RBI Assistant Mains Online Test Series 3, RBI Assistant Mains Quiz
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RBI Assistant Mains Online Test Series 3, RBI Assistant Mains Quiz Series 3. RBI Exam Online Test 2024, RBI Assistant Mains Free Mock Test Exam 2024. RBI Assistant Mains Exam Free Online Quiz 2024, RBI Assistant Mains Full Online Mock Test Series 3rd in English.The RBI Assistant Exam pattern is same as other bank clerk exams followed by Preliminary, Main and Language Proficiency Test. Last year new changes were introduced in the exam pattern. The RBI Assistant Mains Full online mock test paper is free for all students. RBI Assistant Mains Question and Answers in English and Hindi Series 3. Here we are providing RBI Assistant Mains Full Mock Test Paper in English. RBI Assistant Mains Mock Test Series 3rd 2024. Now Test your self for RBI Assistant Mains Exam by using below quiz…
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Question 1 of 50
1. Question
A gardener wants to plant trees in his garden in such a way that the number of trees in each row should be the same. If there are 4 rows or 5 rows or 6 rows, then no tree will be left. Find the least number of trees required.
Correct
The least number of trees that are required = LCM(4, 5, 6) = 60.
Incorrect
The least number of trees that are required = LCM(4, 5, 6) = 60.
-
Question 2 of 50
2. Question
In how many ways can six members be selected from a group of ten members?
Correct
Six members can be selected from ten members in
¹⁰C₆ = ¹⁰C₄ ways.Incorrect
Six members can be selected from ten members in
¹⁰C₆ = ¹⁰C₄ ways. -
Question 3 of 50
3. Question
If cost of sugar increases by 25%. How much percent consumption of sugar should be decreased in order to keep expenditure fixed?
Correct
100
125
—–
125 —– 25
100 —— ? => 20%Incorrect
100
125
—–
125 —– 25
100 —— ? => 20% -
Question 4 of 50
4. Question
The sum of five numbers is 655. The average of the first two numbers is 85 and the third number is 125. Find the average of the two numbers?
Correct
Let the five numbers be P, Q, R, S and T.
=> P + Q + R + S + T = 655.
(P + Q)/2 = 85 and R = 125
P + Q = 170 and R = 125
P + Q + R = 295
S + T = 655 – (P + Q + R) = 360
Average of the last two numbers = (S + T)/2 = 180.Incorrect
Let the five numbers be P, Q, R, S and T.
=> P + Q + R + S + T = 655.
(P + Q)/2 = 85 and R = 125
P + Q = 170 and R = 125
P + Q + R = 295
S + T = 655 – (P + Q + R) = 360
Average of the last two numbers = (S + T)/2 = 180. -
Question 5 of 50
5. Question
The average of 11 results is 50, if the average of first six results is 49 and that of the last six is 52. Find the sixth result?
Correct
1 to 11 = 11 * 50 = 550
1 to 6 = 6 * 49 = 294
6 to 11 = 6 * 52 = 312
6th = 294 + 312 – 550 = 56Incorrect
1 to 11 = 11 * 50 = 550
1 to 6 = 6 * 49 = 294
6 to 11 = 6 * 52 = 312
6th = 294 + 312 – 550 = 56 -
Question 6 of 50
6. Question
Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is :
Correct
Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F of 551 and 1073 = 29;
First number = 551/29 = 19
Third number = 1073/29 = 37.
Required sum = 19 + 29 + 37 = 85.Incorrect
Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F of 551 and 1073 = 29;
First number = 551/29 = 19
Third number = 1073/29 = 37.
Required sum = 19 + 29 + 37 = 85. -
Question 7 of 50
7. Question
The smallest number when increased by ” 1 ” is exactly divisible by 12, 18, 24, 32 and 40 is:
Correct
LCM = 1440
1440 – 1 = 1439Incorrect
LCM = 1440
1440 – 1 = 1439 -
Question 8 of 50
8. Question
Find the roots of the quadratic equation: 2x2 + 3x – 9 = 0?
Correct
2x2 + 6x – 3x – 9 = 0
2x(x + 3) – 3(x + 3) = 0
(x + 3)(2x – 3) = 0
=> x = -3 or x = 3/2.Incorrect
2x2 + 6x – 3x – 9 = 0
2x(x + 3) – 3(x + 3) = 0
(x + 3)(2x – 3) = 0
=> x = -3 or x = 3/2. -
Question 9 of 50
9. Question
A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that all the four bulbs are defective.
Correct
Out of nine, five are good and four are defective. Required probability = ⁴C₄/⁹C₄ = 1/126
Incorrect
Out of nine, five are good and four are defective. Required probability = ⁴C₄/⁹C₄ = 1/126
-
Question 10 of 50
10. Question
There were two candidates in an election. Winner candidate received 62% of votes and won the election by 288 votes. Find the number of votes casted to the winning candidate?
Correct
W = 62% L = 38%
62% – 38% = 24%
24% ——– 288
62% ——– ? => 744Incorrect
W = 62% L = 38%
62% – 38% = 24%
24% ——– 288
62% ——– ? => 744 -
Question 11 of 50
11. Question
The average age of 15 students of a class is 15 years. Out of these, the average age of 5 students is 14 years and that of the other 9 students is 16 years. Tee age of the 15th student is:
Correct
Age of the 15th student=
[15 * 15 – (14 * 5 + 16 * 9)] = (225 – 214) = 11 years.Incorrect
Age of the 15th student=
[15 * 15 – (14 * 5 + 16 * 9)] = (225 – 214) = 11 years. -
Question 12 of 50
12. 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 13 of 50
13. Question
Two cars P and Q start at the same time from A and B which are 120 km apart. If the two cars travel in opposite directions, they meet after one hour and if they travel in same direction (from A towards B), then P meets Q after 6 hours. What is the speed of car P?
Correct
Let their speed be x km/hr and y km/he respectively.
Then, 120/(x + y) = 1 => x + y = 120 — (i)
Now, when they move in same direction:
(Distance traveled by P in 6 hrs) – (Distance traveled by Q in 6 hrs) = 120 km
6x – 6y = 120 => x – y = 20 — (ii)
Sloving (i) and (ii), we get x = 70, y = 50
P’s speed = 70 km/hr.Incorrect
Let their speed be x km/hr and y km/he respectively.
Then, 120/(x + y) = 1 => x + y = 120 — (i)
Now, when they move in same direction:
(Distance traveled by P in 6 hrs) – (Distance traveled by Q in 6 hrs) = 120 km
6x – 6y = 120 => x – y = 20 — (ii)
Sloving (i) and (ii), we get x = 70, y = 50
P’s speed = 70 km/hr. -
Question 14 of 50
14. Question
An article is bought for Rs.675 and sold for Rs.900, find the gain percent?
Correct
675 —- 225
100 —- ? => 33 1/3%Incorrect
675 —- 225
100 —- ? => 33 1/3% -
Question 15 of 50
15. Question
Sreenivas sells a table to Shiva at 10% profit and Shiva sells it to Mahesh at 10% loss. At what price did Sreenivas purchase the table if Mahesh paid Rs. 2178?
Correct
Let the cost price of table for Sreenivas be Rs. x and given that, cost price of table for Mahesh = Rs. 2178.
=> (90%) of (110%) of x = Rs. 2178.
=> (90/100)(110/100)x = 2178
=> x = (2178 * 100)/(9 * 11)
=> x = Rs. 2200Incorrect
Let the cost price of table for Sreenivas be Rs. x and given that, cost price of table for Mahesh = Rs. 2178.
=> (90%) of (110%) of x = Rs. 2178.
=> (90/100)(110/100)x = 2178
=> x = (2178 * 100)/(9 * 11)
=> x = Rs. 2200 -
Question 16 of 50
16. Question
If 5% more is gained by selling an article for Rs. 350 than by selling it for Rs. 340, the cost of the article is:
Correct
Let C.P. be Rs. x.
Then, 5% of x = 350 – 340 = 10
x/20 = 10 => x = 200Incorrect
Let C.P. be Rs. x.
Then, 5% of x = 350 – 340 = 10
x/20 = 10 => x = 200 -
Question 17 of 50
17. Question
The length of a rectangle is two – fifths of the radius of a circle. The radius of the circle is equal to the side of the square, whose area is 1225 sq.units. What is the area (in sq.units) of the rectangle if the rectangle if the breadth is 10 units?
Correct
Given that the area of the square = 1225 sq.units
=> Side of square = √1225 = 35 units
The radius of the circle = side of the square = 35 units Length of the rectangle = 2/5 * 35 = 14 units
Given that breadth = 10 units
Area of the rectangle = lb = 14 * 10 = 140 sq.unitsIncorrect
Given that the area of the square = 1225 sq.units
=> Side of square = √1225 = 35 units
The radius of the circle = side of the square = 35 units Length of the rectangle = 2/5 * 35 = 14 units
Given that breadth = 10 units
Area of the rectangle = lb = 14 * 10 = 140 sq.units -
Question 18 of 50
18. Question
A bag contains a certain number of 50 paise coins, 20 paise coins and 10 paise coins inthe ratio 2:3:4. If the total value of all the coins in the bag is Rs.400, find the number of coins of each kind?
Correct
50*2k + 20*3k + 10*4k = 40000
200k = 40000 => k = 200
50p coins = 2k = 2*200 = 400
20p coins = 3k = 3*200 = 600
10p coins = 4k = 4*200 = 800Incorrect
50*2k + 20*3k + 10*4k = 40000
200k = 40000 => k = 200
50p coins = 2k = 2*200 = 400
20p coins = 3k = 3*200 = 600
10p coins = 4k = 4*200 = 800 -
Question 19 of 50
19. Question
Solve the equation for x : 19(x + y) + 17 = 19(-x + y) – 21
Correct
19x + 19y + 17 = -19x + 19y – 21
38x = -38 => x = -1Incorrect
19x + 19y + 17 = -19x + 19y – 21
38x = -38 => x = -1 -
Question 20 of 50
20. Question
Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time which they take to cross each other is?
Correct
Relative speed = 60 + 40 = 100 km/hr.
= 100 * 5/18 = 250/9 m/sec.
Distance covered in crossing each other = 140 + 160 = 300 m.
Required time = 300 * 9/250 = 54/5 = 10.8 sec.Incorrect
Relative speed = 60 + 40 = 100 km/hr.
= 100 * 5/18 = 250/9 m/sec.
Distance covered in crossing each other = 140 + 160 = 300 m.
Required time = 300 * 9/250 = 54/5 = 10.8 sec. -
Question 21 of 50
21. Question
Subtracting 10% from X is the same as multiplying X by what number?
Correct
X – (10/100) X = X * ?
? = 90%Incorrect
X – (10/100) X = X * ?
? = 90% -
Question 22 of 50
22. Question
A and B go around a circular track of length 600 m on a cycle at speeds of 36 kmph and 54 kmph. After how much time will they meet for the first time at the starting point?
Correct
Time taken to meet for the first time at the starting point
= LCM { length of the track / speed of A , length of the track / speed of B}
= LCM { 600/ (36 * 5/18) , 600/ (54 * 5 /18) }
= LCM (60, 40) = 120 sec.Incorrect
Time taken to meet for the first time at the starting point
= LCM { length of the track / speed of A , length of the track / speed of B}
= LCM { 600/ (36 * 5/18) , 600/ (54 * 5 /18) }
= LCM (60, 40) = 120 sec. -
Question 23 of 50
23. Question
Anil can do a work in 15 days while Sunil can do it in 25 days. How long will they take if both work together?
Correct
1/15 + 1/25 = 8/75
75/8 = 9 3/8 daysIncorrect
1/15 + 1/25 = 8/75
75/8 = 9 3/8 days -
Question 24 of 50
24. Question
A man can row 6 kmph in still water. When the river is running at 1.2 kmph, it takes him 1 hour to row to a place and black. How far is the place?
Correct
M = 6
S = 1.2
DS = 6 + 1.2 = 7.2
US = 6 – 1.2 = 4.8
x/7.2 + x/4.8 = 1
x = 2.88Incorrect
M = 6
S = 1.2
DS = 6 + 1.2 = 7.2
US = 6 – 1.2 = 4.8
x/7.2 + x/4.8 = 1
x = 2.88 -
Question 25 of 50
25. Question
LCM of 455, 117, 338 is:
Correct
Incorrect
-
Question 26 of 50
26. Question
If the wheel is 14 cm then the number of revolutions to cover a distance of 1056 cm is?
Correct
2 * 22/7 * 14 * x = 1056 => x = 12
Incorrect
2 * 22/7 * 14 * x = 1056 => x = 12
-
Question 27 of 50
27. Question
I. x2 + 9x + 20 = 0,
II. y2 + 5y + 6 = 0 to solve both the equations to find the values of x and y?Correct
I. x2 + 4x + 5x + 20 = 0
=>(x + 4)(x + 5) = 0 => x = -4, -5
II. y2 + 3y + 2y + 6 = 0
=>(y + 3)(y + 2) = 0 => y = -3, -2
= x < y.Incorrect
I. x2 + 4x + 5x + 20 = 0
=>(x + 4)(x + 5) = 0 => x = -4, -5
II. y2 + 3y + 2y + 6 = 0
=>(y + 3)(y + 2) = 0 => y = -3, -2
= x < y. -
Question 28 of 50
28. Question
Two trains of length 100 m and 200 m are 100 m apart. They start moving towards each other on parallel tracks, at speeds 54 kmph and 72 kmph. After how much time will the trains meet?
Correct
They are moving in opposite directions, relative speed is equal to the sum of their speeds.
Relative speed = (54 + 72)*5/18 = 7*5 = 35 mps.
The time required = d/s = 100/35 = 20/7 sec.Incorrect
They are moving in opposite directions, relative speed is equal to the sum of their speeds.
Relative speed = (54 + 72)*5/18 = 7*5 = 35 mps.
The time required = d/s = 100/35 = 20/7 sec. -
Question 29 of 50
29. Question
By selling 50 meters of cloth. I gain the selling price of 15 meters. Find the gain percent?
Correct
SP = CP + g
50 SP = 50 CP + 15 SP
35 SP = 50 CP
35 — 15 CP gain
100 — ? => 42 6/7%Incorrect
SP = CP + g
50 SP = 50 CP + 15 SP
35 SP = 50 CP
35 — 15 CP gain
100 — ? => 42 6/7% -
Question 30 of 50
30. Question
Ram’s age and Shyam’s age are in the ratio 3 : 4. Seven years ago the ratio of their ages was 2: 3. Find the ratio of their ages five years hence?
Correct
Let ages of Ramu and Shyamu be x and y respectively.
x/y = 3/4 => x = 3/4 y
Also (x – 7) / (y – 7) = 2/3
=> 3x – 21 = 2y – 14
3x = 2y + 7
But x = 3/4 y
3 * 3/4 y = 2y + 7
9y = 8y + 28 => y = 28 years
Ratio of their ages five years hence
= (21 + 5) / (28 + 5) = 26/33.Incorrect
Let ages of Ramu and Shyamu be x and y respectively.
x/y = 3/4 => x = 3/4 y
Also (x – 7) / (y – 7) = 2/3
=> 3x – 21 = 2y – 14
3x = 2y + 7
But x = 3/4 y
3 * 3/4 y = 2y + 7
9y = 8y + 28 => y = 28 years
Ratio of their ages five years hence
= (21 + 5) / (28 + 5) = 26/33. -
Question 31 of 50
31. Question
A person purchased a TV set for Rs. 16000 and a DVD player for Rs. 6250. He sold both the items together for Rs. 31150. What percentage of profit did he make?
Correct
The total CP = Rs. 16000 + Rs. 6250 = Rs. 22250 and SP = Rs. 31150
Profit(%) = (31150 – 22250)/22250 * 100 = 40%Incorrect
The total CP = Rs. 16000 + Rs. 6250 = Rs. 22250 and SP = Rs. 31150
Profit(%) = (31150 – 22250)/22250 * 100 = 40% -
Question 32 of 50
32. Question
A boy rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. His average speed for the entire trip is approximately?
Correct
Total distance traveled = 10 + 12 = 22 km /hr.
Total time taken = 10/12 + 12/10 = 61/30 hrs.
Average speed = 22 * 30/61 = 10.8 km/hr.Incorrect
Total distance traveled = 10 + 12 = 22 km /hr.
Total time taken = 10/12 + 12/10 = 61/30 hrs.
Average speed = 22 * 30/61 = 10.8 km/hr. -
Question 33 of 50
33. Question
The average age of a husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:
Correct
Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) = 57 years.
Required average = 57/3 = 19 years.Incorrect
Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) = 57 years.
Required average = 57/3 = 19 years. -
Question 34 of 50
34. Question
If x = √8 – √7, y = √6 – √5 and z = √10 – 3, which of the following is true?
Correct
x = √8 – √7 = 1/(√8 + √7) and y = √6 – √5 = 1/(√6 + √5) and z = √10 – 3 = 1/(√10 + 3)
As (√8 + √7) > (√6 + √5), (√8 – √7) < (√6 - √5) As (√10 + 3) > (√8 + √7) > (√6 + √5)
z < x < y.Incorrect
x = √8 – √7 = 1/(√8 + √7) and y = √6 – √5 = 1/(√6 + √5) and z = √10 – 3 = 1/(√10 + 3)
As (√8 + √7) > (√6 + √5), (√8 – √7) < (√6 - √5) As (√10 + 3) > (√8 + √7) > (√6 + √5)
z < x < y. -
Question 35 of 50
35. Question
A brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 25 m * 2 m * 0.75 m?
Correct
25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x
25 = 1/100 * x => x = 25000Incorrect
25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x
25 = 1/100 * x => x = 25000 -
Question 36 of 50
36. Question
The difference between a two-digit number and the number obtained by interchanging the digit is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1:2?
Correct
Since the number is greater than the number obtained on reversing the digits, so the ten’s digit is greater than the unit’s digit.
Let the ten’s and unit’s digits be 2x and x respectively.
Then, (10 * 2x + x) – (10x + 2x) = 36
9x = 36
x = 4
Required difference = (2x + x) – (2x – x) = 2x = 8.Incorrect
Since the number is greater than the number obtained on reversing the digits, so the ten’s digit is greater than the unit’s digit.
Let the ten’s and unit’s digits be 2x and x respectively.
Then, (10 * 2x + x) – (10x + 2x) = 36
9x = 36
x = 4
Required difference = (2x + x) – (2x – x) = 2x = 8. -
Question 37 of 50
37. Question
A gets 3 times as much money as B gets, B gets only Rs.25 more then what C gets. The three gets Rs.675 in all. Find the share of B?
Correct
A+B+C = 675
A = 3B
3B+B+B-25 = 675
5B = 700
B = 140Incorrect
A+B+C = 675
A = 3B
3B+B+B-25 = 675
5B = 700
B = 140 -
Question 38 of 50
38. Question
4500 * ? = 3375
Correct
4500 * x = 3375
= x = 3375/4500 = 3/4Incorrect
4500 * x = 3375
= x = 3375/4500 = 3/4 -
Question 39 of 50
39. Question
The sum of three consecutive integers is 102. Find the lowest of the three?
Correct
Three consecutive numbers can be taken as (P – 1), P, (P + 1).
So, (P – 1) + P + (P + 1) = 102
3P = 102 => P = 34.
The lowest of the three = (P – 1) = 34 – 1 = 33.Incorrect
Three consecutive numbers can be taken as (P – 1), P, (P + 1).
So, (P – 1) + P + (P + 1) = 102
3P = 102 => P = 34.
The lowest of the three = (P – 1) = 34 – 1 = 33. -
Question 40 of 50
40. Question
The sector of a circle has radius of 21 cm and central angle 135o. Find its perimeter?
Correct
Perimeter of the sector = length of the arc + 2(radius)
= (135/360 * 2 * 22/7 * 21) + 2(21)
= 49.5 + 42 = 91.5 cmIncorrect
Perimeter of the sector = length of the arc + 2(radius)
= (135/360 * 2 * 22/7 * 21) + 2(21)
= 49.5 + 42 = 91.5 cm -
Question 41 of 50
41. Question
Rajan borrowed Rs.4000 at 5% p.a compound interest. After 2 years, he repaid Rs.2210 and after 2 more year, the balance with interest. What was the total amount that he paid as interest?
Correct
4000
200 —- I
200
10 —- II
—————
4410
2210
——–
2000
110 —- III
110
5.50 —- IV
———–
2425.50
2210
———–
4635.50
4000
———-
635.50Incorrect
4000
200 —- I
200
10 —- II
—————
4410
2210
——–
2000
110 —- III
110
5.50 —- IV
———–
2425.50
2210
———–
4635.50
4000
———-
635.50 -
Question 42 of 50
42. Question
At what rate percent on simple interest will a sum of money double itself in 30 years?
Correct
P = (P*30*R)/100
R = 3 1/3%Incorrect
P = (P*30*R)/100
R = 3 1/3% -
Question 43 of 50
43. Question
A group consists of 4 men, 6 women and 5 children. In how many ways can 2 men , 3 women and 1 child selected from the given group?
Correct
Two men, three women and one child can be selected in ⁴C₂ * ⁶C₃ * ⁵C₁ ways
= (4 * 3)/(2 * 1) * (6 * 5 * 4)/(3 * 2) * 5
= 600 ways.Incorrect
Two men, three women and one child can be selected in ⁴C₂ * ⁶C₃ * ⁵C₁ ways
= (4 * 3)/(2 * 1) * (6 * 5 * 4)/(3 * 2) * 5
= 600 ways. -
Question 44 of 50
44. Question
Some pens are divided among A, B, C and D. A gets twice the number of pens that B gets. C gets the same number of pens as D gets. If A gets 25 pens more than D and the ratio of the number of pens that B and C get is 2:3, then find the number of pens that D gets?
Correct
Let the number of pens that A, B, C and D get be a, b, c and d respectively.
a:b = 2:1
a = c + 25
b:c = 2:3
a:b:c:d = 4:2:3:3
a, d get 4p, 3p pens
=> 4p – 3p = 25 => p = 25
=> D gets 3p = 3 * 25 = 75 pens.Incorrect
Let the number of pens that A, B, C and D get be a, b, c and d respectively.
a:b = 2:1
a = c + 25
b:c = 2:3
a:b:c:d = 4:2:3:3
a, d get 4p, 3p pens
=> 4p – 3p = 25 => p = 25
=> D gets 3p = 3 * 25 = 75 pens. -
Question 45 of 50
45. Question
Thirty men can do a work in 24 days. In how many days can 20 men can do the work, given that the time spent per day is increased by one-third of the previous time?
Correct
Let the number of hours working per day initially be x. we have M1 D1 H1= M2 D2 H2
30 * 24 * x = 20 * d2 * (4x)/3 => d2 = (30 * 24 * 3)/(24 * 4) = 27 days.Incorrect
Let the number of hours working per day initially be x. we have M1 D1 H1= M2 D2 H2
30 * 24 * x = 20 * d2 * (4x)/3 => d2 = (30 * 24 * 3)/(24 * 4) = 27 days. -
Question 46 of 50
46. Question
In a class of 140 students, 60% of them passed. By what percent is the number of students who passed more than the number of failed students?
Correct
Number of students passed = 60% of 140 = 60/100 * 140 = 84
Number of students failed = 140 – 84 = 56.
Required percentage = 28/56 * 100 = 50%.Incorrect
Number of students passed = 60% of 140 = 60/100 * 140 = 84
Number of students failed = 140 – 84 = 56.
Required percentage = 28/56 * 100 = 50%. -
Question 47 of 50
47. Question
The cost of 16 pens and 8 pencils is Rs.352 and the cost of 4 pens and 4 pencils is Rs.96. Find the cost of each pen?
Correct
Let the cost of each pen and pencil be ‘p’ and ‘q’ respectively.
16p + 8q = 352 — (1)
4p + 4q = 96
8p + 8q = 192 — (2)
(1) – (2) => 8p = 160
=> p = 20Incorrect
Let the cost of each pen and pencil be ‘p’ and ‘q’ respectively.
16p + 8q = 352 — (1)
4p + 4q = 96
8p + 8q = 192 — (2)
(1) – (2) => 8p = 160
=> p = 20 -
Question 48 of 50
48. Question
Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 15 cm.
Correct
Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm2
Incorrect
Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm2
-
Question 49 of 50
49. Question
Sushil got thrice as many marks in English as in Science. His total marks in English, Science and Maths are 162. If the ratio of his marks in English and Maths is 3:5, find his marks in Science?
Correct
S:E = 1:3
E:M = 3:5
————
S:E:M = 3:9:15
3/27 * 162 = 18Incorrect
S:E = 1:3
E:M = 3:5
————
S:E:M = 3:9:15
3/27 * 162 = 18 -
Question 50 of 50
50. Question
A started a business with an investment of Rs. 70000 and after 6 months B joined him investing Rs. 120000. If the profit at the end of a year is Rs. 52000, then the share of B is?
Correct
Ratio of investments of A and B is (70000 * 12) : (120000 * 6) = 7 : 6
Total profit = Rs. 52000
Share of B = 6/13 (52000) = Rs. 24000Incorrect
Ratio of investments of A and B is (70000 * 12) : (120000 * 6) = 7 : 6
Total profit = Rs. 52000
Share of B = 6/13 (52000) = Rs. 24000