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

Exams > Cat > Quantitaitve Aptitude

BASIC ARITHMETIC MCQs

Total Questions : 150 | Page 4 of 15 pages

Question 31. In a college

50 %

students are boys out of which

40 %

have got

80 %

and above marks. Overall,

45 %

of all the students have got

80 %

and above marks. Find the percentage of girls who have got less than

80 %

marks.

33%
60%
50%
40%
Cannot be determined

Discuss Question

Answer: Option C. -> 50%
:
C
Suppose 100 students are in the college: 50 boys and 50 girls. Now

40 %

of the boys i.e. 20 boys have got

80 %

and above marks. And Total 45 students have got

80 %

and above marks. So, number of girls who have got

80 %

and above = 25. Total number of girls = 50. Hence

50 %

of the girls have got less than

80 %

marks.

Question 32. On a certain sum, the difference between the compound interest and the simple interest for the second year is Rs.4500 and the same for the third year is Rs.14175. What is the sum?

200000
216000
196000
212000
156000

Discuss Question

Answer: Option A. -> 200000
:
A
The difference in the SI and CI for the second year

= \frac{P r^{2}}{100^{2}} = 4500.................... (i)

The difference in the SI and CI for the third year

= \frac{P r^{2}}{100^{2}} (3 + \frac{r}{100}) = 14175........... (i i)

From the equation (i) and (ii).

4500 \times (3 + \frac{r}{100}) = 14175

\frac{r}{100} = (\frac{14175}{4500} - 3)

r = \frac{675}{45} .

Now ,from equation (i)

P = \frac{4500}{r^{2}} \times 100^{2} \Rightarrow P = \frac{4500}{675 \times 675} \times 100^{2} \times 45 \times 45 \Rightarrow P = 2, 00, 000.

Question 33. To build a house at a faster pace, the contractor increases the wages/ hour of each worker by

20 %

and the working hours by

25 %

. By how much percent has the daily earnings of each labourer increased?

Discuss Question

Answer: Option D. -> 50%
:
D
Solve using assumption
Lets say the number of hours / day initially = 10 and the wages = Rs 100 for 10 hours
Wages / hour =

\frac{100}{10}

= 10/hr
Earnings initially = 100 for 10 hours
New wages =

1.2 \times 10

= 12 / hr
New number of hours =

1.25 \times 10

= 12.5 hours
New earnings =

12 \times 12.5

= 150

%

change

= \frac{50}{100} \times 100 = 50 %

, Option (4).
2nd method:-

W a g e s (w) = n o . o f h o u r s (h) \times w a g e s / h o u r (p)

\frac{W_{2}}{W_{1}} = \frac{h_{2} \times p_{2}}{h_{1} \times p_{1}} = \frac{1.25 h_{1} \times 1.2 p_{1}}{h_{1} \times p_{1}} = 1.2 \times 1.25 = 1.5

F.O.M = 1.5 i.e

%

increase =

50 %

. Option (4) .

Question 34.

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?

Discuss Question

Answer: Option E. -> 53
:
E
Let the larger number and smaller number is L and S respectively.

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

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

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

Question 35. In a parliamentary debate, a bill was being passed. 600 votes were cast initially but after some discussion, the opposition to the bill went up by

150 % .

The bill was then rejected by a majority 2 times as great as that by which it was passed previously. How many people rejected the bill initially?

Discuss Question

Answer: Option D. -> 200
:
D
There are totally 600 voters, so after the number of people who initially rejected increases by

150 %

, it must still be below 600. That is true only for option (d).
Verifying the answer chosen:
Initially,
Against = 200, for = 400. Hence, majority = 400-200 = 200
After discussions,
Against

= 200 + 1.5 \times 200 = 500

, for = 600-500 = 100. Hence, majority = 500-100 = 400, which is twice the initial majority. Thus, this is correct answer.
Alternatively:
There are totally 600 voters. Let number of peole, who were inAgainst = A, and who were in favour = F. Hence, majority = F - A
F + A = 600
After discussions,
The opposition to the bill went

150 %

. Aganist = 2.5 A, Favour

= 600 - 2.5 A

According to given question,

2 [600 - 2.5 A] = [600 - 2 A] \Rightarrow A = 200 a n d F = 400.

Question 36. Anil and Binod started a business by investing Rs.35,000 and Rs.13,000 respectively. At the end of every month, Anil withdraws certain amount from his investment and Binod invests the same amount as Anil has withdrawn. At the end of the year they share the profits in the ratio

1 : 1

. Find the amount invested by Binod every month.

1000
2000
2500
3000
None of these

Discuss Question

Answer: Option B. -> 2000
:
B
Let x be the Amount that Anil withdraws and Binod invests at the end of every month.
Then,

35, 000 - 11 x = 13, 000.

11 x = 22, 000 & x = 2000.

Alternatively:-

\begin{matrix} 1^{s t} Y e a r & 2^{n d} Y e a r & 3^{r d} Y e a r & ............. & 12^{t h} Y e a r Anil & 35, 000 & (35, 000 - x) & (35, 000 - 2 x) & ............. & (35, 000 - 11 x) Binod & 13, 000 & (13, 000 - x) & (13, 000 - 2 x) & ............. & (13, 000 - 11 x) \end{matrix}

35, 000 \times 12 - 66 x = 13, 000 \times 12 + 66 x

22, 000 \times 12 = 132 x

x = 2000.

Question 37. A certain amount of money has been invested to earn a simple interest of

10 %

per annum. The interest earned every year is again invested at the same rate. After how many years will the amount become the square of itself?

12
8
7
10
can't be determined

Discuss Question

Answer: Option E. -> can't be determined
:
E
Unless we know the amount we cannot derive the answer.
Hence option (e)

Question 38. Laxman had 100 cows at the beginning of the year 1992 and the number of cows each year increases by

10 %

. At the end of each year he doubles the present no. of cows by buying extra cows. What is the number of cows he has at the beginning of 1994?

Discuss Question

Answer: Option E. -> 484
:
E
Solution:
At the beginning of the year 1992, He was having 100 cows.
At the beginning of the year 1993, Number of cows