Graduate Trainee Quantitative Ability Access Bank Job
Graduate Trainee Quantitative Ability Access Bank Job
Question 31
A box contains 90 nuts each of 100 gms and 100 bolts each of 150 gms.
If the entire box weighs 35.5 kg, then the weight of the empty box is:
If the entire box weighs 35.5 kg, then the weight of the empty box is:
Weight of nuts = 90 × 100 g = 9000 g = 9 kg.
Weight of bolts = 100 × 150 g = 15000 g = 15 kg.
Total weight of nuts and bolts = 9 + 15 = 24 kg.
Total box weight = 35.5 kg.
Weight of empty box = 35.5 − 24 = 11.5 kg.
Weight of bolts = 100 × 150 g = 15000 g = 15 kg.
Total weight of nuts and bolts = 9 + 15 = 24 kg.
Total box weight = 35.5 kg.
Weight of empty box = 35.5 − 24 = 11.5 kg.
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Question 32
If the radius of a circle is increased by 20%, then the area is increased by:
Original area \( A = \pi r^2 \).
New radius = 1.2r, so new area \( = \pi (1.2r)^2 = 1.44 \pi r^2 \).
Percentage increase = (1.44 − 1) × 100 = 44%.
New radius = 1.2r, so new area \( = \pi (1.2r)^2 = 1.44 \pi r^2 \).
Percentage increase = (1.44 − 1) × 100 = 44%.
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Question 33
Tom, Dick and Harry went for lunch to a restaurant. Tom had \$100 with him, Dick had \$60 and Harry had \$40. They got a bill for \$104 and decided to give a tip of \$16. They further decided to share the total expenses in the ratio of the amounts of money each carried. The amount of money which Tom paid more than what Harry paid is:
Total money carried = 100 + 60 + 40 = 200.
Total bill with tip = 104 + 16 = 120.
Tom’s share = (100/200) × 120 = 60.
Dick’s share = (60/200) × 120 = 36.
Harry’s share = (40/200) × 120 = 24.
Difference between Tom and Harry = 60 − 24 = 36.
Total bill with tip = 104 + 16 = 120.
Tom’s share = (100/200) × 120 = 60.
Dick’s share = (60/200) × 120 = 36.
Harry’s share = (40/200) × 120 = 24.
Difference between Tom and Harry = 60 − 24 = 36.
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Question 34
A motorcycle stunt man belonging to a fair rides over the vertical walls of a circular well at an average speed of 54 kph for 5 minutes. If the radius of the well is 5 metres, then the distance travelled is:
Speed = 54 km/h = 15 m/s.
Time = 5 minutes = 300 s.
Distance = speed × time = 15 × 300 = 4500 m = 4.5 km.
Time = 5 minutes = 300 s.
Distance = speed × time = 15 × 300 = 4500 m = 4.5 km.
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Question 35
A zookeeper counted the heads of the animals in a zoo and found it to be 80. When he counted the legs of the animals, he found it to be 260. If the zoo had either pigeons or horses, how many horses were there in the zoo?
Let the number of horses be \( x \), and pigeons be \( 80 - x \).
Each horse has 4 legs, each pigeon has 2 legs.
Total legs = 4x + 2(80 − x) = 260.
\(4x + 160 − 2x = 260 \Rightarrow 2x = 100 \Rightarrow x = 50.\)
So, there are 50 horses.
Each horse has 4 legs, each pigeon has 2 legs.
Total legs = 4x + 2(80 − x) = 260.
\(4x + 160 − 2x = 260 \Rightarrow 2x = 100 \Rightarrow x = 50.\)
So, there are 50 horses.
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Question 36
A runs \( \dfrac{13}{5} \) times as fast as B. If A gives a start of 240 m, how far must the post be so that A and B might reach at the same time?
Let the distance be \( d \).
Time taken by A = \( \dfrac{d}{13} \), time taken by B = \( \dfrac{d - 240}{5} \).
Equating: \( \dfrac{d}{13} = \dfrac{d - 240}{5} \).
Cross-multiplying: \( 5d = 13(d - 240) \Rightarrow 5d = 13d - 3120 \Rightarrow 8d = 3120 \Rightarrow d = 390. \)
Time taken by A = \( \dfrac{d}{13} \), time taken by B = \( \dfrac{d - 240}{5} \).
Equating: \( \dfrac{d}{13} = \dfrac{d - 240}{5} \).
Cross-multiplying: \( 5d = 13(d - 240) \Rightarrow 5d = 13d - 3120 \Rightarrow 8d = 3120 \Rightarrow d = 390. \)
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Question 37
A gives B a start of 30 seconds in a 1 km race and still beats him by 20 m. 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 km?
Let A’s speed be \( v_A \) m/s and B’s speed be \( v_B \) m/s.
From the first condition: when A covers 1000 m, B covers 980 m.
\( \dfrac{v_B}{v_A} = \dfrac{980}{1000} = 0.98. \)
From the second condition: A’s 35 s head start means \( 35v_B = 1000 - 1000\times0.98 = 20v_A \).
Solving gives \( v_A = 4 \text{ m/s} \).
Time = \( \dfrac{1000}{4} = 250 \text{ s}. \)
From the first condition: when A covers 1000 m, B covers 980 m.
\( \dfrac{v_B}{v_A} = \dfrac{980}{1000} = 0.98. \)
From the second condition: A’s 35 s head start means \( 35v_B = 1000 - 1000\times0.98 = 20v_A \).
Solving gives \( v_A = 4 \text{ m/s} \).
Time = \( \dfrac{1000}{4} = 250 \text{ s}. \)
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Question 38
A father is three times as old as his son. After fifteen years the father will be twice as old as his son's age at that time. Hence the father's present age is:
Let the son's present age be \( x \). Then the father's age is \( 3x \).
After 15 years: father = \( 3x + 15 \), son = \( x + 15 \).
Given: \( 3x + 15 = 2(x + 15) \Rightarrow 3x + 15 = 2x + 30 \Rightarrow x = 15. \)
Hence father’s age = \( 3x = 45. \)
After 15 years: father = \( 3x + 15 \), son = \( x + 15 \).
Given: \( 3x + 15 = 2(x + 15) \Rightarrow 3x + 15 = 2x + 30 \Rightarrow x = 15. \)
Hence father’s age = \( 3x = 45. \)
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Question 39
A portion of \$7200 is invested at a 4% annual return, while the remainder is invested at a 5% annual return. If the annual income from both portions is the same, what is the total income from the two investments?
Let \( x \) be the amount invested at 4%, and \( 7200 - x \) at 5%.
Given both incomes are equal: \( 0.04x = 0.05(7200 - x) \).
\( 0.04x = 360 - 0.05x \Rightarrow 0.09x = 360 \Rightarrow x = 4000. \)
Income from each = \( 0.04 × 4000 = 160. \)
Total income = \( 160 + 160 = 320. \)
Given both incomes are equal: \( 0.04x = 0.05(7200 - x) \).
\( 0.04x = 360 - 0.05x \Rightarrow 0.09x = 360 \Rightarrow x = 4000. \)
Income from each = \( 0.04 × 4000 = 160. \)
Total income = \( 160 + 160 = 320. \)
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Question 40
The sum of the first 100 numbers, 1 to 100, is divisible by:
Sum of first \( n \) natural numbers = \( \dfrac{n(n + 1)}{2} \).
For \( n = 100 \): \( \dfrac{100 \times 101}{2} = 5050. \)
5050 ÷ 2 = 2525 (divisible by 2), but not divisible by 4 or 8.
Hence, it is divisible by 2 only.
For \( n = 100 \): \( \dfrac{100 \times 101}{2} = 5050. \)
5050 ÷ 2 = 2525 (divisible by 2), but not divisible by 4 or 8.
Hence, it is divisible by 2 only.