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Data Interpertation & airthmatic

Data Interpertation & airthmatic

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Two bags A and B contain different colors of balls. In bag A, there are 2 red balls, 3 green balls and 5 white balls while bag B contain 4 red balls, β€˜x’ green balls and 3 white balls. If the probability of drawing two green balls from bag A is 7/330 more than the probability of drawing 3 green balls from bag B, then find the value of β€˜x’. a) 4 b) 6 c) 5 d) 7 e) None of these

43.05% of 799.89 + 56.89% of 899.89 =? (a) 857 (b) 785 (c) 587 (d) 875 (e) 890

4649.98 Γ· 99.9 + 3749.9 Γ· 9.98 – 3659.98 Γ· 39.98 =? (a) 430 (b) 303 (c) 330 (d) 230 (e) 130

7429.98 Γ· 300.02 Γ— 10.004 =? (a) 256 (b) 251 (c) 258 (d) 248 (e) 266

48.01 Γ— 34.9 Γ· 2.95 = ? (a) 576 (b) 591 (c) 564 (d) 553 (e) 560

√5183.98 ÷ √36 = ? (a) 21 (b) 6 (c) 12 (d) 18 (e) 26

1299 Γ· 19.99 Γ— 25.01 + 400.01 =? (a) 2025 (b) 2300 (c) 1925 (d) 2200 (e) 1700

151.011 – 419.999 + 649.991 =? (a) 380 (b) 420 (c) 350 (d) 410 (e) 360

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#missing #series 1. 18,19,22,38,?,1459 a.96 b.157 c.148 d.77 e.163 2. 18,10,12,22,?,146 a.42 b.52 c.62 d.32 e.None of these 3. 7,25,46,?,105,149 a.66 b.72 c.78 d.84 e.62 4. 972,162,54,27,18,? a.27 b.10 c.9 d.36 e.15 5. 15,105,213,?,483,645 a.339 b.342 c.250 d.327 e.357 6. 9,?,26,90,115,331 a.13 b.17 c.19 d.15 e.21 7. 27,?,141,369,1281 a.57 b.61 c.65 d.69 e.73 8. 98,139,?,229,282 a.152 b.162 c.192 d.172 e.182 9. 35,88,176,847,? a.1595 b.1321 c.1496 d.1231 e.1121 10.11,12,23,134,?,12356 a.1260 b.1245 c.1235 d.1255 e.1225

#wrong #series 1. 15,17,20,29,56,137 a.15 b.20 c.29 d.56 e.137 2. 14,60,152,292,474,704 a.14 b.152 c.292 d.474 e.704 3. 19,49,115,247,469,819 a.49 b.115 c.247 d.469 e.819 4. 45,52,63,72,81,90 a.63 b.45 c.90 d.81 e.52 5. 441,435,423,402,373,331 a.331 b.441 c.435 d.402 e.373 6. 9871,8870,9760,8983,9649 a.8983 b.9871 c.9760 d.9649 e.8870 7. 78,105,138,175,216,259 a.105 b.78 c.175 d.216 e.259 8. 98,109.4,122.7,137.9,156 a.98 b.109.4 c.122.7 d.137.9 e.156 9. 800,400,200,1500,5250,23625 a.800 b.200 c.1500 d.5250 e.23625 10.13,17,31,69,133,233 a.69 b.233 c.31 d.133 e.17

10.I.π’™πŸ + πŸπŸŽπ’™ βˆ’ πŸπŸ’ = 𝟎 II.π’šπŸ βˆ’ πŸπŸπ’š + πŸπŸ– = 𝟎 a. 𝒙 > π’š b. 𝒙 < π’š c. 𝒙 β‰₯ π’š d. 𝒙 ≀ π’š e. 𝐱 = 𝐲 𝐨𝐫 𝐫𝐞π₯𝐚𝐭𝐒𝐨𝐧 𝐜𝐚𝐧𝐭 π›πž πžπ¬π­πšπ›π₯𝐒𝐬𝐑𝐞𝐝

9. I.π’™πŸ‘ = πŸπŸπŸπŸ”πŸ• II.π’šπŸ = πŸ“πŸπŸ— a. 𝒙 > π’š b. 𝒙 < π’š c. 𝒙 β‰₯ π’š d. 𝒙 ≀ π’š e. 𝐱 = 𝐲 𝐨𝐫 𝐫𝐞π₯𝐚𝐭𝐒𝐨𝐧 𝐜𝐚𝐧𝐭 π›πž πžπ¬π­πšπ›π₯𝐒𝐬𝐑𝐞𝐝

8. I.π’™πŸ = πŸπŸ“πŸπŸ II.π’šπŸ’ = πŸπŸπŸ—πŸ” a. 𝒙 > π’š b. 𝒙 < π’š c. 𝒙 β‰₯ π’š d. 𝒙 ≀ π’š e. 𝐱 = 𝐲 𝐨𝐫 𝐫𝐞π₯𝐚𝐭𝐒𝐨𝐧 𝐜𝐚𝐧𝐭 π›πž πžπ¬π­πšπ›π₯𝐒𝐬𝐑𝐞𝐝

7. I.πŸπ’™πŸ βˆ’ πŸπŸ‘π’™ + 𝟐𝟏 = 𝟎 II.π’šπŸ βˆ’ πŸπŸ•π’š + πŸ”πŸŽ = 𝟎 a. 𝒙 > π’š b. 𝒙 < π’š c. 𝒙 β‰₯ π’š d. 𝒙 ≀ π’š e. 𝐱 = 𝐲 𝐨𝐫 𝐫𝐞π₯𝐚𝐭𝐒𝐨𝐧 𝐜𝐚𝐧𝐭 π›πž πžπ¬π­πšπ›π₯𝐒𝐬𝐑𝐞𝐝

6. I.π’™πŸ + πŸ•π’™ βˆ’ πŸπŸ•πŸŽ = 𝟎 II.πŸ—π’šπŸ βˆ’ πŸπŸ”π’š + πŸπŸ” = 𝟎 a. 𝒙 > π’š b. 𝒙 < π’š c. 𝒙 β‰₯ π’š d. 𝒙 ≀ π’š e. 𝐱 = 𝐲 𝐨𝐫 𝐫𝐞π₯𝐚𝐭𝐒𝐨𝐧 𝐜𝐚𝐧𝐭 π›πž πžπ¬π­πšπ›π₯𝐒𝐬𝐑𝐞𝐝

5. I.πŸπŸπ’™πŸ + πŸπŸ•π’™ + 𝟏𝟎 = 𝟎 II.π’šπŸ βˆ’ πŸπŸπ’š + πŸ‘πŸ“ = 𝟎 a. 𝒙 > π’š b. 𝒙 < π’š c. 𝒙 β‰₯ π’š d. 𝒙 ≀ π’š e. 𝐱 = 𝐲 𝐨𝐫 𝐫𝐞π₯𝐚𝐭𝐒𝐨𝐧 𝐜𝐚𝐧𝐭 π›πž πžπ¬π­πšπ›π₯𝐒𝐬𝐑𝐞𝐝

4. I.π’™πŸ + 𝒙 βˆ’ πŸπŸ‘πŸ = 𝟎 II.π’šπŸ βˆ’ πŸπŸ‘π’š + πŸπŸ‘πŸ = 𝟎 a. 𝒙 > π’š b. 𝒙 < π’š c. 𝒙 β‰₯ π’š d. 𝒙 ≀ π’š e. 𝐱 = 𝐲 𝐨𝐫 𝐫𝐞π₯𝐚𝐭𝐒𝐨𝐧 𝐜𝐚𝐧𝐭 π›πž πžπ¬π­πšπ›π₯𝐒𝐬𝐑𝐞𝐝

3. I.π’™πŸ + πŸ–π’™ + 𝟏𝟐 = 𝟎 II.π’šπŸ + πŸπŸ”π’š + πŸ‘πŸ— = 𝟎 a. 𝒙 > π’š b. 𝒙 < π’š c. 𝒙 β‰₯ π’š d. 𝒙 ≀ π’š e. 𝐱 = 𝐲 𝐨𝐫 𝐫𝐞π₯𝐚𝐭𝐒𝐨𝐧 𝐜𝐚𝐧𝐭 π›πž πžπ¬π­πšπ›π₯𝐒𝐬𝐑𝐞𝐝

2. I.πŸ’π’™πŸ + πŸπŸ–π’™ + πŸ’πŸŽ = 𝟎 II.π’šπŸ + πŸ’π’š βˆ’ πŸ‘πŸ = 𝟎 a. 𝒙 > π’š b. 𝒙 < π’š c. 𝒙 β‰₯ π’š d. 𝒙 ≀ π’š e. 𝐱 = 𝐲 𝐨𝐫 𝐫𝐞π₯𝐚𝐭𝐒𝐨𝐧 𝐜𝐚𝐧𝐭 π›πž πžπ¬π­πšπ›π₯𝐒𝐬𝐑𝐞𝐝