Competive Exam Math 1



Q. In a bag containing Blue, Yellow, and Orange tokens, the ratio of Blue to Yellow tokens is 5 : 2, while the ratio of Orange to Blue tokens is 11 : 18. What is the ratio of Orange to Yellow tokens ? (āĻĒā§ā§°āĻļā§āύ: āĻāϟāĻž āĻŦ⧇āĻ—āϤ āύ⧀āϞāĻž (Blue), āĻšāĻžāϞāϧ⧀āϝāĻŧāĻž (Yellow) āφ⧰⧁ āĻ•āĻŽāϞāĻž (Orange) āĻŸā§‹āϕ⧇āύ āφāϛ⧇āĨ¤ āύ⧀āϞāĻž āφ⧰⧁ āĻšāĻžāϞāϧ⧀āϝāĻŧāĻž āĻŸā§‹āϕ⧇āύ⧰ āĻ…āύ⧁āĻĒāĻžāϤ 5 : 2, āφ⧰⧁ āĻ•āĻŽāϞāĻž āφ⧰⧁ āύ⧀āϞāĻž āĻŸā§‹āϕ⧇āύ⧰ āĻ…āύ⧁āĻĒāĻžāϤ 11 : 18āĨ¤ āϤ⧇āĻ¨ā§āϤ⧇ āĻ•āĻŽāϞāĻž āφ⧰⧁ āĻšāĻžāϞāϧ⧀āϝāĻŧāĻž āĻŸā§‹āϕ⧇āύ⧰ āĻ…āύ⧁āĻĒāĻžāϤ āĻ•āĻŋāĻŽāĻžāύ ?)


Options / āĻŦāĻŋāĻ•āĻ˛ā§āĻĒāϏāĻŽā§‚āĻš: A. 55 : 36   B. 63 : 39   C. 59 : 40   D. 51 : 32


Soln


Blue : Yellow = 5 : 2


Multiply by 18: Blue : Yellow = 90 : 36


Orange : Blue = 11 : 18


Multiply by 5: Orange : Blue = 55 : 90


Now Blue is common (90).


So, Orange : Yellow = 55 : 36


Ans: A. 55 : 36


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Q. A shopkeeper has ₹10 and ₹5 notes whose total value is ₹350. If he has 45 notes in total, how many ₹10 notes does he have ? (Q.47 āĻāϜāύ āĻĻā§‹āĻ•āĻžāύāĻĻāĻžā§°ā§° āĻ“āϚ⧰āϤ ₹10 āφ⧰⧁ ₹5-ā§° āύ⧋āϟ āφāϛ⧇āĨ¤ āύ⧋āϟāϏāĻŽā§‚āĻšā§° āĻŽā§āĻ  āĻŽā§‚āĻ˛ā§āϝ ₹350āĨ¤ āϝāĻĻāĻŋ āĻŽā§āĻ  āύ⧋āϟ⧰ āϏāĻ‚āĻ–ā§āϝāĻž 45 āĻšāϝāĻŧ, āϤ⧇āĻ¨ā§āϤ⧇ ₹10-ā§° āύ⧋āϟ āĻ•āĻŋāĻŽāĻžāύ āφāϛ⧇ ?)


Options / āĻŦāĻŋāĻ•āĻ˛ā§āĻĒāϏāĻŽā§‚āĻš: A. 29  B. 31  C. 27  D. 25


Soln / āϏāĻŽāĻžāϧāĻžāύ

₹10 = x and ₹5 = y


10x + 5y = 350 .................... (1)


x + y = 45 


y = 45 - x


10x + 5(45 - x) = 350 


10x + 225 + 5x = 350 


5x = 125


x = 25


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Q. Which of the following decimal expansions is not a rational number ? (āϤāϞ⧰ āϕ⧋āύāĻŸā§‹ āĻĻāĻļāĻŽāĻŋāĻ• āĻĒā§ā§°āϏāĻžā§°āĻŖ āĻĒā§°āĻŋāĻŽā§‡āϝāĻŧ āϏāĻ‚āĻ–ā§āϝāĻž āύāĻšāϝāĻŧ ?)


Options: (A) 43.123456789  (B) 0.12012001200012000...   (C) 43.123456789  (D) More than one of the above / āĻ“āĻĒā§°ā§° āĻāϟāĻžāϤāĻ•ā§ˆ āĻ…āϧāĻŋāĻ•  (E) None of the above / āĻ“āĻĒā§°ā§° āĻāϟāĻžāĻ“ āύāĻšāϝāĻŧ


Rule Rational Number (āĻĒā§°āĻŋāĻŽā§‡āϝāĻŧ āϏāĻ‚āĻ–ā§āϝāĻž): Click Here

Explanation
(A) 43.123456789 : i. Terminating decimal. ii. Rational Number. (āϏāϏ⧀āĻŽ āĻĻāĻļāĻŽāĻŋāĻ•āĨ¤ āϏ⧇āϝāĻŧ⧇āĻšā§‡ āχ āĻĒā§°āĻŋāĻŽā§‡āϝāĻŧ āϏāĻ‚āĻ–ā§āϝāĻžāĨ¤)(B) 0.12012001200012000... : i. Non-terminating.,  ii. Digits do not repeat in a fixed pattern. , iii. Irrational Number. (āĻĻāĻļāĻŽāĻŋāĻ•āĻŸā§‹ āĻ…āϏ⧀āĻŽ āφ⧰⧁ āϕ⧋āύ⧋ āύāĻŋā§°ā§āĻĻāĻŋāĻˇā§āϟ āϧ⧰āϪ⧇ āĻĒ⧁āύ⧰āĻžāĻŦ⧃āĻ¤ā§āϤāĻŋ āĻšā§‹ā§ąāĻž āύāĻžāχāĨ¤ āϏ⧇āϝāĻŧ⧇āĻšā§‡ āχ āĻ…āĻĒā§°āĻŋāĻŽā§‡āϝāĻŧ āϏāĻ‚āĻ–ā§āϝāĻžāĨ¤)

(C) 43.123456789 : i. Terminating decimal.  ii. Rational Number. (āϏāϏ⧀āĻŽ āĻĻāĻļāĻŽāĻŋāĻ•āĨ¤ āϏ⧇āϝāĻŧ⧇āĻšā§‡ āχ āĻĒā§°āĻŋāĻŽā§‡āϝāĻŧ āϏāĻ‚āĻ–ā§āϝāĻžāĨ¤)

Ans / āωāĻ¤ā§āϤ⧰ : (B) 0.12012001200012000...

Remember / āĻŽāύāϤ ā§°āĻžāĻ–āĻŋāĻŦāĻž : Non-terminating + Non-recurring = Irrational Number (āĻ…āϏ⧀āĻŽ + āĻ…āĻĒ⧁āύ⧰āĻžāĻŦ⧃āĻ¤ā§āϤāĻŋāĻŽā§‚āϞāĻ• = āĻ…āĻĒā§°āĻŋāĻŽā§‡āϝāĻŧ āϏāĻ‚āĻ–ā§āϝāĻž)

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Q. For dividend a and divisor b, we have: a = bq + r, Which relation is correct for the remainder r ? (āĻ­āĻžāĻœā§āϝ a āφ⧰⧁ āĻ­āĻžāϜāĻ• b ā§° āĻŦāĻžāĻŦ⧇ a = bq + r āϤ āĻ…ā§ąāĻļāĻŋāĻˇā§āϟ ā§° āĻŦāĻžāĻŦ⧇ āϕ⧋āύāĻŸā§‹ āϏāĻŽā§āĻĒā§°ā§āĻ• āϏāĻ āĻŋāĻ• ?

Options: (A) a ≤ r ≤ b   (B) 0 > r ≤ b  (C) 0 ≤ r < b   (D) More than one of the above  (E) None of the above

 

Explanation: According to the Division Algorithm, a = bq + r where (āϝ'āϤ),: a = āĻ­āĻžāĻœā§āϝ (Dividend), b = āĻ­āĻžāϜāĻ• (Divisor), q = āĻ­āĻžāĻ—āĻĢāϞ (Quotient), r = āĻ…ā§ąāĻļāĻŋāĻˇā§āϟ (Remainder)

The remainder must always satisfy: 0 ≤ r < b

That means: āĻ…ā§°ā§āĻĨāĻžā§Ž,



  • Remainder can be 0.(āĻ…ā§ąāĻļāĻŋāĻˇā§āϟ 0 āĻš'āĻŦ āĻĒāĻžā§°ā§‡āĨ¤

  • Remainder cannot be negative. (āĻ‹āĻŖāĻžāĻ¤ā§āĻŽāĻ• āĻš'āĻŦ āĻ¨ā§‹ā§ąāĻžā§°ā§‡āĨ¤)

  • Remainder must be smaller than the divisor. (āĻ­āĻžāϜāĻ•āϤāĻ•ā§ˆ āϏ⧰⧁ āĻš'āĻŦ āϞāĻžāĻ—āĻŋāĻŦāĨ¤)


Ans / āωāĻ¤ā§āϤ⧰ : (C) 0 ≤ r < b


Example: 17 = 5 × 3 + 2, Here, r = 2 and 0 ≤ 2 < 5


Tip"Remainder is always smaller than the divisor." ("āĻ…ā§ąāĻļāĻŋāĻˇā§āϟ āϏāĻĻāĻžāϝāĻŧ āĻ­āĻžāϜāĻ•āϤāĻ•ā§ˆ āϏ⧰⧁ āĻšāϝāĻŧāĨ¤")


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Q. For any integer , the form of an odd number is / āϝāĻŋāϕ⧋āύ⧋ āĻĒā§‚ā§°ā§āĻŖāϏāĻ‚āĻ–ā§āϝāĻž ā§° āĻŦāĻžāĻŦ⧇ āĻŦāĻŋāώāĻŽ āϏāĻ‚āĻ–ā§āϝāĻžā§° ā§°ā§‚āĻĒ āĻš'āϞ -


(A) 2p + 1 | (B) 2p | (C) p | (D) More than one of the above / āĻ“āĻĒā§°ā§° āĻāϟāĻžāϤāĻ•ā§ˆ āĻ…āϧāĻŋāĻ• | (E) None of the above / āĻ“āĻĒā§°ā§° āĻāϟāĻžāĻ“ āύāĻšāϝāĻŧ


Ans / āωāĻ¤ā§āϤ⧰: (A) 2p + 1


Explanation / āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: An odd number is not divisible by 2, and its general form is 2p+1, where is any integer. / āĻŦāĻŋāώāĻŽ āϏāĻ‚āĻ–ā§āϝāĻžāĻ• ⧍-⧰⧇ āϏāĻŽā§āĻĒā§‚ā§°ā§āĻŖāĻ­āĻžā§ąā§‡ āĻ­āĻžāĻ— āϕ⧰āĻŋāĻŦ āĻ¨ā§‹ā§ąāĻžā§°āĻŋ, āφ⧰⧁ āχāϝāĻŧāĻžā§° āϏāĻžāϧāĻžā§°āĻŖ ā§°ā§‚āĻĒ 2p+1, āϝ'āϤ p āϝāĻŋāϕ⧋āύ⧋ āĻĒā§‚ā§°ā§āĻŖāϏāĻ‚āĻ–ā§āϝāĻžāĨ¤


Examples / āωāĻĻāĻžāĻšā§°āĻŖ: i. p = 1 → 2(1) + 1 = 3, ii. p = 2p → 2(2)+1 = 5


So, 2p + 1 always gives an odd number. / āϏ⧇āϝāĻŧ⧇āĻšā§‡ 2p + 1 āϏāĻĻāĻžāϝāĻŧ āĻāϟāĻž āĻŦāĻŋāώāĻŽ āϏāĻ‚āĻ–ā§āϝāĻž āĻĻāĻŋāϝāĻŧ⧇āĨ¤


Remember / āĻŽāύāϤ ā§°āĻžāĻ–āĻŋāĻŦāĻž: i. Even Number (āϝ⧁āĻ—ā§āĻŽ āϏāĻ‚āĻ–ā§āϝāĻž)2p,  ii. Odd Number (āĻŦāĻŋāώāĻŽ āϏāĻ‚āĻ–ā§āϝāĻž)2p + 1


Ans / āωāĻ¤ā§āϤ⧰: (A) 2p+1


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Q. The number 3.24636363... is / āϏāĻ‚āĻ–ā§āϝāĻž 3.24636363..... āĻšā§ˆāϛ⧇ -


(A) Natural Number (āĻĒā§ā§°āĻžāĻ•ā§ƒāϤ āϏāĻ‚āĻ–ā§āϝāĻž)  (B) Rational Number (āĻĒā§°āĻŋāĻŽā§‡āϝāĻŧ āϏāĻ‚āĻ–ā§āϝāĻž)  (C) Irrational Number (āĻ…āĻĒā§°āĻŋāĻŽā§‡āϝāĻŧ āϏāĻ‚āĻ–ā§āϝāĻž)
(D) More than one of the above (āĻ“āĻĒā§°ā§° āĻāϟāĻžāϤāĻ•ā§ˆ āĻ…āϧāĻŋāĻ•)  (E) None of the above (āĻ“āĻĒā§°ā§° āĻāϟāĻžāĻ“ āύāĻšāϝāĻŧ)


Ans / āωāĻ¤ā§āϤ⧰: (B) Rational Number (āĻĒā§°āĻŋāĻŽā§‡āϝāĻŧ āϏāĻ‚āĻ–ā§āϝāĻž)


Explanation / āĻŦā§āϝāĻžāĻ–ā§āϝāĻž: The decimal number 3.24636363... is non-terminating but recurring, because 63 repeats again and again. Therefore, it is a Rational Number. / āĻĻāĻļāĻŽāĻŋāĻ• āϏāĻ‚āĻ–ā§āϝāĻž 3.24636363... āĻ…āϏ⧀āĻŽ (non-terminating) āĻ•āĻŋāĻ¨ā§āϤ⧁ 63 āĻ…āĻ‚āĻļāĻŸā§‹ āĻĒ⧁āύ⧰āĻžāĻŦ⧃āĻ¤ā§āϤāĻŋ (recurring) āĻšā§ˆ āφāϛ⧇āĨ¤ āϏ⧇āϝāĻŧ⧇āĻšā§‡ āχ āĻāϟāĻž āĻĒā§°āĻŋāĻŽā§‡āϝāĻŧ āϏāĻ‚āĻ–ā§āϝāĻžāĨ¤


Rule / āύāĻŋāϝāĻŧāĻŽ: i. Terminating decimal = Rational Number (āϏāϏ⧀āĻŽ āĻĻāĻļāĻŽāĻŋāĻ• = āĻĒā§°āĻŋāĻŽā§‡āϝāĻŧ āϏāĻ‚āĻ–ā§āϝāĻž)


ii. Non-terminating Recurring decimal = Rational Number (āĻ…āϏ⧀āĻŽ āĻ•āĻŋāĻ¨ā§āϤ⧁ āĻĒ⧁āύ⧰āĻžāĻŦ⧃āĻ¤ā§āϤāĻŋāĻŽā§‚āϞāĻ• āĻĻāĻļāĻŽāĻŋāĻ• = āĻĒā§°āĻŋāĻŽā§‡āϝāĻŧ āϏāĻ‚āĻ–ā§āϝāĻž)


iii. Non-terminating Non-recurring decimal = Irrational Number (āĻ…āϏ⧀āĻŽ āφ⧰⧁ āĻ…āĻĒ⧁āύ⧰āĻžāĻŦ⧃āĻ¤ā§āϤāĻŋāĻŽā§‚āϞāĻ• āĻĻāĻļāĻŽāĻŋāĻ• = āĻ…āĻĒā§°āĻŋāĻŽā§‡āϝāĻŧ āϏāĻ‚āĻ–ā§āϝāĻž)


Example / āωāĻĻāĻžāĻšā§°āĻŖ: 3.24636363.... Since 63 repeats, the number is Rational (āĻĒā§°āĻŋāĻŽā§‡āϝāĻŧ).


Ans / āωāĻ¤ā§āϤ⧰: (B) Rational Number (āĻĒā§°āĻŋāĻŽā§‡āϝāĻŧ āϏāĻ‚āĻ–ā§āϝāĻž).


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Q. The mean (average) of the first 50 odd natural numbers is / āĻĒā§ā§°āĻĨāĻŽ ā§Ģā§ĻāϟāĻž āĻŦāĻŋāώāĻŽ āĻĒā§ā§°āĻžāĻ•ā§ƒāϤ āϏāĻ‚āĻ–ā§āϝāĻžā§° āĻ—āĻĄāĻŧ (Mean) āĻ•āĻŋāĻŽāĻžāύ ?


(A) 30 | (B) 40 | (C) 50 | (D) More than one of the above / āĻ“āĻĒā§°ā§° āĻāϟāĻžāϤāĻ•ā§ˆ āĻ…āϧāĻŋāĻ• | (E) None of the above / āĻ“āĻĒā§°ā§° āĻāϟāĻžāĻ“ āύāĻšāϝāĻŧ


Ans / āωāĻ¤ā§āϤ⧰: (C) 50


Explanation / āĻŦā§āϝāĻžāĻ–ā§āϝāĻž:


The first 50 odd natural numbers are: 1, 3, 5, 7, ..., 99 / āĻĒā§ā§°āĻĨāĻŽ ā§Ģā§ĻāϟāĻž āĻŦāĻŋāώāĻŽ āϏāĻ‚āĻ–ā§āϝāĻž āĻš'āϞ: 1, 3, 5, ..., 99


For numbers in an Arithmetic Progression (A.P.),


Mean = First Term + Last Term / 2 = 1 + 99 / 2 = 50


Shortcut / āϚāĻŽā§ āĻ•ā§ŒāĻļāϞ: The mean of the first n odd natural numbers is always . (āĻĒā§ā§°āĻĨāĻŽ n āϟāĻž āĻŦāĻŋāώāĻŽ āĻĒā§ā§°āĻžāĻ•ā§ƒāϤ āϏāĻ‚āĻ–ā§āϝāĻžā§° āĻ—āĻĄāĻŧ āϏāĻĻāĻžāϝāĻŧ āĻšāϝāĻŧāĨ¤)


So, for the first 50 odd numbers: 50


Ans / āωāĻ¤ā§āϤ⧰: (C) 50.


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Q. The sum of the numerator and denominator of a fraction is 11. If 1 is added to the numerator and 2 is subtracted from the denominator, the fraction becomes 2/3​. Find the fraction. āĻāϟāĻž āĻ­āĻ—ā§āύāĻžāĻ‚āĻļā§° āϞāĻŦ āφ⧰⧁ āĻšā§°ā§° āϝ⧋āĻ—āĻĢāϞ 11āĨ¤ āϝāĻĻāĻŋ āϞāĻŦāϤ 1 āϝ⧋āĻ— āϕ⧰āĻž āĻšāϝāĻŧ āφ⧰⧁ āĻšā§°ā§° āĻĒā§°āĻž 2 āĻŦāĻŋāϝāĻŧā§‹āĻ— āϕ⧰āĻž āĻšāϝāĻŧ, āϤ⧇āĻ¨ā§āϤ⧇ āĻ­āĻ—ā§āύāĻžāĻ‚āĻļāĻŸā§‹ 2/3 āĻšāϝāĻŧāĨ¤ āĻ­āĻ—ā§āύāĻžāĻ‚āĻļāĻŸā§‹ āύāĻŋā§°ā§āĻŖāϝāĻŧ āϕ⧰āĻžāĨ¤


Options: (A) 5/6(B) 6/5(C) 3/8(D) More than one of the above / āĻ“āĻĒā§°ā§° āĻāϟāĻžāϤāĻ•ā§ˆ āĻ…āϧāĻŋāĻ•  (E) None of the above / āĻ“āĻĒā§°ā§° āĻāϟāĻžāĻ“ āύāĻšāϝāĻŧ


Solution / āϏāĻŽāĻžāϧāĻžāύ


Given: Numerator + 1 / Denominator - 2 = 2/3


(C) 3/8 



Correct


Also, 


Sum is 11.


Ans / āωāĻ¤ā§āϤ⧰: 


Fraction = 3/8 ( Need to Check)



Q. The area of a triangle is 61.5 m². If one side (base) is 12.3 m, find the length of the perpendicular (height) drawn from the opposite vertex to that side. āĻāϟāĻž āĻ¤ā§ā§°āĻŋāϭ⧁āϜ⧰ āĻ•ā§āώ⧇āĻ¤ā§ā§°āĻĢāϞ 61.5 m²āĨ¤ āϝāĻĻāĻŋ āχāϝāĻŧāĻžā§° āĻāϟāĻž āĻŦāĻžāĻšā§ (āĻ­āĻŋāĻ¤ā§āϤāĻŋ) 12.3 m āĻšāϝāĻŧ, āϤ⧇āĻ¨ā§āϤ⧇ āĻŦāĻŋāĻĒā§°ā§€āϤ āĻļā§€ā§°ā§āώāĻŦāĻŋāĻ¨ā§āĻĻ⧁⧰ āĻĒā§°āĻž āϏ⧇āχ āĻŦāĻžāĻšā§āϞ⧈ āĻ…āĻ‚āĻ•āĻŋāϤ āϞāĻŽā§āĻŦā§° āĻĻā§ˆā§°ā§āĻ˜ā§āϝ āύāĻŋā§°ā§āĻŖāϝāĻŧ āϕ⧰āĻžāĨ¤


Options: (A) 11.5 m (B) 10.5 m (C) 11 m (D) 10 m


 Solution / āϏāĻŽāĻžāϧāĻžāύ



 61.5 = 12 × 12.3 × h


h = 2 × 61.5 / 12.3 = 123 / 12.3 = 10 m 


Ans / āωāĻ¤ā§āϤ⧰: (D) 10 m