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: āĻāĻāĻž āĻŦā§āĻāϤ āύā§āϞāĻž (Blue), āĻšāĻžāϞāϧā§āϝāĻŧāĻž (Yellow) āĻā§°ā§ āĻāĻŽāϞāĻž (Orange) āĻā§āĻā§āύ āĻāĻā§āĨ¤ āύā§āϞāĻž āĻā§°ā§ āĻšāĻžāϞāϧā§āϝāĻŧāĻž āĻā§āĻā§āύ⧰ āĻ āύā§āĻĒāĻžāϤ 5 : 2, āĻā§°ā§ āĻāĻŽāϞāĻž āĻā§°ā§ āύā§āϞāĻž āĻā§āĻā§āύ⧰ āĻ āύā§āĻĒāĻžāϤ 11 : 18āĨ¤ āϤā§āύā§āϤ⧠āĻāĻŽāϞāĻž āĻā§°ā§ āĻšāĻžāϞāϧā§āϝāĻŧāĻž āĻā§āĻā§āύ⧰ āĻ āύā§āĻĒāĻžāϤ āĻāĻŋāĻŽāĻžāύ ?)
Options / āĻŦāĻŋāĻāϞā§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 âš10 and âš5 notes whose total value is âš350. If he has 45 notes in total, how many âš10 notes does he have ? (Q.47 āĻāĻāύ āĻĻā§āĻāĻžāύāĻĻāĻžā§°ā§° āĻāĻā§°āϤ âš10 āĻā§°ā§ âš5-ā§° āύā§āĻ āĻāĻā§āĨ¤ āύā§āĻāϏāĻŽā§āĻšā§° āĻŽā§āĻ āĻŽā§āϞā§āϝ âš350āĨ¤ āϝāĻĻāĻŋ āĻŽā§āĻ āύā§āĻā§° āϏāĻāĻā§āϝāĻž 45 āĻšāϝāĻŧ, āϤā§āύā§āϤ⧠âš10-ā§° āύā§āĻ āĻāĻŋāĻŽāĻžāύ āĻāĻā§ ?)
Options / āĻŦāĻŋāĻāϞā§āĻĒāϏāĻŽā§āĻš: A. 29 B. 31 C. 27 D. 25
Soln / āϏāĻŽāĻžāϧāĻžāύ
âš10 = x and âš5 = 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⧰ āĻā§āύāĻā§ āĻĻāĻļāĻŽāĻŋāĻ āĻĒā§ā§°āϏāĻžā§°āĻŖ āĻĒā§°āĻŋāĻŽā§āϝāĻŧ āϏāĻāĻā§āϝāĻž āύāĻšāϝāĻŧ ?)
Options: (A) 43.123456789 (B) 0.12012001200012000... (C) 43.123456789 (D) More than one of the above / āĻāĻĒā§°ā§° āĻāĻāĻžāϤāĻā§ āĻ āϧāĻŋāĻ (E) None of the above / āĻāĻĒā§°ā§° āĻāĻāĻžāĻ āύāĻšāϝāĻŧ
Rule Rational Number (āĻĒā§°āĻŋāĻŽā§āϝāĻŧ āϏāĻāĻā§āϝāĻž): Click Here
(A) 43.123456789 : i. Terminating decimal. ii. Rational Number. (āϏāϏā§āĻŽ āĻĻāĻļāĻŽāĻŋāĻāĨ¤ āϏā§āϝāĻŧā§āĻšā§ āĻ āĻĒā§°āĻŋāĻŽā§āϝāĻŧ āϏāĻāĻā§āϝāĻžāĨ¤)(B) 0.12012001200012000... : i. Non-terminating., ii. Digits do not repeat in a fixed pattern. , iii. Irrational Number. (āĻĻāĻļāĻŽāĻŋāĻāĻā§ āĻ āϏā§āĻŽ āĻā§°ā§ āĻā§āύ⧠āύāĻŋā§°ā§āĻĻāĻŋāώā§āĻ āϧ⧰āĻŖā§ āĻĒā§āύ⧰āĻžāĻŦā§āϤā§āϤāĻŋ āĻšā§ā§ąāĻž āύāĻžāĻāĨ¤ āϏā§āϝāĻŧā§āĻšā§ āĻ āĻ āĻĒā§°āĻŋāĻŽā§āϝāĻŧ āϏāĻāĻā§āϝāĻžāĨ¤)
- Remainder can be 0.(āĻ ā§ąāĻļāĻŋāώā§āĻ 0 āĻš'āĻŦ āĻĒāĻžā§°ā§āĨ¤
- Remainder cannot be negative. (āĻāĻŖāĻžāϤā§āĻŽāĻ āĻš'āĻŦ āύā§ā§ąāĻžā§°ā§āĨ¤)
- Remainder must be smaller than the divisor. (āĻāĻžāĻāĻāϤāĻā§ āϏ⧰⧠āĻš'āĻŦ āϞāĻžāĻāĻŋāĻŦāĨ¤)
Ans / āĻāϤā§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§ąāĻļāĻŋāώā§āĻ āϏāĻĻāĻžāϝāĻŧ āĻāĻžāĻāĻāϤāĻā§ āϏ⧰⧠āĻšāϝāĻŧāĨ¤")
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Q. For any integer , the form of an odd number is / āϝāĻŋāĻā§āύ⧠āĻĒā§ā§°ā§āĻŖāϏāĻāĻā§āϝāĻž ā§° āĻŦāĻžāĻŦā§ āĻŦāĻŋāώāĻŽ āϏāĻāĻā§āϝāĻžā§° ā§°ā§āĻĒ āĻš'āϞ -
(A) 2p + 1 | (B) 2p | (C) p | (D) More than one of the above / āĻāĻĒā§°ā§° āĻāĻāĻžāϤāĻā§ āĻ āϧāĻŋāĻ | (E) None of the above / āĻāĻĒā§°ā§° āĻāĻāĻžāĻ āύāĻšāϝāĻŧ
Ans / āĻāϤā§āϤ⧰: (A) 2p + 1
Explanation / āĻŦā§āϝāĻžāĻā§āϝāĻž: An odd number is not divisible by 2, and its general form is 2p+1, where is any integer. / āĻŦāĻŋāώāĻŽ āϏāĻāĻā§āϝāĻžāĻ ā§¨-ā§°ā§ āϏāĻŽā§āĻĒā§ā§°ā§āĻŖāĻāĻžā§ąā§ āĻāĻžāĻ āĻā§°āĻŋāĻŦ āύā§ā§ąāĻžā§°āĻŋ, āĻā§°ā§ āĻāϝāĻŧāĻžā§° āϏāĻžāϧāĻžā§°āĻŖ ā§°ā§āĻĒ 2p+1, āϝ'āϤ p āϝāĻŋāĻā§āύ⧠āĻĒā§ā§°ā§āĻŖāϏāĻāĻā§āϝāĻžāĨ¤
Examples / āĻāĻĻāĻžāĻšā§°āĻŖ: i. p = 1 → 2(1) + 1 = 3, ii. p = 2p → 2(2)+1 = 5
So, 2p + 1 always gives an odd number. / āϏā§āϝāĻŧā§āĻšā§ 2p + 1 āϏāĻĻāĻžāϝāĻŧ āĻāĻāĻž āĻŦāĻŋāώāĻŽ āϏāĻāĻā§āϝāĻž āĻĻāĻŋāϝāĻŧā§āĨ¤
Remember / āĻŽāύāϤ ā§°āĻžāĻāĻŋāĻŦāĻž: i. Even Number (āϝā§āĻā§āĻŽ āϏāĻāĻā§āϝāĻž) → 2p, ii. Odd Number (āĻŦāĻŋāώāĻŽ āϏāĻāĻā§āϝāĻž) → 2p + 1
Ans / āĻāϤā§āϤ⧰: (A) 2p+1
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Q. The number 3.24636363... is / āϏāĻāĻā§āϝāĻž 3.24636363..... āĻšā§āĻā§ -
(A) Natural Number (āĻĒā§ā§°āĻžāĻā§āϤ āϏāĻāĻā§āϝāĻž) (B) Rational Number (āĻĒā§°āĻŋāĻŽā§āϝāĻŧ āϏāĻāĻā§āϝāĻž) (C) Irrational Number (āĻ
āĻĒā§°āĻŋāĻŽā§āϝāĻŧ āϏāĻāĻā§āϝāĻž)
(D) More than one of the above (āĻāĻĒā§°ā§° āĻāĻāĻžāϤāĻā§ āĻ
āϧāĻŋāĻ) (E) None of the above (āĻāĻĒā§°ā§° āĻāĻāĻžāĻ āύāĻšāϝāĻŧ)
Ans / āĻāϤā§āϤ⧰: (B) Rational Number (āĻĒā§°āĻŋāĻŽā§āϝāĻŧ āϏāĻāĻā§āϝāĻž)
Explanation / āĻŦā§āϝāĻžāĻā§āϝāĻž: The decimal number 3.24636363... is non-terminating but recurring, because 63 repeats again and again. Therefore, it is a Rational Number. / āĻĻāĻļāĻŽāĻŋāĻ āϏāĻāĻā§āϝāĻž 3.24636363... āĻ āϏā§āĻŽ (non-terminating) āĻāĻŋāύā§āϤ⧠63 āĻ āĻāĻļāĻā§ āĻĒā§āύ⧰āĻžāĻŦā§āϤā§āϤāĻŋ (recurring) āĻšā§ āĻāĻā§āĨ¤ āϏā§āϝāĻŧā§āĻšā§ āĻ āĻāĻāĻž āĻĒā§°āĻŋāĻŽā§āϝāĻŧ āϏāĻāĻā§āϝāĻžāĨ¤
Rule / āύāĻŋāϝāĻŧāĻŽ: i. Terminating decimal = Rational Number (āϏāϏā§āĻŽ āĻĻāĻļāĻŽāĻŋāĻ = āĻĒā§°āĻŋāĻŽā§āϝāĻŧ āϏāĻāĻā§āϝāĻž)
ii. Non-terminating Recurring decimal = Rational Number (āĻ āϏā§āĻŽ āĻāĻŋāύā§āϤ⧠āĻĒā§āύ⧰āĻžāĻŦā§āϤā§āϤāĻŋāĻŽā§āϞāĻ āĻĻāĻļāĻŽāĻŋāĻ = āĻĒā§°āĻŋāĻŽā§āϝāĻŧ āϏāĻāĻā§āϝāĻž)
iii. Non-terminating Non-recurring decimal = Irrational Number (āĻ āϏā§āĻŽ āĻā§°ā§ āĻ āĻĒā§āύ⧰āĻžāĻŦā§āϤā§āϤāĻŋāĻŽā§āϞāĻ āĻĻāĻļāĻŽāĻŋāĻ = āĻ āĻĒā§°āĻŋāĻŽā§āϝāĻŧ āϏāĻāĻā§āϝāĻž)
Example / āĻāĻĻāĻžāĻšā§°āĻŖ: 3.24636363.... Since 63 repeats, the number is Rational (āĻĒā§°āĻŋāĻŽā§āϝāĻŧ).
Ans / āĻāϤā§āϤ⧰: (B) Rational Number (āĻĒā§°āĻŋāĻŽā§āϝāĻŧ āϏāĻāĻā§āϝāĻž).
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Q. The mean (average) of the first 50 odd natural numbers is / āĻĒā§ā§°āĻĨāĻŽ ā§Ģā§ĻāĻāĻž āĻŦāĻŋāώāĻŽ āĻĒā§ā§°āĻžāĻā§āϤ āϏāĻāĻā§āϝāĻžā§° āĻāĻĄāĻŧ (Mean) āĻāĻŋāĻŽāĻžāύ ?
(A) 30 | (B) 40 | (C) 50 | (D) More than one of the above / āĻāĻĒā§°ā§° āĻāĻāĻžāϤāĻā§ āĻ āϧāĻŋāĻ | (E) None of the above / āĻāĻĒā§°ā§° āĻāĻāĻžāĻ āύāĻšāϝāĻŧ
Ans / āĻāϤā§āϤ⧰: (C) 50
Explanation / āĻŦā§āϝāĻžāĻā§āϝāĻž:
The first 50 odd natural numbers are: 1, 3, 5, 7, ..., 99 / āĻĒā§ā§°āĻĨāĻŽ ā§Ģā§ĻāĻāĻž āĻŦāĻŋāώāĻŽ āϏāĻāĻā§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 / āĻāĻŽā§ āĻā§āĻļāϞ: The mean of the first n odd natural numbers is always . (āĻĒā§ā§°āĻĨāĻŽ n āĻāĻž āĻŦāĻŋāώāĻŽ āĻĒā§ā§°āĻžāĻā§āϤ āϏāĻāĻā§āϝāĻžā§° āĻāĻĄāĻŧ āϏāĻĻāĻžāϝāĻŧ āĻšāϝāĻŧāĨ¤)
So, for the first 50 odd numbers: 50
Ans / āĻāϤā§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 2/3â. Find the fraction. āĻāĻāĻž āĻāĻā§āύāĻžāĻāĻļā§° āϞāĻŦ āĻā§°ā§ āĻšā§°ā§° āϝā§āĻāĻĢāϞ 11āĨ¤ āϝāĻĻāĻŋ āϞāĻŦāϤ 1 āϝā§āĻ āĻā§°āĻž āĻšāϝāĻŧ āĻā§°ā§ āĻšā§°ā§° āĻĒā§°āĻž 2 āĻŦāĻŋāϝāĻŧā§āĻ āĻā§°āĻž āĻšāϝāĻŧ, āϤā§āύā§āϤ⧠āĻāĻā§āύāĻžāĻāĻļāĻā§ 2/3 āĻšāϝāĻŧāĨ¤ āĻāĻā§āύāĻžāĻāĻļāĻā§ āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻžāĨ¤
Options: (A) 5/6(B) 6/5(C) 3/8(D) More than one of the above / āĻāĻĒā§°ā§° āĻāĻāĻžāϤāĻā§ āĻ āϧāĻŋāĻ (E) None of the above / āĻāĻĒā§°ā§° āĻāĻāĻžāĻ āύāĻšāϝāĻŧ
Solution / āϏāĻŽāĻžāϧāĻžāύ
Given: Numerator + 1 / Denominator - 2 = 2/3
(C) 3/8
Correct
Also,
Sum is 11.
Ans / āĻāϤā§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. āĻāĻāĻž āϤā§ā§°āĻŋāĻā§āĻā§° āĻā§āώā§āϤā§ā§°āĻĢāϞ 61.5 m²āĨ¤ āϝāĻĻāĻŋ āĻāϝāĻŧāĻžā§° āĻāĻāĻž āĻŦāĻžāĻšā§ (āĻāĻŋāϤā§āϤāĻŋ) 12.3 m āĻšāϝāĻŧ, āϤā§āύā§āϤ⧠āĻŦāĻŋāĻĒā§°ā§āϤ āĻļā§ā§°ā§āώāĻŦāĻŋāύā§āĻĻā§ā§° āĻĒā§°āĻž āϏā§āĻ āĻŦāĻžāĻšā§āϞ⧠āĻ āĻāĻāĻŋāϤ āϞāĻŽā§āĻŦā§° āĻĻā§ā§°ā§āĻā§āϝ āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°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 / āĻāϤā§āϤ⧰: (D) 10 m