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

Exams > Cat > Quantitaitve Aptitude

BASIC ARITHMETIC MCQs

Total Questions : 150 | Page 10 of 15 pages

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?

$20 %$
$25 %$
$30 %$
$50 %$

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 92.

For construction, a certain amount of wire was being shipped, out of which $10 %$ was stolen. After using $85 %$ of the remainder, 472.5 m of wire was left. How much wire was initially shipped?

5000 m
4000 m
3250 m
3500 m

Discuss Question

Answer: Option D. -> 3500 m
:
D

Take a middle answer option and verify the answers.

Initial = 3500

Amount left after being stolen $= 0.9 \times 3500 = 3150$

Amount left $= 15 % o f 3150 = 472.5$

2nd method:- Let the total length of original wire = x

Remaining wire $= 100 % - 10 % (S t o l e n w i r e) = 90 %$ . After using $85 %$ of remaining wire, $15 %$ of remaining wire was left.

$15 % o f 90 % 0 f x = 472.5$

$0.135 o f x = 472.5 \Rightarrow x = 3500.$

Question 93.

If the side of a square is increased by $25 %$ , then its area is increased by how many percent?

$50 %$
$58 %$
$56.25 %$
$> 60 %$

Discuss Question

Answer: Option C. ->

56.25 %

:
C

Let the side be 10 cm. Then the area will be $100 c m^{2}$

New side $= 125 % o f 10 = 12.5 c m;$

$A r e a = (12.5)^{2} = 156.25$

Percentage increase in area $= 56.25 % .$

Alternatively: If x is the percentage increase in the side of a square, then increase in area is given by
$x + x + \frac{x \times x}{100} = 2 x + \frac{x^{2}}{100} = 25 + 25 + \frac{25 \times 25}{100} = 56.25 %$

Question 94.

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

Answer: Option C. ->

56.25 %

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 95.

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 96.

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 in Against = 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 97.

$20 %$ 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 98.

Rita scored $25 %$ less than Nita. How much percent more did Nita score than Rita?

$20 %$
$25 %$
$33.33 %$
$16.66 %$
53

Discuss Question

Answer: Option C. ->

33.33 %

:
C

Approach 1:
$%$ Nita scored more than Gita $= \frac{25}{100 - 25} \times 100 % = 33.33 %$
Approach 2:
Rita’s score is lesser by $25 %$ $(\frac{1}{4})$ . Nita’s score will be greater by $\frac{1}{x - 1} % = (\frac{1}{3}) = 33.33 %$

Question 99.

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 $= 2 (1.1 \times 100) = 220.$

At the beginning of the year 1994, Number of cows $= 2 (1.1 \times 220) = 484.$

Question 100.

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.