Test: Elementary Mathematics - 1 - CDS MCQ

In ΔPQR,
By Pythagoras theorem,
PR² = PQ² + QR²
As S is the mid-point of QR,
QS = SR ... (1)
This means QR = 2SR
So, PR² = PQ² + 4SR² … (2)

In ΔPQS,
By Pythagoras theorem,
PS² = PQ² + QS²
Also,
PS² = PQ² + SR²(because QS = SR) … (3)

Subtracting (3) from (2), we get
PR² - PS² = PQ² + 4SR² - (PQ² + SR²)
PR² - PS² = 3SR²
PS² = PR² - 3SR²

We know that,
7³ = 343
(7³)²¹ = (343)²¹
Rem[] = Rem[] = Rem[1/342] = 1

x = 111...1 (20 digits), y = 333...3 (10 digits) and z = 222...2 (10 digits)
Therefore,

Thus, is equal to 1.

Statement I is true, but Statement II is false.
This is because there can be two or more readings with the same frequencies.

Lead in the mixture = 1/3 x 25 + 2/5 x 125 = 175/3kg
Tin in the mixture = 2/3 x 25 + 3/5 x 125 = 275/3kg
Required ratio =
⇒ 7 : 11

Consider statement 1:
y² = sec²x + cosec²x

Hence, statement 1 is true.

Now, let us consider statement 2.

Now, cotx =
Or, cot²(x/2) - 2cotx cot(x/2) - 1 = 0
Or, cot(x/2) = (2cotx + (4cot²x + 4)^1/2)/2 (Taking only positive sign as 0° < x < 45°)
Now, the minimum value of cot(x/2) for the given range of x will be when x = 45°.
So, plugging in x as 45°, we get
cot(x/2) = y > (2 + (4 + 4)^1/2)/2
Or, y > (2 + 2(2)^1/2)/2 or y > 2
Hence, statement 2 is correct as well.
Let us now consider statement 3.
LHS of statement 3 =
=
=
=
= cosx + sinx = a say
Now, (cosx + sinx)² = a²
Or, cos²x + sin²x + 2sinxcosx = a²
Or, 1 + sin²x = a²
Now, as sin2x < 1, (as x < 45°), we have a² < 2
Thus, a < 2 as well
Hence, the third statement is false.
Thus, only 2 statements are true.
Hence, answer option 3 is correct.

a^x = b
b^y = c ⇒ a^xy = c
c^z = a ⇒ a^xyz = a¹
⇒ xyz = 1
Hence, this option is not the right answer.

Let α and β be the roots of the equation ax² + bx + c = 0.
So, α + β = -b/a; α​​​​​​​β = c/a
Let 1/α and 1/β be the roots of equation px² + qx + r = 0.
So,

Squaring both sides, we get

On solving, we get b²p² = q²c² ... (1)
Now, from αβ = c/a​​​ and 1/αβ = r/p, we get (c/a) = (p/r) ⇒ ap = cr  ...(2)
Dividing equation (1) by (2), we get: acq² = b²pr

Number of eatables produced by company A : Number of eatables produced by company C
= 15% of 10 : 35% of 7
= 1.5 : 2.45 = 3 : 5

Required percentage = (3/5 × 16% + 3/4 × 21% + 3/8 × 8%) × 100 = x 100 = (567/20) x 100 = 28.35%.

Total number illiterate people in AP and MP = ((7/9) × 27% + (4/5) × 21%) × 32,76,000 = (21 + 16.8) × 32,760 = 12,38,328

Required ratio = = 15/14

Price is raised by 12%.
So, % reduction in consumption so as not to increase the expenditure
= [% Price increase/(100 + % Price Increase)] x 100
= (12/112) x 100
= (1200/112)
= 75/7
= 10(5/7)

Hence, option 4 is the correct answer.

cosθ₁ + cos θ₁ + cosθ₁ = 3 = 1 + 1 + 1
⇒ cosθ₁ = 1 = 0°
⇒ θ₁ = 0°
sin θ₁ = sin 0° = 0

In triangle ABC, BC || PQ.

∴ By BPT (Basic Proportionality Theorem),
AP/PB = AQ/QC
So, =
4x(5x - 3) - 3(5x - 3) = 8x(3x - 1) - 7(3x - 1)
20x² - 12x - 15x + 9 = 24x² - 8x - 21x + 7
0 = 4x² - 2x - 2
2x² - x - 1 = 0
2x² - 2x + x - 1 = 0
2x(x - 1) + 1(x - 1) = 0
x = 1, - (1/2)
Negative value of x is not possible as it would make the sides negative.
So, x = 1 is the correct answer.

Speed of A = 6 km/hr
Speed of A in m/s = (5/3) m/s
Time taken by A in 100 m race = 60 seconds
Let speed of B = x m/s
Time taken by B in (100 − 8) m race = (92/x) seconds
A.T.Q.
(92/x) - 60 = 9

⇒ (92/x) = 69

⇒​​​​​​​ x = (92/69) m/s
Speed of B in km/hr:
4.8 km/hr
This is the required answer.

Area of triangle = 1/2 x Base x Altitude
Given: Base = 6 units & Area of triangle = 12 Sq units
Hence, Altitude = 4 units
Now,Area of quadrilateral ABCD = 1/2 x (sum of parallel sides) x altitude = 1/2 (14 + 8) × 4 = 44 sq units.
Hence, 44 sq units is the correct answer.

x = a cosθ+ b sinθ … (1)
y = a cosθ– b sinθ … (2)
Squaring x and y, we get
⇒ x² = a² cos²θ+ b² sin²θ+ 2ab cosθ sinθ
⇒ y² = a² sin²θ+ b² cos²θ– 2ab cosθ sinθ
⇒ x² + y² = a²(cos²θ + sin²θ) + b²(cos²θ + sin²θ)

Let the number of men who left the job be x.

Clearly, 1-day work of 30 men = 1/40

1-day work of 1 man = 1/(30 x 40)

Now, 24-day work of 30 men = 24/40

Remaining people = (30 - x)

And 40-day work of (30 - x) men =

The work is complete in 64 days.

Hence,


⇒ x = 18
Thus, 18 men left the job.

f(x) = x³ - 4x + p
f(0) = p
f(1) = 1 - 4 + p = p - 3
f(0) and f(1) have opposite signs
By multiplying f(0) and f(1), we get result less than zero.
⇒ p(p - 3) < 0
⇒ 0 < p < 3

(3 tanθ – 2 cotθ)² = (3 tanθ)² + (2 cotθ)² – 2(3 tanθ) (2 cotθ)
= 9 tan²θ + 4 cot²θ - 12
9 tan²θ + 4 cot²θ = (3 tanθ – 2 cotθ)² + 12
= 0 + 12
Minimum value will be 12.

Let x be the value of each installment




x = Rs. 4840

Since CI = P
⇒ 832 = P…(i)
Also, SI =
⇒ 800 =
⇒ P = 40,000/R
∴ From equation (i), 832 =
Or, 832 = 4R + 800
⇒ R = 32/4 = 8%

=
6.5 =
26x = 24x + 10
x = 5