Quantitative Aptitude (QA) Test

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Quantitative Aptitude (QA) Test - 4

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Weekly Quiz Competition

Question 1
3 / -1

If the sum of the roots of the quadratic equations mx² + (2m - 1)x + 4 and (3m + 1)x² - 6x + (2m - 3) are equal, then find the sum of values of m.

A
5/6

B
1/6

C
-5/6

D
Cannot be determined

Solution

The sum of the roots of a quadratic equation ax² + bx + c is - b / a
⇒ -(2m - 1)/m = 6/(3m + 1)
⇒ 6m² - m -1 = -6m
⇒ 6m² + 5m -1 = 0
⇒ m = 1/6 or -1
Thus, the sum of the values of m: 1/6 -1 = -5/6
Question 2
3 / -1

If the last day of the year 1899 is second to the last day of the week, then find the day of the week on which the date 21^st April, 1904 falls. Assume that a week starts on Sunday.

A
Monday

B
Thursday

C
Tuesday

D
Wednesday

Solution

The number of days between 31^st
December 1899 and 31^st December 1903 = 365 x 4 = 1460
The number of days between 31^stDecember 1903 and 21^st April 1904 = 31 + 29 + 31+21 = 112
Total number of days = 1572
Dividing 1572 by 7 the remainder = 4
Hence, 21^st April 1904 falls on Thursday.
Hence, option 2.
Alternatively,
We know that in a non-leap year there are 52 weeks and 1 odd day.
There is no leap year between 1899 to 1903. Hence, there are 4 odd days.
Number of odd days between 31^st December 1903 and 21^st April 1904 = 3 + 1 + 3 + 0 = 7
Thus, there are total 4 odd days between the given time frame. 21^st April 1904 will be a Tuesday.
Hence, 21^st April 1904 falls on Thursday. Hence, option 2.
Question 3
3 / -1

What will be the maximum volume that can be obtained by rotating a right angled triangle of dimensions 6, 8 and 10 units.

A
76.8π

B
32π

C
96π

D
128π

Solution

The maximum volume will be obtained when it is rotated about the smaller side rather than the hypotenuse.
Radius = 8 units, Height = 6 units
Thus, the volume = (1/3) x (π x 64 x 6) = 128π units.
Hence, option 4.
Question 4
3 / -1

The price of 1 pencil is subtracted from the price of 5 pens is equal to one fifth of the price of 3 pencils and 7 pens. Find the price of pencil as a percentage of the price of pen.

A
225

B
226

C
227

D
228

Solution

Let the price of one pencil be x and the price of one pen be y.

⇒ (5y - x) = (3x + 7y)/5
⇒ 25y - 5x = 3x + 7y
⇒ x = 2.25y

Thus, the price of the pencil as a percentage of the price of the pen = (2.25y /y) x 100 = 225%
Question 5
3 / -1

How many subsets of S = (1, 2, 3 , . . . , 400) have the product of their elements an even number?

A
2⁴⁰⁰ - 2²⁰⁰

B
2²⁰⁰- 2¹⁵⁰

C
100

D
225

Solution

For a given set containing N elements, the number of non-zero subsets = 2^N - 1
Number of non-zero subsets in S = 2⁴⁰⁰ - 1
The product of the elements of a subset will be odd when all the elements of the subset are odd.
The odd elements in S are S_odd = (1, 3, 5 , ..., 399)
Number of non-zero subsets in S_odd = 2²⁰⁰-1
Number of subsets where product of its elements is even = (2⁴⁰⁰ - 1) -(2²⁰⁰-1) = 2⁴⁰⁰ - 2²⁰⁰
Hence, option 1.
Question 6
3 / -1

Find the maximum area of the isosceles trapezium (in units²) whose unequal sides are 4 units and 6 units. Given that it is inscribed in a circle whose radius is 2√3

A
5(2√2 - √3)

B
5(2√2 + √3)

C
10(2√2 - √3)

D
10(2√2 + √3)

Solution

The maximum area of trapezium will be obtained when the two sides are on the opposite sides of the diameter.
The height of the trapezium = OE + OF
OE = √OC² - √EC² = √3
OF = √OB² - √FB² = 2√2
∴ the area of the trapezium = (1 / 2) x (2√2 + √3) x (6 + 4) = 5(2√2 + √3)
Hence, option 2.
Question 7
3 / -1

The year on year increase in the price of a product can be represented as the sum of two different constants A and B which has variable co-efficients. If the price of the product in 2004 was Rs. 80 and the coeffiicients of A and B for the session 2004-05 are 2 and 5 respectively and for the session 2005-06 are 4 and 1 respectively, then find the relation between A and B if the price increases at a constant rate of 20% per year.

A
2A = 5B

B
4A = 7B

C
A = 3.125B

D
2A = -5B

Solution

The increase in price in 2004-05 session = 80 x 0.2 = Rs.16.
2A + 5B = 16 ...(i)
The increase in price in 2005-06 session = 96 x 0.2 = Rs.19.2
4A + B = 19.2 ...(ii)
Solving (i) and (ii),
A = 3.125B
Hence, option 3
Question 8
3 / -1

The average speed of Mumbai metro train which travels from Ghatkopar to Versova is 30 km/hr and the distance between the two stations is 32 km. Which function represents the distance 'd(t)' remaining in a trip from Ghatkopar to Versova after a certain time 't' from the start?

A
d(t) = 30t - 32

B
d(t) = 32 - 30t

C
d(t) = 32t / 30

D
d(t) = 30 / 32t

Solution

We know that,
Distance remaining = Total distance - Distance travelled
⇒ d(t) = 32 - 30t
Question 9
3 / -1

In a bag there are 3 blue coloured balls, 6 green coloured balls, 2 white coloured balls and 5 yellow coloured balls. If 2 balls are selected at random from the bag, then what is the probability that none of them is blue? (Enter the correct option number)

A
13/120

B
13/240

C
78/120

D
156/120

Solution

Other than blue any other can be selected.
Hence, 2 out of 13 can be selected in ¹³C₂ ways and total ways of selecting 2 balls out of 16 is ¹⁶C₂ ways.
The required probability = ¹³C₂/¹⁶C₂= 13/20 = 78/120
Question 10
3 / -1

Find the value of M in the figure given below,

A
6.8

B
6.7

C
6.6

D
6.5

Solution

⇒ 3M + 9 + M + 5 + 140= 180
⇒ 4M = 26
⇒ M = 6.5°
Question 11
3 / -1

What is the area of an obtuse angled triangle having two sides as 18 m and 24 m and the angle between them as 150°?

A
78 sq.m

B
108 sq.m

C
84 sq.m

D
96 sq.m

Solution

In a triangle ABC, if AB, AC and the included angle A are known, Area = (1/2) x AB x AC x sin A
= (1/2) x 18 x 24 x sin 150 = 216 x 0.5 = 108 sq.m
Question 12
3 / -1

A finance company keeps revising the interest rate it offers every year with a constant percentage increase in rate of interest over the previous year. However, the rate of interest remains same for a certain investment over the years of investment. If an investment of one lakh rupees in 2010 at 6% rate of simple interest and an investment of one lakh rupees in 2011 amounts to the same sum in 2014, then find the rate of simple interest offered in 2015.

A
25.28

B
25.29

C
25.27

D
25.25

Solution

The amount obtained in 2014 by investing 100000 in 2010 = 100000 x (1 + 0.24) = 124000

Let the rate of interest offered in 2011 be r.
∴ 100000 x (1 + 3r/100) = 124000
⇒ r = 8%

The increase percentage in the offered rate = (2/6) x 100 = 33.33 %

The rate of interest offered in 2015 = 8 x (1 + 1/3)⁴= 25.28 %
Question 13
3 / -1

6 equally spaced girls are standing along the circumference of a circle of radius 10 m and facing the center. What is the shortest distance between any two girls facing each other?

A
20

B
21

C
19

D
23

Solution

If we draw a figure according to the information provided in the question then we can note that:
In the question it is good that each girl is facing the center of the circle.

Radius of circle =10m, Diameter of circle =20m

We need to find the shortest distance between any two girls faces each other.

We can see from the image that oppositely standing girls will only face each other so the shortest distance between any two girls facing each other will be equal to the diameter of the circle ie 20 m.
Question 14
3 / -1

What is the sum of the two middle terms of the arithmetic progression {-17, - 1 3 , . . . , 19}?

A
2

B
4

C
3

D
5

Solution

Since the sum of two middle term s is asked, the series has an even number of terms.
Now for even number of terms we know that, Sum of the middle two term of an A.P. will be equal to the sum of the first and last terms of the A.P.
i.e. - 17 + 19 = 2

Alternatively,

First term = a = -17
Commond difference = d = -13 - (-17) = 4
Last term = 19 = -17 + (n - 1)4
⇒ n = 10

Middle term will be 5th and 6th terms.
Their sum will be = a + 4d + a + 5d = 2a + 9d
i.e. 2 (-17) + 9 (4) = 2
Question 15
3 / -1

In how many ways can you arrange 3 girls and 4 boys in a row such that no boys sit next to each other?

A
147

B
148

C
142

D
144

Solution

3 girls can be arranged in 3! Ways.

Now there are 4 gaps remaining where 4 boys can be arranged in 4! ways.

Total number of ways = 3! x 4! = 6 x 24 = 144
Question 16
3 / -1

Of the three numbers, second is twice the first and is also thrice the third. If the average of the three numbers is 77, find the largest number.

A
115

B
126

C
84

D
72

Solution

Let the numbers be x, y and z.
y = 2x and y = 3z
As the average = 77
⇒ x + y + z = 231
⇒ y/2 + y + y/3 = 231
⇒ y = 126
Hence, option 2.
Question 17
3 / -1

Establish relation between x and y if:
1. x² + 13x + 40 = 0
2. y² + 8y + 15 = 0

A
x > y

B
x < y

C
x ≥ y

D
x ≤ y

Solution

Solving the quadratic equations we get,
x = -8 or -5 and y = -5 or -3.
⇒ x ≤ y
Hence, option 4.
Question 18
3 / -1

A shopkeeper gives a discount of 10% on the marked price and sells it for Rs. 540 to earn a profit of Rs. 90. Find the profit percentage if the product is marked Rs. 200 above its cost price for the same absolute value of discount.

A
30%

B
31.11%

C
40%

D
Cannot be determined

Solution

Case 1:
Let the marked price be m.
The selling price = 0.9m
⇒ 0.9m = 540
⇒ m = 600
Cost price = 540 -90 = Rs. 450
Case 2:
Discount = Rs. 60, Marked price = Rs. 650
Selling price = 650 - 60 = Rs. 590
Profit = 590 - 450 = Rs. 140
Profit percentage = (140/450) x 100 = 31.11%
Question 19
3 / -1

A thief is running on a circular track of radius 5 m at 1.5 m / s . A policeman whose speed is twice that of the thief arrived at the starting point of the track 4 seconds after the thief. The policeman can either run along the circular track in the direction of the thief or go to the centre and then go to any point on the circle from the centre. Find the minimum distance covered by the thief after the policeman starts chasing him.

A
6 meters

B
4 meters

C
5 meters

D
4.5 meters

Solution

When the policeman arrives, the thief has traveled 6 meters.
So, the time taken by the policeman to catch the thief along the circular track = 6/1.5 = 4 seconds.
The total distance covered by the policeman = 12 meters.
Thus, the minimum distance is when the policeman takes the route via the center i.e. he travels a length of 2 x 5 = 10 meters.
The minimum time taken = 10/3 seconds.
The minimum distance covered by the thief = 1.5 x (10/3) = 5 meters
Hence, option 3.
Question 20
3 / -1

A dishonest shopkeeper mixes pure sugar with some sugar like substance to gain extra profit. The cost price of sugar like substance is 2/3 times the pure sugar. If the shopkeeper wants to gain a profit of 30 % on selling the mixture at the cost price of pure sugar, then find the ratio of sugar like substance to pure sugar in the mixture.

A
5:6

B
7:5

C
9:4

D
5:9

Solution

Let the cost price of 1 kg of pure sugar be Rs. x.
The cost price of 1 kg sugar-like substance = Rs. 2x / 3
The selling price of one kg of mixture = Rs. x.
Let the amount of pure sugar in 1 kg of mixture be y.
The amount of sugar like substance = (1 - y)
The cost price of mixture = xy + (2x / 3) x (1 - y) = (x / 3) x (2 + y)
⇒ x / [(x / 3) x (2 + y)] = 1.3
⇒ y = 4/13
Thus, the ratio of sugar-like substance to pure sugar = (1-y):y = 9:4
Hence, option 3.
Question 21
3 / -1

If x = 4⁴ x 5⁶ x 6⁵ x 7¹⁰ x 8⁸, then how many factors of x are of the form 6 A²?

A
1368

B
1365

C
1367

D
1369

Solution

x = 4⁴ x 5⁶ x 6⁵ x 7¹⁰ x 8⁸
⇒ x = 2³⁷ x 3⁵ x 5⁶ x 7¹⁰ = (2 x 3) (2³⁶ x 3⁴ x 5⁶ x 7¹⁰)

Power of 2 can take values of 0, 2 , 4 , . . . , 36 = 19
Power of 3 can take values of 0, 2,4 = 3
Power of 5 can take values of 0, 2,4, 6 = 4
Power of 7 can take values of 0, 2,4, 6, 8, 10 = 6

Total number of factors = 19 x 3 x 4 x 6 = 1368
Question 22
3 / -1

In an examination, 65% of the candidates passed in Mathematics, 55% passed in History, 22% failed in both the subjects and 48909 passed in both the subjects. Find the total number of candidates.

A
116450

B
116454

C
116452

D
116457

Solution

Let the percentage total number of candidates who passed in both the subjects be x.

Let the sets M and H represent students who passed in Mathematics and History respectively.

65 - x + x + 55 - x = 100 - 22
⇒ x = 42%

Now, 42% i.e. 48909 students passed in both the subjects.
∴ Total number of students = 48909 x (100/42) = 116450
Question 23
3 / -1

x is a positive integer with value at most equal to 110. How many values of x are possible if x is not a factor of (x - 1)!?

A
25

B
12

C
28

D
None of these

Solution

If x is a prime number greater than 1, then x is not a factor of (x - 1)!
There are 29 primes up to 110. 4 is also one such number.
Hence, option 4.
Question 24
3 / -1

A person starts travelling from a point P towards R passes through Q which is between P and R. The time taken to travel from P to Q is half that of the time taken from Q to R. However, after passing through any of the given points, he travels at a speed which is two-thirds that of the previous journey. Find the time taken, in seconds, to travel from P to R and back to P, if he always passes through Q in the journey and takes 15 minutes to go from P to Q initially.

A
8000

B
8438.25

C
8437

D
8437.5

Solution

Let the initial speed of the person be x and the time taken to go from P to Q be t.
The distance from Q to R = 4xt / 3
The time taken to go from R to Q = (4xt/3) / (4x/9) = 3t
The time taken to go from Q to P = (xt) / (8x/27) = (27/8)t
Hence, the total time taken = t + 2t + 3t + (27 / 8)t = (75 / 8)t = 9.375t
∵ t = 15 minutes = 900 seconds
The total time taken = 8437.5 seconds.
Hence, option 4.
Question 25
3 / -1

x²- 3x + 2 > 0, y² - 4y + 3 > 0. If x, y are positive integers and xy = 16, then find the value of Ix-yI.

A
1

B
0

C
3

D
2

Solution

x²- 3x + 2 > 0 is true when x < 1, x > 2 ... (i)
y² - 4y + 3 > 0 is true wheny < 1, y > 3 ... (ii)
∵ xy = 16
The possible values of (x, y) are (1, 16), (2, 8), (4, 4), (8, 2) and (16, 1).
From (i) and (ii), we can see that only (4,4) satisifies both the conditions.
⇒ |x - y | = |4 - 4 | = 0
Hence, option 2.
Question 26
3 / -1

Find the minimum value of the quadratic expression 3x² + 11x - 23.

A
-23

B
0

C
17/12

D
-397/12

Solution

The minimum value of a quadratic expression ax² + bx + c (when a > 0) occurs at x = -b / 2a and its minimum value is -(b² - 4ac) / 4a
The minimum value of the equation is = [4(3)(-23) - (11)²]/12 = -397 / 12
Hence, option 4.
Question 27
3 / -1

50 square stone slabs of equal size were needed to cover, a floor area of 72 sq. m. The length of each stone slab is :

A
102 cm

B
120 cm

C
201 cm

D
210 cm

Solution

Area of each slab = 72 / 50 m²= 1.44 m²
Length of each slab =√1.44 =1.2m =120cm
Question 28
3 / -1

A and B start working on a project. After 10 days B is replaced by C who can complete the project alone in 20 days. If A and B together can complete the project in 30 days, then find the day on which the project will be completed after C replaces B, if the efficiency of B is twice that of A.

A
15^th day

B
11^th day

C
9^th day

D
7^th day

Solution

Let the number of days required by B alone to complete the project be x
The number of days required by A alone to complete it = 2x
1/x + l/2x = 1/30
⇒ x = 45
The part of work done in 10 days =1/3
Work done by A and C in 1 day = (1/90) + (1/20) = (11/180)
∴ The number of days taken by A and C = (2/3) / (11/180) = 120/11 = (10*12) / 11 days.
Hence, option 2.
Question 29
3 / -1

37% of the total number of people in a city read newspaper ABC and exceed the number of people who read XYZ by 8000. If 27% of the total population do not read any newspaper then find the population of the city if there are only two newspapers in the city. [Assume that no one reads more than one newspaper].

A
8 lakhs

B
10 lakhs

C
12 lakhs

D
14 lakhs

Solution

Let the total population of the town be x.
The number of people who read XYZ = 0.37x - 8000
0.37x+0.37x - 8000 + 0.27x = x
⇒ x = 8,00,000
Hence, option 1.
Question 30
3 / -1

If x² + x - 2 < 0; then which of the following statement is true?
1. x - 1 > 0
2. x - 1 < 0
3. x + 2 > 0
4. x - 2 < 0

A
Only 1 and 3

B
Only 2 and 4

C
Only 2 and 3

D
Only 1 and 4

Solution

x² + x - 2 < 0
⇒ (x - 1) (x + 2) < 0
⇒ (x - 1) < 0 and (x + 2) > 0 or (x - 1) > 0 and (x + 2) < 0
⇒ x < 1, x > -2 or x > 1, x < -2 which is not possible.
∴ -2 < x < 1
⇒ (x - 1) < 0 and (x + 2) > 0
Hence, option 3.
Question 31
3 / -1

At 12:00 AM both the hour and a minute hand were overlapping each other. How many more minutes does the minute hand travel than the hour hand in the next 54 minutes?

A
48

B
49

C
49.5

D
50

Solution

When the minute hand travels 60 minutes, the hour hand travels 5 minutes.
The minute hand travels 55 minutes more than the hour hand in an hour.
In 54 minutes, the Minute hand travels = (54 x 55/ 60) = 49.5
Hence, option 3.
Question 32
3 / -1

A and B start working together on a project and both have the same efficiency in the beginning. However the efficiency of A decreases to 0.6 times the usual after working for 2 hours. If the project can be finished in 120 man hours, then find the minimum number of days required to finish the work if both do an equal manhours of work each day and the sum of the total number of hours of work by both each day is 12. [ Initial efficiency of A = Efficiency of B = 1 man hour].

A
10 days

B
12 days

C
16 days

D
Cannot be determined

Solution

Let the number of hours worked by B be x.
The hours worked by A = (12 - x)
The manhours of work finished by A = (12 - x - 2) x 0.6 + 2
∵ (12 - x - 2) x 0.6 + 2 = x
⇒ x = 5
The total manhours of work done by them in a day = 2 * x = 2 * 5 = 10
Thus, the number of days required to finish the work = 120/10 =12 days
Hence, option 2.
Question 33
3 / -1

The average number of apples per carton for three cartons of apple is 90. The average price per carton is Rs.1500. The average price of apples in the two cartons which do not contain the highest number of apples is 20 and 25 respectively. The price of the carton with the highest number of apples is 1900. If the highest difference in the number of apples in any two cartons is 100, then find the lowest number of apples in a carton.

A
50

B
45

C
55

D
48

Solution

Let the number of apples in the carton with the least number of apples be x and the second least number of apples be y.
∵ x +y + x + 100 = 270
⇒ 2x +y = 170...(i)
Also, the total amount spent on the two cartons with x and y apples = 4500 - 1900 = 2600
⇒ 20x + 25y = 2600
⇒ 4x + 5y = 520.. .(ii)
From (i) and (ii)
x = 55,y = 60
The lowest number of apples in a cartoon = 55.
Hence, option 3.
Question 34
3 / -1

A saree is one-fourth by length green, two-fifth by length red and the remaining 3850 cm by length is black. What is the length of the saree in meters?

A
110

B
111

C
113

D
115

Solution

Let the length of the saree be ‘x’ cm.

x - (1/4) x - (2/5) x = 3850
⇒ (7/20) x = 3850
⇒ x = 11000 cm = 110 m
Question 35
3 / -1

There are two sections A and B of a class, consisting of 36 and 44 students respectively. If the average weight of section A is 40 Kg and that of section B is 35 Kg, find the average weight of the whole class.

A
30 kg

B
35 kg

C
42.5 kg

D
37.25 kg

Solution

The total weight of (36 + 44) Students of A and B = (36×40) + (44×35) = 2980 kg
⇒ Average weight of the whole class = 2980/80 = 37.25kg

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Quantitative Aptitude (QA) Test - 4

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SHARING IS CARING

Question 1

5/6

1/6

-5/6

Cannot be determined

Solution

Question 2

Monday

Thursday

Tuesday

Wednesday

Solution

Question 3

76.8π

32π

96π

128π

Solution

Question 4

225

226

227

228

Solution

Question 5

2400 - 2200

2200- 2150

100

225

Solution

Question 6

5(2√2 - √3)

5(2√2 + √3)

10(2√2 - √3)

10(2√2 + √3)

Solution

Question 7

2A = 5B

4A = 7B

A = 3.125B

2A = -5B

Solution

Question 8

d(t) = 30t - 32

d(t) = 32 - 30t

d(t) = 32t / 30

d(t) = 30 / 32t

Solution

Question 9

13/120

13/240

78/120

156/120

Solution

Question 10

6.8

6.7

6.6

6.5

Solution

Question 11

78 sq.m

108 sq.m

84 sq.m

96 sq.m

Solution

Question 12

25.28

25.29

25.27

25.25

Solution

2⁴⁰⁰ - 2²⁰⁰

2²⁰⁰- 2¹⁵⁰