Read the given data carefully and answer the questions based on it. Three stores (X, Y and Z) sold two different types of Mouses (Wired +Wireless) . Number of wired mouses sold by X is 240. Ratio of number of wireless mouses to the wired mouses sold by X is 5: 3. Total number of wireless mouses sold by all the stores are 960. Total mouse sold by store Y are 25% more than that by store X. Number of wireless mouses sold by store X is 80% of number of wired mouses sold by Y. Average number of mouse sold by all the three stores is 690.
Wired mouse sold by X and Y together is what percentage more or less than wireless mouse sold by Y.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (C) ¹⁴⁶⁄₃⅔%
Explanation
For Store X, wired mouses = 240 and wireless mouses = (5 / 3) × 240 = 400, so total mouses sold by X = 640.
Total mouses sold by Y = 1.25 × 640 = 800.
Wired mouses sold by Y = 400 / 0.80 = 500.
Wireless mouses sold by Y = 800 - 500 = 300.
Wired mouses sold by X and Y together = 240 + 500 = 740.
Percentage = ((740 - 300) / 300) × 100 = (440 / 300) × 100 = 146 ⅔ %.
2Quantitative Aptitude
Read the given data carefully and answer the questions based on it. Three stores (X, Y and Z) sold two different types of Mouses (Wired +Wireless) . Number of wired mouses sold by X is 240. Ratio of number of wireless mouses to the wired mouses sold by X is 5: 3. Total number of wireless mouses sold by all the stores are 960. Total mouse sold by store Y are 25% more than that by store X. Number of wireless mouses sold by store X is 80% of number of wired mouses sold by Y. Average number of mouse sold by all the three stores is 690.
Find the ratio of total mouse sold by Z to total wired mouse sold by all the three stores.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (A) 21:37
Explanation
Total mouses sold by all three stores = 690 × 3 = 2070.
Total mouses sold by Z = 2070 - (640 + 800) = 630.
Wireless mouses sold by Z = 960 - (400 + 300) = 260.
Wired mouses sold by Z = 630 - 260 = 370.
Total wired mouses sold by X, Y, and Z = 240 + 500 + 370 = 1110.
Required ratio = 630 : 1110 = 21 : 37.
3Quantitative Aptitude
Read the given data carefully and answer the questions based on it. Three stores (X, Y and Z) sold two different types of Mouses (Wired +Wireless) . Number of wired mouses sold by X is 240. Ratio of number of wireless mouses to the wired mouses sold by X is 5: 3. Total number of wireless mouses sold by all the stores are 960. Total mouse sold by store Y are 25% more than that by store X. Number of wireless mouses sold by store X is 80% of number of wired mouses sold by Y. Average number of mouse sold by all the three stores is 690.
If number of wireless mouses sold by store A is 40% more than wired mouse sold by store X and average number of mouses sold by X and A is 637, then find the number of wired mouses sold by store A.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (D) 298
Explanation
Wireless mouses sold by store A = 1.40 × 240 = 336.
Combined total mouses sold by stores X and A = 637 × 2 = 1274.
Total mouses sold by store X = 640.
Total mouses sold by store A = 1274 - 640 = 634.
Wired mouses sold by store A = 634 - 336 = 298.
4Quantitative Aptitude
Read the given data carefully and answer the questions based on it. Three stores (X, Y and Z) sold two different types of Mouses (Wired +Wireless) . Number of wired mouses sold by X is 240. Ratio of number of wireless mouses to the wired mouses sold by X is 5: 3. Total number of wireless mouses sold by all the stores are 960. Total mouse sold by store Y are 25% more than that by store X. Number of wireless mouses sold by store X is 80% of number of wired mouses sold by Y. Average number of mouse sold by all the three stores is 690.
Price of wired mouse is 2 / 3ʳᵈ of price of wireless mouse. If store X earned revenue of Rs.67200 by selling all the mouse, then find the price of wireless mouse.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (B) Rs.120
Explanation
Let the price of a wireless mouse be P rupees.
Price of a wired mouse = (2 / 3) P rupees.
Total revenue of store X = (240 × (2 / 3) P) + (400 × P) = 160 P + 400 P = 560 P.
Given revenue = 67200.
560 P = 67200
P = 67200 / 560 = 120.
5Quantitative Aptitude
Read the given data carefully and answer the questions based on it.Three stores (X, Y and Z) sold two different types of Mouses (Wired +Wireless) . Number of wired mouses sold by X is 240. Ratio of number of wireless mouses to the wired mouses sold by X is 5: 3. Total number of wireless mouses sold by all the stores are 960. Total mouse sold by store Y are 25% more than that by store X. Number of wireless mouses sold by store X is 80% of number of wired mouses sold by Y. Average number of mouse sold by all the three stores is 690.
If 30% of the total mouse sold by store Z are faulty and ratio of faulty wired mouse and faulty wireless mouse is 5:4, then find the difference between non- faulty wired mouse and non- faulty wireless mouse.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (E) 89
Explanation
Total mouses sold by store Z = 630.
Total faulty mouses = 30% of 630 = 189.
Faulty wired mouses = (5 / 9) × 189 = 105.
Faulty wireless mouses = (4 / 9) × 189 = 84.
Non-faulty wired mouses = 370 - 105 = 265.
Non-faulty wireless mouses = 260 - 84 = 176.
Difference = 265 - 176 = 89.
6Quantitative Aptitude
Find the wrong number in the following number series:
6,13,27,48,76,112,153
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (D) 112
Explanation
The pattern of differences between consecutive terms adds consecutive multiples of 7:
13 - 6 = 7
27 - 13 = 14
48 - 27 = 21
76 - 48 = 28
76 + 35 = 111
111 + 42 = 153
The term 112 is incorrect and should be replaced by 111.
7Quantitative Aptitude
Find the wrong number in the following number series:
8,5,7,13.5,31,82.5,254
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (B) 254
Explanation
The pattern followed in the series is:
8 × 0.5 + 1 = 5
5 × 1 + 2 = 7
7 × 1.5 + 3 = 13.5
13.5 × 2 + 4 = 31
31 × 2.5 + 5 = 82.5
82.5 × 3 + 6 = 253.5
The term 254 is incorrect and should be 253.5.
8Quantitative Aptitude
Find the wrong number in the following number series:
7,14,22,36,58,94,152
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (D) 7
Explanation
Each term starting from the third term is the sum of the preceding two terms:
8 + 14 = 22
14 + 22 = 36
22 + 36 = 58
36 + 58 = 94
58 + 94 = 152
The first term 7 is incorrect and should be 8.
9Quantitative Aptitude
Find the wrong number in the following number series:
2,6,14,25,38,55,74
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (E) 6
Explanation
The differences between consecutive terms are consecutive prime numbers starting from 5:
2 + 5 = 7
7 + 7 = 14
14 + 11 = 25
25 + 13 = 38
38 + 17 = 55
55 + 19 = 74
The term 6 is incorrect and should be 7.
10Quantitative Aptitude
Find the wrong number in the following number series:
11,15,42,60,183,219,562
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (A) 60
Explanation
The pattern of addition in the series consists of alternating squares and cubes:
11 + 2² = 11 + 4 = 15
15 + 3³ = 15 + 27 = 42
42 + 4² = 42 + 16 = 58
58 + 5³ = 58 + 125 = 183
183 + 6² = 183 + 36 = 219
219 + 7³ = 219 + 343 = 562
The term 60 is incorrect and should be 58.
11Quantitative Aptitude
A boat covers a distance of 120km in 6 hours in downstream. The ratio of speed of boat in still water to speed of stream is 3: 1. Find in how much time boat cover 84 km in upstream if speed of stream is increased by 80% .
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (B) 14 hours
Explanation
Downstream speed = 120 / 6 = 20 km/h.
Given the ratio of boat speed in still water to stream speed is 3 : 1, let boat speed = 3x and stream speed = x.
3x + x = 20, which gives x = 5 km/h.
Boat speed in still water = 15 km/h and stream speed = 5 km/h.
Increased stream speed = 5 × 1.80 = 9 km/h.
New upstream speed = 15 - 9 = 6 km/h.
Time taken to cover 84 km upstream = 84 / 6 = 14 hours.
12Quantitative Aptitude
Harry spent 20% of his monthly income on House rent. 20% of the remaining on Food and remaining amount he spent on clothing and investment in the ratio of 7: 9. If the difference between expenditure on Food and clothing is Rs.10800, then find his annual income.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (A) Rs.1080000
Explanation
Let monthly income = 100x.
Expenditure on house rent = 20% of 100x = 20x, leaving 80x.
Expenditure on food = 20% of 80x = 16x, leaving 64x.
The remaining 64x is divided between clothing and investment in ratio 7 : 9, so clothing expenditure = (7 / 16) × 64x = 28x.
Difference between clothing and food expenditure = 28x - 16x = 12x = 10800, giving x = 900.
Monthly income = 100 × 900 = Rs. 90000.
Annual income = 12 × 90000 = Rs. 1080000.
13Quantitative Aptitude
A and B started a business with investment of Rs.24000 and Rs.30000 respectively. After 4 months, A withdraws 16⅔% of investment, B withdraws all his investment and C joined the business. If at the end of the year ratio of profit share of A, B and C is 32: 15: 20, then find the initial investment of C.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (B) Rs.20000
Explanation
A's total equivalent investment = (24000 × 4) + (24000 × (5/6) × 8) = 96000 + 160000 = 256000.
B's total equivalent investment = 30000 × 4 = 120000.
Let C's initial investment be C.
C joins for 8 months, so C's equivalent investment = C × 8.
Given ratio of profit shares B : C = 15 : 20 = 3 : 4.
120000 / (8 × C) = 3 / 4.
8 × C = 160000, giving C = Rs. 20000.
14Quantitative Aptitude
6 inlet pipes can fill 60% of the tank in 6 minutes. 4 outlet pipes can empty the same tank in 20 min. find the time taken by 2 inlet and 2 outlet pipe to fill/empty the tank when open together.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (B) 120 min
Explanation
6 inlet pipes fill 60% of the tank in 6 minutes, so 6 inlet pipes fill 100% in 10 minutes.
Rate of 1 inlet pipe = 1 / 60 of tank per minute.
4 outlet pipes empty 100% of the tank in 20 minutes.
Rate of 1 outlet pipe = 1 / 80 of tank per minute.
Combined rate of 2 inlet and 2 outlet pipes = 2 × (1 / 60) - 2 × (1 / 80) = (1 / 30) - (1 / 40) = 1 / 120 of tank per minute.
Hence, 2 inlet and 2 outlet pipes together fill the tank in 120 minutes.
15Quantitative Aptitude
In a mixture A, quantity of milk is 18 liter which is 120% of quantity of water. In mixture B, quantity of milk is 24 liter which is 80% of quantity of water. If both mixture is mixed to form another mixture C, then find the difference between quantity of milk and water in the final mixture.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (B) 3 liters
Explanation
In mixture A, quantity of water = 18 / 1.20 = 15 liters.
In mixture B, quantity of water = 24 / 0.80 = 30 liters.
Total milk in mixture C = 18 + 24 = 42 liters.
Total water in mixture C = 15 + 30 = 45 liters.
Difference between quantity of milk and water = 45 - 42 = 3 liters.
16Quantitative Aptitude
4 years ago, age of Ayush was equal to average of present age of Ayush, Rahul and Deepak. Deepak is 6 years older than Rahul and 3 years younger than Ayush. Find the present age of Deepak.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (E) can't be determined
Explanation
Let present ages of Ayush, Rahul, and Deepak be A, R, and D.
Deepak is 6 years older than Rahul, so R = D - 6.
Deepak is 3 years younger than Ayush, so A = D + 3.
The average of their present ages = (A + R + D) / 3 = ((D + 3) + (D - 6) + D) / 3 = (3D - 3) / 3 = D - 1.
Four years ago, Ayush's age = A - 4 = (D + 3) - 4 = D - 1.
Equating Ayush's age 4 years ago to the average gives D - 1 = D - 1, which holds true for all values of D.
Since no independent equation remains to solve for D, the present age cannot be determined.
17Quantitative Aptitude
A shopkeeper mixed 20 kg of rice which costs Rs.90 per kg with P kg of soyabean which costs Rs.60 per kg and sold the mixture at Rs.63 kg per kg due to which he incurred a loss of 12.5% . Find the value of P.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (D) 30 kg
Explanation
Total cost price = (20 × 90) + (P × 60) = 1800 + 60P.
Total selling price = 63 × (20 + P) = 1260 + 63P.
Loss is 12.5%, so selling price = 87.5% of cost price = (7/8) × (1800 + 60P).
1260 + 63P = 1575 + 52.5P.
10.5P = 315.
P = 30 kg.
18Quantitative Aptitude
There are 20 balls in a bag in which 6 are red balls, x are yellow balls and remaining are black balls. If a ball is picked, the probability of getting a yellow ball is 0.5. Find the difference between no. of yellow balls and black ball.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (C) 6
Explanation
Probability of picking a yellow ball = x / 20 = 0.5, giving x = 10 yellow balls.
Number of red balls = 6.
Number of black balls = 20 - 6 - 10 = 4.
Difference between yellow balls and black balls = 10 - 4 = 6.
19Quantitative Aptitude
A person shares an amount of money in between his friend, son and brother in the ratio of 3: 2: 1. His son invest his share in a scheme which offers simple interest at 40% per annum for 4 years. He earned total interest Rs.640. Find the share of his brother.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (A) Rs.200
Explanation
Let shares of friend, son, and brother be 3x, 2x, and 1x.
Son's investment = 2x.
Simple interest = (P × R × T) / 100 = (2x × 40 × 4) / 100 = 3.2x.
3.2x = 640, so x = 200.
Brother's share = 1x = Rs. 200.
20Quantitative Aptitude
The curved surface area of a cylinder is 1760 cm² and total surface area of cylinder is 2992 cm² . Find the radius of the cylinder.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (C) 14 cm
Explanation
Total surface area - Curved surface area = 2πr² = 2992 - 1760 = 1232 cm².
2 × (22 / 7) × r² = 1232.
(44 / 7) × r² = 1232.
r² = (1232 × 7) / 44 = 196.
r = 14 cm.
21Quantitative Aptitude
In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer:
I.x² + x - 6 = 0 II.15y² - 11y + 2 = 0
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (E) if x = y or no relation can be established between x and y .
Comparing the roots:
When x = -3, -3 < 1/3 and -3 < 2/5, so x < y.
When x = 2, 2 > 1/3 and 2 > 2/5, so x > y.
Since x is both smaller and larger than y depending on the root taken, no relation can be established between x and y.
22Quantitative Aptitude
In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer:
I.x² + 21x + 108 = 0 II.2y² + 53y + 350 = 0
Comparing the roots:
-9 > -14 and -9 > -12.5
-12 > -14 and -12 > -12.5
Since every root of x is greater than every root of y, x > y.
23Quantitative Aptitude
In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer:
I.x = ³√(216) II.y⁴ = 1296
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (B) if x ≥ y
Explanation
Solving equation I:
x = ∛216 = 6
Solving equation II:
y⁴ = 1296
y = 6, -6
Comparing the roots:
When y = 6, x = y.
When y = -6, x > y.
Combining both cases gives x ≥ y.
24Quantitative Aptitude
In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer:
I.6x + 7y = 101 II.13y - 10x = 29
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (C) if x < y
Explanation
Multiplying equation I by 5 and equation II by 3:
30x + 35y = 505
-30x + 39y = 87
Adding the two equations:
74y = 592
y = 8
Substituting y = 8 into equation I:
6x + 7(8) = 101
6x = 45
x = 7.5
Comparing the values: 7.5 < 8, so x < y.
25Quantitative Aptitude
In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer:
I. 12x² - 7x + 1 = 0 II. 10y² - 37y + 7 = 0
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (E) if x = y or no relation can be established between x and y .
Comparing the roots:
1/4 > 1/5, but 1/4 < 7/2.
1/3 > 1/5, but 1/3 < 7/2.
Because x can be greater than y or less than y, no relation can be established between x and y.
26Quantitative Aptitude
Table Given below shows total number of boys in four different (P, Q, R and S) academies and it also shows percentage of girls in these four academies. Read the data carefully and answer the questions.
Note: Total students in an academy = Number of boys + Number of girls
| Academy | Total boys | Percentage of girls |
|---|---|---|
| P | 240 | 76% |
| Q | 560 | 30% |
| R | 1000 | 37.5% |
| S | 840 | 40% |
Total number of girls in academy P & Q together is how much more or less than total number of girls in academy R & S together.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (E) 160
Explanation
Total students in academy P = 240 / (1 - 0.76) = 1000, so number of girls in P = 760.
Total students in academy Q = 560 / (1 - 0.30) = 800, so number of girls in Q = 240.
Total girls in P and Q together = 760 + 240 = 1000.
Total students in academy R = 1000 / (1 - 0.375) = 1600, so number of girls in R = 600.
Total students in academy S = 840 / (1 - 0.40) = 1400, so number of girls in S = 560.
Total girls in R and S together = 600 + 560 = 1160.
Difference = 1160 - 1000 = 160.
27Quantitative Aptitude
Table Given below shows total number of boys in four different (P, Q, R and S) academies and it also shows percentage of girls in these four academies. Read the data carefully and answer the questions.
Note: Total students in an academy = Number of boys + Number of girls
| Academy | Total boys | Percentage of girls |
|---|---|---|
| P | 240 | 76% |
| Q | 560 | 30% |
| R | 1000 | 37.5% |
| S | 840 | 40% |
Total number of students in S are what percent more than total number of students in P?
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (C) 40%
Explanation
Total students in academy S = 840 / 0.60 = 1400.
Total students in academy P = 240 / 0.24 = 1000.
Percentage by which S is more than P = ((1400 - 1000) / 1000) × 100 = 40%.
28Quantitative Aptitude
Table Given below shows total number of boys in four different (P, Q, R and S) academies and it also shows percentage of girls in these four academies. Read the data carefully and answer the questions.
Note: Total students in an academy = Number of boys + Number of girls
| Academy | Total boys | Percentage of girls |
|---|---|---|
| P | 240 | 76% |
| Q | 560 | 30% |
| R | 1000 | 37.5% |
| S | 840 | 40% |
Total number of boys in academy T are 25% more than total number of boys in Q, while total number of girls in T are 40% more than total number of students in R. Find total number of students in T.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (A) 2940
Explanation
Boys in academy T = 560 × 1.25 = 700.
Total students in academy R = 1000 / 0.625 = 1600.
Girls in academy T = 1600 × 1.40 = 2240.
Total students in academy T = 700 + 2240 = 2940.
29Quantitative Aptitude
Table Given below shows total number of boys in four different (P, Q, R and S) academies and it also shows percentage of girls in these four academies. Read the data carefully and answer the questions.
Note: Total students in an academy = Number of boys + Number of girls
| Academy | Total boys | Percentage of girls |
|---|---|---|
| P | 240 | 76% |
| Q | 560 | 30% |
| R | 1000 | 37.5% |
| S | 840 | 40% |
Total number of students in S is how much more than total number of girls in P and Q together.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (B) 400
Explanation
Total students in academy S = 840 / 0.60 = 1400.
Girls in academy P = 1000 - 240 = 760.
Girls in academy Q = 800 - 560 = 240.
Total girls in P and Q together = 760 + 240 = 1000.
Difference = 1400 - 1000 = 400.
30Quantitative Aptitude
Table Given below shows total number of boys in four different (P, Q, R and S) academies and it also shows percentage of girls in these four academies. Read the data carefully and answer the questions.
Note: Total students in an academy = Number of boys + Number of girls
| Academy | Total boys | Percentage of girls |
|---|---|---|
| P | 240 | 76% |
| Q | 560 | 30% |
| R | 1000 | 37.5% |
| S | 840 | 40% |
If there is one coach for per 100 students in each academy, then find total no. of coaches required for all students in these four academies.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (B) 48
Explanation
Total students in P = 240 / 0.24 = 1000.
Total students in Q = 560 / 0.70 = 800.
Total students in R = 1000 / 0.625 = 1600.
Total students in S = 840 / 0.60 = 1400.
Sum of total students in P, Q, R, and S = 1000 + 800 + 1600 + 1400 = 4800.
Coaches required = 4800 / 100 = 48.
31Quantitative Aptitude
In each of these questions, two equation (I) and (II) are given. You have to solve both the equations and give answer
I.121x² - 110 + 21 = 0 II.169y² - 208y + 63 = 0
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (E) If x = y or no relation can be established between x and y
Explanation
Solving equation I: 121x² - 110x + 21 = 0
(11x - 3)(11x - 7) = 0
x = 3/11 = 39/143, and x = 7/11 = 91/143.
Solving equation II: 169y² - 208y + 63 = 0
(13y - 7)(13y - 9) = 0
y = 7/13 = 77/143, and y = 9/13 = 99/143.
Comparing root values:
39/143 < 77/143, but 91/143 > 77/143.
Since x is less than y in some cases and greater than y in others, no relation can be established between x and y.
32Quantitative Aptitude
In each of these questions, two equation (I) and (II) are given. You have to solve both the equations and give answer
I.x² - 6x + 12 = 4 II.y² + 4y - 10 = -13
Since 2 > -1, 2 > -3, 4 > -1, and 4 > -3, x > y holds for all values.
33Quantitative Aptitude
In each of these questions, two equation (I) and (II) are given. You have to solve both the equations and give answer
I.12x² - 7x + 1 = 0 II.20y² - 9y + 1 = 0
Comparing values:
1/3 > 1/4, 1/3 > 1/5, 1/4 = 1/4, and 1/4 > 1/5.
Thus, x ≥ y holds for all root comparisons.
34Quantitative Aptitude
In each of these questions, two equation (I) and (II) are given. You have to solve both the equations and give answer
I. x² + 26x + 165 = 0 II. y² + 23y + 132 = 0
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (E) If x = y or no relation can be established between x and y
Comparing values:
-11 > -12, but -15 < -12.
Since x is greater than y in one pair and less than y in another, no relation can be established between x and y.
35Quantitative Aptitude
In each of these questions, two equation (I) and (II) are given. You have to solve both the equations and give answer
I. 1 + ³³⁄ₓ + (270)/(x²) = 0 II. y² + 27y + 180 = 0
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (D) If x ≤ y
Explanation
Multiplying equation I by x² gives: x² + 33x + 270 = 0
(x + 15)(x + 18) = 0
x = -15, -18.
Comparing values:
-15 = -15, -15 < -12, -18 < -15, and -18 < -12.
Thus, x ≤ y holds for all comparisons.
36Quantitative Aptitude
A boat travels in upstream. If the speed of boat in upstream is decreased by 25% then it is equal to the speed of current and speed of boat in still water is given as 70 km/hr , then find downstream speed of boat? (In km/hr ).
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (B) 100
Explanation
Boat speed in still water B = 70 km/hr and speed of current = C km/hr.
Upstream speed U = B - C = 70 - C.
Decreasing upstream speed by 25% gives 0.75 × (70 - C) = C.
Solving this gives 3 × (70 - C) = 4C, which simplifies to 210 - 3C = 4C, so 7C = 210, yielding C = 30 km/hr.
Downstream speed = B + C = 70 + 30 = 100 km/hr.
37Quantitative Aptitude
A man invested Rs.2P and Rs. P at 8% compound interest (compounding annually) and 15% simple interest for two years respectively. Difference between compound interest and simple interest received after two years is Rs.410. Find the value of P.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (A) Rs.12500
Explanation
For an investment of 2P at 8% compound interest per annum for 2 years, the effective interest rate is 8 + 8 + (8 × 8)/100 = 16.64%.
Compound interest received = 2P × 0.1664 = 0.3328P.
For an investment of P at 15% simple interest per annum for 2 years, simple interest received = P × 0.15 × 2 = 0.30P.
The difference between compound interest and simple interest is 0.3328P - 0.30P = 0.0328P.
Equating this to the given difference gives 0.0328P = 410, so P = 410 / 0.0328 = 12500.
38Quantitative Aptitude
Length of train X is 200 meters and length of train Y is 'a' meter more than train X. If speed of both train X & Y is equal and they cross a signal in 8 sec and 20 secs respectively, then in what time train 'Y' will cross 400 m long platform.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (C) 36 sec
Explanation
Speed of train X is distance divided by time = 200 / 8 = 25 m/s.
Since train Y has the same speed, its total length is speed × time = 25 × 20 = 500 meters.
To cross a 400-meter platform, train Y must cover a total distance equal to its own length plus the platform length, which is 500 + 400 = 900 meters.
Time taken by train Y to cross the platform = 900 / 25 = 36 seconds.
39Quantitative Aptitude
Amit is 66⅔% more efficient than Vivek who can do the same work in 25% more time than Sameer. If Amit and Sameer together can complete 70% of work in 15 days, then in how many days 90% of work can be completed by Vivek and Sameer together.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (B) 25 days
Explanation
Let efficiency of Sameer be E_S. Vivek takes 25% more time than Sameer, so the ratio of efficiencies of Vivek and Sameer is 4 : 5.
Amit is 66⅔% more efficient than Vivek, so the ratio of efficiencies of Amit and Vivek is 5 : 3 = 20 : 12.
Combining these ratios, efficiency ratio of Amit, Vivek, and Sameer is 20 : 12 : 15.
Amit and Sameer together have a combined efficiency of 20 + 15 = 35 units/day and complete 70% of total work in 15 days, meaning 70% of work = 35 × 15 = 525 units, so total work = 750 units.
To complete 90% of work (675 units), Vivek and Sameer together with a combined efficiency of 12 + 15 = 27 units/day will take 675 / 27 = 25 days.
40Quantitative Aptitude
Area of square is 576 cm² and perimeter of square is equal to perimeter of a rectangle. If length of rectangle is 37.5% more than the side of the square, then find the area of rectangle.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (D) 495 cm²
Explanation
Side of the square s = √576 = 24 cm.
Perimeter of the square = 4 × 24 = 96 cm.
Since the perimeter of the rectangle equals the perimeter of the square, 2(l + b) = 96, so l + b = 48 cm.
Length of rectangle l = 24 × (1 + 0.375) = 33 cm.
Breadth of rectangle b = 48 - 33 = 15 cm.
Area of rectangle = l × b = 33 × 15 = 495 cm².
41Quantitative Aptitude
A shopkeeper has two types of watches (Analog and Digital). Cost price of an analog watch is 32.5% more than cost price of a digital watch. If cost price of analog watch is increased by 30% and sold in Rs.2067 at the profit of 25% . If digital watch marked up 15% above cost price, then find the marked price of digital watch (in Rs.).
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (B) 1104
Explanation
Let the cost price of a digital watch be D.
Cost price of an analog watch A = 1.325D.
After a 30% increase, the new cost price of the analog watch becomes 1.30 × 1.325D.
Selling price of the analog watch at 25% profit is 1.25 × 1.30 × 1.325D = 2067.
Solving for D gives 2.03125D = 2067, which yields D = 960.
The digital watch is marked up by 15%, so its marked price is 960 × 1.15 = 1104.
42Quantitative Aptitude
In a vessel P, quantity of spirit was 12 liters more than that of water. After adding 80 liter of mixture having ratio of water and spirit 3: 5 in the vessel P, ratio of water and spirit in the final mixture becomes 9: 13. Find the total quantity of water in the final mixture.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (C) 72 ltr
Explanation
Let initial quantity of water in vessel P be w liters, so initial spirit is (w + 12) liters.
Adding 80 liters of mixture with water to spirit ratio 3 : 5 adds 30 liters of water and 50 liters of spirit.
Total water becomes (w + 30) liters and total spirit becomes (w + 62) liters.
Given final ratio of water to spirit is 9 : 13, setting up (w + 30) / (w + 62) = 9 / 13 gives 13w + 390 = 9w + 558.
Solving gives 4w = 168, so w = 42 liters.
Total quantity of water in the final mixture = 42 + 30 = 72 liters.
43Quantitative Aptitude
P and Q started a business together with capital of Rs.1600 and Rs.2000 respectively. P left the business after 5 months. P gets 20% of total profit for managing business and remaining is distributed in ratio of their profit sharing. If after a year, P gets total profit of Rs.500 then, find the total profit earned by them.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (A) Rs.1250
Explanation
Ratio of investment-time products of P and Q is (1600 × 5) : (2000 × 12) = 8000 : 24000 = 1 : 3.
Let total profit be T.
P receives 20% of T for managing the business, leaving 80% of T to be split in the ratio 1 : 3.
P's share of the remaining profit is (1/4) × 0.80T = 0.20T.
Total profit received by P = 0.20T + 0.20T = 0.40T.
Equating P's total profit to 500 gives 0.40T = 500, so total profit T = 500 / 0.40 = 1250.
44Quantitative Aptitude
The monthly income of a person is Rs.25000 from which he spends 20% on rent, 10% of remaining on food and 20% of remaining on clothing. The remaining amount, he shares between his son and daughter in the ratio of 5: 4. Find the difference between amount spent on rent and share of his daughter.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (B) Rs.1400
Explanation
Total income = ₹25000.
Amount spent on rent = 20% of 25000 = ₹5000, leaving ₹20000.
Amount spent on food = 10% of 20000 = ₹2000, leaving ₹18000.
Amount spent on clothing = 20% of 18000 = ₹3600, leaving ₹14400.
The remaining ₹14400 is divided between son and daughter in ratio 5 : 4, so daughter's share = (4/9) × 14400 = ₹6400.
Difference between amount spent on rent and daughter's share = 6400 - 5000 = 1400.
45Quantitative Aptitude
A bag has 7 pink, 4 black and 3 yellow balls while another bag has 5 pink, 6 yellow and 3 blue balls. A bag is selected at random and a ball drawn out of it, then find the probability that the ball drawn is pink.
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (B) 3/7
Explanation
Probability of selecting either bag is 1/2.
First bag contains 14 balls total (7 pink, 4 black, 3 yellow), so probability of drawing a pink ball from the first bag is 7/14 = 1/2.
Second bag contains 14 balls total (5 pink, 6 yellow, 3 blue), so probability of drawing a pink ball from the second bag is 5/14.
Total probability of drawing a pink ball = (1/2 × 1/2) + (1/2 × 5/14) = 1/4 + 5/28 = 12/28 = 3/7.
46Quantitative Aptitude
Directions (51-55) : What approximate value should come in the place of question (?) mark:
?³ × 17.98 + 12.03 % of 450.03 = (14.02)² + ⁴√(15.99)
● Correct
● Not quite — the correct answer is highlighted
Show answer & explanationHide answer⌄
Answer: (B) 2
Explanation
Approximating the given values to their nearest integers:
17.98 ≈ 18
12.03% of 450.03 ≈ 12% of 450 = (12 × 450) / 100 = 54
(14.02)² ≈ 14² = 196
⁴√15.99 ≈ ⁴√16 = 2
