Quantitative Ability Test

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Quantitative Ability Test - 3

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Weekly Quiz Competition

Question 1
4 / -1

A sum of money becomes 5 times at simple interest in 16 years. What is the rate of interest?

A
33.33%

B
25%

C
20%

D
30%

Solution

Given

The Sum of money becomes 5 times of itself in 16 years

Formula used:

(i) SI = PRT / 100

Where P = principal

R = rate

T = time

(ii) A = SI + P

Where, A = Amount

SI = simple interest

Let the principal be P.

For 16 years,

SI = (P × R × 16) / 100

= 16PR / 100

A = SI + P

⇒ 5P = 16PR / 100 + P

⇒ 4P = 16PR / 100

⇒ R = 400 / 16

⇒ R = 25%

∴ The rate of interest is 25%.

Hence, the correct option is (B).
Question 2
4 / -1

If x² − 2pxy − y² = 0 and x² − 2qxy − y² = 0 bisect angles between each other, then:

A
p + q = 1

B
pq = 1

C
pq + 1 = 0

D
p² + pq + q² = 0

Solution

Equation of angle bisector for ax² + 2hxy + by² = 0 is

x2 − y²a − b = xy/h

For x² − 2pxy − y² = 0, equation of angle bisector will be

⇒ x² − y² / 1 − (−1) = xy / − p

⇒ x² − y² + 2xy / p = 0 .....(i)

Given equation of angle bisector is:

x² − 2qxy + y² = 0 .....(ii)

Comparing both the equations:

⇒ 2 / p = −2q

⇒ pq = −1

⇒ pq + 1 = 0

Hence, the correct option is (C).
Question 3
4 / -1

If A = {1, 4}, B = {2, 3}, C = {3, 5} then (A × B) ∩ (A × C) is equal to:

A
{(1, 3), (4, 3)}

B
{(1, 3), (2, 5)}

C
{(1, 3), (1, 5), (2, 5)}

D
None of these

Solution

Given: A = {1, 4}, B = {2, 3}, C = {3, 5}

Here, we have to find (A × B) ∩ (A × C)

As we know that, A × B = {(a, b) ∣ a ∈ A and b ∈ B}

⇒ A × B = {(1, 2), (1, 3), (4, 2), (4, 3)} ......(i)

⇒ A × C = {(1, 3), (1,5 ), (4, 3), (4, 5)} ......(ii)

From (i) and (ii) we can say that,

⇒ (A × B) ∩ (A × C) = {(1, 3), (4, 3)}

Hence, the correct option is (A).
Question 4
4 / -1

The value of 5^1/4 × (125)^0.25 is =?

A
√5

B
5

C
5√5

D
25

Solution

Given:

5^1/4× (125)^0.25

= 5^0.25 × (5³)^0.25

= 5^0.25× 5^(3×0.25)

= 5^0.25 × 5^0.75

= 5^{(0.25 + 0.75)}

= 5¹

= 5

Hence, the correct option is (B)
Question 5
4 / -1

Find the remainder when 67⁹⁹ is divided by 7.

A
4

B
6

C
1

D
2

Solution

If we check for more power we will find that the remainder start repeating themselves as 4, 2, 1, 4, 2, 1 and so on. So when we get A number having greater power and to be divided by the other number B, we will break power in (4n + x) and the final remainder will depend on x i.e.

A^x / B.

Hence, the correct option is (C).
Question 6
4 / -1

Which of the following is not a finite set?

A
{x : x ∈ N and x² < 36}

B
{x ∈ Z : 0 < x < 10}

C
{x : x∈N and x² = x}

D
{x ∈ N : x is even}

Solution

A set that has the finite number of elements is said to be a finite set.

Option (D): {x ∈ N : x is even}

As we can see that, elements of {x∈N:x is even} are: 2, 4, 6, 8,…….

So, there are infinitely many elements in the set {x∈N:x is even}

Thus, {x ∈ N : x is even } is not a finite set.

Option (A) {x : x ∈ N and x² < 36}

As we can see that, elements of {x:x∈N and x² < 36} are: 1,2,3,4,5

So, there are 5 elements in the set {x:x∈N and x² < 36}

Thus, the set {x : x ∈ N and x² < 36} is a finite set.

Option (B): {x ∈ z : 0 < x < 10}

As we can see that, elements of {x ∈ z : 0 < x < 10} are: 1, 2, 3,………,9

So, there are 9 elements in the set {x ∈ z : 0 < x < 10}

Thus, the set {x ∈ z : 0 < x < 10} is a finite set.

Option (C): {x : x ∈ N and x² = x}

∵ x²= x

⇒ x² − x = 0

⇒ x(x − 1) = 0

⇒ x = 0 or 1

But since x∈N.

So, x = 0 ∉ {x : x ∈ N and x² = x}

Therefore, only x = 1 ∈ {x : x ∈ N and x² = x}

Thus, {x : x ∈ N and x²= x} is a finite set.

Hence, the correct option is (D).
Question 7
4 / -1

The relation R on the set of integer is given by R = {(a, b) : a − b is divisible by 7, where a, b ∈ Z}, then R is a/an:

A
Reflexive and symmetric

B
Reflexive and transitive

C
Only reflexive

D
Equivalence

Solution

Given that

R is a relation on Z and is defined as:

R = {(a, b) : a − b is divisible by 7 where a, b ∈ Z}

We know that:

A relation R on a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive.

Therefore,

R is reflexive as:

a − a = 0 is divisible by 7 for all a ∈ Z.

Suppose, if (a, b) ∈ R

⇒ 7 divides a − b i.e.,

⇒ a − b = 7m, where m ∈ Z

⇒ b−a = 7n, where n = −m

⇒ 7 divides b−a too, which implies that:

(b, a) ∈ R.

Therefore, R is symmetric.

Suppose, if (a, b) ∈ R and (b, c) ∈ R, then:

a − b and b − c are divisible by 7.

⇒ a − b = 7m and b − c = 7n, where m, n ∈ Z

⇒ a − c = 7q, where q = m + n

⇒ (a, c) ∈ R

Therefore, R is transitive.

Hence, the correct option is (D).
Question 8
4 / -1

The distance between two points A and B is 600 km. When they start moving towards each other they meet in 12 hours. If A started moving 5 hours after B, then they meet in 10 hours. Taking these into account find the speed of B.

A
20 km/hr

B
25 km/hr

C
30 km/hr

D
15 km/hr

Solution

Given

Distance between A and B = 600 km

Time taken to meet each other (when they start together) = 12 hours

Distance = Speed × Time

Relative speed of two bodies when they travel towards each other = Sum of their speed

When they start moving towards each other, they meet in 12 hours,

Let the speed of A be ‘A’ km/hr and the speed of B be ‘B’ km/hr

⇒ 600 / (A + B) = 12

⇒ A + B = 50......(1)

If A started moving 5 hours after B, then they meet in 10 hours,

Distance covered by A and B in 10 hours = 50 × 10 = 500

Remaining distance (600 - 500 = 100 km) was already covered by B in 5 hours

Speed of B = 100 / 5 = 20 km/hr

∴ The speed of B is 20 km/hr

Hence, the correct option is (A).
Question 9
4 / -1

A
3 × 1

B
2 × 1

C
2 × 3

D
3 × 3

Solution

Multiplication of matrices is possible if the order of the new matrix is a row of the first matrix and column of the second matrix.

i.e., m₁ × n₂

Now,

Order of new matrix formed by multiplication = m₁ × n₂= 2×1

Hence, the correct option is (B).
Question 10
4 / -1

car covers a distance of 4 km in 6 minutes. If it's speed is decreased by 2 km/hr, then find the time taken by the car to cover same distance.

A
120/19 min

B
2/19 min

C
45 min

D
60/19 min

Solution

Given

Distance = 4 km

Time = 6 minutes

Distance = Speed × Time

Speed = Distance / Time

Time = 6 minutes

⇒ Time = ( 6/60) hr = (1/10) hr

⇒ Speed = DistanceTime = 4(1/10)

⇒ Speed = 40 km/hr

According to the question,

If its speed is decreased by 2 km/hr

⇒ New speed = (40 - 2) km/hr = 38 km/hr

Time taken by the car to cover same distance,

Time = Distance / Speed

⇒ Time = (4/38) hr = (4/38) × 60 = 120/19 minutes

∴ The time taken by the car to cover same distance is 120/19 minutes.

Hence, the correct option is (A).

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Quantitative Ability Test - 3

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Quantitative Ability Test - 3

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Question 1

33.33%

25%

20%

30%

Solution

Question 2

p + q = 1

pq = 1

pq + 1 = 0

p2 + pq + q2 = 0

Solution

Question 3

{(1, 3), (4, 3)}

{(1, 3), (2, 5)}

{(1, 3), (1, 5), (2, 5)}

None of these

Solution

Question 4

√5

5

5√5

25

Solution

Question 5

4

6

1

2

Solution

Question 6

{x : x ∈ N and x2 < 36}

{x ∈ Z : 0 < x < 10}

{x : x∈N and x2 = x}

{x ∈ N : x is even}

Solution

Question 7

Reflexive and symmetric

Reflexive and transitive

Only reflexive

Equivalence

Solution

Question 8

20 km/hr

25 km/hr

30 km/hr

15 km/hr

Solution

Question 9

3 × 1

2 × 1

2 × 3

3 × 3

Solution

Question 10

120/19 min

2/19 min

45 min

60/19 min

Solution

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p² + pq + q² = 0

{x : x ∈ N and x² < 36}

{x : x∈N and x² = x}