Mathematics Mock Test - 10 - CDS MCQ

The sum of the first 100 natural numbers is:

=  (n * (n + 1)) / 2
=  (100 * 101) / 2
=  50 * 101

101 is an odd number and 50 is divisible by 2.
Hence, 50 * 101 will be divisible by 2.

The factors of 1080 which are perfect square:

1080 → 2³ × 3³ × 5

For, a number to be a perfect square, all the powers of numbers should be even number.

Power of 2 → 0 or 2
Power of 3 → 0 or 2
Power of 5 → 0 

So, the factors which are perfect square are 1, 4, 9, 36.
Hence, Option B is correct.

Given:

73 × 75 × 78 × 57 × 197 × 37 is divided by 34

Calculation:

73 × 75 × 78 × 57 × 197 × 3734

We have taken individual remainder like

When 73 is divided by 34 gives remainder is 5

Similarly

Assume one person is born every day. In 100 years, there will be 25 leap years. So 25*1 additional people will be born on these days.
So, total people born will be = 365 x 100 x 1 + 25 x 1
And people born on 29th february = 25 x 1
Hence percentage will be 

Seema saved Rs. 900 in the first 3 months. She must saved Rs. (11400 – 900) = Rs. 10500 in the subsequent months.
The sequence will be of the form: 350 + 400 +……….. n terms = 10500

Solving, we get n = 15
The savings of Rs. 10500 is done in 15 months. Seema saved Rs. 11400 in 15+3 = 18 months.
Hence, option A is the correct answer.

24220 / 1211 = 20

Let the score of Om =x
Total marks of all the students = 21*62 +x
As per the question, The average score of all the 22 students is one more than the average score of the 21 students other than Dev,
Or (21*62 +x)/22 -1 = (21*62 +x – 82.5)/21
(21*62 + x -22)/22 = (21*62 +x – 82.5)/21. Therefore, x =51

At present the total age of the family = 5 × 20 =100
The total age of the family at the time of the birth of the youngest member,
= 100 - 10 - (10 × 4)
= 50
Therefore, average age of the family at the time of birth of the youngest member,
= 50/4 =12.5

► If new values of x, y, z are x′, y′ and z′, and respectively then x′ :  y′ = 4 : 5, y′ :  z′ = 3 : 4

⇒ x′ :  y′ :  z′ = 12 : 15 : 20
⇒ x + y + z = 5000
⇒ x′ + 50 + y′ + 100 + z′ + 150 = 5000 x′ + y′ + z′ = 4700
⇒ 12k + 15k + 20k = 4700 k = 100

► x = 1200 + 50 = 1250
► y = 1500 + 100 = 1600 z = 2000 + 150 = 2150
► x + y = 1250 + 1600 = 2850

Correct option is C
Time taken by first sprinter 
= 100/10 = 10sec
Time taken by second sprinter 
= 100/80 = 12.5sec
Difference = 12.5 - 10 = 2.5 sec

Since the second rabbit covers 7/120 of the distance in 2 hours 30 minutes
⇒ it covers 8.4 / 120 = 7% of the distance in 3 hours.

Thus, in 3 hours both rabbits together cover 15% of the distance which means 5% per hour so they will meet in 20 hours.

The ratio of speeds = 8 : 7.
⇒ the second rabbit would cover 700 ft to the meeting point in 20 hours and its speed would be 35 feet/hr.

So Option D is correct

Sum of the roots and the product of the roots are -20 and 3 respectively.

In f(x) = ax² + bx + c
When a > 0 and D < 0
Then f(x) is always positive.
x² + 2ax + 10 − 3a > 0, ∀x ∈ R

⇒ D < 0
⇒ 4a² − 4(10 − 3a) < 0
⇒ a² + 3a − 10 < 0
⇒ (a+5)(a−2) < 0
⇒ a ∈ (−5,2)

Let's consider x-5 as 'p'

Case 1: p ≥ 0

|x - 5| |2 + 5|x - 5| - 24 = 0

p² +5p - 24 = 0

p² + 8p - 3p - 24 = 0

p(p + 8) -3 (p + 8) = 0

(p + 8) (p - 3) = 0

p = -8 and p = 3

x - 5 = 3,x = 8 This is a real root since x is greater than 5.

x - 5 = -8, x = -3. This root can be negated because x is not greater than 5.

Case 2: p < 0

p² - 5p - 24 = 0

p² - 8p + 3p - 24 = 0

p=8, -3

x - 5 = 8, x = 13. This root can be negated because x is not less than 5

x - 5 = -3, x = 2. This is a real root because x is less than 5.

The sum of the real roots = 8 + 2 = 10

|a| ≠ |b| ≠|c| and -10 ≤ a, b, c ≤ 10

Expression: [abc - (a + b + c)]

For maximum value, two of a,b and c should be negative, as all three negative will make abc negative.

Thus, max value will occur if a= -10, b = -9, c = 8

⇒ Max value = [(-10 × 9 × 8)-(-10-9+8)]

= 720 + 11 = 731

Cos A = 1 - Cos²A
⇒ Cos A = Sin²A
⇒ Cos²A = Sin⁴A
⇒ 1 – Sin²A = Sin⁴A
⇒ 1 = Sin⁴A + Sin²A
⇒ 1³ = (Sin⁴A + Sin²A)³
⇒ 1 = Sin¹²A + Sin⁶A + 3 Sin⁸A + 3 Sin¹⁰A
⇒ Sin¹²A + Sin⁶A + 3 Sin⁸A + 3 Sin¹⁰A – 1 = 0

On comparing,
a = 1, b = 3 , c = 3 , d = 1
⇒ (a+b)/(c+d) = 1

Hence, the answer is 1

L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
∴ Required number = (90 x 4) + 4 = 364.

Greatest number of 4−digits is 9999.

Now, 15=3×5

25 = 5×5

40 = 2×2×2×5

and 75 = 3×5×5

L.C.M. of 15,25,40 and 75 is 2×2×2×3×5×5 = 600.

On dividing 9999 by 600, the remainder is 399.

Required number = (9999−399) = 9600.

The cost per toffee = 75 / 125 = Rs. 0.6 = 60 paise.
Cost of 1 million toffees = 600000.
∵ There is a discount of 40% offered on this quantity.
Thus, the total cost for 1 million toffees is 60% of 600000 = 360000 

So option C is correct

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x)

⇒ 28x - 22x = 350800 - (13900 * 22)
⇒ 6x = 45000
⇒ x = 7500
So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400

SI = PRT / 100
⇒ (SI x 100) / RT = P
⇒ P = (6200*100) / 32
⇒ P = 19375

• Total number of outcomes possible when a die is rolled, n(S) = 6 (? 1 or 2 or 3 or 4 or 5 or 6)
• E = Event that the number shown in the dice is divisible by 3 = {3, 6}
  Hence, n(E) = 2

Let S be the sample space.

• n(S) = Total number of ways of selecting 3 students from 25 students = ²⁵C₃

Let E = Event of selecting 1 girl and 2 boys

• n(E) = Number of ways of selecting 1 girl and 2 boys

15 boys and 10 girls are there in a class. We need to select 2 boys from 15 boys and 1 girl from 10 girls

Number of ways in which this can be done: 
¹⁵C₂ × ¹⁰C₁
Hence n(E) = ¹⁵C₂ × ¹⁰C₁

Given expression = 1/log₆₀ 3 + 1/log₆₀ 4 + 1/log₆₀ 5
= log₆₀ (3 x 4 x 5)
= log₆₀ 60
= 1.

Explanation:

I.(a - 3)(a - 4) = 0
=> a = 3, 4

II. (b - 2)(b - 1) = 0
=> b = 1, 2
=> a > b

I. x2 + 3x + 2x + 6 = 0
=> (x + 3)(x + 2) = 0 => x = -3 or -2
II. y2 + 7y + 2y + 14 = 0
=> (y + 7)(y + 2) = 0 => y = -7 or -2
No relationship can be established between x and y.