SNAP 2024 Quantitative Aptitude & Data Interpretation
Q1–10 of 20If N= a*b where a,b are prime numbers, then what are the number of factors of N3 /b2
Given: N = a × b, where a and b are prime numbers.
Substitute: We need to find factors of N³ / b².
Substitute
N:(a × b)³ / b².Expand numerator:
(a³ × b³) / b².
Simplify: Using exponent rules (x^m / x^n = x^(m-n)):
b³ / b² = b¹.The expression simplifies to:
a³ × b¹.
Formula for Factors: If a number is x^p × y^q (where x, y are primes), the total number of factors is (p + 1)(q + 1).
Calculation:
Power of
ais 3.(3 + 1) = 4.Power of
bis 1.(1 + 1) = 2.Total Factors =
4 × 2 = 8
X is a single digit number such that the LCM of the two digit number X6 and 54 is 108. What is the value of
X?
We are looking for a digit X for the number X6 (a two-digit number ending in 6) such that LCM(X6, 54) = 108.
Analyze 54: Prime factorization is
2 × 3³(2 × 27).Analyze 108: Prime factorization is
2² × 3³(4 × 27).Deduce X6:
For the LCM to be
2² × 3³, the numberX6must contribute the2²factor (since 54 only has2¹).So,
X6must be divisible by 4.Let's test the options. If
X=3, the number is 36.
Verification:
Factors of 36:
2² × 3².Factors of 54:
2¹ × 3³.LCM = Max power of primes =
2² × 3³=4 × 27= 108.This matches the condition. Therefore,
X = 3.
If one of the roots of ax2 + bx = c = 0is half of the other root, then which of the following conditions
must be true?
Let the equation be ax² + bx + c = 0.
Let the roots be α and β.
Condition: One root is half the other. So, let α = k and β = 2k.
Sum of Roots:
α + β = -b/ak + 2k = 3k.3k = -b/a=>k = -b / 3a.
Product of Roots:
α × β = c/ak × 2k = 2k².2k² = c/a.
Substitution: Substitute the value of
kfrom Step 1 into the equation in Step 2.2(-b / 3a)² = c/a2(b² / 9a²) = c/a
Simplify:
Multiply both sides by
9a².2b² = (c/a) × 9a²2b² = 9ac.
In a two-digit number, the difference between the digits is 4, and 10 times the number is equal to 14 times
the sum of the number obtained by adding the digits of the original and the number formed by reversing the
digits. Find the number.
Check Condition 1: "Difference between digits is 4".
Option A (84):
8 - 4 = 4. (Correct)
Check Condition 2: "10 times the number equals 14 times the sum of (Sum of Digits + Reversed Number)".
Original Number: 84.
Sum of Digits:
8 + 4 = 12.Reversed Number: 48.
Sum of (Digits + Reversed):
12 + 48 = 60.LHS (10 × Number):
10 × 84 = 840.RHS (14 × Sum):
14 × 60 = 840.
Conclusion: Since LHS = RHS, 84 is the correct number.
. A large cube of volume 64 cubic metre is broken down into small cubes of volume 8 cubic metre each. The
percentage increase in the surface area is
Large Cube:
Volume = 64. Side = ∛64 = 4.
Surface Area =
6 × side²=6 × 4²=6 × 16= 96.
Small Cubes:
Volume = 8. Side = ∛8 = 2.
Number of Cubes = Total Volume / Small Volume =
64 / 8= 8 cubes.Surface Area of ONE small cube =
6 × 2²=6 × 4= 24.Total Surface Area =
8 cubes × 24= 192.
Percentage Increase:
Increase = New Area - Old Area =
192 - 96= 96.% Increase =
(Increase / Old Area) × 100% Increase =
(96 / 96) × 100= 100%.
A, B and C are employed to do a piece of work for Rs.700. A and B are supposed to finish 17/28th of work
together. Then, Amount that is paid to C is?
Total Work: Represented as 1 unit.
Work by A and B: They finish 17/28 of the work.
Work by C:
Remaining Work =
1 - 17/28=11/28.
Calculate Wage: Wages are distributed proportional to work done.
Total Wage = Rs. 700.
C's Share =
(11/28) × 700.Calculation:
700 / 28 = 25.11 × 25= Rs. 275.
A tank is fitted with 3 pipes A, B and C. A takes twice as much time to fill the tank as B takes to empty the
tank. B takes twice as much time to empty the tank as C takes to fill the tank. If it is known that C takes 10
hours to fill the tank, in how much time three quarters of the tank will be filled provided that all the taps
were opened simultaneously?
Rate of Pipe C (Fill):
Time = 10 hours.
Rate =
1/10(or 10%) per hour.
Rate of Pipe B (Empty):
Takes twice as long as C. Time =
2 × 10 = 20hours.Rate =
-1/20(or -5%) per hour (negative because it empties).
Rate of Pipe A (Fill):
Takes twice as long as B. Time =
2 × 20 = 40hours.Rate =
1/40(or 2.5%) per hour.
Combined Rate:
1/40 - 1/20 + 1/10.Common denominator (40):
(1 - 2 + 4) / 40=3/40.Alternatively in %:
2.5% - 5% + 10%= 7.5% per hour.
Calculate Time:
Target: Fill 3/4 of the tank (75%).
Time =
Target / Rate=0.75 / 0.075(or75% / 7.5%).Time = 10 hours
Sam and Ram run a 100 m race. Sam beats Ram by 20 m. Sam gives Ram a head start such that both Ram
and Sam will be meeting each other at the 75 m mark. What will be the result of the race?
Original Ratio:
Sam runs 100m, Ram runs 80m (since Sam beats him by 20m).
Speed Ratio Sam:Ram = 5:4.
Scenario 2:
They meet at the 75m mark. This means Sam ran 75m.
In the time Sam runs 75m, Ram runs
(4/5) × 75= 60m.For them to meet, Ram must have had a head start of
75 - 60= 15m.
Race Conclusion:
The race is 100m. Sam has 25m left to run (
100 - 75).In the time Sam runs 25m, Ram runs
(4/5) × 25= 20m.Ram was at 75m. He runs 20m more. He reaches
75 + 20= 95m.Sam finishes at 100m. Ram is at 95m.
Result: Ram loses by 5 meters.
If cosec(30°) + cot(60°) = y. What is the value of y?
Values:
cosec(30°)=1 / sin(30°)=1 / (1/2)= 2.cot(60°)=1 / tan(60°)=1 / √3.
Expression:
2 + (1 / √3).
Combine:
Multiply 2 by
√3/√3to get common denominator.(2√3 / √3) + (1 / √3).(2√3 + 1) / √3.
If a2 = b3 = c4,
b2 = d5 ,
then find the value of loga b x logcd.
Given: a² = b³.
This implies
a = b^(3/2).Therefore,
logₐ(b)=1 / log_b(a)=1 / (3/2)= 2/3.
Given: c⁴ = b³ and b² = d⁵.
c = b^(3/4).d = b^(2/5).
Calculate logc(d):
log_c(d)=log(d) / log(c)(Change of base).log(b^(2/5)) / log(b^(3/4)).(2/5) / (3/4)=(2/5) × (4/3)= 8/15.
Final Product:
logₐ(b) × log_c(d)=(2/3) × (8/15).16 / 45.