Math, Science and Pedagogy MCQ Question with Answer | ||||
Quiz-1 | Quiz-2 | Quiz-3 | Quiz-4 | Quiz-5 |
Quiz-6 | Quiz-7 | Quiz-8 | Quiz-9 | Quiz-10 |
Quiz-11 | Quiz-12 | Quiz-13 | Quiz-14 | Quiz-15 |
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 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 5^{m}?
(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 = 5^{6} × 3^{4} × 11 × 12 × 124 Since N is is divisible by 5^{m} ∴ 5^{m} = 5^{6} 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 is 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, … f(k) = 8k+1 For k=1, f(k) = 9= 3^{2}; for k=2, f(k) = 17; for k=3, f(k) = 25 = 5^{2} for k=4, f(k) = 33; for k=5, f(k) = 41 Values of f(k) are square of an odd integer only for some values of k So, the 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 2^{3} x 3 × 5 and 2^{4} x 5 x 7 is
(a) 2^{12} x 3 x 5^{2} x 7
(b) 2^{4} x 5 x 7 x 9
(c) 2^{4} x 3 x 5 x 7
(d) 2^{3} X3 X5 X7
Answer: (c) 2^{4} x 3 x 5 x 7 Solution: LCM = prime factor with highest powe = 2^{4} 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 = $\frac{7 +\,\,2 +3}{3}=4$ |
Q11. In Δ DEF and APQR, if PQ = DE, EF = PR and FD = QR, then
(a) Δ DEF ≅ Δ RPQ
(b) Δ DEF ≅ Δ QPR
(c) Δ DEF ≅ Δ QRP
(d) Δ DEF ≅ Δ PRQ
Answer: (b) Δ DEF ≅ Δ QPR Solution: From figure, PQ = DE, EF = PR and FD = QR ( all given) ∴ DEF ≅ Δ QPR |
Q12. In a quadrilateral ABCD, ∠D = 60° and ∠C = 100 °. The bisectors of ∠A and ∠B meet at point P. The measures of ∠APB is
(a) 80°
(b) 70°
(c) 100°
(d) 60°
Answer: (a) 80° Solution: Let ∠A = 2x and ∠B = 2y ∴ 60° + 100° + 2x + 2y = 360 ( sum of all interior angle of quadilateral is 360°) or x + y = 100 From Δ APB, x + y + ∠APB = 180° ( sum of all interior angle of traiangle is 180°) Or, APB = 180 – 100 (since, x + y = 100°) ∠APB = 80° |
Q13. A train T leaves a place P at 5:00 a.m. and reaches another place Q at 9:00 a.m. another train T2 leaves place Q at 7:00 a.m. and reaches place P at 10:30 a.m. The time at which the two trains cross each other is
(a) 8:26 a.m.
(b) 7:56 a.m.
(c) 8:15 a.m.
(d) 8:00 a.m.
Answer: (b) 7:56 a.m. Solution: Let Distance between P and Q = x km. And the train meet each other t hour after 7:00 am Speed of first train x/4 km/hr and distance coveer when they meet = $\frac{\left( 2+\mathrm{t} \right) \mathrm{x}}{4}$ km Speed of 2^{nd} train = $\frac{2\mathrm{x}}{7}$ km/hr and distance coveer when they meet = $\frac{2\mathrm{xt}}{7}$ Now, $\frac{\left( 2+\mathrm{t} \right) \mathrm{x}}{4}+\frac{2\mathrm{xt}}{7}\,\,=\mathrm{x}$ ⇒ 14 + 7t + 8t = 28 ⇒ 15t = 14 ⇒ t = 14/15 hr = 56 min ∴ They meet each other at 7:56 am |
Q14. X can do $\frac{3}{4}$ of work in 12 days. In how many days X can finish the ½ work?
(a) 8 days
(b) 16 days
(c) 12 days
(d) 24 days
Answer: (a) 8 days Solution: X can do $\frac{3}{4}$ of work in 12 days X can do $\frac{1}{2}$ of a work in $\frac{12\times 4}{3\times 2}=8$ days |
Q15. If after 24 % of wastage the net output of a coal mine is 68400 quintals. Then, the total output of the coal-mine in quintals is
(a) 70000
(b) 90000
(c) 80000
(d) 89000
Answer: (b) 90000 Solution: the total output of the coal-mine = 100x quintals Wastage = 24x and use full output = 76 76x = 68400 ⇒ x = 900 ∴ the total output of the coal-mine = 90000 quintals |
Q16. A man purchased a watch for 400 and sold it at a gain of 20 % of the selling price. The selling price of the watch is
(a) 300
(b) 320
(c) 440
(d) 500
Answer: (d) 500 Solution: Let selling price = 100x Then, 400 + 20x = 100x ⇒ 80x = 400 ⇒x = 5 ∴ Selling price = 500 |
Q17. The ratio of ages of A and B is 2: 5 and the ratio of ages of B and C is 3: 4. What is the ratio of ages of A, B and C?
(a) 6:15:20
(b) 8: 5: 3
(c) 6: 5: 4
(d) 2: 15: 4
Answer: (a) 6:15:20 Solution: Using Ratio and Portion trick |
Q18. Divide the number 26244 by the smallest number, so that the quotient is a perfect cube, so the smallest number is
(a) 4
(b) 6
(c) 36
(d) 16
Answer: (c) 36 Solution: $26244 =2\times 2\times 3\times 3\times \overline{3\times 3\times 3}\times \overline{3\times 3\times 3}$ ∴ Required the smallest number = 2×3 |
Q19. When in two linear equations a_{1}x + b_{1}y = c_{1} and a_{2}x + b_{2}y = C_{2}, $\frac{\mathrm{a}_1}{\mathrm{a}_2}\ne \frac{\mathrm{b}_1}{\mathrm{b}_2}$ then the graph is
(a) parallel
(b) intersection at one point
(c) coincident
(d) None of the above
Answer: (b) intersection at one point Solution: $\frac{\mathrm{a}_1}{\mathrm{a}_2}\ne \frac{\mathrm{b}_1}{\mathrm{b}_2}$ this case is unique solution. So the graph of equation will interect in a point. |
Q20. The radii of the bases of two cylinders are in the ratio of 2: 3 and their heights are in the ratio of 5:3. The ratio of their volumes is
(a) 7: 6
(b) 10: 9
(c) 4: 9
(d) 20:27
Answer: (d) 20:27 Solution: Let their ratio and height respectively = 2r, 3r and 5h, 3h Ratio of their volume = $\frac{\frac{4}{3}\Pi \left( 2\mathrm{r} \right) ^25\mathrm{h}}{\frac{4}{3}\Pi \left( 3\mathrm{r} \right) ^23\mathrm{h}}=\frac{20}{27}=20:27$ |
Q21. The nature of Mathematics is
(a) logical
(b) ornamental
(c) difficult
(d) not for common people
Answer: (a) logical |
Q22. The term, ‘ Mathematical tools ‘ refers to
(a) calculators, rulers, tape measures, protractors, compass, etc.
(b) charts based on formulae and concepts, graph papers, dotted sheets, etc
(c) all types of materials including language, written symbols, meaningful instructions to establish their purpose
(d) physical material like geo-board and 3D models, cubic rods, etc
Answer: (c) all types of materials including language, written symbols, meaningful instructions to establish their purpose |
Q23. Read the following question from the class VI textbook
” Write a pair of integers whose sum gives a negative integer. ”
The above question refers to
(a) reflective question
(b) closed-ended question
(c) multi-disciplinary question
(d) open-ended question
Answer: (d) open-ended question |
Q24. A learner exhibiting difficulty in sorting, recognizing patterns, orienting numbers and shapes, telling time and measurement may have dyscalculia with difficulty in
(a) language processing
(b) visual-memory
(c) visual-motor coordination
(d) visual-spatial skills
Answer: (d) visual-spatial skills |
Q25. In class VI-VI, in the unit of Understanding Quadrilaterals, important results related to the angle-sum property of quadrilaterals are introduced using paper folding activity followed by the exercise based on these properties. At this level proof of the angle, property is not given, as the students of class VI are at Van Hiele level of
(a) Level 2-Informal Deduction
(b) Level 1-analysis
(c) Level 3-Deduction
(d) Level 0-Visualization
Answer: (a) Level 2-Informal Deduction |
Q26. CBSE announced the celebration of ‘ GANIT Week in schools to commemorate the birth anniversary of the legendary mathematician, Srinivasa Ramanujan. GANIT stands for
(a) Growing Ability in Numerical Innovations and Techniques
(b) Growing aptitude in Numerical Innovations and Training
(c) Growing Ability in Numerical Innovations and Training
(d) Growing aptitude in Numerical Innovations and Techniques
Answer: (b) Growing aptitude in Numerical Innovations and Training |
Q27. The teaching stage does not have the
(a) proactive stage
(b) interactive stage
(c) post active stage
(d) middle active stage
Answer: (d) middle active stage |
Q28. The role of intervening variables in the teaching process of mathematics is
(a) diagnostic activities
(b) remedial activities
(c) evaluation-related activities
(d) All of the above
Answer: (d) All of the above |
Q29. The main aim of drill work in mathematics is
(a) to understand the new formula
(b) to enhance computation skill
(c) utilise the knowledge in the circumstance
(d) to clear the new concept
Answer: (b) to enhance computation skill |
Q30. An appropriate method for the establishment of the formulae in Mathematics is
(a) induction
(b) planning
(c) synthesis
(d) None of these
Answer: (a) induction |