## Twelfth SSC CGL level Solution Set, topic Arithmetic mixed topics

This is the twelfth solution set of 10 practice problem exercise for SSC CGL exam on Arithmetic mixed topics. Students must complete the corresponding question set in prescribed time first and then only refer to this solution set.

### Twelfth solution set- 10 problems for SSC CGL exam: Arithmetic mixed topics - time 12 mins

**Q1. **If 9 men working for 9 hours a day for each of the 9 days produce 108 units of work then, 6 men working 6 hours a day for each of the 6 days will produce units of work,

- 32
- 64
- 72
- 24

** Solution:**

Total Man Hours is $9\times{9}\times{9}=9^3$ MH and this produces 108 work units.

So in 1 MH, work done $= \displaystyle\frac{108}{9^3}$ work units.

In the second case we have, number of Man Hours $= 6\times{6}\times{6}=6^3$.

Thus by following Unitary method we can be certain that this amount of Man hours will produce an amount of work,

$=\displaystyle\frac{108}{9^3}\times{6^3}=\displaystyle\frac{108}{3^3}\times{8}=\displaystyle\frac{12}{3}\times{8}=32$ units of work.

**Answer:** Option a: 32.

**Key concepts used:** *Breakthrough concept* used, Man hours producing equivalent amount of work produced -- unitary method. By using a single concept of man hours (it could have been man days), number of men, number of hours per day and number of days all could be combined into one value.

* Aside*:

*This is the* **Mandays technique***that we have elaborately used in our article on* * Solving Time and work problems in a few simple steps part 2*.

*With the use of this technique derivation becomes much simpler and clearer because of number of men and number of days merge together into one variable that is the amount of work.*

**Q2.** A swimmer swims from his starting point against current for 10 minutes and then swims backwards for next 10 minutes to reach a point distant 1000 meters from the starting point. The speed of the current then is,

- 2 km/hr
- 3 km/hr
- 4 km/hr
- 6 km/hr

**Solution:**

On the way back in favor of the current the swimmer overshoots his starting point by 1000 meters.

Let us imagine this case as equivalent to two objects moving in the same direction (by reversing the way back swim) where they started from the same point but the faster one overtook the slower one at their relative speed.

The * relative speed in this case is the difference of downstream and upstream speeds* which

*.*

**equals twice the stream speed****Downstream speed** = Still water speed $+$ Stream speed, and

**Upstream speed** = Still water speed $-$ Stream speed.

So, **Relative speed** = Downstream speed $-$ Upstream speed = 2$\times{}$(Stream speed)

At this speed then the faster object covers 1000 meters in 10 minutes of movement, that is, in one sixth of an hour. So the relative speed is 6 km/hr which is twice the stream speed. Thus stream speed is 3 km/hr.

**Answer:** Option b : 3 km/hr.

**Key concepts used:** Basic concepts of time and distance -- separating out the two swims, forward and backward, enabled reversing the direction of the backward swim and conversion of the two swims into a race in the same direction of two moving objects moving at different speeds -- knowledge that relative speed will be difference of downstream and upstream speed which always is twice the stream speed, still water swimmer speed canceling out.

No equation, no variables, just a **quick conceptual solution.**

**Q3.** A tank has three pipes, the pipes A and B together can empty the tank in half the time that C, the third pipe, takes to empty it alone. Pipe A working alone takes half the time taken by pipe B to empty the tank alone. Together the three pipes take 6 hours 40 minutes to empty the tank. The time that the pipe A takes to empty the tank alone in hrs is,

- 7
- 15
- 10
- 30

**Solution:**

Let us assume the rate of emptying the tank per hour (fraction of tank volume per hour) of the three pipes be $A$, $B$ and $C$.

From first statement,

$\displaystyle\frac{1}{2}(A + B)=C$.

From second statement,

$\displaystyle\frac{1}{2}A = B$.

Combining the two,

$\displaystyle\frac{1}{2}\left(A + \displaystyle\frac{1}{2}A\right)=\displaystyle\frac{3}{4}A=C$

From third statement, noting that the three pipes worked together for 6 hours 40 minutes or $6\displaystyle\frac{2}{3}=\displaystyle\frac{20}{3}$ hours to do the whole work $W$ of emptying the tank, we have,

$\displaystyle\frac{20}{3}(A + B + C) $

$= \displaystyle\frac{20}{3}\left(A + \displaystyle\frac{1}{2}A + \displaystyle\frac{3}{4}A\right)$

$=15A$

$=W$

So pipe A alone will take 15 hours to empty the tank.

**Answer:** Option b: 15.

**Key concepts used:** This is by far the quickest and easiest approach to the solution by * assuming the variables as the rates of work per hour for each of the worker*. This simplifies the whole deduction of such types of

**pipe and tank**or

**time and work problems**. The

**only difference between the two types of problems**is,

*in time and work problems, there usually are no destructive (read tank emptying pipe) worker in any problem.*

*Aside:*

*For detailed reference of use of this powerful simplifying concept of Definition tuning approach refer to our article on simplification of Time and Work problems in a few steps part 1 and part 2.*

**Q4. **If A and B can finish a job together in 20 days, B and C together in 10 days and C and A in 12 days then, A, B and C together will finish the work in days,

- $8\displaystyle\frac{4}{7}$
- $30$
- $\displaystyle\frac{7}{60}$
- $4\displaystyle\frac{2}{7}$

**Solution:**

Just like before, assuming rate of work of the three persons as A, B and C in terms of work portion done per day, we have by the three statements,

$20(A + B)=W$,

Or, $(A + B)=\displaystyle\frac{1}{20}W$.

$10(B + C) = W$,

Or, $(B + C) = \displaystyle\frac{1}{10}W$.

$12(C + A)=W$,

Or, $(C + A) =\displaystyle\frac{1}{12}W$.

Adding the three we get,

$2(A + B + C) = W\left(\displaystyle\frac{1}{20} + \displaystyle\frac{1}{10} + \displaystyle\frac{1}{12}\right)$

$\hspace{10mm} = \displaystyle\frac{28}{120}$,

Or, $\displaystyle\frac{120}{14}\left(A + B + C\right) = W$

So the three persons will take together $8\displaystyle\frac{4}{7}$ days to finish the job.

**Answer:** Option a: $8\displaystyle\frac{4}{7}$.

**Key concepts used:** Assuming variables as rate of work (in terms of work portion) per day for each person enabled us to avoid inverse variables and directly assess the effect of more than one of them working together.

*Aside:*

*For detailed reference of use of this powerful simplifying concept of Definition tuning approach refer to our article on simplification of Time and Work problems in a few steps part 1 and part 2.*

**Q5. **A seller gives 4% discount on his marked price. What should be his marked price when the cost price is Rs. 200 and he wants to make a profit of 20%?

- Rs. 192
- Rs. 240
- Rs. 250
- Rs. 260

**Solution:**

Profit is to be Rs. 40.

So, Sale Price SP must be

= Rs. 240 = 96% of $MP$ (because of 4% discount on Marked Price MP). Thus,

$MP = \displaystyle\frac{240}{.96} = \displaystyle\frac{10}{.04} = $Rs. 250.

**Answer:** Option c: Rs. 250.

**Key concepts used:** *Discount percentage is on Marked Price* while *profit percentage is on Cost Price*. Also, Profit = Sale Price $-$ Cost Price.

**Q6.** If a mobile telephone bill is paid before due date, the company gives a rebate of 4%. A person saves Rs. 13 by paying his bill before due date. What was his billed amount (in Rs.)?

- 225
- 125
- 325
- 425

**Solution:**

Rebate or discount of 4% of total amount = Rs. 13,

Or, 100% of total amount = 13 times 25 = Rs. 325.

**Answer:** Option c : 325.

**Key concepts used:** Basic percentage concepts. * Discount or rebate is always on final total amout*, it may be Marked Price or Billed Amount.

** Q7.** The compound interest on a certain amount of money at a certain rate per year for two years is Rs. 2050 and the simple interest on same amount at same rate for three years is Rs. 3000. The amount of money is,

- Rs. 20,000
- Rs. 21,000
- Rs. 25,000
- Rs. 18,000

**Solution: **

Simple interest for 3 years = Rs. 3000. As * simple interest remains same for every year*,

Simple interest per year is Rs. 1000. For two years the simple interest would then be Rs. 2000.

The **extra Rs. 50 of compound interest** **for the second year **is** on Rs. 1000**.

So compound interest rate is $\displaystyle\frac{50}{1000} = 5$%.

This is same as the simple interest rate.

So, simple interest for one year Rs. 1000 = 5% of total amount (this is the rule of simple interest).

Thus total amount is, 100% = Rs. 20,000.

**Answer:** Option a: Rs. 20,000.

** Key concepts used:** Isolating the extra interest of Rs. 50 accrued over the second year -- deducing that this extra interest is on the interest at the end of second year produced at the rate of interest -- from amount of simple interest in a year and the rate of interest finding out the actual amount of money.

** Q8.** A tank is normally filled in 8 hours but takes another 2 hours more because of a leak in the tank. How long would it take for the full tank to get empty because of the leak?

- 25 hours
- 16 hours
- 40 hours
- 20 hours

** Solution:**

Let $A$ and $L$ be the filling and leaking rates in terms of portion of tank per hour.

**Aside:**

*Here we have used the Rate of work (emptying tank in this case) assumption technique, Definition tuning approach, to avoid dealing with inverse of variables. We have used this technique extensively in our articles on*

**Solving Time and Work problems in a few simple steps****part 1***and*

*.*

**part 2***Time and work problems and Pipes and cisterns problems are exactly similar except that tank draining (or emptying) is equivalent to work destruction which we haven't encountered yet in Time and work problems.*

When both filling and the leak worked together to fill the tank in 10 hours, we have,

$10(A - L) = Full$,

Or, $(A - L) = \displaystyle\frac{1}{10}Full$

Again, $8A = Full$,

Or, $A = \displaystyle\frac{1}{8}Full$.

Subtracting the first from the second,

$L = \left(\displaystyle\frac{1}{8} - \displaystyle\frac{1}{10}\right)\times{Full} = \displaystyle\frac{1}{40}Full$.

So, the leak will take 40 hours to empty the full tank.

**Answer:** Option c: 40 hours.

**Key concepts used:** Assuming per hour rates (in terms of portion of tank) of filling and emptying by the leak enabled formation of simple linear equations in two variables to eliminate filling rate. By this approach, inverse of variables is totally avoided.

**Q9.** If the rate of interest is 4% per annum in the first year and 3% per annum for the second year, the compound interest on Rs. 2000 in 2 years will be,

- Rs. 140.40
- Rs. 142.40
- Rs. 143.40
- Rs. 141.40

**Solution:**

First year interest = 4% of Rs. 2000 = Rs. 80.

Second year interest = 3% of Rs. 2000 + 3% of Rs. 80 = Rs. 62.40.

Total interest = Rs. 142.40.

**Answer:** Option b: Rs. 142.40.

**Key concepts used:** Basic concept of compound interest that, interest on each period is the sum of interest (at the given rate) on the original amount and the interest (at the given rate) on the amount of interest earned till the last period.

** Q10.** The marked price of trousers and shirts are in the ratio of 2 : 1. The seller gives 40% discount on the shirt. If the total discount on the set of trousers and shirt is 30%, the discount offered on the trousers is,

- 30%
- 25%
- 20%
- 15%

**Solution:**

Let us apply the * Portions use technique* with the knowledge that

**marked price of shirt**is 1 portion out total 3 and that of trousers is 2 portions out of total 3.

So total discount is = 30% on total **3 price portions** = 40% on **1 price portion** (for shirt) + $x$% on **2 price portions** (for trousers).

Or, $3\times{0.3} = 0.4 + 2x$, price portion, whatever be the amount, cancels out.

Or, $x = \frac{0.9 - 0.4}{2}=0.25 = 25$%.

**Answer:** Option b: 25%.

**Key concepts used:** In problems involving ratios, * Portions use technique* simplifies derivation significantly.

*Aside:*

*The Portion use technique is a powerful rich ratio concept that simplifies many of the tough ratio problems. You can refer to detailed application of this technique in the articles on*

**Simplifying liquid mixing problems in a few steps part 1***and*

**part 2.**