**INDUSTRY AND COMPANY AWARENESS (ICA)**

### Exams > Cat > Quantitaitve Aptitude

## BASIC ARITHMETIC MCQs

Total Questions : 150
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**1**of 15 pages**Answer: Option C. ->**Rs. 20,000

:

C

Profit's share of A and B =(52,000×12)(39,000×8)=2:1

Let the profit be Rs. x, then B receives 25% as commission for managing business, the remaining 75% of the total profit x is shared between A and B in the ratio 2:1. Hence B will get 13rd part of this in addition to his commission. Hence his total earning

=0.25x+13×0.75x

=0.5x=20,000

x=40,000

So,the remaining profit goes to A,hence the profit of A is Rs.20,000.

**There are 2 baskets with n identical balls. Half the balls from basket 1 are put into basket 2. The ratio of balls in Basket 1 to Basket 2 gets reversed. Next 13rd and 23rd of the balls from the 1st basket and 2nd basket respectively are exchanged. The ratio of balls in basket 2 to basket1 is?**

__Question 2.__**Answer: Option D. ->**1:2

:

D

Option (d)

=23b+23a13b+13a=2:1

Required ratio(basket 2: basket 1) = 1:2

**Answer: Option C. ->**7

:

C

Option (c )

Total = 65 pages

Go from answer options

Option (a)→F=6

A= 4

Together 1024in 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)

:

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= 72required time = x – 9 = 63

**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?**

__Question 5.__**Answer: Option B. ->**8.3 mins

:

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=30and Ds=20which 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).

**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___**

__Question 6.__:

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

Given that, 70+120+x3=60100×Y

190+x3=3Y5

70+120+x3−40=40100×Y

190+x3−40=2Y5

3Y5−40=2Y5

Y = 200 and X = 170

:

1999 =5 crores

2000=5×2=10crores

2001=10×3=30crores

2002=30×1.5=45crores

**Answer: Option D. ->**30 ltr

:

D

Suppose the total quantity is x litres.

So, the required ratio 2: 3

3x−62x+6=23⇒x=10⇒3x=30

So, the quantity of the mixture was 30 ltr.

**Answer: Option C. ->**16

:

C

**Method 1:**Using man-days

Amount of work in terms of man days =24×20=480 man days

So, number of men required to complete the work in 12 days =48012=40men.

16 more men are required. Hence option (c)

**Method 2:**Using constant product rule

Decrease in number of days =1(208)=12.5

Increase in number of men =11.5=16 men to finish the same work.

**Answer: Option B. ->**5 days

:

B

Arun can finish 100% of work in 12 days he can finish (10012)=8.33%of the work in a day.

Ajit can finish 100% of work in 15 days he can finish (10015)=6.67% of the work in a day.

Amit can finish 100% of the work in 20 days he can finish(10020)=5% of the work in a day.

So, if all three work together, then they can finish (8.33+6.67+5)%=20% of the work in a day.

So, they can complete the work in (10020)= 5 days. Hence option (B).

**2nd method:-**W= 12A=15B=20C .

n(A+B+C)=12A,n(A+34A+35A)=12A,n=5days.

Option (2).