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- Question 1 of 35
1. Question
Quantitative AptitudeWhat approximate value should come in the place of question mark (?) in the following question?
428.88 / 12.88 [425.25 + 99.99 / 49.99 (54.88 – 33.33) ] = ?
CorrectWe have to find the approximate value So 428.88/12.88[425.25 + 99.99/49.99(54.88 – 33.33)] = ?
can be written
⇒ 429/13[425 + 100/50(55 – 33)] = ?
Now using BODMAS rule ⇒ 429/13[425 + 100/50(22)] = ? ⇒ 33 × [425 + 2(22)] = ?
⇒ 33 × [425 + 44] = ?
⇒ 33 × [469] = ?
⇒ ? = 15477 IncorrectWe have to find the approximate value So 428.88/12.88[425.25 + 99.99/49.99(54.88 – 33.33)] = ?
can be written
⇒ 429/13[425 + 100/50(55 – 33)] = ?
Now using BODMAS rule ⇒ 429/13[425 + 100/50(22)] = ? ⇒ 33 × [425 + 2(22)] = ?
⇒ 33 × [425 + 44] = ?
⇒ 33 × [469] = ?
⇒ ? = 15477 UnattemptedWe have to find the approximate value So 428.88/12.88[425.25 + 99.99/49.99(54.88 – 33.33)] = ?
can be written
⇒ 429/13[425 + 100/50(55 – 33)] = ?
Now using BODMAS rule ⇒ 429/13[425 + 100/50(22)] = ? ⇒ 33 × [425 + 2(22)] = ?
⇒ 33 × [425 + 44] = ?
⇒ 33 × [469] = ?
⇒ ? = 15477 - Question 2 of 35
2. Question
Quantitative AptitudeWhat approximate value should come in the place of question mark (?) in the following question?
845.89 – 1078.88 + 23.33 × 15.22 + ? = 576.88
CorrectWe have to find the approximate value So 845.89 – 1078.88 + 23.33 × 15.22 + ? = 576.88
can be written
⇒ 846 – 1079 + 23 × 15 + ? = 577
Now using BODMAS rule ⇒ 846 – 1079 + 345 + ? = 577 ⇒ 1191 – 1079 + ? = 577
⇒ 112 + ? = 577
⇒ ? = 577 – 112
⇒ ? = 465 IncorrectWe have to find the approximate value So 845.89 – 1078.88 + 23.33 × 15.22 + ? = 576.88
can be written
⇒ 846 – 1079 + 23 × 15 + ? = 577
Now using BODMAS rule ⇒ 846 – 1079 + 345 + ? = 577 ⇒ 1191 – 1079 + ? = 577
⇒ 112 + ? = 577
⇒ ? = 577 – 112
⇒ ? = 465 UnattemptedWe have to find the approximate value So 845.89 – 1078.88 + 23.33 × 15.22 + ? = 576.88
can be written
⇒ 846 – 1079 + 23 × 15 + ? = 577
Now using BODMAS rule ⇒ 846 – 1079 + 345 + ? = 577 ⇒ 1191 – 1079 + ? = 577
⇒ 112 + ? = 577
⇒ ? = 577 – 112
⇒ ? = 465 - Question 3 of 35
3. Question
Quantitative AptitudeWhat approximate value should come in the place of question mark (?) in the following question?
974.99 / 194.88 + 1403.99 / 77.77 – 33.33 × 42.22 / 11.11 = ? – 976.44
CorrectWe have to find the approximate value So 974.99/194.88 + 1403.99/77.77 – 33.33 × 42.22/11.11 = ? – 976.44
can be written
⇒ 975/195 + 1404/78 – 33 × 42/11 = ? – 976
Now using BODMAS rule ⇒ 5 + 18 – 3 × 42 = ? – 976 ⇒ -103 = ? – 976
⇒ ? = 976 – 103
⇒ ? = 873 IncorrectWe have to find the approximate value So 974.99/194.88 + 1403.99/77.77 – 33.33 × 42.22/11.11 = ? – 976.44
can be written
⇒ 975/195 + 1404/78 – 33 × 42/11 = ? – 976
Now using BODMAS rule ⇒ 5 + 18 – 3 × 42 = ? – 976 ⇒ -103 = ? – 976
⇒ ? = 976 – 103
⇒ ? = 873 UnattemptedWe have to find the approximate value So 974.99/194.88 + 1403.99/77.77 – 33.33 × 42.22/11.11 = ? – 976.44
can be written
⇒ 975/195 + 1404/78 – 33 × 42/11 = ? – 976
Now using BODMAS rule ⇒ 5 + 18 – 3 × 42 = ? – 976 ⇒ -103 = ? – 976
⇒ ? = 976 – 103
⇒ ? = 873 - Question 4 of 35
4. Question
Quantitative AptitudeWhat approximate value should come in the place of question mark (?) in the following question?
8.88 / 3.33 [1296.062/4 + {146411/4 – (16.44 – 12.44)}] = ?
CorrectWe have to find the approximate value So, 8.88/3.33[1296.062/4 + {146411/4 – (16.44 – 12.44)}] = ?
can be written
⇒ 9/3[12962/4 + {146411/4 – (16 – 12)}] = ?
Now using BODMAS rule ⇒ 3[12962/4 + {146411/4 – (16 – 12)}] = ? ⇒ 3[(64)2/4 + {(114)1/4 – (16 – 12)}] = ?
⇒ 3[(6)2+ {11 – 4}] = ?
⇒ 3 × [36 + 7] = ?
⇒ 3 × [43] = ?
⇒ ? = 129 IncorrectWe have to find the approximate value So, 8.88/3.33[1296.062/4 + {146411/4 – (16.44 – 12.44)}] = ?
can be written
⇒ 9/3[12962/4 + {146411/4 – (16 – 12)}] = ?
Now using BODMAS rule ⇒ 3[12962/4 + {146411/4 – (16 – 12)}] = ? ⇒ 3[(64)2/4 + {(114)1/4 – (16 – 12)}] = ?
⇒ 3[(6)2+ {11 – 4}] = ?
⇒ 3 × [36 + 7] = ?
⇒ 3 × [43] = ?
⇒ ? = 129 UnattemptedWe have to find the approximate value So, 8.88/3.33[1296.062/4 + {146411/4 – (16.44 – 12.44)}] = ?
can be written
⇒ 9/3[12962/4 + {146411/4 – (16 – 12)}] = ?
Now using BODMAS rule ⇒ 3[12962/4 + {146411/4 – (16 – 12)}] = ? ⇒ 3[(64)2/4 + {(114)1/4 – (16 – 12)}] = ?
⇒ 3[(6)2+ {11 – 4}] = ?
⇒ 3 × [36 + 7] = ?
⇒ 3 × [43] = ?
⇒ ? = 129 - Question 5 of 35
5. Question
Quantitative AptitudeWhat approximate value should come in the place of question mark (?) in the following question?
2196.98 2/3 + 9260.991/3 – 65535.993/4 – ? = 4096.065/6 + 243.333/5
CorrectThe correct answer is option 4 i.e. – 4957
Understanding Application We have to find the approximate value So, 2196.98 2/3 + 9260.991/3 – 65535.993/4 – ? = 4096.065/6 + 243.333/5
can be written
⇒ 2197 2/3 + 92611/3 – 655363/4 – ? = 40965/6 + 2433/5
Now using BODMAS rule ⇒ (133)2/3 + (213)1/3 – (164)3/4 – ? = (46)5/6 + (35)3/5 ⇒ (13)2 + (21) – (16)3 – ? = (4)5 + (3)3
⇒ 169 + (21) – 4096 – ? = 1024 + 27
⇒ 190 – 4096 – ? = 1051
⇒ – 3906 – ? = 1051
Hence ⇒ – 4957 = ? IncorrectThe correct answer is option 4 i.e. – 4957
Understanding Application We have to find the approximate value So, 2196.98 2/3 + 9260.991/3 – 65535.993/4 – ? = 4096.065/6 + 243.333/5
can be written
⇒ 2197 2/3 + 92611/3 – 655363/4 – ? = 40965/6 + 2433/5
Now using BODMAS rule ⇒ (133)2/3 + (213)1/3 – (164)3/4 – ? = (46)5/6 + (35)3/5 ⇒ (13)2 + (21) – (16)3 – ? = (4)5 + (3)3
⇒ 169 + (21) – 4096 – ? = 1024 + 27
⇒ 190 – 4096 – ? = 1051
⇒ – 3906 – ? = 1051
Hence ⇒ – 4957 = ? UnattemptedThe correct answer is option 4 i.e. – 4957
Understanding Application We have to find the approximate value So, 2196.98 2/3 + 9260.991/3 – 65535.993/4 – ? = 4096.065/6 + 243.333/5
can be written
⇒ 2197 2/3 + 92611/3 – 655363/4 – ? = 40965/6 + 2433/5
Now using BODMAS rule ⇒ (133)2/3 + (213)1/3 – (164)3/4 – ? = (46)5/6 + (35)3/5 ⇒ (13)2 + (21) – (16)3 – ? = (4)5 + (3)3
⇒ 169 + (21) – 4096 – ? = 1024 + 27
⇒ 190 – 4096 – ? = 1051
⇒ – 3906 – ? = 1051
Hence ⇒ – 4957 = ? - Question 6 of 35
6. Question
Quantitative AptitudeWhat approximate value should come in the place of question mark (?) in the following question?
740.99 of 5.05 / 38.99 + 3135.98 ÷ 55.99 – 58.08 = ? of 3.33 / 4.44
CorrectWe have to find the approximate value So 740.99 of 5.05/38.99 + 3135.98 ÷ 55.99 – 58.08 = ? of 3.33/4.44
can be written
⇒ 741 of 5/39 + 3136 ÷ 56 – 58 = ? of 3/4
Now using BODMAS rule ⇒ 741 × 5/39 + 3136 ÷ 56 – 58 = ? × 3/4 ⇒ 19 × 5 + 56 – 58 = ? × 3/4
⇒ 95 + 56 – 58 = ? × 3/4
⇒ 151 – 58 = ? × 3/4
⇒ 93 = ? × 3/4
⇒ 93/3 = ?/4
⇒ 31 = ?/4
⇒ ? = 124 IncorrectWe have to find the approximate value So 740.99 of 5.05/38.99 + 3135.98 ÷ 55.99 – 58.08 = ? of 3.33/4.44
can be written
⇒ 741 of 5/39 + 3136 ÷ 56 – 58 = ? of 3/4
Now using BODMAS rule ⇒ 741 × 5/39 + 3136 ÷ 56 – 58 = ? × 3/4 ⇒ 19 × 5 + 56 – 58 = ? × 3/4
⇒ 95 + 56 – 58 = ? × 3/4
⇒ 151 – 58 = ? × 3/4
⇒ 93 = ? × 3/4
⇒ 93/3 = ?/4
⇒ 31 = ?/4
⇒ ? = 124 UnattemptedWe have to find the approximate value So 740.99 of 5.05/38.99 + 3135.98 ÷ 55.99 – 58.08 = ? of 3.33/4.44
can be written
⇒ 741 of 5/39 + 3136 ÷ 56 – 58 = ? of 3/4
Now using BODMAS rule ⇒ 741 × 5/39 + 3136 ÷ 56 – 58 = ? × 3/4 ⇒ 19 × 5 + 56 – 58 = ? × 3/4
⇒ 95 + 56 – 58 = ? × 3/4
⇒ 151 – 58 = ? × 3/4
⇒ 93 = ? × 3/4
⇒ 93/3 = ?/4
⇒ 31 = ?/4
⇒ ? = 124 - Question 7 of 35
7. Question
Quantitative AptitudeWhat approximate value should come in the place of question mark (?) in the following question?
(17)3 of 5.05 / 288.99 – 6560.98 ÷ 27.07 / 6.99 + 5929.09 of 0.99 / 10.88 = ?
CorrectWe have to find the approximate value So, (17)3 of 5.05/288.99 – 6560.98 ÷ 27.07/6.99 + 5929.09 of 0.99/10.88 = ?
can be written
⇒ (17)3 of 5/289 – 6561 ÷ 27/7 + 5929 of 1/11 = ?
Now using BODMAS rule ⇒ (17)3 × 5/289 – 6561 ÷ 27/7 + 5929 × 1/11 = ? ⇒ 4913 × 5/289 – 6561 × 7/27 + 5929 × 1/11 = ?
⇒ 17 × 5 – 243 × 7 + 539 = ?
⇒ 85 – 1701 + 539 = ?
⇒ 624 – 1701 = ?
⇒ ? = 1077 IncorrectWe have to find the approximate value So, (17)3 of 5.05/288.99 – 6560.98 ÷ 27.07/6.99 + 5929.09 of 0.99/10.88 = ?
can be written
⇒ (17)3 of 5/289 – 6561 ÷ 27/7 + 5929 of 1/11 = ?
Now using BODMAS rule ⇒ (17)3 × 5/289 – 6561 ÷ 27/7 + 5929 × 1/11 = ? ⇒ 4913 × 5/289 – 6561 × 7/27 + 5929 × 1/11 = ?
⇒ 17 × 5 – 243 × 7 + 539 = ?
⇒ 85 – 1701 + 539 = ?
⇒ 624 – 1701 = ?
⇒ ? = 1077 UnattemptedWe have to find the approximate value So, (17)3 of 5.05/288.99 – 6560.98 ÷ 27.07/6.99 + 5929.09 of 0.99/10.88 = ?
can be written
⇒ (17)3 of 5/289 – 6561 ÷ 27/7 + 5929 of 1/11 = ?
Now using BODMAS rule ⇒ (17)3 × 5/289 – 6561 ÷ 27/7 + 5929 × 1/11 = ? ⇒ 4913 × 5/289 – 6561 × 7/27 + 5929 × 1/11 = ?
⇒ 17 × 5 – 243 × 7 + 539 = ?
⇒ 85 – 1701 + 539 = ?
⇒ 624 – 1701 = ?
⇒ ? = 1077 - Question 8 of 35
8. Question
Quantitative AptitudeWhat approximate value should come in the place of question mark (?) in the following question?
1045.89 – 1078.08 of 1 / 1.99 + 23.33 of 15.22 + ? = 976.88
CorrectWe have to find the approximate value So 1045.89 – 1078.08 of 1/1.99 + 23.33 of 15.22 + ? = 976.88
can be written
⇒ 1046 – 1078 of 1/2 + 23 of 15 + ? = 977
Now using BODMAS rule ⇒ 1046 – 1078 × 1/2 + 23 × 15 + ? = 977 ⇒ 1046 – 539 + 345 + ? = 977
⇒ 1391 – 539 + ? = 977
⇒ 852 + ? = 977
⇒ ? = 977 – 852
⇒ ? = 125 IncorrectWe have to find the approximate value So 1045.89 – 1078.08 of 1/1.99 + 23.33 of 15.22 + ? = 976.88
can be written
⇒ 1046 – 1078 of 1/2 + 23 of 15 + ? = 977
Now using BODMAS rule ⇒ 1046 – 1078 × 1/2 + 23 × 15 + ? = 977 ⇒ 1046 – 539 + 345 + ? = 977
⇒ 1391 – 539 + ? = 977
⇒ 852 + ? = 977
⇒ ? = 977 – 852
⇒ ? = 125 UnattemptedWe have to find the approximate value So 1045.89 – 1078.08 of 1/1.99 + 23.33 of 15.22 + ? = 976.88
can be written
⇒ 1046 – 1078 of 1/2 + 23 of 15 + ? = 977
Now using BODMAS rule ⇒ 1046 – 1078 × 1/2 + 23 × 15 + ? = 977 ⇒ 1046 – 539 + 345 + ? = 977
⇒ 1391 – 539 + ? = 977
⇒ 852 + ? = 977
⇒ ? = 977 – 852
⇒ ? = 125 - Question 9 of 35
9. Question
Quantitative AptitudeWhat approximate value should come in the place of question mark (?) in the following question?
12.88 [414.44 + 199.99 / 49.99 {(54.88 – 33.33) – 24.44}] = ?
CorrectWe have to find the approximate value So 12.88[414.44 + 199.99/49.99{(54.88 – 33.33) – 24.44}] = ?
can be written
⇒ 13[414 + 200/50{(55 – 33) – 24}] = ?
Now using BODMAS rule ⇒ 13[414 + 4{(22) – 24}] = ? ⇒ 13[414 + 4{22 – 24}] = ?
⇒ 13[414 + 4{-2}] = ?
⇒ 13[414 – 8] = ?
⇒ 13 × 406 = ?
⇒ ? = 5278 IncorrectWe have to find the approximate value So 12.88[414.44 + 199.99/49.99{(54.88 – 33.33) – 24.44}] = ?
can be written
⇒ 13[414 + 200/50{(55 – 33) – 24}] = ?
Now using BODMAS rule ⇒ 13[414 + 4{(22) – 24}] = ? ⇒ 13[414 + 4{22 – 24}] = ?
⇒ 13[414 + 4{-2}] = ?
⇒ 13[414 – 8] = ?
⇒ 13 × 406 = ?
⇒ ? = 5278 UnattemptedWe have to find the approximate value So 12.88[414.44 + 199.99/49.99{(54.88 – 33.33) – 24.44}] = ?
can be written
⇒ 13[414 + 200/50{(55 – 33) – 24}] = ?
Now using BODMAS rule ⇒ 13[414 + 4{(22) – 24}] = ? ⇒ 13[414 + 4{22 – 24}] = ?
⇒ 13[414 + 4{-2}] = ?
⇒ 13[414 – 8] = ?
⇒ 13 × 406 = ?
⇒ ? = 5278 - Question 10 of 35
10. Question
Quantitative AptitudeWhat approximate value should come in the place of question mark (?) in the following question?
3√2743.99 + 5√7776.06 – 4√4095.89 = ? + 6√4095.89
CorrectWe have to find the approximate value So 3√2743.99 + 5√7776.06 – 4√4095.89 = ? + 6√4095.89
can be written
⇒ 3√2744 + 5√7776 – 4√4096 = ? + 6√4096
Now using BODMAS rule ⇒ (143)1/3 + (65)1/5 – (84)1/4 = ? + (46)1/6 ⇒ 14 + 6 – 8 = ? + 4
⇒ 20 – 8 = ? + 4
⇒ 12 = ? + 4
⇒ ? = 8 IncorrectWe have to find the approximate value So 3√2743.99 + 5√7776.06 – 4√4095.89 = ? + 6√4095.89
can be written
⇒ 3√2744 + 5√7776 – 4√4096 = ? + 6√4096
Now using BODMAS rule ⇒ (143)1/3 + (65)1/5 – (84)1/4 = ? + (46)1/6 ⇒ 14 + 6 – 8 = ? + 4
⇒ 20 – 8 = ? + 4
⇒ 12 = ? + 4
⇒ ? = 8 UnattemptedWe have to find the approximate value So 3√2743.99 + 5√7776.06 – 4√4095.89 = ? + 6√4095.89
can be written
⇒ 3√2744 + 5√7776 – 4√4096 = ? + 6√4096
Now using BODMAS rule ⇒ (143)1/3 + (65)1/5 – (84)1/4 = ? + (46)1/6 ⇒ 14 + 6 – 8 = ? + 4
⇒ 20 – 8 = ? + 4
⇒ 12 = ? + 4
⇒ ? = 8 - Question 11 of 35
11. Question
Quantitative AptitudeThe presence of men, women and children in the annual meeting at Bhawnathpur panchayat are given below
Find the average presence of men, women and children together in march and december month
CorrectTotal presence in march and december. March: 17 + 12 + 12 = 41 December: 28 + 23 + 16 = 67
Total = 41 + 67 = 108
Average 108/3 = 36 IncorrectTotal presence in march and december. March: 17 + 12 + 12 = 41 December: 28 + 23 + 16 = 67
Total = 41 + 67 = 108
Average 108/3 = 36 UnattemptedTotal presence in march and december. March: 17 + 12 + 12 = 41 December: 28 + 23 + 16 = 67
Total = 41 + 67 = 108
Average 108/3 = 36 - Question 12 of 35
12. Question
Quantitative AptitudeIf A is 25% of B and B is 30% of C. Then C is what percent of sum of 80% of (A + B)?
CorrectGiven A = 25% of B B = 30% of C
Now ⇒ A = 25/100 B ⇒ A/B = 1/4 …….(i)
⇒ B = 30/100 C
⇒ B/C = 3/10 ………(ii)
Balancing the equation ⇒ A/B = 1/4 × 3/3 ⇒ B/C = 3/10 × 4/4
So A = 3 B = 12
C = 40
Now ⇒ 80% of (A + B) = 15/100 × 80 = 12
⇒ 40/100 × 12 = 4.8% IncorrectGiven A = 25% of B B = 30% of C
Now ⇒ A = 25/100 B ⇒ A/B = 1/4 …….(i)
⇒ B = 30/100 C
⇒ B/C = 3/10 ………(ii)
Balancing the equation ⇒ A/B = 1/4 × 3/3 ⇒ B/C = 3/10 × 4/4
So A = 3 B = 12
C = 40
Now ⇒ 80% of (A + B) = 15/100 × 80 = 12
⇒ 40/100 × 12 = 4.8% UnattemptedGiven A = 25% of B B = 30% of C
Now ⇒ A = 25/100 B ⇒ A/B = 1/4 …….(i)
⇒ B = 30/100 C
⇒ B/C = 3/10 ………(ii)
Balancing the equation ⇒ A/B = 1/4 × 3/3 ⇒ B/C = 3/10 × 4/4
So A = 3 B = 12
C = 40
Now ⇒ 80% of (A + B) = 15/100 × 80 = 12
⇒ 40/100 × 12 = 4.8% - Question 13 of 35
13. Question
Quantitative AptitudeVishal marked 30% above the price of his bicycle and gives a 10% discount to the customer. Find what percent he has profit or loss at the last.
CorrectTricks during successive discount ± Profit/Loss = ±1st% ± 2nd% ± (1st% × 2nd%)/100 So ⇒ +30 – 10 + (30 × -10)/100 ⇒ +20 – (300/100)
⇒ +20 – 3
⇒ +17
17% profit IncorrectTricks during successive discount ± Profit/Loss = ±1st% ± 2nd% ± (1st% × 2nd%)/100 So ⇒ +30 – 10 + (30 × -10)/100 ⇒ +20 – (300/100)
⇒ +20 – 3
⇒ +17
17% profit UnattemptedTricks during successive discount ± Profit/Loss = ±1st% ± 2nd% ± (1st% × 2nd%)/100 So ⇒ +30 – 10 + (30 × -10)/100 ⇒ +20 – (300/100)
⇒ +20 – 3
⇒ +17
17% profit - Question 14 of 35
14. Question
Quantitative AptitudeA man has started work alone, he can complete the work in 15 days but he left the work after 10 days. After it two boys complete the remaining work in 5 days. If the capacity of the man and two boys is 1 : 1 then find how much man earned. Construction charge of this work is 75000 rupees.
CorrectMan can complete in 15 days But he worked only 10 days If we take capacity of men is x unit per day Then total work = 15x Man 15x = 75000 rupees
Now ⇒ 10x = (75000/15x) × 10x = 50000
IncorrectMan can complete in 15 days But he worked only 10 days If we take capacity of men is x unit per day Then total work = 15x Man 15x = 75000 rupees
Now ⇒ 10x = (75000/15x) × 10x = 50000
UnattemptedMan can complete in 15 days But he worked only 10 days If we take capacity of men is x unit per day Then total work = 15x Man 15x = 75000 rupees
Now ⇒ 10x = (75000/15x) × 10x = 50000
- Question 15 of 35
15. Question
Quantitative AptitudeIn a building, a cat and a dog live together. They are playing together the speed of the cat is 5 m/s and the speed of the dog is 2 m/s. If both stand at two corners distance between them is 200 m. If the first 14 sec they run toward each other direction, after it the only dog runs the opposite direction. So find the time when both are met.
CorrectThe actual speed at first 14 sec ⇒ 5 + 2 = 7 m/s So
Distance travelled in 14 sec = 14 × 7 = 98 m
Remaining distance ⇒ 200 – 98 = 102 m Actual speed after 14 sec ⇒ 5 – 2 = 3 m/s Hence ⇒ time = 102/3 = 34 sec IncorrectThe actual speed at first 14 sec ⇒ 5 + 2 = 7 m/s So
Distance travelled in 14 sec = 14 × 7 = 98 m
Remaining distance ⇒ 200 – 98 = 102 m Actual speed after 14 sec ⇒ 5 – 2 = 3 m/s Hence ⇒ time = 102/3 = 34 sec UnattemptedThe actual speed at first 14 sec ⇒ 5 + 2 = 7 m/s So
Distance travelled in 14 sec = 14 × 7 = 98 m
Remaining distance ⇒ 200 – 98 = 102 m Actual speed after 14 sec ⇒ 5 – 2 = 3 m/s Hence ⇒ time = 102/3 = 34 sec - Question 16 of 35
16. Question
Quantitative AptitudeA sum of Rs. 50000 is given to the customer at a rate of 15% for 2 years, compounded yearly. calculate the amount which is paid by the customer.
CorrectTricks Now Total CI = 7500 + 7500 + 1125 = 16125 Hence Amount = P + CI ⇒ A = 50000 + 16125
= 66125
IncorrectTricks Now Total CI = 7500 + 7500 + 1125 = 16125 Hence Amount = P + CI ⇒ A = 50000 + 16125
= 66125
UnattemptedTricks Now Total CI = 7500 + 7500 + 1125 = 16125 Hence Amount = P + CI ⇒ A = 50000 + 16125
= 66125
- Question 17 of 35
17. Question
Quantitative AptitudeFind the fourth proportion.
4 : 13 : : 64 : ?CorrectProportion a/b = c/d ⇒ 4/13 = 64/? ⇒ ? = (13 × 64)/4
⇒ ? = 208
IncorrectProportion a/b = c/d ⇒ 4/13 = 64/? ⇒ ? = (13 × 64)/4
⇒ ? = 208
UnattemptedProportion a/b = c/d ⇒ 4/13 = 64/? ⇒ ? = (13 × 64)/4
⇒ ? = 208
- Question 18 of 35
18. Question
Quantitative AptitudeThree friends Chintu, Mintu and Pintu, start a business with the capital Rs. 65000, Rs. 70000 and Rs. 85000 respectively. Mintu has left the business after 5 months. If the profit is Rs. 64500 at the end of the year then find the profit of Mintu.
CorrectTricks Chintu : Mintu : Pintu (65000 × 12) : (70000 × 5) : (85000 × 12)
⇒ (65 × 12) : (70 × 5) : (85 × 12)
⇒ 780 : 350 : 1020
⇒ 78 : 35 : 102
Now ⇒ 215 unit = 64500 So
35 unit = (64500/215) × 35
= 10500
Mintu profit = 10500 IncorrectTricks Chintu : Mintu : Pintu (65000 × 12) : (70000 × 5) : (85000 × 12)
⇒ (65 × 12) : (70 × 5) : (85 × 12)
⇒ 780 : 350 : 1020
⇒ 78 : 35 : 102
Now ⇒ 215 unit = 64500 So
35 unit = (64500/215) × 35
= 10500
Mintu profit = 10500 UnattemptedTricks Chintu : Mintu : Pintu (65000 × 12) : (70000 × 5) : (85000 × 12)
⇒ (65 × 12) : (70 × 5) : (85 × 12)
⇒ 780 : 350 : 1020
⇒ 78 : 35 : 102
Now ⇒ 215 unit = 64500 So
35 unit = (64500/215) × 35
= 10500
Mintu profit = 10500 - Question 19 of 35
19. Question
Quantitative AptitudeIn a park, there are different slots for parking. two-wheeler and four-wheeler vans are separated in the park. If count only the number of the vans then the total number of vans is 90 and if the count number of wheels then the total number of wheels is 224. Find the total number of two-wheeler van.
CorrectLet two-wheeler van = X and four-wheeler van = Y
X + Y = 90 ………(i) and
2X + 4Y = 224 …….(ii)
Now eq(i) × 4 – eq(ii)
4X + 4Y = 360 2X + 4Y = 224
⇒ 2X = 136
⇒ X = 68
Two-wheeler van = 68 IncorrectLet two-wheeler van = X and four-wheeler van = Y
X + Y = 90 ………(i) and
2X + 4Y = 224 …….(ii)
Now eq(i) × 4 – eq(ii)
4X + 4Y = 360 2X + 4Y = 224
⇒ 2X = 136
⇒ X = 68
Two-wheeler van = 68 UnattemptedLet two-wheeler van = X and four-wheeler van = Y
X + Y = 90 ………(i) and
2X + 4Y = 224 …….(ii)
Now eq(i) × 4 – eq(ii)
4X + 4Y = 360 2X + 4Y = 224
⇒ 2X = 136
⇒ X = 68
Two-wheeler van = 68 - Question 20 of 35
20. Question
Quantitative AptitudeFind the total number of 10 letter words, with or without meaning, which can be formed out of the letters of the word EXCEPTIONAL, where the repetition of the letters is not allowed.
CorrectHere Given letter EXCEPTIONAL Total letter having 10 types When makes 10 digit number Because repetition is not allowed then
the first place of this number can be 10 types
2nd place of this number can be 9 types
so far..
total number 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3628800 IncorrectHere Given letter EXCEPTIONAL Total letter having 10 types When makes 10 digit number Because repetition is not allowed then
the first place of this number can be 10 types
2nd place of this number can be 9 types
so far..
total number 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3628800 UnattemptedHere Given letter EXCEPTIONAL Total letter having 10 types When makes 10 digit number Because repetition is not allowed then
the first place of this number can be 10 types
2nd place of this number can be 9 types
so far..
total number 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3628800 - Question 21 of 35
21. Question
Quantitative AptitudeFind the missing number in the following series.
95, 91, 82, ?, 41, 5
CorrectThe series follows the pattern – 22, – 32, – 42, – 52, – 62 95 – 22 = 91
91 – 32 = 82
82 – 42 = 66
66 – 52 = 41
41 – 62 = 5
Hence, the missing term is 66. IncorrectThe series follows the pattern – 22, – 32, – 42, – 52, – 62 95 – 22 = 91
91 – 32 = 82
82 – 42 = 66
66 – 52 = 41
41 – 62 = 5
Hence, the missing term is 66. UnattemptedThe series follows the pattern – 22, – 32, – 42, – 52, – 62 95 – 22 = 91
91 – 32 = 82
82 – 42 = 66
66 – 52 = 41
41 – 62 = 5
Hence, the missing term is 66. - Question 22 of 35
22. Question
Quantitative AptitudeFind the missing number in the following series.
1, 1, 1/2, 3/2, ?, 15/8
CorrectThe series follows the pattern x 1, / 2, x 3, / 4, x 5, /6 1 x 1 = 1
1/2 = 1/2
(1/2) x 3 = 3/2
(3/2)/4 = 3/8
(3/8) x 5 = 15/8
Hence, the missing term is 3/8. IncorrectThe series follows the pattern x 1, / 2, x 3, / 4, x 5, /6 1 x 1 = 1
1/2 = 1/2
(1/2) x 3 = 3/2
(3/2)/4 = 3/8
(3/8) x 5 = 15/8
Hence, the missing term is 3/8. UnattemptedThe series follows the pattern x 1, / 2, x 3, / 4, x 5, /6 1 x 1 = 1
1/2 = 1/2
(1/2) x 3 = 3/2
(3/2)/4 = 3/8
(3/8) x 5 = 15/8
Hence, the missing term is 3/8. - Question 23 of 35
23. Question
Quantitative AptitudeFind the missing number in the following series.
– 10, – 11, ?, 2, 40, 159
CorrectThe series follows the pattern – 10, -11, – 9, 2, 40, 159
Difference between terms – 1, 2, 11, 38, 119
Difference between terms 3, 9, 27, 81
Hence, the missing term is – 9.
IncorrectThe series follows the pattern – 10, -11, – 9, 2, 40, 159
Difference between terms – 1, 2, 11, 38, 119
Difference between terms 3, 9, 27, 81
Hence, the missing term is – 9.
UnattemptedThe series follows the pattern – 10, -11, – 9, 2, 40, 159
Difference between terms – 1, 2, 11, 38, 119
Difference between terms 3, 9, 27, 81
Hence, the missing term is – 9.
- Question 24 of 35
24. Question
Quantitative AptitudeFind the missing number in the following series.
– 2, – 3, – 8, ?, – 272, – 2991
CorrectThe series follows the pattern x 2 + 1 , x 3 + 1, x 5 + 1, x 7 + 1, x 11 + 1
– 2 x 2 + 1 = – 3
– 3 x 3 + 1= – 8
– 8 x 5 + 1 = – 39
– 39 x 7 + 1 = – 272
– 272 x 11 + 1 = – 2991
Hence, the missing term is -39
IncorrectThe series follows the pattern x 2 + 1 , x 3 + 1, x 5 + 1, x 7 + 1, x 11 + 1
– 2 x 2 + 1 = – 3
– 3 x 3 + 1= – 8
– 8 x 5 + 1 = – 39
– 39 x 7 + 1 = – 272
– 272 x 11 + 1 = – 2991
Hence, the missing term is -39
UnattemptedThe series follows the pattern x 2 + 1 , x 3 + 1, x 5 + 1, x 7 + 1, x 11 + 1
– 2 x 2 + 1 = – 3
– 3 x 3 + 1= – 8
– 8 x 5 + 1 = – 39
– 39 x 7 + 1 = – 272
– 272 x 11 + 1 = – 2991
Hence, the missing term is -39
- Question 25 of 35
25. Question
Quantitative AptitudeFind the missing number in the following series.
63, 24, ?, 20/3, 47/9, 128/27
CorrectThe series follows the pattern x (1/3) + 3, x (1/3) + 3, x (1/3) + 3
63 x (1/3) + 3 = 24
24 x (1/3) + 3 = 11
11 x (1/3) + 3 = 20/3
20/3 x (1/3) + 3 = 47/9
47/9 x (1/3) + 3 = 128/27
Hence, the missing term is 11
IncorrectThe series follows the pattern x (1/3) + 3, x (1/3) + 3, x (1/3) + 3
63 x (1/3) + 3 = 24
24 x (1/3) + 3 = 11
11 x (1/3) + 3 = 20/3
20/3 x (1/3) + 3 = 47/9
47/9 x (1/3) + 3 = 128/27
Hence, the missing term is 11
UnattemptedThe series follows the pattern x (1/3) + 3, x (1/3) + 3, x (1/3) + 3
63 x (1/3) + 3 = 24
24 x (1/3) + 3 = 11
11 x (1/3) + 3 = 20/3
20/3 x (1/3) + 3 = 47/9
47/9 x (1/3) + 3 = 128/27
Hence, the missing term is 11
- Question 26 of 35
26. Question
Quantitative AptitudeQuantity I : Value of ( x + y )3 + xy + ( x + y)(4/3) when x = 3 and y = 5
Quantity II : Value of ( a + b )2 + a2 – b2 when a = 4 and b = 3CorrectQuantity I Quantity II ( x + y )3 + xy + ( x + y)(4/3) ( a + b )2 + a2 – b2 x = 3 and y = 5 a = 4, b = 3 83 + 15 + 84/3 72 + 42 – 32 543 56 IncorrectQuantity I Quantity II ( x + y )3 + xy + ( x + y)(4/3) ( a + b )2 + a2 – b2 x = 3 and y = 5 a = 4, b = 3 83 + 15 + 84/3 72 + 42 – 32 543 56 UnattemptedQuantity I Quantity II ( x + y )3 + xy + ( x + y)(4/3) ( a + b )2 + a2 – b2 x = 3 and y = 5 a = 4, b = 3 83 + 15 + 84/3 72 + 42 – 32 543 56 - Question 27 of 35
27. Question
Quantitative AptitudeBrass consists of copper and zinc. In 700gm of brass the ratio of copper to zinc is 5 : 2.
Quantity I : The amount of copper added to make the ratio 11 : 4
Quantity II : The amount of copper added to make the ratio 13 : 4CorrectAmount of copper in brass when ratio is 5 : 2 = 700 x (5/7) = 500gm
Let the amount of copper added in first case be x and second case be y
Quantity I Quantity II (500 + x)/200 = 11/4 (500 + y)/ 200 = 13/4 2000 + 4x = 2200 2000 + 4y = 2600 x = 50 y = 150 Therefore, y > x
IncorrectAmount of copper in brass when ratio is 5 : 2 = 700 x (5/7) = 500gm
Let the amount of copper added in first case be x and second case be y
Quantity I Quantity II (500 + x)/200 = 11/4 (500 + y)/ 200 = 13/4 2000 + 4x = 2200 2000 + 4y = 2600 x = 50 y = 150 Therefore, y > x
UnattemptedAmount of copper in brass when ratio is 5 : 2 = 700 x (5/7) = 500gm
Let the amount of copper added in first case be x and second case be y
Quantity I Quantity II (500 + x)/200 = 11/4 (500 + y)/ 200 = 13/4 2000 + 4x = 2200 2000 + 4y = 2600 x = 50 y = 150 Therefore, y > x
- Question 28 of 35
28. Question
Quantitative AptitudeA shop has two offers.
Quantity I : The percent discount on a offer of buy 2 get 1 free.
Quantity II : The percent discount on a offer of buy 4 get 2 free.CorrectLet the cost of each unit be x
Quantity I Quantity II Cost without offer = 3x Cost without offer = 6x Discounted cost = 2x Discounted cost = 4x Discount = 3x – 2x = x Discount = 6x – 4x = 2x Discount% = (x/3x)100 Discount% = (2x/6x)100 = 33.33% = 33.33% Both the values are same.
IncorrectLet the cost of each unit be x
Quantity I Quantity II Cost without offer = 3x Cost without offer = 6x Discounted cost = 2x Discounted cost = 4x Discount = 3x – 2x = x Discount = 6x – 4x = 2x Discount% = (x/3x)100 Discount% = (2x/6x)100 = 33.33% = 33.33% Both the values are same.
UnattemptedLet the cost of each unit be x
Quantity I Quantity II Cost without offer = 3x Cost without offer = 6x Discounted cost = 2x Discounted cost = 4x Discount = 3x – 2x = x Discount = 6x – 4x = 2x Discount% = (x/3x)100 Discount% = (2x/6x)100 = 33.33% = 33.33% Both the values are same.
- Question 29 of 35
29. Question
Quantitative AptitudeQuantity I: The chance of a person winning if he rolls 2 dice and gets a sum of 5
Quantity II: The chance of a person winning if the second dice gets a number greater than the first dice.CorrectThe total number of outcomes when two dice are rolled = 6 x 6 = 36
Favourable outcomes for first case = {1, 4}, {2, 3}, {3, 2}, {4, 1}
Favourable outcomes for second case = {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {3, 4}, {3, 5}, {3, 6}, {4, 5}, {4, 6}, {5, 6}
Quantity I Quantity II Favourable outcomes = 4 Favourable outcomes = 15 Chance of winning = 4/36 Chance of winning = 15/36 = 1/9 = 5/12 Quantity II > Quantity I
IncorrectThe total number of outcomes when two dice are rolled = 6 x 6 = 36
Favourable outcomes for first case = {1, 4}, {2, 3}, {3, 2}, {4, 1}
Favourable outcomes for second case = {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {3, 4}, {3, 5}, {3, 6}, {4, 5}, {4, 6}, {5, 6}
Quantity I Quantity II Favourable outcomes = 4 Favourable outcomes = 15 Chance of winning = 4/36 Chance of winning = 15/36 = 1/9 = 5/12 Quantity II > Quantity I
UnattemptedThe total number of outcomes when two dice are rolled = 6 x 6 = 36
Favourable outcomes for first case = {1, 4}, {2, 3}, {3, 2}, {4, 1}
Favourable outcomes for second case = {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {3, 4}, {3, 5}, {3, 6}, {4, 5}, {4, 6}, {5, 6}
Quantity I Quantity II Favourable outcomes = 4 Favourable outcomes = 15 Chance of winning = 4/36 Chance of winning = 15/36 = 1/9 = 5/12 Quantity II > Quantity I
- Question 30 of 35
30. Question
Quantitative AptitudeQuantity I: The average speed when a person covers 3/8 of the total distance with a speed of 3km/hr, 1/4 of the total distance with a speed of 4km/hr and the rest with 2km/hr.
Quantity II: The average speed when a person covers 1/3 of the total distance with a speed of 4km/hr, 1/3 of the total distance with 3km/hr and the rest with 6km/hr
CorrectLet the total distance be x
Avg speed = total distance/total time taken
t = d/s
Quantity I Quantity II Av1 = x /[(3x/8 x 3) + (x/4 x 4) + (3x/8 x 2)] Av2 = x/[(x/3 x 4) + (x/3 x 3) + (x/3 x 6)] Av1 = x/[(3x/24) + (x/16) + (3x/16)] Av2 = x/[(x/12) + (x/9) + (x/18)] Av1 = 4/[(3/6) + (1/4) + (3/4)] Av2 = 3/[(1/4) + (1/3) + (1/6)] Av1 = 8/3 Av2 = 4 Av2 > Av1
IncorrectLet the total distance be x
Avg speed = total distance/total time taken
t = d/s
Quantity I Quantity II Av1 = x /[(3x/8 x 3) + (x/4 x 4) + (3x/8 x 2)] Av2 = x/[(x/3 x 4) + (x/3 x 3) + (x/3 x 6)] Av1 = x/[(3x/24) + (x/16) + (3x/16)] Av2 = x/[(x/12) + (x/9) + (x/18)] Av1 = 4/[(3/6) + (1/4) + (3/4)] Av2 = 3/[(1/4) + (1/3) + (1/6)] Av1 = 8/3 Av2 = 4 Av2 > Av1
UnattemptedLet the total distance be x
Avg speed = total distance/total time taken
t = d/s
Quantity I Quantity II Av1 = x /[(3x/8 x 3) + (x/4 x 4) + (3x/8 x 2)] Av2 = x/[(x/3 x 4) + (x/3 x 3) + (x/3 x 6)] Av1 = x/[(3x/24) + (x/16) + (3x/16)] Av2 = x/[(x/12) + (x/9) + (x/18)] Av1 = 4/[(3/6) + (1/4) + (3/4)] Av2 = 3/[(1/4) + (1/3) + (1/6)] Av1 = 8/3 Av2 = 4 Av2 > Av1
- Question 31 of 35
31. Question
Quantitative AptitudeDirection: The line graph below shows the number of hours in which 5 persons A, B, C, D and E complete a piece of work. Study the graph carefully and answer the questions that follow.
Find the difference of time taken by A and B together to complete the work and the time taken by C and E together to complete the work.
CorrectFrom the line graph: A and B can complete the work in 18 and 12 hours respectively.
Suppose total work = 108 units (LCM of 18 and 12)
So,
Efficiency of A = 108/18 = 6
Efficiency of B = 108/12 = 9
So, time taken by A and B together to complete the work = 108/(6 + 9) = 108/15 = 7.2 hours
From the line graph:
C and E can complete the work in 10 and 15 hours respectively.
Suppose total work = 30 units (LCM of 10 and 15)
So, Efficiency of C = 30/10 = 3
Efficiency of E = 30/15 = 2
So, time taken by C and E together to complete the work = 30/(3 + 2) = 6 hours
Hence, required difference = 7.2 – 6 = 1.2 hours
IncorrectFrom the line graph: A and B can complete the work in 18 and 12 hours respectively.
Suppose total work = 108 units (LCM of 18 and 12)
So,
Efficiency of A = 108/18 = 6
Efficiency of B = 108/12 = 9
So, time taken by A and B together to complete the work = 108/(6 + 9) = 108/15 = 7.2 hours
From the line graph:
C and E can complete the work in 10 and 15 hours respectively.
Suppose total work = 30 units (LCM of 10 and 15)
So, Efficiency of C = 30/10 = 3
Efficiency of E = 30/15 = 2
So, time taken by C and E together to complete the work = 30/(3 + 2) = 6 hours
Hence, required difference = 7.2 – 6 = 1.2 hours
UnattemptedFrom the line graph: A and B can complete the work in 18 and 12 hours respectively.
Suppose total work = 108 units (LCM of 18 and 12)
So,
Efficiency of A = 108/18 = 6
Efficiency of B = 108/12 = 9
So, time taken by A and B together to complete the work = 108/(6 + 9) = 108/15 = 7.2 hours
From the line graph:
C and E can complete the work in 10 and 15 hours respectively.
Suppose total work = 30 units (LCM of 10 and 15)
So, Efficiency of C = 30/10 = 3
Efficiency of E = 30/15 = 2
So, time taken by C and E together to complete the work = 30/(3 + 2) = 6 hours
Hence, required difference = 7.2 – 6 = 1.2 hours
- Question 32 of 35
32. Question
Quantitative AptitudeDirection: The line graph below shows the number of hours in which 5 persons A, B, C, D and E complete a piece of work. Study the graph carefully and answer the questions that follow.
What is the time taken by B, D and E to complete the work together?
CorrectFrom the line graph: B, D and E can complete the work in 12, 24 and 15 hours respectively.
Suppose total work = 120 units (LCM of 12, 24 and 15)
So, Efficiency of B = 120/12 = 10
Efficiency of D = 120/24 = 5
Efficiency of E = 120/15 = 8
So, time taken by B, D and E together to complete the work = 120/(10 + 5 + 8) = 120/23 hours
IncorrectFrom the line graph: B, D and E can complete the work in 12, 24 and 15 hours respectively.
Suppose total work = 120 units (LCM of 12, 24 and 15)
So, Efficiency of B = 120/12 = 10
Efficiency of D = 120/24 = 5
Efficiency of E = 120/15 = 8
So, time taken by B, D and E together to complete the work = 120/(10 + 5 + 8) = 120/23 hours
UnattemptedFrom the line graph: B, D and E can complete the work in 12, 24 and 15 hours respectively.
Suppose total work = 120 units (LCM of 12, 24 and 15)
So, Efficiency of B = 120/12 = 10
Efficiency of D = 120/24 = 5
Efficiency of E = 120/15 = 8
So, time taken by B, D and E together to complete the work = 120/(10 + 5 + 8) = 120/23 hours
- Question 33 of 35
33. Question
Quantitative AptitudeDirection: The line graph below shows the number of hours in which 5 persons A, B, C, D and E complete a piece of work. Study the graph carefully and answer the questions that follow.
A and C started the work working alternatively (One hours each) starting with C. After how much time, work will be finished?
CorrectFrom the line graph: A and C can complete the work in 18 and 10 hours respectively.
Suppose total work = 90 units (LCM of 18 and 10)
So,
Efficiency of A = 90/18 = 5
Efficiency of C = 90/10 = 9
Hence, Work done in first 2 hours = 9 + 5 = 14 units
So, work done in 12 hours (= 2 × 6) = 14 × 6 = 84 units
Now, remaining 6 units will be done by C in 6/9 = 2/3 hours
Hence, total time = 12 + 2/3 = 38/3 hours
IncorrectFrom the line graph: A and C can complete the work in 18 and 10 hours respectively.
Suppose total work = 90 units (LCM of 18 and 10)
So,
Efficiency of A = 90/18 = 5
Efficiency of C = 90/10 = 9
Hence, Work done in first 2 hours = 9 + 5 = 14 units
So, work done in 12 hours (= 2 × 6) = 14 × 6 = 84 units
Now, remaining 6 units will be done by C in 6/9 = 2/3 hours
Hence, total time = 12 + 2/3 = 38/3 hours
UnattemptedFrom the line graph: A and C can complete the work in 18 and 10 hours respectively.
Suppose total work = 90 units (LCM of 18 and 10)
So,
Efficiency of A = 90/18 = 5
Efficiency of C = 90/10 = 9
Hence, Work done in first 2 hours = 9 + 5 = 14 units
So, work done in 12 hours (= 2 × 6) = 14 × 6 = 84 units
Now, remaining 6 units will be done by C in 6/9 = 2/3 hours
Hence, total time = 12 + 2/3 = 38/3 hours
- Question 34 of 35
34. Question
Quantitative AptitudeDirection: The line graph below shows the number of hours in which 5 persons A, B, C, D and E complete a piece of work. Study the graph carefully and answer the questions that follow.
B started the work alone and left after 4 hours, who can complete the remaining work in 10 hours?
CorrectSuppose total work = 360 units (LCM of 18, 12, 10, 24, 15) So,
Efficiency of A = 360/18 = 20
Efficiency of B = 360/12 = 30
Efficiency of C = 360/10 = 36
Efficiency of D = 360/24 = 15
Efficiency of E = 360/15 = 24
B worked for 4 hours, so work done by B = 30 × 4 = 120 units
So, Remaining work = 360 – 120 = 240 units
240 units will be done in 10 hours by E only as his efficiency is 10.
IncorrectSuppose total work = 360 units (LCM of 18, 12, 10, 24, 15) So,
Efficiency of A = 360/18 = 20
Efficiency of B = 360/12 = 30
Efficiency of C = 360/10 = 36
Efficiency of D = 360/24 = 15
Efficiency of E = 360/15 = 24
B worked for 4 hours, so work done by B = 30 × 4 = 120 units
So, Remaining work = 360 – 120 = 240 units
240 units will be done in 10 hours by E only as his efficiency is 10.
UnattemptedSuppose total work = 360 units (LCM of 18, 12, 10, 24, 15) So,
Efficiency of A = 360/18 = 20
Efficiency of B = 360/12 = 30
Efficiency of C = 360/10 = 36
Efficiency of D = 360/24 = 15
Efficiency of E = 360/15 = 24
B worked for 4 hours, so work done by B = 30 × 4 = 120 units
So, Remaining work = 360 – 120 = 240 units
240 units will be done in 10 hours by E only as his efficiency is 10.
- Question 35 of 35
35. Question
Quantitative AptitudeDirection: The line graph below shows the number of hours in which 5 persons A, B, C, D and E complete a piece of work. Study the graph carefully and answer the questions that follow.
C and D together started the work but C is working with 1/3 of his efficiency and D is working with full efficiency. In how much time the work will be finished?
CorrectFrom the line graph: C and D can complete the work in 10 and 24 hours respectively.
Suppose total work = 120 units (LCM of 10 and 24)
So, Efficiency of C = 120/10 = 12
Efficiency of D = 120/24 = 5
Since, C is working with 1/3 of his efficiency and D is working with full efficiency
So, time taken by C and D together to complete the work = 120/(4 + 5) = 40/3 hours
IncorrectFrom the line graph: C and D can complete the work in 10 and 24 hours respectively.
Suppose total work = 120 units (LCM of 10 and 24)
So, Efficiency of C = 120/10 = 12
Efficiency of D = 120/24 = 5
Since, C is working with 1/3 of his efficiency and D is working with full efficiency
So, time taken by C and D together to complete the work = 120/(4 + 5) = 40/3 hours
UnattemptedFrom the line graph: C and D can complete the work in 10 and 24 hours respectively.
Suppose total work = 120 units (LCM of 10 and 24)
So, Efficiency of C = 120/10 = 12
Efficiency of D = 120/24 = 5
Since, C is working with 1/3 of his efficiency and D is working with full efficiency
So, time taken by C and D together to complete the work = 120/(4 + 5) = 40/3 hours