CAT Quantitative Aptitude- 1 - Quant Aptitude Free MCQ Test with solutions

We can use the formula

Let the present ages of Akhil and Akash be x and y respectively.

⇒ 5x - 40 = y - 8
⇒ 5x - y = 32
⇒ y = 5x - 32

⇒ 15x + 180 = 7y + 84
⇒ 15x + 180 = 7(5x - 32) + 84
⇒ 15x + 96 = 35x - 224
⇒ 20x = 320
⇒ x = 16
⇒ y = 48
x:y = 16:48 = 1:3
Sum of antecedent and consequent = 1 + 3 = 4

1. Let the marked price of the fridge be Rs f.
   Then marked price of the television is Rs 4f.

2. Intended (correct) discounts:

• Fridge: 20% off ⇒ pay 80% of f ⇒ 0.8f.

• Television: 25% off ⇒ pay 75% of 4f ⇒ 0.75×4f = 3f.

• Correct total payment = 0.8f + 3f = 3.8f.

3. Interchanged (actual) discounts:

• Fridge: 25% off
  ⇒ pay 75% of f
  ⇒ 0.75f.

• Television: 20% off
  ⇒ pay 80% of 4f
  ⇒ 0.8×4f = 3.2f.

• Actual total payment = 0.75f + 3.2f = 3.95f.

4. Required ratio (Actual : Correct) = 3.95f : 3.8f = 3.95 : 3.8

5. Simplify:

• Multiply by 20 to clear decimals: 79 : 76.

The ratio of the books with them is 
LCM of 3, 5, 7 and 11 is 1155
So, they must be having books in the ratio
which is equivalent to 
385 : 231 : 165 : 105
Thus, the minimum number of books they must be having is
385 + 231 + 165 + 105 = 886
Hence, 886 is the correct answer.

We set y = −k with k = 0,1,2,... So the inequality becomes
7x + 9k ≤ 63,
with x = 0,1,2,... .

For a fixed x, compute R = 63 − 7x. We need 9k ≤ R, so k can be any integer from 0 up to the largest k whose 9k is ≤ R. The number of k-values equals that largest k plus 1.

Compute for x = 0 to 9 (because when k = 0 we get 7x ≤ 63 → x ≤ 9):

• x = 0: R = 63. The largest multiple of 9 ≤ 63 is 63 = 9×7, so largest k = 7 → number of k = 7 + 1 = 8.

• x = 1: R = 56. Largest multiple of 9 ≤ 56 is 54 = 9×6, so largest k = 6 → number = 7.

• x = 2: R = 49. Largest multiple of 9 ≤ 49 is 45 = 9×5, so largest k = 5 → number = 6.

• x = 3: R = 42. Largest multiple of 9 ≤ 42 is 36 = 9×4, so largest k = 4 → number = 5.

• x = 4: R = 35. Largest multiple of 9 ≤ 35 is 27 = 9×3, so largest k = 3 → number = 4.

• x = 5: R = 28. Largest multiple of 9 ≤ 28 is 27 = 9×3, so largest k = 3 → number = 4.

• x = 6: R = 21. Largest multiple of 9 ≤ 21 is 18 = 9×2, so largest k = 2 → number = 3.

• x = 7: R = 14. Largest multiple of 9 ≤ 14 is 9 = 9×1, so largest k = 1 → number = 2.

• x = 8: R = 7. Largest multiple of 9 ≤ 7 is 0 = 9×0, so largest k = 0 → number = 1.

• x = 9: R = 0. Largest multiple of 9 ≤ 0 is 0 = 9×0, so largest k = 0 → number = 1.

Now add the counts: 8 + 7 + 6 + 5 + 4 + 4 + 3 + 2 + 1 + 1 = 41.

Consider radius of each smaller circle be r and that of the larger circle be R = 6 cm.
Construct PD such that PD is perpendicular to AB.
In right triangle ADP, AP = DP/cos(APD) = DP/ cos(60) = 2DP = 2r
In right triangle OFA, OF/cos(AOF) = AO ⇒ R/(1/2) = AP+OP  ⇒ 2R=2r+r+R   ⇒ R = 3r
 In right triangle OEP, EP=OPcos(EPO) = (R + r)((√3)/2 = (4R/3)((√3) = 2R/(√3)
PQ=2EP = 4R/√3)
Perimeter = 12R/(√3) = 24√3 cm (R = 6cm)

Let us draw perpendicular from centre of the circle to AB and CD.
​

We are given that ∠OED = 30°, therefore ∠AOE = 90° – 30° = 60°.

Let R be the radius of the given circle. Then, in right-angle triangle ODF:
OD² = OF² + FD²
⇒ OF² = R² – 5²
⇒ OF = √(R² – 25) cm

In right-angle triangle OFE:
sin(OEF) = OF / OE
⇒ 1/2 = OF / OE
⇒ OE = 2 × √(R² – 25) … (1)

Similarly, in right-angle triangle OGA:
OG² = OA² – AG²
⇒ OG = √(R² – 9)

In right-angle triangle OGE:
sin(OEG) = OG / OE
⇒ √3/2 = OG / OE
⇒ OE = (2/√3) × √(R² – 9) … (2)

By equating (1) and (2):
2 × √(R² – 25) = (2/√3) × √(R² – 9)

⇒ 3 × (R² – 25) = R² – 9
⇒ 2R² = 66
⇒ R² = 33
⇒ R = √33 cm

The number of triangles that can be formed for a given perimeter 'p' is given by when p is even and  when p is odd, where [] is the nearest integer function.
The number of scalene triangles that can be formed for a given perimeter 'p' is given by when p is even and if p is odd.
Number of scalene triangles that can be formed with a perimeter of 45 cm = [42² /48] = [36.75]= 37
Therefore, 37 is the correct answer.

In the triangle BOA, tan 
Therefore, θ=60 degrees.
The shaded area = Area of two triangles - area of section of circle.
As the radius is perpendicular to a tangent, triangle OAB is a right angled triangle, with right angle at B, with area = 1/2 ∗ 5 ∗ 5√3​.
Hence area of both triangles= 2 ∗ 1/2 ∗ 5 ∗ 5√3​=25√3​.
The angle BOA = 
Hence, angle BOC = 2*BOA = 120°
Hence the area of the section = area of circle/3 
Hence area of shaded portion= 

Let us draw a line from point E that is parallel to AG and intersects DB at H.

Let f(x) = |x+3|+|x-3|+|x-6|+|x-5|+|x+5|
f(x) is a linear equation in x at all times. It will be a union of straight lines with inflection points at -3, 3, 6, 5 and -5. Hence, the minimum value of the expression occurs at one of these inflection points. We will calculate the value of f(x) at each of these points and then find out the least possible value of f(x)
f(-3) = 25
f(3) = 19
f(6) = 24
f(5) = 21
f(-5) = 31
19 is the least value.

Step 1: Use the equation to express one variable

• From x + 2y = 20, write x = 20 − 2y.

Step 2: Plug into the inequality

• 2x + y > 16 ⇒ 2(20 − 2y) + y > 16

• 40 − 4y + y > 16 ⇒ 40 − 3y > 16 ⇒ 3y < 24 ⇒ y < 8.

Step 3: Apply non-negative condition

• y must be a non-negative integer less than 8: y = 0,1,2,3,4,5,6,7 (8 values).

• For each y, x = 20 − 2y is automatically a non-negative integer (since y ≤ 7 ⇒ x ≥ 6).

Answer

• There are 8 ordered pairs (x,y).

3x + 4y + z = a --> (1)
2x + 6y + 4z = b --> (2)
x - y - 2z = c --> (3)
Eliminating z from (1) and (2) , (1)x4 - (2) : 12x + 16y + 4z - (2x + 6y + 4z) = 4a - b
10x + 10 y = 4a - b
⇒ x + y = (4a - b)/10 - (4)
Eliminating z from (1) and (3), (1)x2 + 3 = 6x + 8y + 2z + (x - y - 2z) = 2a + c
⇒ x + y = (2a + c)/7 - (5)
Equating equations (4) and (5), we get,
(4a - b)/10 = (2a + c)/7
28a - 7b = 20a + 10c
Thus, 8a = 7b + 10c

The equation can be written as follows:


For the logarithms in the question to be defined, x - 1 > 0 ⇒ x > 1
So, the only possible value of x = 5

∣log _(6x+4)(3x−2)∣=1
3x−2>0
x>2/3........(1)
log _(6x+4) (3x−2)=±1
Case 1
log _(6x+4) (3x−2)=1
∴ 6x+4 = 3x-2
∴ 6x+4=3x−2
x=-2
which is not possible according to the equation 1
Case 2

(6x+4)(3x−2)=1
18x² =9

Check domain for each:

x = +1/√2 ≈ 0.7071: base 6x+4 ≈ 8.2426 (>0 and ≠1), argument 3x−2 ≈ 0.1213 (>0). Valid.

x = −1/√2 ≈ −0.7071: base 6x+4 ≈ −0.2426 (negative) and argument negative — invalid.

So only x = +1/√2 is acceptable.

Number of possible values = 1.

For the equation to be defined, 0 < 18 - 3^x
So, the maximum value of x can be 2.
​
Taking 3^x as t, then:
​
Let us take the two roots to be A and B. A and B would be values of 3^x  that satisfy the equation.
However, we need to find the product of 3^x+2 that satisfy the equation. So, we must find the value of (A+2)*(B+2) = AB + 2*(A+B) + 4.
From the quadratic equation t^2-18t+27=0, we know that A+B = 18 and AB=27
So, the required product will be 27+36+4 = 67
So the answer is Option C

Step 1: Find the repeating cycle
The LCM of 3 and 11 is 33. Hence, in every block of 33 numbers, the count of numbers satisfying the condition will be the same.

Multiples of 3 less than 33 that leave odd remainders when divided by 11 are:
3, 9, 12, 18, 27
So, there are 5 such numbers in each block of 33.

Step 2: Extend the sequence
In the next block (34–66), the valid numbers are:
36, 42, 45, 51, 60 (which are 33+3, 33+9, 33+12, 33+18, 33+27)

Step 3: Count number of blocks
Below 1000,
1000 ÷ 33 = 30
Thus, there will be 30 complete blocks.
Between 990 and 1000, the extra valid numbers are 993 and 999.

Step 4: First block sum
S1 = 3 + 9 + 12 + 18 + 27 = 69

Step 5: General series sum

• Second block: S₂ = 69 + (5 × 33) = 69 + 165

• Third block: S₃ = 69 + (2 × 165)

• General block: Sn = 69 + 165 × (n – 1)

Step 6: Total sum of 30 blocks
S = (69 × 30) + [165 × (1 + 2 + 3 + … + 29)]
S = 2070 + 165 × (29 × 30 ÷ 2)
S = 2070 + 71775 = 73845

Step 7: Add the last two terms
S' = 73845 + 993 + 999 = 75837

Minimum time to reach the hypotenuse from A

• Right triangle ABC with right angle at A; legs:
  AB=15 km,
  AC=20 km;
  hypotenuse BC opposite A.

• “Reach the hypotenuse from A” means go from A to the line BC along the shortest path, i.e., the perpendicular from A to BC.

• The perpendicular distance from the right-angle vertex to the hypotenuse equals (product of legs)/hypotenuse.

Compute:

• Hypotenuse BC = √(15²+20²)
  = √(225+400)
  = √625
  = 25 km.

• Distance d from A to BC
  = (AB·AC)/BC
  = (15·20)/25
  = 300/25
  = 12 km.

Speed = 30 km/h.

• Time t = distance /speed = 12/30 h = 0.4 h = 24 minutes.

Answer: 24 minutes.

Minimum of (a−b)²+(a−c)²+(a−d)² given a+b+c+d=30, a,b,c,d∈Z

• Let s=30−a=b+c+d. For fixed a, b²+c²+d² is minimized when b,c,d are as equal as possible (integers differ by at most 1).
• Choose a so that b,c,d cluster around a to shrink (a−b),(a−c),(a−d).
• Take a=8 and (b,c,d)=(7,7,8) (sum 30). Then
• (a−b)²+(a−c)²+(a−d)²
  =(8−7)²+(8−7)²+(8−8)²
  =1+1+0=2.

Time taken by a person to complete 1 job = 120
Work done by a person in 1 day = 1/120
Work done on 2nd day by 2 persons of same efficiency = 1/120 + 2/120 = 3/120
Work done on 3rd day by 3 persons . of same efficiency = 1/120 + 2/120 + 3/120 = 6/120
Work done on nth day by n persons of same efficiency = (1+2+3+4....... + n)/120
1 Job is completed on nth day. So the work done will be equal to 1:
(1+2+3+4....... + n)/120 = 1
n(n+1)/2 = 120 ( Sum of 1st n natural numbers = n(n+1)/2 )
Substituting the values from the options, we'll get n=15