Bihar STET Mathematics Questions Solved Problems with Detailed Solutions [Free PDF]

Formula Used:

Average = Sum of observations/Total no. of observations

Calculation:

Using the above formula -

⇒ Sum of marks in 4 subjects = 91 × 4 = 364

⇒ Average of 6 subjects = \(\frac{364+176}{6}\) = \(\frac{540}{6}\) = 90

∴ The correct answer is 90

Given:

The diameter of a sphere is 3.5 cm

Concept used:

Total surface area = 4πr2

Here r = radius

Calculation:

Total surface area = 4 × 22/7 × (3.5/2)²

⇒ 4 × 22/7 × 12.25/4

⇒ 22/7 × 12.25

⇒ 38.5

∴ The total surface area of the sphere is 38.5 cm2.

The number when divided by 16 or 12 or 8 or 6 or 4 leaves a remainder of 2.

∴ The number when divided by LCM of 16 or 12 or 8 or 6 or 4 that is 48 should also leave a remainder of 2

⇒ When number is divided by 48, then remainder = 2

∴ Number = 48n + 2, where n ϵ N

∴ Smallest number which satisfies the given conditions = 48 × 1 + 2 = 48 + 2 = 50.

Concept:

We know that composite numbers are the numbers which are not prime.

Calculation:

By identifying the prime numbers lying between 67 and 101 we can find the number of composite numbers lying between 67 and 101.

⇒ Prime numbers lying between 67 and 101 are 71, 73, 79, 83, 89 and 97 i.e., 6 prime numbers.

Total numbers lying between 67 and 101 = 33

⇒ Number of composite numbers lying between 67 and 101 = 33 – 6 = 27

Calculations:

Let the number of questions attempted correctly be X and those attempted wrong be Y.

Total question attempted= 75

⇒ X + Y = 75      ---(1)

Also total score is 125. Since 4 marks is awarded for every correct answer and 1 marks is lost for every wrong answer. Thus

 ⇒(X × 4) +(Y × (-1)) = 125

⇒ 4X – Y = 125      ---(2)

Adding (1) and (2) both sides, we get

⇒ 5X = 200

⇒ X = 40

Given:

The given number is 486

Concept:

Expressing any number as factors of prime numbers means reducing the number to its least possible multiple

Prime number = Number which is divisible by only 1 and itself

Calculation:

486 = 2 × 3 × 3 × 3 × 3 × 3

⇒ 486 = 2 × 3⁵

Explanation:

Given numbers

15, 20 and 35

The smallest number will be the LCM (Least Common Multiple) of these three numbers.

Let us find the prime factors for each number

15 = 3 × 5

20 = 2 × 2 × 5

35 = 5 × 7

The LCM of the three numbers is given by

LCM = 2 × 2 × 5 × 3 × 7 = 420

420 is exactly divisible by 15, 20, and 35.

Since we have a remainder of 8 in each case, the smallest number in each case will be given by

= 420 + 8 = 428.

Concept:

Percentage profit is represented by

\(\% {\bf{Profit}} = \frac{{{\bf{SP}} - {\bf{CP}}}}{{{\bf{CP}}}}\), where, SP = Selling Price, CP = Cost Price

Percentage Loss is represented by

\(\% {\bf{Loss}} = \frac{{{\bf{CP}} - {\bf{SP}}}}{{{\bf{CP}}}}\)

Calculation:

Given:

For 1^st Pipe: SP = Rs.12 per pipe, %Profit = 20% = \(\frac{1}{5}\)

For 2^nd Pipe: SP = Rs.12 per pipe, %Los s= 20% = \(\frac{1}{5}\)

Now, we now that

For 1^st Pipe:

\(\% {\bf{Profit}} = \frac{{{\bf{SP}} - {\bf{CP}}}}{{{\bf{CP}}}}\)⇒ \(\frac{1}{5}\)= \(\frac{12 - CP_1 }{CP_1 }\)⇒ CP₁ = Rs. 10

For 2^nd Pipe:

\(\% {\bf{Loss}} = \frac{{{\bf{CP}} - {\bf{SP}}}}{{{\bf{CP}}}}\)⇒ \(\frac{1}{5}\) = \(\frac{CP_2 - 12}{CP_2 }\)⇒ CP₂ = Rs. 15

For the whole transaction

Total cost price = 10 + 15 = Rs. 25

Total selling price = 12 + 12 = Rs. 24

∴ there is a loss for the man and that total % loss is given by

\(\% {\bf{Loss}} = \frac{{{\bf{Total}}\;{\bf{CP}} - {\bf{Total}}\;{\bf{SP}}}}{{{\bf{Total}}\;{\bf{CP}}}} = \frac{{25 - 24}}{{25}} = \frac{1}{{25}} \times 100 = 4\% \) Loss

Concept:

When two bodies run in the opposite direction, then their relative velocities get added.

1 km/hr = 5/18 m/s.

1 m/s = 18/5 km/hr.

Calculation:

Given:

Length of train = 220 m, Speed of train = 54 km/hr ⇒ 15 m/s and Speed of man = 12 km/hr ⇒ 10/3 m/s [ 1 km/hr = 5/18 m/s]

\(Time = \frac{Distance}{Speed}\)

⇒ \(\frac{220}{\frac{10}{3}\;+\;15} (Speed\;gets\;added\;due\;to\;different\;direction)\)

\(⇒ \frac{220\;\times\;3}{55}\)

⇒ 12 sec.

Calculation:

Given:

33% of A is equal to 55% of B

\(⇒ \frac{33}{100}\times A=\frac{55}{100}\times B\)

\(⇒ \frac{A}{B}=\frac{55}{100}\times \frac{100}{33}\)

⇒ A : B = 5 : 3.

Given,

Ratio of three numbers = 3 : 4 : 5

Sum of first two numbers = 21

Calculation:

Ratio of three numbers = 3x : 4x : 5x

3x + 4x = 21

⇒ 7x = 21

⇒ x = 21/7

⇒ x = 3

Second number = 4 × 3 = 12

Third number = 5 × 3 = 15

∴ Difference of last two numbers = 15 – 12 = 3