CAT 2021 Quant Question
A box has 450 balls, each either white or black, there being as many metallic white balls as metallic black balls. If 40% of the white balls and 50% of the black balls are metallic, then the number of non-metallic balls in the box is
Answer:
The box contains 450 balls
Let’s consider the number of white and black balls to be ‘x’ and ‘y’ respectively.
The number of white metallic balls is the same as black metallic balls.
40% of the white balls and 50% of the black balls are metallic.
So, 0.4x =0.5y
x : y = 5 : 4.
Hence the number of White balls is 250 (100 - metallic and 150 - Non-metallic)
Hence the number of Black balls is 200 (100 - metallic and 100 - Non-metallic)
The total number of non-metallic balls is 250.
The answer is '250’
CAT 2021 Quant Question
Anil, Bobby and Chintu jointly invest in a business and agree to share the overall profit in proportion to their investments. Anil’s share of investment is 70%. His share of profit decreases by ₹ 420 if the overall profit goes down from 18% to 15%. Chintu’s share of profit increases by ₹ 80 if the overall profit goes up from 15% to 17%. The amount, in INR, invested by Bobby is
A. 2400
B. 2200
C. 2000
D. 1800
Answer:
Anil’s share of investment is 70%. His share of profit decreases by ₹ 420 if the overall profit goes down from 18% to 15%.
Let‘s consider the total amount invested by the three to be ‘x’
So, 70% of (18% of x - 15% of x) = 420.
70% of 3% of x = 420
x = 20,000.
Chintu’s share of profit increases by ₹ 80 if the overall profit goes up from 15% to 17%.
Let’s consider the chintu’s percentage share to be ‘c’.
So, c% of 2% of 20,000 = 80.
c = 20%.
Therefore, the percentage of the amount invested by Bobby is 10%
Hence, 10% of 20,000 is 2000.
The answer is '2000’
Choice C is the correct answer.
CAT 2021 Quant Question
If a rhombus has area 12 sq cm and side length 5 cm, then the length, in cm, of its longer diagonal is
Answer:
Let’s consider the length of the longer and shorter diagonal of the rhombus to be ‘2a’ and ‘2b’ respectively.
The area of the rhombus = 12 sq cm
ab = 6.
The side of the rhombus = 5 cm.
By solving eq(1) & (2) we get
Long diagonal = 2a = √(37) + √(13)
Hence, the answer is '√(37) + √(13)'
Choice C is the correct answer.
CAT 2021 Quant Question
Three positive integers x, y and z are in arithmetic progression. If y − x > 2 and xyz = 5(x + y + z), then z − x equals
A. 8
B. 10
C. 14
D. 12
Answer:
Given that, x, y, z are in arithmetic progression.
Also given that xyz = 5(x + y + z).
xyz = 5(3y). Since x, y, z are in arithmetic progression.
xz = 15.
The possible combinations for x and z are (3, 5) (1, 15).
z - x > 4 (Since y - x > 2).
Therefore,
x = 1, y = 8, z = 15
Hence, z - x = 14.
Hence, the answer is '14’
Choice C is the correct answer.
CAT 2021 Quant Question:
Let D and E be points on sides AB and AC, respectively, of a triangle ABC, such that AD : BD = 2 : 1 and AE : CE = 2 : 3. If the area of the triangle ADE is 8 sq cm, then the area of the triangle ABC, in sq cm, is
Answer
The area of the triangle ADE = 8 sqcm.
xy × sin(A) = 4
The area of the triangle ABC = 
= 30 sq cm
The answer is 30
CAT 2021 Quant Question:
For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4-digit number satisfying the above conditions is
Answer
For a 4-digit number we can make 4 placeholders to represent the four digits
Now, we know that the sum of the thousands, hundreds and the tens place digit is 14, and the sum of the hundreds, tens and the units place digit is 15.
So we can conclude that the unit’s digit is one unit greater than the thousandth place.
To maximize the 4-digit number the value of a should be as large as possible. And the value of the tens digit can be 9 at most. So the maximum value of a is 4.
To satisfy the sum of the thousands, hundreds and the tens place digit, the value of the hundreds place should be equal to 1.
Hence the largest possible number is 4195
The answer is 4195
CAT 2021 Quant Question:
Answer:
We have a function f(x) = 
We can complete the squares in the numerator as
The minimum value of k = 3, as the square part is always positive and another positive number is added to the equation.
So the fraction will have the minimum value when k = 3
And, the maximum value when k approaches its maximum value. We can observe that the fraction has a 2k + 1 in the denominator and so it will always be one more than twice the numerator. As the value of k increases the value of the addition of one gets less and less of a value to the overall denominator. The fraction approaches 1/2 , but never 1/2 itself.
Hence the range becomes 
Choice C is the correct answer.
CAT 2021 Quant Question:
In a football tournament, a player has played a certain number of matches and 10 more matches are to be played. If he scores a total of one goal over the next 10 matches, his overall average will be 0.15 goals per match. On the other hand, if he scores a total of two goals over the next 10 matches, his overall average will be 0.2 goals per match. The number of matches he has played is
Answer:
Let the number of matches already played be x, then over the next ten matches if one more goal is scored, the average goals become 0.15. Further if the player scores two goals the average becomes 0.2.
From this we can see that one extra goal is increasing the average by 0.05 (0.2 - 0.15). Hence it can be said that the total number of matches played till the end is 20.
Therefore, x + 10 = 20
Hence, x = 10
The answer is '10’
Answer:
2.25 ≤ 2 + 
has to be either lesser than or equal to 202.
Hence, the value of n can be 5 at most.
Further, subtracting 2 from the first two parts of the inequality,
Hence, n ≥ -4
So from the inequality we can say that the integral value of n can be [-4, 5].
For 3 + 
have integral values, n but be greater than or equal to -1 and go up till 5.
Hence the number of values of n can be -1, 0, 1, 2, 3, 4, 5.
7 values
Hence, the answer is 7
CAT 2021 Quant Question:
For all real numbers x the condition |3x - 20| + |3x - 40| = 20 necessarily holds if
A. 6 < x < 11
B. 7 < x < 12
C. 10 < x < 15
D. 9 < x < 14
Answer:
Let’s assume y to be 3x,
Now, the equation becomes | y - 20 | + | y - 40 | = 20
Let’s comprehend the meaning of | y - 20 |,
| y - 20 | simply means, the distance between y and 20
Let’s jot the three terms, y, 20 and 40 on a number line.
Now, y could be anywhere on the number line.
Observe that if y is on the left of 20, the distance between y and ‘40’ will be greater than 20 and the condition | y - 20 | + | y - 40 | = 20 fails.
If y is on the right of 40, the distance between y and ‘20’ will be greater than 20 and the condition | y - 20 | + | y - 40 | = 20 fails.
Hence the point y needs to be between 20 and 40.
Will any point between 20 and 40 qualify to be ‘y’ ?
Yes, Since the distance between 20 and 40 is exactly 20, for any ‘y’ between 20 and 40, the distance between 20 and ‘y’ plus the distance between ‘y’ and 40 is exactly 20.
So, 20 < y < 40
20 < 3x < 40
20/3 < x < 40/3
Option B alone holds good.
Alternate Solution:
We know that the minimum value of the modulus function can be zero.
Thus value occurs in the two modulus functions at 20/3 and 40/3 respectively.
So, for numbers less than 20/3 the value of both the functions increases and similarly, the value of the sum of the functions increases when x is greater than 40/3.
Between 20/3 and 40/3, the value of the functions remains the same as one of the inequality increases by the same amount as the decrease in the other.
So, the value of the sum of the modulus remains at 20 for the range of values [20/3 , 40/3].
Which can be rewritten as [6.66,13.33]
The option which satisfies this condition is 7 < x < 12.
Hence, the answer is 7 < x < 12
Choice B is the correct answer.
