ARITHMETIC REVIEW TEST - 3
Q1–10 of 25There is a leak in the bottom of a tank. This leak can empty a full tank in 8 hours. When the tank is full, a tap is opened into the tank which admits 6 litres per hour and the tank is now emptied in 12 hours. What is the capacity of the tank?
Calculation:
Leak alone empties the tank in $8\text{ hours}$ $\implies \text{Rate} = -\frac{1}{8}$ tank/hour.
Leak + Tap together empty the tank in $12\text{ hours}$ $\implies \text{Rate} = -\frac{1}{12}$ tank/hour.
Tap filling rate $= -\frac{1}{12} - \left(-\frac{1}{8}\right) = \frac{1}{24}$ tank/hour.
Tap alone takes $24\text{ hours}$ to fill the tank.
Given tap rate $= 6\text{ litres/hour}$.
Capacity $= 24 \text{ hours} \times 6\text{ litres/hour} = 144\text{ litres}$.
Correct Answer: (c) 144 litres
The winning relay team in a high school sports competition clocked 48 minutes for a distance of 13.2 km. Its runners A, B, C and D maintained speeds of 15 kmph, 16, 17 kmph and 18 kmph respectively. What is the ratio of the time taken by B to that taken by D?
Calculation:
Speed of B $= 16\text{ kmph}$.
Speed of D $= 18\text{ kmph}$.
Since time taken is inversely proportional to speed ($\text{Time} \propto \frac{1}{\text{Speed}}$):
$$\frac{\text{Time}_B}{\text{Time}_D} = \frac{\text{Speed}_D}{\text{Speed}_B} = \frac{18}{16} = \frac{9}{8}$$
Correct Answer: (c) 9:8
Three bells chime at intervals of 18,24 and 32 minutes respectively. At a certain time they begin to chime together. What length of time will elapse before they chime together again?
Calculation:
The interval is the LCM of $18, 24,$ and $32\text{ minutes}$.
$18 = 2 \times 3^2$, $24 = 2^3 \times 3$, $32 = 2^5$.
$\text{LCM} = 2^5 \times 3^2 = 32 \times 9 = 288\text{ minutes}$.
$288\text{ minutes} = 4\text{ hours } 48\text{ minutes}$.
Correct Answer: (b) 4 hours 48 minutes
In a race of 200 meters run, Ashish beats Sunil by 20 metres and Nalin by 40 metres. If Sunil and Nalin are running a race of 100 metres with exactly the same speeds as before, then by how many metres will Sunil beat Nalin?
Calculation:
In a $200\text{ m}$ race: when Ashish runs $200\text{ m}$, Sunil runs $180\text{ m}$ ($200 - 20$) and Nalin runs $160\text{ m}$ ($200 - 40$).
Ratio of speeds of Sunil to Nalin $= \frac{180}{160} = \frac{9}{8}$.
In a $100\text{ m}$ race: when Sunil runs $100\text{ m}$, Nalin runs $100 \times \frac{8}{9} = \frac{800}{9} = 88.89\text{ m}$.
Sunil beats Nalin by $100 - 88.89 = 11.11\text{ metres}$.
Correct Answer: (a) 11.11 metres
A man invests ₹ 3000 at a rate of 5% per annum. How much more should he invest at a rate of 8%, so that he can earn a total of 6% per annum?
Calculation:
Let additional amount $= ₹x$ invested at $8\%$.
Total investment $= 3000 + x$.
Overall interest condition:
$$3000 \times 5\% + x \times 8\% = (3000 + x) \times 6\%$$$$150 + 0.08x = 180 + 0.06x \implies 0.02x = 30 \implies x = ₹1500$$
Correct Answer: (c) ₹1500
Use the following data for Questions 6 to 9:
Q. What is the shortest distance between M₁ and M₂?
Course details: Circular course radius $r = 14\text{ km}$ $\implies \text{Circumference } C = 2 \pi r = 2 \times \frac{22}{7} \times 14 = 88\text{ km}$.
Initial speeds: $v_H = 3v$, $v_R = v$.
Relative speed $= 3v + v = 4v$.
Time to first meeting $M_1$: Helitabh covers $\frac{3}{4}C = 66\text{ km}$ clockwise; Ruk Ruk covers $\frac{1}{4}C = 22\text{ km}$ counter-clockwise.
At $M_1$ (outside start), they reverse directions and swap speeds: Helitabh now has speed $v$, Ruk Ruk has speed $3v$.
Time to second meeting $M_2$: Helitabh covers $\frac{1}{4}C = 22\text{ km}$ back; Ruk Ruk covers $\frac{3}{4}C = 66\text{ km}$ back. They meet at $M_2$, which is located at $22\text{ km} + 22\text{ km} = 44\text{ km}$ from start (diametrically opposite to starting point $M_0$).
At $M_2$, they swap again. Helitabh (speed $3v$) covers $\frac{3}{4}C = 66\text{ km}$ to reach $M_3$. $M_3$ is at $44 + 66 = 110 \equiv 22\text{ km}$ mark from start.
At $M_3$, they swap again. Helitabh (speed $v$) covers $\frac{1}{4}C = 22\text{ km}$ to reach $M_4$, which is at $22 - 22 = 0\text{ km}$ mark ($M_0$).
Q6: Shortest distance between $M_1$ and $M_2$
$M_1$ is at $66\text{ km}$ mark ($270^\circ$ angle) and $M_2$ is at $44\text{ km}$ mark ($180^\circ$ angle).
Angle difference $= 90^\circ$.
Shortest straight-line distance $= \sqrt{r^2 + r^2} = r\sqrt{2} = 14\sqrt{2}\text{ km}$.
Correct Answer: (b) $14\sqrt{2}$ km
Use the following data for Questions 6 to 9:
Course details: Circular course radius $r = 14\text{ km}$ $\implies \text{Circumference } C = 2 \pi r = 2 \times \frac{22}{7} \times 14 = 88\text{ km}$.
Initial speeds: $v_H = 3v$, $v_R = v$.
Relative speed $= 3v + v = 4v$.
Time to first meeting $M_1$: Helitabh covers $\frac{3}{4}C = 66\text{ km}$ clockwise; Ruk Ruk covers $\frac{1}{4}C = 22\text{ km}$ counter-clockwise.
At $M_1$ (outside start), they reverse directions and swap speeds: Helitabh now has speed $v$, Ruk Ruk has speed $3v$.
Time to second meeting $M_2$: Helitabh covers $\frac{1}{4}C = 22\text{ km}$ back; Ruk Ruk covers $\frac{3}{4}C = 66\text{ km}$ back. They meet at $M_2$, which is located at $22\text{ km} + 22\text{ km} = 44\text{ km}$ from start (diametrically opposite to starting point $M_0$).
At $M_2$, they swap again. Helitabh (speed $3v$) covers $\frac{3}{4}C = 66\text{ km}$ to reach $M_3$. $M_3$ is at $44 + 66 = 110 \equiv 22\text{ km}$ mark from start.
At $M_3$, they swap again. Helitabh (speed $v$) covers $\frac{1}{4}C = 22\text{ km}$ to reach $M_4$, which is at $22 - 22 = 0\text{ km}$ mark ($M_0$).
Q7: Shortest distance between $M_1$ and $M_3$ along the course
$M_1$ is at $66\text{ km}$ mark, $M_3$ is at $22\text{ km}$ mark.
Distance along circumference $= 66 - 22 = 44\text{ km}$.
Correct Answer: (a) 44 km
Use the following data for Questions 6 to 9:
Course details: Circular course radius $r = 14\text{ km}$ $\implies \text{Circumference } C = 2 \pi r = 2 \times \frac{22}{7} \times 14 = 88\text{ km}$.
Initial speeds: $v_H = 3v$, $v_R = v$.
Relative speed $= 3v + v = 4v$.
Time to first meeting $M_1$: Helitabh covers $\frac{3}{4}C = 66\text{ km}$ clockwise; Ruk Ruk covers $\frac{1}{4}C = 22\text{ km}$ counter-clockwise.
At $M_1$ (outside start), they reverse directions and swap speeds: Helitabh now has speed $v$, Ruk Ruk has speed $3v$.
Time to second meeting $M_2$: Helitabh covers $\frac{1}{4}C = 22\text{ km}$ back; Ruk Ruk covers $\frac{3}{4}C = 66\text{ km}$ back. They meet at $M_2$, which is located at $22\text{ km} + 22\text{ km} = 44\text{ km}$ from start (diametrically opposite to starting point $M_0$).
At $M_2$, they swap again. Helitabh (speed $3v$) covers $\frac{3}{4}C = 66\text{ km}$ to reach $M_3$. $M_3$ is at $44 + 66 = 110 \equiv 22\text{ km}$ mark from start.
At $M_3$, they swap again. Helitabh (speed $v$) covers $\frac{1}{4}C = 22\text{ km}$ to reach $M_4$, which is at $22 - 22 = 0\text{ km}$ mark ($M_0$).
Q8: Point that coincides with $M_0$
$M_4$ is at $0\text{ km}$ mark, which coincides with $M_0$.
Correct Answer: (d) $M_4$
Use the following data for Questions 6 to 9:
Course details: Circular course radius $r = 14\text{ km}$ $\implies \text{Circumference } C = 2 \pi r = 2 \times \frac{22}{7} \times 14 = 88\text{ km}$.
Initial speeds: $v_H = 3v$, $v_R = v$.
Relative speed $= 3v + v = 4v$.
Time to first meeting $M_1$: Helitabh covers $\frac{3}{4}C = 66\text{ km}$ clockwise; Ruk Ruk covers $\frac{1}{4}C = 22\text{ km}$ counter-clockwise.
At $M_1$ (outside start), they reverse directions and swap speeds: Helitabh now has speed $v$, Ruk Ruk has speed $3v$.
Time to second meeting $M_2$: Helitabh covers $\frac{1}{4}C = 22\text{ km}$ back; Ruk Ruk covers $\frac{3}{4}C = 66\text{ km}$ back. They meet at $M_2$, which is located at $22\text{ km} + 22\text{ km} = 44\text{ km}$ from start (diametrically opposite to starting point $M_0$).
At $M_2$, they swap again. Helitabh (speed $3v$) covers $\frac{3}{4}C = 66\text{ km}$ to reach $M_3$. $M_3$ is at $44 + 66 = 110 \equiv 22\text{ km}$ mark from start.
At $M_3$, they swap again. Helitabh (speed $v$) covers $\frac{1}{4}C = 22\text{ km}$ to reach $M_4$, which is at $22 - 22 = 0\text{ km}$ mark ($M_0$).
Segment 1 (to $M_1$) $= 66\text{ km}$.
Segment 2 (to $M_2$) $= 22\text{ km}$.
Segment 3 (to $M_3$) $= 66\text{ km}$.
Total distance $= 66 + 22 + 66 = 154\text{ km}$.
Correct Answer: (a) 154 km
Directions for Questions 10 to 12: A certain race is made up of three stretches A, B and C, each 4 km long, and to be covered by a certain mode of transport. The following table gives these modes of transport for the stretches, and the minimum and maximum possible speeds (in kmph) over these stretches. The speed over a particular stretch is assumed to be constant. The previous record for the race is ten minutes.
Stretch | Mode of transport | Min. Speed | Max Speed |
| A | Car | 80 | 120 |
| B | Motor-cycle | 60 | 100 |
| C | Bicycle | 20 | 40 |
Q. Anshuman travels at minimum speed by car over A and completes stretch B at the fastest possible speed. At what speed should he cover stretch C in order to break the previous record?
Distance per stretch: $4\text{ km}$. Previous record $= 10\text{ minutes} = \frac{1}{6}\text{ hour}$.
Q10:
Time for stretch A (Min speed $80\text{ kmph}$) $= \frac{4}{80} = \frac{1}{20}\text{ hr} = 3\text{ min}$.
Time for stretch B (Max speed $100\text{ kmph}$) $= \frac{4}{100} = \frac{1}{25}\text{ hr} = 2.4\text{ min}$.
Time used $= 3 + 2.4 = 5.4\text{ min}$.
Time remaining to beat record $= 10 - 5.4 = 4.6\text{ min} = \frac{4.6}{60}\text{ hr}$.
Required speed for C $= \frac{4}{4.6/60} \approx 52.17\text{ kmph}$.
Since maximum speed for stretch C is $40\text{ kmph}$, achieving $>52.17\text{ kmph}$ is impossible.
Correct Answer: (c) This is not possible