Mathematics Mock Test - 7 - CDS MCQ

A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. So red light fashes after every 1/3 min and green light flashes every 2/5 min. LCM of both the fractions is 2 min.
Hence, they flash together after every 2 min. So in an hour they flash together 30 times.

The Correct Answer is C: 10

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The size of each row would be the HCF of 363, 429 and 693. Difference between 363 and 429 =66.

Factors of 66 are 66, 33, 22, 11, 6, 3, 2, 1.

66 need not to be checked as it is even and 363 is odd. 33 divides 363, hence would automatically divide

429 and also divides 693. Hence, 33 is the correct answer for the size of each row.

For how many rows would be required we need to follow the following process:

Minimum number of rows required = 363/33 + 429/33 + 693/33 = 11 + 13 + 21 = 45 rows.

Therefore, the correct answer is A

Let total sales be ‘x’

The commission that Sailesh will get is x – 265000

He gets 6% on sales upto 100000 and 5% on sales greater than that.

Calculating his commission on total sales:

0.06*100000 + 0.05(x-100000)

Equating,

0.05x + 1000 = x – 265000

0.95x = 266000

x= 280000

Hence, his sales were worth 280,000

Let the CP of the each toy be “x”. CP of 12 toys will be “12x”. Now the shopkeeper made a 10% profit on CP. This means that

12x(1.1)= 2112 or x=160 . Hence the CP of each toy is ₹160.

Now let the SP of each toy be “m”. Now he sold 8 toys at 20% discount. This means that 8m(0.8) or 6.4m

He sold 4 toys at an additional 25% discount. 4m(0.8)(0.75)=2.4m  Now 6.4m+2.4m=8.8m=2112 or m=240

Hence CP= 160 and SP=240. Hence profit percentage is 50%.

5616 ÷ 18 = 312
312 / 8 = 39

Change in total weight of 10 students = difference in weight of the student who joined and the student

=> weigth of first student who left = 30 + (10×2) = 50

weight of the student who joined last = 50 – 10 = 40...
Thus change in average weight = (40 – 30)/10 = 1...
 

The first ten composite numbers are: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18. 

Required average:

= (4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18) / 10
= 112 / 10
= 11.2

Let the incomes be 4x, 5x, 6x and the spending be 6y, 7y, 8y and savings are (4x–6y), (5x–7y) & (6x–8y)
Sheldon saves 1/4^th of his income.

Therefore:

⇒ 4x – 6y = 4x / 4
⇒ 4x – 6y = x
⇒ 3x = 6y
⇒ x / y = 2
∴ y = x / 2

Ratio of Sheldon’s Leonard’s & Howard’s savings:

= 4x – 6y : 5x – 7y : 6x – 8y
= x : 5x – 7y : 6x – 8y
= x : 5x – 7x / 2 : 6x – 8x / 2
= x : 3x / 2 : 2x
= 2 : 3 : 4 

Since P to R is double the distance of P to Q,
Therefore, it is evident that the time taken from P to R and back would be double the time taken from P to Q and back (i.e. double of 6.5 hours = 13 hours).

Since going from P to R takes 9 hours, coming back from R to P would take 4 hours i.e. 13- 9 = 4

So Option A is correct

Initial distance = 25 dog leaps
Per-minute dog makes 5 dog leaps and cat makes 6 cat leaps = 3 dog leaps
⇒  Relative speed = 2 dog leaps / minutes
⇒  An initial distance of 25 dog leaps would get covered in 12.5 minutes.

So Option D is correct

(a² + b²)x² − 2b(a + c)x + (b²+c²) = 0
Roots are real and equal ∴ D = 0
D = b² − 4ac = 0

⇒ [−2b(a+c)]² − 4(a² + b²)(b² + c²) = 0
⇒ b²(a² + c² + 2ac) −(a²b² + a²c² + b⁴ + c²c²) = 0
⇒ b²a² + b²c² + 2acb² − a²b² − a²c² − b⁴ − b²c² = 0
⇒ 2acb² − a²c² − 2acb² = 0
⇒ (b² − ac)² = 0
⇒ b² = ac

Let the two consecutive positive integers be x and x + 1.

⇒ x² + (x + 1)² - x(x + 1) = 91
⇒ x² + x - 90 = 0
⇒ (x + 10)(x - 9) = 0
⇒ x = -10 or 9.
x = 9 [∵ x is positive]

Hence the two consecutive positive integers are 9 and 10.

Case 1: x ≥ 40/3

we get 3x-20 +3x-40 = 20

6x=80

Case 2
we get 3x - 20 + 40 - 3x = 20
we get 20 = 20
So we get x 
Case 3x  < 20/3
we get 20-3x+40-3x =20
40=6x
x = 20/3
but this is not possible
so we get from case 1,2 and 3

Now looking at options
we can say only option C satisfies for all x .
Hence 7<x<12.

The critical points of the function are -4, -6, -8, … , -98 ( 48 points).

For all integers less than -98  and greater than -4 f(x) > 0 always .

for x= -5, f(x) < 0

Similarly, for x= -9, -13, …., -97 (This is an AP with common difference -4)

Hence, in total there are 24 such integers satisfying f(x)< 0.

We know that,
h² = p² + b² Given, p and b are positive integer, so h2 will be sum of two perfect squares.

We see 
a) 7² + 5² = 74
b) 6² + 4² = 52
c) 3² + 2² = 13
d) Can’t be expressed as a sum of two perfect squares

Therefore the answer is Option D.

Given numbers are 43, 91, and 183.
Subtract smallest number from both the highest numbers.
We have three cases:
183 > 43; 183 > 91 and 91 > 43
Therrefore,
183 – 43 = 140
183 – 91 = 92
91 – 43 = 48
Now, we have three new numbers: 140, 48 and 92.

HCF (140, 48 and 92) = 4
The highest number that divides 183, 91, and 43 and leaves the same remainder is 4.

A loss of 25% means a cost price of 100 corresponding to a selling price of 75. CP as a percentage of the SP would then be 133.33%

Alternatively,

Cost price = 1025
Loss% = 25%
SP = 1025 - (1025 * 25/100) = 768.75
⇒ 768.75 * (x/100) = 1025
⇒ 768.75x = 102500
⇒ x = 102500/768.75 = 133.33%

So option D is correct

R sold the table at 20% profit at Rs. 90. Thus R's cost price x 1.2 = 90
R’s Cost price = Rs. 75

We also know that E sold it to R at 25% profit.
Thus, E’s Cost price x 1.25 = 75
⇒ E’s cost price = 60

So option D is correct

Simple Interest (SI) = P N R / 100

P is the Principal loan amount = Rs.1400

N is the number of years of deposit

R is the rate of interest

It is given that the loan period is as many years as the rate of interest.

So, N = R

Interest at the end of the loan period (SI ) = Rs.686

So,

686 = 1400 * R * R /100

R² = 686*100 /1400

R² = 49

R = 7%

Simple Interest = PRT / 100
Here P = Principal, R = Rate of interest and T = time period
Hence, Simple Interest ∝ T
Required Ratio = (Simple Interest for 5 years) / (Simple Interest for 15 years)
⇒ T₁/ T₂ = 5 / 15
⇒ 1 / 3 = 1 : 3

• Let P(QA) = passing QA’s section
• P(V) = passing Verbal section
• So, P(QA/V) = P(QA∩V)/P(V) = 0.6/0.8 = 0.75
  Hence, the correct answer is option A.

Total number of outcomes possible when a coin is tossed = 2 (∵ Head or Tail)

• Hence, total number of outcomes possible when 3 coins are tossed, n(S) = 2 x 2 x 2 = 8
  ​(∵ S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH})

E = event of getting at most two Tails = {TTH, THT, HTT, THH, HTH, HHT, HHH}

• Hence, n(E) = 7

Explanation:

I. (a + 6)(a + 5) = 0
=> a = -6, -5
II. (b + 5)(b + 1) = 0
=> b = -5, -1 => a ≤ b