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

Exams > Cat > Quantitaitve Aptitude

BASIC ARITHMETIC MCQs

Total Questions : 150 | Page 5 of 15 pages

Question 41. Deepak invested Rs. 50000 in a bond which was giving him

10 %

interest per annum at compound interest. Find the money received by Deepak after 3 years if he has to pay

20 %

tax on the compound interest every year.

Discuss Question

Answer: Option E. -> Rs. 62985.6
:
E
Deepak was getting

10 %

interest,

20 %

of the interest i.e.

20 %

10 %

(= 2 %)

was the tax which he had to pay, so the net interest rate (after tax)

= 10 - 2 = 8 %

So, amount received by him after 3 years

= 50000 {(1 + (\frac{8}{100}))}^{3} = R s .62985 .6

Hence option (5)
2nd method:-By, pascal triangles for

n = 3 & r = 8 %

. Option (5).

Question 42. Pankaj takes a loan of Rs. 10000 at

10 %

per annum compounded annually which is to be repaid in two equal annual instalments: one at the end of the first year and second at the end of the second year. The value of each instalment is:

Discuss Question

Answer: Option A. -> Rs. 5760
:
A
Let the value of each installment is X, Total amount is P

P = X (K + K^{2} + K^{3} + . . . . . . . . . . . . . . . . . . .), W h e r e K = \frac{100}{100 + r}

Here only two years and r is

10 %

then

P = X (K + K^{2})

10000 = x {\frac{10}{11} + {(\frac{10}{11})}^{2}}

X = 5761.9
Alternatively

10, 000 (1.1)^{2} = x [1 + (1.1)]

x = 5761.9

Question 43. In what ratio should three varieties of chemical A, B and C at the rate Rs. 120, Rs. 144 and Rs. 174 per litre be mixed to come up with a mixture worth Rs. 141 per litre?

Discuss Question

Answer: Option C. -> 11: 77: 7
:
C
First mix varieties

A & C

in a ratio to get a mixture at the cost Rs. 141 per litre.
Applying Alligation:

In What Ratio Should Three Varieties Of Chemical A, B And C ...

Now mix varieties

A & B

to get a mixture at the cost of Rs. 141 per litre.

So, A: C in ratio 11: 7 and A: B in ratio 1: 7.The required ratio A: B: C = 11: 77: 7. Hence option (c).
2nd method:- By, observation, 141 is nearest to B (144) then A(120) and then C(174). When we mix all three then highest contribution will be according the above, i.e

B \to A \to C

, Now only option (c).

Question 44. 4 men and 2 boys can do a job in

6 \frac{2}{3}

days. 3 women and 4 boys can finish the same job in 5 days. Also 2 men and 3 women can finish the same job in 4 days. In how many days can 1 man, 1 woman and 1 boy finish the work at their double efficiency___

Discuss Question

:
In 1 day 4 men and 2 boys can do

[\frac{100}{(\frac{20}{3})}] % = 15 %

of the job. ... (i)
In 1 day 3 women and 4 boys can do

\frac{100}{5} = 20 %

of the job.... (ii)
In 1 day 2 men and 3 women can do

\frac{100}{4} = 25 %

of the job.... (iii)
So, adding (i), (ii), (iii)
6 men, 6 women and 6 boys can do 15 + 20 + 25 =

60 %

of the job in 1 day.
So, 1 man, 1 woman and 1 boys can do

10 %

of the job in a day.
All double their efficiency, they can do

20 %

of the job in a day.
They can complete the job in

\frac{100}{20}

= 5 days.

Question 45. A man can finish a work in 12 days while B can finish the same work in 15 days. If both work together, then calculate the time taken to complete the work.

6(23) days
6 days
8 days
6(34) days

Discuss Question

Answer: Option A. -> 6(23) days
:
A
A can finish the work in 12 days.
He can finish

100 %

of the work in 12 days.
In 1 day he can finish

(\frac{100}{12}) % = 8.33 %

of the work.
Similarly B can finish

(\frac{100}{15}) % = 6.67 %

of the work in 1 day.
When they both work together, they can finish

(8.33 + 6.67) % = 15 %

of the work in 1 day
So, to complete

100 %

of the work, they will take

(\frac{100}{15}) e = 6 (\frac{2}{3})

days. Hence option (a).
2nd method:-Total work (W) = 12A = 15B ,

B = \frac{4}{5} A

n(A+B) = W = 12A or 15B (Here, we take 12A).

n (A + B) = 12 A, n (A + \frac{4}{5} A), n = \frac{20}{3} = 6 (\frac{2}{3})

days. Option (a).

Question 46. From a container 6 litres milk was drawn out and was replaced by water. Again 6 litres of mixture was drawn out and replaced by water. Thus the quantity of milk in the container after these two operations is 9: 16. Calculate the quantity of the milk initially in the container:

Discuss Question

Answer: Option B. -> 15 ltr
:
B
After these two transfers = milk is left in the container in ratio 9: 25

\frac{(k - 6)^{2}}{k^{2}} = \frac{9}{25 k} = 15.

Hence option (b)

Question 47. A can finish a work in 16 days while B can finish the work in 8 days. After A started the work alone, B joined him after 4 days. The total time taken to finish the work is___

Discuss Question

:
A can finish the work in 16 days. In one day he can finish

(\frac{100}{16}) %

of the work.
In 4 days, working alone he will finish

[(\frac{100}{16}) \times 4] = 25 %

of the work.
Now B joins him.
Amount of work left = 100 – 25 =

75 %

B can do

(\frac{100}{8}) %

of the work.
A and B together can do

(\frac{100}{8}) % + (\frac{100}{16}) % = (\frac{300}{16}) %

of the work.
Work left

= 75 %

. Hence time taken

= [\frac{(75 \times 16)}{300}] = 4

days.
So, total time taken = 4 + 4 = 8 days.
2nd method:- In 4 days A will finish

25 %

of work. Remaining work is

75 %

, Efficiency of B is double than A i.e

50 %

work will done by B and

25 %

will done by A. Thus it will take 4 days. Total 8 days.

Question 48. As a worker, Ajit is thrice as good as Dev. If both working together can finish the work in 5 days, then determine the time taken by Ajit to complete the work alone.

103 days
5 days
203 days
7 days

Discuss Question

Answer: Option C. -> 203 days
:
C
Suppose Ajit can finish the work in ‘x’ days. In one day he can do

(\frac{100}{x}) %

of the work.
As Ajit is thrice as good as Dev, Dev will do

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

of what Ajit can do in a day.
So, Dev can do

(\frac{100}{3 x}) %

of work in a day.
Now they complete the work in 5 days, so in 1 day they must be doing

\frac{100}{5} = 20 %

of the work.
So,

(\frac{100}{x}) + (\frac{100}{3 x}) = 20

x = (\frac{20}{3})

So, Ajit can complete the work alone in

(\frac{20}{3})

days. Hence option (d).
2nd method:- Efficiency

A = 3 D, 5 (A + D) = n A (i . e .,) 5 \times \frac{4}{3} A = n A, n = \frac{20}{3}

days.

Question 49. Pipe A can fill a tank in 4 hours while Pipe B can empty the same tank completely in 5 hours. If both the pipes are working simultaneously, then determine the time in which an empty tank will be completely filled.

Discuss Question

Answer: Option C. -> 20
:
C
Pipe A can fill a tank completely in 4 hours, Thereforein 1 hour, it can fill

\frac{100}{4} = 25 %

of the tank.
Pipe B can empty the tank in 5 hours, Therefore in 1 hour, it can empty

\frac{100}{5} = 20 %

of the tank.
So, when both are simultaneously working, then in 1 hour

5 %

of the tank will be filled.
For filling the tank completely, it will take

\frac{100}{5}

= 20 hours.
2nd method:-

(\frac{1}{4}) - (\frac{1}{5}) = \frac{1}{20}

. Thus tank will fill in 20 days.

Question 50. A and B can do a work in 12 days, B and C can do the same work in 15 days and A and C can complete the work in 20 days.Days required by A, B and C working together is (in days) ___

Discuss Question

:
Let a, b and c be the % of the work done by A, B and C in one day respectively.