CTET Practice Test: Mathematics - 5 - CTET & State TET MCQ

The length of each bar gives the required information.

The prime factorization of 36 is: 2 x 2 x 3 x 3

The prime factorization of 72 is: 2 x 2 x 2 x 3 x 3

Eliminate the duplicate factors of the two lists, then multiply them once with the remaining factors of the lists to get LCM = 2 x 2 x 2 x 3 x 3 = 72 

Let 3 consecutive terms A.P is a –d, a , a + d. and the sum is 51

so, (a –d) + a + (a + d) = 51
⇒ 3a –d + d = 51
⇒ 3a = 51
⇒ a = 17

The product of first and third terms is 273

So  it stand for ( a –d) (a + d) = 273
⇒ a² –d² = 273
⇒ 172 –d ² = 273
⇒ 289 –d ² = 273
⇒  d ² = 289 –273
⇒ d ² = 16
⇒ d = 4

Hence the 3rd terms ( a+d ) is 21

The Roman numeral XXX is equivalent to the Hindu-Arabic numeral 30.

Explanation:

• X represents 10 in Roman numerals.
• XXX means 10 + 10 + 10 = 30.

Since op is perpendicular to PQ, the ∠OPQ = 90°
Now, in right angled triangle OPQ,

The Number is 428721.
2 is in place of tenth and ten thousand.
Those two numbers are 20 and 20,000
So,
The product of those two numbers are :
=> 20 × 20,000
=> 4,00,000

Step 1: Question statement and Inferences

It is given that

 x+y=−10 ....(i)

xy=16  ....(ii)

We need to find which of the given statements must be true.

Step 2 & 3: Finding required values and calculating the final answer

Squaring (i) on both sides, we get:

(x+y)²=100

x²+y²+2xy=100

....(iii)

We know that

(x−y)²=x²+y²−2xy

(x−y)²=x²+y²+2xy−4xy

(x−y)²=(x+y)²−4xy

Using (ii) and (iii) we get:

(x−y)²=100−64

(x−y)²=36

⇒(x−y)=±6

Therefore (I) is not a must be true statement.

Now let us look at (II).

(iii) – 2*(ii) gives us:

(x²+y²+2xy)−2xy=100−32

⇒x²+y²=68

Therefore statement II is true.

We already observed above that

(x−y)²=36

We know that

36=35+136=7∗5+1

Therefore

(x−y)²

 leaves a remainder of 1 when divided by 7.

Answer: Option (E)

Since Tangents from an external point to a circle are equal.
∴ PA = PB = 4 cm.
BR = CR = 5 cm CQ = AQ = 6 cm ⇒ Perimeter of ΔPQR = PQ + QR + RP
⇒ Perimeter of ΔPQR = PA + AQ + QC + CR + BR + PB  ⇒ Perimeter of ΔPQR = 4 + 6 + 6 + 5 + 5 + 4
⇒ Perimeter of Δ PQR = 30 cm 

To solve this problem, we need to find a number that satisfies the following conditions:

1. When divided by 2, 3, 4, 5, or 6, the remainder is 1.
2. The number is divisible by 7.
3. The number lies between 250 and 350.

Let's start by finding the least common multiple (LCM) of 2, 3, 4, 5, and 6, which is the smallest number divisible by all of these numbers.

LCM(2, 3, 4, 5, 6) = 60

We need to find a number of the form 7k, where k is an integer, that leaves a remainder of 1 when divided by 60. The numbers in this sequence can be expressed as 60n + 1, where n is an integer.

Now, let's find the first few numbers of the form 60n + 1 that are divisible by 7 and lie between 250 and 350:

• For n = 4: 60(4) + 1 = 241 (not divisible by 7)
• For n = 5: 60(5) + 1 = 301 (divisible by 7)

So, the number we're looking for is 301.

Now, let's find the sum of its digits: 3 + 0 + 1 = 4

Therefore, the sum of the digits of the number is 4.

From the above figure we can see that the line segments joining the centre of circular faces is perpendicular to the base.

Given equation is

And, we need to find the value of x, given that x is an integer.

Step 2 & 3: Finding required values and calculating the final answer

Let us rewriting the last term of (I) to get x-1 in the denominator:

Let us now bring the terms with the same denominator together:

• (x + 1)² = {2(x – 1)}²
• [(x+1)²-{2(x-1)}² ]=0 ………….. (II)

• Observe that this is of the form  a² – b² = (a+b)(a-b)

   

  So let us say:

  a = x + 1             

  b = 2(x – 1) = 2x - 2

   

  Therefore:

  a + b = x + 1 + 2x – 2 = 3x -1

  a – b = x + 1 - 2x + 2 = -x + 3

   

  Therefore we can rewrite (II) as:

  (3x-1)(3-x)=0

• Putting each factor to zero:

Since it is given that x is an integer,

Thus, x = 3 is the required answer.

Answer: Option (E)

In ∠BOC

⇒ x + y = 80°

⇒ 2x + 2y. = 160°

Also, 2x + 2y + 2z = 180°

⇒ 160° + 2z = 180°

⇒ ∠BAC = 2z = 20°

2,3,5,7,11,13,17,19,23,29,31,37,41,43,47

A = 11 and
B = 17
17-11 = 6

Given, BC = 6 cm and ∠B= 45°
We know that, the construction of a triangle is not possible, if sum of two sides is less than or equal to the third side of the triangle.
i.e., AB + BC < AC
⇒ BC < AC – AB
⇒  6 < AC-AB
So, if AC – AB= 6.9 cm, then construction of ΔABC with given conditions is not possible.

Lets break the figure -
It has 6 surfaces.
Area of bottom surface = L * B
Area of Top Surface = L * B
Area of Right side surface = B * H
Area of Left Side surface  = B * H
Area of front Surface = H * L
Area of Back surface = H * L 
Total surface area of Cuboid = sum of all the surfaces.
=> (Bottom surface area + Top surface area + Right side surface area + Left side surface area + front side surface area
back side surface area)
=> (LB + LB + BH + BH + HL + HL)
=> (2LB + 2BH + 2HL)
=> 2(LB + BH + HL)
hence option B is correct.

Given :

The ratio of the income of X and Y is 4 : 3.

The ratio of monthly expenses of X and Y is 3 : 2. 

X and Y save 6000 rupees each month.

Concept used :

Savings = Income - expense

Calculations :

Let the ratio of monthly income of X and Y be 4a and 3a respectively. 

Let the ratio of monthly expenses of X and Y be 3b and 2b respectively. 

Savings of X = 4a - 3b

4a - 3b = 6000      ....(1) 

Savings of Y = 3a - 2b 

3a - 2b = 6000      ....(2) 

Solving equation 1 and 2 

We get a = 6000 and b = 6000

Total monthly income of X and Y = 4a + 3a = 7a 

⇒ 7 × 6000 

⇒ 42000 rupees 

∴ Option 2 is the correct answer.

• Step 1: Find the prime factors of each number




• 17 = 17 (prime number)


• 5 = 5 (prime number)




• Step 2: Identify the common and uncommon prime factors




• Common prime factors: None


• Uncommon prime factors: 17, 5




• Step 3: Multiply all the prime factors to get the L.C.M.




• L.C.M. = 17 x 5 = 85

Volume of cube = (side of cube)³
V = 4 * 4 * 4 cm³ 
V = 64 cm³ 
Hence the option B is correct.

A data is a collection of numbers gathered to give some information.

The alphabet 'B' has horizontal reflection because half a figure is a mirror image of the other half.