SBI PO

Quantity Comparison Practice Questions & Answers with Detailed Solutions

As far as banking exams are concerned, quantity comparison is one important concept and the questions related to this can either be numerical or analytical. This is the concept in which the value of the quantities are being calculated and then comparison is made between quantity 1 and quantity 2. In this article we will be discussing Quantity Comparison Practice Questions & Answers with detailed Solutions. Mentioned below are different types of quantity comparisons:

  • Based on three quantities
  • Quadratic equation based questions
  • Two quantities of different topic
  • Two quantities of same topic
  • When one quantity has numerical value

General Tips as How to go for Quantity Comparison questions

  • Elaborate calculations are not required in most of the quantity comparison questions
  • Try to compare the quantities based on non-graphical data provided and not on the basis of geometric figures or any other visual aid.
  • To solve quantity comparison questions you are only required to determine which quantity (if either) is greater, not how much greater the quantity is.

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Quantity Comparison Practice Questions & Answers with Detailed Solutions

Question 1: In the question, two Quantities I and II are given. You have to solve both the Quantities to establish the correct relation between Quantity-I and Quantity-II and choose the correct option.

A set contains possible three digit numbers whose sum of digits is 24. A number is picked at random.

Quantity-I: Probability that the number is divisible by 2.

Quantity-II: Probability that the number is divisible by 3.

A) Quantity-I > Quantity-II

B) Quantity-I < Quantity-II

C) Quantity-I ≤ Quantity-II

D) Quantity-I = Quantity-II or No relation

E) Quantity-I ≥ Quantity-II

Question 2: In the question, two Quantities I and II are given. You have to solve both the Quantities to establish the correct relation between Quantity-I and Quantity-II and choose the correct option.

The age of three friends Deepak, Dixit and Parul is (x – 2) years, (x + 1) years, and (x + 4) years respectively.The average age of Deepak, Dixit and Parul taken together is (x – 12.5) years less than the average age of Dixit and Parul taken together.

Quantity-I: Average age of Deepak and Parul (in years).

Quantity-II: Age of Dixit (in years)

A) Quantity-I > Quantity-II

B) Quantity-I < Quantity-II

C) Quantity-I ≤ Quantity-II

D) Quantity-I = Quantity-II or No relation

E) Quantity-I ≥ Quantity-II

Question 3: In the question, two quantities I and II are given. You have to solve both the quantities to establish the correct relation between Quantity-I and Quantity-II and choose the correct option.

Quantity I: A is 50% more efficient than B, and thrice as efficient as C. If B and C work together, then the work is completed in 20 days. In how many days A alone can complete the work?

Quantity II: 20 days

A) Quantity-I < Quantity-II

B) Quantity-I > Quantity-II

C) Quantity-I ≤ Quantity-II

D) Quantity-I = Quantity-II or No relation

E) Quantity-I ≥ Quantity-II

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Question 4: In the question, two quantities I and II are given. You have to solve both the quantities to establish the correct relation between Quantity-I and Quantity-II and choose the correct option.

Quantity-I: A box contains 6 red, 4 white, and 5 black balls. Priyam draws 2 balls at random from the bag. What will be the probability that at least 1 of them is red in colour?

Quantity-II: 0.6

A) Quantity-I > Quantity-II

B) Quantity-I < Quantity-II

C) Quantity-I ≤ Quantity-II

D) Quantity-I = Quantity-II or No relation

E) Quantity-I ≥ Quantity-II

Question 5: In the question, two quantities I and II are given. You have to solve both the quantities to establish the correct relation between Quantity-I and Quantity-II and choose the correct option.

Quantity-I: Income of Gaurav and Saurav is Rs. 28000 and Rs. 30000, respectively. If Gaurav saves Rs. 8000 which is 80% of Saurav’s savings, then find the difference between expenditure of Saurav and Gaurav.

Quantity-II: Rs. 500

A) Quantity-I > Quantity-II

B) Quantity-I < Quantity-II

C) Quantity-I ≤ Quantity-II

D) Quantity-I = Quantity-II or No relation

E) Quantity-I ≥ Quantity-II

Question 6: In the question, two quantities I and II are given. You have to solve both the quantities to establish the correct relation between Quantity-I and Quantity-II and choose the correct option.

Quantity-I: ‘A’ can complete a piece of work in (x + 49) days while ‘B’ can complete the same work in (x + 36) days. If both of them working together can complete the whole work in ‘x’ days then find the value of x.

Quantity-II: Find the value of ‘x’ if (x – 44)2 = 172 – 4x.

A) Quantity-I > Quantity-II

B) Quantity-I < Quantity-II

C) Quantity-I ≤ Quantity-II

D) Quantity-I = Quantity-II or No relation

E) Quantity-I ≥ Quantity-II

Question 7: In the question, two quantities I and II are given. You have to solve both the quantities to establish the correct relation between Quantity-I and Quantity-II and choose the correct option.

Quantity-I: A shopkeeper bought three articles of Rs. 6000 each. He sold first article at a profit of 15%, second article at a profit of 12% and third article at a loss of 15%. Find overall profit/loss percentage faced by the shopkeeper.

Quantity-II: 6%

A) Quantity-I > Quantity-II

B) Quantity-I < Quantity-II

C) Quantity-I ≤ Quantity-II

D) Quantity-I = Quantity-II or No relation

E) Quantity-I ≥ Quantity-II

Question 8: In the question, two quantities I and II are given. You have to solve both the quantities to establish the correct relation between Quantity-I and Quantity-II and choose the correct option.

Quantity-I: ‘A’ and ‘B’ together can complete a piece of work in 16 days while ‘B’ and ‘C’ together can complete the same work in 12 days. ‘A’ started the work and worked on it for 9 days and left then ‘B’ joined the work and worked on it for 15 days and left. At last ‘C’ finished the remaining work in 5 days. In how many days ‘C’ alone will complete the whole work.

Quantity-II: 16 days

A) Quantity-I > Quantity-II

B) Quantity-I < Quantity-II

C) Quantity-I ≤ Quantity-II

D) Quantity-I = Quantity-II or No relation

E) Quantity-I ≥ Quantity-II

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Question 9: Quantity-I: A mixture contains milk and water in the ratio of 3:2 respectively. If 64 litres of milk and 64 litres of water are added to this mixture, then the ratio of milk and water changes to 10:7, respectively. Find the initial quantity of water in the mixture.

Quantity-II: 240 litres of a mixture contains 30% wine and 70% whisky. A certain quantity of wine is added to this mixture such that the new mixture contains 40% wine and rest whisky. Find the quantity of wine added.

A) Quantity-I > Quantity-II

B) Quantity-I < Quantity-II

C) Quantity-I ≤ Quantity-II

D) Quantity-I = Quantity-II or No relation

E) Quantity-I ≥ Quantity-II

Question 10: In the question, two quantities I and II are given. You have to solve both the quantities to establish the correct relation between Quantity-I and Quantity-II and choose the correct option.

Quantity I: Five balls are to be drawn from a bag containing 7 red balls and ‘x’ green balls. Find the value of ‘x’ if the number of ways in which the balls can be drawn such that exactly three red balls are drawn is 525.

Quantity II: A bag contains red and blue balls in the ratio of 3:2, respectively. Two balls are drawn randomly from the bag and the probability that both the balls are of different colors is 18/35. Find the number of red balls in the bag.

A) Quantity-I < Quantity-II

B) Quantity-I > Quantity-II

C) Quantity-I ≤ Quantity-II

D) Quantity-I = Quantity-II or No relation

E) Quantity-I ≥ Quantity-II

प्रश्न 1: प्रश्न में, दो मात्राएँ I और II दी गई हैं। आपको मात्रा- I और मात्रा – II के बीच सही संबंध स्थापित करने के लिए दोनों मात्राओं को हल करना है और सही विकल्प चुनना है।

एक सेट में संभावित तीन अंकों की संख्या है, जिनके अंकों का जोड़ 24 है। एक संख्या को यादृच्छिक रूप से चुना जाता है।

मात्रा- I: प्रायिकता है कि संख्या 2 से विभाज्य है।

मात्रा- II: प्रायिकता है कि संख्या 3 से विभाज्य है।

A) मात्रा- I> मात्रा- II

B) मात्रा- I < मात्रा- II

C) मात्रा- I≤ मात्रा- II

D) मात्रा- I = मात्रा- II या कोई संबंध नहीं

E) मात्रा- I ≥ मात्रा- II

प्रश्न 2: प्रश्न में, दो मात्राएँ I और II दी गई हैं। आपको मात्रा- I और मात्रा – II के बीच सही संबंध स्थापित करने के लिए दोनों मात्राओं को हल करना है और सही विकल्प चुनना है।

तीन दोस्तों Deepak, Dixit और Parul की आयु क्रमशः (x – 2) वर्ष, (x + 1) वर्ष और (x + 4) वर्ष है। Deepak, Dixit और Parul की औसत आयु , Dixit और Parul की औसत आयु की तुलना में (x – 12.5%) वर्ष कम है।

मात्रा- I: Deepak और Parul की औसत आयु (वर्षों में)।

मात्रा- II: Dixit की आयु (वर्षों में)

A) मात्रा- I> मात्रा- II

B) मात्रा- I < मात्रा- II

C) मात्रा- I ≤ मात्रा- II

D) मात्रा- I = मात्रा- II या कोई संबंध नहीं है

E) मात्रा- I≥ मात्रा- II

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प्रश्न 3: प्रश्न में, दो मात्राएँ I और II दी गई हैं। आपको मात्रा- I और मात्रा- II के बीच सही संबंध स्थापित करने के लिए दोनों संबंधों को हल करना है और सही विकल्प चुनना है।

मात्रा I: A, B की तुलना में 50% अधिक कार्यकुशल है, और C से तीन गुणा कार्यकुशल है। यदि B और C एक साथ काम करते हैं, तो कार्य 20 दिनों में पूरा हो जाता है। A अकेले कितने दिनों में काम पूरा कर सकता है?

मात्रा II: 20 दिन

A) मात्रा- I <मात्रा- II

B) मात्रा- I >मात्रा- II

C) मात्रा- I ≤ मात्रा- II

D) मात्रा- I = मात्रा- II या कोई संबंध नहीं

E) मात्रा- I ≥ मात्रा- II

प्रश्न 4: प्रश्न में, दो मात्राएँ I और II दी गई हैं। आपको मात्रा- I और मात्रा- II के बीच सही संबंध स्थापित करने के लिए दोनों संबंधों को हल करना है और सही विकल्प चुनना है।

मात्रा- I: एक box में 6 red, 4 white और 5 black balls हैं। Priyam ने बैग से यादृच्छिक रूप से 2 balls निकाली। ऐसी कितनी प्रायिकता होगी कि उनमें से कम से कम 1 का red colour हो?

मात्रा- II: 0.6

A) मात्रा- I> मात्रा- II

B) मात्रा- I < मात्रा- II

C) मात्रा- I ≤ मात्रा- II

D) मात्रा- I = मात्रा- II या कोई संबंध नहीं

E) मात्रा- I ≥ मात्रा- II

प्रश्न 5: प्रश्न में, दो मात्राएँ I और II दी गई हैं। आपको मात्रा- I और मात्रा- II के बीच सही संबंध स्थापित करने के लिए दोनों संबंधों को हल करना है और सही विकल्प चुनना है।

मात्रा- I: यदि Gaurav और Saurav की आय क्रमशः Rs. 28000 और Rs. 30000 है | यदि Gaurav Rs. 8000 बचाता है, जो Saurav की बचत का 80% है, तो Saurav और Gaurav के खर्च के बीच का अंतर ज्ञात करें ?

मात्रा- II: Rs. 500

A) मात्रा- I > मात्रा- II

B) मात्रा- I < मात्रा- II

C) मात्रा- I ≤ मात्रा- II

D) मात्रा- I = मात्रा- II या कोई संबंध नहीं

E) मात्रा- I ≥ मात्रा- II

प्रश्न 6: प्रश्न में, दो मात्राएँ I और II दी गई हैं। आपको मात्रा- I और मात्रा- II के बीच सही संबंध स्थापित करने के लिए दोनों संबंधों को हल करना है और सही विकल्प चुनना है।

मात्रा- I: ’A’ एक काम को (x + 49) दिनों में पूरा कर सकता है, जबकि’ B’ उसी काम को (x + 36) दिनों में पूरा कर सकता है। यदि दोनों एक साथ काम करते हैं, तो पूरा काम ‘x’ दिनों में पूरा कर सकते हैं, फिर x का मान ज्ञात करें।

मात्रा- II:- यदि (x – 44)2 = 172 – 4x है तो ‘x’ का मान ज्ञात करें?

A) मात्रा- I> मात्रा- II

B) मात्रा- I <मात्रा- II

C) मात्रा- I≤मात्रा- II

D) मात्रा- I = मात्रा- II या कोई संबंध नहीं

E) मात्रा- I≥ मात्रा- II

प्रश्न 7: प्रश्न में, दो मात्राएँ I और II दी गई हैं। आपको मात्रा- I और मात्रा- II के बीच सही संबंध स्थापित करने के लिए दोनों संबंधों को हल करना है और सही विकल्प चुनना है।

मात्रा- I: एक दुकानदार ने Rs. 6000 प्रत्येक में तीन वस्तुएं खरीदी। उसने पहली वस्तु को 15% के लाभ पर बेचा, दूसरी वस्तु को 12% के लाभ पर और तीसरी वस्तु को 15% की हानि पर बेचा।दुकानदार द्वारा हुए समग्र लाभ / हानि प्रतिशत ज्ञात करें?

मात्रा- II: 6%

A) मात्रा- I> मात्रा- II

B) मात्रा- I <मात्रा- II

C) मात्रा- I≤मात्रा- II

D) मात्रा- I = मात्रा- II या कोई संबंध नहीं

E) मात्रा- I≥ मात्रा- II

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प्रश्न 8: प्रश्न में, दो मात्राएँ I और II दी गई हैं। आपको मात्रा- I और मात्रा- II के बीच सही संबंध स्थापित करने के लिए दोनों संबंधों को हल करना है और सही विकल्प चुनना है।

मात्रा- I: ’A’ और ‘B’ साथ में 16 दिनों में एक काम पूरा कर सकते हैं, जबकि ‘B’ और ‘C’ साथ में 12 दिनों में उसी काम पूरा कर सकते हैं। ‘A’ ने काम शुरू किया और 9 दिनों तक इस पर काम किया और फिर कार्य छोड़ दिया और फिर ‘B’ काम में शामिल हो गया और 15 दिनों तक इस पर काम किया और काम छोड़ दिया।अंत में ‘C’ ने शेष कार्य को 5 दिनों में समाप्त कर दिया।कितने दिनों में अकेले ’C’ पूरे काम को समाप्त करेगा?

मात्रा- II: 16 दिन

A) मात्रा- I> मात्रा- II

B) मात्रा- I <मात्रा- II

C) मात्रा- I≤मात्रा- II

D) मात्रा- I = मात्रा- II या कोई संबंध नहीं

E) मात्रा- I≥ मात्रा- II

प्रश्न 9: मात्रा- I: मिश्रण में दूध और पानी का अनुपात क्रमशः 3: 2 है। यदि इस मिश्रण में 64 litres दूध और 64 litres पानी मिलाया जाए, तो दूध और पानी का अनुपात क्रमशः 10: 7 में बदल जाता है। मिश्रण में पानी की प्रारंभिक मात्रा ज्ञात करें?

मात्रा- II: 240 litres मिश्रण में 30% wine और 70% whisky है। इस मिश्रण में एक निश्चित मात्रा में wine मिलाई जाती है, ताकि नए मिश्रण में 40% wine और बाकि whisky हो। डाले गए wine की मात्रा ज्ञात करें ?

A) Quantity-I > Quantity-II

B) Quantity-I < Quantity-II

C) Quantity-I ≤ Quantity-II

D) Quantity-I = Quantity-II or No relation

E) Quantity-I ≥ Quantity-II

प्रश्न 10: प्रश्न में, दो मात्राएँ I और II दी गई हैं। आपको मात्रा- I और मात्रा- II के बीच सही संबंध स्थापित करने के लिए दोनों संबंधों को हल करना है और सही विकल्प चुनना है।

मात्रा I: एक बैग में 7 red balls और ‘x’ green balls है और बैग में से पांच balls को निकाला जाता है।‘x’ का मान ज्ञात करें यदि ठीक 3 red balls को निकालने के तरीकों की संख्या 525 हैं।

मात्रा II: एक बैग में क्रमशः 3: 2 के अनुपात में red और blue balls हैं। बैग से दो balls को निकाला जाए और दोनों balls के अलग-अलग colors के होने की प्रायिकता 18/35 है । बैग में red balls की संख्या ज्ञात करें ?

A) मात्रा-I < मात्रा-II

B) मात्रा-I > मात्रा-II

C) मात्रा-I ≤ मात्रा-II

D) मात्रा-I = मात्रा-II और कोई सम्बन्ध नहीं

E) मात्रा-I ≥ मात्रा-II

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ANSWER KEYS and SOLUTIONS:

1) – B)2) – D)3) – D)4) – A)5) – B)6) – D)
7) – B)8) – D)9) – A)10) – A)

Solution 1: B)

Digits of three digits number that will give sum as 24 will be

(9, 9, 6), (9, 8, 7), and (8, 8, 8)

Three digits number formed by (9, 9, 6) = (3!)/(2!) = 3

Three digits number formed by (9, 8, 7) = (3!) = 6

Three digits number formed by (8, 8, 8) = (3!)/(3!) = 1

Total required three digit number = 3 + 6 + 1 = 10

Quantity-II:

So, Three digits number formed by (9, 9, 6) that will be divisible by 2 = 1

So, Three digits number formed by (9, 8, 7) that will be divisible by 2 = 2

So, Three digits number formed by (8, 8, 8) that will be divisible by 2 = 1

Therefore, required probability = (1 + 2 + 1)/10 = 4/10 = 2/5

Quantity-II:

As sum of digits is 24, therefore all such numbers will be divisible by 3.

Therefore, required probability = 1

Hence, option b.

Solution 2: D)

According to question,

[(x – 2) + (x + 1) + (x + 4)]/3 = [(x + 1) + (x + 4)]/2 – (x – 12.5)

(x + 1) = (2x + 5)/2 – (x – 12.5)

2x – 11.5 = x + 2.5

2x – x = 11.5 + 2.5

x = 14

So, age of Deepak, Dixit and Parul is 12 years, 15 years, and 18 years.

Quantity-II:

Average age of Deepak and Parul = (12 + 18)/2 = 15 years

Quantity-II:

Age of Dixit = 15 years

Hence, option d.

Solution 3: D)

Quantity I:

Let the time taken by A complete the work be 2x days.

Then, time taken by B alone complete the work = 3x days

Time taken by C alone to complete the work = 6x days

According to question: 1/3x + 1/6x = 1/20

3/6x = 1/20, x = 10

So, the time taken by A alone to complete the work = 20 days

So, Quantity I = Quantity II.

Hence, option d.

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Solution 4: A)

Quantity-II:

Required Probability = {(6C1×9C1) + 6C2}/15C2 = {(6 ×9) + 15}/105 = 69/105 ~ 0.65

Quantity-II: 0.6

Hence, option a.

Solution 5: B)

Quantity-II:

Saving of Saurav = 8000/0.8 = Rs. 10000

Expenditure of Gaurav = 28000 – 8000 = Rs. 20000

Expenditure of Saurav = 30000 – 10000 = Rs. 20000

Required difference = 20000 – 20000 = Rs. 0

Quantity-II: Rs. 500

Hence, option b.

Solution 6: D)

Quantity I:

According to question;

1/x + 1/(x + 49) + 1/(x + 36)

Or, 1/x = {x + 36 + x + 49}/{(x + 36)(x + 49)}

Or, x{2×2 + 85} = x2 + 85x + 36 × 49

Or, x2 = 36 × 49

Or, x = 6 × 7 = 42

So, Quantity I = 42

Quantity II:

(x – 44)2 = 172 – 4x

x2 – 84x + 1764 = 0

Or, x2 – 42x – 42x + 1764 = 0

Or, x(x – 42) – 42(x – 42) = 0

Or, (x – 42)(x – 42) = 0

Or, x = 42

So, Quantity II = 42

Therefore, Quantity I = Quantity II

Hence, option d.

Solution 7: B)

Quantity I:

Total cost price of three articles = 6000 × 3 = Rs. 18000

Selling price of first article = 1.15 × 6000 = Rs. 6900

Selling price of second article = 1.12 × 6000 = Rs. 6720

Selling price of third article = 0.85 × 6000 = Rs. 5100

Total selling price of all three articles = 6900 + 6720 + 5100 = Rs. 18720

Desired profit = 18720 – 18000 = Rs. 720

Desired profit percent = 720/18000 × 100 = 4%

So, Quantity I = 4%

Quantity II = 6%

Therefore Quantity II > Quantity I

Hence, option b.

Solution 8: D)

Quantity I:

Let total amount of work = 48 units (LCM of 16 and 12)

Amount of work done by A and B together in one day = 48/16 = 3 units

Amount of work done by B and C together in one day = 48/12 = 4 units

Let efficiencies of A, B and C be a units/day, b units/day and c units/day respectively

So, 9a + 15b + 5c = 48

9a + 9b + 5b + 5c + b = 48

9(a + b) + 5(b + c) + b = 48

9 × 3 + 5 × 4 + b = 48

b = 48 – 47 = 1 unit/day

Efficiency of C = 4 – 1 = 3

Desired Time = 48/3 = 16 days.

So, Quantity I = 16 days

Quantity II = 16 days

Therefore Quantity I = Quantity II

Hence, option d.

Solution 9: A)

Quantity-II:

Let, amount of milk and water in mixture be 3x litres and 2x litres respectively.

According to question,

(3x + 64): (2x + 64) = 10: 7

21x + 448 = 20x + 640

21x – 20x = 640 – 448

x = 192

So, initial quantity of water in mixture = 192 × 2 = 384 litres

Quantity-II:

Quantity of whisky in initial mixture = 70% of 240 = 168 litres

According to question,

60% of (240 + x) = 168

144 + 0.6x = 168

0.6x = 24

x = 24/0.6

x = 40 litres

Hence, option a.

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Solution 10: A)

Quantity I:

Number of ways = 7C3 × xC2 = 525

35 × x(x – 1)/2 = 525

x2 – x = 30

x2 – x – 30 = 0

x2 – 6x + 5x – 30 = 0

x(x – 6) + 5(x – 6) = 0

(x – 6)(x + 5) = 0

x = 6, -5

Number balls can’t be negative, so the value of ‘x’ = 6

Quantity II:

Let the bag contains 3x red balls and 2x blue balls.

So, the probability that both the balls drawn are of different colour = (3xC1 × 2xC1)/5xC1 = 18/35

(3x × 2x × 2)/{5x(5x – 1)} = 18/35

2x/(5x – 1) = 3/7

14x = 15x – 3

x = 3

So, the number of red balls in the bag = 3 × 3 = 9

So, Quantity I < Quantity II

Hence, option a.

Cheena Sawhney

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