Free CAT Quiz series 3 | CAT Online Test Series 3, CAT Free Mock Test
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Free CAT Quiz series 3 | CAT Online Test Series 3, CAT Free Mock Test. Common Admission Test (CAT) is one of the most challenging and competitive MBA entrance test in our country. Check your level of preparation for CAT with free All India Mock test 2024. CAT Exam Online Test 2024, CAT Free Mock Test Exam 2024. CAT Exam Free Online Quiz 2024, CAT Full Online Mock Test Series 3rd in English. CAT Online Test for All Subjects, CAT Free Mock Test Series in English. CAT Free Mock Test Series 3. CAT English Language Online Test in English Series 3rd. Here we are providing CAT Full Mock Test Paper in English. CAT Mock Test Series 3rd 2024. Now Test your self for CAT Exam by using below quiz…
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Question 1 of 50
1. Question
Find the least number by which (59)(36)(17) has to be multiplied to obtain a perfect square?
Correct
For any perfect square the index of the power of each of its prime factors is even. in the given number the power of 5 and 17 have odd indices, while 3 has an even index.
The least number = (5)(17) = 85.Incorrect
For any perfect square the index of the power of each of its prime factors is even. in the given number the power of 5 and 17 have odd indices, while 3 has an even index.
The least number = (5)(17) = 85. -
Question 2 of 50
2. Question
Find the least square number which is divisible by 10, 12, 15 and 18?
Correct
Incorrect
-
Question 3 of 50
3. Question
The sum of the two numbers is 12 and their product is 35. What is the sum of the reciprocals of these numbers?
Correct
Let the numbers be a and b. Then,
a + b = 12 and ab = 35.
(a + b)/ab = 12/35
= (1/b + 1/a) = 12/35
Sum of reciprocals of given numbers = 12/35.Incorrect
Let the numbers be a and b. Then,
a + b = 12 and ab = 35.
(a + b)/ab = 12/35
= (1/b + 1/a) = 12/35
Sum of reciprocals of given numbers = 12/35. -
Question 4 of 50
4. Question
How long does a train 110 m long running at the speed of 72 km/hr takes to cross a bridge 132 m length?
Correct
Speed = 72 * 5/18 = 20 m/sec
Total distance covered = 110 + 132 = 242 m.
Required time = 242/20 = 12.1 sec.Incorrect
Speed = 72 * 5/18 = 20 m/sec
Total distance covered = 110 + 132 = 242 m.
Required time = 242/20 = 12.1 sec. -
Question 5 of 50
5. Question
A car covers a distance of 624 km in 6 ½ hours. Find its speed?
Correct
624/6 = 104 kmph
Incorrect
624/6 = 104 kmph
-
Question 6 of 50
6. Question
Find the roots of quadratic equation: 2x2 + 5x + 2 = 0?
Correct
2x2 + 4x + x + 2 = 0
2x(x + 2) + 1(x + 2) = 0
(x + 2)(2x + 1) = 0 => x = -2, -1/2Incorrect
2x2 + 4x + x + 2 = 0
2x(x + 2) + 1(x + 2) = 0
(x + 2)(2x + 1) = 0 => x = -2, -1/2 -
Question 7 of 50
7. Question
Three pipes of same capacity can fill a tank in 8 hours. If there are only two pipes of same capacity, the tank can be filled in.
Correct
The part of the tank filled by three pipes in one hour = 1/8
=> The part of the tank filled by two pipes in 1 hour = 2/3 * 1/8 = 1/12.
The tank can be filled in 12 hours.Incorrect
The part of the tank filled by three pipes in one hour = 1/8
=> The part of the tank filled by two pipes in 1 hour = 2/3 * 1/8 = 1/12.
The tank can be filled in 12 hours. -
Question 8 of 50
8. Question
A shopkeeper purchased 70 kg of potatoes for Rs. 420 and sold the whole lot at the rate of Rs. 6.50 per kg. What will be his gain percent?
Correct
C.P. of 1 kg = 420/7 = Rs. 6
S.P. of 1 kg = Rs. 6.50
Gain % = 0.50/6 * 100 = 25/3 = 8 1/3 %Incorrect
C.P. of 1 kg = 420/7 = Rs. 6
S.P. of 1 kg = Rs. 6.50
Gain % = 0.50/6 * 100 = 25/3 = 8 1/3 % -
Question 9 of 50
9. Question
Which of the following is the greatest?
Correct
5 > 4 and 1/2 > 1/4
51/2 > 41/4
41/4 cannot be the greatest. Hence 31/4 also cannot be the greatest.
37/10 = (37)1/10 = (2187)1/10
51/2 = 55/10 = (55)1/10 = (3125)1/10
61/5 = 62/10 = (62)1/10 = (36)1/10
As (3125)1/10 > (2187)1/10 > (36)1/10 , 51/2 is the greatest.Incorrect
5 > 4 and 1/2 > 1/4
51/2 > 41/4
41/4 cannot be the greatest. Hence 31/4 also cannot be the greatest.
37/10 = (37)1/10 = (2187)1/10
51/2 = 55/10 = (55)1/10 = (3125)1/10
61/5 = 62/10 = (62)1/10 = (36)1/10
As (3125)1/10 > (2187)1/10 > (36)1/10 , 51/2 is the greatest. -
Question 10 of 50
10. Question
36 * 48 ÷ 64 + 36 ÷ 12 = ?
Correct
36 * 48 / 64 + 36/12 = 27 + 3 = 30
Incorrect
36 * 48 / 64 + 36/12 = 27 + 3 = 30
-
Question 11 of 50
11. Question
Twelve men can complete a piece of work in 32 days. The same work can be completed by 16 women in 36 days and by 48 boys in 16 days. Find the time taken by one man, one woman and one boy working together to complete the work?
Correct
12 men take 32 days to complete the work.
One man will take (12 * 32) days to complete it.
Similarly one woman will take (16 * 36) days to complete it and one boy will take (48 * 16) days to complete it.
One man, one woman and one boy working together will complete = 1/(12 * 32) + 1/(16 * 36) + 1/(48 * 16)
= 1/(4 * 3 * 16 * 2) + 1/(16 * 4 * 9) + 1/(16 * 3 * 4 * 4)
= 1/64(1/6 + 1/9 + 1/12) = 1/64 * 13/36 of the work in a day.
They will take (64 * 36)/13 days to complete the work working together.Incorrect
12 men take 32 days to complete the work.
One man will take (12 * 32) days to complete it.
Similarly one woman will take (16 * 36) days to complete it and one boy will take (48 * 16) days to complete it.
One man, one woman and one boy working together will complete = 1/(12 * 32) + 1/(16 * 36) + 1/(48 * 16)
= 1/(4 * 3 * 16 * 2) + 1/(16 * 4 * 9) + 1/(16 * 3 * 4 * 4)
= 1/64(1/6 + 1/9 + 1/12) = 1/64 * 13/36 of the work in a day.
They will take (64 * 36)/13 days to complete the work working together. -
Question 12 of 50
12. Question
There are some pigeons and hares in a zoo. If heads are counted, there are 200. If legs are counted, there are 580. The number of hares in the zoo is?
Correct
200*2 = 400
580
—–
180
1—-2
?—-180 = 90Incorrect
200*2 = 400
580
—–
180
1—-2
?—-180 = 90 -
Question 13 of 50
13. Question
Two trains of equal lengths are running at speeds of 30 kmph and 60 kmph. The two trains crossed each other in 30 seconds when travelling in opposite direction. In what time will they cross each other when travelling in the same direction?
Correct
Let the length of each train be x m.
(x + x) / (60 + 30)5/18 = (750 * 18) / (30 * 5) = 90 sec.Incorrect
Let the length of each train be x m.
(x + x) / (60 + 30)5/18 = (750 * 18) / (30 * 5) = 90 sec. -
Question 14 of 50
14. Question
[50 – {20 + 60% of (26/13 + 24/12 + 21) – 2} + 8] = 625/?
Correct
[50 – {20 + 60% of (26/13 + 24/12 + 21) – 2} + 8] = 625/x => [50 – {20 + 3/5 of (25) – 2} + 8] = 625/x
=> [50 – 33 + 8] = 625/x => x = 25Incorrect
[50 – {20 + 60% of (26/13 + 24/12 + 21) – 2} + 8] = 625/x => [50 – {20 + 3/5 of (25) – 2} + 8] = 625/x
=> [50 – 33 + 8] = 625/x => x = 25 -
Question 15 of 50
15. Question
The circumferences of two circles are 264 meters and 352 meters. Find the difference between the areas of the larger and the smaller circles.
Correct
Let the radii of the smaller and the larger circles be s m and l m respectively.
2∏s = 264 and 2∏l = 352
s = 264/2∏ and l = 352/2∏
Difference between the areas = ∏l2 – ∏s2
= ∏{1762/∏2 – 1322/∏2}
= 1762/∏ – 1322/∏
= (176 – 132)(176 + 132)/∏
= (44)(308)/(22/7) = (2)(308)(7) = 4312 sq mIncorrect
Let the radii of the smaller and the larger circles be s m and l m respectively.
2∏s = 264 and 2∏l = 352
s = 264/2∏ and l = 352/2∏
Difference between the areas = ∏l2 – ∏s2
= ∏{1762/∏2 – 1322/∏2}
= 1762/∏ – 1322/∏
= (176 – 132)(176 + 132)/∏
= (44)(308)/(22/7) = (2)(308)(7) = 4312 sq m -
Question 16 of 50
16. Question
What is the sum of all the prime numbers from 20 to 40?
Correct
23 + 29 + 31 + 37 = 120
Incorrect
23 + 29 + 31 + 37 = 120
-
Question 17 of 50
17. Question
The volume of a cube is 1728 cc. Find its surface.
Correct
a3 = 1728 => a = 12
6a2 = 6 * 12 * 12 = 864Incorrect
a3 = 1728 => a = 12
6a2 = 6 * 12 * 12 = 864 -
Question 18 of 50
18. Question
The average of 35 numbers is 25. If each number is multiplied by 5, find the new average?
Correct
Sum of the 35 numbers = 35 * 25 = 875
If each number is multiplied by 5, the sum also gets multiplied by 5 and the average also gets multiplied by 5.
Thus, the new average = 25 * 5 = 125.Incorrect
Sum of the 35 numbers = 35 * 25 = 875
If each number is multiplied by 5, the sum also gets multiplied by 5 and the average also gets multiplied by 5.
Thus, the new average = 25 * 5 = 125. -
Question 19 of 50
19. Question
There is food for 760 men for 22 days. How many more men should join after two days so that the same food may last for 19 days more?
Correct
760 —- 22
760 —- 20
x —– 19
x*19 = 760*20
x = 800
760
——-
40Incorrect
760 —- 22
760 —- 20
x —– 19
x*19 = 760*20
x = 800
760
——-
40 -
Question 20 of 50
20. Question
A and B invests Rs.8000 and Rs.9000 in a business. After 4 months, A withdraws half of his capital and 2 months later, B withdraws one-third of his capital. In what ratio should they share the profits at the end of the year?
Correct
A : B
(8000*4)+(4000*8) : (9000*6)+(6000*6)
64000 : 90000
32 : 45Incorrect
A : B
(8000*4)+(4000*8) : (9000*6)+(6000*6)
64000 : 90000
32 : 45 -
Question 21 of 50
21. Question
I. a2 – 13a + 42 = 0,
II. b2 – 15b + 56 = 0 to solve both the equations to find the values of a and b?Correct
I. a2 – 13a + 42 = 0
=>(a – 6)(a – 7) = 0 => a = 6, 7
II. b2 – 15b + 56 = 0
=>(b – 7)(b – 8) = 0 => b = 7, 8
=>a ≤ bIncorrect
I. a2 – 13a + 42 = 0
=>(a – 6)(a – 7) = 0 => a = 6, 7
II. b2 – 15b + 56 = 0
=>(b – 7)(b – 8) = 0 => b = 7, 8
=>a ≤ b -
Question 22 of 50
22. Question
Krishan and Nandan jointly started a business. Krishan invested three times as Nandan did and invested his money for double time as compared to Nandan. Nandan earned Rs. 4000. If the gain is proportional to the money invested and the time for which the money is invested then the total gain was?
Correct
3:1
2:1
——
6:1
1 —– 4000
7 —– ? => Rs.28,000Incorrect
3:1
2:1
——
6:1
1 —– 4000
7 —– ? => Rs.28,000 -
Question 23 of 50
23. Question
6 – [5/6 + {3 7/8 – 2 1/3 + 1 7/9}] = ?
Correct
6 – [5/6 + {3 7/8 – 2 1/3} + 1 7/9] => 6 – [5/6 + {3 1/8 – 7/3} + 16/9]
=> 6 – [5/6 + {(279 – 168 + 128)/72}] => 6 – 5/6 + 239/72
=> 6 – (60 + 239)/72 => 6 – 46 – 11/72
=> 2 – 11/72 => 1 + 1 – 11/72 => 1 + 61/72 = 1 61/72Incorrect
6 – [5/6 + {3 7/8 – 2 1/3} + 1 7/9] => 6 – [5/6 + {3 1/8 – 7/3} + 16/9]
=> 6 – [5/6 + {(279 – 168 + 128)/72}] => 6 – 5/6 + 239/72
=> 6 – (60 + 239)/72 => 6 – 46 – 11/72
=> 2 – 11/72 => 1 + 1 – 11/72 => 1 + 61/72 = 1 61/72 -
Question 24 of 50
24. Question
The difference between simple and compound interest on Rs. 1200 for one year at 10% per annum reckoned half-yearly is?
Correct
S.I. = (1200 * 10 * 1)/100 = Rs. 120
C.I. = [1200 * (1 + 5/100)2 – 1200] = Rs. 123 Difference = (123 – 120) = Rs. 3.Incorrect
S.I. = (1200 * 10 * 1)/100 = Rs. 120
C.I. = [1200 * (1 + 5/100)2 – 1200] = Rs. 123 Difference = (123 – 120) = Rs. 3. -
Question 25 of 50
25. Question
(0.9 * 0.9 – 0.8 * 0.8)/1.7 = ?
Correct
As numerator is of the form a2 – b2 = (a + b) (a – b), so the given expression simplifying becomes
(0.92 – 0.82)/(0.9 + 0.8) (0.9 + 0.8) (0.9 – 0.8)/0.9 + 0.8 = 0.1Incorrect
As numerator is of the form a2 – b2 = (a + b) (a – b), so the given expression simplifying becomes
(0.92 – 0.82)/(0.9 + 0.8) (0.9 + 0.8) (0.9 – 0.8)/0.9 + 0.8 = 0.1 -
Question 26 of 50
26. Question
Find out the C.I on Rs.5000 at 4% p.a. compound half-yearly for 1 1/2 years.
Correct
A = 5000(51/50)3
= 5306.04
5000
———–
306.04Incorrect
A = 5000(51/50)3
= 5306.04
5000
———–
306.04 -
Question 27 of 50
27. Question
If three eighth of a number is 141. What will be the approximately value of 32.08% of this number?
Correct
x * 3/8 = 141 => x= 376
376 * 32.08/100 = 120Incorrect
x * 3/8 = 141 => x= 376
376 * 32.08/100 = 120 -
Question 28 of 50
28. Question
A team of eight entered for a shooting competition. The best marks man scored 85 points. If he had scored 92 points, the average scores for. The team would have been 84. How many points altogether did the team score?
Correct
8 * 84 = 672 – 7 = 665
Incorrect
8 * 84 = 672 – 7 = 665
-
Question 29 of 50
29. Question
A motorcyclist goes from Bombay to Pune, a distance of 192 kms at an average of 32 kmph speed. Another man starts from Bombay by car 2 ½ hours after the first, and reaches Pune ½ hour earlier. What is the ratio of the speed of the motorcycle and the car?
Correct
T = 192/32 = 6 h
T = 6 – 3 = 3
Time Ratio = 6:3 = 2:1
Speed Ratio = 1:2Incorrect
T = 192/32 = 6 h
T = 6 – 3 = 3
Time Ratio = 6:3 = 2:1
Speed Ratio = 1:2 -
Question 30 of 50
30. Question
A, B and C play a cricket match. The ratio of the runs scored by them in the match is A:B = 2:3 and B:C = 2:5. If the total runs scored by all of them are 75, the runs scored by B are?
Correct
A:B = 2:3
B:C = 2:5
A:B:C = 4:6:15
6/25 * 75 = 18Incorrect
A:B = 2:3
B:C = 2:5
A:B:C = 4:6:15
6/25 * 75 = 18 -
Question 31 of 50
31. Question
I. a3 – 988 = 343,
II. b2 – 72 = 49 to solve both the equations to find the values of a and b?Correct
a3 = 1331 => a = 11
b2 = 121 => b = ± 11
a ≥ bIncorrect
a3 = 1331 => a = 11
b2 = 121 => b = ± 11
a ≥ b -
Question 32 of 50
32. Question
If the price has fallen by 10% what percent of its consumption be: increased so that the expenditure may be the same as before?
Correct
100 – 10 = 90
90——10
100——? => 11 1/9%Incorrect
100 – 10 = 90
90——10
100——? => 11 1/9% -
Question 33 of 50
33. Question
Visitors to show were charged Rs.15 each on the first day. Rs.7.50 on the second day, Rs.2.50 on the third day and total attendance on the three days were in ratio 2:5:13 respectively. The average charge per person for the whole show is?
Correct
2: 5: 13
2x 5x 13x
15 7.5 2.5
30x + 37.5x + 32.5x = 100x/20x
Average = 5Incorrect
2: 5: 13
2x 5x 13x
15 7.5 2.5
30x + 37.5x + 32.5x = 100x/20x
Average = 5 -
Question 34 of 50
34. Question
The difference between the compound interest compounded annually and simple interest for 2 years at 20% per annum is Rs.144. Find the principal?
Correct
P = 144(100/5)2 => P = 3600
Incorrect
P = 144(100/5)2 => P = 3600
-
Question 35 of 50
35. Question
150 men consume 1050 kg of rice in 30 days. In how many days will 70 men consume 980 kg of rice?
Correct
Rate of consumption of each man = 1050/(150 * 30) = 7/30 kg/day
Let us say 70 men take x days to consume 150 kg.
Quantity consumed by each item in x days = (7x/30) kg
Quantity consumed by 70 men in x days = (7/30 x)(70) kg
= (7/30 x) * (70) = 960
x = 60 days.Incorrect
Rate of consumption of each man = 1050/(150 * 30) = 7/30 kg/day
Let us say 70 men take x days to consume 150 kg.
Quantity consumed by each item in x days = (7x/30) kg
Quantity consumed by 70 men in x days = (7/30 x)(70) kg
= (7/30 x) * (70) = 960
x = 60 days. -
Question 36 of 50
36. Question
Siddharth wants to borrow Rs.6000 at rate of interest 6% p.a. at S.I and lend the same amount at C.I at same rate of interest for two years. What would be his income in the above transaction?
Correct
Amount of money Siddharth borrowed at S.I at 6% p.a. for two years = Rs.6,000
He lend the same amount for C.I at 6% p.a. for two years.
=> Siddharth’s income = C.I – S.I
= p[1 + r/ 100]n – p – pnr/100
= p{ [1 + r/ 100]2 – 1 – nr/100
= 6,000{ [1 + 6/100]2 – 1 – 12/100}
= 6,000 {(1.06)2– 1 – 0.12} = 6,000(1.1236 – 1 – 0.12)
= 6,000 (0.0036) = 6 * 3.6 = Rs.21.60Incorrect
Amount of money Siddharth borrowed at S.I at 6% p.a. for two years = Rs.6,000
He lend the same amount for C.I at 6% p.a. for two years.
=> Siddharth’s income = C.I – S.I
= p[1 + r/ 100]n – p – pnr/100
= p{ [1 + r/ 100]2 – 1 – nr/100
= 6,000{ [1 + 6/100]2 – 1 – 12/100}
= 6,000 {(1.06)2– 1 – 0.12} = 6,000(1.1236 – 1 – 0.12)
= 6,000 (0.0036) = 6 * 3.6 = Rs.21.60 -
Question 37 of 50
37. Question
The proportion of copper and zinc in the brass is 13:7. How much zinc will there be in 100 kg of brass?
Correct
7/20 * 100 = 35
Incorrect
7/20 * 100 = 35
-
Question 38 of 50
38. Question
What sum of money put at C.I amounts in 2 years to Rs.8820 and in 3 years to Rs.9261?
Correct
8820 —- 441
100 —- ? => 5%
x *105/100 * 105/100 = 8820
x*1.1025=8820
x=8820/1.1025 => 8000Incorrect
8820 —- 441
100 —- ? => 5%
x *105/100 * 105/100 = 8820
x*1.1025=8820
x=8820/1.1025 => 8000 -
Question 39 of 50
39. Question
The ratio of the number of ladies to gents at a party was 1:2 but when 2 ladies and 2 gents left, the ratio became 1:3. How many people were at the party originally?
Correct
x, 2x
(x-2):(2x-2) = 1:3
3x-6 = 2x-2
x = 4
x+2x = 3x
=> 3*4 = 12Incorrect
x, 2x
(x-2):(2x-2) = 1:3
3x-6 = 2x-2
x = 4
x+2x = 3x
=> 3*4 = 12 -
Question 40 of 50
40. Question
4000 was divided into two parts such a way that when first part was invested at 3% and the second at 5%, the whole annual interest from both the investments is Rs.144, how much was put at 3%?
Correct
(x*3*1)/100 + [(4000 – x)*5*1]/100 = 144
3x/100 + 200 – 5x/100 = 144
2x/100 = 56 → x = 2800Incorrect
(x*3*1)/100 + [(4000 – x)*5*1]/100 = 144
3x/100 + 200 – 5x/100 = 144
2x/100 = 56 → x = 2800 -
Question 41 of 50
41. Question
A car after covering ½ of a journey of 100 km develops engine trouble and later travels at ½ of its original speed. As a result, it arrives 2 hours late than its normal time. What is the normal speed of the car is?
Correct
[50/x + 50/(x/2)] – 100/x = 2
x = 25Incorrect
[50/x + 50/(x/2)] – 100/x = 2
x = 25 -
Question 42 of 50
42. Question
Four car rental agencies A, B, C and D rented a plot for parking their cars during the night. A parked 15 cars for 12 days, B parked 12 cars for 20 days, C parked 18 cars for 18 days and D parked 16 cars for 15 days. If A paid Rs. 1125 as rent for parking his cars, what is the total rent paid by all the four agencies?
Correct
The ratio in which the four agencies will be paying the rents = 15 * 12 : 12 * 20 : 18 * 18 : 16 * 15
= 180 : 240 : 324 : 240 = 45 : 60 : 81 : 60
Let us consider the four amounts to be 45k, 60k, 81k and 60k respectively.
The total rent paid by the four agencies = 45k + 60k + 81k + 60k= 246k
It is given that A paid Rs. 1125
45k = 1125 => k = 25
246k = 246(25) = Rs. 6150
Thus the total rent paid by all the four agencies is Rs. 6150.Incorrect
The ratio in which the four agencies will be paying the rents = 15 * 12 : 12 * 20 : 18 * 18 : 16 * 15
= 180 : 240 : 324 : 240 = 45 : 60 : 81 : 60
Let us consider the four amounts to be 45k, 60k, 81k and 60k respectively.
The total rent paid by the four agencies = 45k + 60k + 81k + 60k= 246k
It is given that A paid Rs. 1125
45k = 1125 => k = 25
246k = 246(25) = Rs. 6150
Thus the total rent paid by all the four agencies is Rs. 6150. -
Question 43 of 50
43. Question
What is the sum of the local values of the digits 2, 3, 4, 5 in the number 2345?
Correct
2000 + 300 + 40 + 5 = 2345
Incorrect
2000 + 300 + 40 + 5 = 2345
-
Question 44 of 50
44. Question
A train 100 meters long completely crosses a 300 meters long bridge in 45 seconds. What is the speed of the train is?
Correct
S = (100 + 300)/45 = 400/45 * 18/5 = 32
Incorrect
S = (100 + 300)/45 = 400/45 * 18/5 = 32
-
Question 45 of 50
45. Question
A man engaged a servant on the condition that he would pay him Rs.710 and a uniform after a year service. He served only for 8 months and got Rs.460 and a uniform. Find the price of the uniform?
Correct
8/12 = 2/3 * 710 = 473.33
460.00
—————
13.33
1/3 uniform 13.33
1 ————— ? => Rs.40Incorrect
8/12 = 2/3 * 710 = 473.33
460.00
—————
13.33
1/3 uniform 13.33
1 ————— ? => Rs.40 -
Question 46 of 50
46. Question
Train P crosses a pole in 30 seconds and train Q crosses the same pole in one minute and 15 seconds. The length of train P is three-fourths the length of train Q. What is the ratio of the speed of train P to that of train Q?
Correct
Given that train P crosses a pole in 30 seconds and train Q crosses the same pole in one minute and 15 seconds.
Let the length of train P be LP and that of train Q be LQ
given that LP = 3/4 LQ
As the train P and Q crosses the pole in 30 seconds and 75 seconds respectively,
=> Speed of train P = VP = LP/30
Speed of train Q = VQ = LQ/75
LP = 3/4 LQ
=> VP = 3/4 LQ/(30) = LQ/40
Ratio of their speeds = VP : VQ
= LQ/40 : LQ/75 => 1/40 : 1/75 = 15 : 8Incorrect
Given that train P crosses a pole in 30 seconds and train Q crosses the same pole in one minute and 15 seconds.
Let the length of train P be LP and that of train Q be LQ
given that LP = 3/4 LQ
As the train P and Q crosses the pole in 30 seconds and 75 seconds respectively,
=> Speed of train P = VP = LP/30
Speed of train Q = VQ = LQ/75
LP = 3/4 LQ
=> VP = 3/4 LQ/(30) = LQ/40
Ratio of their speeds = VP : VQ
= LQ/40 : LQ/75 => 1/40 : 1/75 = 15 : 8 -
Question 47 of 50
47. Question
The cash difference between the selling prices of an article at a profit of 4% and 6% is Rs. 3. The ratio of the two selling prices is:
Correct
Let C.P. of the article be Rs. x.
Then, required ratio = 104% of x / 106% of x
= 104/106 = 52/53 = 52:53Incorrect
Let C.P. of the article be Rs. x.
Then, required ratio = 104% of x / 106% of x
= 104/106 = 52/53 = 52:53 -
Question 48 of 50
48. Question
A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected, if it should have 5 seniors and 5 juniors?
Correct
Here, five seniors out of 12 seniors can be selected in ¹²C₅ ways. Also, five juniors out of ten juniors can be selected ¹⁰C₅ ways. Hence the total number of different ways of selection = ¹²C₅ * ¹⁰C₅ = ¹²C₇ * ¹⁰C₅
= ¹²C₅ = ¹²C₇Incorrect
Here, five seniors out of 12 seniors can be selected in ¹²C₅ ways. Also, five juniors out of ten juniors can be selected ¹⁰C₅ ways. Hence the total number of different ways of selection = ¹²C₅ * ¹⁰C₅ = ¹²C₇ * ¹⁰C₅
= ¹²C₅ = ¹²C₇ -
Question 49 of 50
49. Question
A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is:
Correct
Let the ten’s and unit’s digit be x and 8/x respectively.
Then,
(10x + 8/x) + 18 = 10 * 8/x + x
9x2 + 18x – 72 = 0
x2 + 2x – 8 = 0
(x + 4)(x – 2) = 0
x = 2
So, ten’s digit = 2 and unit’s digit = 4.
Hence, required number = 24.Incorrect
Let the ten’s and unit’s digit be x and 8/x respectively.
Then,
(10x + 8/x) + 18 = 10 * 8/x + x
9x2 + 18x – 72 = 0
x2 + 2x – 8 = 0
(x + 4)(x – 2) = 0
x = 2
So, ten’s digit = 2 and unit’s digit = 4.
Hence, required number = 24. -
Question 50 of 50
50. Question
A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by:
Correct
Let the ten’s digit be x and unit’s digit be y.
Then, number 10x + y.
Number obtained by interchanging the digits = 10y + x.
(10x + y) + (10y + x) = 11(x + y)
which is divisible by 11.Incorrect
Let the ten’s digit be x and unit’s digit be y.
Then, number 10x + y.
Number obtained by interchanging the digits = 10y + x.
(10x + y) + (10y + x) = 11(x + y)
which is divisible by 11.