Quantitative Aptitude (QA) Test

Result

Quantitative Aptitude (QA) Test - 27

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Weekly Quiz Competition

Question 1
3 / -1

Find the remainder of 25! when divided by 29.

A
2

B
3

C
4

D
5

Solution
Question 2
3 / -1

α and β are distinct roots of a quadratic equation where α² + 1 = 3α and β² = 3β - 1. Find the equation whose roots are [(1/α) + 2] and [(1/β) + 2].

A
x²-7x-11=0

B
x²-7x+11=0

C
x²+7x+11=0

D
x²+7x-11=0

Solution

given: α² + 1 = 3α and β2 = 3β - 1

⇒ α²-3α+1 =0 and β2-3β+1 =0

both α and β satisfy the equation : x²-3x+1=0

thus, α+β = -(-3)/1 = 3 and α×β = 1/1 = 1

now,

the new roots are [(1/α) + 2] and [(1/β) + 2]

⇒ sum = [(1/α) + 2] + [(1/β) + 2] = (1/α) + (1/β) + 4 = (α+β)/(αβ) + 4 = (3)(1) + 4 = 7

⇒ product = [(1/α) + 2]×[(1/β) + 2] = 1/(αβ) + 2/α +2/β + 4 = 1/(1) + 2(α+β)/(αβ) + 4 = 1 + 2(3)(1) + 4 = 11

Thus required equation = x²-7x+11
Question 3
3 / -1

In the Year 2024, Abhijeet spent 60% of what he earned. His income increased by 100rs. every month as compared to the preceding month. if he saved 48000rs. in that year, find his income in the second last month.

A
10450

B
9450

C
10550

D
10500

Solution
Question 4
3 / -1

Two positive integers p and q are in the ratio 7:13. If 14 and 26 are subtracted respectively from p and q, the ratio of the resulting numbers, which are positive, taken in the same order, is again 7:13. what is the second least value of (p+q).

A
60

B
80

C
70

D
100

Solution
Question 5
3 / -1

Roger and Misha attempted to solve a quadratic equation. Roger made a mistake in writing down the constant term. He ended up with the roots (7, 3). Misha made a mistake in writing down the coefficient of x. She got the roots as (3, 3). Find the exact roots of the original quadratic equation.

A
(9, 1)

B
(-9, 1)

C
(5, 4)

D
(5, 4)

Solution

Calculation:

Since Roger made a mistake in the constant term, so his product of roots will be wrong but his sum of roots will be correct.

∴ The sum of roots = 7 + 3

⇒ 10.

Misha made a mistake in the coefficient of ‘x’, so her sum of roots will be wrong but the product of roots will be correct.

Hence, the product of roots = 3 × 3

⇒ 9.

∴ The required equation is x² – 10x + 9 = 0.

∴ The roots of this equation are 9 and 1.
Question 6
3 / -1

In how many ways can eight letters be selected from the word CONTINUATION such that at least five distinct letters are included in it?

A
94

B
92

C
95

D
91

Solution

The letters of the word CONTINUATION are:

C ; U ; A ; O,O ; T,T ; I,I ; N, N, N

Now, we have to select 8 letters out of 12 letters such that at least 5 distinct letters. The possibilities are as follows:

Total number of ways = 4 + 6 + 30 + 30 + 24 = 94
Question 7
3 / -1

The ratio of men and women in shift I workers is 1 : 3 and ratio between the women and men in shift II workers is 2 : 3. When both shift workers are present then ratio between the men and women is 23 : 27. Find the ratio between the men of shift II and shift I.

A
5 : 3

B
2 : 5

C
18 : 5

D
5 : 2

Solution

Calculation:

Let number of men and women workers of shift I be a and 3a respectively.

Let number of men and women workers of shift II be 3b and 2b respectively.

⇒ (a + 3b)/(3a + 2b) = 23 : 27

⇒ 27a + 81b = 69a + 46b

⇒ 42a = 35b

⇒ 6a = 5b

⇒ a/b = 5/6

Ratio of men of shift II and shift I = 3b : a

= 3 × 6 : 5 = 18 : 5

∴ Ratio of men of shift II and shift I = 18 : 5.

Important Points

Read the question properly and ratio in which two group mixed to form third group.
Question 8
3 / -1

The number x is a five-digit number with no Os and the last 3 digits of x form a perfect cube divisible by 8. When the digits of x are considered in the reverse order, the resulting
number also shows the same property. How many such numbers are possible?

A
1

B
2

C
3

D
4

Solution

Cubes divisible by 8 and with no Os are 216 and 512. So, the last 3 digits that are divisible by 8 can be either 216 or 512.
When the number is reversed, the same property holds. Hence the number should be a palindrome.

So, the number could be 21512 or 61216. i.e., 2 numbers are possible.
Question 9
3 / -1

If the vertices of an octagon are used to form a triangle, what is the probability that the triangle doesn't have any side common with the octagon?

A
2/7

B
3/7

C
1/7

D
4/7

Solution

Step 1: Calculate the total number of triangles that can be formed from the vertices of the octagon.

Step 2: Calculate the number of triangles that have at least one side in common with the octagon.

Triangles with exactly one common side:

8 sides × 4 choices for the third vertex = 32 triangles

Triangles with exactly two common sides (adjacent vertices):

8 triangles

Total triangles with at least one side in common:

32 + 8 = 40 triangles

Step 3: Calculate the number of favorable triangles that do not share any side with the octagon.

Favorable triangles = 56 - 40 = 16.

Step 4: Calculate the probability.

Final Answer: The probability that the triangle doesn't have any side in common with the octagon is 2/7.
Question 10
3 / -1

A trader under weighs rice and gets a profit of 40%. Had he used the correct weight he would have still got a profit of 19%. What weight does he use for 1 kg?

A
800gm

B
900gm

C
880gm

D
850gm

Solution

Given:

Profit when false weight is given = 40% and Profit when actual weight is given = 19%

let, the the cost of 1kg rice = 100rs

then, the selling price should 1.19 × 100rs = 119rs

now, at this selling price when false weight is given he gets 40% profit.

⇒ 1.4 × X = 119

⇒ X = 119/1.4 = 85rs

85rs is correspondent to 850gm. Thus, answer is 850gm.