Basic Arithmetic(Exams > Cat > Quantitaitve Aptitude ) Questions and answers for exam Preparation

Exams > Cat > Quantitaitve Aptitude

BASIC ARITHMETIC MCQs

Total Questions : 150 | Page 1 of 15 pages

Question 1. A started a business with Rs. 52,000 and after 4 months B joined him with Rs.39,000. At the end of the year, out of the total profits B received total Rs. 20,000 including

25 %

of the profits as commission for managing the business. What amount did A receive?

Rs. 70,000
Rs. 60,000
Rs. 20,000
Rs. 40,000
Rs. 2,00,000

Discuss Question

Answer: Option C. -> Rs. 20,000
:
C
Profit's share of A and B

= \frac{(52, 000 \times 12)}{(39, 000 \times 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

{\frac{1}{3}}^{r d}

part of this in addition to his commission. Hence his total earning

= 0.25 x + \frac{1}{3} \times 0.75 x

= 0.5 x = 20, 000

x = 40, 000

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

Question 2. 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

{\frac{1}{3}}^{r d}

and

{\frac{2}{3}}^{r d}

of the balls from the

1^{s t}

basket and

2^{n d}

basket respectively are exchanged. The ratio of balls in basket 2 to basket1 is?

Discuss Question

Answer: Option D. -> 1:2
:
D
Option (d)

= \frac{\frac{2}{3} b + \frac{2}{3} a}{\frac{1}{3} b + \frac{1}{3} a} = 2 : 1

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

Question 3. 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)

Discuss Question

Answer: Option C. -> 7
:
C
Option (c )
Total = 65 pages
Go from answer options
Option

(a) \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)

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___

Discuss Question

:
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

Question 5. 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?

3.33 mins
8.3 mins
5.76 mins
8.25 mins

Discuss Question

Answer: Option B. -> 8.3 mins
:
B
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).

Question 6. 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___

Discuss Question

:
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{3 Y}{5}

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

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

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

Y = 200 and X = 170

Question 7. Fresco has doubled its turnover in 2000 from 5 crores in 1999, tripled its turnover in 2001 and grew by

50 %

in 2002.The turnover at the end of 2002 is___

Discuss Question

:
1999 =5 crores

2000 = 5 \times 2 = 10 c r o r e s

2001 = 10 \times 3 = 30 c r o r e s

2002 = 30 \times 1.5 = 45 c r o r e s

Question 8. The ratio of milk and water in the container is 3: 2 and when 10 litres of the mixture is taken out and is replaced by water the ratio becomes 2: 3. The total quantity of the mixture in the container is.

15 ltr
10 ltr
12 ltr
30 ltr
18 ltr

Discuss Question

Answer: Option D. -> 30 ltr
:
D
Suppose the total quantity is x litres.
So, the required ratio 2: 3

\frac{3 x - 6}{2 x + 6} = \frac{2}{3} \Rightarrow x = 10 \Rightarrow 3 x = 30

So, the quantity of the mixture was 30 ltr.

Question 9. 24 men can complete a work in 20 days. How many more men are required to complete the work in 12 days?

Discuss Question

Answer: Option C. -> 16
:
C
Method 1:Using man-days
Amount of work in terms of man days

= 24 \times 20 = 480

man days
So, number of men required to complete the work in 12 days

= \frac{480}{12} = 40

men.
16 more men are required. Hence option (c)
Method 2: Using constant product rule
Decrease in number of days

= \frac{1}{(\frac{20}{8})} = \frac{1}{2.5}

Increase in number of men

= \frac{1}{1.5} =

16 men to finish the same work.

Question 10. Arun can finish a work in 12 days; Ajit can finish the same work in 15 days while Amit can finish the same work in 20 days. Find the total time taken when all three work together to complete the work.