Free Basic Arithmetic - 05 Practice Test - CAT 

Let total work is LCM of (15, 20) = 60

Work done by Anil in a day = 4 units

Work done by Sunil in a day = 3 units

Anil worked for (t – 6) days and Sunil worked for t days.

4(t – 6) + 3t = 60

t = 12; Option (c).  

2nd method:- During last 6 days Sunil alone will work i.e  $30\%$  of work will be done by Sunil alone. Remaining  $70\%$ of work will be done together. In one day both will do  $\left(\frac{100}{15}\right)+\left(\frac{100}{20}\right)=11.67\%$ , then $70\%$ will complete in 6 days. Total 6+6= 12 days.Option (c).

 The average of the 5 temperatures is:  $\frac{a+b+c+d+e}{5}=50$

One of these temps is 45: $\frac{a+b+c+d+45}{5}=50$.

Solving for the variables: a + b + c + d = 205

In order to find the greatest range of temps, we minimize all temps but one. Remember, though, that 45 is the lowest temp possible, so: 45 + 45 + 45 + d = 205 Solving for the variable: d = 70

70 - 45 = 25 Option (b).

2nd method:- For finding maximum range, we will take 4 days minimum out of 5 days i.e 4 days temp. will take 45, then total negative deviations will be 20. Thus for average 50, positive deviation should be also 20 . Thus highest temp will be (50+20)=70 degrees. Then range =70-45 = 25. Option (b).

Total Cost Price = Rs 5000 . Total profit =  $10\%$ . Total Selling Price  $=1.1\times 5000=5500$

Now go from answer options

If the number of footballs sold at profit is 25, then revenue for these 25 balls $=1.3\times 25\times 100=3250$

And the number of footballs sold at loss will be 25; revenue for these balls $=0.9\times 25\times 100=2250$

Sum= 5500 =  $10\%$ profit on 5000

Thus, the assumption is right

As the distance is kept constant, time is inversely proportional to speed and hence the time taken is in HM. Let the time required for rowing in still water be x, then rowing time down with stream will be x-9; x  is the HM of 84 and x-9. Therefore x= 72 required time = x – 9 = 63

Let Ajay’s score be x and maximum marks be Y.

Given that, $\frac{70+120+x}{3}=\frac{60}{100}\times Y$

$\frac{190+x}{3}=\frac{3Y}{5}$

$\frac{70+120+x}{3}-40=\frac{40}{100}\times Y$

$\frac{190+x}{3}-40=\frac{2Y}{5}$

$\frac{3Y}{5}-40=\frac{2Y}{5}$

Y = 200 and X = 170

Let amount borrowed be Rs. X

Amount to be paid back = Rs. 1.04X

Amount received due to lending at 6% compound interest payable half -yearly = $x(1+\frac{6}{200}{)}^2=1.0609x$

Profit = 1.0609X – 1.04X = 209

0.0209X = 209, X = 10,000; ‘A’ borrowed Rs.10, 000 initially. Option(a)

Alternate Solution

We see that the answer options are far apart and hence the exact calculations are not required. There is an increase of 2% and he is gaining 209 rupees in one year.Hence the amount has to be close to 10000 and hence the answer.

Let the average score after ${18}^{th}$  Mock CSAT be x.

$1^{st}$  Method:- Weighted Average:  $\frac{18x+98}{19}=x+4;x=22$ answer is  $x+4=26$, option (c).

$2^{nd}$  Method:- Using Alligation

Use the reverse gear approach now.

Put X = 22; we will get 18: 1; answer is x + 4 = 26, option (c).

Let speed of son be s,and speed of father be f,then
$\frac{s}{f}=\frac{3}{2}$ (inverse ratio of the time) ( If D is the distance ,then  $\frac{D}{f}=30$ and $\frac{D}{s}=20$ which implies $\frac{s}{f}=\frac{3}{2}$)

Let speed of dad be 20 and son be 30m/s

Now in 5 mins, the father covers 100 m

And son covers 150 m

The son covers 100 m in 3.33 min,

Hence the son will be able to catch up with the father after 8.3 mins from the start. Option (b).

 Let total work be 100 units.

Ghulam’s one day work = $\frac{100}{25}$ = 4 units.

Pranab’s one day work = 12 units.

Krishna’s one day work = 2 units.

Combined work of the trio = 4 + 12 + 2 = 18 units.

Hence, They will take  $\frac{100}{18}=\frac{50}{9}$ days. Option(a).  

2nd method:- In place of unit, we can take in percentage form also.In one day, $G=4\%,P=12\%,K=2\%$

 Total in one day= $18\%$ , Then  $100\%$ in $\frac{100}{8}$ days i.e $\frac{50}{9}$ days. Option(a).

Average Speed  $=\frac{2\times 50\times 100}{50+100}=\frac{10,000}{150}=66.67$ . Distance is kept constant, hence Speed and Time are inversely proportional, hence, we can use Harmonic Mean.  $H.M=\left(\frac{2ab}{a+b}\right)$   Option(c).

Alternative,   Average  speed= total distance / total time= 66.67. Option (c).

The LCM of 18, 24 and 36 is 72. Let total work is be 72 units.

Sushilkumar Shinde and, Dr. Farooq’s 1 day work  $=\frac{1}{18}\times 72=4units$

Dr. Farooq and S. Jaipal Reddy’s 1 day work   $=\frac{1}{24}\times 72=3units$

S. Jaipal Reddy and Sushilkumar Shinde’s 1 day work   $=\frac{1}{36}\times 72=2units$

Combined work of Ministers  $=\frac{4+3+2}{2}=\frac{9}{2}$  units

Hence, they will take  $\frac{72}{9}\times 2=16$  days;

Option (d).

Option (d)

Conventional approach :

$FinalValue=InitialValue{(1+\frac{r}{100})}^t$

Thus, rate  $={\left(\frac{FV}{IV}\right)}^{\frac{1}{t}}-1$

$\Rightarrow \frac{FV}{IV}=\frac{48}{36}=1.33$

Going from answer options

If 10 is the answer

${1.1}^3={11}^2\times 11=12\times 11\approx 1.33$

Therefore r is approximately  $10\%.$

Option (c )

Total = 65 pages

Go from answer options

Option $\left(a\right)\to F=6$

A= 4

Together  $\frac{10}{24}$  in one hour . This multiplied by (65-4=61),

should be an integer which is not true. Hence, answer a is eliminated Similarly, options (b),(d)are eliminated.

Option (c)  $=\frac{35}{25}\times 60,$which is an integer

Reverse gear method 2

If Francis writes 35 pages, then Amit writes 30 pages. Use answer options, to find each person’s individual time the time difference should be one hour. This happens only with option (c)

Let the larger number and smaller number is L and S respectively.

$0.2L=0.3S-2.3\&2L-3S=-23$

$L-S=10\&2L-2S=20$

Solving the above two equations we get L = 53. Hence option (e)