91. In an examination, the success to failure ratio was 5 : 2, if the number of failures been 14 more, then the success to failure ratio would have been 9 : 5. The total number of candidates, who appeared for the examination, was:

एक परीक्षा में, सफलता का विफलता से अनुपात 5 : 2 था, यदि विफलताओं की संख्या 14 अधिक होती, तो सफलता का विफलता से अनुपात 9 : 5 होता। परीक्षा में उपस्थित होने वाले छात्रों की कुल संख्या कितनी थी:


Option”B” is correct

The success to failure ratio = 5 : 2

When the number of failures are 14 more, then the success to failure ratio would have been 9 : 5

Calculation:

Let the successful candidates be 5x and the failure candidates be 2x.

According to Question,

(5x – 14)/(2x + 14) = 9/5

⇒ 25x – 70 = 18x + 126

⇒ x = 196/7 = 28

Total number of candidates who appeared for the examination = 5x + 2x = 7x = 7 × 28 = 196

92. A jar contains a mixture of pure juice And water in a ratio of 5 : 2. From this, 21 lit juice mixture is taken out and 11 lit water is mixed in remaining mixture due to which ratio of pure juice to water becomes 7 : 5. Find the initial quantity of pure juice in the mixture.

एक जार में 5 : 2 के अनुपात में शुद्ध रस और पानी का मिश्रण है। इसमें से 21 लीटर रस का मिश्रण निकाला जाता है और शेष मिश्रण में 11 लीटर पानी मिलाया जाता है, जिसके कारण शुद्ध रस और पानी का अनुपात 7 : 5 हो जाता है। मिश्रण में शुद्ध रस की प्रारंभिक मात्रा ज्ञात कीजिये।


Option”C” is correct
The ratio of different components remain the same always unless any particular component is added to it

Suppose the quantity of pure juice And pure water in the mixture after 21 lit is taken out is 5x And 2x respectively

According to the question,

5x2x+11=75⇒5x2x+11=75

⇒ 25x = 14x + 77

⇒ x = 7

Initial quantity of pure juice in mixture = 5x + 21 × (5/7)

= 35 + 15 = 50 lit

Short Method :

∵ the ratio of different components remain the same always unless any particular component is added to it

∵ no pure juice is added, only water is added so the ratio of pure juice must be the same

⇒ Pure juice : water

 

⇒ 11 units = 11 lit

⇒ 1 unit = 1 lit

Initial quantity of pure juice = 35 units + 21 × (5/7)

= 50 lit

93. Two numbers are such that the square of one is 224 less than 8 times the square of the other. If the numbers are in the ratio of 3 : 4, find the numbers.

दो संख्याएँ ऐसी हैं, जिनमें से एक का वर्ग दूसरे के वर्ग के 8 गुना से 224 कम है। यदि संख्या 3: 4 के अनुपात में हैं, तो संख्याएँ ज्ञात कीजिए।


Option”D” is correct

The two numbers are in the ratio = 3x : 4x

According to the question

(4x)2 + 224 = 8 × (3x)2

⇒ 16x2 + 224 = 8 × 9x2

⇒ 72x2 – 16x2 = 224

⇒ 56x2 = 224

⇒ x2 = 224/56

⇒ x2 = 4

⇒ x = √4

⇒ x = 2

∴ The number are 3 × 2 = 6 and 4 × 2 = 8.

94. Divide 500 into two parts such that the ratio of one to the other are in 5 : 3?

500 को दो भागों में विभाजित कीजिये जैसे कि एक से दूसरे का अनुपात 5 : 3 है?


Option”A” is correct

Let the ratio of two parts be 5x and 3x

Sum of Ratios = 5x + 3x = 8x

⇒ 8x = 500

Value of first part = (5x/8x) × 500 = 312.5

Value of second part = (3x/8x) × 500 = 187.5

∴ The value of two parts is 312.5 and 187.5 respectively.

95. The price of two articles are in the ratio of 3 : 2 respectively. The price of first article is increased by 40% and the price of second article is decreased by x%. If the new ratio of the price of the two articles is 7 : 3 respectively, then what is the value of x?

दो वस्तुओं का मूल्य क्रमशः 3 : 2 के अनुपात में है। पहली वस्तु के मूल्य में 40% की वृद्धि हुई है और दूसरी वस्तु के मूल्य में x% की कमी हुई है। यदि दो वस्तुओं के मूल्य का नया अनुपात क्रमशः 7 : 3 है, तो x का मान क्या है?


Option”D” is correct

The price of two articles are in the ratio of 3 : 2 respectively

Formula Used:

Selling price = Cost price + profit

Selling price = Cost price – loss

Calculation:

Let the cost price of two articles be 3y and 2y respectively, then

When the price of first article is increased by 40%

Price of first article = 3y + (40/100) × 3y = 4.2y

When the price of second article is decreased by x%

Price of second article = 2y – (x/100) × 2y = 2y – 2xy/100

Now,

⇒ (4.2y)/(2y – 2xy/100) = 7 : 3

⇒ 12.6y = 14y – 14xy/100

⇒ 14xy/100 = 1.4y

⇒ x = 10%

∴ Value of x is 10

96. If (4a + 8b) ∶ (3a – b) = 5 ∶ 2 and (3c + 2d) ∶ (6c – 2d) = 11 ∶ 7. Find the compound ratio of (a2 + b2) ∶ (2a2 + 3b2) and (8c + 2d) ∶ (3c + 4d).

यदि (4a + 8b) ∶ (3a – b) = 5 ∶ 2 और (3c + 2d) ∶ (6c – 2d) = 11 ∶ 7 है। (a2 + b2) ∶ (2a2 + 3b2) and (8c + 2d) ∶ (3c + 4d) का मिश्रित अनुपात ज्ञात कीजिए।


Option”C” is correct

Now, (4a + 8b) ∶ (3a – b) = 5 ∶ 2

∴ 2(4a + 8b) = 5(3a – b)

∴ 8a + 16b = 15a – 5b

∴ 7a = 21b

∴ a ∶ b = 3 ∶ 1

Now, (3c + 2d) ∶ (6c – 2d) = 11 ∶ 7

∴ 7(3c + 2d) = 11(6c – 2d)

∴ 21c + 14d = 66c – 22d

∴ 45c = 36d

∴ c ∶ d = 4 ∶ 5

Now, (a2 + b2) ∶ (2a2 + 3b2) = (32 + 12) ∶ (2(32) + 3(12)) = 10 ∶ 21

Now, (8c + 2d) ∶ (3c + 4d) = (8(4) + 2(5)) ∶ (3(4) + 4(5)) = 21 ∶ 16

Thus, compound ratio of (a2 + b2) ∶ (2a2 + 3b2) and (8c + 2d) ∶ (3c + 4d) = (10)(21) ∶ (21)(16) = 210 ∶ 336

5 : 8

97. Rs. 500 is to be divided among X, Y And Z in such a way that Rs. 16 more than 2/5th of X’s share, Rs. 70 less than 3/4th of Y’s share and Rs. 4 less than 3/5th of Z’s share are equal. Find the share of Z.

X, Y और Z के बीच 500 रुपए को इस प्रकार बांटा जाना है जिससे X के हिस्से के 2/5 गुना से 16 रुपए अधिक, Y के हिस्से के 3/4 गुना से 70 रुपए कम और Z के हिस्से का 3/5 गुना से 4 रूपये कम बराबर हो जाए। तो Z का हिस्सा ज्ञात कीजिए।


Option”D” is correct

As per the given statement:

[X × 2/5] + 16 = [Y × 3/4] – 70 = [Z × 3/5] – 4 = 6K (Suppose)

∴ X = (6K – 16) × 5/2 = 15K – 40

Y = (6K + 70) × 4/3 = 8K + 280/3

Z = (6K + 4) × 5/3 = 10K + 20/3

Since total amount = Rs. 500

∴ 15K – 40 + 8K + 280/3 + 10K + 20/3 = 500

⇒ 33K = 440

⇒ K = 40/3

∴ Share of Z = 10 × 40/3 + 20/3 = 420/3 = Rs. 140

98. Three vessels each of 10 litre capacity contain a mixture of milk and water in the ratio 2 ∶ 1, 3 ∶ 1 and 3 ∶ 2. If all the three vessels are emptied into a large vessel, then find the ratio of milk and water in the new mixture.

2 ∶ 1, 3 ∶ 1 और 3 ∶ 2 के अनुपात में दूध और पानी के मिश्रण वाले प्रत्येक 10 लीटर क्षमता वाले तीन बर्तन हैं। यदि तीनों बर्तनों को एक बड़े बर्तन में खाली किया जाता है, तब नए मिश्रण में दूध और पानी का अनुपात ज्ञात कीजिये।


Option”B” is correct

Quantity of total milk = 10 × 2/3 + 10 × 3/4 + 10 × 3/5 = 20/3 + 15/2 + 6 = 121/6

Quantity of total water = 10 × 1/3 + 10 × 1/4 + 10 × 2/5 = 10/3 + 5/2 + 4 = 59/6

∴ Ratio of milk and water in the new mixture = 121 ∶ 59 

99. The price of a pen and a marker is in the ratio 4 ∶ 5. A shopkeeper bought some pens and markers by paying for them in the ratio 3 ∶ 2. If he bought 45 pens, how many markers did he buy?

एक पेन और एक मार्कर की कीमत 4 ∶ 5 के अनुपात में है। एक दुकानदार ने 3 ∶ 2 के अनुपात में कीमत चुकाकर कुछ पेन और मार्कर खरीदे। यदि उसने 45 पेन खरीदे, तो उसने कितने मार्कर खरीदे?


Option”B” is correct

Let he buy ‘x’ markers

⇒ Ratio of pens and markers bought = 45 ∶ x

Now,

⇒ Total price paid = No. of items bought × Price of item

Also,

⇒ Compound ratio of two ratios (a ∶ b) and (c ∶ d) = (ac ∶ bd)

⇒ Ratio of price paid = Ratio of pens and markers bought × Ratio of prices of pen and marker

⇒ 3 ∶ 2 = (45 ∶ x) × (4 ∶ 5)

⇒ 3 ∶ 2 = 180 ∶ 5x

⇒ 3 ∶ 2 = 36 ∶ x

⇒ x = (2/3) × 36 = 24

∴ He bought 24 markers

100. There are 24700 students in 4 schools A, B, C and D. If half students of A, two-third of B, three-fourth of C and four-fifth of D are same number of students, then find the ratio of students in school A and D.

4 स्कूल A, B, C और D में 24700 छात्र हैं। यदि A के आधे, B के दो-तिहाई, C के तीन-चौथाई और D के 4/5वें भाग में छात्रों की संख्या समान है, तो स्कूल A और D में छात्रों का अनुपात ज्ञात कीजिए।


Option”D” is correct

According to question,

A/2 = 2B/3 = 3C/4 = 4D/5

∴ A/B = 4/3

⇒ B/C = 9/8

⇒ C/D = 16/15

⇒ A ∶ B = 4 ∶ 3

⇒ B ∶ C = 9 ∶ 8

⇒ C ∶ D = 16 ∶ 15

⇒ A ∶ B ∶ C ∶ D = 24 : 18 : 16 : 15

∴ The ratio of number of students in school A and D = 8 : 5

7 thoughts on “Ratio & Proportion Questions”

  1. Nice application 👍👍 9
    Thank so much 🥰🥰
    Very helpful for my pripresion

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