2015 27th AUG 1st SHIFT]fa:b 5:12 and b : ca) 3:28:5, then a : c is(b) 2:3(d) 4 5 |
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Answer» a:b = 5:12 b:c = 8:5 multiplying a:b by 2 and b:c by 3, we get a:b = 10:24 b:c = 24:15 So a:b:c = 10:24:15 and a:c = 10:15 = 2:3 |
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| 24. |
en pr bveOC3 |
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| 25. |
at bor) |
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| 26. |
tobe placed ineach bor can hold aer ct i-hous naany usays Can we place thobll so tt no bon emau empty? |
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| 27. |
en or 18 women can do a piece ofIf 12 mwork in 14 days. How long will 8 menand 16 women take to finish the work?41RRB 2007 |
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Answer» One man one day work = 1/12*14 One woman one day work = 1/18*14 Then,8 men one day work = 8/12*14= 1/3*7 = 1/2116 women one day work = 16/18*14 = 4/63 Therefore,8 men and 16 women one day work = 1/21 + 4/63= (3 + 4)/63 = 7/63 = 1/9 Therefore 8 men and 16 women together will complete work in 9 days |
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| 28. |
9 z ^ { 3 } - 27 ^ { 2 } z - 100 z + 300 \text { if } ( 3 z + 10 ) |
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| 29. |
Quis = guer- gsodb |
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| 30. |
A dome of a building is in the form of a hemisphere. From inside, it was white-washedat the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square metre, findthe |
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| 31. |
Find the compound interest on6400 for 2 years at 55% per annum. |
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Answer» For compound interest amount after n years A = P(1 + R/100)^n Where P = 6400, R = 11/2, n = 2 A = 6400(1 + 11/200)^2 = 6400(211/200)^2 = 6400/4*(2.11)^2 = 1600*4.45 = 7120 CI = A - P = 7120 - 6400 = Rs 720 |
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| 32. |
Doyt[ st 9“7{3& |
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Answer» Prime factorisation of 74297429 = 17×19×23 |
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| 33. |
ST LRSS TR A N& यूंक्लिड विभाजन प्रमेयिका का प्रयोग करके दर्शाइए कि किसी धनात्मक पृर्णाक का97, क्र + 1 या 9#+8 के रूप का होता है।-.3 ञ2 4 e ol A i A |
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Answer» आज्ञा देना किसी भी सकारात्मक पूर्णांक और b = 3 a = 3q + r, जहाँ q 3 0 और 0 3 r <3 इसलिए, प्रत्येक संख्या को इन तीन रूपों के रूप में दर्शाया जा सकता है। तीन मामले हैं। केस 1: जब a = 3q, जहाँ m एक पूर्णांक है जैसे कि m = केस 2: जब a = 3q + 1, a ³= (3q +1)³ a³ = 27q³ + 27q² + 9q + 1 a³= 9 (3q³ + 3q ²+ q) + 1 a³= 9 m + 1 जहाँ m एक पूर्णांक है जैसे कि m = (3q ³+ 3q ²+ q) केस 3: जब a = 3q + 2, a ³ = (3q +2)³ एक 3 = 27q ³+ 54q²+ 36q + 8 एक 3 = 9 (3q ³+ 6q ²+ 4q) + 8 एक 3 = 9 मी + 8 जहाँ m एक पूर्णांक है जैसे कि m = (3q ³ + 6q ²+ 4q) इसलिए, किसी भी धनात्मक पूर्णांक का घन 9m, 9m + 1, या 9m + 8 का है। |
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| 34. |
2.Mrs.Kaushikhas a square plot with the (ameasurement as shown in the figure. She wants toconstruct a house in the middle of the plot. A garden is developedaround the house. Find the total cost of developing a garden aroundthe house at the rate of 55 per m2. |
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Answer» tqsm |
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| 35. |
, Mrs. Kaushik has a square plot with the(a)measurement as shown in the figure. She wants toconstruct a house in the middle of the plot. A garden is developedaround the house. Find the total cost of developing a garden aroundthe house at the rate of t 55 per nm2 |
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| 36. |
2. Mrs. Kaushik has a square plot with themeasurement as shown in the figure. She wants toconstruct a house in the middle of the plot. A garden is developedaround the house. Find the total cost of developing a garden aroundthe house at the rate of 55 per m2. |
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| 37. |
2012 Planner10 A Foruits bay Contains is of Oranges,of Pomegranates, 5 of the applesand rest of the Foruits are MangoeseThe total fornits in the bag were240 find.© The fraction of Mangoes in the bag. |
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Answer» total fruits 240mangoes = total fruits - rem fruitsm = 240 - ( 240/3 +240/4 +240/5)m =240 - ( 80 +60+48)m =240 - 188m =52fraction =52/240= 13/60 answer |
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| 38. |
Kaushik has a square plot as shown in Fig. 20.15. She wants to constructse in the middle of the plot. A garden is developed around the house.a houseind the total cost of developing a garden around the house, if cost ofdeveloping a garden is Rs 55 per square metre. |
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| 39. |
2. Mrs. Kaushik has a square plot with themeasurement as shown in the figure. She wants toconstruct a house in the middle of the plot. A garden is developedaround the house. Find the total cost of developing a garden aroundthe house at the rate of 55 per m2lor 25 |
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Answer» Side of a square plot = 25 m Area of square plot = = = 625 m2 Length of the house = 20 m and Breadth of the house = 15 m Area of the house = length breadth = 20 15 = 300 m2 Area of garden = Area of square plot – Area of house = 625 – 300 = 325 m2 Cost of developing the garden per sq. m = ` 55 Cost of developing the garden 325 sq. m = = ` 17,875 |
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| 40. |
iasaal-60 m2. Mrs. Kaushik has a square plot with the(a)measurement as shown in the figure. She wants toconstruct a house in the middle of the plot. A garden is developedaround the house. Find the total cost of developing a garden aroundthe house at the rate of 55 per m2. |
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| 41. |
boaMrs. Kaushik has a square plot with themeasurement as shown in the figure. She wants to2.(a)nstruct a house in the middle of the plot. A garden is developedcoaround the house. Find the total cost of developing a garden aroundthe house at the rate of55 per m2 |
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| 42. |
. +1 = gsod guisy |o ST - 1] 2 i |
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Answer» 4sin(x)cos(x)+2sin(x)-2cos(x)-1=0 2sin(x)(2cos(x)+1)-1(2cos(x)+1)=0 (2sin(x)-1)(2cos(x)+1)=0 sin(x)=1/2 or cos(x)=-1/2 x=2npi+pi/6 or x=2pi/3,4pi/3,8pi/3 etc |
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| 43. |
There is a basket of fruit with 23 apples and 17 oranges. What is the minimum number ofchanges you can make so that the ratio of apples to oranges is 3 to 1?3 |
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Answer» If we remove 2 apples and 10 oranges then,Number of apples = 21Number of oranges = 7Ratio=21 : 7 = 3 : 1 |
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| 44. |
ST 1. Write the next three natural numbers after 10999 |
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Answer» 11000,11001,11002 is the answer |
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| 45. |
3. In figure, if PQR is thetangent to a circle at Qwhose centre is O, AB isa chord parallel to PR andzBQR 70°, then findZAQB-- \ |
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| 46. |
\left. \begin{array} { l } { \operatorname { cosec } ( - 1410 ^ { \circ } ) } \\ { \operatorname { sin } ( - \frac { 11 \pi } { 3 } ) } \end{array} \right. |
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| 47. |
The denominator of a rational number is greater than its numerator by 8. If thenumerator is increased by 17 and the denominator is decreased by 1, the numberobtained isFind the rational number. |
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Answer» answer is simple th answer is 7/15 answer is 7/15 is the correct answer |
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| 48. |
14. AB and CD are two parallel chords of a circle lying on the opposite sides of the centre such thatAB-24 cm and CD = 10 cm. If the chord AB is at a distance of 5 cmn from the centre, find thedistance of chord CD from the centre of the circle. |
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| 49. |
Sanya has a piece of land which is in the shape of a Rhombus. Shewants to divide it equally in to two parts one for her son and other forher daughter.If the perimeter of the land is 400 m and one of its diagonalis 160m27.Find How much area eÄ ch of them will get? |
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Answer» As the diagonals of rhombus bisect each other: Therefore OD = 1/2.BD = 1/2 x 160 = 80 m OC = 1/2.AC In△OCD, we have, CD^2= OC^2+ OD^2 100^2 =OC^2+ 80^2 ⇒ OC^2= 10000 - 6400 ⇒OC^2 = 3600⇒OC = 60 m Therefore, area of△BCD = 1/2(BD x OC) = 1/2 x 160 x 60 = 4800 m2 ∴Each of them will get 4800 m2of area for their crops. |
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| 50. |
o the given equation.130 |
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Answer» 8y=19+58y=24y=24/8 =3 |
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