Graduate Trainee Quantitative Ability Access Bank Job
Graduate Trainee Quantitative Ability Access Bank Job
Question 51
If apples are bought at the rate of 30 for a rupee, how many apples must be sold for a rupee so as to gain 20%?
Cost price of 1 apple = \( 1/30 \) rupee.
To gain 20%, selling price of 1 apple = \( 1/30 \times 1.2 = 1/25 \) rupee.
Therefore, number of apples sold for 1 rupee = 25.
Thus, 25 apples must be sold for 1 rupee to gain 20%.
To gain 20%, selling price of 1 apple = \( 1/30 \times 1.2 = 1/25 \) rupee.
Therefore, number of apples sold for 1 rupee = 25.
Thus, 25 apples must be sold for 1 rupee to gain 20%.
Not clear enough?
Question 52
A merchant marks his goods in such a way that the profit on sale of 50 articles is equal to the selling price of 25 articles. What is his profit margin?
Let the cost price of each article be \( \$1 \).
Total cost price for 50 articles = \( \$50 \).
Profit = Selling price of 25 articles = \( 25S/50 = 0.5S \).
Thus, profit = \( 25/50 = 50\% \).
Hence, the profit margin is 50%.
Total cost price for 50 articles = \( \$50 \).
Profit = Selling price of 25 articles = \( 25S/50 = 0.5S \).
Thus, profit = \( 25/50 = 50\% \).
Hence, the profit margin is 50%.
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Question 53
A predator is chasing its prey. The predator takes 4 leaps for every 6 leaps of the prey and the predator covers as much distance in 2 leaps as 3 leaps of the prey. Will the predator succeed in getting its food?
Let the prey's one leap = \( x \).
In 6 leaps, prey covers \( 6x \).
Predator’s 1 leap = \( 1.5x \), hence in 4 leaps predator covers \( 6x \).
Both cover the same distance, meaning the predator will never overtake the prey.
Therefore, the predator will never succeed.
In 6 leaps, prey covers \( 6x \).
Predator’s 1 leap = \( 1.5x \), hence in 4 leaps predator covers \( 6x \).
Both cover the same distance, meaning the predator will never overtake the prey.
Therefore, the predator will never succeed.
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Question 54
Two boys begin together to write out a booklet containing 535 lines. The first boy starts with the first line, writing at the rate of 100 lines an hour; and the second starts with the last line, writing line 534 and so on, backwards proceeding at the rate of 50 lines an hour. At what line will they meet?
Let time = \( t \) hours when they meet.
First boy writes \( 100t \) lines, second boy writes \( 50t \) lines.
Total = 535 lines ⇒ \( 100t + 50t = 535 \).
\( t = 535/150 = 3.566 \) hours.
Lines written by first boy = \( 100 \times 3.566 = 356.6 \approx 357 \).
They meet at line 357.
First boy writes \( 100t \) lines, second boy writes \( 50t \) lines.
Total = 535 lines ⇒ \( 100t + 50t = 535 \).
\( t = 535/150 = 3.566 \) hours.
Lines written by first boy = \( 100 \times 3.566 = 356.6 \approx 357 \).
They meet at line 357.
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Question 55
A gives B a start of 30 seconds in a kilometre race and still beats him by 20 metres. However, when he gives B a start of 35 seconds, they finish the race in a dead heat. How long does A take to run the kilometre?
Let A's speed = \( v_A \) m/s and B's speed = \( v_B \) m/s.
When B starts 30 s earlier, A beats him by 20 m:
\( 1000/v_A = 1000/v_B - 30 \).
When B starts 35 s earlier, they tie:
\( 1000/v_A = 1000/v_B - 35 \).
Subtracting, \( 5 = 20/v_B \) ⇒ \( v_B = 4 \) m/s.
Substitute into first: \( 1000/v_A = 1000/4 - 30 = 220 \) ⇒ \( v_A = 1000/220 = 4.545 \) m/s.
Time for A = \( 1000 / 4.545 = 220 \) s.
When B starts 30 s earlier, A beats him by 20 m:
\( 1000/v_A = 1000/v_B - 30 \).
When B starts 35 s earlier, they tie:
\( 1000/v_A = 1000/v_B - 35 \).
Subtracting, \( 5 = 20/v_B \) ⇒ \( v_B = 4 \) m/s.
Substitute into first: \( 1000/v_A = 1000/4 - 30 = 220 \) ⇒ \( v_A = 1000/220 = 4.545 \) m/s.
Time for A = \( 1000 / 4.545 = 220 \) s.
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Question 56
A man and a woman, 81 miles apart from each other, start travelling towards each other at the same time. If the man covers 5 miles per hour to the woman's 4 miles per hour, how far will the woman have travelled when they meet?
Relative speed = \( 5 + 4 = 9 \) mph.
Time to meet = \( 81 / 9 = 9 \) hours.
Distance covered by woman = \( 9 \times 4 = 36 \) miles.
Hence, she travels 36 miles when they meet.
Time to meet = \( 81 / 9 = 9 \) hours.
Distance covered by woman = \( 9 \times 4 = 36 \) miles.
Hence, she travels 36 miles when they meet.
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Question 57
The speed of a motor boat itself is 20 km/h and the rate of flow of the river is 4 km/h. Moving with the stream the boat went 120 km. What distance will the boat cover during the same time going against the stream?
Downstream speed = \( 20 + 4 = 24 \) km/h.
Upstream speed = \( 20 - 4 = 16 \) km/h.
Time taken downstream = \( 120 / 24 = 5 \) hours.
Distance covered upstream in same time = \( 16 \times 5 = 80 \) km.
Hence, the boat covers 80 km upstream.
Upstream speed = \( 20 - 4 = 16 \) km/h.
Time taken downstream = \( 120 / 24 = 5 \) hours.
Distance covered upstream in same time = \( 16 \times 5 = 80 \) km.
Hence, the boat covers 80 km upstream.
Not clear enough?
Question 58
A can give B 20 points, A can give C 32 points and B can give C 15 points. How many points make the game?
Let the game be of \( x \) points.
A : B = \( x : (x - 20) \), A : C = \( x : (x - 32) \).
B : C = \( (x - 20) : (x - 32) = 15 \) points difference.
\( (x - 20) - (x - 32) = 15 \) ⇒ \( 12 = 15 \times (x - 32)/(x - 20) \) ⇒ \( x = 200 \).
Hence, the game is of 200 points.
A : B = \( x : (x - 20) \), A : C = \( x : (x - 32) \).
B : C = \( (x - 20) : (x - 32) = 15 \) points difference.
\( (x - 20) - (x - 32) = 15 \) ⇒ \( 12 = 15 \times (x - 32)/(x - 20) \) ⇒ \( x = 200 \).
Hence, the game is of 200 points.
Not clear enough?
Question 59
The average age of a group of 12 students is 20 years. If 4 more students join the group, the average age increases by 1 year. The average age of the new students is _______.
Total age of 12 students = \( 12 \times 20 = 240 \).
New average = 21 ⇒ total age of 16 students = \( 16 \times 21 = 336 \).
Age of 4 new students = \( 336 - 240 = 96 \).
Average age of 4 new students = \( 96 / 4 = 24 \).
Hence, their average age is 24 years.
New average = 21 ⇒ total age of 16 students = \( 16 \times 21 = 336 \).
Age of 4 new students = \( 336 - 240 = 96 \).
Average age of 4 new students = \( 96 / 4 = 24 \).
Hence, their average age is 24 years.
Not clear enough?
Question 60
I travel the first part of my journey at 40 km/h and the second part at 60 km/h and cover the total distance of 240 km to my destination in 5 hours. How long did the first part of my journey last?
Let the time taken for the first part be \( t \) hours.
Then, distance covered in first part = \( 40t \) and in the second part = \( 60(5 - t) \).
Total distance = 240 km.
\( 40t + 60(5 - t) = 240 \) ⇒ \( 40t + 300 - 60t = 240 \).
\( 20t = 60 \Rightarrow t = 3 \).
Hence, the first part of the journey lasted 3 hours.
Then, distance covered in first part = \( 40t \) and in the second part = \( 60(5 - t) \).
Total distance = 240 km.
\( 40t + 60(5 - t) = 240 \) ⇒ \( 40t + 300 - 60t = 240 \).
\( 20t = 60 \Rightarrow t = 3 \).
Hence, the first part of the journey lasted 3 hours.
Not clear enough?
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Line 2: A rainbow in the sky
Line 3: so was it when my life began
Line 4: so is it now
Line 5: so is it when I have grown
Line 6: the child is the father of the man.
The object of the extract is____?
The primary stress is indicated by capital letters. Choose the one that has the correct stress:
presidency
