XAT Quant Based LR Questions

XAT Quant Based LR Questions is one of the moderately important topics for XAT. These questions typically involve data interpretation, mathematical puzzles, and logical caselets, requiring candidates to quickly process numerical data and arrive at logical conclusions. To excel, it's important to strengthen your fundamentals in arithmetic, algebra, and geometry while practicing different problem types like tables, graphs, and arrangement-based puzzles. Checking with the XAT Syllabus will help you to know similar other topics for XAT.

For focused preparation, enrolling in an XAT crash course can be a game-changer. These courses help you to get yourself familiar and becomes helpful in mastering Quant-based LR questions. With the right strategy and approach, you can tackle even the trickiest LR questions.

XAT Quant Based LR Questions

Instructions

Read the following scenario and answer the THREE questions that follow.

Comprehension:
The diagram below represents a road network connecting five towns, namely Meereen, Lannisport, Winterfell, Oldtown, and Gulltown. The maximum speed limits along any stretch of road are as shown in the diagram. The straight road that connects Meereen to Gulltown passes through Oldtown. Another straight road, running west to east, connecting Meereen to Winterfell, passes through Lannisport. Further, two straight roads, one from Lannisport to Oldtown and another from Winterfell to Gulltown, are perpendicular to the road joining Meereen to Winterfell, and run from south to north.

Consider a car always travelling at the maximum permissible speed, and always taking the shortest route. It takes 1 hour to reach Oldtown from Meereen, 2 hours to reach Gulltown from Oldtown, and 45 minutes to reach Winterfell from Gulltown. (For this problem, always consider the shortest route in terms of distance.)

Question 1

Tyrion Lannister drove from Meereen to Oldtown, then from Oldtown to Lannisport, and finally from Lannisport to Winterfell, always taking the shortest paths. He always drove at a speed
10 km/hr below the maximum speed limits for the stretches he took.
What was his total driving time closest to?

3 hrs 19 mins

4 hrs 20 mins

7 hrs 28 mins

6 hrs 42 mins

7 hrs 50 mins

Video Solution

Solution

We are told that a car always travelling at the maximum permissible speed, and always taking the shortest route takes 1 hour to reach Oldtown from Meereen, 2 hours to reach Gulltown from Oldtown, and 45 minutes to reach Winterfell from Gulltown.

So, distance between Oldtown and Meereen is 100*1 = 100km

Distance between Gulltown and Oldtown is 25*2 = 50km.

Distance between Winterfell and Gulltown. is 120*3/4 = 90km.

Let the distance between Meereen and Winterfall be x.

Using the pythagoras theorem, $$(100+50)^2=90^2+x^2$$

x = 120 km

For convience, we are refering the vertices as the starting letters of towns.

Using the similar triangle rule, MOL is similar to MGW.

So, $$\frac{MO}{ML}=\frac{MG}{MW}$$

$$\frac{100}{ML}=\frac{150}{120}$$

ML = 80 km.

LW = 40 km.

$$\frac{MO}{LO}=\frac{MG}{WG}$$

$$\frac{100}{LO}=\frac{150}{90}$$

LO = 60 km

The path Tyrion Lannister took is MO, OL, LW.

Distances of these path are 100, 60, 40. The respective speeds are 90, 10, 110.

So, the total time taken will be $$\frac{100}{90}+\frac{60}{10}+\frac{40}{110}\ =\ 1.11+6+0.36\ \ =7.47$$

0.47 of an hour is 28 mins.

So, answer will be 7 hr 28 mins

Instructions

Read the following scenario and answer the THREE questions that follow.

Question 2

Missandei starts from Gulltown towards Oldtown by the shortest path, driving at the maximum permissible speed. From Oldtown, she drives at a speed of 10 km/hr towards Lannisport. When Missandei starts from Gulltown, Varys starts at the same time from Lannisport to Oldtown along the shortest path, always driving at the maximum permissible speed.
If they don’t stop anywhere, at what point will they meet?

Approximately 13.33 km south of Oldtown.

Approximately 54.33 km north of Lannisport.

Approximately 57.33 km north of Lannisport.

Approximately 6.67 km south of Oldtown.

Cannot be uniquely determined from the given information.

Video Solution

Solution

So, distance between Oldtown and Meereen is 100*1 = 100km

Distance between Gulltown and Oldtown is 25*2 = 50km.

Distance between Winterfell and Gulltown. is 120*3/4 = 90km.

Let the distance between Meereen and Winterfall be x.

Using the pythagoras theorem, $$(100+50)^2=90^2+x^2$$

x = 120 km

For convience, we are refering the vertices as the starting letters of towns.

Using the similar triangle rule, MOL is similar to MGW.

So, $$\frac{MO}{ML}=\frac{MG}{MW}$$

$$\frac{100}{ML}=\frac{150}{120}$$

ML = 80 km.

LW = 40 km.

$$\frac{MO}{LO}=\frac{MG}{WG}$$

$$\frac{100}{LO}=\frac{150}{90}$$

LO = 60 km

The path taken by Missandei is GO, OL at a speed of 25 kmph and 10 kmph. The path taken by Varys is LO at a speed of 20 kmph.

Missandei reaches O 2 hours after he starts. In the meantime, Varys travels 20 * 2 = 40km.

Now, both are travelling in opposite directions and 20 km away.

The speed of Varys is 20 kmph, and that of Missandei is 10 kmph.

The time taken for them to meet is 20/30 = 2/3 hr.

In 2/3 hr, Missandei travels 6.67 km, and Varys travels 13.33 km.

So, they will meet at 6.67 km south of Oldtown.

Instructions

Read the following scenario and answer the THREE questions that follow.

Question 3

The capital city, King’s Landing, located 40 km to the south of Gulltown on the road connecting Gulltown to Winterfell, did not have a straight road, connecting to Meereen. Now, a new expressway is being built to connect these two towns by a straight road.

What should be the maximum speed limit allowed on this expressway so that it cuts down the travel time, from Meeren to King’s Landing, from the fastest possible route through the road network shown in the diagram, by 20 minutes?

100 km/hr

130 km/hr

139 km/hr

120 km/hr

157 km/hr

Video Solution

Solution

So, distance between Oldtown and Meereen is 100*1 = 100km

Distance between Gulltown and Oldtown is 25*2 = 50km.

Distance between Winterfell and Gulltown. Is 120*3/4 = 90km.

Let the distance between Meereen and Winterfall be x.

Using the Pythagoras theorem, $$(100+50)^2=90^2+x^2$$

x = 120 km

For convenience, we are referring to the vertices as the starting letters of towns.

Using the similar triangle rule, MOL is similar to MGW.

So, $$\frac{MO}{ML}=\frac{MG}{MW}$$

$$\frac{100}{ML}=\frac{150}{120}$$

ML = 80 km.

LW = 40 km.

$$\frac{MO}{LO}=\frac{MG}{WG}$$

$$\frac{100}{LO}=\frac{150}{90}$$

LO = 60 km

The direct path from M to K will be $$\sqrt{\ 120^2+50^2}=\ 130\ km$$.

Time taken for the path MOGK,= $$\frac{100}{100}+\frac{50}{25}+\frac{40}{120}\ =1+2+\frac{1}{3}=3.33$$

Time taken for the path MWK = $$\frac{120}{120}+\frac{50}{120}\ =1.42$$

So, the second case is MWK, which is the fastest route. The time taken will be 1 hr 25 mins.

If the travel time is cut down by 20 mins, the new time will be 1 hr 5 mins. So, it should take 13/12 hrs.

speed = $$\dfrac{130}{\frac{13}{12}}=120 kmph$$

Instructions

Read the following scenario and answer the THREE questions that follow.

A store offers a choice of five different discount coupons to its customers, described as follows:
Coupon A: A flat discount of Rs. 250 on a minimum spend of Rs. 1200 in one transaction.
Coupon B: A 15% discount on a minimum spend of Rs. 500 in one transaction, up to a maximum discount of Rs. 300.
Coupon C: A flat discount of Rs. 100 on a minimum spend of Rs. 600 in one transaction.
Coupon D: A 10% discount on a minimum spend of Rs. 250 in one transaction, up to a maximum discount of Rs. 100.
Coupon E: A flat discount of Rs. 50 on a minimum spend of Rs. 200 in one transaction.

The customers are allowed to use at most one coupon in one transaction, i.e., two or more coupons cannot be combined for the same transaction.

Question 1

Four customers used four different discount coupons for their respective transactions in such a way that they obtained a total discount of Rs. 710. Which discount coupon was not used?

Coupon A

Coupon B

Coupon C

Coupon D

Coupon E

Video Solution

Solution

Tabulating the coupons and the respective minimum & maximum discounts on minimum spend as below-

From the table, we can infer that

Coupon A can be availed only when someone spends a minimum amount of Rs. 1200 and he will obtain a flat discount of Rs. 250.

Coupon B can be availed only when someone spends a minimum amount of Rs. 500 such that he will get a discount of 15% on it which is Rs. 75 and a maximum discount of Rs. 300 can be given when he spends amount greater than or equal to Rs. 2000.

Coupon C can be availed only when someone spends a minimum amount of Rs. 600 and he will obtain a flat discount of Rs. 100.

Coupon D can be availed only when someone spends a minimum amount of Rs. 250 such that he will get a discount of 10% on it which is Rs. 25 and a maximum discount of Rs. 100 can be given when he spends amount greater than or equal to Rs. 1000.

Coupon E can be availed only when someone spends a minimum amount of Rs. 200 and he will obtain a flat discount of Rs. 50.

In this question, it is given that four customers used four different discount coupons for their respective transactions in such a way that they obtained a total discount of Rs. 710.

Let us assume that coupon A was not used, then the four customers can get a maximum discount of (300+100+100+50) = Rs. 550

From the observation, we see that the sum of maximum discounts if all 5 coupons are used is (250+300+100+100+50) = Rs. 800

Therefore, to get a discount of Rs. 710 only one coupon that we can skip is Coupon E because in all other cases, the total discount will be less than Rs. 710

Hence, Option E is the correct answer.

Instructions

Read the following scenario and answer the THREE questions that follow.

The customers are allowed to use at most one coupon in one transaction, i.e., two or more coupons cannot be combined for the same transaction.

Question 2

Four customers used four different discount coupons for their respective transactions in such a way that nobody used any discount coupon sub-optimally. (A discount coupon is used sub-optimally if using another discount coupon could have resulted in a higher discount for the same transaction.) What was the minimum combined spend (before application of any discount)?

Rs. 2250

Rs. 2500

Rs. 2350

Rs. 2300

Rs. 1550

Video Solution

Solution

Tabulating the coupons and the respective minimum & maximum discounts on minimum spend as below-

From the table, we can infer that

Coupon A can be availed only when someone spends a minimum amount of Rs. 1200 and he will obtain a flat discount of Rs. 250.

Coupon C can be availed only when someone spends a minimum amount of Rs. 600 and he will obtain a flat discount of Rs. 100.

Coupon E can be availed only when someone spends a minimum amount of Rs. 200 and he will obtain a flat discount of Rs. 50.

In this question,it is given that four customers used four different discount coupons for their respective transactions in such a way that nobody used any discount coupon sub-optimally. (A discount coupon is used sub-optimally if using another discount coupon could have resulted in a higher discount for the same transaction.)

We need to find the the minimum combined spend (before applying of any discount)

For that, we starts checking for the minimum amount that needs to be spend.

From the table, we know that we can avail coupon E by spending Rs. 200 so we use this and get flat discount of Rs. 50 and we cannot get more discount by spending this amount.

Next, we check for coupon D(as, it requires 2nd lowest minimum spend)

If we spend Rs. 250 and use coupon D, we get only Rs. 25 as a discount, but if we use coupon E for the same amount, we can get Rs. 50 flat discount. (using a discount sub-optimally) so we have to spend a minimum of Rs. 500 so that we get a discount of Rs.50, but now again, we see that by spending Rs. 500, we can get a better discount by using Coupon B.

So, now we will use coupon B for Rs. 500 spending.

Till now, we have used coupons E and B (we can't use these coupons now, also we cannot use coupon D)

We need to use 4 coupons so using remaining 2 coupons (A and C) by spending a amount of Rs. 1200 and Rs. 600 for coupon A and coupon C respectively.

Hence, the minimum combined spend is 200 + 500 + 1200 + 600 = Rs. 2500

Option B is the correct answer.

Instructions

Read the following scenario and answer the THREE questions that follow.

The customers are allowed to use at most one coupon in one transaction, i.e., two or more coupons cannot be combined for the same transaction.

Question 3

A family wanted to purchase four products worth Rs. 1000 each, and another product worth Rs. 300. They were told that they could:
I) pay for the five products through one or more transactions in any way they wanted, as long as the purchase amount of any one product would not get split into different transactions, and
II) use the same discount coupon repeatedly for separate transactions, if they opt for more than one transaction.
What was the maximum discount that they could obtain for their purchase?

Rs. 600

Rs. 645

Rs. 650

Rs. 700

None of the remaining options is correct.

Video Solution

Solution

Tabulating the coupons and the respective minimum & maximum discounts on minimum spend as below-

From the table, we can infer that

Coupon A can be availed only when someone spends a minimum amount of Rs. 1200 and he will obtain a flat discount of Rs. 250.

Coupon C can be availed only when someone spends a minimum amount of Rs. 600 and he will obtain a flat discount of Rs. 100.

Coupon E can be availed only when someone spends a minimum amount of Rs. 200 and he will obtain a flat discount of Rs. 50.

For this question, the maximum discount can be obtained using I) pay for the five products through one or more transactions in any way they wanted, as long as the purchase amount of any one product would not be split into different transactions.

They can split the transactions as
(a) Rs. 2000 with Coupon B: A 15% discount giving Rs. 300 discount

(b) Rs. 1000 again with Coupon B: A 15% discount giving Rs. 150 discount

So, the total max discount they can get is Rs. 300 + 150 + 250 = Rs. 700

Hence, the answer is Rs. 700

XAT Quant Based LR Questions

XAT Quant Based LR Questions

Join CAT 2026 course by 5-Time CAT 100%iler

Download resources