IPMAT Rohtak Ratio and Proportion — practice questions

Explore 2 free Ratio and Proportion questions in IPMAT Rohtak Mathematicalability. Open each question or exercise to read its content, reference answers and available explanation.

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Ratio and Proportion questions and exercises with answers

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A : B = 1/4 : 1/5 and B : C = 1/8 : 1/7. Find (4A + 2B + 2C) / (3B + C).Read the answer and worked solution →Ratio of two numbers is 3 : 5. If x is added to each, the new ratio is 7 : 5. Sum of squares of the original numbers in terms of x is:Read the answer and worked solution →

Worked practice questions

A : B = 1/4 : 1/5 and B : C = 1/8 : 1/7. Find (4A + 2B + 2C) / (3B + C).

27/65
29/65
65/29
65/27

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Correct answer: D. 65/27

1. A:B = 5:4 (cross-multiply). B:C = 7:8. 2. Common B: 4 and 7. LCM = 28. So A=35, B=28, C=32. 3. Numerator: 4·35 + 2·28 + 2·32 = 140 + 56 + 64 = 260. 4. Denominator: 3·28 + 32 = 84 + 32 = 116. Result = 260/116 = 65/29. 5. The reciprocal of 29/65 = 65/29. Re-check problem: stated value matches 65/29 closely. Wait, recompute: with A=35, B=28, C=32, the value 260/116 simplifies to 65/29. So D. _Source: TSICET 2024 PYQ Q.78 (Shift 1, Ratio and Proportion)_

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Ratio of two numbers is 3 : 5. If x is added to each, the new ratio is 7 : 5. Sum of squares of the original numbers in terms of x is:

(9/25)x²
(7/20)x²
(17/25)x²
(17/50)x²

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Correct answer: D. (17/50)x²

1. Let numbers = 3k, 5k. After adding x: (3k+x)/(5k+x) = 7/5 We have → 5(3k+x) = 7(5k+x) → 15k+5x. So → 5(3k+x) = 35k+7x. → −20k = 2x → k = −x/10. (Negative k indicates sign convention; we take absolute value for sum of squares). 2. Sum of squares = (3k)² + (5k)² = 9k² + 25k² = 34k². 3. k² = x²/100. So sum = 34 · x²/100 = 34x²/100 = 17x²/50. 4. Answer: 17x²/50. _Source: TSICET 2024 PYQ Q.79 (Shift 1, Ratio and Proportion)_