DCCB Mock Test in English Series 3, DCCB Online Test Series 3
Finish Quiz
0 of 50 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
Information
DCCB Mock Test in English Series 3, DCCB Online Test Series 3, DCCB Free Online Test Series 3. RRB Exam Online Test 2019, DCCB Free Mock Test Exam 2019. DCCB Exam Free Online Quiz 2019, DCCB Full Online Mock Test Series 3rd in English. RRB Online Test for All Subjects, DCCB Free Mock Test Series in English. DCCB Free Mock Test Series 3. DCCB English Language Online Test in English Series 3rd. Our DCCB Staff Assistant Mock Test follows the exact Exam Pattern.
This paper has 50 questions.
Time allowed is 50 minutes.
The DCCB Online Test Series 3rd, DCCB Free Online Test Exam is Very helpful for all students. Now Scroll down below n click on “Start Quiz” or “Start Test” and Test yourself.
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 50 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Average score 

Your score 

Categories
 Not categorized 0%
Pos.  Name  Entered on  Points  Result 

Table is loading  
No data available  
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
 Answered
 Review

Question 1 of 50
1. Question
If the sides of a cube are in the ratio 4:3. What is the ratio of their diagonals?
Correct
a_{1}:a_{2} = 4:3
d_{1}:d_{2} = 4:3Incorrect
a_{1}:a_{2} = 4:3
d_{1}:d_{2} = 4:3 
Question 2 of 50
2. Question
Find the one which does not belong to that group ?
Correct
Y^{+4}C^{2}A^{+4}E^{2}C, K^{+4}O^{2}M^{+4}Q^{2}O, P^{+4}T^{2}R^{+3}U^{1}T, G^{+4}K^{2}I^{+4}M^{2}K and D^{+4}H^{2}F^{+4}J^{2}H.
Except PTRUT, all other groups follow similar pattern.Incorrect
Y^{+4}C^{2}A^{+4}E^{2}C, K^{+4}O^{2}M^{+4}Q^{2}O, P^{+4}T^{2}R^{+3}U^{1}T, G^{+4}K^{2}I^{+4}M^{2}K and D^{+4}H^{2}F^{+4}J^{2}H.
Except PTRUT, all other groups follow similar pattern. 
Question 3 of 50
3. Question
A man sells two articles for Rs.3600 each and he gains 30% on the first and loses 30% on the next. Find his total gain or loss?
Correct
(30*30)/100 = 9%loss
Incorrect
(30*30)/100 = 9%loss

Question 4 of 50
4. Question
How much time will a train of length 200 m moving at a speed of 72 kmph take to cross another train of length 300 m, moving at 36 kmph in the same direction?
Correct
The distance to be covered = Sum of their lengths = 200 + 300 = 500 m.
Relative speed = 72 36 = 36 kmph = 36 * 5/18 = 10 mps.
Time required = d/s = 500/10 = 50 sec.Incorrect
The distance to be covered = Sum of their lengths = 200 + 300 = 500 m.
Relative speed = 72 36 = 36 kmph = 36 * 5/18 = 10 mps.
Time required = d/s = 500/10 = 50 sec. 
Question 5 of 50
5. Question
I. a^{2} – 7a + 12 = 0,
II. b^{2} – 3b + 2 = 0 to solve both the equations to find the values of a and b?Correct
I.(a – 3)(a – 4) = 0
=> a = 3, 4
II. (b – 2)(b – 1) = 0
=> b = 1, 2
=> a > bIncorrect
I.(a – 3)(a – 4) = 0
=> a = 3, 4
II. (b – 2)(b – 1) = 0
=> b = 1, 2
=> a > b 
Question 6 of 50
6. Question
If an article is sold at 19% profit instead of 12% profit, then the profit would be Rs. 105 more. What is the cost price?
Correct
Let the cost price of an article be Rs. x.
(19% of x) – (12% of x) = 105
19x/100 – 12x/100 = 105 => 7x = 105 * 100
=> x = 1500
Cost price = Rs. 1500Incorrect
Let the cost price of an article be Rs. x.
(19% of x) – (12% of x) = 105
19x/100 – 12x/100 = 105 => 7x = 105 * 100
=> x = 1500
Cost price = Rs. 1500 
Question 7 of 50
7. Question
(64 + 9 + 9) / (2 * 20 + 1) = ?
Correct
(64 + 9 + 9) / (2 * 20 + 1) = 82/41 = 2
Incorrect
(64 + 9 + 9) / (2 * 20 + 1) = 82/41 = 2

Question 8 of 50
8. Question
All the water in container A which was filled to its brim was poured into two containers B and C. The quantity of water in container B was 62.5% less than the capacity of container A. If 148 liters was now transferred from C to B, then both the containers would have equal quantities of water. What was the initial quantity of water in container A?
Correct
B has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.
Quantity of water in B = 8k – 5k = 3k.
Quantity of water in container C = 8k – 3k = 5k
Container: A B C
Quantity of water: 8k 3k 5k
It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.
5k – 148 = 3k + 148 => 2k = 296 => k = 148
The initial quantity of water in A = 8k = 8 * 148 = 1184 liters.Incorrect
B has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.
Quantity of water in B = 8k – 5k = 3k.
Quantity of water in container C = 8k – 3k = 5k
Container: A B C
Quantity of water: 8k 3k 5k
It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.
5k – 148 = 3k + 148 => 2k = 296 => k = 148
The initial quantity of water in A = 8k = 8 * 148 = 1184 liters. 
Question 9 of 50
9. Question
The length of rectangle is thrice its breadth and its perimeter is 96 m, find the area of the rectangle?
Correct
2(3x + x) = 96
l = 36 b = 12
lb = 36 * 12 = 432Incorrect
2(3x + x) = 96
l = 36 b = 12
lb = 36 * 12 = 432 
Question 10 of 50
10. Question
The least number of five digits that have 144 their HCF is?
Correct
144) 10000 (69
9936
——–
64
10000 + 144 – 64 = 10080Incorrect
144) 10000 (69
9936
——–
64
10000 + 144 – 64 = 10080 
Question 11 of 50
11. Question
x varies inversely as square of y. Given that y = 2 for x = 1. The value of x for y = 6 will be equal to:
Correct
Given x = k/y^{2}, where k is a constant.
Now, y = 2 and x = 1 gives k = 4.
x = 4/y^{2} => x = 4/6^{2}, when
y = 6 => x = 4/36 = 1/9.Incorrect
Given x = k/y^{2}, where k is a constant.
Now, y = 2 and x = 1 gives k = 4.
x = 4/y^{2} => x = 4/6^{2}, when
y = 6 => x = 4/36 = 1/9. 
Question 12 of 50
12. Question
If the area of circle is 616 sq cm then its circumference?
Correct
22/7 r^{2} = 616 => r = 14
2 * 22/7 * 14 = 88Incorrect
22/7 r^{2} = 616 => r = 14
2 * 22/7 * 14 = 88 
Question 13 of 50
13. Question
In a bag there are coins of 50 paisa, 25 paisa and one rupee in the proportion 5:6:2. If there are in all Rs.42, the number of 25 paisa coins is?
Correct
5x 6x 2x
50 25 100
250x + 150x + 200x = 4200
600x = 4200
x = 7 => 6x = 42Incorrect
5x 6x 2x
50 25 100
250x + 150x + 200x = 4200
600x = 4200
x = 7 => 6x = 42 
Question 14 of 50
14. Question
A can finish a work in 18 days B can do the same work in 15 days. B worked for 10 days and left the job. In how many days, A alone can finish the remaining work?
Correct
B’s 10 day’s work = 1/15 * 10 = 2/3
Remaining work = (1 – 2/3) = 1/3
Now, 1/18 work is done by A in 1 day.
1/3 work is done by A in (18 * 1/3) = 6 days.Incorrect
B’s 10 day’s work = 1/15 * 10 = 2/3
Remaining work = (1 – 2/3) = 1/3
Now, 1/18 work is done by A in 1 day.
1/3 work is done by A in (18 * 1/3) = 6 days. 
Question 15 of 50
15. Question
A, B and C completed a piece of work, A worked for 6 days, B for 9 days and C for 4 days. Their daily wages were in the ratio of 3:4:5. Find the daily wages of C, if their total earning was Rs.1480?
Correct
3x 4x 5x
6 9 4
18x + 36x + 20x = 1480
74x = 1480 => x = 20
5x = 100 Rs.Incorrect
3x 4x 5x
6 9 4
18x + 36x + 20x = 1480
74x = 1480 => x = 20
5x = 100 Rs. 
Question 16 of 50
16. Question
7394 – 1236 – 4652 = ?
Correct
Incorrect

Question 17 of 50
17. Question
If x;Y = 5:2, then (8x + 9y):(8x + 2y) is :
Correct
Let x = 5k and y = 2k.
Then, (8x + 9y)/(8x + 2y)
= [(8 * 5k) + (9 * 2k)] / [(8 * 5k) + (2 * 2k)] = 58k/44k = 29/22
(8x + 9y):(8x + 2y) = 29:22Incorrect
Let x = 5k and y = 2k.
Then, (8x + 9y)/(8x + 2y)
= [(8 * 5k) + (9 * 2k)] / [(8 * 5k) + (2 * 2k)] = 58k/44k = 29/22
(8x + 9y):(8x + 2y) = 29:22 
Question 18 of 50
18. Question
A student scored an average of 80 marks in 3 subjects: Physics, Chemistry and Mathematics. If the average marks in Physics and Mathematics is 90 and that in Physics and Chemistry is 70, what are the marks in Physics?
Correct
Given M + P + C = 80 * 3 = 240 — (1)
M + P = 90 * 2 = 180 — (2)
P + C = 70 * 2 = 140 — (3)
Where M, P and C are marks obtained by the student in Mathematics, Physics and Chemistry.
P = (2) + (3) – (1) = 180 + 140 – 240 = 80Incorrect
Given M + P + C = 80 * 3 = 240 — (1)
M + P = 90 * 2 = 180 — (2)
P + C = 70 * 2 = 140 — (3)
Where M, P and C are marks obtained by the student in Mathematics, Physics and Chemistry.
P = (2) + (3) – (1) = 180 + 140 – 240 = 80 
Question 19 of 50
19. Question
Find the nearest to 25268 which is exactly divisible by 467?
Correct
Incorrect

Question 20 of 50
20. Question
What is the difference between the compound interest on Rs.12000 at 20% p.a. for one year when compounded yearly and half yearly?
Correct
When compounded annually, interest
= 12000[1 + 20/100]^{1} – 12000 = Rs.2400
When compounded semiannually, interest
= 12000[1 + 10/100]^{2} – 12000 = Rs.2520
Required difference = 2520 – 2400 = Rs.120Incorrect
When compounded annually, interest
= 12000[1 + 20/100]^{1} – 12000 = Rs.2400
When compounded semiannually, interest
= 12000[1 + 10/100]^{2} – 12000 = Rs.2520
Required difference = 2520 – 2400 = Rs.120 
Question 21 of 50
21. Question
How many three letter words are formed using the letters of the word TIME?
Correct
The number of letters in the given word is four.
The number of three letter words that can be formed using these four letters is ⁴P₃ = 4 * 3 * 2 = 24.Incorrect
The number of letters in the given word is four.
The number of three letter words that can be formed using these four letters is ⁴P₃ = 4 * 3 * 2 = 24. 
Question 22 of 50
22. Question
If p, q and r are positive integers and satisfy x = (p + q r)/r = (p – q + r)/q = (q + r – p)/p, then the value of x is?
Correct
When two or more ratios are equal, each of the ratios are equal to sum of the numerators divided by the sum of the denominators, provided sum of the denominators is nonzero.
Hence, x = (p + q r)/r = (p – q + r)/q = (q + r – p)/p
=> x = (p + q – r + p – q + r + q + r – p) / (r + q + p)
=> x = (r + q + p) / (r + q + p) = 1
p + q + r is nonzero.Incorrect
When two or more ratios are equal, each of the ratios are equal to sum of the numerators divided by the sum of the denominators, provided sum of the denominators is nonzero.
Hence, x = (p + q r)/r = (p – q + r)/q = (q + r – p)/p
=> x = (p + q – r + p – q + r + q + r – p) / (r + q + p)
=> x = (r + q + p) / (r + q + p) = 1
p + q + r is nonzero. 
Question 23 of 50
23. Question
If a: b = 3:4, b:c = 7:9, c:d = 5:7, find a:d?
Correct
a/d = (3/4)*(7/9)*(5/7) => 5/12
Incorrect
a/d = (3/4)*(7/9)*(5/7) => 5/12

Question 24 of 50
24. Question
Solve for m if 36 (6^{m}) = (216)^{3m+4}
Correct
36 (6^{m}) = (216)^{3m+4} => 6^{2} 6^{m} = (6^{3})^{3m+4} => 6^{2+m} = 6^{9m+12} Equating powers of 6 on both sides.
m + 2 = 9m + 12 => 10 = 8m => m = 5/4Incorrect
36 (6^{m}) = (216)^{3m+4} => 6^{2} 6^{m} = (6^{3})^{3m+4} => 6^{2+m} = 6^{9m+12} Equating powers of 6 on both sides.
m + 2 = 9m + 12 => 10 = 8m => m = 5/4 
Question 25 of 50
25. Question
X men can do a work in 120 days. If there were 20 men less, the work would have taken 60 days more. What is the value of X?
Correct
We have M_{1} D_{1} = M_{2} D_{2}
120X = (X – 20)180
=> 2X = (X – 20) 3 => 2X = 3X – 60
=> X = 60Incorrect
We have M_{1} D_{1} = M_{2} D_{2}
120X = (X – 20)180
=> 2X = (X – 20) 3 => 2X = 3X – 60
=> X = 60 
Question 26 of 50
26. Question
Which of the following groups of fractions is in descending order?
Correct
The fractions considered are 8/15 9/13 6/11
To compare them we make the denominators the same. So the fractions are
(8 * 13 * 11)/2145, (9 * 15 * 11)/2145 and (6 * 15 * 13)/2145
1144/2145, 1485/2145 and 1170/2145
so in descending order the fractions will be
1485/2145, 1170/2145 and 1144/2145 i.e., 9/13 , 6/11 , 8/15Incorrect
The fractions considered are 8/15 9/13 6/11
To compare them we make the denominators the same. So the fractions are
(8 * 13 * 11)/2145, (9 * 15 * 11)/2145 and (6 * 15 * 13)/2145
1144/2145, 1485/2145 and 1170/2145
so in descending order the fractions will be
1485/2145, 1170/2145 and 1144/2145 i.e., 9/13 , 6/11 , 8/15 
Question 27 of 50
27. Question
The least number which when diminished by 7 is divisible by 21, 28, 36 and 45 is?
Correct
LCM = 1260
1260 + 7 = 1267Incorrect
LCM = 1260
1260 + 7 = 1267 
Question 28 of 50
28. Question
A can give B 100 meters start and C 200 meters start in a kilometer race. How much start can B give C in a kilometer race?
Correct
A runs 1000 m while B runs 900 m and C runs 800 m.
The number of meters that C runs when B runs 1000 m,
= (1000 * 800)/900 = 8000/9 = 888.88 m.
B can give C = 1000 – 888.88 = 111.12 m.Incorrect
A runs 1000 m while B runs 900 m and C runs 800 m.
The number of meters that C runs when B runs 1000 m,
= (1000 * 800)/900 = 8000/9 = 888.88 m.
B can give C = 1000 – 888.88 = 111.12 m. 
Question 29 of 50
29. Question
64 boys and 40 girls form a group for social work. During their membership drive, same number of boys and girls joined the group. How many members does the group have now, if the ratio of boys to girls is 4 : 3?
Correct
Let us say x boys and x girls joined the group.
(64 + x) / (40 + x) = 4/3
192 + 3x = 160 + 4x => x = 32
Number of members in the group = 64 + x + 40 + x
= 104 + 2x = 168.Incorrect
Let us say x boys and x girls joined the group.
(64 + x) / (40 + x) = 4/3
192 + 3x = 160 + 4x => x = 32
Number of members in the group = 64 + x + 40 + x
= 104 + 2x = 168. 
Question 30 of 50
30. 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 31 of 50
31. Question
A train running at a speed of 36 kmph crosses an electric pole in 12 seconds. In how much time will it cross a 350 m long platform?
Correct
Let the length of the train be x m.
When a train crosses an electric pole, the distance covered is its own length.
So, x = 12 * 36 * 5 /18 m = 120 m.
Time taken to cross the platform = (120 +350)/ 36 * 5/18 = 47 min.Incorrect
Let the length of the train be x m.
When a train crosses an electric pole, the distance covered is its own length.
So, x = 12 * 36 * 5 /18 m = 120 m.
Time taken to cross the platform = (120 +350)/ 36 * 5/18 = 47 min. 
Question 32 of 50
32. Question
Two trains of equal length, running with the speeds of 60 and 40 kmph, take 50 seconds to cross each other while they are running in the same direction. What time will they take to cross each other if they are running in opposite directions?
Correct
RS = 60 40 = 20 * 5/18 = 100/18
T = 50
D = 50 * 100/18 = 2500/9
RS = 60 + 40 = 100 * 5/18
T = 2500/9 * 18/500 = 10 secIncorrect
RS = 60 40 = 20 * 5/18 = 100/18
T = 50
D = 50 * 100/18 = 2500/9
RS = 60 + 40 = 100 * 5/18
T = 2500/9 * 18/500 = 10 sec 
Question 33 of 50
33. Question
The average salary of workers in an industry is Rs.200 the average salary of technicians being Rs.400 and that of nontechnicians being Rs.125. What is the total number of workers?
Correct
Incorrect

Question 34 of 50
34. Question
A can do a half of certain work in 70 days and B one third of the same in 35 days. They together will do the whole work in.
Correct
A = 140 days
B = 105 days
1/140 + 1/105 = 7/420 = 1/60
=>60 daysIncorrect
A = 140 days
B = 105 days
1/140 + 1/105 = 7/420 = 1/60
=>60 days 
Question 35 of 50
35. Question
It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the car is?
Correct
Let the speed of the train be x km/hr and that of the car be y km/hr.
Then, 120/x + 480/y = 8 or 1/x + 4/y = 1/15 — (i)
And, 200/x + 400/y = 25/3 or 1/x + 2/y = 1/24 — (ii)
Solving (i) and (ii), we get x = 60 and y = 80
Ratio of speeds = 60:80 = 3:4Incorrect
Let the speed of the train be x km/hr and that of the car be y km/hr.
Then, 120/x + 480/y = 8 or 1/x + 4/y = 1/15 — (i)
And, 200/x + 400/y = 25/3 or 1/x + 2/y = 1/24 — (ii)
Solving (i) and (ii), we get x = 60 and y = 80
Ratio of speeds = 60:80 = 3:4 
Question 36 of 50
36. Question
A man purchases 8 pens for Rs.9 and sells 9 pens for Rs.8, how much profit or loss does he make?
Correct
81 — 17
100 —– ? è 20.98%lossIncorrect
81 — 17
100 —– ? è 20.98%loss 
Question 37 of 50
37. Question
A and B can do a piece of work in 3 days, B and C in 4 days, C and A in 6 days. How long will C take to do it?
Correct
2c = ¼ + 1/6 – 1/3 = 1/12
c = 1/24 => 24 daysIncorrect
2c = ¼ + 1/6 – 1/3 = 1/12
c = 1/24 => 24 days 
Question 38 of 50
38. Question
Find the greatest number that, while dividing 47, 215 and 365, gives the same remainder in each case?
Correct
Calculate the differences, taking two numbers at a time as follows:
(21547) = 168
(365215) = 150
(36547) = 318
HCF of 168, 150 and 318 we get 6, which is the greatest number, which while dividing 47, 215 and 365 gives the same remainder in each cases is 5.Incorrect
Calculate the differences, taking two numbers at a time as follows:
(21547) = 168
(365215) = 150
(36547) = 318
HCF of 168, 150 and 318 we get 6, which is the greatest number, which while dividing 47, 215 and 365 gives the same remainder in each cases is 5. 
Question 39 of 50
39. Question
Find the expenditure on digging a well 14m deep and of 3m diameter at Rs.15 per cubic meter?
Correct
22/7 * 14 * 3/2 * 3/2 = 99 m^{2}
99 * 15 = 1485Incorrect
22/7 * 14 * 3/2 * 3/2 = 99 m^{2}
99 * 15 = 1485 
Question 40 of 50
40. 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 41 of 50
41. Question
The average marks in mathematics scored by the pupils of a school at the public examination were 39. If four of these pupils who actually scored 5, 12, 15 and 19 marks at the examination had not been sent up, the average marks for the school would have been 44. Find the number of pupils sent up for examination from the school?
Correct
39x = 5 + 12 + 15 + 19 + (x – 4)44
x = 25Incorrect
39x = 5 + 12 + 15 + 19 + (x – 4)44
x = 25 
Question 42 of 50
42. Question
Surface area of two spheres are in the ratio 1:4 what is the ratio of their volumes?
Correct
Incorrect

Question 43 of 50
43. Question
A tank 3 m long, 2 m wide and 1.5 m deep is dug in a field 20 m long and 14 m wide. If the earth dug out is evenly spread out over the field, the level of the field will raise by nearly?
Correct
Incorrect

Question 44 of 50
44. Question
The average of five consecutive odd numbers is 61. What is the difference between the highest and lowest numbers?
Correct
Let the numbers x, x + 2, x + 4, x + 6 and x + 8
Then, [x + (x + 2) + (x + 4) + (x + 6) + (x + 8)]5 = 61
or 5x + 20 = 305 => x = 57
So, required difference = (57 + 8) – 57 = 8.Incorrect
Let the numbers x, x + 2, x + 4, x + 6 and x + 8
Then, [x + (x + 2) + (x + 4) + (x + 6) + (x + 8)]5 = 61
or 5x + 20 = 305 => x = 57
So, required difference = (57 + 8) – 57 = 8. 
Question 45 of 50
45. 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 46 of 50
46. Question
The least number which when divided by 8, 12 and 16, leave in each remainder 3, is?
Correct
LCM = 48 + 3 = 51
Incorrect
LCM = 48 + 3 = 51

Question 47 of 50
47. Question
(0.9 * 0.9 – 0.8 * 0.8)/1.7 = ?
Correct
As numerator is of the form a^{2} – b^{2} = (a + b) (a – b), so the given expression simplifying becomes
(0.9^{2} – 0.8^{2})/(0.9 + 0.8) (0.9 + 0.8) (0.9 – 0.8)/0.9 + 0.8 = 0.1Incorrect
As numerator is of the form a^{2} – b^{2} = (a + b) (a – b), so the given expression simplifying becomes
(0.9^{2} – 0.8^{2})/(0.9 + 0.8) (0.9 + 0.8) (0.9 – 0.8)/0.9 + 0.8 = 0.1 
Question 48 of 50
48. Question
If a number is chosen at random from the set {1, 2, 3, …., 100}, then the probability that the chosen number is a perfect cube is .
Correct
We have 1, 8, 27 and 64 as perfect cubes from 1 to 100.
Thus, the probability of picking a perfect cube is
4/100 = 1/25.Incorrect
We have 1, 8, 27 and 64 as perfect cubes from 1 to 100.
Thus, the probability of picking a perfect cube is
4/100 = 1/25. 
Question 49 of 50
49. 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 50 of 50
50. Question
Ravi can do a piece of work in 30 days while Prakash can do it in 40 days. In how many days will they finish it together?
Correct
1/30 + 1/40 = 7/120
120/7 = 17 1/7 daysIncorrect
1/30 + 1/40 = 7/120
120/7 = 17 1/7 days