# Free Basic Arithmetic - 05 Practice Test - CAT

### Question 1

Anil and Sunil can complete a job in 15 days and 20 days respectively. Both of them started it together. Anil left 6 days before the work was completed. Find the time taken (in days) to complete the work.

#### SOLUTION

Solution :C

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 (10015)+(10020)=11.67% , then 70% will complete in 6 days. Total 6+6= 12 days.Option (c).

### Question 2

The average of temperatures at noontime from Tuesday to Saturday is 50; the lowest one is 45, what can be the possible maximum range of the temperatures.

#### SOLUTION

Solution :B

The average of the 5 temperatures is: a+b+c+d+e5 = 50

One of these temps is 45: a+b+c+d+455 = 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).

### Question 3

A shopkeeper buys 50 footballs at Rs.100 per football. He sells a part of the footballs at a profit of 30% and on the remaining; he incurs a loss of 10% . If the overall profit is 10%, then the number of footballs sold at profit is?

#### SOLUTION

Solution :A

Total Cost Price = Rs 5000 . Total profit = 10% . Total Selling Price =1.1×5000=5500

Now go from answer options

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

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

Sum= 5500 = 10% profit on 5000

Thus, the assumption is right

### Question 4

A crew can row a certain course upstream in 84 minutes; they can row the same course downstream in 9 minutes less than they can row in still water. The time they would take to row down with the stream

#### SOLUTION

Solution :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

### Question 5

Three students Ankit, Abhash and Ajay appeared for an exam. If Ankit and Abhash secured 70 marks and 120 marks respectively, the average marks secured by all of them were 60% of the maximum marks. The passing marks in it were 40 less than their average marks and 40% of the maximum marks. The marks secured by Ajay

#### SOLUTION

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

Given that, 70+120+x3=60100×Y

190+x3=3Y570+120+x3−40=40100×Y

190+x3−40=2Y53Y5−40=2Y5

Y = 200 and X = 170

### Question 6

‘A’ borrowed money at 4% per annum simple interest payable yearly and lent it immediately at 6% per annum compound interest payable half-yearly and gained thereby 209 rupees at the end of a year. What sum of money did he borrow?

#### SOLUTION

Solution :A

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+6200)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.

### Question 7

An Aspirant scored 98 in his 19th Mock CSAT and thus increases his average by 4. What is his average after the 19th Mock CSAT.

#### SOLUTION

Solution :C

Let the average score after 18th Mock CSAT be x.

1st Method:- Weighted Average: 18x+9819=x+4;x=22 answer is x+4=26, option (c).

2nd Method:- Using AlligationUse the reverse gear approach now.

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

### Question 8

Next door lives a man with his son. They both work in the same factory. I watch them going to work through my window. The father leaves for work 10 min earlier than his son. One day I asked him about it and he told me he takes 30 min to walk to his factory ,whereas his son is able to cover the distance in only 20 min. I wondered if the father were to leave the house 5 min earlier than his son, how soon would the son catch up with the father?

#### SOLUTION

Solution :B

Let speed of son be s,and speed of father be f,then

sf=32 (inverse ratio of the time) ( If D is the distance ,then Df=30 and Ds=20 which implies sf=32)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).

### Question 9

Pranab is thrice as efficient as Ghulam, and Ghulam is twice as efficient as Krishna. If the trio work together, how long will they take to complete a job which Ghulam completes in 25 days.

#### SOLUTION

Solution :A

Let total work be 100 units.

Ghulam’s one day work = 10025 = 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 10018=509 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 1008 days i.e 509 days. Option(a).

### Question 10

Minister of Railways, Shri DineshTrivedi travels from Bangalore to Delhi at 50 km/hr and returns at 100 km/hr. What was his average speed?

#### SOLUTION

Solution :C

Average Speed =2×50×10050+100=10,000150=66.67 . Distance is kept constant, hence Speed and Time are inversely proportional, hence, we can use Harmonic Mean. H.M=(2aba+b) Option(c).

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

### Question 11

Minister of Power, Shri Sushilkumar Shinde and Minister of New and Renewable Energy, Dr. Farooq working together can do a piece of work in 18 days, Minister of New and Renewable Energy, Dr. Farooq and Minister of Petroleum and Natural Gas, Shri S. Jaipal Reddy in 24 days, Minister of Petroleum and Natural Gas and Minister of Power in 36 days. If all of them work together, how many days will it take?

#### SOLUTION

Solution :D

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 =118×72=4 units

Dr. Farooq and S. Jaipal Reddy’s 1 day work =124×72=3 units

S. Jaipal Reddy and Sushilkumar Shinde’s 1 day work =136×72=2 units

Combined work of Ministers =4+3+22=92 units

Hence, they will take 729×2=16 days;

Option (d).

### Question 12

The population of a town was 3600 three years back. It is 4800 right now. . What is the rate of growth of population, if it has been constant over the years and has been compounding annually?

#### SOLUTION

Solution :D

Option (d)

Conventional approach :

Final Value=Initial Value(1+r100)t

Thus, rate =(FVIV)1t−1

⇒FVIV=4836=1.33

Going from answer options

If 10 is the answer

1.13=112×11=12×11≈1.33

Therefore r is approximately 10%.

### Question 13

Amit and Francis are on detention and they need to write 65 pages together, Amit writing for an hour extra than Francis. Francis can write 2 pages/hour more than Amit and therefore, he did 5 pages more than Amit.What is the speed per hour of Francis? (pages/hr)

#### SOLUTION

Solution :C

Option (c )

Total = 65 pages

Go from answer options

Option (a)→F=6

A= 4

Together 1024 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) =3525×60,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)

### Question 14

20% of a larger number is 2.3 less than 30% of a smaller number. The larger number also exceeds the smaller number by 10. What is the value of the larger number?

#### SOLUTION

Solution :D

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)

### Question 15

Ram bought two cars and C.P of 1st car = 150000, C.P of 2nd car = 175000. He sold first car at a profit of 30% and second car at a loss of 10%. What is his profit?

#### SOLUTION

Solution :1st car:

C.P = 150000

S.P = 195000

Profit = 45000

2nd car:

C.P = 175000

S.P = 157500

loss = 17500

Overall profit = 45000 - 17500 = 27500