ARITHMETIC REVIEW TEST - 2
Q1–10 of 25A car after travelling 18 km from a point A developed some problem in the engine and the speed became 4/5th of its original speed. As a result, the car reached point B 45 minutes late. If the engine had developed the same problem after travelling 30 km from A, the car would have reached B only 36 minutes late. The original speed of the car (in km per hour) and the distance between the points A and B (in km) are
Concept: Reduced speed is $\frac{4}{5}$ of original speed $\implies$ Time taken is $\frac{5}{4}$ of original time (an extra $\frac{1}{4}$ of the usual time).
Calculation:
In the extra $12\text{ km}$ ($30\text{ km} - 18\text{ km}$), the delay was reduced by $9\text{ min}$ ($45\text{ min} - 36\text{ min}$).
If $\frac{1}{4}$ of usual time for $12\text{ km} = 9\text{ min}$, then the usual time to cover $12\text{ km} = 9 \times 4 = 36\text{ min} = 0.6\text{ hours}$.
Original Speed $= \frac{12\text{ km}}{0.6\text{ hours}} = 20\text{ km/hr}$.
For the first $18\text{ km}$, time taken at normal speed $= \frac{18}{20} = 0.9\text{ hours} = 54\text{ min}$.
A delay of $45\text{ min}$ means the remaining distance usually takes $45 \times 4 = 180\text{ min} = 3\text{ hours}$.
Distance after $18\text{ km} = 20 \times 3 = 60\text{ km}$.
Total Distance $= 18 + 60 = 78\text{ km}$.
Correct Answer: (d) None of these (Speed $= 20\text{ km/hr}$, Distance $= 78\text{ km}$)
A, B and C individually can finish a work in 6, 8 and 15 hours respectively. They started the work together and after completing the work got ₹ 94.60. when they divide the money among themselves, A, B and C will get respectively (in ₹)
Calculation:
Time taken ratio $= 6 : 8 : 15$.
Efficiency / Work done ratio $= \frac{1}{6} : \frac{1}{8} : \frac{1}{15} = 20 : 15 : 8$.
Total units $= 20 + 15 + 8 = 43$.
Share of A $= 94.60 \times \frac{20}{43} = ₹44.00$.
Share of B $= 94.60 \times \frac{15}{43} = ₹33.00$.
Share of C $= 94.60 \times \frac{8}{43} = ₹17.60$.
Correct Answer: (a) 44, 33, 17.60
Two trains are traveling in opposite direction at uniform speed 60 and 50 km per hour respectively. They take 5 seconds to cross each other. If the two trains had traveled in the same direction, then a passenger sitting in the faster moving train would have overtaken the other train in 18 seconds. The length of the trains in metres are
Relative speed in opposite direction $= 60 + 50 = 110\text{ km/hr} = 110 \times \frac{5}{18} = \frac{275}{9}\text{ m/s}$.
Total length of both trains ($L_1 + L_2$) $= \frac{275}{9} \times 5 = \frac{1375}{9} \approx 152.78\text{ metres}$.
Relative speed in same direction $= 60 - 50 = 10\text{ km/hr} = 10 \times \frac{5}{18} = \frac{25}{9}\text{ m/s}$.
Length of slower train ($L_2$) $= \text{Relative Speed} \times 18\text{ s} = \frac{25}{9} \times 18 = 50\text{ metres}$.
Length of faster train ($L_1$) $= 152.78 - 50 = 102.78\text{ metres}$.
Correct Answer: (c) 102.78, 50
Assume that an equal number of people are born on each day. Find approximately the percentage of the people whose birthday will fall on 29th February.
Calculation:
Over a 4-year period (1461 days), 29th February occurs once.
Probability $= \frac{1}{1461} \approx 0.00068447$.
Percentage $= 0.00068447 \times 100 \approx 0.0684\%$.
Correct Answer: (c) 0.0684
A sum of money compounded annually becomes ₹ 625 in two years and ₹ 675 in three years. The rate of interest per annum is
Calculation:
Amount after 2 years $= ₹625$, Amount after 3 years $= ₹675$.
Interest earned in 3rd year $= 675 - 625 = ₹50$.
Rate of Interest $= \frac{50}{625} \times 100 = 8\%$.
Correct Answer: (b) 8%
Every day Asha’s husband meets her at the city railway station at 6:00 p.m. and drives her to their residence. One day she left early from the office and reached the railway station at 5:00 p.m. She started walking towards her home, met her husband coming from their residence on the way and they reached home 10 minutes earlier than the usual time. For how long did she walk?
Calculation:
They reached home $10\text{ minutes}$ earlier, which means the husband saved $10\text{ minutes}$ of total round-trip driving time.
Thus, he saved $5\text{ minutes}$ of driving in one direction.
He usually meets her at the station at 6:00 p.m., so he met her $5\text{ minutes}$ earlier on the road at 5:55 p.m.
Since Asha reached the station at 5:00 p.m. and walked until 5:55 p.m., she walked for 55 minutes.
Correct Answer: (d) 55 minutes
Three machines, A, B and C can be used to produce a product. Machine A will take 60 hours to produce a million units. Machine B is twice as fast as Machine A. Machine C will take the same amount of time to produce a million units as A and B running together. How much time will be required to produce a million units if all the three machines are used simultaneously?
Calculation:
Time by A $= 60\text{ hrs}$ $\implies \text{Rate } A = \frac{1}{60}$.
Machine B is twice as fast as A $\implies \text{Rate } B = \frac{2}{60} = \frac{1}{30}$.
Machine C rate $= A + B = \frac{1}{60} + \frac{1}{30} = \frac{3}{60} = \frac{1}{20}$.
Combined Rate $= A + B + C = \frac{1}{20} + \frac{1}{20} = \frac{1}{10}$.
Time needed $= 10\text{ hours}$.
Correct Answer: (b) 10 hours
Mr. and Mrs. Shah travel from City A to City B and break journey at City C in between. Somewhere between City A and City C, Mrs. Shah asks “How far have we travelled?” Mr. Shah replies, “Half as far as the distance from here to City C”. Somewhere between City C and City B, exactly 200 km from the point where she asked the first question, Mrs. Shah asks “How far do we have to go?” Mr. Shah replies “Half as far as the distance from City C to here.” The distance between Cities A and B in km. is
Calculation:
Let distance from $A$ to first stop point $= x$. Distance from first stop point to $C = 2x$. So distance $AC = 3x$.
Let distance from second stop point to $B = y$. Distance from $C$ to second stop point $= 2y$. So distance $CB = 3y$.
Distance between the two stop points $= 2x + 2y = 200\text{ km}$ $\implies x + y = 100\text{ km}$.
Total Distance $AB = 3x + 3y = 3(x + y) = 3 \times 100 = 300\text{ km}$.
Correct Answer: (d) 300
A shop sells ball point pen refills. It used to sell refills for 50 paise each, and there were hardly any takers. When he reduced the price, the remaining refills were sold out enabling the shopkeeper to realize ₹ 35.89. How many refills were sold at the reduced price?
Calculation:
Total revenue $= ₹35.89 = 3589\text{ paise}$.
The price per refill must be an integer in paise and strictly less than $50\text{ paise}$.
$3589 = 37 \times 97$.
Since the reduced price must be $< 50\text{ paise}$, the price per refill is $37\text{ paise}$.
Number of refills sold $= 97$.
Correct Answer: (d) 97
Anand and Bharat can cut 5 kg of wood in 20 min, Bharat and Chandra can cut 5 kg of wood in 40 min. Chandra and Anand can cut 5 kg. of wood in 30 min. How much time Chandra will take to cut 5 kg of wood alone?
Calculation: Rates to cut $5\text{ kg}$ of wood:
$A + B = \frac{1}{20}$ per minute
$B + C = \frac{1}{40}$ per minute
$C + A = \frac{1}{30}$ per minute
Sum: $2(A + B + C) = \frac{1}{20} + \frac{1}{40} + \frac{1}{30} = \frac{13}{120} \implies A + B + C = \frac{13}{240}$
Rate of Chandra ($C$) $= (A + B + C) - (A + B) = \frac{13}{240} - \frac{1}{20} = \frac{1}{240}$
Time taken by Chandra alone $= 240\text{ minutes}$.
Correct Answer: (c) 240 minutes