Mathematics and Pedagogy MCQ Questions | |
Quiz-1 | Quiz-2 |
Q1. If $\frac{\mathrm{x}}{\mathrm{y}}=\frac{3}{5}$ then the value of is $\frac{\mathrm{x}-\mathrm{y}}{\mathrm{x}+\mathrm{y}}$
(a) $-\frac{1}{4}$
(b) $\frac{1}{4}$
(c) $-\frac{1}{6}$
(d) -6
Answer: (a) $-\frac{1}{4}$ Solution: $\frac{\mathrm{x}}{\mathrm{y}}=\frac{3}{5}$ Using Dividendo and Componendo $\Longrightarrow \frac{\mathrm{x}-\mathrm{y}}{\mathrm{x}=\mathrm{y}}=\frac{3-5}{3+5}=-\frac{1}{4}\,\,$ |
Q2. If x is a negative real number, then
(a) |xl = x
(b) | x| = –x
(c) | x| = $\frac{1}{\mathrm{x}}$
(d) |x| = $-\frac{1}{\mathrm{x}}$
Answer: (b) | x| = –x Solution: Clearly, defined absolute value i.e | x| = – x |
Q3. What is the maximum value of m, if the number N = 35 x 45 x 55 x 60 x 124 x 75 is divisible by 5m?
(a) 4
(b) 5
(c) 6
(d) 7
Answer: (c) 6 Solution: N = 35 x 45 x 55 x 60 x 124 x 75 = 3 × 5 × 3 × 3 × 5 × 5 × 11 × 5 × 12 × 124 × 5 × 5 × 3 = 56 × 34 × 11 × 12 × 124 Since N is is divisible by 5m ∴ 5m = 56 or m = 6 |
Q4. A student was asked to multiply a number by 25. He instead multiplied the number by 52 and got the answer 324 more than the correct answer. The number to be multiplied was
(a) 12
(b) 15
(c) 25
(d) 32
Answer: (a) 12 Solution: Let the number = x 52x – 25x = 324 ⇒ 27x = 324 ⇒x = 12 |
Q5. Consider the following statements:
I. No integer of the form 4k + 3, where k an integer, can be expressed as the sum of two squares.
II. The Square of an odd integer can express in the form 8k +1, where k is an integer.
Which of the statement (s) given above is/ are correct?
(a) Only I
(b) Only II
(c) Both I and II
(d) Neither I nor II
Answer: (a) Only I Solution: I. f(k) = 4k+3 For k=1, f(k) = 7; for k=2, f(k) = 11; for k=3, f(k) = 15 Values of f(k) cannot be expressed as sum of two squares for k = 1, 2, 3, … II. f(k) = 8k+1 For k=1, f(k) = 9= 32; for k=2, f(k) = 17; for k=3, f(k) = 25 = 52 for k=4, f(k) = 33; for k=5, f(k) = 41 Values of f(k) is square of an odd integer only for some values of k So, only statement I is correct |
Q6. If a =$\sqrt{\left( 2013 \right) ^2+2013+2014}$, then the value of a is
(a) 1002
(b) 1007
(c) 2013
(d) 2014
Answer: (d) 2014 Solution: $\sqrt{\left( 2013 \right) ^2+2013+2014}$ = $\sqrt{\left( 2013 \right) ^2+2.2013.1+1}$ = $\sqrt{\left( 2013+1 \right) ^2}$ = 2014 |
Q7. LCM of 23 x 3 × 5 and 24 x 5 x 7 is
(a) 212 x 3 x 52 x 7
(b) 24 x 5 x 7 x 9
(c) 24 x 3 x 5 x 7
(d) 23 ×3 ×5 ×7
Answer: (c) 24 x 3 x 5 x 7 Solution: LCM = prime factor with highest powe = 24 x 3 x 5 x 7 |
Q8. The value of is equal to $\frac{\left( 0.5 \right) ^4-\left( 0.4 \right) ^4}{\left( 0.5 \right) ^2+\left( 0.4 \right) ^2}$
(a) 0.9
(b) 0.09
(c) 9
(d) 0.009
Answer: (b) 0.09 Solution: $\frac{\left( 0.5 \right) ^4-\left( 0.4 \right) ^4}{\left( 0.5 \right) ^2+\left( 0.4 \right) ^2}$ = $\frac{\left( \left( 0.5 \right) ^2+\left( 0.4 \right) ^2 \right) .\left( \left( 0.5 \right) ^2-\left( 0.4 \right) ^2 \right)}{\left( \left( 0.5 \right) ^2+\left( 0.4 \right) ^2 \right)}$ = (0.5 + 0.4) (0.5 – 0.4) = 0.09 |
Q9. The sum of all interior angles of a regular convex polygon is 1080°. The measure of each of its interior angles is
(a) 108°
(b) 135°
(c) 72°
(d) 120°
Answer: (b) 135° Solution: sum of all interior angles of a regular convex polygon = (2n – 4)×90° ∴ (2n – 4)×90 = 1080 ⇒ 2n – 4 = 12 ⇒2n = 16 ⇒ n = 8 ∴ each interior angles of a regular convex polygon = $\frac{1080\degree}{8}=135\degree$ |
Q10.The mean of range, mode, and median of the data 4, 3, 2, 2, 7, 2, 2, 0, 3, 4, 4 is
(a) 4
(b) 3
(c) 5
(d) 2
Answer: (a) 4 Solution: Arranging the given data in a table with their frequency:
From the table Range = 7 – 0 = 7 Mode = 2 ( frequency 4) Median = 3 ∴ mean of range, mode and median = $\frac{7 +\,\,2 +3}{3}=4$ |