IBPS CLERK Quantitative MATHS 01 Free Mock Test

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Question 1 of 35

1. Question

Quantitative Aptitude

What 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) ] = ?

14480
16720
15477
81924
11980

Correct

We have to find the approximate value

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

Incorrect

We have to find the approximate value

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

Unattempted

We have to find the approximate value

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 Aptitude

What approximate value should come in the place of question mark (?) in the following question?

845.89 – 1078.88 + 23.33 × 15.22 + ? = 576.88

Correct

We have to find the approximate value

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

Incorrect

We have to find the approximate value

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

Unattempted

We have to find the approximate value

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 Aptitude

What 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

1080
932
663
890
873

Correct

We have to find the approximate value

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

Incorrect

We have to find the approximate value

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

Unattempted

We have to find the approximate value

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 Aptitude

What approximate value should come in the place of question mark (?) in the following question?

8.88 / 3.33 [1296.06^2/4+ {14641^1/4 – (16.44 – 12.44)}] = ?

Correct

We have to find the approximate value

So,

8.88/3.33[1296.06^2/4+ {14641^1/4– (16.44 – 12.44)}] = ?

can be written

⇒ 9/3[1296^2/4+ {14641^1/4 – (16 – 12)}] = ?

Now using BODMAS rule

⇒ 3[1296^2/4+ {14641^1/4 – (16 – 12)}] = ?

⇒ 3[(6⁴)^2/4+ {(11⁴)^1/4 – (16 – 12)}] = ?

⇒ 3[(6)²+ {11 – 4}] = ?

⇒ 3 × [36 + 7] = ?

⇒ 3 × [43] = ?

⇒ ? = 129

Incorrect

We have to find the approximate value

So,

8.88/3.33[1296.06^2/4+ {14641^1/4– (16.44 – 12.44)}] = ?

can be written

⇒ 9/3[1296^2/4+ {14641^1/4 – (16 – 12)}] = ?

Now using BODMAS rule

⇒ 3[1296^2/4+ {14641^1/4 – (16 – 12)}] = ?

⇒ 3[(6⁴)^2/4+ {(11⁴)^1/4 – (16 – 12)}] = ?

⇒ 3[(6)²+ {11 – 4}] = ?

⇒ 3 × [36 + 7] = ?

⇒ 3 × [43] = ?

⇒ ? = 129

Unattempted

We have to find the approximate value

So,

8.88/3.33[1296.06^2/4+ {14641^1/4– (16.44 – 12.44)}] = ?

can be written

⇒ 9/3[1296^2/4+ {14641^1/4 – (16 – 12)}] = ?

Now using BODMAS rule

⇒ 3[1296^2/4+ {14641^1/4 – (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 Aptitude

What approximate value should come in the place of question mark (?) in the following question?

2196.98 ^2/3+ 9260.99^1/3 – 65535.99^3/4 – ? = 4096.06^5/6+ 243.33^3/5

4957
6375
5042
- 4957
- 5042

Correct

The correct answer is option 4 i.e. – 4957

Understanding	Application
We have to find the approximate value	So, 2196.98^2/3 + 9260.99^1/3– 65535.99^3/4– ? = 4096.06^5/6+ 243.33^3/5 can be written ⇒ 2197^2/3+ 9261^1/3– 65536^3/4– ? = 4096^5/6+ 243^3/5
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 = ?

Incorrect

The correct answer is option 4 i.e. – 4957

Understanding	Application
We have to find the approximate value	So, 2196.98^2/3 + 9260.99^1/3– 65535.99^3/4– ? = 4096.06^5/6+ 243.33^3/5 can be written ⇒ 2197^2/3+ 9261^1/3– 65536^3/4– ? = 4096^5/6+ 243^3/5
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 = ?

Unattempted

The correct answer is option 4 i.e. – 4957

Understanding	Application
We have to find the approximate value	So, 2196.98^2/3 + 9260.99^1/3– 65535.99^3/4– ? = 4096.06^5/6+ 243.33^3/5 can be written ⇒ 2197^2/3+ 9261^1/3– 65536^3/4– ? = 4096^5/6+ 243^3/5
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 Aptitude

What 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

Correct

We have to find the approximate value

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

Incorrect

We have to find the approximate value

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

Unattempted

We have to find the approximate value

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 Aptitude

What 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 = ?

1077
1050
999
1175
1277

Correct

We have to find the approximate value

So,

(17)³of 5.05/288.99 – 6560.98 ÷ 27.07/6.99 + 5929.09 of 0.99/10.88 = ?

can be written

⇒ (17)³of 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

Incorrect

We have to find the approximate value

So,

(17)³of 5.05/288.99 – 6560.98 ÷ 27.07/6.99 + 5929.09 of 0.99/10.88 = ?

can be written

⇒ (17)³of 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

Unattempted

We have to find the approximate value

So,

(17)³of 5.05/288.99 – 6560.98 ÷ 27.07/6.99 + 5929.09 of 0.99/10.88 = ?

can be written

⇒ (17)³of 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 Aptitude

What 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

Correct

We have to find the approximate value

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

Incorrect

We have to find the approximate value

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

Unattempted

We have to find the approximate value

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 Aptitude

What 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}] = ?

5478
5608
5278
5076
6091

Correct

We have to find the approximate value

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

Incorrect

We have to find the approximate value

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

Unattempted

We have to find the approximate value

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 Aptitude

What 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

Correct

We have to find the approximate value

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

Incorrect

We have to find the approximate value

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

Unattempted

We have to find the approximate value

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 Aptitude
The 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
- 36
- 24
- 38
- 40
- 32
Correct
Total presence in march and december. March: 17 + 12 + 12 = 41
December: 28 + 23 + 16 = 67

Total = 41 + 67 = 108
Average 108/3 = 36
Incorrect
Total presence in march and december. March: 17 + 12 + 12 = 41
December: 28 + 23 + 16 = 67

Total = 41 + 67 = 108
Average 108/3 = 36
Unattempted
Total 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 Aptitude

If A is 25% of B and B is 30% of C. Then C is what percent of sum of 80% of (A + B)?

10%
5.5%
4.8%
6.8%
3.5%

Correct

Given	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%

Incorrect

Given	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%

Unattempted

Given	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 Aptitude
Vishal 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.
- 17% loss
- 20% profit
- 17% profit
- 20% loss
- No profit No loss
Correct
Tricks during successive discount ± Profit/Loss = ±1st% ± 2nd% ± (1st% × 2nd%)/100
So ⇒ +30 – 10 + (30 × -10)/100
⇒ +20 – (300/100)
⇒ +20 – 3
⇒ +17
17% profit
Incorrect
Tricks during successive discount ± Profit/Loss = ±1st% ± 2nd% ± (1st% × 2nd%)/100
So ⇒ +30 – 10 + (30 × -10)/100
⇒ +20 – (300/100)
⇒ +20 – 3
⇒ +17
17% profit
Unattempted
Tricks 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 Aptitude
A 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.
- Rs. 35000
- Rs. 40000
- Rs. 45000
- Rs. 50000
- Rs. 55000
Correct
Man 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
Incorrect
Man 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
Unattempted
Man 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 Aptitude

In 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.

34 sec
40 sec
20 sec
30 sec
44 sec

Correct

The 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

Incorrect

The 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

Unattempted

The 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 Aptitude

A 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.

Rs. 60625
Rs. 55125
Rs. 60125
Rs. 66125
Rs. 50125

Correct

Tricks

Now

Total CI = 7500 + 7500 + 1125 = 16125

Hence

Amount = P + CI

⇒ A = 50000 + 16125

= 66125

Incorrect

Tricks

Now

Total CI = 7500 + 7500 + 1125 = 16125

Hence

Amount = P + CI

⇒ A = 50000 + 16125

= 66125

Unattempted

Tricks

Now

Total CI = 7500 + 7500 + 1125 = 16125

Hence

Amount = P + CI

⇒ A = 50000 + 16125

= 66125

Question 17 of 35
17. Question
Quantitative Aptitude
Find the fourth proportion.
4 : 13 : : 64 : ?
- 104
- 250
- 310
- 405
- 208
Correct
Proportion a/b = c/d
⇒ 4/13 = 64/?
⇒ ? = (13 × 64)/4
⇒ ? = 208

Incorrect
Proportion a/b = c/d
⇒ 4/13 = 64/?
⇒ ? = (13 × 64)/4
⇒ ? = 208

Unattempted
Proportion a/b = c/d
⇒ 4/13 = 64/?
⇒ ? = (13 × 64)/4
⇒ ? = 208

Question 18 of 35

18. Question

Quantitative Aptitude

Three 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.

17000
14000
16500
10500
15000

Correct

Tricks

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

35 unit = (64500/215) × 35

= 10500

Mintu profit = 10500

Incorrect

Tricks

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

35 unit = (64500/215) × 35

= 10500

Mintu profit = 10500

Unattempted

Tricks

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

35 unit = (64500/215) × 35

= 10500

Mintu profit = 10500

Question 19 of 35
19. Question
Quantitative Aptitude
In 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.
- 68
- 70
- 22
- 20
- 80
Correct
Let 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
Incorrect
Let 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
Unattempted
Let 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 Aptitude

Find 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.

3628800
3528800
1535800
2516820
3568410

Correct

Here 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

Incorrect

Here 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

Unattempted

Here 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 Aptitude
Find the missing number in the following series.
95, 91, 82, ?, 41, 5
- 66
- 72
- 73
- 69
- None of these
Correct
The 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.
Incorrect
The 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.
Unattempted
The 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 Aptitude
Find the missing number in the following series.
1, 1, 1/2, 3/2, ?, 15/8
- 2
- 3/8
- 5/8
- 3
- None of these
Correct
The 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.
Incorrect
The 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.
Unattempted
The 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 Aptitude

Find the missing number in the following series.

– 10, – 11, ?, 2, 40, 159

1
- 9
- 3
- 7
None of these

Correct

The 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.

Incorrect

The 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.

Unattempted

The 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 Aptitude
Find the missing number in the following series.
– 2, – 3, – 8, ?, – 272, – 2991
- - 30
- - 30
- - 39
- - 40
- None of these
Correct
The 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
Incorrect
The 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
Unattempted
The 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 Aptitude
Find the missing number in the following series.
63, 24, ?, 20/3, 47/9, 128/27
- 12
- 6
- 9
- 11
- None of these
Correct
The 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
20/3x (1/3) + 3 = 47/9
47/9 x (1/3) + 3 = 128/27
Hence, the missing term is 11
Incorrect
The 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
20/3x (1/3) + 3 = 47/9
47/9 x (1/3) + 3 = 128/27
Hence, the missing term is 11
Unattempted
The 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
20/3x (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 Aptitude
Quantity 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 = 3
- Quantity I > Quantity II
- Quantity I > Quantity II
- Quantity I < Quantity II
- Quantity II > Quantity I
- Quantity II > Quantity I
Correct
Quantity 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 + 8^4/3 7² + 4² – 3²
543 56
Incorrect
Quantity 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 + 8^4/3 7² + 4² – 3²
543 56
Unattempted
Quantity 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 + 8^4/3 7² + 4² – 3²
543 56
Question 27 of 35
27. Question
Quantitative Aptitude
Brass 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 : 4
- Quantity I > Quantity II
- Quantity I > Quantity II
- Quantity I < Quantity II
- Quantity II > Quantity I
- Quantity I = Quantity II or relation can't be established
Correct
Amount 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
Incorrect
Amount 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
Unattempted
Amount 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 Aptitude

A 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.

Quantity I > Quantity II
Quantity I > Quantity II
Quantity I < Quantity II
Quantity II > Quantity I
Quantity I = Quantity II or relation can't be established

Correct

Let 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.

Incorrect

Let 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.

Unattempted

Let 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 Aptitude
Quantity 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.
- Quantity I > Quantity II
- Quantity I > Quantity II
- Quantity I < Quantity II
- Quantity II > Quantity I
- Quantity I = Quantity II or relation can't be established
Correct
The 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
Incorrect
The 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
Unattempted
The 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 Aptitude

Quantity 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

Quantity I > Quantity II
Quantity I > Quantity II
Quantity I < Quantity II
Quantity II > Quantity I
Quantity I = Quantity II or relation can't be established

Correct

Let 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

Incorrect

Let 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

Unattempted

Let 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 Aptitude

Direction: 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.

1.5 hours
1.2 hours
1 hours
1.8 hours
2 hours

Correct

From 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

Incorrect

From 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

Unattempted

From 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 Aptitude

Direction: 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?

90/13 hours
140/27 hours
120/23 hours
108/19 hours
None of these

Correct

From 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

Incorrect

From 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

Unattempted

From 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 Aptitude

Direction: 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?

13 hours
41/3 hours
35/3 hours
39/4 hours
38/3 hours

Correct

From 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

Incorrect

From 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

Unattempted

From 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 Aptitude

Direction: 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?

E
A
C
D
None of these

Correct

Suppose 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.

Incorrect

Suppose 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.

Unattempted

Suppose 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 Aptitude

Direction: 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?

13 hours
41/3 hours
40/3 hours
14 hours
38/3 hours

Correct

From 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

Incorrect

From 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

Unattempted

From 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

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IBPS CLERK Quantitative MATHS 01 Free Mock Test

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IBPS CLERK : Quantitative Maths

Quantity 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 + 8^4/3	7² + 4² – 3²
543	56

Quantity I	Quantity II
(500 + x)/200 = 11/4	(500 + y)/ 200 = 13/4
2000 + 4x = 2200	2000 + 4y = 2600
x = 50	y = 150

Quantity I	Quantity II
Favourable outcomes = 4	Favourable outcomes = 15
Chance of winning = 4/36	Chance of winning = 15/36
= 1/9	= 5/12