IBPS RRB Office Assistant Quantitative Aptitude Free Online Mock Test - 1
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Quantitative Aptitude Questions for IBPS RRB Office Assistant Online Test, IBPS RRB Free Mock Test Series. IBPS Free Mock Test for Office Assistant. IBPS RRB is conducted every year for selection to the post of both IBPS RRB Assistant and IBPS RRB Officer Cadre in Regional Rural Banks spread across the country. IBPS Quantitative Aptitude Online Test in English. IBPS RRB Quantitative Aptitude Quiz 2024. The IBPS Full online mock test paper is free for all students. IBPS Question and Answers in Hindi and English. IBPS Mock test for Quantitative Aptitude Subject Via Online Mode. Here we are providing Quantitative Aptitude Question for IBPS RRB Exam in English. IBPS RRB Office Assistant Mock Test Series 2024. Now Test your self for IBPS Exam by using below quiz…
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Question 1 of 40
1. Question
Madhan was 3 times as old as his daughter 10 years ago. After 10 years, Madhan will be twice as old as his daughter. Find out the present age of Madhan.
Correct
Let age of the daughter before 10 years = x
Then, age of Madhan before 10 years = 3x
After 10 years, Madhan will be twice as old as his daughter.
? 3x + 20 = 2( x+20)
? x = 20
? Present age of Madhan = 3x + 10
= 3× 20+ 10
= 70 years.
Incorrect
Let age of the daughter before 10 years = x
Then, age of Madhan before 10 years = 3x
After 10 years, Madhan will be twice as old as his daughter.
? 3x + 20 = 2( x+20)
? x = 20
? Present age of Madhan = 3x + 10
= 3× 20+ 10
= 70 years.
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Question 2 of 40
2. Question
P walks around a circular field at the rate of two rounds per hour while Q runs around it at the rate of eight rounds per hour. They start in the opposite direction from the same point at 6.45 a.m. They shall first cross each other at?
Correct
Since P and Q move in the opposite direction along the circular field, so they will first meet each other when there is a difference of one round between the two.
Relative speed of P and Q = 8+2 = 10 rounds per hour.
Time taken to complete one round at this speed = 1/10 hr = 6 min.
6 Hrs 45 min + 6 min = 6 hrs 51 min i.e, 6.51 a.m
Incorrect
Since P and Q move in the opposite direction along the circular field, so they will first meet each other when there is a difference of one round between the two.
Relative speed of P and Q = 8+2 = 10 rounds per hour.
Time taken to complete one round at this speed = 1/10 hr = 6 min.
6 Hrs 45 min + 6 min = 6 hrs 51 min i.e, 6.51 a.m
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Question 3 of 40
3. Question
Three pipes P, Q and R can fill a tank from empty to full in 40 minutes, 15 minutes, and 30 minutes respectively. When the tank is empty, all the three pipes are opened. P, Q and R discharge chemical solutions X, Y and Z respectively. What is the proportion of the solution Y in the liquid in the tank after 5 minutes?
Correct
Part of the tank filled by pipe P in 1 minute = 1 / 40
Part of the tank filled by pipe Q in 1 minute = 1 / 15
Part of the tank filled by pipe R in 1 minute = 1/ 30
Here we have to find the proportion of the solution Y.
Pipe Q discharges chemical solution Y.
Part of the tank filled by pipe Q in 5 minutes
= 5 × 1 / 15 = 1 / 3
Part of the tank filled by pipe P, Q, R together in 1 minute
= 1/40 + 1/15 + 1/30
= (3+8+4)/120 = 15 / 120 = 1 / 8.
Part of the tank filled by pipe P, Q, R together in 5 minute
= 5 × 1/8 = 5/ 8
Required proportion = (1/3) / ( 5/8) = 8 / 15
Incorrect
Part of the tank filled by pipe P in 1 minute = 1 / 40
Part of the tank filled by pipe Q in 1 minute = 1 / 15
Part of the tank filled by pipe R in 1 minute = 1/ 30
Here we have to find the proportion of the solution Y.
Pipe Q discharges chemical solution Y.
Part of the tank filled by pipe Q in 5 minutes
= 5 × 1 / 15 = 1 / 3
Part of the tank filled by pipe P, Q, R together in 1 minute
= 1/40 + 1/15 + 1/30
= (3+8+4)/120 = 15 / 120 = 1 / 8.
Part of the tank filled by pipe P, Q, R together in 5 minute
= 5 × 1/8 = 5/ 8
Required proportion = (1/3) / ( 5/8) = 8 / 15
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Question 4 of 40
4. Question
A boat running downstream covers a distance of 20 km in 5 hours while for covering the same distance upstream, it takes 10 hours. What is the speed of the stream?
Correct
Rate of downstream=(20 / 5 ) kmph= 4kmph
Rate of upstream =( 20/10) kmph= 2kmph
Therefore Speed of the stream= (1/2)(4 – 2) kmph= 1 kmph
Incorrect
Rate of downstream=(20 / 5 ) kmph= 4kmph
Rate of upstream =( 20/10) kmph= 2kmph
Therefore Speed of the stream= (1/2)(4 – 2) kmph= 1 kmph
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Question 5 of 40
5. Question
Two stations X and Y are 170 km apart on a straight line. One train starts from X at 6 a.m. and travels towards Y at 25 kmph. Another train starts from Y at 8 a.m. and travels towards X at a speed of 35 kmph. At what time will they meet?
Correct
Suppose they meet z hours after 6 a.m.
Distance covered by X in z hours = 25× z km.
Distance covered by Y in (z – 2) hours = 35(z – 2) km.
Therefore 25z + 35(z- 2) = 170
60z = 240
Z = 4.
So, they meet at 10 a.m.
Incorrect
Suppose they meet z hours after 6 a.m.
Distance covered by X in z hours = 25× z km.
Distance covered by Y in (z – 2) hours = 35(z – 2) km.
Therefore 25z + 35(z- 2) = 170
60z = 240
Z = 4.
So, they meet at 10 a.m.
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Question 6 of 40
6. Question
Simple interest on a certain sum of money for 4 years at 5% per annum is half the compound interest on Rs. 8000 for 2 years at 10% per annum. The sum placed on simple interest is:
Correct
C.I = 8000 * (1+(10/100))2 –8000
= 8000 *110/100 * 110/100 – 8000 = 9680 – 8000 = Rs.1680
Therefore, S.I = C.I / 2 = 1680 / 2=Rs. 840
840 = sum * 4 * 5 / 100
:. Sum = Rs.(840 * 100)/ (4 *5) = Rs.4200
Incorrect
C.I = 8000 * (1+(10/100))2 –8000
= 8000 *110/100 * 110/100 – 8000 = 9680 – 8000 = Rs.1680
Therefore, S.I = C.I / 2 = 1680 / 2=Rs. 840
840 = sum * 4 * 5 / 100
:. Sum = Rs.(840 * 100)/ (4 *5) = Rs.4200
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Question 7 of 40
7. Question
On selling 25 balls at Rs. 1015, there is a profit equal to the cost price of 10 balls. What is the cost price of a ball?
Correct
Profit = (S.P of 25 balls) – (C.P of 25 balls) = 1015 – (C.P of 25 balls)
Given that Profit = (C.P of 10 balls)
=> 1015 – (C.P of 25 balls) = (C.P of 10 balls)
=> (C.P of 25 balls) + (C.P of 10 balls) = 1015
=> C.P of 35 balls = 1015
=> C.P of 1 ball=1015 / 35 = Rs. 29
Incorrect
Profit = (S.P of 25 balls) – (C.P of 25 balls) = 1015 – (C.P of 25 balls)
Given that Profit = (C.P of 10 balls)
=> 1015 – (C.P of 25 balls) = (C.P of 10 balls)
=> (C.P of 25 balls) + (C.P of 10 balls) = 1015
=> C.P of 35 balls = 1015
=> C.P of 1 ball=1015 / 35 = Rs. 29
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Question 8 of 40
8. Question
A letter lock consists of four rings each marked with five different letters. The number of distinct unsuccessful attempts to open the lock is at the most -.
Correct
Since each ring consists of five different letters, the total number of attempts possible with the four rings is =5 * 5 * 5 * 5 = 625. Of these attempts, one of them is a successful attempt.
Maximum number of unsuccessful attempts = 625 – 1 = 624.
Incorrect
Since each ring consists of five different letters, the total number of attempts possible with the four rings is =5 * 5 * 5 * 5 = 625. Of these attempts, one of them is a successful attempt.
Maximum number of unsuccessful attempts = 625 – 1 = 624.
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Question 9 of 40
9. Question
Arun started a business with Rs. 40,000 and is joined afterwards by Arul with Rs.60, 000. After how many months did Arul join if the profits at the end of the year are divided equally?
Correct
Suppose, Arul joined after x months.
Then, 40000 * 12 = 60000 * (12 – x) => 48 = 72 – 6x => 6x = 24 → x = 4
Incorrect
Suppose, Arul joined after x months.
Then, 40000 * 12 = 60000 * (12 – x) => 48 = 72 – 6x => 6x = 24 → x = 4
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Question 10 of 40
10. Question
Kathir and Anand can do a work in 20 days and 30 days respectively. Kathir started the work and left after 4 days. Anand took over and completed the work. In how many days was the total work completed?
Correct
Kathir’s one day’s work= 1/20
Kathir’s 4 day’s work =4* (1/20) = 1/5
Work left= 1-1/5 = 4 /5
Anand’s one day’s work= 1/30
Anand can do work in = (4/5) *(30/ 1) = 24 days
So total days = 24+4 = 28 days
Incorrect
Kathir’s one day’s work= 1/20
Kathir’s 4 day’s work =4* (1/20) = 1/5
Work left= 1-1/5 = 4 /5
Anand’s one day’s work= 1/30
Anand can do work in = (4/5) *(30/ 1) = 24 days
So total days = 24+4 = 28 days
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Question 11 of 40
11. Question
Population of a city is 1.2 lakh. If the population of male increases by 5% and the female by 10%, the population will be 1.2835 lakh. What is the number of female in the city ?
Correct
Let the population of female is ‘x’.
Population of male = 1.2 – x
= (110x/100) + [(105/100) (1.2 – x)] = 1.2835
110 x + 126 – 105x = 128.35
5x = 128.35 – 126 = 2.35
x = (2.35/5) = 0.47lakh = 47000
Incorrect
Let the population of female is ‘x’.
Population of male = 1.2 – x
= (110x/100) + [(105/100) (1.2 – x)] = 1.2835
110 x + 126 – 105x = 128.35
5x = 128.35 – 126 = 2.35
x = (2.35/5) = 0.47lakh = 47000
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Question 12 of 40
12. Question
A shopkeeper marks his goods 20% above the cost price but give 11% discount on it. If he sells the article for Rs.1575.30 then what is the cost price ?
Correct
Let the cost price is ‘x’
x×(120/ 100) × (89/100) = 1575.3
x = (157530 × 100) / (120×89) = 1475
Incorrect
Let the cost price is ‘x’
x×(120/ 100) × (89/100) = 1575.3
x = (157530 × 100) / (120×89) = 1475
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Question 13 of 40
13. Question
If Rs. 6200 amounts to Rs. 8804 in 3 years 6 months, what will Rs. 7800 amount to in 4 years 6 months at the same rate percent per annum ?
Correct
S.I. = 8804 – 6200 = 2604
r = (2604 × 100) / (6200 × 3.5) = 12% p.a
Now for Rs. 7800, S.I. = (7800 × 4.5 × 12)/100 = 4212
Req. amount = 7800 + 4212 = 12012
Incorrect
S.I. = 8804 – 6200 = 2604
r = (2604 × 100) / (6200 × 3.5) = 12% p.a
Now for Rs. 7800, S.I. = (7800 × 4.5 × 12)/100 = 4212
Req. amount = 7800 + 4212 = 12012
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Question 14 of 40
14. Question
The compound interest on a certain sum of money for two years at 8% p.a. is Rs. 499.20. What will be the simple interest at the same rate and for the same time period ?
Correct
P[1 +( 8/100)]2 – P = 499.20
P (27/25)2 – P = 499.20
P (729-625) / 625 = 499.20
P = (499.20 × 625) / 104 = 3000
S.I = (3000 × 8 × 2) / 100 = 480
Incorrect
P[1 +( 8/100)]2 – P = 499.20
P (27/25)2 – P = 499.20
P (729-625) / 625 = 499.20
P = (499.20 × 625) / 104 = 3000
S.I = (3000 × 8 × 2) / 100 = 480
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Question 15 of 40
15. Question
Certain number of persons can do a work in 50 days. If there were 7 persons more the work could be finished in 14 days less. How many persons were there initially ?
Correct
Let the original number of men ‘x’
7 person (50 – 14 = 36) days work = x persons 14 days work
x = (7 × 36)/14 = 18
Incorrect
Let the original number of men ‘x’
7 person (50 – 14 = 36) days work = x persons 14 days work
x = (7 × 36)/14 = 18
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Question 16 of 40
16. Question
[7.995 + 25.96 + 13.02] ÷ 2.490 = ? What approximate value will come in place of question mark (?) in the following questions? (You are not expected to calculate the exact value.)
Correct
? = (7.995 + 25.96 + 13.02) ÷ 2.490
= (8 + 26 + 13) ÷ 2.5 = 47 ÷ 2.5 ≈ 19
Incorrect
? = (7.995 + 25.96 + 13.02) ÷ 2.490
= (8 + 26 + 13) ÷ 2.5 = 47 ÷ 2.5 ≈ 19
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Question 17 of 40
17. Question
(5.95 x 8.06) ÷ (8.03 x 2.04) = ? – 4.95 What approximate value will come in place of question mark (?) in the following questions? (You are not expected to calculate the exact value.)
Correct
? -. 4 95 = (5.95 × 8.06) ÷ (8.03 × 2.04)
- ? – 5 = (6 × 8) / (8 × 2) = 3
? ≈ 3 + 5 = 8
Incorrect
? -. 4 95 = (5.95 × 8.06) ÷ (8.03 × 2.04)
- ? – 5 = (6 × 8) / (8 × 2) = 3
? ≈ 3 + 5 = 8
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Question 18 of 40
18. Question
4.014 ÷ 1.894 x 15.011 ÷ 8.971= (?) What approximate value will come in place of question mark (?) in the following questions? (You are not expected to calculate the exact value.)
Correct
Answer: c
(?) = 4.014 ÷ 1.894 × 15.011 ÷ 8.971
≈ 4 ÷ 2 × 15 ÷ 9 = 2 × 15 ÷ 9 = (30/9) ≈ 3
Incorrect
Answer: c
(?) = 4.014 ÷ 1.894 × 15.011 ÷ 8.971
≈ 4 ÷ 2 × 15 ÷ 9 = 2 × 15 ÷ 9 = (30/9) ≈ 3
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Question 19 of 40
19. Question
(6.90)3 – (2.99)3 – (6.02)3 = (?)2 What approximate value will come in place of question mark (?) in the following questions? (You are not expected to calculate the exact value.)
Correct
( ?)2 = (6.90)3 – (2.99)3 – (6.02)3
≈ (7)3 – (3)3 – (6)3
= 343 – 27 – 216 = 343 – 243 = 100
? = √100 = 10
Incorrect
( ?)2 = (6.90)3 – (2.99)3 – (6.02)3
≈ (7)3 – (3)3 – (6)3
= 343 – 27 – 216 = 343 – 243 = 100
? = √100 = 10
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Question 20 of 40
20. Question
[(6.03 x 2.98) – (1.99 x 2.012)] / (2.053 —360.5) = ? What approximate value will come in place of question mark (?) in the following questions? (You are not expected to calculate the exact value.)
Correct
? = [ (6.03 x 2.98) – (1.99 x 2.012) ] / [ (2.05)3 -360.5]
= (6 × 3 – 2 × 22) / [(2)3 – (36)1/2] = 10 / (8-6) = 10/2 ≈ 5
Incorrect
? = [ (6.03 x 2.98) – (1.99 x 2.012) ] / [ (2.05)3 -360.5]
= (6 × 3 – 2 × 22) / [(2)3 – (36)1/2] = 10 / (8-6) = 10/2 ≈ 5
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Question 21 of 40
21. Question
The students of ABC Institute take part in various extra-curricular activities, the distribution of which is as shown below. Total number of 800
What the difference between the, total number of students taking part in Volleyball and Swimming together and the total number of students talking part in Football?
Correct
Reqd difference = (26 + 16 – 17)% of 800 = 25% of 800 = 200
Incorrect
Reqd difference = (26 + 16 – 17)% of 800 = 25% of 800 = 200
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Question 22 of 40
22. Question
The students of ABC Institute take part in various extra-curricular activities, the distribution of which is as shown below. Total number of 800
What is the ratio of the total number of students taking part in Tennis and Swimming together to that taking part in Cricket and Football together?
Correct
Required ratio = (22 + 16) / (19+17) = 38/36 = 19:18
Incorrect
Required ratio = (22 + 16) / (19+17) = 38/36 = 19:18
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Question 23 of 40
23. Question
The students of ABC Institute take part in various extra-curricular activities, the distribution of which is as shown below. Total number of 800
What is the number of girls who take part in Volleyball, if the ratio of boys to girls is 5 : 11?
Correct
Number of students taking part in Volleyball = (26 × 800)/100 = 208
Let the number of boys be 5x and girls be I I x.
Now, 5x + 11x = 208
or, I6x =.208
x = 208 /16 =13
Number of girl students who take part in Volleyball = I I x 13 = 143
Incorrect
Number of students taking part in Volleyball = (26 × 800)/100 = 208
Let the number of boys be 5x and girls be I I x.
Now, 5x + 11x = 208
or, I6x =.208
x = 208 /16 =13
Number of girl students who take part in Volleyball = I I x 13 = 143
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Question 24 of 40
24. Question
The students of ABC Institute take part in various extra-curricular activities, the distribution of which is as shown below. Total number of
What is the average number of students taking part in Tennis, Cricket, Football and Swimming?
Correct
Reqd average [ (22+19+17+16)%of 800 ] / 4
= (74% of 800) / 4= (74 x 800) / (100×4) = 148
Incorrect
Reqd average [ (22+19+17+16)%of 800 ] / 4
= (74% of 800) / 4= (74 x 800) / (100×4) = 148
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Question 25 of 40
25. Question
The students of ABC Institute take part in various extra-curricular activities, the distribution of which is as shown below. Total number of 800
If out of the number of students taking part in Tennis, 79 are girls, what is the difference between the number of boys and that of girls taking part in Tennis?
Correct
Total number of students taking part in Tennis = 22% of 800 = (22 x 800) /100 = 176
Out of the number of students taking part in tennis, 79 are girls.
The number of boys who take part in Tennis = 176 — 79 = 97
Difference = 97 — 79 = 18
Incorrect
Total number of students taking part in Tennis = 22% of 800 = (22 x 800) /100 = 176
Out of the number of students taking part in tennis, 79 are girls.
The number of boys who take part in Tennis = 176 — 79 = 97
Difference = 97 — 79 = 18
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Question 26 of 40
26. Question
15, 16, 10, 33, ?, 66.25 What will come in place of the question mark :
Correct
x1+1, ÷2+2, x3+3, ÷4+4, x5+5
Incorrect
x1+1, ÷2+2, x3+3, ÷4+4, x5+5
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Question 27 of 40
27. Question
2, 9, 82, ?, 32826 What will come in place of the question mark :
Correct
x2^2-2+3, x3^2-3+4, x4^2-4+5, x5^2-5+6
Incorrect
x2^2-2+3, x3^2-3+4, x4^2-4+5, x5^2-5+6
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Question 28 of 40
28. Question
32, 321, 682, 1123, ?, 2277 What will come in place of the question mark :
Correct
+17^2, +19^2, +21^2, +23^2, +25^2
Incorrect
+17^2, +19^2, +21^2, +23^2, +25^2
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Question 29 of 40
29. Question
84, ?, 78, 109, 72, 113, 70 What will come in place of the question mark :
Correct
+23, -29, +31, -37, +41, -43
Incorrect
+23, -29, +31, -37, +41, -43
-
Question 30 of 40
30. Question
11, 15, ?, 71, 333 What will come in place of the question mark :
Correct
Diff of diff is multiples of 6
Incorrect
Diff of diff is multiples of 6
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Question 31 of 40
31. Question
33 ÷ 37 × (27)2 × 11.25 + 75% of 45 = ? What should come in place of the question mark (?) in the following questions?
Correct
? = 33 ÷ 37 × (33)2 × 11.25 + [(75 × 45) / 100]
= (3)3 + 6 – 7 × 11.25 + 33.75
= 9 × 11.25 + 33.75
= 101.25 + 33.75 = 135
Incorrect
? = 33 ÷ 37 × (33)2 × 11.25 + [(75 × 45) / 100]
= (3)3 + 6 – 7 × 11.25 + 33.75
= 9 × 11.25 + 33.75
= 101.25 + 33.75 = 135
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Question 32 of 40
32. Question
144% of 185 – 44% of 85 = 200 + ? What should come in place of the question mark (?) in the following questions?
Correct
[(144 × 185) / 100] – [(44 × 85) / 100]
= 266.7 – 37.4 = 229 = 200 + 29
Incorrect
[(144 × 185) / 100] – [(44 × 85) / 100]
= 266.7 – 37.4 = 229 = 200 + 29
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Question 33 of 40
33. Question
(17.35)2 – (8.85)2 = 200 + ? What should come in place of the question mark (?) in the following questions?
Correct
(17.35)2 – (8.85)2 = (17.35 + 8.85) (17.35 – 8.85)
= 2.62 × 8.5 = 222.7
Incorrect
(17.35)2 – (8.85)2 = (17.35 + 8.85) (17.35 – 8.85)
= 2.62 × 8.5 = 222.7
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Question 34 of 40
34. Question
(1 / 13) × 3237 + (3 / 14) × 5362 + 200% of 1 = ? + 1335 What should come in place of the question mark (?) in the following questions?
Correct
(3237 / 13) + [(3 × 5362) / 14] + [ (200 × 1) / 100]
= 249 + 1149 + 2 = 1400
? = 1400 – 1335 = 65
Incorrect
(3237 / 13) + [(3 × 5362) / 14] + [ (200 × 1) / 100]
= 249 + 1149 + 2 = 1400
? = 1400 – 1335 = 65
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Question 35 of 40
35. Question
(11 / 7) of (5 / 8) of (13 / 9) of 8568 = ? What should come in place of the question mark (?) in the following questions?
Correct
Incorrect
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Question 36 of 40
36. Question
200.1 × 9.9 – 25 × 62.5 + 12 × 144 = ?– 26.49 What should come in place of the question mark (?) in the following questions?
Correct
200.1 × 9.9 – 25 × 62.5 + 12 × 144 = ?– 26.49
Or, 1980.99 – 1562.5 + 1728 = ?– 26.49
Or, ? = 1980.99 + 1728 + 26.49 – 1562.5
= 3735.48 – 1562.5 = 2172.98
Incorrect
200.1 × 9.9 – 25 × 62.5 + 12 × 144 = ?– 26.49
Or, 1980.99 – 1562.5 + 1728 = ?– 26.49
Or, ? = 1980.99 + 1728 + 26.49 – 1562.5
= 3735.48 – 1562.5 = 2172.98
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Question 37 of 40
37. Question
√48 +√80 + √176 + √324 – √121 = ? + 7 + 4√11 What should come in place of the question mark (?) in the following questions?
Correct
? + 7 + 4√11 =√48 + √80 + √176 + √324 – √121
Or, ? = 4√3 + 4√5 + 4√ + 7 – 7 – 4√11
= 4 (√3 + √5 )
Incorrect
? + 7 + 4√11 =√48 + √80 + √176 + √324 – √121
Or, ? = 4√3 + 4√5 + 4√ + 7 – 7 – 4√11
= 4 (√3 + √5 )
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Question 38 of 40
38. Question
1265 ÷ 25.3 + 102 × 98 – (23)2 = ? What should come in place of the question mark (?) in the following questions?
Correct
1265 / 25.3 + 102 × 98 – (23)2 = ?
= 50 + 9996 – 529 = 9517
Incorrect
1265 / 25.3 + 102 × 98 – (23)2 = ?
= 50 + 9996 – 529 = 9517
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Question 39 of 40
39. Question
3√12167 + 3√21952% of 280 – 3√704969% of 56 = ? What should come in place of the question mark (?) in the following questions?
Correct
3√12167 + 3√21952% of 280 – 3√704969% of 56 = ?
= 23 + [ (28 × 280) / 100 ] – [ (89 × 56) / 100 ]
= 23 + 78.4 – 49.84 = 51.56
Incorrect
3√12167 + 3√21952% of 280 – 3√704969% of 56 = ?
= 23 + [ (28 × 280) / 100 ] – [ (89 × 56) / 100 ]
= 23 + 78.4 – 49.84 = 51.56
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Question 40 of 40
40. Question
{ [ 33 (17 / 25) ] × [ 34 (22 / 37) ] } + { [ 35 (28 / 52) ] × [ 36 (37 / 57) ] } = ? What should come in place of the question mark (?) in the following questions?
Correct
= ?{ [ 33 (17 / 25) ] × [ 34 (22 / 37) ] } + { [ 35 (28 / 52) ] × [ 36 (37 / 57) ] }
= [ (842 / 25) × (1280 / 37) ] + [ (1848 / 52) × (2089 / 57) ]
= [ (842 / 5) × (256 / 37) ] + [ (616 / 52) × (2089 / 19) ]
= (215552 / 185) + (1286824 / 988)
= 1165.14 + 1302.45 = 2467.59
Incorrect
= ?{ [ 33 (17 / 25) ] × [ 34 (22 / 37) ] } + { [ 35 (28 / 52) ] × [ 36 (37 / 57) ] }
= [ (842 / 25) × (1280 / 37) ] + [ (1848 / 52) × (2089 / 57) ]
= [ (842 / 5) × (256 / 37) ] + [ (616 / 52) × (2089 / 19) ]
= (215552 / 185) + (1286824 / 988)
= 1165.14 + 1302.45 = 2467.59
