Quantitative Aptitude Test

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Quantitative Aptitude Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

In a class there is 39 students, the average weight decreases by 4 kg, when a new students is joining the class. If the average weight of students is 64 kg. What is the weight of new student?

A
90 kg

B
96 kg

C
110 kg

D
85 kg

E
80 kg

Solution

Given:

A total number of students = 39.

The average weight of students is 64 kg.

The average weight decreases by 4 kg.

Total weight of class = 39 × 64 kg = 2496 kg

Total weight after new students joins = (39 + 1)(64 - 4) = 2400 kg

The weight of new student = Total weight of class - Total weight after new students joins.

⇒ 2496 kg - 2400 kg

⇒ 96 kg

∴ The weight of new student is 96 kg.
Question 2
1 / -0

In a college election, two candidates A and B contested. If in the election 30% votes were declared invalid and A got 1000 votes more than B. Calculate the total votes in the election if votes received by A is 1375.

A
3000

B
1500

C
2500

D
2200

E
2800

Solution

Given:

Votes received by A = 1375

A got 1000 votes more than B.

⇒ Votes received by B = 1375 – 1000 = 375

∴ Total valid votes in the election = 1375 + 375 = 1750

Suppose total votes in the election be 100x.

∴ Total valid votes in the election = $(\frac{70}{100})$ × 100x = 70x

∴ 70x = 1750

⇒ x = 25

∴ Total number of votes in the election = 25 × 100 = 2500
Question 3
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Directions For Questions

Direction: In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

...view full instructions

I. x³ = 8

II. y²= 4

A
x > y

B
x < y

C
x ≥ y

D
x ≤ y

E
x = y or the relationship between x and y cannot be established.

Solution

Given

I.x³= 8

⇒ x³= 2³

⇒ x= 2

II. y²= 4

⇒ y²- 4= 0

⇒ (y+ 2)(y- 2) =0

⇒ y= 2, -2

Comparison between x and y (via Tabulation):

value of x value of y Relation

2 2 x = y

2 -2 x> y

∴x ≥ y
Question 4
1 / -0

Directions For Questions

Direction: In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

...view full instructions

I. x² - 3x + 2 = 0

II. y² = 1

A
x > y

B
x < y

C
x ≥ y

D
x ≤ y

E
x = y or the relationship between x and y cannot be established.

Solution

Given

I. x²- 3x + 2 = 0

⇒ x²- 2x- 1x+2 =0

⇒ x(x- 2) - 1(x- 2) =0

⇒ (x-1)(x- 2) =0

⇒ x= 1 , 2

II. y²= 1

⇒ y²- 1= 0

⇒ (y+ 1)(y- 1) =0

⇒ y= 1, -1

Comparison between x and y (via Tabulation):

Value of x Value of y Relation

1 1 x = y

1 -1 x> y

2 1 x> y

2 -1 x > y

∴x ≥ y
Question 5
1 / -0

Directions For Questions

Direction: In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

...view full instructions

I. $x^{\frac{5}{2}} = \sqrt{243}$

II. $2 y^{\frac{3}{2}} = \sqrt{256}$

A
x > y

B
x < y

C
x ≥ y

D
x ≤ y

E
x = y or relationship between x and y cannot be established.

Solution

Given

I. $x^{\frac{5}{2}} = \sqrt{243}$

⇒ $x^{\frac{5}{2}} = {(243)}^{\frac{1}{2}}$

⇒ $x^{\frac{5}{2}} = {(3^{5})}^{\frac{1}{2}}$

⇒ $x^{\frac{5}{2}} = 3^{\frac{5}{2}}$

⇒ x = 3

II. $2 y^{\frac{3}{2}} = \sqrt{256}$

⇒ $y^{\frac{3}{2}} = (\frac{1}{2}) \times \sqrt{256}$

⇒ $y^{\frac{3}{2}} = \sqrt{(\frac{256}{4})}$

⇒ $y^{\frac{3}{2}} = \sqrt{64}$

⇒ $y^{\frac{3}{2}} = 4^{\frac{3}{2}}$

⇒ y = 4

∴, x < y

Question 6

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Directions For Questions

Direction: In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

...view full instructions

I: x² – 2x – 35 = 0

II: 2y² + 72y + 70 = 0

x > y

x ≥ y

x < y

x ≤ y

x = y or the relationship between x and y cannot be established.

Solution

Given

I: x²– 2x – 35 = 0

⇒ x²– 7x + 5x – 35 = 0

⇒ x × (x – 7) + 5 × (x – 7) = 0

⇒ (x – 7) × (x + 5) = 0

⇒ x = 7, -5

II: 2y²+ 72y + 70 = 0

⇒ 2y²+ 2y + 70y + 70 = 0

⇒ 2y × (y + 1) + 70 × (y + 1) = 0

⇒ (y + 1) × (2y + 70) = 0

⇒ y = -1, $\frac{- 70}{2}$

⇒ y = (-1), (-35)

Comparison between x and y (via Tabulation):

Value of x	Relation	Value of y
7	>	-1
7	>	-35
-5	<	-1
-5	>	-35

∴ Relationship between x and y cannot be established.

Question 7
1 / -0

The difference between simple interest and compound interest on a certain sum of money for 2 years at 8% per annum is Rs 120. Find the sum.

A
Rs. 19,750

B
Rs. 17,570

C
Rs. 21,570

D
Rs. 18,750

E
Rs. 20,050

Solution

Given:

Difference between simple interest and compound interest on a certain sum of money = Rs. $120$

Time $=2$ years

Rate of interest $=8 \%$

We know that:

Simple interest $=\frac{(P \times r \times t)}{100} \quad[\because P=$ principle, $t=$ time, $r=$ rate of interest $]$

Compound Interest $=\left[P(1+\frac{r}{100})^{ t}-P\right] \quad[\because P=$ principle amount, $t=$ time, $r=$ rate of interest, $n=$ number of times interest applied per time periods]

Let, the sum i.e., Principal amount $=$ Rs $100 x$

$\therefore$ Simple interest earned = Rs.$\frac{(100 x \times 2 \times 8)}{100}=$ Rs. $16 x$

$\therefore$ Compound interest earned $=$ Rs. $\left[100 x \times(1+\frac{8}{100})^{2}-100 x\right]$

$=$ Rs. $(\frac{2916 x}{25}-100 x)$

According to the question,

$(\frac{2916 x}{25}-100 x)-16 x=120$

$\Rightarrow\frac{(2916 x-2900 x)}{25}=120$

$\Rightarrow 16 {x}=3000$

$\Rightarrow {x}=\frac{3000}{16}$

$\Rightarrow {x}=\frac{375}{2}$

$\Rightarrow$ Total Sum $=$ Rs. $(\frac{100 \times 375}{2})=$ Rs. 18,750

$\therefore$ Total sum is Rs. 18,750.
Question 8
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Directions For Questions

Direction: What should come in place of the question mark '?' in the following number series?

...view full instructions

4, 5, 4, 9, ?, 52, 155

A
19

B
20

C
17

D
18

E
21

Solution

The series follows following pattern:

4 × 1 + 1 = 5

5 × 1 - 1 = 4

4 × 2 + 1 = 9

9 × 2 - 1 = 17

17 × 3 + 1 = 52

52 × 3 - 1 = 155

∴ The value of ? is 17.
Question 9
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A number is increased by 84, it becomes 107% of itself. What is the number?

A
600

B
900

C
1500

D
1200

E
1400

Solution

Let the number be yAccording to the problem statement,

⇒ y + 84 = 107 × $\frac{y}{100}$

⇒ 84 = 1.07y – y = 0.07y

⇒ y = 1200

∴ the number is 1200.
Question 10
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Directions For Questions

Direction: What should come in place of the question mark '?' in the following number series?

...view full instructions

14, 14, 28, 84, 336, ?

A
1680

B
1682

C
2016

D
1344

E
None of these

Solution

The series follows the following pattern:

14 × 1 = 14

14 × 2 = 28

28 × 3 = 84

84 × 4 = 336

336 × 5 = 1680

∴ The value of ? is 1680.

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Re-Attempt Weekly Quiz Competition

Quantitative Aptitude Test - 10

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Quantitative Aptitude Test - 10

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SHARING IS CARING

Question 1

90 kg

96 kg

110 kg

85 kg

80 kg

Solution

Question 2

3000

1500

2500

2200

2800

Solution

Question 3

x > y

x < y

x ≥ y

x ≤ y

x = y or the relationship between x and y cannot be established.

Solution

Question 4

x > y

x < y

x ≥ y

x ≤ y

x = y or the relationship between x and y cannot be established.

Solution

Question 5

x > y

x < y

x ≥ y

x ≤ y

x = y or relationship between x and y cannot be established.

Solution

Question 6

x > y

x ≥ y

x < y

x ≤ y

x = y or the relationship between x and y cannot be established.

Solution

Question 7

Rs. 19,750

Rs. 17,570

Rs. 21,570

Rs. 18,750

Rs. 20,050

Solution

Question 8

19

20

17

18

21

Solution

Question 9

600

900

1500

1200

1400

Solution

Question 10

1680

1682

2016

1344

None of these

Solution

User

Question Analysis