Arithmetics
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Arithmetics: Series
Find out K in following series
(1) | 400 | 394 | 388 | 382 | 376 | K | ||||
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Answer | K = C =370 | |||||||||
Take Home | Descending series, from 376, we have increment towards left by 6, so k is by decreasing 6. |
(2) | 15 | 28 | 27 | 50 | 39 | 62 | K | |||
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Answer | K= A = 51 | |||||||||
Take Home | Increase ,decrease then increase like this . Mixture of two sequences, one is 15,27,39,K and this is at deviation of constant 12. So the K = 39+12=51 And second series is 28,50,62; which we need not to explore. |
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(3) | 97 | 89 | 83 | 79 | 73 | K | ||||
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Answer | K=D=71 | |||||||||
Take Home | Decreasing series, put like this 97. 89, 83, 79, 73, K 97. 89, 83, 79, 73, K ----------------------------------- ,. 8,. 6,. 4,. 6, ----------------------------------- Thus 97. (-8) 89,. (-6), 83, (-4), 79, (-6), 73, [-2] 71= K. |
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(4) | 15 | 16 | 20 | 29 | 45 | K | ||||
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Answer | K=B= 70 | |||||||||
Take Home | Increasing series Arranging like 15, 16, 20, 29, 45, K 15, 16, 20, 29, 45, K ----------------------------------------------- , 1,. 4,. 9,. 16 1^0, 2^2,3^2, 4^2, [5^2] 25 +45 = 70 =. K ---------------------------------------------------- |
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(5) | 4 | 1335 | 2335 | 3064 | 3576 | K | ||||
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Answer | K = A = 3919 | |||||||||
Take Home | Increasing series Arranging like 4,. 1335,. 2335,. 3064,. 3576. K 4,. 1335,. 2335,. 3064,. 3576. K ______________________________________________ ,. 1331,. 1000,. 729,. 512. We get a pattern in difference series. 11^3,. 10^3,. 9^3,. 8^3. [ 7^3] = 343. Add +3576 to get K ________________________________________________ 3919 = K = A. |
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(6) | 10 | 14 | 22 | 38 | 70 | K | ||||
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Answer | K = B= 134 | |||||||||
Take Home | Increasing sequence Let us find difference series 10. 14. 22. 38. 70. K 10. 14. 22. 38. 70. K Diff. 4. 8. 16. 32 Inspection 4x1,. 4x2,. 8x2,. 16x2 [32x2] = 64+70 = 134 = B = K |
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(7) | 3 | 3 | 6 | 18 | 72 | 360 | K | |||
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Answer | K= D= 2160 | |||||||||
Take Home | Increasing series, sudden jump after 72, seems to be the role of multiplication By inspection we see 3. 3. 6. 18. 72. 360. K 3×1. 3x2. 6x3. 18x4. 72x5. [360x6] = 2160 = D = K |
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**(8) | 4 | 8 | 24 | 120 | 840 | K | ||||
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Answer | K = C = 9240 | |||||||||
Take Home | Increasing, sudden jump after 24, role of multiplication The Series 4,. 8,. 24,. 120,. 840,. K Inspection. 8/4,. 24/8,. 120/24,. 840/120,. K/840 2,. 3,. 5,. 7,. [ 11 , the next prime] K/840 = 11 <=> K = 840x11 = 9240 = C |
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(9) | 5 | 10 | 40 | 240 | 1920 | K | ||||
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Answer | K = B = 19200 | |||||||||
Take Home | Increasing series, sudden jump after 40 , role of multiplication The series 5. 10. 40. 240. 1920. K Inspection. 10/5. 40/10. 240/40. 1920/240. K/1920 2. 4. 6. 8. [10 even ] K/1920 = 10 ⇔ K = 19200 = B |
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**(10) | 4 | 5 | 12 | 39 | 160 | K | ||||
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Answer | K = A = 805 | |||||||||
Take Home | Increasing series, jump at 39, role of multiplication, but also successive terms not dividing evenly, so arithmetico geometric series might be. The Series. 4,. 5,. 12,. 39,. 160,. K Inspection. 4x1+1. 5x2+2. 12x3+3. 39x4+4. 160x5+5= 805 = A |
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**(11) | 6 | 12 | 15 | 75 | 82 | K | ||||
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Answer | K= C = 902 | |||||||||
Take Home | Series is increasing, not much jump, pattern not smooth. The series. 6,. 12,. 15,. 75,. 82,. K Inspection. 6. 6x2. 12+3. 15x5. 75+7 Observations Two types of series intermixed 2) either multiplied or added ( alternately ) by a prime number Therefore K = [ Last value] [ x ] [ next prime ] = 82 x 11 = 902 = C. |
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(12) | 8 | 4 | 4 | 6 | 12 | 30 | K | |||
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Answer | K = C= 90 | |||||||||
Take Home | Series not jump, decreasing and increasing mixed type The Series. 8,. 4,. 4,. 6,. 12,. 30,. K Inspection by ratio 4/8. 4/4. 6/4. 12/6. 30/12. K/30 Values. ½ 1. 1+2/4 2. 2+6/12 Or equal to. ½ 1. 1+ ½ 2. 2+½ [3] as there are two mixed series one goes by ½, 1+½ ans 2+½ etc and another goes by 1,2,3 etc So K/30 = 3 ⇔ K= 90 = C. |
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(13) | 6 | 8 | 19 | 61 | 249 | K | ||||
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Answer | K= C = 1251 | |||||||||
Take Home | Series increasing,jumps at 19 , effect of multiplication, division not giving even distribution, let us try representing terms with respect to previous terms The Series 6,. 8,. 19,. 61,. 249,. K Inspection by representing with respect to previous term ,. 6+2. 8x2+3. 19x3+4. 61x4+5. [249x5+6] K = 249x5+6 = 1245+6 = 1251 = C |
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(14) | 289 | 256 | 225 | 196 | 169 | K | ||||
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Answer | K = D= 144 | |||||||||
Take Home | Two answers are there for this series, one is given here, so we took 144 as answer. How 144 is the answer. Must learn by heart squares of some nos. The series 289,. 256,. 225,. 196,. 169,. K Inspection. 17x17. 16x16. 15x15. 14x14. 13x13. 12x12 K = 12x12= 144 = D Another Solution , but the option is not there to support this answer. The series is in decreasing order. The series 289,. 256,. 225,. 196,. 169,. K 289,. 256,. 225,. 196,. 169,. K _________________________________________________________ Subtracting. X. 33. 31. 29. 25. 169 - K 33. 31. 29. 25. 169 - K _________________________________________________________ Inspection. 2. 3. 4 [5] So K is given by Equation : 25 - (169 - K) = 5 ⇔ 25 - 5 -169 = - K ⇔ K = 169 -25 +5 = 169 -20 = 149 Another plausible solution is 149, but it is not given in the option. |
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**(15) | 361 | 529 | 841 | 961 | K | |||||
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Answer | K = D = 1369 | |||||||||
Take Home | The problem is little difficult, one should be able to recognise squares of big numbers. The series is. 361. 529. 841. 961,. K Inspection. 19x19. 23x23. 29x29. 31x31. Square of some no > 31 Then one should be able to recognise that these numbers are successive prime numbers. Hence, K = square of successive prime number after 31 = 37 x 37 = 1369 = D |
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(16) | 9 | 64 | 25 | 216 | 49 | K | ||||
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Take Home | The series is increasing and decreasing mixed type It is based on squares and cubes The series is. 9,. 64,. 25,. 216,. 49,. K Inspection by sq and cubes 3x3. 4x4x4. 5x5. 6x6x6. 7x7. [8x8x8] K = 8x8x8 = 64x8 = 512 |
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(17) | 121 | 265 | 434 | 630 | K | 1111 | ||||
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Answer | K = D = 855 | |||||||||
Take Home | This series is increasing. The series is. 121. 265. 434. 630. K. 1111 Shift series 121. 265. 434. 630. K. 1111 __________________________________________________________________ Subtracting. X. 144. 169. 196. Recognising 12x12. 13x13. 14x14. [15x15] Thus K - 630 = 15x15 ⇔ K = 630 + 225 = 855 = D |
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(18) | 3 | 9 | 27 | 81 | K | |||||
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Answer | K= B = 243 | |||||||||
Take Home | Series is increasing and have some kind of multiplication The series is 3,. 9,. 27,. 81,. K Representing from previous term 3,. 3x3. 9x3. 27x3. [81x3] K = 81x3 = 243 = B |
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(19) | 1 | 2 | 3 | 5 | 8 | K | ||||
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Answer | K = C = 13 | |||||||||
Take Home | It is a fibonacci series, succeeding term is sum of two previous term The series is. 1,. 2,. 3,. 5,. 8,. K Representing from previous two terms 1,. 2,. 1+2,. 2+3,. 3+5,. [5+8] Therefore K = 5+8 = 13 = C |
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(20) | 4 | 5 | 9 | 18 | 34 | K | ||||
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Answer | K=D=59 | |||||||||
Take Home | The series is increasing. Not sudden jump The series is 4,. 5,. 9,. 18,. 34,. K The shift series 4,. 5,. 9,. 18,. 34,. K ___________________________________________________________ By subtracting. 1. 4. 9. 16. K-34 Recognising. 1. 2x2. 3x3. 4x4. [ 5x5 ] Therefore, K -34 = 5x5 ⇔ K = 34+25 = 59 = D. |
Time & Work
Time & Work (1)Karan completes ⅔ of his work in 10 days. What amount of time, he will take to complete ⅗ of the same work ? : |
Options:(A)- 4 (B)- 8 (C)- 6 (D)- 9 |
Answer:Ans: (D) = 9 |
Calculations:⅔ of work is done in 10 days => complete (1) work will be done in 10 / (⅔) or (10*3)/2 = 15 days => ⅗ of work will be done in 15* (⅗) = 9 days. |
Take Home:Unitary method is used. First find out the unit amount of work is done in what time and then find out time for required work |
Time & Work (2)"A" can cultivate ⅖ of a land in 6 days, and "B" can cultivate ⅓ of same land in 10 days. Working together A and B can cultivate ⅘ of the land in how many days. ? : |
Options:(A)- 4 (B)- 5 (C)- 8 (D)- 10 |
Answer:Ans: ( C ) = 8 |
Calculations:A:: ⅖ Of land is cultivated in 6 days, So in one day A will cultivate (⅖)/(6) = 2/30. B:: ⅓ Of land is cultivated in 10 days, so in one day B will cultivate (⅓)/(10) = 1/30 Therefore, A and B combinedly in one day will cultivate (2/30) + (1/30) = 3/30 = 1/10 of land => A & B will complete the work in 10 days. => A and B combinedly will cultivate ⅘ land of work 10*(⅘) = 8 days |
Take Home:Works done combinedly by A and B together in one day are additive. Therefore they are totalled to get further result of ⅘ of work done by both of them together |
Time & Work (3)"A" and "B" can do a work in 12 days, B and C can do in 15 days and C and A can do the work in 20 days. If A,B,C work together, how many days they will take to complete the work ? : |
Options:(A)- 5 (B)- 7+ 5/6 (C)- 10 (D)- 15+ 2/3 |
Answer:Ans: ( C ) = 10 |
Calculations:A&B doing work in 12 days , so in one day they do 1/12 work. B&C doing work in 15 days , so in one day they do 1/15 work. C&A doing work in 20 days , so in one day they do 1/20 work. Therefore, A&B,B&C, and C&A will do work in one day = 1/12 + 1/15 + 1/20 = (5+4+3)/60 = 12/60 = ⅕ => 2(A,B,C) have done in one day the work =⅕ => A,B,C have done in one day the work (½)* (⅕) = 1/10 => A,B,C combinedly will complete work in 10 days. |
Take Home:Works done combinedly by A,B, and C together in one day are additive. If in one day, the part of work done by an identity is a/b, then work will be completed by the identity in b/a days. |
Time & Work (4)A can do a piece of work in 20 days B can do the same work in 30 days. They finished the total work in 8 days with the help of C. If they earned Rs. 5550, then what is share of C ? : |
Options:(A)- 1800 (B)- 1850 (C)- 1900 (D)- 1950 |
Answer:Ans: (B) = 1850 |
Calculations:A can do a piece of work in 20 days => A does in one day 1/20 B can do the same work in 30 days => B does in one day 1/30 Given A+B+C does work in 8 days => A+B+C does in 1 day ⅛ If C completes work in c days then in one day he would be doing 1/c work Therefore, 1/20 + 1/30 + 1/c = ⅛ => 1/c = ⅛ - 1/20 - 1/30 = (15 - 6 - 4) / 120 => 1/c = 5/120 Therefore, in one day A,B,C have done work as 1/20, 1/30, 5/120 The proportion of work done in one day by A,B,C is 6:4:5 Thus C 's share in total money will be [5/(6+4+5)]* 5550 = (5/15)* 5550= 1850 |
Take Home:Actually this payment should be made in proportion to the work done by all the three A,B, and C, but since they have done work from beginning to end, therefore, money is divided in proportion to the daily work done (rate of work done per day). |
Time & Work (5)The ratios of efficiencies of P,Q,R is 2:3:5. The total wages of P,Q,R working for 14, 24, and 20 days respectively are Rs 6000. Find the total wages of all the three, when P works for 9 days, Q works for 14 days, and R works for 8 days ? |
Options:(A)- 3000 (B)- 2860 (C)- 2450 (D)- 3240 |
Answer:Ans: (A) = ₹ 3000 |
Calculations:Given K* ( 14*(2/(2+3+5))+24*(3/(2+3+5))+20*(5/(2+3+5))) = 6000, where K is an unknown constant for efficiency. => K* ((14*2/10) + (24*3/10) + (20*5/10)) = 6000 … (1) => K* ( (28+72+100)/10) =6000 => K *(200/10) = 6000 => K = 300. ….(2) Now we are to find K* ( 9*(2/(10)+ 14*(3/(10))+8*(5/(10))) where K = 300 from (2). = 300* (18/10 + 42/10 + 40/10) = 300* 100/10 = 3000 = option (A) |
Take Home:Payment is being made in terms of efficiency-days. A twist question could be: What is one day payment of P? To find out one day payment of P: We have P,Q and R is paid Rs 6000 for 14 days of work of P. P's efficiency is 2*K/10, where K is given by (1). Therefore, Ps payment for 14 days will be: = 14*300*(2/10) [ the first term of equation in (1) ] =14*60 . That means P's pay for one day will be 14*60/14 = 60. …(3) Similarly Q's pay for 24 days will be 24*300*(3/10)= 24*90. Q's pay for one day will be 24*90/24 = 90. …(4) R's pay for one day will be 20*300*(5/10)*(1/20) = 150 …(5) Required payment 9days P + 14days Q + 8days R = 9*60+14*90+8*150 = 540+1260+1200= 1800+1200= 3000. = option (A) |
Time & Work (6)To do any work C alone takes twice as long as A and B takes together; A takes 3 times as long as B and C takes together, All the three together completes the work in 5 days. How long will each take to complete separately ? |
Options:(A)- 20, 12, 15 (B)- 18, 20, 22 (C)- 15, 16, 18 (D)- 12, 15, 17 |
Answer:Ans: (A) = 20,12,15 |
Calculations:Let us assume A, B, C do daily work as a,b, and c. Then, given is c= (a+b)/2 [ only then C will take twice as long as A and B take together.] Also a= (b+c)/3 , and (a+b+c) = ⅕ [ because A, B, C together takes 5days We have (a+b+c) *1/2 = ⅕*½ = 1/10 => c+c/2 = 1/10 => c= (1/10)*(⅔)= 1/15 => C will do work in 15 days Further (a+b+c) *⅓ = ⅕*⅓ = 1/15 => a/3 + a = 1/15 => a= (1/15)*(3/4)= 1/20 => A will do work in 20 days Further (a+b+c) = ⅕ => (1/20)+b+(1/15) = 1/5 => b= ⅕ - 1/15 - 1/20 => b= 12/60 - 4/60 - 3/60 = 5/60 = 1/12 => B will do work in 12 days Therefore, option (A) is correct. |
Take Home:Per day works are additive |
Time & Work (7)A, B, C all together can do a piece of work in 20 days, in which B takes twice as long as A and C together do the work, and C takes twice as long as A and B together take to do the work. In how many days B alone can do the work ? |
Options:(A)- 40 (B)- 35 (C)- 60 (D)- 45 |
Answer:Ans: (C) = 60 |
Calculations:Let us assume A, B, C do daily work as a,b, and c. Then, given is b = (a+c)/2 [ only then B will take twice as long as A and C take together.] Also c = (a+b)/2 , and (a+b+c) = 1/20 [ because A, B, C together takes 20days ] We have (a+b+c) *1/2 = 1/20*½ = 1/40 => b+b/2 = 1/40 => b= (1/40)*(⅔)= 1/60 => B will do work in 60 days Therefore, option (C) is correct. |
Take Home:Per day works are additive. Secondly B will take twice as long as A and C take together iff B is slower in doing work as compared to A & C together. Thus b = (a + c)/2 , where a, b, and care per day works done by A,B,and C respectively |
Time & Work (8)A, B each working alone can do the work in 12 and 36 days. They started work together, but A left the work after some time, and B finished the remaining work in four days. After how many days from the start did A leave the work ? |
Options:(A)- 10 (B)- 8 (C)- 12 (D)- 3 (E)- None |
Answer:Ans: (B) = 8 |
Calculations:Let us assume A, B do daily work as a,b. Let, after x days from beginning, A has left the work. Then, given is : a= 1/12, b = 1/36 x days together A and B and 4 days only B completes the work. => x*(a+b) + 4*b = 1 => x*(( 1/12)+(1/36)) + 4*(1/36) = 1 => x *((3+1)/36) = 1-(1/9) = 8/9 => x*(4/36) = 8/9. => x = (36/4)*(8/9)= 8 Therefore, option (B) is correct. |
Take Home:Per day works are additive. |
Time & Work (9)A man "A" can do a piece of work in 10 days, man "B" can do the same work in 12 days, while "C" can do the same work in 15 days. They started work together, but after 2 days, A left the work, and remaining work was completed by "B" and "C". Find out in how many days the work will be completed.? |
Options:(A)- 5 days (B)- 7 days (C)- 2 ⅙ days (D)- 3 ⅓ days (E)- None |
Answer:Ans: (D) = 3 ⅓ |
Calculations:Let us assume A, B, and C do daily work as a,b, and c respectively. Given that after 2 days from beginning, A has left the work. Then, given is : a= 1/10, b = 1/12, and c = 1/15 2 days together A, B, and C and say x days the B, and C do the work to complete it. => 2*(a+b+c) + x*(b + c) = 1 => 2*(( 1/10)+(1/12)+(1/15)) + x*((1/12)+(1/15)) = 1 => 2 *((6+5+4)/60) + x*((5+4)/60)= 1 => 2*15/60 + x*9/60 = 1 => x = (60/9)* ( 1-(30/60)) = (60/9)*(½) = 30/9 = 10/3 = 3 ⅓ Therefore, option (D) is correct. |
Take Home:Per day works are additive. |
Time & Work (10)Sita can do a work in 12 days working 4 hours daily.Gita can finish the same in 15 days working 3 hours daily. In how many days they can finish this work working 4.5 hours daily? |
Options:(A)- 5 5/31 days (B)- 6 days (C)- 6 4/31days (D)- 5 3/31 days (E)- None |
Answer:Ans: (A) = 5 5/31 |
Calculations:Let us assume Sita and Gita do per hour work as s, and g respectively. Then, given is : s = 1/(12*4)=1/48, g = 1/(15*3) = 1/45 Let x days are needed to complete the work by Sita and Gita working each day 4.5 hours. Then, (x*4.5)*(s+g) = 1 => x*(9/2)*((1/48)+(1/45)) = 1 => x*(9/2)*((15+16)/720) = 1 => x*(9/2)*(31/720) = 1 => x = (2/9)*(720/31) = 2*80/31 = 160/ 31= 5+5/31 = 5 5/31 = option (A) |
Take Home:Per day works are additive. |
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