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

Exams > Cat > Quantitaitve Aptitude

BASIC ARITHMETIC MCQs

Total Questions : 150 | Page 8 of 15 pages

Question 71. Speed of A is twice that of B. Both start from a point X towards a point Y. If B took 3 hours more than A, what is the time taken by B?

4 hr
6 hr
8 hr
2 hr

Discuss Question

Answer: Option B. -> 6 hr
:
B
Distance is constant, speed and time are inversely proportional. Therefore, if ratio of speed of A and B is 2:1, ratio of time taken will be 1:2
Time taken by A= X
Time taken by B= 2x
2X-X=3
x=3. B takes 6 hours

Question 72. Which is the least number, which when subtracted from both the numerator and the denominator of the ratio

\frac{80}{70}

, gives

\frac{13}{11}

Discuss Question

Answer: Option C. -> 15
:
C
Solving conventionally,
Let x be the required number

\frac{80 - x}{70 - x} = \frac{13}{11}

880 – 11x = 910 – 13x
2x = 30
x= 15
Shortcut:- Use answer options. Always start with the middle answer option

Question 73. Till the age of 21, Arun grows such that his height varies as the square root of this age. When Arun is 9 years old, his height is 4 feet. What is his height at 16 years?

5 feet
5 feet 2 inches
5 feet 4 inches
none of these

Discuss Question

Answer: Option C. -> 5 feet 4 inches
:
C
Option (c)
Relationship given:-

4 = K \sqrt{9} \to K = \frac{4}{3}

H α \sqrt{A H} = K \sqrt{A}

Height of Arun at 16 years

H = \frac{4}{3} \times 4 = \frac{16}{3}

= 5 feet 4 inches

Question 74. A sum of Rs 530 is divided between P,Q,R such that P gets

\frac{1}{2}

that of Q and Q gets twice that of R. What is the ratio of Q to that of the sum of P and Q?

Discuss Question

Answer: Option B. -> 2:3
:
B

P = \frac{1}{2} Q; Q = 2 R o r \frac{P}{Q} = \frac{1}{2}; \frac{Q}{R} = \frac{2}{1}

Shares of P, Q, R = x, 2x,x

\frac{Q}{P + Q} = \frac{2 x}{3 x} = 2 : 3

Question 75. ‘a’ and ‘b’ are any two distinct numbers belonging to the set {1,2,3,4,5,6}. How many ratios of the form

\frac{a}{b}

can be formed which are not integers.

Discuss Question

Answer: Option C. -> 22
:
C
For each A selected there are five ways of selecting a B.. There are total 30 ratios that can be formed. Now for ‘1’ there are 5 numbers that can come in the numerator to form an integer.
For 2 there are 2 numbers and for 3 there is only one.
Therefore 30-5-2-1= 22 is the answer.

Question 76.

Anil and Sunil can complete a job in 15 days and 20 days respectively. Both of them started it together. Anil left 6 days before the work was completed. Find the time taken (in days) to complete the work.

Discuss Question

Answer: Option C. -> 12
:
C

Let total work is LCM of (15, 20) = 60

Work done by Anil in a day = 4 units

Work done by Sunil in a day = 3 units

Anil worked for (t – 6) days and Sunil worked for t days.

4(t – 6) + 3t = 60

t = 12; Option (c).

2nd method:- During last 6 days Sunil alone will work i.e $30 %$ of work will be done by Sunil alone. Remaining $70 %$ of work will be done together. In one day both will do $(\frac{100}{15}) + (\frac{100}{20}) = 11.67 %$ , then $70 %$ will complete in 6 days. Total 6+6= 12 days.Option (c).

Question 77.

The average of temperatures at noontime from Tuesday to Saturday is 50; the lowest one is 45, what can be the possible maximum range of the temperatures.

Discuss Question

Answer: Option B. -> 25
:
B

The average of the 5 temperatures is: $\frac{a + b + c + d + e}{5} = 50$

One of these temps is 45: $\frac{a + b + c + d + 45}{5} = 50$ .

Solving for the variables: a + b + c + d = 205

In order to find the greatest range of temps, we minimize all temps but one. Remember, though, that 45 is the lowest temp possible, so: 45 + 45 + 45 + d = 205 Solving for the variable: d = 70

70 - 45 = 25 Option (b).

2nd method:- For finding maximum range, we will take 4 days minimum out of 5 days i.e 4 days temp. will take 45, then total negative deviations will be 20. Thus for average 50, positive deviation should be also 20 . Thus highest temp will be (50+20)=70 degrees. Then range =70-45 = 25. Option (b).

Question 78.

A shopkeeper buys 50 footballs at Rs.100 per football. He sells a part of the footballs at a profit of $30 %$ and on the remaining; he incurs a loss of $10 %$ . If the overall profit is $10 %$ , then the number of footballs sold at profit is?

Discuss Question

Answer: Option A. -> 25
:
A

Total Cost Price = Rs 5000 . Total profit = $10 %$ . Total Selling Price $= 1.1 \times 5000 = 5500$

Now go from answer options

If the number of footballs sold at profit is 25, then revenue for these 25 balls $= 1.3 \times 25 \times 100 = 3250$

And the number of footballs sold at loss will be 25; revenue for these balls $= 0.9 \times 25 \times 100 = 2250$

Sum= 5500 = $10 %$ profit on 5000

Thus, the assumption is right

Question 79.

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

Answer: Option A. -> 25
:

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 80.

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

Answer: Option A. -> 25
:

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= 72 required time = x – 9 = 63