CAT Online Test Series 4  Free CAT Quiz Series 4  CAT Free Mock Test
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CAT Online Test Series 4  Free CAT Quiz Series 4  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 2021. CAT Exam Online Test 2021, CAT Free Mock Test Exam 2021. CAT Exam Free Online Quiz 2021, CAT Full Online Mock Test Series 4th in English. CAT Online Test for All Subjects, CAT Free Mock Test Series in English. CAT Free Mock Test Series 4. CAT English Language Online Test in English Series 4th. Here we are providing CAT Full Mock Test Paper in English. CAT Mock Test Series 4th 2021. Now Test your self for CAT Exam by using below quiz…
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Question 1 of 50
1. Question
A batsman in his 17^{th} innings makes a score of 85 and their by increasing his average by 3. What is his average after the 17^{th}innings?
Correct
16x + 85 = 17(x + 3)
x = 34 + 3 = 37Incorrect
16x + 85 = 17(x + 3)
x = 34 + 3 = 37 
Question 2 of 50
2. Question
The price of a VCR is marked at Rs. 12,000. If successive discounts of 15%, 10% and 5% be allowed, then at what price does a customer buy it?
Correct
Actual price = 95% of 90% of 85% of Rs. 12000
= 95/100 * 90/100 * 85/100 * 12000 = Rs. 8721.Incorrect
Actual price = 95% of 90% of 85% of Rs. 12000
= 95/100 * 90/100 * 85/100 * 12000 = Rs. 8721. 
Question 3 of 50
3. Question
The base of a right triangle is 8 and hypotenuse is 10. Its area is?
Correct
Incorrect

Question 4 of 50
4. Question
108 * 107 * 96 = ? (to the nearest hundred)
Correct
108 * 107 * 96 ≡ 1109400
Incorrect
108 * 107 * 96 ≡ 1109400

Question 5 of 50
5. Question
Two pipes can separately fill a tank in 20 and 30 hours respectively. Both the pipes are opened to fill the tank but when the tank is full, a leak develops in the tank through which onethird of water supplied by both the pipes goes out. What is the total time taken to fill the tank?
Correct
1/20 + 1/30 = 1/12
1 + 1/3 = 4/3
1 — 12
4/3 — ?
4/3 * 12 = 16 hrsIncorrect
1/20 + 1/30 = 1/12
1 + 1/3 = 4/3
1 — 12
4/3 — ?
4/3 * 12 = 16 hrs 
Question 6 of 50
6. Question
A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
Correct
Speed of the train relative to man = (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr = 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x – 5) km/hr. x – 5 = 45 ==> x = 50 km/hr.
Incorrect
Speed of the train relative to man = (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr = 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x – 5) km/hr. x – 5 = 45 ==> x = 50 km/hr.

Question 7 of 50
7. Question
5 * 5 ÷ 5 + 5 ÷ 5 = ?
Correct
(5 * 5)/5 + 1/5 * 5 = 5 + 1 = 6
Incorrect
(5 * 5)/5 + 1/5 * 5 = 5 + 1 = 6

Question 8 of 50
8. Question
I. a^{2} + 11a + 30 = 0,
II. b^{2} + 6b + 5 = 0 to solve both the equations to find the values of a and b?Correct
I. (a + 6)(a + 5) = 0
=> a = 6, 5
II. (b + 5)(b + 1) = 0
=> b = 5, 1 => a ≤ bIncorrect
I. (a + 6)(a + 5) = 0
=> a = 6, 5
II. (b + 5)(b + 1) = 0
=> b = 5, 1 => a ≤ b 
Question 9 of 50
9. Question
9000 + 16 2/3 % of ? = 10500
Correct
9000 + 16 2/3 % of ? = 10500 => 9000 + 50/3 % of ? = 10500
50/(3 * 100) of ? = 1500 => ? = 1500 * 6
? = 9000Incorrect
9000 + 16 2/3 % of ? = 10500 => 9000 + 50/3 % of ? = 10500
50/(3 * 100) of ? = 1500 => ? = 1500 * 6
? = 9000 
Question 10 of 50
10. Question
9 3/4 + 7 2/17 – 9 1/15 = ?
Correct
Given sum = 9 + 3/4 + 7 + 2/17 – (9 + 1/15)
= (9 + 7 – 9) + (3/4 + 2/17 – 1/15)
= 7 + (765 + 120 – 68)/1020 = 7 817/1020Incorrect
Given sum = 9 + 3/4 + 7 + 2/17 – (9 + 1/15)
= (9 + 7 – 9) + (3/4 + 2/17 – 1/15)
= 7 + (765 + 120 – 68)/1020 = 7 817/1020 
Question 11 of 50
11. Question
The number of new words that can be formed by rearranging the letters of the word ‘ALIVE’ is .
Correct
Number of words which can be formed = 5! – 1 = 120 – 1 = 119.
Incorrect
Number of words which can be formed = 5! – 1 = 120 – 1 = 119.

Question 12 of 50
12. Question
(84 ÷ 14 ÷ 2 ÷ 9) * ? = 33⅓
Correct
(84 ÷ 14 ÷ 2 ÷ 9) * ? = 33⅓
=> (6 ÷ 2 ÷ 9) * ? = 100 / 3
=> 3 / 9 * ? = 100 / 3
=> ? = 100Incorrect
(84 ÷ 14 ÷ 2 ÷ 9) * ? = 33⅓
=> (6 ÷ 2 ÷ 9) * ? = 100 / 3
=> 3 / 9 * ? = 100 / 3
=> ? = 100 
Question 13 of 50
13. Question
A train 800 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min, then the length of the tunnel is?
Correct
Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (800 + x)/60 = 65/3
x = 500 m.Incorrect
Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (800 + x)/60 = 65/3
x = 500 m. 
Question 14 of 50
14. Question
If 35% of a number is 12 less than 50% of that number, then the number is?
Correct
Let the number be x. Then,
50% of x – 35% of x = 12
50/100 x – 35/100 x = 12
x = (12 * 100)/15 = 80.Incorrect
Let the number be x. Then,
50% of x – 35% of x = 12
50/100 x – 35/100 x = 12
x = (12 * 100)/15 = 80. 
Question 15 of 50
15. Question
The sum of the digits of a twodigit number is 12. The difference of the digits is 6. Find the number?
Correct
Let the twodigit number be 10a + b
a + b = 12 — (1)
If a>b, a – b = 6
If b>a, b – a = 6
If a – b = 6, adding it to equation (1), we get
2a = 18 => a =9
so b = 12 – a = 3
Number would be 93.
if b – a = 6, adding it to the equation (1), we get
2b = 18 => b = 9
a = 12 – b = 3.
Number would be 39.
There fore, Number would be 39 or 93.Incorrect
Let the twodigit number be 10a + b
a + b = 12 — (1)
If a>b, a – b = 6
If b>a, b – a = 6
If a – b = 6, adding it to equation (1), we get
2a = 18 => a =9
so b = 12 – a = 3
Number would be 93.
if b – a = 6, adding it to the equation (1), we get
2b = 18 => b = 9
a = 12 – b = 3.
Number would be 39.
There fore, Number would be 39 or 93. 
Question 16 of 50
16. Question
The ratio of the prices of three articles X, Y and Z is 8 : 5 : 3. If the prices of X , Y and Z are increased by 25%, 20% and 33 1/3% respectively, then what would be the ratio of the new prices of X, Y and Z?
Correct
Let the prices of X, Y and Z be 8k, 5k and 3k respectively.
After increase
Price of X = 8k * 125/100 = 10k
Price of Y = 5k * 120/100 = 6k
Price of Z = 3k * (133 1/3)/100 = 4k
Required ratio = 10k : 6k : 4k = 5 : 3 : 2.Incorrect
Let the prices of X, Y and Z be 8k, 5k and 3k respectively.
After increase
Price of X = 8k * 125/100 = 10k
Price of Y = 5k * 120/100 = 6k
Price of Z = 3k * (133 1/3)/100 = 4k
Required ratio = 10k : 6k : 4k = 5 : 3 : 2. 
Question 17 of 50
17. Question
Find the least multiple of 13 which when divided by 6, 8 and 12 leaves 5, 7 and 11 as remainders respectively?
Correct
Incorrect

Question 18 of 50
18. Question
A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in?
Correct
(A + B + C)’s 1 day work = 1/6;
(A + B)’s 1 day work = 1/8
(B + C)’s 1 day work = 1/12
(A + C)’s 1 day work = (2 * 1/6) – (1/8 + 1/12)
= (1/3 – 5/24) = 1/8
So, A and C together will do the work in 8 days.Incorrect
(A + B + C)’s 1 day work = 1/6;
(A + B)’s 1 day work = 1/8
(B + C)’s 1 day work = 1/12
(A + C)’s 1 day work = (2 * 1/6) – (1/8 + 1/12)
= (1/3 – 5/24) = 1/8
So, A and C together will do the work in 8 days. 
Question 19 of 50
19. Question
If the height of a cone is increased by 100% then its volume is increased by?
Correct
Incorrect

Question 20 of 50
20. Question
The ratio of the radius of two circles is 1: 3, and then the ratio of their areas is?
Correct
r_{1}: r_{2} = 1: 3
Πr_{1}^{2}: Πr_{2}^{2}
r_{1}^{2}: r_{2}^{2 }= 1: 9Incorrect
r_{1}: r_{2} = 1: 3
Πr_{1}^{2}: Πr_{2}^{2}
r_{1}^{2}: r_{2}^{2 }= 1: 9 
Question 21 of 50
21. Question
There is a 30% increase in the price of an article in the first year, a 20% decrease in the second year and a 10% increase in the next year. If the final price of the article is Rs. 2288, then what was the price of the article initially?
Correct
Let the price of the article, four years age be Rs. 100 in the 1st year, price of the article = 100 + 30 = Rs. 130. In the 2nd year, price = 130 – 20% of 130 = 130 – 26 = Rs. 104.
In the 3rd year, price = 104 + 10% of 104 = 104 + 10.4 = Rs. 114.40.
But present price of the article is Rs. 2288
for 114.4 —> 100 ; 2288 —> ?
Required price = (2288 * 100)/114.4 = 20 * 100 = Rs. 2000.Incorrect
Let the price of the article, four years age be Rs. 100 in the 1st year, price of the article = 100 + 30 = Rs. 130. In the 2nd year, price = 130 – 20% of 130 = 130 – 26 = Rs. 104.
In the 3rd year, price = 104 + 10% of 104 = 104 + 10.4 = Rs. 114.40.
But present price of the article is Rs. 2288
for 114.4 —> 100 ; 2288 —> ?
Required price = (2288 * 100)/114.4 = 20 * 100 = Rs. 2000. 
Question 22 of 50
22. Question
Difference between two numbers is 5, six times of the smaller lacks by 6 from the four times of the greater. Find the numbers?
Correct
x – y = 5
4x – 6y = 6
x = 12 y = 7Incorrect
x – y = 5
4x – 6y = 6
x = 12 y = 7 
Question 23 of 50
23. Question
The sum of a number and its reciprocal is oneeighth of 34. What is the product of the number and its square root?
Correct
Let the number be x. Then,
x + 1/x = 34/8
8x^{2} – 34x + 8 = 0
4x^{2} – 17x + 4 = 0
(4x – 1)(x – 4) = 0
x = 4
required number = 4 * √4 = 4 * 2 = 8.Incorrect
Let the number be x. Then,
x + 1/x = 34/8
8x^{2} – 34x + 8 = 0
4x^{2} – 17x + 4 = 0
(4x – 1)(x – 4) = 0
x = 4
required number = 4 * √4 = 4 * 2 = 8. 
Question 24 of 50
24. Question
The sum of the digits of a twodigit number is 15 and the difference between the digits is 3. What is the twodigit number?
Correct
Let the ten’s digit be x and unit’s digit by y
Then, x + y = 15 and x – y = 3 or y – x = 3
Solving x + y = 15 and x – y = 3, we get: x = 9, y = 6
Solving x + y = 15 and y – x = 3, we get: x = 6, y = 9
So, the number is either 96 or 69.
Hence, the number cannot be determined.Incorrect
Let the ten’s digit be x and unit’s digit by y
Then, x + y = 15 and x – y = 3 or y – x = 3
Solving x + y = 15 and x – y = 3, we get: x = 9, y = 6
Solving x + y = 15 and y – x = 3, we get: x = 6, y = 9
So, the number is either 96 or 69.
Hence, the number cannot be determined. 
Question 25 of 50
25. Question
The length of a rectangle is increased by 25% and its breadth is decreased by 20%. What is the effect on its area?
Correct
100 * 100 = 10000
125 * 80 = 10000
No changeIncorrect
100 * 100 = 10000
125 * 80 = 10000
No change 
Question 26 of 50
26. Question
A and B can finish a work in 16 days while A alone can do the same work in 24 days. In how many days B alone will complete the work?
Correct
B = 1/16 – 1/24 = 1/48 => 48 days
Incorrect
B = 1/16 – 1/24 = 1/48 => 48 days

Question 27 of 50
27. Question
The ratio of the earnings of P and Q is 9 : 10. If the earnings of P increases by onefourth and the earnings of Q decreases by onefourth, then find the new ratio of their earnings?
Correct
Let the earnings of P and Q be Rs. 9x and Rs. 10x respectively.
New ration = [9x + 1/4(9x)]/[10x – 1/4(10x)]
=> [9x(1 + 1/4)]/[10x(1 – 1/4)] = 9/10 * (5/4)/(3/4) => 3/2.Incorrect
Let the earnings of P and Q be Rs. 9x and Rs. 10x respectively.
New ration = [9x + 1/4(9x)]/[10x – 1/4(10x)]
=> [9x(1 + 1/4)]/[10x(1 – 1/4)] = 9/10 * (5/4)/(3/4) => 3/2. 
Question 28 of 50
28. Question
The food in a camp lasts for 30 men for 40 days. If ten more men join, how many days will the food last?
Correct
one man can consume the same food in 30*40 = 1200 days.
10 more men join, the total number of men = 40
The number of days the food will last = 1200/40 = 30 days.Incorrect
one man can consume the same food in 30*40 = 1200 days.
10 more men join, the total number of men = 40
The number of days the food will last = 1200/40 = 30 days. 
Question 29 of 50
29. Question
The roots of the equation 3x^{2} – 12x + 10 = 0 are?
Correct
The discriminant of the quadratic equation is (12)^{2} – 4(3)(10) i.e., 24. As this is positive but not a perfect square, the roots are irrational and unequal.
Incorrect
The discriminant of the quadratic equation is (12)^{2} – 4(3)(10) i.e., 24. As this is positive but not a perfect square, the roots are irrational and unequal.

Question 30 of 50
30. Question
If one third of 3/4 of a number is 21. Then find the number?
Correct
x * 1/3 * 3/4 =21 => x = 84
Incorrect
x * 1/3 * 3/4 =21 => x = 84

Question 31 of 50
31. Question
A, B and C can do a work in 6, 8 and 12 days respectively doing the work together and get a payment of Rs.1800. What is B’s share?
Correct
WC = 1/6:1/8:1/12 => 4:3:2
3/9 * 1800 = 600Incorrect
WC = 1/6:1/8:1/12 => 4:3:2
3/9 * 1800 = 600 
Question 32 of 50
32. Question
Two trains of equal lengths take 10 sec and 15 sec respectively to cross a telegraph post. If the length of each train be 120 m, in what time will they cross other travelling in opposite direction?
Correct
Speed of the first train = 120/10 = 12 m/sec.
Speed of the second train = 120/5 = 8 m/sec.
Relative speed = 12 + 8 = 20 m/sec.
Required time = (120 + 120)/20 = 12 sec.Incorrect
Speed of the first train = 120/10 = 12 m/sec.
Speed of the second train = 120/5 = 8 m/sec.
Relative speed = 12 + 8 = 20 m/sec.
Required time = (120 + 120)/20 = 12 sec. 
Question 33 of 50
33. Question
The owner of a furniture shop charges his customer 24% more than the cost price. If a customer paid Rs. 8339 for a computer table, then what was the cost price of the computer table?
Correct
CP = SP * (100/(100 + profit%))
= 8339(100/124) = Rs. 6725.Incorrect
CP = SP * (100/(100 + profit%))
= 8339(100/124) = Rs. 6725. 
Question 34 of 50
34. Question
A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
Correct
Speed = [54 * 5/18] m/sec = 15 m/sec.
Length of the train = (15 * 20) m = 300 m.
Let the length of the platform be x meters.
Then, x + 300 / 36 = 15
x + 300 = 540
x = 240 m.Incorrect
Speed = [54 * 5/18] m/sec = 15 m/sec.
Length of the train = (15 * 20) m = 300 m.
Let the length of the platform be x meters.
Then, x + 300 / 36 = 15
x + 300 = 540
x = 240 m. 
Question 35 of 50
35. Question
Six points are marked on a straight line and five points are marked on another line which is parallel to the first line. How many straight lines, including the first two, can be formed with these points?
Correct
We know that, the number of straight lines that can be formed by the 11 points in which 6 points are collinear and no other set of three points, except those that can be selected out of these 6 points are collinear.
Hence, the required number of straight lines
= ¹¹C₂ – ⁶C₂ – ⁵C₂ + 1 + 1
= 55 – 15 – 10 + 2 = 32Incorrect
We know that, the number of straight lines that can be formed by the 11 points in which 6 points are collinear and no other set of three points, except those that can be selected out of these 6 points are collinear.
Hence, the required number of straight lines
= ¹¹C₂ – ⁶C₂ – ⁵C₂ + 1 + 1
= 55 – 15 – 10 + 2 = 32 
Question 36 of 50
36. Question
The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 10, then their difference is:
Correct
Let the numbers be x and (100 – x).
Then, x(100 – x) = 5 * 495
x^{2} – 100x + 2475 = 0
(x – 55)(x – 45) = 0
x = 55 or 45
The numbers are 45 and 55.
Required difference = 55 – 45 = 10.Incorrect
Let the numbers be x and (100 – x).
Then, x(100 – x) = 5 * 495
x^{2} – 100x + 2475 = 0
(x – 55)(x – 45) = 0
x = 55 or 45
The numbers are 45 and 55.
Required difference = 55 – 45 = 10. 
Question 37 of 50
37. Question
(743.30)^{2} = ?
Correct
(743.30)^{2} = 552500
Incorrect
(743.30)^{2} = 552500

Question 38 of 50
38. Question
9 men and 12 boys finish a job in 12 days, 12 men and 12 boys finish it in 10 days. 10 men and 10 boys shall finish it in how many days?
Correct
9M + 12B —– 12 days
12M + 12B —— 10 days
10M + 10B ——?
108M + 144B = 120M +120B
24B = 12M => 1M = 2B
18B + 12B = 30B — 12 days
20B + 10B = 30B —–? => 12 daysIncorrect
9M + 12B —– 12 days
12M + 12B —— 10 days
10M + 10B ——?
108M + 144B = 120M +120B
24B = 12M => 1M = 2B
18B + 12B = 30B — 12 days
20B + 10B = 30B —–? => 12 days 
Question 39 of 50
39. Question
The difference between a number and its threefifth is 50. What is the number?
Correct
Let the number be x. Then,
x – 3/5 x = 50 => 2/5 x = 50
x = (50 * 5)/2 = 125.Incorrect
Let the number be x. Then,
x – 3/5 x = 50 => 2/5 x = 50
x = (50 * 5)/2 = 125. 
Question 40 of 50
40. Question
If the cost price of 50 articles is equal to the selling price of 40 articles, then the gain or loss percent is?
Correct
Given that, cost price of 50 article is equal to selling price of 40 articles.
Let cost price of one article = Rs. 1
Selling price of 40 articles = Rs. 50
But Cost price of 40 articles = Rs. 40
Therefore, the trader made profit.\Percentage of profit = 10/40 * 100 = 25%Incorrect
Given that, cost price of 50 article is equal to selling price of 40 articles.
Let cost price of one article = Rs. 1
Selling price of 40 articles = Rs. 50
But Cost price of 40 articles = Rs. 40
Therefore, the trader made profit.\Percentage of profit = 10/40 * 100 = 25% 
Question 41 of 50
41. Question
A train running at the speed of 60 km/hr crosses a pole in 9 seconds. Find the length of the train.
Correct
Speed = 60*(5/18) m/sec = 50/3 m/sec
Length of Train (Distance) = Speed * Time
(50/3) * 9 = 150 meterIncorrect
Speed = 60*(5/18) m/sec = 50/3 m/sec
Length of Train (Distance) = Speed * Time
(50/3) * 9 = 150 meter 
Question 42 of 50
42. Question
A watch was sold at a loss of 10%. If it was sold for Rs.140 more, there would have been a gain of 4%. What is the cost price?
Correct
90%
104%
——–
14% — 140
100% — ? => Rs.1000Incorrect
90%
104%
——–
14% — 140
100% — ? => Rs.1000 
Question 43 of 50
43. Question
Sum of two numbers is 80. Greater number exceeds by 5 from four times of the smaller. Find the numbers?
Correct
x + y = 80
x – 4y = 5
x = 65 y = 15Incorrect
x + y = 80
x – 4y = 5
x = 65 y = 15 
Question 44 of 50
44. Question
The compound interest accrued on an amount of Rs.44000 at the end of two years is Rs.1193.60. What would be the simple interest accrued on the same amount at the same rate in the same period?
Correct
Let the rate of interest be R% p.a.
4400{[1 + R/100]^{2} – 1} = 11193.60
[1 + R/100]^{2} = (44000 + 11193.60)/44000
[1 + R/100]^{2} = 1 + 2544/1000 = 1 + 159/625
[1 + R/100]^{2} = 784/625 = (28/25)^{2}
1 + R/100 = 28/25
R/100 = 3/25
herefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100
=Rs.10560Incorrect
Let the rate of interest be R% p.a.
4400{[1 + R/100]^{2} – 1} = 11193.60
[1 + R/100]^{2} = (44000 + 11193.60)/44000
[1 + R/100]^{2} = 1 + 2544/1000 = 1 + 159/625
[1 + R/100]^{2} = 784/625 = (28/25)^{2}
1 + R/100 = 28/25
R/100 = 3/25
herefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100
=Rs.10560 
Question 45 of 50
45. Question
If the L.C.M of two numbers is 750 and their product is 18750, find the H.C.F of the numbers.
Correct
H.C.F = (Product of the numbers) / (Their L.C.M) = 18750/750 = 25.
Incorrect
H.C.F = (Product of the numbers) / (Their L.C.M) = 18750/750 = 25.

Question 46 of 50
46. Question
Albert buys 4 horses and 9 cows for Rs. 13,400. If he sells the horses at 10% profit and the cows at 20% profit, then he earns a total profit of Rs. 1880. The cost of a horse is:
Correct
Let C.P. of each horse be Rs. x and C.P. of each cow be Rs. y.
Then, 4x + 9y = 13400 — (i)
And, 10% of 4x + 20% of 9y = 1880
2/5 x + 9/5 y = 1880 => 2x + 9y = 9400 — (ii)
Solving (i) and (ii), we get : x = 2000 and y = 600.
Cost price of each horse = Rs. 2000.Incorrect
Let C.P. of each horse be Rs. x and C.P. of each cow be Rs. y.
Then, 4x + 9y = 13400 — (i)
And, 10% of 4x + 20% of 9y = 1880
2/5 x + 9/5 y = 1880 => 2x + 9y = 9400 — (ii)
Solving (i) and (ii), we get : x = 2000 and y = 600.
Cost price of each horse = Rs. 2000. 
Question 47 of 50
47. Question
Robert is traveling on his cycle and has calculated to reach point A at 2 p.m. if he travels at 10 km/hr; he will reach there at 12 noon if he travels at 15 km/hr. At what speed must he travel to reach A at 1 p.m.?
Correct
Let the distance traveled be x km.
Then, x/10 – x/15 = 2
3x – 2x = 60 => x = 60 km.
Time taken to travel 60 km at 10 km/hr = 60/10 = 6 hrs.
So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m.
Required speed = 60/5 = 12 kmph.Incorrect
Let the distance traveled be x km.
Then, x/10 – x/15 = 2
3x – 2x = 60 => x = 60 km.
Time taken to travel 60 km at 10 km/hr = 60/10 = 6 hrs.
So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m.
Required speed = 60/5 = 12 kmph. 
Question 48 of 50
48. Question
A certain sum is invested at simple interest at 18% p.a. for two years instead of investing at 12% p.a. for the same time period. Therefore the interest received is more by Rs. 840. Find the sum?
Correct
Let the sum be Rs. x.
(x * 18 * 2)/100 – (x * 12 * 2)/100 = 840 => 36x/100 – 24x/100 =840
=> 12x/100 = 840 => x = 7000.Incorrect
Let the sum be Rs. x.
(x * 18 * 2)/100 – (x * 12 * 2)/100 = 840 => 36x/100 – 24x/100 =840
=> 12x/100 = 840 => x = 7000. 
Question 49 of 50
49. Question
A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained?
Correct
Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 – 6 = 6 liters
Remaining water = 8 – 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 – 8) = 8 liters.
Remaining water = (4 2) = 2 liters.
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1.Incorrect
Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 – 6 = 6 liters
Remaining water = 8 – 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 – 8) = 8 liters.
Remaining water = (4 2) = 2 liters.
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1. 
Question 50 of 50
50. Question
How much interest can a person get on Rs. 8200 at 17.5% p.a. simple interest for a period of two years and six months?
Correct
I = (8200 * 2.5 * 17.5)/100 = (8200 * 5 * 35)/(100 * 2 * 2) = Rs. 3587.50
Incorrect
I = (8200 * 2.5 * 17.5)/100 = (8200 * 5 * 35)/(100 * 2 * 2) = Rs. 3587.50