Competive Exam Math 6


Q. Use the cuboid PQRSTUVW to answer Questions 1 - 5



Solutions / āϏāĻŽāĻžāϧāĻžāύ


1. Face PQRS is parallel to ________. (PQRS āĻŽā§āĻ–āĻ–āύ ________-ā§° āϏāĻŽāĻžāĻ¨ā§āϤ⧰āĻžāϞāĨ¤)


Ans / āωāĻ¤ā§āϤ⧰: TUVW


2. PSWT is parallel to ________. (PSWT āĻŽā§āĻ–āĻ–āύ ________-ā§° āϏāĻŽāĻžāĻ¨ā§āϤ⧰āĻžāϞāĨ¤)


Ans / āωāĻ¤ā§āϤ⧰: QRUV


3. ________ is parallel to PQUT. (PQUT-ā§° āϏāĻŽāĻžāĻ¨ā§āϤ⧰āĻžāϞ āĻŽā§āĻ–āĻ–āύ āĻšā§ˆāϛ⧇ ________āĨ¤)


Ans / āωāĻ¤ā§āϤ⧰: TUVW


4. PSWT is perpendicular to ________. (PSWT āĻŽā§āĻ–āĻ–āύ ________-ā§° āϞāĻŽā§āĻŦāĨ¤)


Ans / āωāĻ¤ā§āϤ⧰: QRUV


5. PQUT is perpendicular to ________. PQUT āĻŽā§āĻ–āĻ–āύ ________-ā§° āϞāĻŽā§āĻŦāĨ¤


Ans / āωāĻ¤ā§āϤ⧰: TUVW


6. A triangle with three unequal sides is called ________. (āϤāĻŋāύāĻŋāĻ“āϟāĻž āĻŦāĻžāĻšā§ āĻ…āϏāĻŽāĻžāύ āĻĨāĻ•āĻž āĻ¤ā§ā§°āĻŋāϭ⧁āϜāĻ• āĻ•āĻŋ āĻŦā§‹āϞāĻž āĻšāϝāĻŧ ?)


(a) Right-angled Triangle | (b) Scalene Triangle | (c) Isosceles Triangle


Ans / āωāĻ¤ā§āϤ⧰: (b) Scalene Triangle


Explanation: A scalene triangle has all three sides of different lengths. āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: Scalene triangle-ā§° āϤāĻŋāύāĻŋāĻ“āϟāĻž āĻŦāĻžāĻšā§ā§° āĻĻā§ˆā§°ā§āĻ˜ā§āϝ āĻŦ⧇āϞ⧇āĻ— āĻŦ⧇āϞ⧇āĻ— āĻšāϝāĻŧāĨ¤


7. The sum of the three angles of a triangle is ________. āĻāϟāĻž āĻ¤ā§ā§°āĻŋāϭ⧁āϜ⧰ āϤāĻŋāύāĻŋāϟāĻž āϕ⧋āĻŖā§° āϝ⧋āĻ—āĻĢāϞ āĻ•āĻŋāĻŽāĻžāύ?


(a) 360° | (b) 180° | (c) 260°


Ans / āωāĻ¤ā§āϤ⧰: (b) 180°


Explanation: The sum of the interior angles of every triangle is 180°. āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āĻĒā§ā§°āϤāĻŋāĻŸā§‹ āĻ¤ā§ā§°āĻŋāϭ⧁āϜ⧰ āĻ­āĻŋāϤ⧰⧰ āϤāĻŋāύāĻŋāϟāĻž āϕ⧋āĻŖā§° āϝ⧋āĻ—āĻĢāϞ ā§§ā§Žā§Ļ°āĨ¤


Data / āϤāĻĨā§āϝ: 7, 5, 2, 9, 5, 8


Arrange in ascending order / āĻŠā§°ā§āĻ§ā§āĻŦāĻ•ā§ā§°āĻŽāϤ āϏāĻžāϜāĻŋāϞ⧇: 2, 5, 5, 7, 8, 9


8. Mean (Average) / āĻ—āĻĄāĻŧ : 7 + 5 + 2 + 9 + 5 + 8 / 6 = 36 / 6 =


Ans / āωāĻ¤ā§āϤ⧰: 6


9. Mode / āĻŦāĻšā§āϞāĻ•


Answer / āωāĻ¤ā§āϤ⧰: 5


Explanation: The number 5 appears most frequently. āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: ā§Ģ āϏāĻ‚āĻ–ā§āϝāĻžāĻŸā§‹ āĻ¸ā§°ā§āĻŦāĻžāϧāĻŋāĻ•āĻŦāĻžā§° āĻĻ⧇āĻ–āĻž āϝāĻžāϝāĻŧāĨ¤


10. Median / āĻŽāĻ§ā§āϝāĻ•


Since there are 6 numbers, Median = Average of the 3rd and 4th numbers.


5 + 7 / 2 = 6


Ans / āωāĻ¤ā§āϤ⧰: 6


Explanation: The correct median is 6, not 5.7 or 12. āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: āϏāĻ āĻŋāĻ• Median = ā§ŦāĨ¤ 


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11. The remainder when -76 is divided by 3 is: (-76 āĻ• 3 ⧰⧇ āĻšā§°āĻŖ āϕ⧰āĻŋāϞ⧇ āĻĒā§ā§°āĻžāĻĒā§āϤ āĻ­āĻžāĻ—āĻļ⧇āώ āĻšā§ˆāϛ⧇:)


Trick:



  • Ignore the negative sign first (āĻĒā§ā§°āĻĨāĻŽā§‡ − āϚāĻŋāĻšā§āύ āφāρāϤ⧰āĻžāχ āϞāĻ“āĻ•āĨ¤): 76 ÷ 3 → remainder(āĻ­āĻžāĻ—āĻļ⧇āώ) = 1

  • For negative numbers (āĻ‹āĻŖāĻžāĻ¤ā§āĻŽāĻ• āϏāĻ‚āĻ–ā§āϝāĻžā§° āĻŦāĻžāĻŦ⧇ -):

  •  Remainder = Divisor − Positive remainder (āĻ­āĻžāĻ—āĻļ⧇āώ = āĻšā§°āĻŖāĻ•āĻžā§°ā§€ āϏāĻ‚āĻ–ā§āϝāĻž − āϧāύāĻžāĻ¤ā§āĻŽāĻ• āĻ­āĻžāĻ—āĻļ⧇āώ) = 3 - 1 = 2


Ans / āωāĻ¤ā§āϤ⧰: (C) 2


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Q. A teacher is preparing 15 identical exam folders. Each folder contains: 8 question sheets 6 answer sheets 1 booklet with [(27-3)/6 + 4]. How many total sheets / pages does the teacher prepare for all 15 folders ? Option: A. 330 B. 270 C. 134 D. 360 (Q. āĻāϜāύ āĻļāĻŋāĻ•ā§āώāϕ⧇ ā§§ā§ĢāϟāĻž āĻāϕ⧇ āϧ⧰āĻŖā§° āĻĒā§°ā§€āĻ•ā§āώāĻžā§° āĻĢ'āĻ˛ā§āĻĄāĻžā§° āĻĒā§ā§°āĻ¸ā§āϤ⧁āϤ āϕ⧰āĻŋ āφāϛ⧇āĨ¤ āĻĒā§ā§°āϤāĻŋāĻŸā§‹ āĻĢ'āĻ˛ā§āĻĄāĻžā§°āϤ āφāϛ⧇ - ā§Ž āĻ–āύ āĻĒā§ā§°āĻļā§āύāĻĒāĻ¤ā§ā§° (Question Sheets), ā§Ŧ āĻ–āύ āωāĻ¤ā§āϤ⧰āĻĒāĻ¤ā§ā§° (Answer Sheets), ā§§āĻ–āύ āĻĒ⧁āĻ¸ā§āϤāĻŋāĻ•āĻž, āϝ'āϤ [(27−3)/6 + 4] āϟāĻž āĻĒ⧃āĻˇā§āĻ āĻž āφāϛ⧇āĨ¤ ā§§ā§ĢāϟāĻž āĻĢ'āĻ˛ā§āĻĄāĻžā§°ā§° āĻŦāĻžāĻŦ⧇ āĻļāĻŋāĻ•ā§āώāĻ•āϜāύ⧇ āĻŽā§āĻ  āĻ•āĻŋāĻŽāĻžāύāĻ–āύ āĻļā§āĻŦā§€āϟ/āĻĒ⧃āĻˇā§āĻ āĻž āĻĒā§ā§°āĻ¸ā§āϤ⧁āϤ āϕ⧰āĻŋāĻŦ ?


Solution


1st: Calculate the booklet pages : (27 − 3)/6 + 4 = 24/6 + 4 = 4 + 4 = 8


So, each folder contains:



  • Question sheets = 8

  • Answer sheets = 6

  • Booklet pages = 8


Total per folder: 8 + 6 + 8 = 22


For 15 folders: 22 × 15 = 330


Ans: A. 330


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Q. If A = {6, 8, 10} and B = {4, 6, 8}, then A = ?āĻĒā§ā§°āĻļā§āύ: āϝāĻĻāĻŋ A = {6, 8, 10} āφ⧰⧁ B = {4, 6, 8}, āϤ⧇āĻ¨ā§āϤ⧇ = ?)


Options / āĻŦāĻŋāĻ•āĻ˛ā§āĻĒāϏāĻŽā§‚āĻš: (A) {4, 10}  (B) {4}  (C) {10}  (D) {10, 4}


Trick / āĻŸā§ā§°āĻŋāĻ•:


B − A = Elements in B but NOT in A. (B-āϤ āĻĨāĻ•āĻž āĻ•āĻŋāĻ¨ā§āϤ⧁ A-āϤ āύāĻĨāĻ•āĻž āωāĻĒāĻžāĻĻāĻžāύāĨ¤)


Cancel the common elements from both sets / āĻĻ⧁āϝāĻŧā§‹āϟāĻž āϏāĻŽāĻˇā§āϟāĻŋāϤ āĻĨāĻ•āĻž āĻāϕ⧇ āωāĻĒāĻžāĻĻāĻžāύ āĻ•āĻžāϟāĻŋ āĻĻāĻŋāϝāĻŧāĻ•: B = {4, 6, 8}, A = {6, 8, 10}


Check B / B āϚāĻžāĻ“āρ:



  • 4 → Not in A / A-āϤ āύāĻžāχ

  • 6 → In A, B / A, B-āϤ āφāϛ⧇ : Cancel 

  • 8 → In A, B / A, B -āϤ āφāϛ⧇ : Cancel


Therefore / āϏ⧇āϝāĻŧ⧇āĻšā§‡, B − A = {4}


Ans / āϏāĻ āĻŋāĻ• āωāĻ¤ā§āϤ⧰: (B) {4}


Shortcut : "Start from B → Cancel common elements → The remaining elements are the answer." ("B-ā§° āĻĒā§°āĻž āφ⧰āĻŽā§āĻ­ āϕ⧰āĻ• → āĻāϕ⧇ āωāĻĒāĻžāĻĻāĻžāύ āĻ•āĻžāϟāĻŋ āĻĻāĻŋāϝāĻŧāĻ• → āϝāĻŋ āĻŦāĻžāϕ⧀ āĻĨāĻžāĻ•āĻŋāĻŦ, āϏ⧇āϝāĻŧāĻžāχ B − AāĨ¤")


QIf A = {6, 8, 10} and B = {4, 6, 8}, then B = ?āĻĒā§ā§°āĻļā§āύ: āϝāĻĻāĻŋ A = {6, 8, 10} āφ⧰⧁ B = {4, 6, 8}, āϤ⧇āĻ¨ā§āϤ⧇ Options / āĻŦāĻŋāĻ•āĻ˛ā§āĻĒāϏāĻŽā§‚āĻš:


(A) {4}  (B) {10}  (C) {6, 8}  (D) {4, 10}


Trick / āĻŸā§ā§°āĻŋāĻ•: A − B = Elements in A but NOT in B. (A − B = A-āϤ āĻĨāĻ•āĻž āĻ•āĻŋāĻ¨ā§āϤ⧁ B-āϤ āύāĻĨāĻ•āĻž āωāĻĒāĻžāĻĻāĻžāύāĨ¤)


Cancel the common elements / āĻĻ⧁āϝāĻŧā§‹āϟāĻž āϏāĻŽāĻˇā§āϟāĻŋāϤ āĻĨāĻ•āĻž āĻāϕ⧇ āωāĻĒāĻžāĻĻāĻžāύ āĻ•āĻžāϟāĻŋ āĻĻāĻŋāϝāĻŧāĻ•: A = {6, 8, 10}, B = {4, 6, 8}


Cancel 6 and 8 (common elements / āĻāϕ⧇ āωāĻĒāĻžāĻĻāĻžāύ)


Remaining in A = {10}


Ans / āωāĻ¤ā§āϤ⧰: (B) {10}


Shortcut: Start from A → Cancel common elements → Remaining elements = A − B. (A-ā§° āĻĒā§°āĻž āφ⧰āĻŽā§āĻ­ āϕ⧰āĻ• → āĻāϕ⧇ āωāĻĒāĻžāĻĻāĻžāύ āĻ•āĻžāϟāĻŋ āĻĻāĻŋāϝāĻŧāĻ• → āϝāĻŋ āĻŦāĻžāϕ⧀ āĻĨāĻžāĻ•āĻŋāĻŦ, āϏ⧇āϝāĻŧāĻžāχ A − B)


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Practice Set: Remainder (āĻ­āĻžāĻ—āĻļ⧇āώ)


Q1. The remainder when -25 is divided by 4 is: (Q1. -25 āĻ• 4 ⧰⧇ āĻšā§°āĻŖ āϕ⧰āĻŋāϞ⧇ āĻĒā§ā§°āĻžāĻĒā§āϤ āĻ­āĻžāĻ—āĻļ⧇āώ āĻšā§ˆāϛ⧇:)


(A) -1 (B) 1 (C) 2 (D) 3


Q2. The remainder when -45 is divided by 7 is: (Q2. -45 āĻ• 7 ⧰⧇ āĻšā§°āĻŖ āϕ⧰āĻŋāϞ⧇ āĻĒā§ā§°āĻžāĻĒā§āϤ āĻ­āĻžāĻ—āĻļ⧇āώ āĻšā§ˆāϛ⧇:)


(A) 4 (B) 5 (C) 6 (D) -3


Q3. The remainder when -81 is divided by 5 is: (Q3. -81 āĻ• 5 ⧰⧇ āĻšā§°āĻŖ āϕ⧰āĻŋāϞ⧇ āĻĒā§ā§°āĻžāĻĒā§āϤ āĻ­āĻžāĻ—āĻļ⧇āώ āĻšā§ˆāϛ⧇:)


(A) 1 (B) 2 (C) 4 (D) -1


Q4. The remainder when -100 is divided by 6 is: (Q4. -100 āĻ• 6 ⧰⧇ āĻšā§°āĻŖ āϕ⧰āĻŋāϞ⧇ āĻĒā§ā§°āĻžāĻĒā§āϤ āĻ­āĻžāĻ—āĻļ⧇āώ āĻšā§ˆāϛ⧇:)


(A) 2 (B) 4 (C) 5 (D) -2


Q5. The remainder when -67 is divided by 8 is: (Q5. -67 āĻ• 8 ⧰⧇ āĻšā§°āĻŖ āϕ⧰āĻŋāϞ⧇ āĻĒā§ā§°āĻžāĻĒā§āϤ āĻ­āĻžāĻ—āĻļ⧇āώ āĻšā§ˆāϛ⧇:)


(A) 1 (B) 3 (C) 5 (D) 7


Shortcut Trick / āϏāĻšāϜ āĻŸā§ā§°āĻŋāĻ• : For negative numbers: Remainder = Divisor − Positive remainder (āĻ‹āĻŖāĻžāĻ¤ā§āĻŽāĻ• āϏāĻ‚āĻ–ā§āϝāĻžā§° āĻŦāĻžāĻŦ⧇: āĻ­āĻžāĻ—āĻļ⧇āώ = āĻšā§°āĻŖāĻ•āĻžā§°ā§€ āϏāĻ‚āĻ–ā§āϝāĻž − āϧāύāĻžāĻ¤ā§āĻŽāĻ• āĻ­āĻžāĻ—āĻļ⧇āώ )













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